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INSTRUCTOR’S MANUAL TO ACCOMPANY
ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES Third Edition
Richard M. Felder Ronald W. Rousseau with assistance from Matthew Burke, Swapnil Chhabra, Jun Gao, Gary Huvard, Concepción Jimenez-Gonzalez, Linda Holm, Norman Kaplan, Brian Keyes, Amit Khandelwal, Stephanie Manfredi, Janette Mendez-Santiago, Amy Michel, Dong Niu, Amitabh Sehgal, James Semler, Kai Wang, Esther Wilcox, Jack Winnick, Tao Wu, Jian Zhou
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INSTRUCTOR’S MANUAL to accompany ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES THIRD EDITION
RICHARD M. FELDER North Carolina State University RONALD W. ROUSSEAU Georgia Institute of Technology
JOHN WILEY & SONS New York Chichester Brisbane
Toronto
Singapore
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CONTENTS Notes to the Instructor
iv
Section/Problem Concordance
vi
Sample Assignment Schedule I
ix
Sample Assignment Schedule II
x
Sample Responses to a Creativity Exercise
xi
Transparency Masters
xvi
Compressibility charts Cox vapor pressure chart Psychrometric chart – SI units Psychrometric chart – American engineering units Enthalpy-concentration chart: H2SO4-H2O Enthalpy-concentration chart: NH3-H2O
xvii xxi xxii xxiii xxiv xxv
Problem Solutions Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 (Case Study 1) Chapter 13 (Case Study 2) Chapter 14 (Case Study 3)
2-1 3-1 4-1 5-1 6-1 7-1 8-1 9-1 10-1 11-1 12-1 13-1 14-1
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NOTES TO THE INSTRUCTOR Problem Assignments To aid in the structuring of the course, we have provided a section/problem concordance on pp. vii–ix and two sample assignment schedules on pp. x–xi. We believe there is far too much material in the textbook to attempt to cover in one semester or two quarters. The sample assignment schedules therefore cover only Chapters 1-9, and within those chapters some topics are omitted (e.g. liquid-liquid equilibrium and adsorption on solid surfaces in Chapter 6 and mechanical energy balances in Chapter 7). The missing sections can substitute for Chapters 2 and 3 in classes where the content of those chapters has been well covered in chemistry and/or physics courses), and Chapters 10 (computer flowsheeting) and 11 (transient balances) may be assigned for extra credit or covered in honors sections or subsequent courses in the curriculum. We will discuss the case studies in Chapters 12-14 separately. In the sample assignment schedules, we have designated a number of “bonus problems” which may be ignored, assigned for extra credit as add-ons to the regular assignments (those will be long assignments), or assigned for extra credit in lieu of some of the problems in the regular assignments. The bonus problems may be assigned as individual exercises or students may work on them in pairs. We have had good experience with the latter approach. Creativity Exercises The creativity exercises in the text are designed to stimulate divergent thinking and to induce the students to think about course material from different perspectives. We have used such exercises both as extra-credit assignments to individuals or pairs of students or as the foci of in-class brainstorming sessions. In all cases, we have found that they invariably lead to innovative, clever, and often amusing ideas; they give students who are by nature creative an opportunity to demonstrate their talent and they help other students develop creative problem-solving skills; and the students usually enjoy doing them. There are no “right answers” to such exercises, and so we have not included solutions in this manual. However, to provide an idea of the kind of things that students come up with, we have included on pp. xi–xv a collection of student responses to a creativity exercise given by one of the authors in a junior course on fluid dynamics. Transparency Masters Several of our colleagues have suggested that we include in the text enlarged versions of some of the figures, such as the psychometric charts, which are difficult to read in reduced format. We have chosen not to do so, since whether they are inserts or fold-outs such charts tend to be ripped off (one way or another) or otherwise lost. Instead, we have included in this manual, beginning on p. xvi, large versions of some of the most commonly used figures. These masters can be used to make transparencies for lectures; they can also be copied and distributed to the students for use in solving problem. Case Studies The case studies comprise Chapters 12 through 14 of the text. In them, we seek to (1) illustrate the development of complex chemical processes from basic principles, and to provide a broad process context for the text material; (2) raise questions that require students to think about topics strictly beyond the scope of the first course, and to seek out sources of information other than the text; (3) accustom the students to team project work. We do not organize the activities of case study
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teams, nor do we assign team leaders, although we suggest to the students that they do so. This is a risk, and sometimes it is necessary to step in and get a laggard group started. However, letting the teams shape their own working relationships and structure their own activities usually is an enlightening experience to the students. Problem Solutions The detailed solutions to 634 of the 635 chapter-end problems constitute the principal content of this manual. (The solution to the last problem of Chapter 10 is left as an exercise for the professor, or for anyone else who wants to do it.) With few exceptions, the conversion factors and physical property data needed to solve the endof-chapter problems are contained in the text. It may be presumed that conversion factors for which sources are not explicitly cited come from the front cover table; densities, latent heats, and critical constants come from Table B.1; heat capacity formulas come from Table B.2; enthalpies of combustion gases come from Tables B.8 and B.9; vapor pressures come from Table B.4 or (for water) Table B.3; and enthalpies, internal energies, and specific volumes of water at different temperatures and pressures come from Tables B.5-B.7. As the reader of the text may have discovered, we believe strongly in the systematic use of the flow charts in the solution of material and energy balance problems. When a student comes to us to ask for help with a problem, we first ask to see the labeled flow chart: no flow chart, no help. Other instructors we know demand fully labeled flow charts and solution outlines at the beginning of every problem solution, before any calculations are performed. In any case, we find that the students who can be persuaded to adopt this approach generally complete their assignments in reasonable periods of time and do well in the course; most of those who continue to resist it find themselves taking hours to do the homework problems, and do poorly in the course. Posting Problem Solutions: An Impassioned Plea It is common practice for instructors to photocopy solutions from the manual and to post them after the assignments are handed in, or, even worse, to distribute the solutions to the students. What happens then, of course, is that the solutions get into circulation and reincarnate with increasing frequency as student solutions. After one or two course offerings, the homework problems consequently lose much of their instructional value and become more exercises in stenography than engineering problem solving. In the stoichiometry course particularly, the concepts are relatively elementary: the main point is to teach the students to set up and solve problems. A great deal of classroom lecturing on concepts should therefore not be necessary, and a good deal of the class time can be spent in outlining problem solutions. The burden should be placed on the students to make sure they know how to do the problems: to ask about them in class, to make notes on solutions outlined on the board, and to fill in omitted calculations. Besides being pedagogically superior to solution-posting, this approach should cut down on the ease with which students can simply copy letter-perfect solutions instead of doing the work themselves. Errors A great deal of time and effort has been expended to make the solutions in this manual as free of errors as possible. Nevertheless, errors undoubtedly still exist. We will be grateful to any of our colleagues who send us corrections, no matter how major or minor they may be; we will provide an errata list on the text Web site (http://www2.ncsu.edu/unity/lockers/users/f/felder/public/EPCP.html) and make the corrections in subsequent printings of the text.
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SECTION/PROBLEM CONCORDANCE Key: i-Routine drill j-Application k-Longer or more challenging *-Computer solution required CHAPTER 2 Section
Problems
2.1-2.3 2.4 2.5 2.6 2.7
1-5 , 6-7 i j k j 8-10 , 11-12 , 13 , 14-15 i j j 16-17 , 18-19 , 20 * j j 21-28 , 29 *
i
j
i j k j k 30-31 , 32-37 , 38 , 39-41 , 42-44 *
CHAPTER 3 Section
Problems
3.1-3-2 3.3 3.4 3.5
1-2 , 3-8 , 9 , 10 , 11 , 12-13 i j k j 14-16 , 17-25 , 26 *, 27-31 i j k 32 , 33-46 , 47 i j k k 48 , 49-52 , 53 , 54 *
i
j
k
j
k
j
CHAPTER 4 Section
Problems
4.1-4.3a 4.3b-e 4.4 4.5 4.6a, b 4.6c 4.7a-e 4.7f 4.7g 4.8
1 , 2-5 j k k j k k 6-20 , 21 , 22 *, 22-25 , 26 , 27 * j k 28-30 , 31 j k 32-34 , 35-38 i k j 39-40 , 41 *, 42-45 k k 46 , 47-48 * j k 49-53 , 54-55 * j k 56-57 , 58-59 k k 60-62 , 63 * i j k j k j k k 64-65 , 66 , 67 *, 68-73 , 74-76 , 77-78 , 79 , 80 *
i
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CHAPTER 5 Section
Problems
5.1 5.2a,b 5.2c 5.3a-c 5.4a,b 5.4c
1 , 2-3 , 4 i j 5-6 , 7-15 j k j k j k 16-21 , 22-23 , 24-30 , 31-34 , 35-46 , 47-54 j k j k k 55-56 , 57 *, 58-60 , 61-62 , 63 * i j k j 64-65 , 66-69 , 70 , 71-73 j k j 74-77 , 78 ,79-83
i
j
k
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CHAPTER 6 Section 6.1 6.2, 6.3 6.4a 6.4b 6.4c 6.4d 6.5a,b 6.5c 6.6 6.7
Problems j k j 1-4 , 5 *, 6-8
j k k j k j k 6.9-29 , 30 , 31 *, 32-34 , 35-36 , 39-41 , 42 j 43-44 i j k 45-46 , 47-50 , 51 j k j k j k 52-57 , 58 *, 59 , 60 , 61-63 , 64 * j k j k 65-67 , 68-69 *, 70 , 71-73 i j k 74 , 75-80 , 81-83 i j k 84-85 , 86 , 87 j k j 88-91 , 92 , 93-97 j k 98-99 , 100-101
CHAPTER 7 Section
Problems
7.1, 7.2 7.3 7.4 7.5 7.6 7.7
1-2 , 3 , 4-6 , 7-8 i j 9 , 10-11 i j i j 12-13 , 14-17 , 18 , 19-23 j k 24-28 , 29 j k j k j k k 30-38 , 39 , 40-41 , 42-44 , 45-48 , 49-50 , 51 * j k 52-56 , 57-58
i
j
i
j
CHAPTER 8 Section
Problems
8.1-8.3b 8.3c 8.3d 8.3e 8.4a 8.4b 8.4c 8.4d 8.4e 8.5
1-4 , 5-16 i 17-18 j k j k 19-25 , 26 , 27-31 , 32 j k 33 , 34 * i j 35 , 36-41 i j 42 , 43-44 j k j k k k 45-53 , 54 , 55-56 , 57-65 , 66 *, 67-68 i j k j 69-71 , 72-73 , 74 , 75 i j k 76-77 , 78-79 , 80 i j k j k k j 81-82 , 83-87 , 88-90 , 91 , 92 , 93-94 *, 95-99
i
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CHAPTER 9 Section
Problems
9.1 9.2 9.3, 9.4 9.5a 9.5b 9.5c 9.6a 9.6b
1-2 , 3-4 i 5-6 I j 7-9 , 10 j k j k j k 11-17 , 18 *, 19-21 , 22-23 , 24 *, 25-30 k k j k 31-34 , 35 *, 36 , 37-38 j k 39-44 , 45-47 j k j k 48-50 , 51 , 52-56 , 57-61 j k k k 62-63 , 64-67 , 68 *, 69-70
i
j
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CHAPTER 10 Section
Problems j
10.11-4 , 5 10.2, 10.3
k k j k 6 , 7 *, 8-14 *
CHAPTER 11 Section
Problems j
k
j
j
j
k
j
11.1, 11.2 1-2 , 3 , 4 , 5 *, 6-9 , 10 , 11-14 , 15-19 j k 11.320-26 , 27-30
k
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SAMPLE ASSIGNMENT SCHEDULE I
Assignment
Read (Ch., Sect.)
Problems Due
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
— 1; 2-2.6 2:3, 6, 9, 12 2.7 2:14, 17, 19, 22 3-3.3 2:30, 33; 3:3, 6 3.4-3.5 3:13, 16, 22, 29 4-4.3a 3:32, 37, 40, 49 4.3b-4.3e 3:47; 4:2, 4, 7 4.4 4:10, 11, 14 4.5 4:18, 25, 28 — 4:26, 29 TEST THROUGH SECTION 4.3 4.6a-b 4:32, 39 4.6c, 4.7a-d 4:37, 43 4.7e-g 4:49, 53 4.8 4:56, 65 5-5.2b 4:69, 73; 5:3 5.2c-5.3 5:7, 12, 15, 19 5.4 5:25, 26, 59 6-6.2 5:35, 40, 66; 6:1 6.3 6:2, 9, 12, 15 6.4a,b 6:23, 33 6.4c 6:37, 47 6.5 6:53, 66, 74 7-7.3 6:36, 70 TEST THROUGH SECTION 6.4 7.4-7.5 7.1, 6, 9, 10 7.6 7:12, 18, 25 8-8.3b 7:24, 28, 30 8.3c-8.3e 7:33, 35; 8:2 — 7:45; 8:5, 6, 8 8.4a-c 8:15, 18, 25 8.4d-e 8:29, 36, 44 — 8:46, 49 8.5 8:51, 55, 73 — 8:61 — 8:79, 81, 86 TEST THROUGH SECTION 8.4d 9-9.4 8: 95, 98; 9:1 9.5a 9:7, 12 9.5b-c 9:14 9.6a-b 9:17 9.6c-d 9:32 — 9:48 — 9:54 — 9:63
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Bonus Problems (*Computer problem)
2.13 2:20*, 36-43 2:38-43; 3.9,11 2:40*; 3.26* 3:53*, 54* 4:22* 4:27* 4:28*, 31, 35-36 4:38, 41* 4:39, 43* 4:47*, 48* 4:54*, 55* 4:58,59 4:60-62, 63*, 67* 4:79-80*; 5:4*, 22-23 5: 31-34, 47-54 5:57*, 61-62, 63*; 6:5* 6:30, 31*, 35, 42 6:51, 58*, 60, 64* 6:68-69, 71-73 6:81, 83, 86-87 6:88-101 7:3, 17 7:29, 39, 42-44 7:49-50, 51* 7:52-58 8:12 8:26, 32, 34* 8:54, 57-59 8:62-68 8:74, 80 8:88-90 8:93-94* 9:9 9:18* 9:22-34 9:35*, 37-38, 45-47 9:51, 57-61, 64-65 9:67-70
SAMPLE ASSIGMENT SCHEDULE II
Assignment
Read (Ch., Sect.)
Problems Due
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
— 1; 2-2.6 2:4, 7, 8, 11 2.7 2:15, 16, 18, 23 3-3.3 2:31, 34; 3:4, 7 3.4-3.5 3:12, 17, 23, 28 4-4.3a 3:32, 39, 43, 50 4.3b-4.3e 3:51; 4:3, 4, 6 4.4 4:9, 12, 15 4.5 4:21, 23, 28 — 4:26, 30 TEST THROUGH SECTION 4.3 4.6a-b 4:33, 40 4.6c, 4.7a-d 4:34, 45 4.7e-g 4:50, 51 4.8 4:57, 64 5-5.2b 4:70, 71; 5:2 5.2c-5.3 5:8, 11, 13, 17 5.4 5:25, 29, 58 6-6.2 5:38, 42, 67; 6:1 6.3 6:6, 8, 10, 17 6.4a,b 6:27, 35 6.4c 6:39, 46 6.5 6:52, 65, 75 7-7.3 6:41, 70 TEST THROUGH SECTION 6.4 7.4-7.5 7:2, 7, 9, 11 7.6 7:13, 18, 22 8-8.3b 7:24, 28, 30 8.3c-8.3e 7:32, 37; 8.1 — 7:47; 8:5, 6, 9 8.4a-c 8:14, 17, 24 8.4d-e 8:30, 37, 43 — 8:45, 50 8.5 8:53, 56, 69 — 8:61 — 8:78, 83, 85 TEST THROUGH SECTION 8.4d 9-9.4 8:96, 97; 9:2 9.5a 9:7, 11 9.5b-c 9:15 9.6a-b 9:16 9.6c-d 9:33 — 9:50 — 9:55 — 9:66
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Bonus Problems (*Computer problem)
2.13 2:20*, 36-43 2:38-43; 3.9,11 2:40*; 3.26* 3:53*, 54* 4.22* 4:26, 27* 4: 28*, 31, 35-36 4:38, 41* 4:39, 43* 4:47*, 48* 4:54*, 55* 4:58, 59 4:60-62, 63*, 67* 4:79-80*; 5:4*, 22-23 5: 31-34, 47-54 5:57*, 61-61, 63*; 6.5* 6:30, 31*, 35-36, 42 6:51, 58*, 60, 64* 6:68-69, 71-73 6:81, 83, 86-87 6:88-101 7:3, 17 7:29, 39, 42-44 7:49-50, 51* 7:52-58 8:12 8:26, 32, 34* 8:54, 57-59 8:62-68 8:74, 80 8:88-90 8:93-94* 9:9 9:18* 9:22-34 9:35*, 37-38, 45-47 9:51, 57-61, 64-65 9:67-70
SAMPLE RESPONSES TO A CREATIVITY EXERCISE The exercise that follows was given to a junior class in fluid dynamics. The students were given a week, and were told to do the exercise either individually or in pairs. The grading system used is explained in the statement of the exercise. Thirty-one individuals and nine pairs submitted responses, for a total of 40 responses from 49 students. Some students found their way to Perry’s Handbook and took ideas from there, which was perfectly acceptable; many were more inventive, and submitted a wide variety of clever, ingenious, and humorous responses. The average number of suggested flow measurement techniques was 26; the high was 53, and the low was 5. A summary of the collected responses with duplicates eliminated follows the exercise statement. Exercise You are faced with the task of measuring the volumetric flow rate of a liquid in a large pipeline. The liquid is in turbulent flow, and a flat velocity profile may be assumed (so that you need only measure the fluid velocity to determine the volumetric flow rate). The line is not equipped with a built-in flowmeter; however, there are taps to permit the injection or suspension of devices or substances and the withdrawal of fluid samples. The pipeline is glass and the liquid is clear. Assume that any device you want to insert in the pipe can be made leakproof if necessary, and that any technique you propose can be calibrated against known flow rates of the fluid. Come up with as many ways as you can think of to perform the measurement that might have a chance of working. (Example: insert a small salmon in the pipe, suspend a lure irresistible to salmon upstream of the insertion point, and time how long it takes the fish to traverse a measured section of the pipe.) You will get 1 point for every 5 techniques you think of (no fractional points awarded), up to a maximum of 10 points. Note, however: The techniques must be substantially different from one another to count. Giving me a pitot tube with 10 different manometer fluids, for example, will get you nowhere. Responses 1. Pitot tube. 2. Hot-wire or hot-film anemometer 3. Pass effluent through a venturi meter or orifice meter or nozzle meter or rotameter or ... (no credit for simply naming the meter if it can’t be easily inserted in the pipeline). 4. Pass effluent into a weir, measure height in notch. 5. Inject dye, measure time for it to traverse a known length. 6. Insert a solid object (such as a balloon, a bucket, a cork, a marble, a bar of Ivory soap, or the 311 book), measure time for it to traverse a known length (or travel alongside it on a bicycle or moped or pickup truck and note your speed, or attach it to a string on a spool and measure the rate of rotation of the spool). 7. Insert a series of solid objects, measure rate at which they pass a point (or frequency of collisions with the pipe wall, or rate of collection on a filter).
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8. Inject dye at fixed rate, shine light on the pipe, measure light absorbance downstream (or angle of refraction or turbidity, or put a sunbather under the pipe and measure his rate of tanning). 9. Measure the energy being consumed by the pump being used to move the fluid. 10. Put a magnet in the flow, measure the magnetic force required to hold it in place (or measure its velocity along the wall, or have two external magnetic switches triggered by its passage and time the interval between events, or measure the rate of motion of a compass needle as the magnet passes). 11. Collect effluent (or a sidestream), measure amount (volume, mass) collected in a known time interval (or the rate at which the level in the container rises, or the time required to fill a known volume or to saturate a sponge, or to water a plant or wash a pulp sample, or to saturate a plot of ground in Ethiopia where they really need it). 12. Direct effluent into a container of salt, see how much dissolves. 13. Direct effluent against a raft in a pond, measure its velocity. 14. Discharge effluent horizontally, letting it fall through a known height, and measure its horizontal displacement. 15. Discharge effluent horizontally, and measure the force it exerts on a plate. 16. Discharge effluent (or a sidestream) upward, measure height of fountain (or suspend a ping pong ball at the top, and measure its height) 17. Discharge effluent downward from a flexible hose, and measure height of nozzle above the ground. 18. Let fluid fill a balloon (or a water bed), measure time required for explosion, or volume increase in a known time interval. 19. Insert a U-tube at each of two points in the line, use as a manometer (either straight pipe between points or insert an obstacle to flow, like an orifice or a solid object). 20. Insert paddle wheel (or a water wheel at the outlet), measure rotational speed. 21. Insert propeller, measure rotational speed by counting or automatically. 22. Insert turbine-generator, measure work output (or intensity of light attached to generator). 23. Suspend solid object on a string, measure angle made by string with vertical (or horizontal displacement of object, or rotation of a lever arm, or angle at which your hand is bent back). 24. Drop in an object denser than the fluid, measure horizontal distance traversed before hitting the bottom of the channel. 25. Inject from below an object lighter than the fluid (e.g. a bubble), measure horizontal distance traversed before hitting the top of the channel.
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26. Inject from below an object heavier than the fluid, measure its horizontal displacement (or follow its trajectory using stop-motion photography). 27. Put a flexible fiber (or membrane or easily deformed plate) in the path of the flow, measure its distension in flow direction, or thickness at which flow is sufficient to break it. 28. Pluck a guitar string in the flow, time its period of vibration. 29. Attach tape to the wall, time its unraveling. 30. Feed effluent into a centrifugal pump or a lobbed-impeller flowmeter, measure rotational speed. 31. Measure height of fluid in a vertical standpipe coming from the top of the pipe. 32. Determine time required for effluent to sink a ship (or to flood out the football coach’s house, hopefully with some of his players in it). 33. Determine time required for effluent to float a duck out of a well of known depth (or to float an object of known weight and displacement). 34. Insert a solid object (e.g. a snowball, or a tootsie pop, or Alka Seltzer, or the Wicked Witch of the West), determine time required to dissolve it (or wear it away, or wash the paint off it). 35. Insert an absorbing object, measure its rate of fluid uptake. 36. Insert solid objects of different sizes, find the one such that the drag force is just sufficient to initiate motion (or measure rate at which a given object is dragged along the pipe). 37. Measure vibration intensity or amplitude of tube (or noise level, or sound of a bell clapper), either naturally occurring or after the pipe is struck. 38. Insert an iron bar, measure rate of corrosion. 39. Propel a bullet (or an arrow, or a torpedo, or Nolan Ryan’s fastball) into the pipe outlet, measure distance it travels before stopping. 40. Determine velocity of immersed submarine moving against (or with) the flow (or Mark Spitz, or a trout approaching a lure, or a seal approaching food, or a squirrel approaching an acorn, or a hungry dog approaching a dead rabbit, or a snake approaching a mouse, or a sailor approaching a mermaid, or a horny male frog approaching erotic pictures of female dancing frogs, or a 311 student after this test approaching free beer). 41. Measure vibrational speed of a trout’s tail swimming against the flow and remaining stationary (or the rate of flapping of a piece of cloth or the rate of wobble of a nutating disk or the magnitude of noise generated by “chatterbox” lure). 42. Insert balloon (or piston-fitted cylinder with piston facing upstream, or closed tube with flexible diaphragm covering opening), measure final gas pressure (or rate of pressure increase or rate of motion of piston or intensity of whistle if piston drives gas through it).
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43. Insert solid object, measure force required to hold it still (or extension or compression of a spring or elastic band, or put the object against razor blade and see how long it takes for the blade to split it). 44. Insert a solid object, determine distance required for downstream wake to disappear. 45. Put in plug, measure force required to hold it in (or distance it travels when it is ejected). 46. Measure the shear force on the pipe wall (with or without a bend in the line), or the extension of the pipe length due to shear. 47. Measure the rate of heat generation or temperature rise due to friction in the pipeline. 48. Measure rate at which air is drawn through a Buchner funnel (or pitch or intensity at which it is drawn through a whistle) by the suction created by the flowing fluid. 49. Add heat, measure temperature rise (or rate of evaporation). 50. Insert a hot object, measure its rate of cooling (or a cold object, and measure its rate of heating). 51. Cool, measure temperature drop (or rate of freezing). 52. Pass effluent through a heat exchanger, measure rate of heat transfer. 53. Add an acid or base at a known rate, measure pH downstream or determine amount needed to change litmus paper color (or add salt and measure change in electrical conductivity, or add a radioisotope and measure change in activity, or add a phosphorescent substance and measure luminescence, or add any chemical and measure its concentration by any means). 54. Add sugar at a known rate, measure rate of formation of rock candy downstream. 55. Add a second liquid of different density, measure resulting density change (or add an immiscible liquid, measure its rate of passage). 56. Add a reactant, determine amount needed to react completely with the fluid (or with another reactant on a permeable membrane in the flow channel, or inject chlorine ions and measure the rate of electroplating on a silver electrode). 57. Get a technician to drink the effluent, measure his weight gain after a fixed time (or the time required for his mouth to fill up, or the time required to drown a rat). 58. Add alcohol (or poison, or salt, or Kool-Aid mix) to the fluid at a known rate, have a technician drink it (or do it yourself), and determine the time required to feel the effects. 59. Immerse pipe outlet in water, find the depth at which the hydrostatic head is just sufficient to stop the flow. 60. Insert air tube facing upstream, determine pressure needed to initiate bubbling. 61. Place pipe in wind tunnel, find wind velocity just adequate to stop the flow.
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62. If pipe is only partially filled, put in sailboat, measure wind force needed to hold it stationary. 63. Put another pipe against outlet, find flow in second pipe that just neutralizes unknown flow. 64. Send sound wave through, time passage over known distance (or use Doppler meter, or time passage of an electrical impulse or a light wave). 65. Put bacteria in line, determine reproduction rate. 66. Put algae in pipe, measure change in COD. 67. Put a spawning fish in the line, measure how far the eggs travel in a given time interval. 68. Measure how long it takes for the effluent to put out a fire of a given size. 69. Pass the fluid spirally into a funnel, measure how long it takes for a drop of dye to disappear. 70. If the fluid is combustible, burn it in a combustion engine and measure the rate of power output. 71. Determine how long you can hold your breath, then jump in and see how far you travel before you have to breathe (or see if an animal can make it out of the pipe before drowning). 72. Add effluent to bubble bath, measure extent of generation of bubbles. 73. Insert a fish with a monitor in its heart, time how long it takes him to die. (Must kill a lot of fish to calibrate-don’t tell “Save the Whales.”) 74. Count rate of passage of molecules. 75. Insert a monkey who can insert pegs in holes at a known rate, and count the number of pegs inserted over a known distance. 76. Install a Japanese flowmeter equipped with lots of flashing lights. 77. Correlate the velocity with the rate at which the student pulls out his hair during the experiment. 78. Hire someone to do it. 79. Break into the pipeline company’s office and steal the flow rate records. 80. Look it up in the Enquirer flow rate tables. 81. Fill a balloon, throw it at your boss, and correlate his anger with the flow rate. (?) 82. Ask your local psychic.
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TRANSPARENCY MASTERS The following pages contain oversized renderings of illustrations taken from the text. The illustration numbers are listed below with their text page numbers. You are granted permission to have these illustrations reproduced as transparencies for your own use in conjunction with the textbook, Felder and Rousseau: ELEMENTARY PRINCIPLES OF CHEMICAL PROCESSES, 3rd Edition, John Wiley & Sons. Resale is expressly prohibited. Transparencies may easily be prepared using either thermal copy (ThermoFax) or electrostatic copy (Xerox) machines. See the operating manual for your particular copy machine. Transparency film can be purchased from your usual copy paper supplier. Other illustrations in the book may also serve as transparency masters. For best results, it may be necessary to enlarge the illustration to fill the sheet of copy film. Many electrostatic copiers have this capability. Figure Description Compressibility charts
Cox vapor pressure chart Psychrometric chart – SI units Psychrometric chart – Am. Engr. units Enthalpy conc. chart – H2SO4/H2O Enthalpy conc. chart – NH3/H2O
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Figure Number
Text Page
5.4-1 5.4-2 5.4-3 5.4-4 6.1-4 8.4-1 8.4-2 8.5-1 8.5-2
208 209 210 211 247 382 383 396 400
CHAPTER TWO 2.1 (a) (b) (c)
2.2 (a)
3 wk
7d
24 h
3600 s 1000 ms
1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s 3.2808
ft
554 m 4 1d 1h d kg 24 h 60 min 760 mi h
1
1 h
u 10 9 ms 18144 . 25.98 mi / h 26.0 mi / h
1 kg 10 8 cm 4 1000 g 1 m 4
m
1 h
0.0006214 mi 3600 s
340 m / s
1 m3 35.3145 ft 3
(b)
921 kg 2.20462 lb m m3 1 kg
(c)
5.37 u 10 3 kJ 1 min 1000 J min 60 s 1 kJ
3.85 u 10 4 cm 4 / min g
57.5 lb m / ft 3
1.34 u 10 -3 hp 119.93 hp 120 hp 1 J/s
2.3 Assume that a golf ball occupies the space equivalent to a 2 in u 2 in u 2 in cube. For a classroom with dimensions 40 ft u 40 ft u 40 ft : 40 u 40 u 40 ft 3 (12) 3 in 3 1 ball n balls 6.48 u 10 6 | 7 million balls 3 3 3 2 in ft The estimate could vary by an order of magnitude or more, depending on the assumptions made. 2.4 4.3 light yr 365 d 24 h 1 yr 1 d
3600 s 1.86 u 105 mi 3.2808 ft 1 step 1 h 1 s 0.0006214 mi 2 ft
2.5 Distance from the earth to the moon = 238857 miles 238857 mi
1
m
1 report
0.0006214 mi
0.001 m
4 u 1011 reports
2.6 19 km 1000 m 0.0006214 mi 1000 L 1 L 1 km 1 m 264.17 gal Calculate the total cost to travel x miles. Total Cost
Total Cost
American
European
$14,500
$21,700
Equate the two costs x
44.7 mi / gal
$1.25 1 gal x (mi) 14,500 0.04464 x gal 28 mi x (mi) $1.25 1 gal gal 44.7 mi
4.3 u 10 5 miles
2-1
21,700 0.02796 x
7 u 1016 steps
2.7 10 6 cm 3 220.83 imp. gal
5320 imp. gal 14 h 365 d plane h 1 d 1 yr ton kerosene u 10 5 1188 . plane yr
0.965 g 1 cm 3
1 kg 1 ton 1000 g 1000 kg
4.02 u 10 9 ton crude oil 1 ton kerosene plane yr 5 u 10 ton kerosene yr 7 ton crude oil 1188 . 4834 planes 5000 planes
2.8 (a) (b) (c)
2.9
2.10 2.11
32.1714 ft / s 2
25.0 lb m 25 N
1 9.8066 m / s 2
10 ton
1 lb m 5 u 10 -4 ton
50 u 15 u 2 m 3
500 lb m
(a) mdisplaced fluid
1 lb f 32.1714 lb m ft / s 2
1 kg m / s 2 1N
2.55 kg 2.6 kg 980.66 cm / s 2
1000 g 2.20462 lb m
35.3145 ft 3 1 m3
25.0 lb f
85.3 lb m 1 ft 3
1 dyne 1 g cm / s 2
1 lb f 32.174 ft 2 32.174 lb m / ft s 2 1 s
FG IJ FG 1 IJ | 25 m H K H 10K
1 m3 1 | 5 u 10 2 2 11.5 kg
1 kg 2.20462 lb m
mcylinder U f V f
9 u 10 9 dynes
U cVc U f hSr 2
4.5 u 10 6 lb f
3
U c HSr 2
Uc
Ufh
(30 cm 14.1 cm)(100 . g / cm 3 ) Uc 0.53 g / cm 3 H 30 cm U c H (30 cm)(0.53 g / cm 3 ) (b) U f 171 . g / cm 3 h (30 cm - 20.7 cm) 2.12
Vs
SR 2 H ; Vf 3
R SR 2 H Sr 2 h ; H 3 3
U fVf
U sVs U f
Uf
Us
H H
h3 H2
2
2
2
3
H3 H 3 h3
h
3
2
H
r
2
Us
s
2
Us
Uf h
R h H
FG IJ SR FG H h IJ H K 3H HK h I SR F SR H H U J G 3 H 3 H K
SR 2 H Sh Rh 3 3 H
Vf
r r h
H
Us
R
1
F hI 1 G J H HK 2-2
3
Uf
2.13
Say h m
depth of liquid y y= 1 dA y= 1– h x
2
A(m )
z
h
1m
x = 1– y
2
1 y 2
dA dy
2 1 y 2 dx
dx
1 y 2
d i
Am
2
E d i
A m2
b g
W N
z z 1
1
dA 2
1 h
1 y 2 dy
1 h
Table of integrals, or trigonometric substitution
S 2 1 h 1 1 h sin 1 1 h 2 1 h 4 m u A( m 2 ) 0.879 g 10 6 cm 2 1 kg 9.81 N 3.45 u 10 4 A cm 3 1 m3 10 3 g kg , y 1 y 2 sin 1 y
1
E Substitute for A LS W b N g 3.45 u 10 M b1 hg N2 4
b g
b g
b g
g g0
b g
1 1 h
2
b gOPQ
sin 1 1 h
32.174 lb m ft / s 2 1 slug = 32.174 lb m 1 lb f 1 poundal = 1 lb m ft / s 2 32.174 (a) (i) On the earth: 175 lb m 1 slug M 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W 5.63 u 10 3 poundals 2 s 1 lb m ft / s 2 (ii) On the moon 175 lb m 1 slug M 5.44 slugs 32.174 lb m 175 lb m 32.174 ft 1 poundal W 938 poundals 2 6 s 1 lb m ft / s 2
2.14 1 lb f
1 slug ft / s 2
2-3
2.14 (cont’d) ma a
F/m
0.135 m / s
2
(b) F
2.15 (a) F
25.0 slugs
ma 1 fern = (1 bung)(32.174 ft / s 2 )
FG 1IJ H 6K
1 slug 32.174 lb m
1m 3.2808 ft
5.3623 bung ft / s 2
1 fern 5.3623 bung ft / s 2
(b) On the moon: W On the earth: W 2.16 (a) | (3)(9)
3 bung 32.174 ft 1 fern 2 6 s 5.3623 bung ft / s 2 (3)(32.174) / 5.3623 = 18 fern
(b)
27
(2.7)(8.632)
23
(c) | 2 125 127
(d)
2.365 125.2 127.5 2.17 R |
1 lb m ft / s 2 1 poundal
355 poundals
3 fern
40 u 10 4 5 u 8 u 10 4 | | 1 u 10 4 45 5u 9 4 (3.600 u 10 ) / 45 9 u 10 5 |
| 50 u 10 3 1 u 10 3 | 49 u 10 3 | 5 u 10 4 4.753 u 10 4 9 u 10 2
5 u 10 4
(7 u 10 1 )(3 u 10 5 )(6)(5 u 10 4 ) | 42 u 10 2 | 4 u 10 3 6 (3)(5 u 10 )
Rexact
3812.5 3800 38 . u 10 3
2.18 (a) A: R
731 . 72.4
0.7 o C
X
72.4 731 . 72.6 72.8 73.0 5
s
(72.4 72.8) 2 (731 . 72.8) 2 (72.6 72.8) 2 (72.8 72.8) 2 (73.0 72.8) 2 51
72.8 o C
0.3o C B: R 1031 . 97.3 58 . oC X
97.3 1014 . 98.7 1031 . 100.4 100.2 o C 5
s
(97.3 100.2) 2 (1014 . 100.2) 2 (98.7 100.2) 2 (1031 . 100.2) 2 (100.4 100.2) 2 51 2.3o C
(b) Thermocouple B exhibits a higher degree of scatter and is also more accurate.
2-4
2.19 (a)
12
12
¦
¦ ( X 735. )
Xi
i 1
X
s
C min=
735 . 12 X 2 s 735 . 2(12 . )
711 .
C max=
X 2 s 735 . 2(12 . )
75.9
2
i 1
12 .
12 1
(b) Joanne is more likely to be the statistician, because she wants to make the control limits stricter. (c) Inadequate cleaning between batches, impurities in raw materials, variations in reactor temperature (failure of reactor control system), problems with the color measurement system, operator carelessness 2.20 (a), (b) 1 2 (a) Run 134 131 X Mean(X) 131.9 Stdev(X) 2.2 127.5 Min 136.4 Max (b) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14
X 128 131 133 130 133 129 133 135 137 133 136 138 135 139
Min 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5 127.5
3 129
4 5 6 7 8 9 10 11 12 13 14 15 133 135 131 134 130 131 136 129 130 133 130 133
Mean 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9 131.9
Max 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4 136.4
140 138 136 134 132 130 128 126 0
5
10
15
(c) Beginning with Run 11, the process has been near or well over the upper quality assurance limit. An overhaul would have been reasonable after Run 12.
2.21 (a) Q'
2.36 u 10 4 kg m 2 h
2.10462 lb 3.2808 2 ft 2 kg m2
(2 u 10 4 )(2)(9) | 12 u 10 ( 4 3) | 12 . u 10 6 lb ft 2 / s 36 u 10 3 2 . u 10 lb ft / s 0.00000148 lb ft 2 / s = 148
(b) Q' approximate | Q' exact
1 h 3600 s
2-5
CpP
2.22 N Pr
k
0.583 J / g o C 1936 lb m 0.286 W / m o C ft h
1 h 3.2808 ft 1000 g 3600 s m 2.20462 lb m
(6 u 10 1 )(2 u 10 3 )(3 u 10 3 ) 3 u 10 3 . u 10 3 . The calculator solution is 163 . u 10 3 | | 15 2 (3 u 10 1 )(4 u 10 3 )(2)
N Pr | 2.23 Re
DuU P
Re |
(5 u 10 1 )(2)(8 u 10 1 )(10 6 ) 5 u 101 ( 3) | | 2 u 10 4 the flow is turbulent 3 4 3 (3)(4 u 10)(10 )(4 u 10 )
0.48 ft 1 m 2.067 in 1 m 3 s 3.2808 ft 0.43 u 10 kg / m s 39.37 in
0.805 g cm 3
1 kg 10 6 cm 3 1000 g 1 m 3
2.24 (a) kgd p y
F P IJ FG d uU IJ 2.00 0.600G D H UD K H P K L 100 . u 10 N s / m 2.00 0.600M . kg / m )(100 . u 10 m N (100 1/ 2
1/ 3
p
5
5
3
44.426
k g (0.00500 m) (0100 . ) 100 . u 10 5 m 2 / s
OP LM (0.00500 m)(10.0 m / s)(100 . kg / m ) O PQ / s) Q N (100 . u 10 N s / m ) 1/ 3
2
2
3
5
44.426 k g
1/ 2
2
0.888 m / s
(b) The diameter of the particles is not uniform, the conditions of the system used to model the equation may differ significantly from the conditions in the reactor (out of the range of empirical data), all of the other variables are subject to measurement or estimation error. (c) dp (m) 0.005 0.010 0.005 0.005 0.005
y 0.1 0.1 0.1 0.1 0.1
D (m2/s) 1.00E-05 1.00E-05 2.00E-05 1.00E-05 1.00E-05
P (N-s/m2) U (kg/m3) u (m/s) 1.00E-05 1 10 1.00E-05 1 10 1.00E-05 1 10 2.00E-05 1 10 1.00E-05 1 20
kg 0.889 0.620 1.427 0.796 1.240
2.25 (a) 200 crystals / min mm; 10 crystals / min mm 2
(b) r
200 crystals 0.050 in 25.4 mm 10 crystals min mm in min mm 2 238 crystals / min
b g
(c) D mm
b g
D c in
60r c
(25.4) 2 mm 2 in 2
238 crystals 1 min 4.0 crystals / s 60 s min
FG H
25.4 mm crystals 25.4 D c ; r min 1 in
b
0.050 2 in 2
g b
200 25.4 D c 10 25.4 D c
g
2
2-6
IJ K
rc
crystals
60 s
s
1 min
b g
r c 84.7 D c 108 D c
2
60r c
2.26 (a) 70.5 lb m / ft 3 ; 8.27 u 10 -7 in 2 / lb f (b) U
L8.27 u 10 / ft ) exp M N 3
(70.5 lb m
35.3145 ft 3
70.57 lb m ft
FG lb IJ H ft K F lb IJ PG H in K
(c) U
3
m
Uc
m 3
f 2
g cm 3
P'
62.43U c
N m2
1
7
in 2 lb f
9 u 10 6 N m2
m3
1000 g
6
3
10 cm
3
2.20462 lb m
1 lb m 28,317 cm 3 g 1 ft 3 453593 .
d
OP Q
113 . g / cm 3
62.43U c
12 m2 39.37 2 in 2
0.2248 lb f 1N
14.696 lb f / in 2 u 10 6 N / m 2 101325 .
145 . u 10 4 P'
id
i
d
P' 9.00 u 10 6 N / m 2 U ' 113 . exp[(120 . u 10 10 )(9.00 u 10 6 )] 113 . g / cm 3 cm d i d i 28,317 1728 in 16.39V ' expb3600t c g V '
2.27 (a) V cm 3
i
70.5 exp 8.27 u 10 7 145 . u 10 4 P ' U c 113 . exp 120 . u 10 10 P '
V ' in 3
3
3
b g 3600t cbhrg 0.06102 expb3600t cg 16.39V ' ; t s
(b) The t in the exponent has a coefficient of s-1. 2.28 (a) 3.00 mol / L, 2.00 min -1 0 C 3.00 exp[(-2.00)(0)] = 3.00 mol / L t = 1 C 3.00 exp[(-2.00)(1)] = 0.406 mol / L 0.406 3.00 Cint For t=0.6 min: (0.6 0) 3.00 14 . mol / L 1 0 Cexact 3.00 exp[(-2.00)(0.6)] = 0.9 mol / L
(b) t
For C=0.10 mol/L:
1 0 (010 . 3.00) 0 112 . min 0.406 3 1 C 1 0.10 =ln = - ln = 1.70 min 2.00 3.00 2 3.00
t int t exact
(c) 3.5 C exact vs. t
3 C (mol/L)
2.5 2
(t=0.6, C =1.4)
1.5 1
(t=1.12, C=0.10)
0.5 0 0
1
2
t (min)
2-7
p*
2.29 (a) (b)
60 20 (185 166.2) 20 42 mm Hg 199.8 166.2
c
MAIN PROGRAM FOR PROBLEM 2.29 IMPLICIT REAL*4(A–H, 0–Z) DIMENSION TD(6), PD(6) DO 1 I = 1, 6 READ (5, *) TD(I), PD(I) 1 CONTINUE WRITE (5, 902) 902 FORMAT (‘0’, 5X, ‘TEMPERATURE VAPOR PRESSURE’/6X, * ‘ (C) (MM HG)’/) DO 2 I = 0, 115, 5 T = 100 + I CALL VAP (T, P, TD, PD) WRITE (6, 903) T, P 903 FORMAT (10X, F5.1, 10X, F5.1) 2 CONTINUE END SUBROUTINE VAP (T, P, TD, PD) DIMENSION TD(6), PD(6) I=1 1 IF (TD(I).LE.T.AND.T.LT.TD(I + 1)) GO TO 2 I=I+1 IF (I.EQ.6) STOP GO TO 1 2 P = PD(I) + (T – TD(I))/(TD(I + 1) – TD(I)) * (PD(I + 1) – PD(I)) RETURN END DATA OUTPUT 98.5 1.0 TEMPERATURE VAPOR PRESSURE 131.8 5.0 (C) (MM HG) 100.0 1.2 215.5 100.0 105.0 1.8 215.0 98.7
2.30 (b) ln y b ln a (c) ln y b ln a
ln a bx y ae bx (ln y 2 ln y1 ) / ( x 2 x1 ) ln y bx
(ln 2 ln 1) / (1 2)
ln 2 0.63(1) a
4.00 y
0.693
4.00e 0.693 x
ln a b ln x y ax b (ln y 2 ln y1 ) / (ln x 2 ln x1 ) (ln 2 ln 1) / (ln 1 ln 2) ln y b ln x ln 2 ( 1) ln(1) a 2 y 2 / x
1
(d) ln( xy ) ln a b( y / x ) xy ae by / x y (a / x )e by / x [can' t get y f ( x )] b [ln( xy ) 2 ln( xy ) 1 ] / [( y / x ) 2 ( y / x ) 1 ] (ln 807.0 ln 40.2) / ( 2.0 10 . ) ln a
ln( xy ) b( y / x )
ln 807.0 3 ln( 2.0) a
2-8
2 xy
2e 3 y / x y
3
(2 / x )e 3 y / x
2.30 (cont’d) (e) ln( y 2 / x )
a ( x 2) b y [ax ( x 2) b ]1/ 2
ln a b ln( x 2) y 2 / x
b [ln( y 2 / x ) 2 ln( y 2 / x ) 1 ] / [ln( x 2) 2 ln( x 2) 1 ] (ln 807.0 ln 40.2) / (ln 2.0 ln 10 . ) 4.33 ln( y 2 / x ) b( x 2)
ln a
y2 / x
ln 807.0 4.33 ln(2.0) a
40.2( x 2) 4.33 y
6.34 x 1/ 2 ( x 2) 2.165
2.31 (b) Plot y 2 vs. x 3 on rectangular axes. Slope
(c)
1 ln( y 3)
(d)
1 ( y 1) 2
1 a b b
x Plot
a ( x 3) 3 Plot
40.2
1 vs. ln( y 3)
m, Intcpt
n
x [rect. axes], slope =
1 a , intercept = b b
1 vs. ( x 3) 3 [rect. axes], slope = a , intercept = 0 ( y 1) 2
OR 2 ln( y 1) ln a 3 ln( x 3) Plot ln( y 1) vs. ln( x 3) [rect.] or (y + 1) vs. (x - 3) [log] 3 ln a slope = , intercept = 2 2 (e) ln y
a x b
Plot ln y vs. (f) log10 ( xy )
x [rect.] or y vs.
x [semilog ], slope = a, intercept = b
a( x 2 y 2 ) b
Plot log10 ( xy ) vs. ( x 2 y 2 ) [rect.] slope = a, intercept = b (g)
1 y OR
ax 1 y
b x x y
ax
ax 2 b Plot
b 1 x xy
a
x vs. x 2 [rect.], slope = a , intercept = b y
b 1 1 Plot vs. 2 [rect.] , slope = b, intercept = a 2 xy x x
2-9
2.32 (a) A plot of y vs. R is a line through ( R 5 , y
0.011 ) and ( R 80 , y
0169 . ).
0.18 0.16 0.14 0.12 y
0.1 0.08 0.06 0.04 0.02 0 0
20
40
60
80
100
R
y
aRb
a b
U| V| W
0.011 . 0169 2.11 u 10 3 80 5 y 0.011 2.11 u 10 3 5 4.50 u 10 4
ib g
d
2.11 u 10 3 R 4.50 u 10 4
d2.11 u 10 ib43g 4.50 u 10 0.092 kg H O kg b1200 kg hgb0.092 kg H O kgg 110 kg H O h
(b) R
3
43 y
4
2
2.33 (a) ln T b ln a (b) T T T T
2
ln a b ln I T aI b (ln T2 ln T1 ) / (ln I 2 ln I 1 ) ln T b ln I
2
(ln 120 ln 210) / (ln 40 ln 25)
ln 210 ( 119 . ) ln( 25) a
9677.6 T
119 .
9677.6I 1.19
b9677.6 / T g 85 C I b9677.6 / 85g 535 . L/s 175 C I b9677.6 / 175g 29.1 L / s 290 C I b9677.6 / 290g 19.0 L / s 1.19
9677.6I 1.19 I
1.19
o
o
1.19
o
1.19
(c) The estimate for T=175°C is probably closest to the real value, because the value of temperature is in the range of the data originally taken to fit the line. The value of T=290°C is probably the least likely to be correct, because it is farthest away from the date range.
2-10
ln ((CA-CAe)/(CA0-CAe))
2.34 (a) Yes, because when ln[(C A C Ae ) / (C A0 C Ae )] is plotted vs. t in rectangular coordinates, the plot is a straight line. 0
50
100
150
200
0 -0.5 -1 -1.5 -2 t (m in)
Slope = -0.0093 k = 9.3 u 10-3 min 1 (b) ln[(C A C Ae ) / (C A0 C Ae )] kt C A 0.0495)e ( 9.3u10 (01823 .
CA
3
)(120)
(C A 0 C Ae )e kt C Ae
0.0495 = 9.300 u 10-2 g / L
9.300 u 10-2 g 30.5 gal 23.317 L 8.8 g L 7.4805 gal
C = m / V m = CV
2.35 (a) ft 3 and h -2 , respectively (b) ln(V) vs. t2 in rectangular coordinates, slope=2 and intercept= ln(353 . u 10 2 ) ; or V(logarithmic axis) vs. t2 in semilog coordinates, slope=2, intercept= 353 . u 10 2 3 3 7 2 (c) V ( m ) 100 . u 10 exp(15 . u 10 t ) 2.36 PV k
C / V k ln P
C P
ln C k lnV
8.5
lnP
8 7.5 7 6.5 6 2.5
3
lnP = -1.573(lnV ) + 12.736
k
slope
3.5 lnV
( 1573 . ) 1573 . (dimensionless)
Intercept = ln C 12.736 C G GL G0 G
G G 1 0 m G GL KLC
e12 .736
3.40 u 105 mm Hg cm4.719
K L C m ln
G0 G G GL
ln K L m ln C
ln ( G 0 -G )/(G -G L )= 2 .4 8 3 5 ln C - 1 0 .0 4 5
3 ln(G 0-G)/(G-G L )
2.37 (a)
4
2 1 0 -1 3 .5
4
4 .5
5
ln C
2-11
5 .5
2.37 (cont’d) m slope 2.483 (dimensionless) 10.045 K L
Intercept = ln K L
4.340 u 10 5 ppm-2.483
G 180 . u 103 u 10 3 . 4.340 u 105 (475) 2.483 G 1806 3.00 u 10 3 G C=475 ppm is well beyond the range of the data. 475
(b) C
2.38 (a) For runs 2, 3 and 4: Z aV b p c ln Z ln a b lnV c ln p b 0.68 ln(35 . ) ln a b ln(102 . ) c ln( 9.1) c 146 . ln(2.58) ln a b ln(102 . ) c ln(112 . ) ln(3.72) ln a b ln(175 . ) c ln(112 . ) a = 86.7 volts kPa 1.46 / (L / s) 0.678 . Slope=b, Intercept= ln a c ln p (b) When P is constant (runs 1 to 4), plot ln Z vs. lnV 2
lnZ
1.5 1 0.5 0 -1
-0.5
0
lnZ = 0.5199lnV + 1.0035
0.5
1
1.5
b
lnV
slope
0.52
Intercept = lna c ln P 10035 . When V is constant (runs 5 to 7), plot lnZ vs. lnP. Slope=c, Intercept= ln a c lnV 2
lnZ
1.5 1 0.5 0 1.5
1.7
lnZ = -0.9972lnP + 3.4551
1.9
2.1
2.3
c
lnP
slope
0.997 10 .
Intercept = lna b lnV
3.4551
Z
Plot Z vs V b P c . Slope=a (no intercept) 7 6 5 4 3 2 1 0.05
Z = 31.096VbPc
0.1
0.15
0.2
Vb P c
a
slope
311 . volt kPa / (L / s) .52
The results in part (b) are more reliable, because more data were used to obtain them.
2-12
2.39 (a) sxy sxx
a b y (b) a
¦x y
1 n
¦x
1 n
sx
n
1 n
[(0.4)(0.3) (2.1)(19 . ) (31 . )(3.2)] / 3 4.677
i i
i 1 n
2 i
(0.32 19 . 2 3.2 2 ) / 3 4.647
i 1 n
¦
(0.3 19 . 3.2) / 3 18 . ; sy
xi
i 1
sxy sx s y
b g
sxx sx
4.677 (18 . )(1.867) 4.647 (18 . )2
2
sxx s y sxy sx
b g
sxx sx
1 n
n
¦y
i
0.936
(4.647)(1867 . ) (4.677)(18 . ) 2 4.647 (18 . )
2
(0.4 2.1 31 . ) / 3 1867 .
i 1
0182 .
0.936 x 0182 . sxy
4.677 4.647
sxx
x y 10065 10065 . .
4
y
3
y = 0.936x + 0.182
2
y = 1.0065x
1 0 0
1
2
3
4
x
2.40 (a) 1/C vs. t. Slope= b, intercept=a slope = 0.477 L / g h;
a
Intercept = 0.082 L / g
3 2.5 2 1.5 1 0.5 0
2 1.5 C
1/C
(b) b
1 0.5 0
0
1
2
3
4
5
6
1
t
1/C = 0.4771t + 0.0823
C
2 C-fitted
3
4
5
t
(c) C 1 / (a bt ) 1 / [0.082 0.477(0)] 12.2 g / L t
(1 / C a ) / b
(1 / 0.01 0.082) / 0.477
209.5 h
(d) t=0 and C=0.01 are out of the range of the experimental data. (e) The concentration of the hazardous substance could be enough to cause damage to the biotic resources in the river; the treatment requires an extremely large period of time; some of the hazardous substances might remain in the tank instead of being converted; the decomposition products might not be harmless.
2-13
2.41 (a) and (c)
y
10
1 0.1
1
10
100
x
ax b ln y
(b) y
ln a b ln x; Slope = b, Intercept = ln a
ln y = 0.1684ln x + 1.1258 2
ln y
1.5 1 0.5 0 -1
0
1
2 ln x
3
4
b slope
5
0168 .
a Intercept = ln a 11258 .
3.08
2.42 (a) ln(1-Cp/CA0) vs. t in rectangular coordinates. Slope=-k, intercept=0 (b)
400
600
ln(1-Cp/Cao)
t
k
200
Lab 1
100
600
800
-2 -4
0
k
Lab 3
0.0063 s-1
500
600
t
200
400
Lab 2
0.0111 s-1
600
800
0 -2 -4
-6 ln(1-Cp/Cao)= -0.0064t
t
400
-4
k
0
-6 ln(1-Cp/Cao) = -0.0063t
300
-2
0.0062 s-1
400
200
0
-6 ln(1-Cp/Cao) = -0.0111t
ln(1-Cp/Cao)
ln(1-Cp/Cao)
0
0
800 ln(1-Cp/Cao)
0 200 0 -1 -2 -3 -4 ln(1-Cp/Cao) = -0.0062t
t
k
Lab 4
0.0064 s-1
(c) Disregarding the value of k that is very different from the other three, k is estimated with the average of the calculated k’s. k 0.0063 s-1 (d) Errors in measurement of concentration, poor temperature control, errors in time measurements, delays in taking the samples, impure reactants, impurities acting as catalysts, inadequate mixing, poor sample handling, clerical errors in the reports, dirty reactor.
2-14
2.43 yi
axi I (a )
n
¦
n
¦by
di2
i
i 1
a
n
¦ i 1
2.44
i 1
n
yi xi /
axi
g
2
dI da
n
0
¦ 2b y
i
i 1
g
axi xi
n
¦y x
i i
i 1
a
n
¦x i 1
¦x
2 i
i 1
DIMENSION X(100), Y(100) READ (5, 1) N C N = NUMBER OF DATA POINTS 1FORMAT (I10) READ (5, 2) (X(J), Y(J), J = 1, N 2FORMAT (8F 10.2) SX = 0.0 SY = 0.0 SXX = 0.0 SXY = 0.0 DO 100J = 1, N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J) ** 2 100SXY = SXY + X(J) * Y(J) AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN CALCULATE SLOPE AND INTERCEPT A = (SXY - SX * SY)/(SXX - SX ** 2) B = SY - A * SX WRITE (6, 3) 3FORMAT (1H1, 20X 'PROBLEM 2-39'/) WRITE (6, 4) A, B 4FORMAT (1H0, 'SLOPEb -- bAb =', F6.3, 3X 'INTERCEPTb -- b8b =', F7.3/) C CALCULATE FITTED VALUES OF Y, AND SUM OF SQUARES OF RESIDUALS SSQ = 0.0 DO 200J = 1, N YC = A * X(J) + B RES = Y(J) - YC WRITE (6, 5) X(J), Y(J), YC, RES 5FORMAT (3X 'Xb =', F5.2, 5X /Yb =', F7.2, 5X 'Y(FITTED)b =', F7.2, 5X * 'RESIDUALb =', F6.3) 200SSQ = SSQ + RES ** 2 WRITE (6, 6) SSQ 6FORMAT (IH0, 'SUM OF SQUARES OF RESIDUALSb =', E10.3) STOP END $DATA 5 1.0 2.35 1.5 5.53 2.0 8.92 2.5 12.15 3.0 15.38 SOLUTION: a 6.536, b 4.206
2-15
2 i
0
2.45 (a) E(cal/mol), D0 (cm2/s) (b) ln D vs. 1/T, Slope=-E/R, intercept=ln D0. (c) Intercept = ln D0 = -3.0151 D0 = 0.05 cm2 / s .
3.0E-03
2.9E-03
2.8E-03
2.7E-03
2.6E-03
2.5E-03
2.4E-03
2.3E-03
2.2E-03
2.1E-03
2.0E-03
Slope = E / R = -3666 K E = (3666 K)(1.987 cal / mol K) = 7284 cal / mol
-10.0 ln D
-11.0 -12.0 -13.0 -14.0
ln D = -3666(1/T) - 3.0151
1/T
(d) Spreadsheet T 347 374.2 396.2 420.7 447.7 471.2
D 1.34E-06 2.50E-06 4.55E-06 8.52E-06 1.41E-05 2.00E-05
1/T 2.88E-03 2.67E-03 2.52E-03 2.38E-03 2.23E-03 2.12E-03 Sx Sy Syx Sxx -E/R ln D0
lnD (1/T)*(lnD) -13.5 -0.03897 -12.9 -0.03447 -12.3 -0.03105 -11.7 -0.02775 -11.2 -0.02495 -10.8 -0.02296 2.47E-03 -12.1 -3.00E-02 6.16E-06 -3666 -3.0151
D0
7284
E
0.05
2-16
(1/T)**2 8.31E-06 7.14E-06 6.37E-06 5.65E-06 4.99E-06 4.50E-06
CHAPTER THREE 3.1
16 u 6 u 2 m3 1000 kg | 2 u 10 5 2 103 | 2 u 105 kg 3 m
b
(a) m
8 oz
(b) m
2s
gb gb gd i
106 cm3 1 qt 1g 4 u 106 | | 1 u 102 g / s 3 3 32 oz 1056.68 qt cm 3 u 10 10
b
gd i
(c) Weight of a boxer | 220 lb m 12 u 220 lb m 1 stone | 220 stones Wmax t 14 lb m (d)
dictionary
SD 2 L 4
V |
314 . 4.52 ft 2 4
d
800 miles 5880 ft 7.4805 gal 1 barrel 1 mile 1 ft 3 31.5 gal
i d
i
3 u 4 u 5 u 8 u 102 u 5 u 103 u 7 4 u 3 u 10
(e) (i)V |
| 1 u 107 barrels
6 ft u 1 ft u 0.5 ft 28,317 cm3 | 3 u 3 u 104 | 1 u 105 cm3 1 ft 3
(ii)V |
150 lb m
1 ft 3 62.4 lb m
28,317 cm3 150 u 3 u 104 | | 1 u 105 cm3 3 60 1 ft
(f) SG | 105 . 3.2
(a) (i)
(ii) (b) U 3.3
(a)
995 kg 1 lb m 0.028317 m3 m3 0.45359 kg 1 ft 3 995 kg / m3
62.43 lb m / ft 3 1000 kg / m3
U H2 O u SG
62.43 lb m / ft 3 u 5.7
50 L
62.12 lb m / ft 3
62.12 lb m / ft 3 360 lb m / ft 3
0.70 u 103 kg 1 m3 35 kg m3 103 L
(b)
1150 kg m3 1000 L 1 min 27 L s 0.7 u 1000 kg 1 m3 60 s min
(c)
10 gal 1 ft 3 0.70 u 62.43 lb m # 29 lb m / min 2 min 7.481 gal 1 ft 3
3-1
3.3 (cont’d) (d) Assuming that 1 cm3 kerosene was mixed with Vg (cm3 ) gasoline
d
i d i 1dcm kerosenei 0.82dg kerosenei d0.70V 0.82idg blendi 0.78 V SG V 1dcm blend i
Vg cm3gasoline 0.70Vg g gasoline 3
0.82 0.78 0.78 0.70
g
3
g
g
Vgasoline Vkerosene
Volumetric ratio
3.4
In France: In U.S.:
3.5
50.0 kg
L
0.50 cm3 1 cm3 5 Fr
3
0.5 0 cm
0.50 cm3 gasoline / cm3 kerosene $1
$68.42 0.7 u 1.0 kg 1L 5.22 Fr 50.0 kg L 1 gal $1.20 $22.64 0.70 u 1.0 kg 3.7854 L 1 gal
VB ( ft 3 / h ), m B ( lb m / h )
V ( ft 3 / h), SG
VH ( ft 3 / h ), m H ( lb m / h )
0.850
700 lb m / h
700 lb m ft 3 1319 . ft 3 / h h 0.850 u 62.43 lb m 3 VB ft 0.879 u 62.43 lb m kg / h m B 54.88V B h ft 3 m H VH 0.659 u 62.43 4114 . VH kg / h
(a) V
d i bg d hb
VB VH
b
g
b
g
g
. ft 3 / h 1319 54.88V 4114 . V
m B m H 700 lb m B H 3 VB 114 . ft / h m B 628 lb m / h benzene VH
. ft 3 / h m H 174
. lb m / h hexane 716
(b) – No buildup of mass in unit. – U B and U H at inlet stream conditions are equal to their tabulated values (which are o
strictly valid at 20 C and 1 atm.) – Volumes of benzene and hexane are additive. – Densitometer gives correct reading.
3-2
3.6
1955 . kg H 2SO 4
(a) V
1 kg solution L u 1000 0.35kg H 2SO 4 12563 . . kg
445 L
(b) 195.5 kg H 2 SO 4
L u 1.00 kg 18255 . 195.5 kg H 2 SO 4 0.65 kg H 2 O L 0.35 kg H 2 SO 4 1.000 kg 470 445 u 100% 5.6% % error 445
Videal
3.7
b g E Weight of block bdowng Mass of oil displaced + Mass of water displaced = Mass of block U b0.542gV U b1 0.542gV U V Buoyant force up
oil
c
H 2O
From Table B.1: U c
2.26 g / cm3 , U w
100 . g / cm3 U oil
moil U oil u V 3.325 g / cm3 u 35.3 cm3 moil + flask 117.4 g 124.8 g 242 g 3.8
470 L
b g
117.4 g
b
Buoyant force up = Weight of block down Wblock ( UVg ) disp. Liq
Wdisplaced liquid
b g
b g
. Ag Expt. 1: U w 15
UB 2A g UB
U w 1.00 g/cm3
bg
UB
Let U w
for b p 1 hb 1
bg
subst. 3 for h p 1 in
b2 g, solve for h
b1
hb1
15 .
soln
density of water. Note: U A ! U w (object sinks) Ab hsi
d
Ab h p1 hb1
i
bi g
i
b
d
Ap h p1 Vd 1 Vw
WA WB h p1 pw g
W WB Vw A Ap pw g
(1)
U wVd 1 g WA WB weight of displaced water
(2)
A p h p1
Vw
15 . g / cm3 SG
d
Volume of pond water: Vw subst. 2
0.75
b g
2U B
WA WB pw gAb
h p1 hb1
B
Subst. (1) for Vd 1 , solve for h p1 hb1
Before object is jettisoned
bg
b g
0.75 g / cm3 SG
Archimedes
hU 1
hb 1
15 . 2
Volume displaced: Vd 1
hs 1
WA + WB
Uw u
U B 2 A g U soln
3.9
g
( UVg ) block
b g
Expt. 2: U soln A g
3.325 g / cm3
Ap h p1 Ab h p1 hb1
Vw WA WB Ap pw gAp
g LM 1 MN A
p
3-3
1 Ab
OP PQ
(3) (4)
i
3.9 (cont’d) hs 2
WB WA
Let V A hU2
h b2
(5)
Ab h p 2 hb 2
(6)
d
Volume displaced by boat:Vd 2
Subst. for Vd 2
E , solve for dh
WB pw gAb
p2
hb 2
i (7)
Ap h p 2 Vd 2 V A
Volume of pond water: Vw
b 5g , b 6 g & b 7 g
Vw
Ap h p 2
WB W A pw g p A g
Vw WB WA Ap pw gAp p A gAp
solve for
hp 2
h p 21
bg
subst. 8
bg
for h p 2 in 7 , solve for hb 2
i
Archimedes U WVd 2 g WB
After object is jettisoned
h p 2 hb 2
WA U Ag
volume of jettisoned object =
(8)
Vw WB WA WB Ap pw gAp p A gAp pw gAb
hb 2
(9)
(a) Change in pond level h p 2 h p1
LM 1 A gNp
b8gb 3g WA p
A
1 pW
OP Q
0 !0
VA VA 0 pW p A pW Ap
IJ F K GH
WA b pW p A g b5gF p A pW gAp
GH
I JK
the pond level falls (b)
Change in boat level h p 2 h p1
LM 1 A g MN p A p
A
the boat rises 3.10 (a) U bulk
2.93 kg CaCO 3 L CaCO 3
L OP F p F A 1I I P ! 0 1 M 1 G P G JK JK P G J p A PQ H A K M H p H A MN PQ ! 0
b9 gb4 g WA
p
1 pW Ap
O b5gF VA I M
W
0.70 L CaCO 3 L total
b
p
!0
A
p
W
b
2.05 kg / L
2.05 kg 50 L 9.807 m / s2 1N 100 . u 103 N L 1 kg m / s2 Neglected the weight of the bag itself and of the air in the filled bag.
(b) Wbag
U bulkVg
(c) The limestone would fall short of filling three bags, because – the powder would pack tighter than the original particles. – you could never recover 100% of what you fed to the mill.
3-4
3.11 (a) Wb
122.5 kg 9.807 m / s2
mb g
Ub
m f mnf
(b)
mf
xf
mb
mb
V f Vnf
b2 g,b3g
mb
Fx GH U
f
mb x f
d
Vb
I JK
U nf
f
1 / U b 1 / U nf
(c) x f
mf
Uf
mf mnf
mb x f mb (1 x f )
mb
mnf
U nf
F GH
mb Ub
I JK
1 1 xf U b U nf
mb Ub
Fx GH U
f
1 x f
f
U nf
I (V JK
lungs
Vother )
mb
b
lungs
nf
b
nf
nf
nf
other
b
lungs
b
1 / U f 1 / U nf
F1 1I GH U U JK
F 1 1 I 1 1 V V GH U U JK U U m F 1 1 I F V V I F 1 1 I F 12. 01. I J G GH U U JK GH m JK GH 103 J 11 . . K H 122.5 K x F1 1I FG 1 1 IJ GH U U JK H 0.9 11. K xf
f
1 / U b 1 / U nf
Vb
Vlungs Vother
U nf
(3)
mb 1 1 xf Ub U f U nf
(d) V f Vnf Vlungs Vother mnf
i
1 / 103 . 1 / 1.1 0.31 1 / 0.9 1 / 1.1
1 / U f 1 / U nf
(2)
mb 1 x f
1 x f
119 L
(1)
mf
(1),(2) mnf
Uf
1202 N
(1202 N - 44.0 N) 1 kg m / s2 Wb WI Uwg 0.996 kg / L u 9.807 m / s2 1N mb 122.5 kg 1.03 kg / L Vb 119 L
Vb
mf
1N 1 kg m / s2
other
b
f
f
nf
3-5
0.25
Conc. (g Ile/100 g H2O)
3.12 (a) 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.987
y = 545.5x - 539.03 R2 = 0.9992
0.989
0.991
0.993
0.995
0.997
Density (g/cm3)
5455 . U 539.03
From the plot above, r (b) For U = 0.9940 g / cm3 , m Ile
150 L h
r
3.197 g Ile / 100g H 2 O
0.994 g 1000 cm3 3.197 g Ile 1 kg 3 cm L 103.197 g sol 1000 g
4.6 kg Ile / h
(c) The density of H2O increases as T decreases, therefore the density was higher than it should have been to use the calibration formula. The valve of r and hence the Ile mass flow rate calculated in part (b) would be too high. 3.13 (a)
Mass Flow Rate (kg/min)
1.20 1.00
y = 0.0743x + 0.1523 R2 = 0.9989
0.80 0.60 0.40 0.20 0.00 0.0
2.0
4.0
6.0
8.0
10.0
12.0
Rotameter Reading
b g
From the plot, R = 5.3 m 0.0743 5.3 01523 . 0.546 l g / min
3-6
3.13 (cont’d) (b) Rotamete Collection Collected r Reading Time Volume (min) (cm3) 2 1 297 2 1 301 4 1 454 4 1 448 6 0.5 300 6 0.5 298 8 0.5 371 8 0.5 377 10 0.5 440 10 0.5 453
Mass Flow Rate (kg/min) 0.297 0.301 0.454 0.448 0.600 0.596 0.742 0.754 0.880 0.906
b
Difference Duplicate (Di)
Mean Di
0.004 0.006 0.004
0.0104
0.012 0.026
g
1 0.004 0.006 0.004 0.012 0.026 0.0104 kg / min 5 . Di ) kg / min 0.610 r 0.018 kg / min 95% confidence limits: (0.610 r 174 Di
There is roughly a 95% probability that the true flow rate is between 0.532 kg / min and 0.628 kg / min . 3.14 (a) (b)
(c) (d) (e) (f) (g) (h)
15.0 kmol C 6 H 6 15.0 kmol C 6 H 6
15,000 mol C 6 H 6
78.114 kg C 6 H 6
117 . u 103 kg C 6 H 6 kmol C 6 H 6 1000 mol 15 . u 104 mol C 6 H 6 kmol lb - mole 453.6 mol
15,000 mol C 6 H 6
6 mol C 1 mol C 6 H 6
15,000 mol C 6 H 6
6 mol H 1 mol C 6 H 6
90,000 mol C 12.011 g C mol C 90,000 mol H 1.008 g H mol H
15,000 mol C 6 H 6
33.07 lb - mole C 6 H 6 90,000 mol C 90,000 mol H
1.08 u 106 g C 9.07 u 104 g H
6.022 u 1023 mol
9.03 u 1027 molecules of C6 H 6
3-7
3.15 (a) m
(b) n
175 m3 h
1000 L m3
2526 kg
1000 mol 1 min
min
0.866 kg 1h L 60 min
92.13 kg
60 s
2526 kg / min
457 mol / s
(c) Assumed density (SG) at T, P of stream is the same as the density at 20oC and 1 atm 3.16 (a)
200.0 kg mix 0.150 kg CH 3OH kg mix
(b) m mix
3.17
M m N 2
100.0 lb - mole MA h 0.25 mol N 2 3000 kg h
kmol CH 3OH 1000 mol 32.04 kg CH 3OH
74.08 lb m MA
m / V
28.02 g N 2
0.75 mol H 2
mol N 2 kmol 0.25 kmol N 2 8.52 kg
kmol feed
936 mol CH 3OH
8715 lb m / h
2.02 g H 2
8.52 g mol mol H 2 28.02 kg N 2 2470 kg N 2 h kmol N 2
215 g 65 g 150 g
500 g / 455 mL 110 . g mL
(c) 150 g CaCO 3 / 500 g suspension 3.19
1 lb m mix
1 lb - mole MA 0.850 lb m MA
3.18 M suspension 565 g 65 g 500 g , M CaCO 3 (a) V 455 mL min , m 500 g min (b) U
1 kmol
0.300 g CaCO 3 g suspension
Assume 100 mol mix. mC2 H 5OH mC4 H 8O 2
10.0 mol C 2 H 5OH
46.07 g C 2 H 5OH
75.0 mol C 4 H 8 O 2
mol C 2 H 5OH 88.1 g C 4 H 8 O 2
461 g C 2 H 5OH
6608 g C 4 H 8 O 2 mol C 4 H 8 O 2 15.0 mol CH 3COOH 60.05 g CH 3COOH mCH 3COOH 901 g CH 3COOH mol CH 3COOH 461 g xC 2 H 5OH 0.0578 g C2 H 5OH / g mix 461 g + 6608 g + 901 g 6608 g xC 4 H 8 O 2 0.8291 g C 4 H 8 O 2 / g mix 461 g + 6608 g + 901 g 901 g xCH 3COOH 0113 . g CH 3COOH / g mix 461 g + 6608 g + 901 g 461 g + 6608 g + 901 g MW 79.7 g / mol 100 mol 25 kmol EA 100 kmol mix 79.7 kg mix m 2660 kg mix 75 kmol EA 1 kmol mix
3-8
3.20 (a) Unit Crystallizer Filter Dryer
(b) m gypsu m
Function Form solid gypsum particles from a solution Separate particles from solution Remove water from filter cake 0.35 kg C aSO 4 2 H 2 O L slurry
1 L slurry
0.35 kg CaSO4 2H2O
Vgypsum
CaSO 4 in gypsum: m
CaSO 4 in soln.: m
% recovery =
L CaSO4 2H2O
0151 . L CaSO4 2H2O 2.32 kg CaSO4 2H2O 0.35 kg gypsum 136.15 kg CaSO 4 0.277 kg CaSO 4 172.18 kg gypsum
. g L sol b1 0151
0.35 kg gypsum
(c) m
0 .35 kg C aSO 4 2 H 2 O
1.05 kg 0.209 kg CaSO 4 L 100.209 kg sol
0.05 kg sol 0.209 g CaSO 4 0.95 kg gypsum 100.209 g sol
0.277 g + 3.84 u 10 -5 g u 100% 0.277 g + 0.00186 g
FB:
45.8 L 0.90 kg min L 55.2 L 0.75 kg min L
3.84 u 10 -5 kg CaSO 4
99.3%
3.21 CSA:
0.00186 kg CaSO 4
kmol 0.5496 75 kg kmol 0.4600 90 kg
U| |V || W
kmol 0.5496 min kmol 0.4600 min
1.2
mol CSA mol FB
She was wrong. The mixer would come to a grinding halt and the motor would overheat. 3.22 (a)
150 mol EtOH 6910 g EtO H
V
SG
(b) V c
46.07 g EtOH
6910 g EtOH mol EtOH 0.600 g H 2 O 10365 g H 2 O 0.400 g EtOH
6910 g EtOH
L 789 g EtOH
(6910 +10365) g 19.1 L
L 1000 g
10365 g H 2 O
1000 g H 2 O
0.903
( 6910 10365) g mix
% error
L
L 935.18 g
(19.123 18.472 ) L u 100% 18.472 L
18.472 L 18.5 L 3.5%
3-9
19.123 L 19.1 L
3.23
0.09 mol CH 4
M
16.04 g
0.91 mol Air
29.0 g Air
27.83 g mol mol mol 700 kg kmol 0.090 kmol CH 4 2.264 kmol CH 4 h h 27.83 kg 1.00 kmol mix 2.264 kmol CH 4 0.91 kmol air 22.89 kmol air h h 0.09 kmol CH 4 5% CH 4
2.264 kmol CH 4
0.95 kmol air
h
0.05 kmol CH 4
b
43.01 kmol air h
g
Dilution air required: 43.01 - 22.89 kmol air 1000 mol 20200 mol air h h
1 kmol
20.20 kmol Air 29 kg Air Product gas: 700 kg 1286 kg h h
h
kmol Air
43.01 kmol Air 0.21 kmol O2 32.00 kg O2 h kg O2 0.225 h 1.00 kmol Air 1 kmol O 2 1286 kg total kg
3.24
mi , Ui M
xi
m m ¦ Mi V i i
¦ xi U i
A: B:
mi M , U= Vi V
xi
¦U
1 U
¦
1 M
mi2 ¦V zU i
mi Vi V 1 1 Vi Correct. ¦ M mi M M U 0.60 0.25 0.15 1.091 U 0.917 g / cm 3 0.791 1.049 1.595
¦
i
xi Ui
3.25 (a) Basis: 100 mol N 2 20 mol CH 4 N total x CO
x CH 4
(b) M
Not helpful.
100 20 64 32
R|20 u 80 25 S |T 20 u 40 25
64 mol CO 2 32 mol CO
216 mol
32 64 0.15 m ol C O / m ol , x CO 2 0.30 m ol C O 2 / m ol 216 216 20 100 0.09 mol CH 4 / mol , x N 2 0.46 mol N 2 / mol 216 216
¦y M i
i
015 . u 28 0.30 u 44 0.09 u 16 0.46 u 28
3-10
32 g / mol
3.26 (a) Samples Species
MW
k
1
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
Peak Area 3.6 2.8 2.4 1.7
Mole Mass Fraction Fraction 0.156 0.062 0.233 0.173 0.324 0.353 0.287 0.412
2
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
7.8 2.4 5.6 0.4
0.249 0.146 0.556 0.050
0.111 0.123 0.685 0.081
1.170 0.689 2.615 0.233
18.767 20.712 115.304 13.554
3
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
3.4 4.5 2.6 0.8
0.146 0.371 0.349 0.134
0.064 0.304 0.419 0.212
0.510 1.292 1.214 0.466
8.180 38.835 53.534 27.107
4
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
4.8 2.5 1.3 0.2
0.333 0.332 0.281 0.054
0.173 0.324 0.401 0.102
0.720 0.718 0.607 0.117
11.549 21.575 26.767 6.777
5
CH4 C2H6 C3H8 C4H10
16.04 30.07 44.09 58.12
0.150 0.287 0.467 0.583
6.4 7.9 4.8 2.3
0.141 0.333 0.329 0.197
0.059 0.262 0.380 0.299
0.960 2.267 2.242 1.341
15.398 68.178 98.832 77.933
(b) REAL A(10), MW(10), K(10), MOL(10), MASS(10), MOLT, MASST INTEGER N, ND, ID, J READ (5, *) N CN-NUMBER OF SPECIES READ (5, *) (MW(J), K(J), J = 1, N) READ (5, *) ND DO 20 ID 1, ND READ (5, *)(A(J), J = 1, N) MOLT 0. 0 MASST 0. 0 DO 10 J = 1, N MOL(J) = MASS(J) = MOL(J) * MW(J) MOLT = MOLT + MOL(J) MASST = MASST + MASS(J) 10 CONTINUE DO 15 J = 1, N MOL(J) = MOL(J)/MOLT MASS(J) = MASS(J)/MASST 15 CONTINUE WRITE (6, 1) ID, (J, MOL(J), MASS (J), J = 1, N) 20 CONTINUE 1 FORMAT (' SAMPLE: `, I3, /,
' SPECIES MOLE FR. MASS FR.', /,
10(3X, I3, 2(5X, F5.3), /), /)
3-11
moles
mass
0.540 0.804 1.121 0.991
8.662 24.164 49.416 57.603
END $DATA
4 16 . 04
0 . 150
30 . 07 0. 287 44 . 09 0. 467 58. 12
0 . 583
5 3. 6 2. 8 2 . 4 1. 7 7. 8 2. 4 5. 6 0. 4 3. 4 4 . 5 2. 6 0 . 8 4 . 8 2. 5 1. 3
0. 2
6. 4 7. 9 4 . 8 2 . 3 [OUTPUT] SAMPLE: 1 SPECIES
MOLE FR MASS FR
1
0.156
0.062
2
0.233
0.173
3
0.324
0.353
0.287
0.412
4 SAMPLE: 2 (ETC.)
3.27 (a)
(8.7 u 10 6 u 0.40) kg C
44 kg CO 2 12 kg C
(11 . u 10 6 u 0.26) kg C 28 kg CO 12 kg C ( 3.8 u 10 5 u 0.10) kg C
m
M
16 kg CH 4 12 kg C
1.28 u 10 7 kg CO 2 2.9 u 105 kmol CO 2
6.67 u 10 5 kg CO 2.38 u 10 4 kmol CO 5.07 u 10 4 kg CH 4 3.17 u 10 3 kmol CH 4
(1.28 u 10 7 6.67 u 10 5 5.07 u 10 4 ) kg 1 metric ton 1000 kg
¦y M i
i
0.915 u 44 0.075 u 28 0.01 u 16
13,500
metric tons yr
42.5 g / mol
3.28 (a) Basis: 1 liter of solution 1000 mL
1.03 g 5 g H 2 SO 4 mL 100 g
mol H 2 SO 4 98.08 g H 2 SO 4
3.28 (cont’d)
3-12
0.525 mol / L 0.525 molar solution
(b) t
V V
55 gal
(c) u
V A
t
L u
55 gal
3.7854 L
min
60 s
gal
87 L
min
3.7854 L 10 3 mL 1.03 g gal 1L mL
144 s
0.0500 g H 2 SO 4 g
87 L m 3 1 min min 1000 L 60 s (S u 0.06 2 / 4 ) m 2 45 m 88 s 0.513 m / s
1 lbm 453.59 g
23.6 lb m H 2 SO 4
0.513 m / s
3.29 (a) n1 (mol/min) 0.180 mol C6H14/mol 0.820 mol N2/mol
n2 (mol/min) 0.050 mol C6H14/mol 0.950 mol N2/mol 1.50 L C6H14(l)/min n 3 (mol C6H14(l)/min)
n3
150 . L 0.659 kg 1000 mol 1147 . mol / min min L 86.17 kg
UV W
|RS |T
Hexane balance: 0.180n1 0050 . n2 1147 . (mol C6 H14 / min) solve n1 838 . mol / min n2 = 72.3 mol / min Nitrogen balance: 0.820n1 0950 . n2 (mol N2 / min) (b) Hexane recovery
3.30
30 mL
n3 1147 . u 100% u 100% 76% n1 0180 . 838 .
b g
1 L 0.030 mol 172 g 0155 . g Nauseum 103 mL lL 1 mol
3-13
3.31 (a) kt is dimensionless k (min -1 ) (b) A semilog plot of C A vs. t is a straight line ln CA
ln(CA)
1 0 -1 -2 -3 -4 -5
y = -0.4137x + 0.2512 R2 = 0.9996
0.0
k
ln CAO kt
5.0 t (min)
10.0
0.414 min 1
CAO 1286 ln CAO 02512 . . lb - moles ft 3
FG 1b - molesIJ Cc mol 28.317 liter H ft K liter 1 ft t cbsg 1 min t bming t c 60 60 s
(c) C A
A
3
CA
0.06243C Ac t 3.32 (a)
(b)
(c)
3
2600 mm Hg
1000 mol
b
g
drop primes
b
g
C A mol / L
b
2.14 exp 0.00693t
g
5.30 mol / L 14.696 psi 50.3 psi 760 mm Hg
275 ft H 2 O 101.325 kPa 33.9 ft H 2 O
822.0 kPa
3.00 atm 101325 . 12 m2 u 105 N m2 1 atm 1002 cm2
30.4 N cm2
280 cm Hg 10 mm 101325 . u 106 dynes cm2 1002 cm2 (d) 1 cm 760 mm Hg 12 m2 (e) 1 atm
0.06243C Ac
C A 0 exp kt
1334 . exp 0.419t c 60
200 s C A
2.26462 lb - moles
20 cm Hg 10 mm 1 cm
1 atm 760 mm Hg
3-14
0.737 atm
3.733 u 1010
dynes m2
3.32 (cont’d) (f)
(g)
b
b25.0 14.696gpsi
760 mm Hg 14.696 psi
(h) 325 mm Hg 760 mm Hg
(i)
g
25.0 psig 760 mm Hg gauge 14.696 psig
35.0 psi 760 mm Hg 14.696 psi
3.33 (a) Pg
Ugh
h (m)
b
g
1293 mm Hg gauge
b g
2053 mm Hg abs
b
435 mm Hg gauge
g
13.546 g Hg cm3
1 cm
1540 cm CCl 4
10 mm 1595 g CCl 4 cm3 .
0.92 u 1000 kg 9.81 m / s2 m3 0111 . Pg (kPa)
h (m)
1N 1 kPa 2 3 1 kg m / s 10 N / m2
h
Pg Pg moil
68 kPa h
UV
(b) Pg Patm
0111 . u 68 7.55 m
FG 0.92 u 1000 kg IJ u FG 7.55 u S u 16 H 4 m K H 3
2
m3
IJ K
14 . u 10 6 kg
Ptop Ugh
b
g b g
68 101 115 0.92 u 1000 u 9.81 / 103 h h
5.98 m
3.34 (a) Weight of block = Sum of weights of displaced liquids U 1h1 U 2 h2 (h1 h2 ) AU b g h1 AU 1 g h2 AU 2 g U b h1 h2 (b)
Ptop Patm U1gh0 , Pbottom Patm U1g(h0 h1) U2 gh2 , Wb Ub (h1 h2 ) A
Fdown ( Patm U1gh0 ) A Ub (h1 h2 ) A , Fup [ Patm U1g(h0 h1) U2 gh2 ]A Fdown Fup Ub (h1 h2 ) A U1gh1 A U2 gh2 A Wblock Wliquid displaced
3-15
'P
3.35
bP
atm
g
Ugh Pinside
1 atm 1 atm
F
m
3.36
P
154 N 65 cm2 2
cm
. g1000 kg b105 m3
. u 104 N u 100
9.8066 m 150 m 12 m2 1N s2 1002 cm2 1 kg m / s2
. lb I FG 022481 H 1 N JK f
2250 lb f
14 . u 62.43 lb m 1 ft 3 2.3 u 106 gal 2.69 u 107 lb m UV 3 ft 7.481 gal P0 Ugh . u 62.43 lb m 32.174 ft 30 ft 1 lb f 12 ft 2 lb f 14 14.7 2 in ft 3 s2 32.174 lb m ft / s2 12 2 in 2 32.9 psi
— Structural flaw in the tank. — Tank strength inadequate for that much force. — Molasses corroded tank wall
S u 24 2 u 3 in 3 1 ft 3 8.0 u 62.43 lb m 392 lb m 3.37 (a) mhead 4 12 3 in 3 ft 3 392 lb m 32.174 ft / s 2 1 lb f W mhead g 392 lb f 32.174 lb m ft / s2 30 14.7 lb f S u 202 in 2 Fnet Finside Fout W 392 lb f 4415 lb f in 2 4 The head would blow off. 2 Fnet 4415 lb f 32.174 lb m ft / s 362 ft / s2 Initial acceleration: a mhead 392 lb m 1 lb f
b
g
(b) Vent the reactor through a valve to the outside or a hood before removing the head. 3.38 (a)
a
2m 1m
b
Pa Ugh Patm , Pb Patm If the inside pressure on the door equaled Pa , the force on the door would be F Adoor ( Pa Pb ) UghAdoor Since the pressure at every point on the door is greater than Pa , Since the pressure at every point on the door is greater than Pa , F >UghAdoor
3-16
3.38 (cont’d) (b) Assume an average bathtub 5 ft long, 2.5 ft wide, and 2 ft high takes about 10 min to fill. V 5 u 25 . u 2 ft 3 Vtub . ft 3 / min V 5 u 25 . 125 . ft 3 / min 25 | t 10 min For a full room, h
(i)
10 m
. m 1N 10 m 2 m2 1000 kg 981 F ! 2.0 u 105 N 3 2 2 m s 1 kg m / s The door will break before the room fills F!
(ii)
d i dP i
3.39 (a) Pg g
If the door holds, it will take 3 35.3145 ft 3 Vroom 5 u 15 u 10 m t fill V 12.5 ft 3 / min 1 m3 He will not have enough time.
b
g
1h 60 min
31 h
25 m H 2 O tap
b
junction
101.3 kPa 245 kPa 10.33 m H 2 O 25 5 m H 2 O 101.3 kPa 294 kPa 10.33 m H 2 O
g
(b) Air in the line. (lowers average density of the water.) (c) The line could be clogged, or there could be a leak between the junction and the tap. 3.40
Pabs Pgauge Patm
800 mm Hg 25 mm Hg 800 25 775 mm Hg
b
3.41 (a) P1 U A g h1 h2 P1 P2
bU
B
g
P2 U B gh1 U C gh2
g
b
g
U A gh1 U C U A gh2
LMb10. 0.792g g 981 cm 30.0 cm b137 . 0.792g g cm s cm N I 123.0 kPa F 1 dyne I F 101325 . kPa uG J G H 1 g cm / s K H 1.01325 u 10 dynes / cm JK
(b) P1 121 kPa +
3
2
2
3
6
2
3-17
981 cm 24.0 cm s2
OP Q
3.42 (a) U T g (500 x )
U W gR R
UT (500 x ) UW
b
g
0.866 u 500 10 424 cm 1.000 0.866 u 500 400 87 cm If h 400 cm , R = 1.000 If h 10 cm , R =
b
g
UT (500 x ) U Hg
For Hg, R c
b
g
0.866 . cm u 500 10 312 13.6 0.866 If h 400 cm , R = . cm u 500 400 637 13.6 If h 10 cm , R =
b
g
Use mercury, because the water manometer would have to be too tall. (b) If the manometer were simply filled with toluene, the level in the glass tube would be at the level in the tank. Advantages of using mercury: smaller manometer; less evaporation. (c) The nitrogen blanket is used to avoid contact between toluene and atmospheric oxygen, minimizing the risk of combustion.
3.43
Patm
b
P g 7.23 g P I U gJ b26 cmg dU U igb26 cmg FGH 7.23 K m
U f g 7.23 m U f
Pa Pb
f
atm
atm
w
w
F756 mmHg 1 m 1000 kg 9.81m/s GH 7.23m 100 cm m
2
3
Pa Pb
3.44 (a) 'h 900 hl
(b) 'h
Ib g JK
760 mmHg 1m 1N 26 cm 2 5 2 1 kg m/s 1.01325u10 N m 100 cm
81 . mm Hg
75 . psi 760 mm Hg 388 mm Hg hl = 900 388=512 mm 14.696 psi
388 25 u 2
338 mm Pg =
338 mm Hg
14.696 psi 760 mm Hg
3-18
6.54 psig
3.45 (a) h = L sin T (b) h
b8.7 cmg sinb15qg
3.46 (a) P
Patm Poil PHg
2.3 cm H 2 O
23 mm H 2 O
920 kg 9.81 m / s2 765 365 m3 393 mm Hg
0.10 m
1N 760 mm Hg 2 1 kg m / s 1.01325 u 105 N / m2
(b) — Nonreactive with the vapor in the apparatus. — Lighter than and immiscible with mercury. — Low rate of evaporation (low volatility).
c
3.47 (a) Let U f
h
manometer fluid density 110 . g cm 3 , U ac
c0.791 g cm h
acetone density
3
Differential manometer formula: 'P
dU
f
i
U ac gh
. 0791 . gg 981 cm h (mm) 1 cm 1 dyne g b110 cm s 10 mm 1 g cm/ s 0.02274 hb mmg V b mL sg 62 87 107 123 138 151 hb mmg 5 10 15 20 25 30 'Pb mm Hgg 0.114 0.227 0.341 0.455 0.568 0.682
b
'P mm Hg
(b) lnV
3
2
2
b g
n ln 'P ln K 6
ln(V)
5.5
y = 0.4979x + 5.2068
5 4.5 4 -2.5
-2
-1.5
-1
-0.5
0
ln( P)
b g
. ln 'P 52068 . From the plot above, ln V 04979
3-19
760 mm Hg 1.01325u106 dyne/ cm2
. . , ln K n = 04979 | 05
5.2068 K
183
ml s
bmm Hgg
0.5
3.47 (cont’d) (c) h
23 'P
b0.02274gb23g
132 mL 0.791 g s
mL
104 g s
0.523 mm Hg V 104 g s
1 mol 58.08 g
b
g
183 0.523
0.5
132 mL s
180 . mol s
. 544q R / 18 . 303 K 273 30q C 3.48 (a) T 85q F 4597 . 474q R 460 14q F (b) T 10q C 273 263 K u 18 (c) 'T
(d)
85q C 10 . qK 85q C 18 . qF 85q C 1.8q R 85q K; 153q F; 153q R 10 . qC 1q C 10 . qC
150q R 1q F 150q R 1.0q C 150q R 1.0$ K 150q F; 83.3q K; 1q R 1.8q R 1.8q R
3.49 (a) T
0.0940 u 1000$ FB 4.00 98.0$ C T = 98.0 u 1.8 + 32 = 208$ F
(b) 'T ($ C) 'T ($ F)
0.0940'T ($ FB)
15$ C 100$ L ; T2 T ($ C) aT ($ L) b
b43 15g$ C b1000 - 100g$ L
T ($ C)
0.94$ C 'T (K)
0.94 K
0.94$ C 1.8$ F 169 . $ F 'T ($ R) 169 . $R $ 1.0 C
(c) T1
a
83.3q C
43$ C 1000$ L
F $ CI ; GH $ L JK
0.0311
b
15 0.0311 u 100
0.0311T ($ L) 119 . and T ($ L)
119 . $C
32.15T ($ C) 382.6
(d) Tbp
. $ FB 3232 $ L 88.6$ C 184.6 K 332.3$ R -127.4$ F 9851
(e) 'T
50.0$ L 1.56$ C 16.6$ FB 156 . K 2.8$ F 2.8$ R
3-20
3.50
bT g (a) V b mVg
bT g
100q C
b H 2O
455q C
m AgCl
b g
aT q C b a 0.05524 mV q C 5.27 100a b b 0.2539 mV 24.88 455a b V mV 0.05524T q C 0.2539
b g
b g
T qC
b g
b g
1810 . V mV 4.596
. mV o136 . mV 1856 . q C o2508 . qC (b) 100 ln K n ln R
3.51 (a) ln T
b
g
b
g
b2508. 1856. gq C 20 s
326 . qC / s
KR n
T
ln 250.0 110.0
n
dT dt
1184 .
ln 40.0 20.0
. ln K ln 1100 . 1184 . (ln 200 . ) 1154 . K 3169 . T 3169 . R1184
FG 320 IJ H 3169 . K
(b) R
1/1.184
49.3
(c) Extrapolation error, thermocouple reading wrong. 3.52 (a) PV
0.08206nT
. g 14696 b g Pcbpsig14696 .
bg d i
, V L V c ft 3 u
P atm
b g b
n mol
g
453.59 mol n c lb - moles u , T($ K) lb moles
b Pc 14.696g u V c u 28.317
14.696
b
g
b
g
P c 14.696 u V c
0.08206 u nc u
28317 . ft 3 L T c ($ F) 32 . 27315 1.8
LM N
OP Q
453.59 (T c 32) . u 27315 1 1.8
0.08206 u 14.696 u 453.59 u n c u T c 459.7 28.317 u 18 .
b
b
g
Pc 14.696 V c 1073 . nc T c 4597 .
3-21
g
3.52 (cont’d)
b500 14.696g u 35. 10.73 u b85 459.7g
(b) ntot c
0308 . lb - mole 0.30 lb - mole CO 28 lbm CO 26 . lb m CO lb - mole lb - mole CO
mCO
(c) T c
b3000 14.696g u 35. 459.7 10.73 u 0.308
b g
3.53 (a) T q C
b
2733$ F
g
a u r ohms b
UV . a bW 33028
. a b 0 23624 100
0.308 lb - mole
a 10634 . b 25122 .
b g
T qC
FG kmol IJ n c (kmol) 1 min n c H s K min 60 s 60 Pcbmm Hgg 1 atm Pc Pbatmg 760 mm Hg 760 F m I m 1 min V c VG J V c H s K min 60 s 60
b g
. r ohms 25122 . 10634
(b) n
. b g T cbq Cg 27316
, TK
3
3
b g d b g
i
. Pc mm Hg V c m3 min 0016034 . Pc V c n c 12186 n c . T c q C 27316 60 . 760 T c 27316 60 (c) T
10.634r 25122 . r1
26159 . T1
26.95q C
r2
26157 . T2
26.93q C
r3
44.789 T3
2251 . qC
P (mm Hg)
h Patm
h (29.76 in Hg)
FG 760 mm Hg IJ H 29.92 in Hg K
h1 232 mm P1 987.9 mm Hg h2 156 mm P2
h3
9119 . mm Hg
74 mm P3 829.9 mm Hg
3-22
h 755.9
3.53 (cont’d)
b0.016034gb987.9gb947 60g 0.8331 kmol CH 26.95 27316 . b0.016034gb9119. gb195g 9.501 kmol air min
(d) n1 n2
min
. 26.93 27316 n1 n2 10.33 kmol min
n3
b
. g g b10.33gb2251. 27316 b0.016034gb829.9g
n3 T2 27316 . 0.016034 P3
(e) V3
(f)
4
0.8331 kmol CH 4 16.04 kg CH 4 min kmol 0.21 u 9.501 kmol O2 32.0 kg O2 xCH4
387 m3 min
kg CH 4 min 0.79 u 9.501 kmol N 2 29.0 kg N 2 13.36
min kmol O2 min 13.36 kg CH 4 min 0.0465 kg CH 4 kg (13.36 274) kg / min
REAL, MW, T, SLOPE, INTCPT, KO, E REAL TIME (100), CA (100), TK (100), X (100), Y(100) INTEGER IT, N, NT, J READ 5, MW, NT DO 10 IT=1, NT READ 5, TC, N TK(IT) = TC + 273.15 READ 5, (TIME (J), CA (J), J 1 , N) DO 1 J=1, N CA J CA J / MW
3.54
b g b g b g
bg bg XbJg TIMEbJg YbJg 1./CAbJg
1
CONTINUE CALL LS (X, Y, N, SLOPE, INTCPT)
b g
K IT
SLOPE
WRITE (E, 2) TK (IT), (TIME (J), CA (J), J 1 , N) WRITE (6, 3) K (IT) 10 CONTINUE DO 4 J=1, NT
3-23
kmol N 2
274
kg air min
bg YbJg XJ
bg LOGcKbJgh
1./TK J
3.54 (cont’d) 4
CONTINUE CALL LS (X, Y, NT, SLOPE, INTCPT) KO EXP INTCPT
b
2
3 5
10
g
E 8.314 SLOPE WRITE (6, 5) KO, E FORMAT (' TEMPERATURE (K): ', F6.2, / * ' TIME CA', /, * ' (MIN) (MOLES)', / * 100 (IX, F5.2, 3X, F7.4, /)) FORMAT (' K (L/MOL – MIN): ', F5.3, //) FORMAT (/, ' KO (L/MOL – MIN) : ', E 12.4, /, ' E (J/MOL): ', E 12.4) END SUBROUTINE LS (X, Y, N, SLOPE, INTCPT) REAL X(100), Y(100), SLOPE, INTCPT, SX, SY, SXX, SXY, AN INTEGER N, J SX=0 SY=0 SXX=0 SXY=0 DO 10 J=1,N SX = SX + X(J) SY = SY + Y(J) SXX = SXX + X(J)**2 SXY = SXY + X(J)*Y(J) CONTINUE AN = N SX = SX/AN SY = SY/AN SXX = SXX/AN SXY = SXY/AN SLOPE = (SXY – SX*SY)/(SXX – SX**2) INTCPT = SY – SLOPE*SX RETURN END $ DATA 65.0 94.0 10.0 20.0
4 6 8.1 4.3
[OUTPUT] TEMPERATURE (K): 367.15 TIME CA (MIN) (MOLS/L) 10.00 0.1246 3-24
30.0 40.0 50.0 3.54 (cont’d)
3.0 2.2 1.8
20.00 0.0662 30.00 0.0462 40.00 0.0338
60.0
1.5
50.00 0.0277 60.00 0.0231
b
g
K L / MOL MIN : 0.707 110. 10.0 20.0 30.0 40.0 50.0 60.0
6 3.5 1.8 1.2 0.92 0.73 0.61
127.
6
ETC
bat 94qCg
TEMPERATURE (K): 383.15
b
g
K L / MOL MIN : 1.758
b g E bJ / MOLg: 0.6690E 05
K0 L / MOL MIN : 0.2329E 10
3-25
CHAPTER FOUR 4.1
a.
Continuous, Transient
b.
Input – Output = Accumulation No reactions Generation = 0, Consumption = 0
6.00 c.
4.2
t
kg kg 3.00 s s
dn dn dt dt
100 . m3 1000 kg 1 s 1 m3 3.00 kg
3.00
333 s
a.
Continuous, Steady State
b.
k
c.
Input – Output – Consumption = 0 Steady state Accumulation = 0 A is a reactant Generation = 0
0 CA
C A0
k
kg s
f CA
0
FG IJ FG IJ V FG m IJ C FG mol IJ kVC FG mol IJ C HsK H K H K H sK Hm K
m3 mol V C A0 s m3
4.3
3
A
b
a.
v kg / h m 100 kg / h 0.550 kg B / kg 0.450 kg T / kg
b
A
CA0 kV 1 V
g
0.850 kg B / kg 0.150 kg T / kg l kg / h m
A
3
g
Input – Output = 0 Steady state Accumulation = 0 No reaction Generation = 0, Consumption = 0
0.106 kg B / kg 0.894 kg T / kg
(1) Total Mass Balance: 100.0 kg / h
v m l m
(2) Benzene Balance: 0.550 u 100.0 kg B / h
v Solve (1) & (2) simultaneously m
v 0106 l 0.850m . m
59.7 kg h, m l
40.3 kg h
b.
The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output).
c.
Possible explanations a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.
4-1
4.4
b.
c.
b
n (mol) 0.500 mol N 2 mol 0.500 mol CH 4 mol 100.0 g / s
b g bg C H g g bg C H g g
xB d.
3
8
4
10
b
b
b
b g U| R|n blb - mole DA sg S| 0.21 lb - moleO lb - mole DA V| T 0.79 lb - mole N lb - mole DAW
g
g
26.45 x E lb - mole C2 H 6 / h
n1 lb - mole H 2 O s
0.014 n kg N 2
100 x E g C2 H 6 1 lb m lb - mole C2 H 6 3600 s s 453593 h . g 30 lb m C2 H 6
n E
x E g C2 H 6 g xP
g
0.500 n mol N 2 28 g N 2 1 kg mol N 2 1000 g
nO 2
b
0.21n 2 lb - mole O 2 / s
g
g
2
2
x H 2O
2
xO 2 e.
b g
n mol
0.400 mol NO mol
b
y NO2 mol NO 2 mol
b
n N 2O 4
g
0.600 y NO2 mol N 2O 4 mol
4.5
a.
FG H
lb - mole H 2 O n1 lb - mole n1 n 2
FG H
0.21n 2 lb - mole O 2 lb - mole n1 n 2
IJ K
IJ K
b
n 0.600 y NO 2 mol N 2 O 4
g
Basis: 1000 lbm C3H8 / h fresh feed (Could also take 1 h operation as basis flow chart would be as below except that all / h would be deleted.) 1000 lb m C3H8 / h
g
b
b
n 6 lb m / h
n 7 lb m / h
g
0.02 lb m C3H 8 / lb m 0.98 lb m C3H 6 / lb m
g
0.97 lb m C3H 8 / lb m 0.03 lb m C3H 6 / lb m
Still
Compressor
Reactor
b n blb n blb n blb
g C H / hg CH / h g H / hg
b b
n1 lb m C3H 8 / h
n1 lb m C3H 8 / h 2
m
3
m
4
m
3
4
2
Note: the compressor and the off gas from the absorber are not mentioned explicitly in the process description, but their presence should be inferred.
b b
n3 lb m CH 4 / h n 4 lb m H 2 / h
g
g b
n5 lb m / h
g
Stripper
Absorber
b b n blb
n1 lb m C3H 8 / h
g g
n 2 lb m C3H 6 / h 5
4-2
g g
n 2 lb m C3H 6 / h
6
m
oil / h
g
4.5 (cont’d) b. Overall objective: To produce C3H6 from C3H8. Preheater function: Raise temperature of the reactants to raise the reaction rate. Reactor function: Convert C3H8 to C3H6. Absorption tower function: Separate the C3H8 and C3H6 in the reactor effluent from the other components. Stripping tower function: Recover the C3H8 and C3H6 from the solvent. Distillation column function: Separate the C3H5 from the C3H8. 4.6
a.
3 independent balances (one for each species)
b.
1 , m 3 , m 5 , x2 , y2 , y4 , z4 ) 7 unknowns ( m – 3 balances – 2 mole fraction summations 2 unknowns must be specified
c.
y2
1 x2
b gb g FGH kghA IJK F kg I F kg I 5300 G J m 1200 m G J Overall Balance: m H hK H hK F kg BIJ 1200 y 0.60m FG kg BIJ 5300 x G B Balance: 0.03m H hK H hK A Balance: 5300 x2
FG kg A IJ H h K
3 1200 0.70 m
1
3
1
z4 4.7
a.
2
4
5
1 0.70 y4
3 independent balances (one for each species)
b. Water Balance:
400 g 0.885 g H 2O min g
Acetic Acid Balance:
bg b g
R g 0.995 g H O m 2 R m min g
F g CH OOH IJ . gG b400gb0115 H min K 3
E m
356 g min
R 0.096m E 0.005m
461g min
FG g CH OOH IJ H min K 3
FG g IJ m m FG g IJ m 417 g min H min K H min K F g IJ b0.096gb461g FG g IJ 44 g min = 44 g min . gb400g b0.005gb356g G b0115 H min K H min K
C 400 Overall Balance: m c.
5
R
4-3
E
C
4.7 (cont’d) d. CH3COOH
H 2O some CH3COOH CH3COOH H 2O C4 H9OH
C4 H 9OH CH 3COOH
Extractor
Distillation Column
C4 H9OH
4.8
a.
X-large: 25 broken eggs/min 35 unbroken eggs/min
120 eggs/min 0.30 broken egg/egg 0.70 unbroken egg/egg
b.
120
n1 n2
b
g
25 35 n1 n2 eggs min n1 n2
b0.30gb120g c.
Large: n 1 broken eggs/min n 2 unbroken eggs/min
25 n1
50
U| n V| n W
1
2
11 39
50 large eggs min
b11 50g 0.22 22% of the large eggs (right hand) and b25 70g 36% of the extra-large eggs (left hand) n1 large eggs broken/50 large eggs
d.
are broken. Since it does not require much strength to break an egg, the left hand is probably poorly controlled (rather than strong) relative to the right. Therefore, Fred is right-handed. 4.9
a.
b
m1 lb m strawberries 015 . lb m S / lb m 0.85 lb m W / lb m
c
m2 lb m S sugar
b.
g
b
m3 lb m W evaporated 1.00 lb m jam 0.667 lb m S / lb m 0.333 lb m W / lb m
h
3 unknowns ( m1 , m2 , m3 ) – 2 balances – 1 feed ratio 0 DF
4-4
g
4.10
a.
300 gal
b g
m1 lb m
0.750 lb m C 2 H 5OH / lb m 0.250 lb m H 2 O / lb m
b g
m3 lb m
0.600 lb m C 2 H 5OH / lb m 0.400 lb m H 2 O / lb m
b g m blb g
V40 gal
4 unknowns ( m1 , m2 ,V40 , m3 ) – 2 balances – 2 specific gravities 0 DF
m
2
0.400 lb m C 2 H 5 OH / lb m 0.600 lb m H 2 O / lb m
b.
300 gal
m1
1ft 3 0.877 u 62.4 lb m 7.4805 gal ft 3
Overall balance: m1 m2 m3 C2H5OH balance: 0.750m1 0.400m2 Solve (1) & (2) simultaneously m2 1646 lb m
V40 4.11
a.
b
n1 mol / s
(1) (2)
0.600m3 1646 lb m, , m3
ft 3 7.4805 gal 0.952 u 62.4lb m 1ft 3
3841 lb m
207 gal 3 unknowns ( n1 , n2 , n3 ) – 2 balances 1 DF
g
0.0403 mol C3H 8 / mol 0.9597 mol air / mol
b
2195 lb m
n 2 mol air / s
b
n3 mol / s
g
0.0205 mol C3H 8 / mol 0.9795 mol air / mol
g
0.21mol O 2 / mol 0.79 mol N 2 / mol
b.
Propane feed rate: 0.0403n1 Propane balance: 0.0403n1 Overall balance: 3722 n2
c.
b
g
150 n1 3722 mol / s 0.0205n3 n3 7317 mol / s 7317 n2 3600 mol / s
b
b
g
g
! . The dilution rate should be greater than the value calculated to ensure that ignition is not possible even if the fuel feed rate increases slightly.
4-5
4.12
a.
b
kg / h m
g
0.960 kg CH3OH / kg 0.040 kg H 2O / kg
1000 kg / h 0.500 kg CH 3OH / kg
,x ) 2 unknowns ( m – 2 balances 0 DF
0.500 kg H 2O / kg
673 kg / h
b
g 1 x b kg H O / kg g x kg CH3OH / kg 2
b.
Overall balance: 1000
673 m 327 kg / h m
b g
Methanol balance: 0.500 1000
b g b g
0.960 327 x 673 x
Molar flow rates of methanol and water: 673 kg 0.276 kg CH3OH 1000 g mol CH3OH h kg kg 32.0 g CH3OH
0.276 kg CH3OH / kg
5.80 u 103 mol CH3OH / h
673 kg 0.724 kg H 2 O 1000 g mol H 2 O 2.71 u 104 mol H 2 O / h h kg kg 18 g H 2O Mole fraction of Methanol: 5.80 u 103 0176 . mol CH3OH / mol 5.80 u 103 2.71 u 104 c. 4.13
Analyzer is wrong, flow rates are wrong, impurities in the feed, a reaction is taking place, the system is not at steady state.
a.
Product Feed
Reactor
2253 kg
Reactor effluent
Purifier
2253 kg R = 388
1239 kg R = 583
W aste
b g
m w kg
R = 140
Analyzer Calibration Data 1
xp
x p = 0.000145R
0.1
1.364546
0.01 100
R
4-6
1000
4.13 (cont’d) b.
Effluent: x p
1.3645
1.3645
Product: x p
1.3645
Waste: x p Efficiency c.
b g 0.494 kg P / kg 0.000145b583g 0.861 kg P / kg . kg P / kg 0.000145b140g 0123 0.861b1239g u 100% 95.8% 0.494b2253g 0.000145 388
Mass balance on purifier: 2253 1239 mw mw P balance on purifier: Input: 0.494 kg P / kg 2253 kg 1113 kg P
1014 kg
b
gb g Output: b0.861 kg P / kggb1239 kgg b0123 . kg P / kggb1014 kgg
1192 kg P
The P balance does not close . Analyzer readings are wrong; impure feed; extrapolation beyond analyzer calibration data is risky -- recalibrate; get data for R > 583; not at steady state; additional reaction occurs in purifier; normal data scatter. 4.14
a.
b
g
n1 lb- mole/ h . 00100 lb- mole H2O/ lb- mole . 09900 lb- mole DA/ lb- mole
b v dft / hi
b
g
h n2 lb- mole HO/ 2 2
g
n3 lb- mole/ h . lb- mole H2O / lb- mole 0100 . lb- mole DA/ lb- mole 0900
3
4 unknowns ( n1 , n2 , n3 , v ) – 2 balances – 1 density – 1 meter reading = 0 DF
aR b 96.9 40.0 1.626 50 15
Assume linear relationship: v
v2 v1 R2 R1
Slope: a
va aR1
Intercept: b
v2
b g
b g
40.0 1.626 15
c
1.626 95 15.61 170 ft / h 3
h
b
3
n 2
170 ft 62 .4 lb m lb - mol h ft 3 18.0 lb m
589 lb - moles H 2 O / h
DA balance: 0.9900n1 0.900n3 Overall balance: n1 n2 n3 Solve (1) & (2) simultaneously n1 b.
15.61
g
5890 lb - moles / h, n 3
(1) (2) 6480 lb - moles / h
Bad calibration data, not at steady state, leaks, 7% value is wrong, v R relationship is not linear, extrapolation of analyzer correlation leads to error.
4-7
4.15
a.
b
kg / s m
100 kg / s 0.600 kg E / kg
g
0.900 kg E / kg 0100 . kg H 2 O / kg
0.050 kg S / kg 0.350 kg H 2 O / kg
b
kg / s m
g
, xE , xS ) 3 unknowns ( m – 3 balances 0 DF
b g x b kgS / kg g 1 x x b kg H O / kg g x E kg E / kg S
E
b.
2
g 0100 . b kgS / kgg
2m m 50.0 kg / s
Overall balance: 100
b g x b50g x
S balance: 0.050 100
b g
E balance: 0.600 100
kg E in bottom stream kg E in feed c.
b
S
S
S
0.300 kg E / kg b g b g 0.300b50g kg E in bottom stream 0.25 0.600b100g kg Ein feed
0.900 50 x E 50 x E
b g lnbag b lnbRg lnb x / x g lnb0.400 / 0100 . g . 1491 b lnb R / R g lnb38 / 15g lnbag lnb x g b lnb R g lnb0100 . g 1491 . lnb15g x aRb ln x 2
1
2
1
1
1
6.340 a 1764 u 103 .
u 103 R1.491 . x 1764 R d.
FG x IJ FG 0.900 IJ H a K H 1764 u 10 K . 1 b
3
1 1.491
. 655
Device not calibrated – recalibrate. Calibration curve deviates from linearity at high mass fractions – measure against known standard. Impurities in the stream – analyze a sample. Mixture is not all liquid – check sample. Calibration data are temperature dependent – check calibration at various temperatures. System is not at steady state – take more measurements. Scatter in data – take more measurements.
4-8
4.16
a.
4.00 mol H 2SO 4 0.098 kg H 2SO 4 L of solution L of solution mol H 2SO 4 1213 . kgsolution
b.
bg
b
0.323 kg H 2SO 4 / kgsolution
5 unknowns ( v1 , v2 , v3 , m2 , m3 ) – 2 balances – 3 specific gravities 0 DF
v1 L
bg m b kg g
100 kg 0.200 kg H 2SO 4 / kg 0.800 kg H 2 O / kg . SG 1139
g
v3 L 3
0.323 kg H 2SO 4 / kg 0.677 kg H 2 O / kg SG 1.213
bg m b kg g v2 L 2
0.600 kg H 2SO 4 / kg 0.400 kg H 2 O / kg SG 1.498
Overall mass balance: 100 m2
b g
UV m m 0.677m W
m3
2
Water balance: 0.800 100 0.400m2 v1 v2 v1 v2 c.
4.17
100 kg
L 1139 . kg
44.4 kg 87.80 29.64
L 1498 . kg 2.96
3
3
44.4 kg 144 kg
87.80 L20%solution 29.64 L 60% solution
L 20%solution L 60% solution
1250 kg P 44.4 kg 60%solution L h 144 kg P 1498 . kgsolution
257 L / h
b g
m1 kg @$18 / kg 0.25 kg P / kg 0.75 kg H2O / kg
100 . kg 017 . kg P/ kg 0.83 kg H2O / kg
b g
m2 kg @$10 / kg 012 . kg P / kg 0.88 kg H 2O / kg Overall balance: m1 m2
100 . . m2 Pigment balance: 0.25m1 012
b g Solve (1) and (2) simultaneously m 0.385 kg 25% paint, m Cost of blend: 0.385b$18.00g 0.615b$10.00g $13.08 per kg . b$13.08g $14.39 per kg Selling price: 110 017 . 100 . 1
4-9
2
(1) (2) 0.615 kg12% paint
4.18
b
a.
gb
m1 kg H 2O 85% of entering water
g
100 kg 0.800 kgS / kg 0.200 kg H 2 O / kg
b g m b kg H Og m2 kgS 3
2
b
85% drying: m1
gb g 17.0 kg H O 0.800b100g 80.0 kgS
0.850 0.200 100
Sugar balance: m2
2
Overall balance: 100 17 80 m3 m3 3 kg H 2 O 0.0361 kg H 2O / kg xw 3 80 kg
b
g
17 kg H 2O 80 3 kg
m1 m2 m3 b.
3 kg H 2O
b
0.205 kg H 2O / kg wet sugar
g
1000 tons wet sugar 3 tons H 2 O 100 tons wet sugar day
30 tons H 2 O / day
1000 tons WS 0.800 tons DS 2000 lb m $0.15 365 days day ton WS ton lb m year c.
b b
g
1 xw1 x w 2 ... xw10 0.0504 kg H 2 O / kg 10 1 2 2 SD 0.00181 kg H 2 O / kg xw1 xw ... xw10 x w 9 Endpoints 0.0504 r 3 0.00181 xw
g
b
b
Lower limit
4.19
$8.8 u 10 7 per year
g
g
0.0450, Upper limit
0.0558
d.
The evaporator is probably not working according to design specifications since x w 0.0361 0.0450 .
a.
v1 m 3
c h m b kg H O g 2
1
1.00
SG
d i m b kg suspension g v3 m 3 3
d i
v2 m
3
SG
1.48
5 unknowns ( v1 , v2 , v3 , m1 , m3 ) – 1 mass balance – 1 volume balance – 3 specific gravities 0 DF
400 kg galena SG
7.44
Total mass balance: m1 400
m3
(1)
4-10
4.19 (cont’d) Assume volume additivity:
b g
m1 kg
m3 400 kg m 3 1000 kg 7440 kg
Solve (1) and (2) simultaneously m1 668 kg
v1
4.20
3
m 1000 kg
b g
m3 kg
m3 (2) 1480 kg
668 kg H 2 O, m3 1068 kg suspension
0.668 m 3 water fed to tank
b.
Specific gravity of coal < 1.48 < Specific gravity of slate
c.
The suspension begins to settle. Stir the suspension. 1.00 < Specific gravity of coal < 1.48
a.
b
n1 mol / h
g
b
n2 mol / h
b
0.040 mol H 2 O / mol 0.960 mol DA / mol
g
x mol H 2O / mol
b
g
1 x mol DA / mol
b
n3 mol H 2 O adsorbed / h
g
g
97% of H 2O in feed
. 3.40g kg b354
Adsorption rate: n3
b
g
0.97 0.04n1 n1
97% adsorbed: 156 .
Total mole balance: n1
b g
. 1556 mol H 2 O / h
401 . mol / h
n2 n3 n 2
. Water balance: 0.040 401
4.21
5h
mol H 2O 0.0180 kg H 2 O . 1556 . 401
b
g
38.54 mol / h
b
x 38.54 x 12 1566 . . u 10 5 mol H 2O / mol
g
b.
The calcium chloride pellets have reached their saturation limit. Eventually the mole fraction will reach that of the inlet stream, i.e. 4%.
a.
300 lb m / h 0.55 lb m H 2SO 4 / lb m
b
0.45 lb m H 2 O / lb m
b
B lb m / h m
C lb m / h m
g
g
0.75 lb m H 2SO 4 / lb m 0.25 lb m H 2 O / lb m
0.90 lb m H 2SO 4 / lb m 010 . lb m H 2 O / lb m
B Overall balance: 300 m
m C B H2SO4 balance: 0.55 300 0.90m
b g
(1)
0.75m C
B Solve (1) and (2) simultaneously m
4-11
C 400 lb m / h, m
(2) 700 lb m / h
4.21 (cont’d) b.
500 150 A 7.78 RA 44.4 RA 25 m 70 25 800 200 B 200 B 15.0 RB 100 m RB 20 m 60 20 ln 100 ln 20 x 6.841e0.2682 Rx ln x ln 20 . Rx 4 ln x 0.2682 Rx 1923 10 4 300 44.4 400 100 44.3, mB 400 RB 333 ., mA 300 RA 7.78 15.0 1 55 ln 7.78 x 55% Rx 0.268 6.841
b b
A 150 m
g g
b
g
IJ K
FG H
c.
A m B Overall balance: m
m C
A 0.90m B H2SO4 balance: 0.01xm
d
0.75 0.01 6.841e 0.2682 Rx
15.0 RB 100 RB
d2.59 0.236e
Check: RA
4.22
a.
44.3, Rx
b
n A kmol / h
0.2682 Rx
iR
b
b
g
0.75 m A m B m B
i b7.78 R
A
b0.75 0.01xgm
g
44.4
135 . e0.2682 Rx 813 .
A
7.78 RB
e2.59 0.236e
b g
0.2682 7.78
. e j44.3 135
b g 813 .
0.2682 7.78
g 100 kg / h
b
n P kmol / h
g
0.20 kmol H 2 / kmol 0.80 kmol N 2 / kmol
g
0.50 kmol H 2 / kmol 0.50 kmol N 2 / kmol MW
b
g
b
A
015 .
. 015
. kmol H 2 / kmol 010 0.90 kmol N 2 / kmol n B kmol / h
0.75m C
g
0.20 2.016 0.80 28.012
22.813 kg / kmol
100 kg kmol 4.38 kmol / h h 22.813 kg Overall balance: n A n B 4.38 H2 balance: 010 . n A 0.50n B 0.20 4.38 n P
(1)
b g
Solve (1) and (2) simultaneously n A
4-12
(2) 3.29 kmol / h, n B
. kmol / h 110
. 333
4.22 (cont’d) b.
n P
m P 22.813
Overall balance: n A n B H2 balance: x A n A x B n B
c.
Trial 1 2 3 4 5 6 7 8 9 10 11 12
n A
b b
m P 22.813 x P m P 22.813
P xB x P m 22.813 x B x A XA 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10
g g
b b
P xP x A m 22.813 x B x A
n B
XB 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50
XP 0.10 0.20 0.30 0.40 0.50 0.60 0.10 0.20 0.30 0.40 0.50 0.60
mP 100 100 100 100 100 100 250 250 250 250 250 250
g g
nA 4.38 3.29 2.19 1.10 0.00 -1.10 10.96 8.22 5.48 2.74 0.00 -2.74
nB 0.00 1.10 2.19 3.29 4.38 5.48 0.00 2.74 5.48 8.22 10.96 13.70
The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. d. 4.23
Results are the same as in part c. Venous blood 195.0 ml / min 175 . mg urea / ml
Arterial blood 200.0 ml / min 190 . mg urea / ml
Dialysate
b b
Dialyzing fluid 1500 ml / min a.
g
Water removal rate: 200.0 195.0 5.0 ml / min
b
g
b
g
Urea removal rate: 190 . 200.0 175 . 195.0 b.
g
v ml / min c mg urea / ml
v 1500 5.0 1505ml / min 38.8 mg urea / min 0.258 mg urea / ml c 1505 ml / min
4-13
38.8 mg urea / min
4.23 (cont’d) c. 2.7 11 . mg removed 1 min 10 3 ml 5.0 L 206 min (3.4 h) ml 38.8 mg removed 1L
b
4.24
a.
g
b
n1 kmol / min
g b
20.0 kg CO 2 / min
b
n 2 kmol / min
n3 kmol / min
g
g
0.023 kmol CO 2 / kmol
0.015 kmol CO 2 / kmol 20.0 kg CO 2 kmol 0.455 kmol CO 2 / min 44.0 kg CO 2 min Overall balance: 0.455 n 2 n3 CO2 balance: 0.455 0.015n 2 0.023n3 Solve (1) and (2) simultaneously n 2 55.6 kmol / min, n3 n1
b.
4.25
u
150 m 18 s
8.33 m / s
A
1 2 SD 4
. kmol 561 1 min s m3 D 108 . m min 0123 . kmol 60 s 8.33 m
Spectrophotometer calibration: C Dye concentration: A
0.60 cm 3
Dye injected
b
g bg
3.0 P g V L 4.26
a.
1000 L B / min
3
1
2
1
1
A 0.9 C 3
. gb018 . g b3333
g
3.333 A
3.0 P g
5.0 L
b g y b kmol SO / kmolg 1 y b kmol A / kmolg b kg / min g m x b kg SO / kgg 1 x b kg B / kg g n3 kmol / min 2
3
3
4
2
4
4
4-14
561 . kmol / min
0.600 Pg / L
1L 5.0 mg 103 P g 1L 103 cm 3 1 mg
0.600 P g / L V
b g V d m / min i n b kmol / min g y b kmol SO / kmolg 1 y b kmol A / kmolg 2 kg B / min m 1
018 . C
b
! C Pg / L
kA
(1) (2)
4.26 (cont’d)
2 , m 4 , x4 , y1 , y3 ) 8 unknowns ( n1 , n3 , v1 , m – 3 material balances – 2 analyzer readings – 1 meter reading – 1 gas density formula – 1 specific gravity 0 DF b.
Orifice meter calibration: A log plot of V vs. h is a line through the points h1 100, V1 142 and h2
d
ln V
b ln h ln a V ah ln V2 V1 ln 290 142 2
d
1
1
2.58 a
1
e 2.58
13.2 V
13.2h 0.515
Analyzer calibration: ln y bR ln a y aebR b ln a
b
ln y 2 y1
h1
R2 R1 ln y1 bR1
E
0.0600
5.00 u 10 4 e 0.0600 R
b g 207.3 m h b12.2g b150 14.7g 14.7 batmg 0.460 mol / L = 0.460 kmol / m b75 460g 18. bKg E
210 mm V1 13.2 210
U feed gas
n1
U| 90 20 || lnb0.00166g 0.0600b20g 7.60V y || |W
. 0.00166g g lnb01107
a 5.00 u 10 4
c.
207.3 m 3 0.460 kmol min m3
0.515
3
. kmol min 9534
b g expb0.0600 u 116 . g
. u 104 exp 0.0600 u 82.4 R1 82.4 y1 500
0.0702 kmol SO2 kmol
. y3 500 . u 104 R3 116
0.00100 kmol SO2 kmol
2 m
400, V2
b
d h b g 0.515 lnbh h g lnb400 100g ln a ln V b ln h lnb142g 0.515 ln 100
b
i
1000 L B 130 . kg 1300 kg / min min LB
4-15
3
i
290 .
4.26 (cont’d) A balance: 1 0.0702 95.34
b
SO2
gb g b1 0.00100gn n 88.7 kmol min x balance: b0.0702gb9534 . g( 64.0 kg / kmol) b0.00100gb88.7g(64) m 3
3
4 4
(1)
4 (1 x4 ) B balance: 1300 = m (2) 4 = 1723 kg / min, x4 = 0.245 kg SO2 absorbed / kg Solve (1) and (2) simultaneously m 4 x4 SO2 removed = m
4.27
422 kg SO 2 / min
d.
Decreasing the bubble size increases the bubble surface-to-volume ratio, which results in a higher rate of transfer of SO2 from the gas to the liquid phase.
a.
V2 m 3 / min
b g y b kmolSO / kmol g 1 y b kmol A / kmolg
d i b kg B / min g m
n 3 kmol / min
3
d i n b kmol / min g y b kmolSO / kmolg 1 y b kmol A / kmol g V1 m 3 / min
R3
b g x b kgSO kg g 1 x b kg B / kg g 4 kg / min m
1
2
1
2
3
2
2
4
1
4
P1 , T1 , R1 , h1
b.
2 , n3 , y3 , R3 , m 4 , x4 ) 14 unknowns ( n1 ,V1 , y1 , P1 , T1 , R1 , h1 ,V2 , m – 3 material balances – 3 analyzer and orifice meter readings – 1 gas density formula (relates n1 and V1 ) 2 and V2 ) – 1 specific gravity (relates m 6 DF
b
g b1 y gn
A balance: 1 y1 n1 SO2 balance: y1n1 2 B balance: m
3
y3n3
b1 x gm 4
Calibration formulas:
(1)
3
4 x4 m 64 kgSO 2 / kmol
(2) (3)
4
y1
5.00 u 104 e0.060 R1
(4)
4 0.060 R3
(5)
y3 5.00 u 10 e V1 13.2h10.515 Gas density formula: n1
b
g
(6)
12.2 P1 14.7 / 14.7
bT 460g / 18. 1
Liquid specific gravity: SG 130 . V2
4-16
V1
(7)
b g
(8)
2 kg m m3 h 1300 kg
4.27 (cont’d) c.
T1
75 °F
y1
0.07 kmol SO2/kmol
P1
150 psig
V1
207 m3/h
h1
210 torr
n1
95.26 kmol/h
R1
82.4 x4 (kg SO2/kg)
Trial 1 2 3 4 5 6 7 8 9 10
0.10 0.10 0.10 0.10 0.10 0.20 0.20 0.20 0.20 0.20
y3 (kmol SO2/kmol) V2 (m3/h) n3 (kmol/h) 0.050 0.89 93.25 0.025 1.95 90.86 0.010 2.56 89.48 0.005 2.76 89.03 0.001 2.92 88.68 0.050 0.39 93.25 0.025 0.87 90.86 0.010 1.14 89.48 0.005 1.23 89.03 0.001 1.30 88.68
m4 (kg/h) m2 (kg/h) 1283.45 1155.11 2813.72 2532.35 3694.78 3325.31 3982.57 3584.31 4210.72 3789.65 641.73 513.38 1406.86 1125.49 1847.39 1477.91 1991.28 1593.03 2105.36 1684.29
3
V 2 ( m /h )
V2 vs. y3 3 .5 0 3 .0 0 2 .5 0 2 .0 0 1 .5 0 1 .0 0 0 .5 0 0 .0 0 0 .0 0 0
0 .0 2 0
0 .0 4 0
0 .0 6 0
y 3 ( k m o l S O 2 / k m o l)
x4 = 0 .1 0
x4 = 0 .2 0
For a given SO2 feed rate removing more SO2 (lower y3) requires a higher solvent feed rate ( V2 ). For a given SO2 removal rate (y3), a higher solvent feed rate ( V2 ) tends to a more dilute SO2 solution at the outlet (lower x4). d. 4.28
Answers are the same as in part c. Maximum balances: Overall - 3, Unit 1 - 2; Unit 2 - 3; Mixing point - 3 3 Overall mass balance m 1 Mass balance - Unit 1 m A balance - Unit 1 x1 2 Mass balance - mixing point m A balance - mixing point x2 C balance - mixing point y2
4-17
4.29
a.
100 mol / h 0.300 mol B / mol 0.250 mol T / mol 0.450 mol X / mol
b
n 2 mol / h
b g Column 1 x b mol T / molg 1 x x b mol X / mol g n b mol / h g x B 2 mol B / mol T2
B2
b
g
n 4 mol / h
0.940 mol B / mol 0.060 mol T / mol
Column 2
T2
3
0.020 mol T / mol
b
n5 mol / h
b g x b mol T / molg 1 x x b mol X / mol g T5
B5
Column 1 4 unknowns ( n2 , n3 , x B 2 , xT 2 ) –3 balances – 1 recovery of X in bot. (96%) 0 DF
b
Total mole balance: 100
b g T balance: 0.250b100g B balance: 0.300 100
Total mole balance: n2
b.
0.98n3
(1)
n 2 n3
(2)
x B 2 n 2
(3)
xT 2 n 2 0.020n3
(4)
0.940n 4
(5)
n4 n5
(6)
B balance: x B 2 n 2
0.940n 4 x B5 n5
(7)
T balance: xT 2 n 2
0.060n 4 xT 5 n5
(8)
(1) n 3 44.1 mol / h (3) x B 2 0.536 mol B / mol (5) n 4 30.95 mol / h (7) x B5 0.036 mol B / mol Overall benzene recovery:
T5
Column 2: 4 unknowns ( n3 , n4 , n5 , y x ) – 3 balances – 1 recovery of B in top (97%) 0 DF
gb g
Column 2 97% B recovery: 0.97 x B 2 n 2
g
x B5 mol B / mol
0.980 mol X / mol
Column 1 96% X recovery: 0.96 0.450 100
b
(2) n 2 55.9 mol / h (4) x T 2 0.431 mol T / mol (6) n5 24.96 mol / h (8) x T 5 0.892 mol T / mol
g
0.940 30.95 u 100% 0.300 100
b g 0.892b24.96g u 100 Overall toluene recovery: 0.250b100g
4-18
g
97%
89%
4.30
a.
g
3 kg / h m
0.035 kg S / kg
x3 kg S / kg
0.965 kg H 2 O / kg
b
w kg H 2O / h m
b.
b b
100 kg / h
1
b
g
1 x3 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
b g x b kg S / kgg 1 x b kg H O / kgg 4 kg / h m
4
g
4
4
g
b
w kg H 2O / h 0100 . m
Overall process 100 kg/h 0.035 kg S/kg 0.965 kg H2O/kg
m 10 ( kg / h) 0.050 kg S/kg 0.950 kg H2O/kg m w ( kg H 2 O / h)
b g
Salt balance: 0.035 100 Overall balance: 100 H2O yield: Yw
b b
0.050m 10
m w m 10
g
w kg H 2O recovered m 96.5 kg H 2 O in fresh feed
g
First 4 evaporators
b b
g
m 4 kg / h x 4 kg S/ kg 1 x4 kg H2 O / kg
100 kg/ h 0.035 kg S/ kg 0.965 kg H2 O / kg
b g
Yw
0.31
x4
0.0398
b
g
4 0100 . m w m 4
Salt balance: 0.035 100 c.
g
b g 4 u 0100 . m b kg H O / hg w
Overall balance: 100
2
x4 m 4
4-19
2
g
10
b
10 kg / h m
g
0.050 kg S / kg 0.950 kg H 2O / kg
b
w kg H 2O / h 0100 . m
g
4.31
b g
a.
2n1 mol
Condenser
0.97 mol B / mol 0.03 mol T / mol
b g
b g
n1 mol (89.2% of Bin feed )
n1 mol
0.97 mol B / mol 0.03 mol T / mol
0.97 mol B / mol 0.03 mol T / mol
100 mol 0.50 mol B / mol
Still
0.50 mol T / mol
b gb y b mol B / molg 1 y b mol T / mol g
n 4 mol 45% of feed to reboiler
g
B
B
b g z b mol B / molg 1 z b mol T / molg n 2 mol B
b g x b mol B / molg 1 x b mol T / molg n3 mol
Reboiler
B
B
B
Overall process:
Condenser:
3 unknowns ( n1 , n3 , x B ) Still: 5 unknowns ( n1 , n2 , n4 , y B , z B ) – 2 balances – 2 balances – 1 relationship (89.2% recovery) 3 DF 0 DF
1 unknown ( n1 ) – 0 balances 1 DF
Reboiler:
6 unknowns ( n2 , n3 , n4 , x B , y B , z B ) – 2 balances – 2 relationships (2.25 ratio & 45% vapor) 3 DF
Begin with overall process. b.
Overall process 89.2% recovery: 0.892 0.50 100
b gb g
0.97 n1
Overall balance: 100
n1 n3
B balance: 0.50 100
0.97 n1 x B n3
b g
Reboiler Composition relationship: Percent vaporized: n 4 Mole balance: n2
e j / b1 x g
yB / 1 yB xB
0.45n 2
(1)
n3 n 4
(2)
(Solve (1) and (2) simultaneously.) B balance: z B n2
2.25
B
x B n3 y B n 4
4-20
4.31 (cont’d) c. B fraction in bottoms: x B Moles of overhead: n1 Recovery of toluene: 4.32
0100 . mol B / mol Moles of bottoms: n3
46.0 mol
. gb54.02g b1 x gn u 100% b1 010 u 100% 0.50b100g 0.50b100g B
3
a.
b
m3 kg H 2O
97%
g
Bypass
Mixing point
b g
Evaporator
m1 kg
100 kg . kg S / kg 012 0.88 kg H2O / kg
54.0 mol
012 . kg S / kg 0.88 kg H2O / kg
b g
b g
m4 kg
m5 kg
0.58 kg S / kg 0.42 kg H 2O / kg
0.42 kg S / kg 0.58 kg H2O / kg
b g
m2 kg
012 . kg S / kg 0.88 kg H2O / kg
Overall process:
Evaporator:
2 unknowns ( m3 , m5 ) – 2 balances 0 DF
3 unknowns ( m1 , m3 , m4 ) – 2 balances 1 DF
b g
. 100 Overall S balance: 012 Overall mass balance: 100
Bypass:
2 unknowns ( m1 , m2 ) – 1 independent balance 1 DF
Mixing point: 3 unknowns ( m2 , m4 , m5 ) – 2 balances 1 DF
0.42m5 m3 m5
Mixing point mass balance: m4 m2
m5
. m2 Mixing point S balance: 0.58m4 012
(1)
0.42m5
(2)
Solve (1) and (2) simultaneously Bypass mass balance: 100 m1 m2 b.
m1
90.05 kg, m2
Bypass fraction: c.
m2 100
9.95 kg, m3
714 . kg, m4
18.65 kg, m5
28.6 kg product
0.095
Over-evaporating could degrade the juice; additional evaporation could be uneconomical; a stream consisting of 90% solids could be hard to transport.
4-21
4.33
a.
b
4 kg Cr / h m
b
1 kg / h m
g
b
2 kg / h m
g
g
b g x b kg Cr / kgg 1 x b kg W / kg g 5 kg / h m
0.0515 kg Cr / kg
0.0515 kg Cr / kg
0.9485 kg W / kg
0.9485 kg W / kg
Treatment Unit
b
3 kg / h m
5
5
b g x b kg Cr / kgg 1 x b kg W / kgg 6 kg / h m 6
6
g
0.0515 kg Cr / kg 0.9485 kg W / kg
b.
1 m
b
2 6000 kg / h m
4500 kg / h maximum allowed value 3 6000 4500 1500 kg / h Bypass point mass balance: m 4 95% Cr removal: m
b
gb
g
g
0.95 0.0515 4500 220.2 kg Cr / h 5 4500 220.2 4279.8 kg / h Mass balance on treatment unit: m 0.0515 4500 220.2 Cr balance on treatment unit: x5 0.002707 kg Cr / kg 4779.8 6 1500 4279.8 5779.8 kg / h Mixing point mass balance: m
b
Mixing point Cr balance: x6
c.
b g
g
b
0.0515 1500 0.0002707 4279.8 5779.8
m 1 (kg/h) m 2 (kg/h) m 3 (kg/h) m 4 (kg/h) m 5 (kg/h) 1000 1000 0 48.9 951 2000 2000 0 97.9 1902 3000 3000 0 147 2853 4000 4000 0 196 3804 5000 4500 500 220 4280 6000 4500 1500 220 4280 7000 4500 2500 220 4280 8000 4500 3500 220 4280 9000 4500 4500 220 4280 10000 4500 5500 220 4280
4-22
g
0.0154 kg Cr / kg
m 6 (kg/h) x5 0.00271 951 0.00271 1902 0.00271 2853 0.00271 3804 0.00271 4780 0.00271 5780 0.00271 6780 0.00271 7780 0.00271 8780 0.00271 9780
x6 0.00271 0.00271 0.00271 0.00271 0.00781 0.0154 0.0207 0.0247 0.0277 0.0301
4.33 (cont’d)
x 6 (kg Cr/kg)
m 1 vs. x 6 0.03500 0.03000 0.02500 0.02000 0.01500 0.01000 0.00500 0.00000 0
2000
4000
6000
8000 10000 12000
m 1 (kg /h )
d.
4.34
Cost of additional capacity – installation and maintenance, revenue from additional recovered Cr, anticipated wastewater production in coming years, capacity of waste lagoon, regulatory limits on Cr emissions.
a.
b
175 kg H2O / s 45% of water fed to evaporator
b
1 kg / s m
g
b b
4 kg K 2SO 4 / s m
0196 . kg K 2SO 4 / kg 0.804 kg H 2O / kg
5 kg H 2O / s m
g
g
b b
6 kg K 2SO 4 / s m
Evaporator
7 kg H 2 O / s m
g
g
g
Crystallizer Filter
Filter cake
b
2 kg K 2SO 4 / s 10m
g
b g RS0.400 kg K SO / kg UV T0.600 kg H O / kg W 2 kgsoln / s m 2
Filtrate
b
3 kg / s m
4
2
g
0.400 kg K 2SO 4 / kg 0.600 kg H 2 O / kg
Let K = K2SO4, W = H2 Basis: 175 kg W evaporated/s 1, m 2) Overall process: 2 unknowns ( m - 2 balances 0 DF
Mixing point:
Evaporator:
Crystallizer:
4, m 5, m 6, m 7 ) 4 unknowns ( m – 2 balances – 1 percent evaporation 1 DF
1, m 2 Strategy: Overall balances m 5 % evaporation m 3, m 4 Balances around mixing point m 6, m 7 Balances around evaporator m
4-23
1, m 3, m 4, m 5 ) 4 unknowns ( m - 2 balances 2 DF
2, m 3, m 6, m 7 ) 4 unknowns ( m – 2 balances 2 DF
U| verify that each |V chosen subsystem involves || no more than two W unknown variables
4.34 (cont’d) Overall mass balance: m 1 175 10m 2 m 2 Overall K balance: . m 1 10m 2 0.400m 2 0196
U| V| W
Production rate of crystals 10m 2 45% evaporation: 175 kg evaporated min
0.450m 5
W balance around mixing point: 0.804m 1 0.600m 3 Mass balance around mixing point: m 1 m 3 K balance around evaporator: m 6
m 4
W balance around evaporator: m 5
175 m 7
m 4 m 5
Mole fraction of K in stream entering evaporator =
b.
1 Fresh feed rate: m
c.
b
g
3 kg recycle s m 1 kg fresh feed s m
b
m 4 m 4 m 5
221 kg / s
bg
2 Production rate of crystals 10m Recycle ratio:
m 5
. kg K s s 416
g
352.3 kg recycle . 160 220.8 kg fresh feed
Scale to 75% of capacity. Flow rate of stream entering evaporator = 0.75(398 kg / s) = 299 kg / s 46.3% K, 537% . W
d.
Drying . Principal costs are likely to be the heating cost for the evaporator and the dryer and the cooling cost for the crystallizer.
4-24
4.35
a.
Overall objective: Separate components of a CH4-CO2 mixture, recover CH4, and discharge CO2 to the atmosphere. Absorber function: Separates CO2 from CH4. Stripper function: Removes dissolved CO2 from CH3OH so that the latter can be reused.
b.
The top streams are liquids while the bottom streams are gases. The liquids are heavier than the gases so the liquids fall through the columns and the gases rise.
c.
b
n1 mol / h
g
b g n b mol CO / h g n 5 mol N 2 / h
0.010 mol CO 2 / mol 0.990 mol CH 4 / mol
100 mol / h 0.300 mol CO 2 / mol
b
Absorber
n 2 mol / h
6
g
Stripper
0.005 mol CO 2 / mol
0.700 mol CH 4 / mol
2
b
n 5 mol N 2 / h
g
0.995 mol CH 3OH / mol
b g n b mol CH OH / h g n 3 mol CO 2 / h 4
Overall:
Stripper:
3
3 unknowns ( n1 , n5 , n 6 ) – 2 balances 1 DF
Absorber:
4 unknowns ( n2 , n3 , n4 , n5 ) – 2 balances – 1 percent removal (90%) 1 DF
b0.700gb100g bmol CH / hg Overall mole balance: 100b mol / h g n n Overall CH4 balance:
4
1
Percent CO2 stripped: 0.90n3 Stripper CO2 balance: n3
n1 n6
70.71 mol / h , n2
0.990n1
6
n 6
n 6 0.005n2
Stripper CH3OH balance: n4 d.
4 unknowns ( n1 , n 2 , n3 , n 4 ) – 3 balances 1 DF
0.995n 2
. mol / h , n3 6510
32.55 mol CO 2 / h, n 4
647.7 mol CH 3OH / h ,
29.29 mol CO 2 / h
Fractional CO2 absorption: f CO 2
30.0 0.010n1 30.0
4-25
0.976 mol CO 2 absorbed / mol fed
4.35 (cont’d) Total molar flow rate of liquid feed to stripper and mole fraction of CO2: n3 n3 n 4 680 mol / h , x3 0.0478 mol CO 2 / mol n3 n 4 e.
Scale up to 1000 kg/h (=106 g/h) of product gas: MW1
b
g
b
0.01 44 g CO 2 / mol 0.99 16 g CH 4 / mol
g
16.28 g / mol
u 10 mol / h . bn g d10. u 10 g / hib16.28 g / molg 6142 u 10 mol / h) / (70.71 mol / h) . bn g b100 mol / hg (6142 6
1 new
4
4
feed new
4.36
8.69 u 104 mol / h
f.
Ta Ts The higher temperature in the stripper will help drive off the gas. Pa ! Ps The higher pressure in the absorber will help dissolve the gas in the liquid.
g.
The methanol must have a high solubility for CO2, a low solubility for CH4, and a low volatility at the stripper temperature.
a.
Basis: 100 kg beans fed
e
m kg C H 5 6 14
e
m kg C H 1 6 14
j
300 kg C 6 H14
Ex
j
Condenser
b g b g y b kg oil / kg g 1 x y b kg C H m2 kg
2
13.0 kg oil 87.0 kg S
2
2
6
b g b g 1 y b kg C H m4 kg
F
x 2 kg S / kg
14 / kg
Ev
y 4 kg oil / kg 4
g
6
14
/ kg
g
g
b g
m3 kg
0.75 kg S / kg
b
y3 kg oil / kg
b
g
0.25 y3 kg C 6 H14 / kg
Overall:
b
m6 kg oil
4 unknowns ( m1 , m3 , m6 , y3 ) – 3 balances 1 DF
Extractor:
g
3 unknowns ( m2 , x2 , y2 ) – 3 balances 0 DF
Evaporator: 4 unknowns ( m4 , m5 , m6 , y4 ) 2 unknowns ( m1 , m5 ) – 1 balance – 2 balances 1 DF 2 DF Filter: 7 unknowns ( m2 , m3 , m4 , x2 , y2 , y3 , y4 ) – 3 balances – 1 oil/hexane ratio 3 DF
Mixing Pt:
Start with extractor (0 degrees of freedom) Extractor mass balance: 300 87.0 13.0 kg
4-26
m2
4.36
(cont’d) Extractor S balance: 87.0 kg S
x2 m2
Extractor oil balance: 13.0 kg oil
y2 m2
Filter S balance: 87.0 kg S 0.75m3
b g
m3 m4 Oil / hexane ratio in filter cake:
Filter mass balance: m2 kg y3
y2 1 x2 y 2
0.25 y3
Filter oil balance: 13.0 kg oil
b
y3m3 y4 m4
g
Evaporator hexane balance: 1 y4 m4 Mixing pt. Hexane balance: m1 m5 Evaporator oil balance: y4 m4 b.
m5 300 kg C6 H14
m6
b
m6 100
g
118 . kg oil 0118 . kg oil / kg beans fed 100 kg beans fed m1 28 kg C6 H14 0.28 kg C6 H14 / kg beans fed Fresh hexanefeed 100 100 kg beans fed m5 272 kg C 6 H14 recycled Recycle ratio 9.71 kg C6 H14 recycled / kg C6 H14 fed m1 28 kg C 6 H14 fed
Yield
b b
c.
g
g
Lower heating cost for the evaporator and lower cooling cost for the condenser.
4.37
b
g
m lb m dirt 1 98 lb m dry shirts 3 lb m Whizzo 100 lb
m 2 lb m dirt 98 lb m dry shirts
b
m lb m Whizzo 2
g
b g
Tub
Filter
b g
m lb m 4 013 . lb m dirt / lb m 0.87 lb m Whizzo / lb m
m lb m 3 0.03 lb m dirt / lb m 0.97 lb m Whizzo / lb m
b g b b
m lb m 4 1 x lb m dirt / lb m 4 x lb m Whizzo / lb m 4
g
b g
m lb m 5 0.92 lb m dirt / lb m 0.08 lb m Whizzo / lb m
g
Strategy 95% dirt removal m1 ( 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) m2 , m5 (solves Part (a))
4-27
4.37
(cont’d)
b g around the filter b m , m , x g , but the tub only involves 2 b m , m g and 2 balances are Balances around the mixing point involve 3 unknowns m3 , m6 , x , as do balances 4
6
3
4
allowed for each subsystem. Balances around tub m3 , m4 Balances around mixing point m6 , x (solves Part (b)) a.
. lb dirt b0.05gb2.0g 010 Overall dirt balance: 2.0 010 . b0.92gm m 2.065 lb dirt Overall Whizzo balance: m 3 b0.08gb2.065g blb Whizzog 95% dirt removal: m1
m
5
5
2
b.
m
m
Tub dirt balance: 2 0.03m3 010 . 013 . m4 Tub Whizzo balance: 0.97m3 3 0.87m4 Solve (1) & (2) simultaneously m3 20.4 lb m , m4 19.3 lb m Mixing pt. mass balance: 317 . m6 20.4 lb m m6 17.3 lb m Mixing pt. Whizzo balance: 317 . x4 19.3 0.97 20.4 x 0.833 lb m Whizzo / lb m 833% . Whizzo, 6.7% dirt
b g b gb g
4.38
317 . lb m Whizzo
a.
2720 kg S mixer 3 Discarded C 3L kg L C 3S kg S 3300 kg S
Filter 3
C 2L kg L C 2S kg S
F 3L kg L F 3S kg S
620 kg L mixer 1 Filter 1 F 1L F 1S
mixer 2 C 1L kg L C 1S kg S kg L kg S
Filter 2 F 2L kg L F 2S kg S To holding tank
b g g U| V| W
F1L 6.2 kg L mixer filter 1: 0.01 620 F1L balance: 620 6.2 C1L C1L 6138 . kg L F2 L . F3 L F2 L 6.2 kg L mixer filter 2: 0.01 6138 balance: 6138 . F3L F2 L C3 L C2 L 613.7 kg L F3 L 61 mixer filter 3: 0.01C2 L F3L . kg L balance: 613.7 = 6.1+ C3L C3 L 607.6 kg L
b
4-28
(1) (2)
4.38 (cont’d) Solvent C1S 495 kg S m f 1: 015 . 3300 C1S F1S 2805 kg S balance: 3300 495 F1S C2 S . 495 F3S C2 S 482.6 kg S m f 2: 015 F2 S 2734.6 kg S balance: 495 F3S C2 S F2 S C3S 480.4 kg S m f 3: 015 C3S . 2720 C2 S F3S 2722.2 kg S balance: 2720 + C2S F3S C3S
b
g
b
U| |V || W
g
b
g
Holding Tank Contents 6.2 6.2 12.4 kg leaf 2805 2734.6 b.
5540 kg solvent
b g
5540 kgS
b g
QR kg
0165 . kg E / kg
Q0 kg
. kg E / kg Extraction 013 0.15kg F / kg Unit
0.835 kg W / kg
b g Q b kg Fg
Steam Stripper
0.855kg W / kg
QD kg D
b g
b g Q b kg Dg Q b kg Fg QE kg E
F
0.200 kg E / kg 0.026 kg F / kg 0.774 kg W / kg QB kg 0.013 kg E / kg 0.987 kg W / kg
D F
b
Q3 kg steam Mass of D in Product:
1 kg D
620 kg leaf
1000 kg leaf
0.62 kg D
QD
b g
Water balance around extraction unit: 0.835 5540 0.855QR QR Ethanol balance around extraction unit: 0165 . 5540 013 . 5410 QE QE 211 kg ethanol in extract
b
c.
g
b
g
b
F balance around stripper 0.015 5410 0.026Q0 Q0
b g
g
b
b g
b
4.39
a.
g
b
g
b
C 2 H 2 2 H 2 o C2 H 6 2 mol H 2 react / mol C2 H 2 react 0.5 kmol C2 H 6 formed / mol H 2 react
4-29
g
6085 kg mass of stripper bottom product
W balance around stripper 0.855 5410 QS 0.774 3121 0.987 6085 QS
b
5410 kg
3121 kg mass of stripper overhead product
E balance around stripper 013 . 5410 0.200 3121 0.013QB QB
b g
g
g
g
3796 kg steam fed to stripper
4.39 (cont’d) b. nH 2 nC2 H 2
15 . 2.0 H 2 is limiting reactant
15 . mol H 2 fed 10 . mol C2 H 2 fed 0.75 mol C2 H 2 required (theoretical) 10 . mol fed 0.75 mol required u 100% 333% % excess C2 H 2 . 0.75 mol required c.
4 u 106 tonnes C2 H 6 1 yr 1 day 1 h 1000 kg 1 kmol C2 H 6 2 kmol H 2 2.00 kg H 2 300 days 24 h 3600 s tonne 30.0 kg C2 H 6 1 kmol C2 H 6 1 kmol H 2 yr 20.6 kg H 2 / s
4.40
d.
The extra cost will be involved in separating the product from the excess reactant.
a.
4 NH 3 5 O 2 o 4 NO 6 H 2 O 5 lb - mole O 2 react 125 . lb - mole O 2 react / lb - mole NO formed 4 lb - mole NO formed
b.
dn i
100 kmol NH3 5 kmol O 2 h 4 kmol NH3
O 2 theoretical
d i
40% excess O 2 nO 2 c.
fed
b
125 kmol O 2
140 . 125 kmol O 2
g
175 kmol O 2
b50.0 kg NH gb1 kmol NH / 17 kg NH g 2.94 kmol NH . kmol O b100.0 kg O gb1 kmol O / 32 kg O g 3125 F n I 3125 F n I 5 125 . G GH n JK 2..94 106 H n JK 4 . 3
3
2
2
2
O2
NH 3
3
3
2
O2
NH 3
fed
stoich
O 2 is the limiting reactant Required NH3:
3125 . kmol O 2 4 kmol NH3 5 kmol O 2
2.50 kmol NH3
2.94 2.50 u 100% 17.6% excess NH3 2.50 nO 2 vO 2 [ 0 3125 . 5 [ [ Extent of reaction: nO 2 % excess NH3
d i
Mass of NO:
4.41
a.
b g
0
3125 . kmol O 2 4 kmol NO 30.0 kg NO 5 kmol O 2 1 kmol NO
0.625 kmol
625 mol
75.0 kg NO
By adding the feeds in stoichometric proportion, all of the H2S and SO2 would be consumed. Automation provides for faster and more accurate response to fluctuations in the feed stream, reducing the risk of release of H2S and SO2. It also may reduce labor costs.
4-30
4.41 (cont’d) b.
3.00 u 10 2 kmol 0.85 kmol H 2 S 1 kmol SO 2 h kmol 2 kmol H 2 S
n c
c.
127.5 kmol SO 2 / h
C a lib r a t io n C u r v e
X ( m o l H2S /m o l)
1 .2 0 1 .0 0 0 .8 0 0 .6 0 0 .4 0 0 .2 0 0 .0 0 0 .0
2 0 .0
4 0 .0
6 0 .0
8 0 .0
1 0 0 .0
R a (m V )
0.0199 Ra 0.0605
X d.
b
n c kmol SO 2 / h
b
n f kmol / h
b
g
x kmol H 2S / kmol
g
Blender
g
Flowmeter calibration:
n f aR f 100 kmol / h , R f
n f
Control valve calibration:
n c nc
UVn 15 mV W
25.0 kmol / h, R c 60.0 kmol / h , Rc
FG H
f
UV W
20 Rf 3
10.0 mV n c 25.0 mV
IJ b K
7 5 Rc 3 3
1 7 5 1 20 n f x Rc R f 0.0119 Ra 0.0605 2 3 3 2 3 5 10 R f 0.0119 Ra 0.0605 Rc 7 7
Stoichiometric feed: n c
b
n f
3.00 u 10 2 kmol / h R f
g
3 n f 20
4-31
45 mV
g
4.41 (cont’d)
b gb b g
e.
gb g
10 5 45 0.0119 76.5 0.0605 53.9 mV 7 7 7 5 n c 53.9 127.4 kmol / h 3 3 Faulty sensors, computer problems, analyzer calibration not linear, extrapolation beyond range of calibration data, system had not reached steady state yet. Rc
4.42 165 mol / s
b
x mol C 2 H 4 / mol
b
b
n mol / s
g
1 x mol HBr / mol
g
0.310 mol C2 H 4 / mol . mol HBr / mol 0173 0.517 mol C 2 H 5Br / mol
g
C 2 H 4 HBr o C 2 H 5 Br C balance:
b
g
b
165 mol x mol C2 H 4 2 mol C s mol mol C2 H 4
gb g b
gb g
n 0.310 2 n 0.517 2
. gb1g b gb g nb0173
Br balance: 165 1 x 1
(2)
Solve (1) and (2) simultaneously n 108.77 mol / s, x
b g
1 x
(1)
0.545 mol C 2 H 4 / mol
0.455 mol HBr / mol
Since the C2H4/HBr feed ratio (0.545/0.455) is greater than the stoichiometric ration (=1), HBr is the limiting reactant .
bn g b165 mol / sgb0.455 mol HBr / molg 75.08 mol HBr . gb108.8g 75.08 b0173 u 100% Fractional conversion of HBr HBr fed
dn i 75.08 mol C H dn i b165 mol / sgb0.545 mol C H C 2 H 4 stoich
2
75.08
0.749 mol HBr react / mol fed
4
C 2 H 4 fed
2
4
/ mol
g
89.93 mol C2 H 4
89.93 75.08 19.8% 75.08 Extent of reaction: n C2 H5Br n C2 H5Br vC2 H5Br[ 108.8 0.517
% excess of C2 H 4
d
i
0
4-32
b
gb
g
bg
0 1[ [
56.2 mol / s
4.43
a.
1 O 2 o Cl 2 H 2 O 2
2HCl
Basis: 100 mol HCl fed to reactor
b g n b mol O g n b mol N g n b mol Cl g n b mol H Og n2 mol HCl
100 mol HCl
b
n1 mol air
g
0.21 mol O 2 / mol
100 mol HCl 0.5 mol O 2
2
b
g
85 mol HCl react
1 mol Cl 2
2
15 mol HCl
42.5 mol Cl 2
2 mol HCl
N 2 balance:
2
135 . u 25 n1 160.7 mol air fed
85% conversion 85 mol HCl react n2
b85gb1 2g
2
25 mol O 2
2 mol HCl
35% excess air: 0.21n1 mol O 2 fed
n6
4
6
35% excess
n5
2
5
0.79 mol N 2 / mol
bO gstoic
3
42.5 mol H 2 O
b160.7gb0.79g
n4 n4
127 mol N 2
O balance:
b160.7gb0.21g mol O
2
2 mol O 1 mol O 2
2 n3
42.5 mol H 2 O
1 mol O n3 1 mol H 2 O
12.5 mol O 2
Total moles: 5
¦nj
239.5 mol
j 2
b.
0.063
mol O 2 mol N 2 mol HCl , 0.052 , 0.530 , mol mol mol
0177 .
mol Cl 2 mol H 2 O , 0177 . mol mol
15 mol HCl 239.5 mol
As before, n1 160.7 mol air fed , n2
15 mol HCl
1 2HCl O 2 o Cl 2 H 2 O 2 ni
E
bn g
i 0
vi [
HCl: 15 100 2[ [
42.5 mol
4-33
4.43 (cont’d)
N 2 : n4
b g 0.79b160.7g
Cl 2 : n5
[
42.5 mol Cl 2
H 2 O: n6
[
42.5 mol H 2 O
c.
4.44
1 [ 12.5 mol O 2 2 127 mol N 2
0.21 160.7
O 2 : n3
These molar quantities are the same as in part (a), so the mole fractions would also be the same. Use of pure O2 would eliminate the need for an extra process to remove the N2 from the product gas, but O2 costs much more than air. The cheaper process will be the process of choice.
b g Fe O 3H SO o Fe bSO g 3H O bTiOgSO 2H O o H TiO bsg H SO H TiO bsg o TiO bsg H O
FeTiO3 2H 2SO 4 o TiO SO 4 FeSO 4 2H 2 O 2
3
2
4
2
4
2
2
2
3
2
4 3
2
3
2
4
2
Basis: 1000 kg TiO2 produced 1000 kg TiO 2
kmol TiO 2
1 kmol FeTiO 3
79.90 kg TiO 2
1 kmol TiO 2
12.52 kmol FeTiO 3 dec.
12.52 kmol FeTiO 3 decomposes
1 kmol FeTiO 3 feed
14.06 kmol FeTiO 3 fed
0.89 kmol FeTiO 3 dec. 14.06 kmol FeTiO 3
b
1 kmol Ti
47.90 kg Ti
1 kmol FeTiO 3
kmol Ti
673.5 kg Ti / M kg ore
g
0.243 M
6735 . kg Ti fed
2772 kg ore fed
b
g
Ore is made up entirely of 14.06 kmol FeTiO 3 + n kmol Fe2 O3 (Assumption!) n
2772 kg ore
638.1 kg Fe 2 O 3
14.06 kmol FeTiO 3 151.74 kg FeTiO 3 kmol FeTiO3 kmol Fe2O 3
. kg Fe2 O 3 6381
4.00 kmol Fe2O 3
159.69 kg Fe2 O 3
14.06 kmol FeTiO3 2 kmol H2SO4 4.00 kmol FeTiO3 3 kmol H2SO4 4012 . kmol H2SO4 1 kmol FeTiO3 1 kmol Fe2O3
b
50% excess: 15 . 4012 . kmol H 2SO 4 Mass of 80% solution: 5902.4 kg H 2 SO 4 / M
g
6018 . kmol H 2SO 4 fed
60.18 kmol H 2SO 4
a
bkg solng
98.08 kg H 2SO 4 1 kmol H 2SO 4
0.80 M
4-34
a
5902.4 kg H 2SO 4
7380 kg 80% H 2 SO 4 feed
4.45
a.
Plot C (log scale) vs. R (linear scale) on semilog paper, get straight line through
dR
i
0.30 g m 3 and
10, C1
1
ln C bR ln a C b
b
g
0.0575 , ln a 48 10 C 0169 . e 0.0575 R
d
i C c(ftlb E
48, C2
2
m)
3
b g
b g
ln 2.67 0.0575 48
453.6 g 35.31 ft 3 1 m3 1 lb m
d
d2867 ft sib60 s ming
178 . a
e 1.78
0169 .
i
u 10 5 e 0.0575 R 1055 .
3
138 ft 3 lb m coal
1250 lb m min R
d
37 C c lb m SO 2 ft 3
i
u 105 eb 1055 .
8.86 u 105 lb m SO 2 138 ft 3 ft 3 1 lb m coal c.
IK
16,020C c
16,020C ' 0169 . e 0.0575 R C c lb m SO 2 ft 3 b.
2.67 g m 3
ae br
ln 2.67 0.30
C g m3
FH R
gb g
0.0575 37
0.012 0.018
8.86 u 10
5
lb m SO 2 ft 3
lb m SO 2 compliance achieved lb m coal
S O 2 o SO 2 1250 lb m coal 0.05 lb m S 64.06 lb m SO 2 min
1 lb m coal
32.06 lb m S
124.9 lb m SO 2 generated min
2867 ft 3 60 s 886 . u 105 lb m SO2 152 . lbm SO2 min in scrubbed gas s 1 min ft3 air 1250 lbm coal/min 62.5 lb m S/min
% removal d.
furnace ash
b124.9 15.2g lb
scrubbing fluid stack gas 124.9 lbm SO2 /min
scrubber scrubbed gas 15.2 lb m SO2 /min liquid effluent (124.9 – 15.2) lb m SO2 (absorbed)/min
SO 2 scrubbed min u 100% 124.9 lb m SO 2 fed to scrubber min m
88%
The regulation was avoided by diluting the stack gas with fresh air before it exited from the stack. The new regulation prevents this since the mass of SO2 emitted per mass of coal burned is independent of the flow rate of air in the stack.
4-35
4.46
a.
A B ===== C + D nA nA [ 0
nB
nB [
yA
nC
nC [
yB
nD
nD [
yC
0 0
0
nI
nI
yD
0
¦ ni
Total nT
At equilibrium:
en en en en
bn bn
yC y D y A yB
C0 A0
gb [ gbn
B0
j [j n [j n [j n
[ nT T
C0 D0
T
g [ g
[ c n D0 [ c c
b
c
A0
B0
T
4.87 ( nT ’s cancel)
c
gh b
387 . [ 2c nC0 nD0 487 . nA0 nB0 [ c nC0nD0 487 . nA0nB0
g
0
[a[ 2c b[ c c 0]
?[ c
b.
a 387 . 1 2 b r b 4ac where b nC0 nD0 487 . nA0 nB0 2a c nC0nD0 487 . nA0nB0
e
b[
b 387 .
F b gH
1 1174 . r 2 387 .
[e e2
1174 c .
. g b1174
nA0 80, nC0 nC nA nB nC nD 4.87
2
4.87
b gb gIK [
4 387 4.87 .
e1
0.496
2.54 is also a solution but leads to a negative conversion
Fractional conversion: X A c.
g
nA0 1 nB0 1 nC0 nB0 nI 0 0
Basis: 1 mol A feed Constants: a
b
j
b
XB
g
n A0 n A n A0
[ e1 n A0
g
0.496
nJ 0 0 nC 0 0 ! [ c 70 mol 70 nC0 [ c n A0 [ c n A0 70 mol n B0 [ c 80 70 10 mol nC 0 [ c 70 mol n D0 [ c 70 mol yC y D y A yB
nD0
b gb g b gb g
70 70 nC n D n AnB n A0 70 10
4-36
4.87 n A0
170.6 mol methanol fed
4.46 (cont’d) Product gas n A nB nC
U| |V || W
yA 170.6 70 100.6 mol yB 10 mol yC 70 mol yD 70 mol
nD ntotal
4.47
0.401 mol CH3OH mol 0.040 mol CH3COOH mol 0.279 mol CH3COOCH 3 mol 0.279 mol H 2 O mol
250.6 mol
d.
Cost of reactants, selling price for product, market for product, rate of reaction, need for heating or cooling, and many other items.
a.
o CO 2 H 2 CO H 2 O m (A)
(B)
(C)
(D)
b g n b mol H Og n b mol CO g n b mol H g n b mol Ig
. mol 100 0.20 mol CO / mol . mol CO 2 / mol 010 0.40 mol H 2 O / mol 0.30 mol I / mol
n A mol CO 2
B
2
C
2
D I
6 unknowns ( n A , nB , nC , nD , n I , [ )
Degree of freedom analysis:
bg
– 4 expressions for ni [ – 1 balance on I – 1 equilibrium relationship 0 DF b.
Since two moles are prodcued for every two moles that react, ntotal out ntotal in 100 . mol
b g b g
b g
n A 0.20 [ nB 0.40 [ nC 010 . [ nD [ n I 0.30 ntot
(1) (2) (3) (4) (5)
100 . mol
At equilibrium: yD c.
nD
[
yC y D y A yB
b
. [ gb[ g b010 b0.20 [ gb0.40 [ g / molg
nC nD n A nB
0110 . mol H 2
The reaction has not reached equilibrium yet.
4-37
0.0247 exp
FG 4020 IJ [ H 1123 K
0110 . mol
4.47
(cont’d) d.
T (K) 1223 1123 1023 923 823 723 623 673 698 688
x (CO) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
x (H2O) 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
1123 1123 1123 1123
0.2 0.4 0.3 0.5
0.4 0.2 0.3 0.4
x (CO2)
Keq Keq (Goal Seek) Extent of Reaction 0.6610 0.6610 0.2242 0.8858 0.8856 0.2424 1.2569 1.2569 0.2643 1.9240 1.9242 0.2905 3.2662 3.2661 0.3219 6.4187 6.4188 0.3585 15.6692 15.6692 0.3992 9.7017 9.7011 0.3785 7.8331 7.8331 0.3684 8.5171 8.5177 0.3724
0 0 0 0 0 0 0 0 0 0 0.1 0.1 0 0
0.8858 0.8858 0.8858 0.8858
0.8863 0.8857 0.8856 0.8867
0.1101 0.1100 0.1454 0.2156
y (H2) 0.224 0.242 0.264 0.291 0.322 0.358 0.399 0.378 0.368 0.372 0.110 0.110 0.145 0.216
The lower the temperature, the higher the extent of reaction. An equimolar feed ratio of carbon monoxide and water also maximizes the extent of reaction. 4.48
a.
A 2B o C
b
ln K e1 / K e 2 1 T1 1 T2
E
4
11458
1 373 1 573
ln K e1 11458 T1
ln A0
b.
b g g lnd10.5 / 2.316 u 10 i
ln A0 E T K
ln K e
ln 10.5 11458 373 28.37 A0
4.79 u 1013
b gh
c
Ke
4.79 u 10 13 exp 11458 T K atm 2 Ke (450K) 0.0548 atm1
nA nB nC
n A0 [ nB 0 2[ nC 0 [
nT
nT 0
U| |V | 2[ |W
b b b
gb gb gb
g
y A n A0 [ nT 0 2[ y B nB 0 2[ nT 0 2[ yC nC 0 [ nT 0 2[ nT 0 n A0 nB 0 nC 0
b
g
g
g
At equilibrium,
yC 1 y A y B2 P2 c.
bn bn
C0 A0
gb [ gb n
g 2[ g
[ e nT 0 2[ e e
B0
2 2
e
bg
1 P2
bg
Ke T (substitute for K T from Part a.) e
Basis: 1 mol A (CO) n A0
1 nB 0
b
[ e 2 2[ e
1 nC 0
g
2
b1 [ gb1 2[ g e
e
2
1 4 atm 2
0 nT 0
2, P
b g
K e 423
2 atm , T
423K
0.278 atm -2 [ 2e [ e 01317 .
4-38
0
4.47 (cont’d) (For this particular set of initial conditions, we get a quadratic equation. In general, the equation will be cubic.)
[e
, 0.844 0156 .
Reject the second solution, since it leads to a negative nB .
. g c2 2b0156 . gh y 0.500 b1 0156 y . gh c2 2b0156 . gh y 0.408 c1 2b0156 y . g c2 2b0156 . gh y 0.092 b0 0156 n n [ Fractional Conversion of CO b Ag n n yA
A
B
B
C
C
A0
A
A0
d.
0156 . mol A reacted / mol A feed
A0
Use the equations from part b. i) ii) iii) iv)
Fractional conversion decreases with increasing fraction of CO. Fractional conversion decreases with increasing fraction of CH3OH. Fractional conversion decreases with increasing temperature. Fractional conversion increases with increasing pressure.
e. *
1
2
REAL TRU, A, E, YA0, YC0, T, P, KE, P2KE, C0, C1, C2, C3, EK, EKPI, FN, FDN, NT, CON, YA, YB, YC INTEGER NIT, INMAX TAU = 0.0001 INMAX = 10 A = 4.79E–13 E = 11458. READ (5, *) YA0, YB0, YC0, T, P KE = A * EXP(E/T) P2KE = P*P*KE C0 = YC0 – P2KE * YA0 * YB0 * YB0 C1 = 1. – 4. * YC0 + P2KE * YB0 * (YB0 + 4. * YA0) C2 = 4. * (YC0 –1. – P2KE * (YA0 + YB0)) C3 = 4. * (1. + P2KE) EK = 0.0 (Assume an initial value [ e 0. 0 ) NIT = 0 FN = C0 + EK * (C1 + EK * (C2 + EK * C3)) FDN = C1 + EK * (2. * C2 + EK * 3. * C3) EKPI = EK - FN/FDN NIT = NIT + 1 IF (NIT.EQ.INMAX) GOTO 4 IF (ABS((EKPI – EK)/EKPI).LT.TAU) GOTO 2 EK = EKPI GOTO 1 NT = 1. – 2. * EKPI YA = (YA0 – EKPI)/NT YB = (YB0 – 2. + EKPI)/NT YC = (YC0 + EKPI)/NT
4-39
4.48 (cont’d) CON = EKPI/YA0 WRITE (6, 3) YA, YB, YC, CON STOP WRITE (6, 5) INMAX, EKPI FORMAT (' YA YB YC CON', 1, 4(F6.3, 1X)) FORMAT ('DID NOT CONVERGE IN', I3, 'ITERATIONS',/, * 'CURRENT VALUE = ', F6.3) END
4 3
$DATA 0.5 RESULTS YA 0.500
0.5
YB 0.408
0.0
423.
YC 0.092
2.
CON 0.156
Note: This will only find one root — there are two others that can only be found by choosing different initial values of [ a 4.49
a.
CH 4 O 2 o HCHO H 2O
(1)
CH 4 2O 2 o CO 2 2H 2O
(2)
100 mol / s
b g n b mol O / sg n b mol HCHO / sg n b mol H O / sg n b mol CO g n1 mol CH 4 / s
0.50 mol CH 4 / mol 0.50 mol O 2 / mol
2
2
3
4
5
2
2
7 unknowns ( n1 , n 2 , n3 , n 4 , n5 , [ 1 , [ 2 ) – 5 equations for n [ , [ i
e
1
2
j
2 DF b.
n1 n2
1
n4
[ 1 [ 2[
n5
[ 2
n3
c.
50 [ 1 [ 2 50 [ 2[
1
(1) (2)
2
(3) (4)
2
(5)
Fractional conversion: Fractional yield:
n3 50
b50 n g 1
50 0.855 n3
0.900 n1
5.00 mol CH / s 4
42.75 mol HCHO / s
4-40
4.49 (cont’d)
U| || V| || W
y CH 0.0500 mol CH 4 / mol Equation 3 [ 1 42.75 4 yO 0.0275 mol O 2 / mol Equation 1 [ 2 2.25 2 Equation 2 n 2 2.75 y HCHO 0.4275 mol HCHO / mol y H O 0.4725 mol H 2 O / mol Equation 4 n 4 47.25 2 y CO 0.0225 mol CO 2 / mol Equation 5 n5 2.25 2 Selectivity:
42.75 mol HCHO / s 2.25 mol CO 2 / s
19.0 mol HCHO / mol CO 2
4-41
4.50
a.
Design for low conversion and feed ethane in excess. Low conversion and excess ethane make the second reaction unlikely.
b.
C2H6 + Cl2 o C2H5Cl + HCl, C2H5Cl + Cl2 o C2H4Cl2 + HCl Basis: 100 mol C2H5Cl produced
c.
n1 (mol C2H6)
100 mol C2H5OH
n2 (mol C2H6)
n3 (mol C2H6) n4 (mol HCl) n5 (mol C2H5Cl2)
5 unknowns –3 atomic balances 2 D.F.
Selectivity: 100 mol C 2 H 5 Cl 14n5 (mol C 2 H 4 Cl 2 ) n5
U| n 714.3 mol C H in g n V 2n 2b100g 2n 2b7.143g|W n 114.3 mol C H out 6b714.3g 5b100g 6b114.3g n 4b7.143g n 607.1 mol HCl 2n 100 607.1 2b7.143g n 114.3 mol Cl b
. n1 13% conversion: 1 015 C balance:
H balance: Cl balance:
7.143 mol C 2 H 4 Cl 2
3
1
3
1
2
6
3
2
6
4
2
4
2
Feed Ratio: 114.3 mol Cl 2 / 714.3 mol C2 H 6
2
016 . mol Cl 2 / mol C2 H 6
Maximum possible amount of C2H5Cl: 114.3 mol Cl 2 1 mol C 2 H 5 Cl n max 114.3 mol C 2 H 5 Cl 1 mol Cl 2 Fractional yield of C2H5Cl:
4.51
nC2 H5Cl n max
100 mol 114.3 mol
0.875
d.
Some of the C2H4Cl2 is further chlorinated in an undesired side reaction: C2H5Cl2 + Cl2 o C2H4Cl3 + HCl
a.
C2H4 + H2O o C2H5OH, 2 C2H5OH o (C2H5)2O + H2O Basis: 100 mol effluent gas 100 mol 0.433 mol C 2 H 4 / mol
n (mol C H ) 1 2 4 n [mol H O (v)] 2 2 n 3 (mol I)
3 unknowns
0.025 mol C 2 H 5 OH / mol
-2 independent atomic balances
0.0014 mol (C H ) O / mol 2 5 2 0.093 mol I / mol
-1 I balance 0 D. F.
0.4476 mol H O (v) / mol 2
(1) C balance: 2n1
b
100 2 0.433 2 0.025 4 0.0014
(2) H balance: 4n1 2n2 (3) O balance: n2
b
b
g
100 4 0.433 6 0.025 10 0.0014 2 0.4476
g
g
100 0.025 0.0014 0.4476
Note; Eq. (1) 2 Eq. (3) 2 (4) I balance: n3 = 9.3
Eq. (2) 2 independent atomic balances
4-42
4.51 (cont'd) b. (1) n1 46.08 mol C 2 H 6 (3) n2 47.4 mol H 2 O Reactor feed contains 44.8% C 2 H 6 , 46.1% H 2 O, 9.1% I (4) n3 9.3 mol I
U| V| W
46.08 43.3 u 100% 6.0% 46.08 If all C2H4 were converted and the second reaction did not occur, nC2 H5OH
% conversion of C2H4:
d
d
Fractional Yield of C2H5OH: nC2 H5OH / nC2 H5OH Selectivity of C2H5OH to (C2H5)2O: 2.5 mol C 2 H 5 OH 0.14 mol (C 2 H 5 ) 2 O c. 4.52
i
max
b2.5 / 46.08g
i
max
46.08 mol
0.054
17.9 mol C 2 H 5OH / mol (C 2 H 5 ) 2 O
Keep conversion low to prevent C2H5OH from being in reactor long enough to form significant amounts of (C2H5)2O. Separate and recycle unreacted C2H4.
bg
bg
bg
bg
CaF2 s H 2 SO 4 l o CaSO 4 s 2HF g
1 metric ton acid
1000 kg acid
0.60 kg HF
1 metric ton acid
1 kg acid
600 kg HF
Basis: 100 kg Ore dissolved (not fed) 100 kg Ore d issolved 0.96 kg CaF 2 /kg 0.04 kg SiO 2/ kg nA (kg 93% H2 SO4 ) 0.93 H2 SO4 kg/ kg 0.07 H2 O kg/ kg
n1 n2 n3 n4
(kg CaSO4) (kg HF) (kg H 4SiF6 ) (kg H 2SO 4)
n5 (kg H2 O)
Atomic balance - Si:
b g
0.04 100 kg SiO 2
28.1 kg Si
n3 (kg H 4 SiF6 )
60.1 kg SiO 2
28.1 kg Si n3 146.1 kg H 4 SiF6
Atomic balance - F:
b g
0.96 100 kg CaF2
38.0 kg F
n2 (kg HF)
78.1 kg CaF2 114.0 kg F 9.72 kg H 4 SiF6 n2 146.1 kg H 4 SiF6 600 kg HF 100 kg ore diss. 41.2 kg HF
1 kg ore feed 0.95 kg ore diss.
4-43
19.0 kg F 20.0 kg HF 412 . kg HF 1533 kg ore
9.72 kg H 4 SiF6
4.53
a.
C 6 H 6 Cl 2 o C 6 H 5 Cl HCl C 6 H 5 Cl Cl 2 o C 6 H 4 Cl 2 HCl C 6 H 4 Cl 2 Cl 2 o C 6 H 3 Cl 3 HCl Convert output wt% to mol%: Basis 100 g output species C6H 6 C 6 H 5 Cl C 6 H 4 Cl 2 C 6 H 3 Cl 3
g 65.0 32.0 2.5 0.5
Mol. Wt. 78.11 112.56 147.01 181.46
mol 0.832 0.284 0.017 0.003
mol % 73.2 25.0 1.5 0.3
total 1.136 Basis: 100 mol output n 4 (mol HCl(g )) n 3 (mol I)
n1 (mol C6 H6 )
65.0 mo l C6 H6 32.0 mo l C6 H 5 Cl 2.5 mo l C 6 H 4 Cl 2 0.5 mo l C6 H 3 Cl 3
n2 (mol Cl 2) n3 (mol I) b.
4 unknowns -3 atomic balances -1 wt% Cl 2 in feed 0 D.F.
b
100 mol C H g H balance: 6b100g 6b65.0g 5b32.0g 4b2.5g 3b0.5g n n 38.5 mol HCl Cl balance: 2n 38.5 32.0 2b2.5g 3b0.5g n 38.5 mol Cl Theoretical C H 38.5 mol Cl b1 mol C H 1 mol Cl g 38.5 mol C H Excess C H : b100 38.5g 38.5 u 100% 160% excess C H Fractional Conversion: b100 65.0g 100 0.350 mol C H react / mol fed
6 65.0 32.0 2.5 0.5 n1
C balance: 6n1
6
4
2
6
6
6
4
2
6
2
6
2
6
2
6
6
6
6
6
6
6
Yield: (32.0 mol C 6 H 5 Cl) (38.5 mol C 6 H 5 Cl maximum) = 0.831
U| V| W
38.5 mol Cl 2 70.91 g Cl 2 1 g gas 2091 g gas g gas 0.357 mole Cl 2 0.98 g Cl 2 g liquid Liquid feed: 100 mol C 6 H 6 78.11 g C 6 H 6 / mol C 6 H 6 7811 g liquid Gas feed:
b
gb
g
c.
Low conversion low residence time in reactor lower chance of 2nd and 3rd reactions occurring. Large excess of C 6 H 6 Cl 2 much more likely to encounter C 6 H 6 than substituted C 6 H 6 higher selectivity.
d.
Dissolve in water to produce hydrochloric acid.
e.
Reagent grade costs much more. Use only if impurities in technical grade mixture affect the reaction rate or desired product yield.
4-44
4.54
a.
2CO 2 2CO O 2
2A 2B C
O 2 N 2 2NO
C D 2E
bn bn bn bn bn
n A0 2[ e1 yA n B 0 2[ e 2 yB nC 0 [ e1 [ e 2 y C nD0 [ e2 yD n E 0 2[ e 2 yE
nA nB nC nD nE
bn
ntotal = nT 0 [ e1
gb gb
g g
2[ e1 nT 0 [ e1 n T 0 [ e1 B 0 2[ e1 nT 0 [ e1 C 0 [ e1 [ e 2 n T 0 [ e1 D 0 1[ e 2 nT 0 [ e1 E 0 2[ e 2 A0
gb gb
gb
g g
g
n A0 n B 0 nC 0 n D 0 n E 0
T0
g
Equilibrium at 3000K and 1 atm
bn
g bn [ [ g 01071 . bn 2[ g bn [ g bn 2[ g bn [ [ gbn [ g 0.01493
y B2 y C y 2A
2[ e1
2
A0
e1
C0 2
e1
T0
e2
e1
2
y E2 yC y D
B0
A0
E f1
B0
b
e1
01071 . n A0 2[ e1 f2
b
e2
e2
g bn
D0
2
T0
e2
g b gbn
g bn [ [ g g bn 2[ g 0
[ e1 n B 0 2[ e1
0.01493 nC 0 [ e1 [ e 2
D0
[ e2
2
C0
e1
e2
2
E0
e2
U| V| b g W b g
0 Defines functions f 1 [ 1 , [ 2 and f2 [1, [ 2
b.
Given all nio’s, solve above equations for [e1 and [e2 nA, nB, nC, nD, nE yA, yB, yC, yD, yE
c.
nA0 = nC0 = nD0 = 0.333, nB0 = nE0 = 0 [e1 =0.0593, [e2 = 0.0208 yA = 0.2027, yB = 0.1120, yC = 0.3510, yD = 0.2950, yE = 0.0393
d.
a11 d 1 a12 d 2 f 1 a12 f 2 a 22 f 1 d1 a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a11 d 1 a12 d 2 f 1 a12 f 2 a 22 f 1 d1 a11 a 22 a12 a 21
b[ g
e1 new
[ e1 d 1
a 21 d 1 a 22 d 2 f 2 a 21 f 1 a11 f 2 d2 a11 a 22 a12 a 21
b[ g
e 2 new
[ e1 d 2
a 21 d 1 a 22 d 2 f 2 a 21 f 1 a11 f 2 d2 a11 a 22 a12 a 21
b[ g
e 2 new
(Solution given following program listing.)
4-45
[ e1 d 2
4.54 (cont’d) . 1 30 2
3
100 4 120
IMPLICIT REAL * 4(N) WRITE (6, 1) FORMAT('1', 30X, 'SOLUTION TO PROBLEM 4.57'///) READ (5, *) NA0, NB0, NC0, ND0, NE0 IF (NA0.LT.0.0)STOP WRITE (6, 2) NA0, NB0, NC0, ND0, NE0 FORMAT('0', 15X, 'NA0, NB0, NC0, ND0, NE0 *', 5F6.2/) NTO = NA0 + NB0 + NC0 + ND0 + NE0 NMAX = 10 X1 = 0.1 X2 = 0.1 DO 100 J = 1, NMAX NA = NA0 – X1 – X1 NB = NB0 + X1 + X1 NC = NC0 + X1 – X2 ND = ND0 – X2 NE = NE0 + X2 + X2 NAS = NA ** 2 NBS = NB ** 2 NES = NE ** 2 NT = NT0 + X1 F1 = 0.1071 * NAS * NT – NBS * NC F2 = 0.01493 * NC * ND – NES A11 = –0.4284 * NA * NT * 0.1071 * NAS – 4.0 * NB * NC – NBS A12 = NBS A21 = 0.01493 * ND A22 = –0.01493 * (NC + ND) – 4.0 * NE DEN = A11 * A22 – A12 * A21 D1 = (A12 * F2 – A22 * F1)/DEN D2 = (A21 * F1 – A11 * F2)/DEN X1C = X1 + D1 X2C = X2 + D2 WRITE (6, 3) J, X1, X2, X1C, X2C FORMAT(20X, 'ITER *', I3, 3X, 'X1A, X2A =', 2F10.5, 6X, 'X1C, X2C =', * 2F10.5) IF (ABS(D1/X1C).LT.1.0E–5.AND.ABS(D2/X2C).LT.1.0E–5) GOTO 120 X1 = X1C X2 = X2C CONTINUE WRITE (6, 4) NMAX FORMAT('0', 10X, 'PROGRAM DID NOT CONVERGE IN', I2, 'ITERATIONS'/) STOP YA = NA/NT YB = NB/NT YC = NC/NT YD = ND/NT YE = NE/NT WRITE (6, 5) YA, YB, YC, YD, YE 5 FORMAT ('0', 15X, 'YA, YB, YC, YD, YE =', 1P5E14.4///) GOTO 30 END $DATA 0.3333 0.00 0.3333 0.3333 0.0 0.50 0.0 0.0 0.50 0.0 0.20 0.20 0.20 0.20 0.20
4-46
4.54 (cont’d) SOLUTION TO PROBLEM 4.54 NA0, NB0, NC0, ND0, NE0 = 0.33 0.00 0.33 ITER = 1 X1A, X2A = 0.10000 0.10000 ITER = 2 X1A, X2A = 0.06418 0.05181 ITER = 3 X1A, X2A = 0.05969 0.02486 ITER = 4 X1A, X2A = 0.05437 0.02213 ITER = 5 X1A, X2A = 0.05931 0.02086 ITER = 6 X1A, X2A = 0.05930 0.02083
YA, YB, YC, YD, YE =
0.33 0.00 X1C, X2C = 0.06418 X1C, X2C = 0.05969 X1C, X2C = 0.05937 X1C, X2C = 0.05931 X1C, X2C = 0.05930 X1C, X2C = 0.05930
2.0270E 01 2.9501E 01
NA0, NB0, NC0, ND0, NE0 = 0.20 0.20 ITER = 1 X1A, X2A = 0.10000 p ITER = 7 X1A, X2A = –0.02244
0.20
1.1197 E 01 3.9319 E 02
0.20 0.10000
3.5100E 01
0.20 X1C, X2C = 0.00012
–0.08339 X1C, X2C = –0.02244
YA, YB, YC, YD, YE = 2.5051E 01 15868 . E 01 2.8989 E 01 3.3991E 02 4.55
0.05181 0.02986 0.02213 0.02086 0.02083 0.02083
0.00037 –0.08339
2.6693E 01
(B)
a.
mB0 (kg A / h) 1 kg B/ kg A fed to reactor ( A) m A 0 (kg A / h) x RA (kg R / kg A)
( P) m A 0 (kg A / h) x RA (kg R / kg A)
m3 (kg A / h)
R o 5
x R 3 (Kg R / kg)
99% conv.
m P (kg P / h) 0.0075 kg R / kg P
f mA0 (kg A / h) x RA (kg R / kg A)
Reactor
Splitting point 4 unknowns (mA0, mB0, f, xRA) -1 independent balance (mass) 3 D.F.
4 unknowns (xRA, mB0, m3, xRA) -1 balance (mass) -1 conversion 2 D.F
Mixing point
Total process
5 unknowns (f, m3, mP, xRA, xR3)
7 unknowns (mA0, xRA, f, mB0, m3, xR3, mP)
-2 balances (mass, R)
-5 relations
3 D.F.
2 D.F.
4-47
4.55 (cont’d) b. Mass balance on splitting point: mA0 = mB0 + f mA0
(1)
Mass balance on reactor: 2 mB0 = m3
(2)
99% conversion of R: xR3 m3 = 0.01 xRA mB0
(3)
Mass balance on mixing point: m3 + f mA0 = mP R balance on mixing point: xR3 m3 + xRA f mA0 = 0.0075 mP
(4) (5)
Given xRA and mP, solve simultaneously for mA0, mB0, f, m3, xR3 c.
mA0 = 2778 kg A/h mB0 = 2072 kg B/h fA = 0.255 kg bypass/kg fresh feed mP 4850 4850 4850 4850 4850 4850 4850 4850 4850
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 3327 3022 2870 2778 2717 2674 2641 2616 2596
mB0 1523 1828 1980 2072 2133 2176 2209 2234 2254
f 0.54 0.40 0.31 0.25 0.21 0.19 0.16 0.15 0.13
mP 2450 2450 2450 2450 2450 2450 2450 2450 2450
xRA 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
mA0 1663 1511 1435 1389 1359 1337 1321 1308 1298
mB0 762 914 990 1036 1066 1088 1104 1117 1127
f 0.54 0.40 0.31 0.25 0.22 0.19 0.16 0.15 0.13
f v s . x RA f (kg bypass/kg fresh feed)
d.
0.60 0.50 0.40 0.30 0.20 0.10 0.00 0.00
0.02
0.04
0.06 x R A (k g R / k g A )
4-48
0.08
0.10
0.12
4.56
a.
900 kg HCHO 1 kmol HCHO h 30.03 kg HCHO
30.0 kmol HCHO / h
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h n 2 (kmol H 2 / h) n 3 (kmol CH 3 OH / h)
% conversion:
30.0 n1
0.60 n1
50.0 kmol CH 3 OH / h
b.
n (kmol CH OH / h) 1 3
30.0 kmol HCHO / h
30.0 kmol HCHO / h
n 2 (kmol H 2 / h)
n 2 (kmol H 2 / h)
n 3 (kmol CH 3 OH / h) n (kmol CH OH / h) 3 3
Overall C balance: n1 (1) = 30.0 (1) n1 = 30.0 kmol CH3OH/h (fresh feed) Single pass conversion:
30.0 n1 n3
0.60 n3
20.0 kmol CH 3OH / h
n1 + n3 = 50.0 kmol CH3OH fed to reactor/h
4.57
c.
Increased xsp will (1) require a larger reactor and so will increase the cost of the reactor and (2) lower the quantities of unreacted methanol and so will decrease the cost of the separation. The plot would resemble a concave upward parabola with a minimum around xsp = 60%.
a.
Convert effluent composition to molar basis. Basis: 100 g effluent: 10.6 g H 2 1 mol H 2 2.02 g H 2 64.0 g CO 1 mol CO 28.01 g CO
5.25 mol H 2 2.28 mol CO
25.4 g CH 3 OH 1 mol CH 3 OH 32.04 g CH 3 OH 0.793 mol CH 3 OH
4-49
H : 0.631 mol H / mol 2 2 CO: 0.274 mol CO / mol CH OH: 0.0953 mol CH OH / mol 3 3
4.57 (cont’d) n4 (mol / min) 0.004 mol CH 3OH(v)/ mol x (mol CO/ mol) (0.896 - x) (mol H 2 / mol) Cond.
Reactor
350 mol/ min n1 (mol CO/ min) n 2 (mol H 2 / min)
0.631 mol CH 3OH(v)/ mol
n 3 (mol CH 3OH(l) / min)
0.274 mol CO/ mol CO H 2 o CH 3OH 0.0953 mol H / mol 2
Condenser
Overall process
3 unknowns (n3, n4, x)
2 unknowns (n1, n2)
-3 balances 0 degrees of freedom
-2 independent atomic balances 0 degrees of freedom
Balances around condenser
U| V| W
n 32.1 mol CH 3 OH(l) / min CO: 350 0.274 n x 3 4 n H : 350 0.631 n ( 0.996 x ) 318.7 mol recycle / min 2 4 4 CH OH: 350 0.0953 n 0.004 n x .301 molCO / mol 3 3 4 Overall balances
U|V |W
n C: n = n 1 3 1 n H: 2n = 4n 2 2 2
32.08 mol / min CO in feed 64.16 mol / min H 2 in feed
Single pass conversion of CO:
32.08 318.72 0.3009 350 0.274 u 100% 32.08 318.72 0.3009
–
32.08 0 u 100% 100% 32.08 Reactor conditions or feed rates drifting. (Recalibrate measurement instruments.)
–
Impurities in feed. (Re-analyze feed.)
–
Leak in methanol outlet pipe before flowmeter. (Check for it.)
Overall conversion of CO: b.
25.07%
4-50
4.58
a.
Basis: 100 kmol reactor feed/hr n3 (kmol CH4 /h) 100 kmol /h
Reactor
n1 (kmol CH4 /h) 80 kmol CH4 /h n2 (kmol Cl2 /h) 20 kmol Cl2 /h
n3 (kmol CH4 /h) n4 (kmol HCl /h) 5n5 (kmol CH3Cl /h) n5 (kmol CH2Cl 2 /h)
Cond.
Solvent
Absorb n3 (kmol CH4 /h) n4 (kmol HCl/h)
n4 (kmol HCl/h)
5n5 (kmol CH3Cl /h) Still 5n5 (kmol CH3Cl /h) n5 (kmol CH2Cl 2 /h)
n5 (kmol CH2Cl 2 /h)
Overall process: 4 unknowns (n1, n2, n4, n5) -3 balances = 1 D.F. Mixing Point: 3 unknowns (n1, n2, n3) -2 balances = 1 D.F. Reactor: 3 unknowns (n3, n4, n5) -3 balances = 0 D.F. Condenser: 3 unknowns (n3, n4, n5) -0 balances = 3 D.F. Absorption column: 2 unknowns (n3, n4) -0 balances = 2 D.F. Distillation Column: 2 unknowns (n4, n5) -0 balances = 2 D.F. Atomic balances around reactor: ½ 1) C balance : 80 n 3 5n 5 n 5 ° 2) H balance : 320 4n 3 n 4 15n 5 2n 5 ¾ Solve for n 3 , n 4 , n 5 ° 3) Cl balance : 40 n 4 5n 5 2n 5 ¿ CH4 balance around mixing point: n1 = (80 – n3) Cl2 balance: n2 = 20 b.
Solve for n1
For a basis of 100 kmol/h into reactor n1 = 17.1 kmol CH4/h
n4 = 20.0 kmol HCl/h
n2 = 20.0 kmol Cl2/h
5n5 = 14.5 kmol CH3Cl/h
n3 = 62.9 kmol CH4/h c.
(1000 kg CH3Cl/h)(1 kmol/50.49 kg) = 19.81 kmol CH3Cl/h Scale factor =
19.81 kmol CH 3 Cl/h 14.5 kmol CH 3Cl/h
1.366
n tot 50.6 kmol/h n (17.1)(1.366) 23.3 kmol CH 4 /h ½ Fresh feed: 1 ¾ n 2 (20.0)(1.366) 27.3 kmol Cl 2 /h ¿ 46.0 mol% CH 4 , 54.0 mole% Cl 2 Recycle: n3 = (62.9)(1.366) = 85.9 kmol CH4 recycled/h
4-51
4.59
a.
Basis: 100 mol fed to reactor/h 25 mol O2/h, 75 mol C2H4/h n1 (mol C 2H 4 //h) n2 (mol O 2 /h)
Seperator
reactor nC2H4 ( mol C 2H 4 /h) nO2 (mol O 2 /h)
75 mol C 2H 4 //h 25 mol O2 /h
n1 (mol C 2H 4 //h) n2 (mol O 2 /h) n3 (mol C 2H 4O /h) n4 (mol CO 2 /h) n5 (mol H 2O /h)
n3 (mol C 2H 4O /h)
n4 (mol CO 2 /h) n5 (mol H 2O /h)
Reactor 5 unknowns (n1 - n5) -3 atomic balances -1 - % yield -1 - % conversion 0 D.F. Strategy: 1. Solve balances around reactor to find n1- n5 2. Solve balances around mixing point to find nO2, nC2H4 (1) % Conversion n1 = .800 * 75 (2) % yield: (.200)(75) mol C 2 H 4 u
90 mol C 2 H 4 O 100 mol C 2 H 4
n 3 (production rate of C 2 H 4 O)
(3) C balance (reactor): 150 = 2 n1 + 2 n3 + n4 (4) H balance (reactor): 300 = 4 n1 + 4 n3 + 2 n5 (5) O balance (reactor): 50 = 2 n2 + n3 + 2 n4 + n5 (6) O2 balance (mix pt): nO2 = 25 – n2 (7) C2H4 balance (mix pt): nC2H4 = 75 – n1 Overall conversion of C2H4: 100% b.
c.
n1 = 60.0 mol C2H4/h
n5 = 3.00 mol H2O/h
n2 = 13.75 mol O2 /h
nO2 = 11.25 mol O2/h
n3 = 13.5 mol C2H4O/h n4 = 3.00 mol CO2/h
nC2H4 = 15.0 mol C2H4/h 100% conversion of C2H4
Scale factor =
2000 lbm C 2 H 4 O 1 lb - mole C 2 H 4 O h h 44.05 lbm C 2 H 4 O 13.5 mol C 2 H 4 O
nC2H4 = (3.363)(15.0) = 50.4 lb-mol C2H4/h nO2 = (3.363)(11.25) = 37.8 lb-mol O2/h
4-52
3.363
lb mol / h mol / h
4.60
a.
Basis: 100 mol feed/h 100 mol/h
n1 (mol /h)
32 mol CO/h 64 mol H 2 / h 4 mol N 2 / h
.13 mol N 2 /mol
reactor
n3 (mol CH 3 OH / h)
cond.
500 mol / h x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x2 (mol H 2 / h)
n3 (mol / h) x1 (mol N 2 /mol) x2 (mol CO / mol) 1-x1-x2 (mol H 2 / h) Purge
Mixing point balances: total: (100) + 500 = n1 n1 = 600 mol/h N2: 4 + x1 * 500 = .13 * 600 x1 = 0.148 mol N2/mol Overall system balances: N2: 4 = .148 * n3 n3 = 27 mol/h Atomic C: 32 = n2 + x2*27 n2 = 24.3 mol CH3OH/h Atomic H: 2 * 64 = 4*24.3 + 2*(1-0.148-x2)*27 x2 = 0.284 mol CO/mol Overall CO conversion: 100*[32-0.284(27)]/32 = 76% Single pass CO conversion: 24.3/ (32+.284*500) = 14% b.
Recycle: To recover unconsumed CO and H2 and get a better overall conversion. Purge: to prevent buildup of N2.
4.61
a.
2N2 + 3H2 -> NH3 (1-yp) (1-fsp) n1 (mol N2) (1-yp) (1-fsp) 3n1 (mol H2) (1-yp) n2 (mol I)
1 mol (1-XI0)/4 (mol N2 / mol) 3/4 (1-XI0) (mol H2 / mol) XI0 (mol I / mol)
nr (mol) n1 (mol N2) 3n1 (mol H2) n2 (mol I)
Reactor
4-53
(1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I)
nr (mol) (1-fsp) n1 (mol N2) (1-fsp) 3n1 (mol H2) n2 (mol I) 2 fsp n1 (mol NH3)
yp (1-fsp) n1 (mol N2) yp (1-fsp) 3n1 (mol H2) yp n2 (mol I)
Condenser
np (mol) 2 fsp n1 (mol NH3)
4.61 (cont’d) At mixing point: N2: (1-XI0)/4 + (1-yp)(1-fsp) n1 = n1 I: XI0 + (1-yp) n2 = n2 Total moles fed to reactor: nr = 4n1 + n2 Moles of NH3 produced: np = 2fspn1 Overall N2 conversion:
b.
(1 X I0 ) / 4 y p (1 f sp )n 1 (1 X I0 ) / 4
u 100%
XI0 = 0.01 fsp = 0.20 yp = 0.10 n1 = 0.884 mol N2
nr = 3.636 mol fed
n2 = 0.1 mol I
np = 0.3536 mol NH3 produced N2 conversion = 71.4%
c.
Recycle: recover and reuse unconsumed reactants. Purge: avoid accumulation of I in the system.
d.
Increasing XI0 results in increasing nr, decreasing np, and has no effect on fov. Increasing fsp results in decreasing nr, increasing np, and increasing fov. Increasing yp results in decreasing nr, decreasing np, and decreasing fov. Optimal values would result in a low value of nr and fsp, and a high value of np, this would give the highest profit. XI0 0.01 0.05 0.10 0.01 0.01 0.01 0.10 0.10 0.10
fsp 0.20 0.20 0.20 0.30 0.40 0.50 0.20 0.20 0.20
yp 0.10 0.10 0.10 0.10 0.10 0.10 0.20 0.30 0.40
nr 3.636 3.893 4.214 2.776 2.252 1.900 3.000 2.379 1.981
4-54
np 0.354 0.339 0.321 0.401 0.430 0.450 0.250 0.205 0.173
fov 71.4% 71.4% 71.4% 81.1% 87.0% 90.9% 55.6% 45.5% 38.5%
4.62
a.
i - C 4 H 10 C 4 H 8
Basis: 1-hour operation
C 8 H 18
n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10) n 1 (C 8 H18) m 4 (91% H 2 SO4 )
D
P decanter
E
Units of n : kmol Units of m: kg
n 1 (C 8 H18) n 2 (n-C 4 H10 ) n 3 (i-C 4 H 10)
still
n 5 (n-C 4 H10) n 6 (i-C 4 H 10) n 7 (C 8 H18) m 8 (91% H 2 SO 4 )
reactor
C B
F
n 1 (C 8 H18) n 2 (n-C 4 H10 )
m 4 (kg 91% H 2 SO4 )
40000 kg A n 0 kmol 0.25 i-C4 H10 0.50 n-C4 H10 0.25 C4 H 8
n 3 (i-C 4 H 10)
Calculate moles of feed M
. g b0.25gb5610 . g b0.75gb5812
0.25 M L C4 H10 0.50 M n C4 H10 0.25 M C4 H 8 57.6 kg kmol
n0
b40000 kggb1 kmol 57.6 kgg
Overall n - C 4 H 10 balance: n2
694 kmol
b0.50gb694g
347 kmol n - C 4 H 10 in product
C 8 H 18 balance: n1
b0.25gb694g kmol C H 4
8
react 1 mol C 8 H 18 1 mol C 4 H 8
. kmol C 8 H 8 in product 1735
b
At (A), 5 mol i - C 4 H 10 1 mole C 4 H 8 n mol i - C 4 H 10
b
Note: n mol C 4 H 8
g
694g g b 5gb 0 .25 gb A
moles C 4 H 8 at A=173.5
173.5 at (A), (B) and (C) and in feed
b gb g
i - C 4 H 10 balance around first mixing point 0.25 694 n3 n3
867.5
694 kmol i - C 4 H 10 recycled from still
At C, 200 mol i - C4 H10 mol C4 H 8
b
n mol i - C4 H10
g b200gb1735. g C
4-55
34,700 kmol i - C4 H10
867.5 kmol i -C 4 H 10 at
bA g and b Bg
4.62 (cont’d) i - C 4 H 10 balance around second mixing point 867.5 n6 n6
34,700
33,800 kmol C 4 H 10 in recycle E
Recycle E: Since Streams (D) and (E) have the same composition,
b g bmoles n - C H g bmoles C H g n bmoles C H g n
n5 moles n - C 4 H 10 n2 n7 n1
4
b g bmoles i - C H g
n6 moles i - C 4 H 10
E
n3
10 D
8
18 E
6
8
18 D
3
4
n7
E
n5
16,900 kmol n - C 4 H 10
10 D
8460 kmol C 4 H 18
Hydrocarbons entering reactor:
kg I b347 16900gbkmol n - C H g FGH 58.12 kmol JK kg I F . kg IJ 1735. kmol C H FG 5610 b867.5 33800gb kmol i - C H g G 5812 . H kmol K H kmol JK kg I F 8460 kmol C H G 114.22 J 4.00 u 10 kg . H kmol K H SO solution entering reactor 4.00 u 10 kg HC 2 kg H SO baq g band leaving reactor g 1 kg HC 8.00 u 10 kg H SO baq g m b H SO in recycleg n b n - C H in recycleg 8.00 u 10 b H SO leaving reactor g n n b n - C H leaving reactor g m 7.84 u 10 kg H SO baq g in recycle E 4
10
4
8
2
10
4
6
18
6
4
6
8
2
6
4
2
m4
2
4
6
8
8
2
2
2
4
4
5
4
10
5
4
10
4
H 2 SO 4 entering reactor H 2 SO 4 in E
b g
16 . u 10 5 kg H 2 SO 4 aq recycled from decanter
b g 1480 kmol H SO in recycle d ib g d16. u 10 ib0.09gkg H O b1 kmol 18.02 kgg 799 kmol H O from decanter
16 . u 10 5 0.91 kg H 2 SO 4 1 kmol 98.08 kg 5
2
2
4
2
Summary: (Change amounts to flow rates) Product: 173.5 kmol C 8 H 18 h , 347 kmol n - C 4 H 10 h Recycle from still: 694 kmol i - C 4 H 10 h Acid recycle: 1480 kmol H 2 SO 4 h , 799 kmol H 2 O h Recycle E: 16,900 kmol n - C 4 H 10 h , 33,800 kmol L - C 4 H 10 h , 8460 kmol C 8 H 18 h, 7.84 u 10 6 kg h 91% H 2 SO 4 72,740 kmol H 2 SO 4 h , 39,150 kmol H 2 O h
4-56
4.63
a.
A balance on ith tank (input = output + consumption) Ai kC Ai C Bi mol liter min V L v L min C A , i 1 mol L vC
b
g
b
g
b
E y v, note V / v
gbg
W
C Ai kW C Ai C Bi
C A , i 1 B balance. By analogy, C B , i 1
C Bi kW C Ai C Bi
Subtract equations C Bi C Ai
C B , i 1 C A, i 1
A from balances on bi 1g tank
C B , i 2 C A, i 2 C B 0 C A 0
st
b.
C Bi C Ai
C A , i 1 2 C Ai
C B 0 C A 0 C Bi
b
C Ai C B 0 C A0 . Substitute in A balance from part (a).
g
2 0 C Ai kW C Ai C Ai C B 0 C A0 . Collect terms in C Ai , C 1Ai , C Ai .
b
g
kW C AL 1 kW C B 0 C A0 C A, i 1 2 D C AL E C AL J
0 where D
0 kW , E
b
g
1 kW C B 0 C A0 , J
C A , i 1
E E 2 4DJ
(Only + rather than r: since DJ is negative and the 2D negative solution would yield a negative concentration.)
Solution: C Ai
c. k= v= V= CA0 = CB0 = alpha = beta =
36.2 5000 2000 0.0567 0.1000 14.48 1.6270
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14
gamma -5.670E-02 -2.791E-02 -1.512E-02 -8.631E-03 -5.076E-03 -3.038E-03 -1.837E-03 -1.118E-03 -6.830E-04 -4.182E-04 -2.565E-04 -1.574E-04 -9.667E-05 -5.939E-05
CA(N) 2.791E-02 1.512E-02 8.631E-03 5.076E-03 3.038E-03 1.837E-03 1.118E-03 6.830E-04 4.182E-04 2.565E-04 1.574E-04 9.667E-05 5.939E-05 3.649E-05
xA(N) 0.5077 0.7333 0.8478 0.9105 0.9464 0.9676 0.9803 0.9880 0.9926 0.9955 0.9972 0.9983 0.9990 0.9994
(xmin = 0.50, N = 1), (xmin = 0.80, N = 3), (xmin = 0.90, N = 4), (xmin = 0.95, N = 6), (xmin = 0.99, N = 9), (xmin = 0.999, N = 13). As xmin o 1, the required number of tanks and hence the process cost becomes infinite. d.
(i) k increases N decreases (faster reaction fewer tanks) (ii) v increases N increases (faster throughput less time spent in reactor lower conversion per reactor)
(iii) V increases N decreases (larger reactor more time spent in reactor higher conversion per reactor)
4-57
4.64
a.
Basis: 1000 g gas Species
m (g)
MW
n (mol)
mole % (wet)
mole % (dry)
C3H8
800
44.09
18.145
77.2%
87.5%
C4H10
150
58.12
2.581
11.0%
12.5%
H2O
50
18.02
2.775
11.8%
Total
1000
23.501
100%
100%
Total moles = 23.50 mol, Total moles (dry) = 20.74 mol Ratio: 2.775 / 20.726 = 0.134 mol H 2 O / mol dry gas b.
C3H8 + 5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Theoretical O2: C3H8: C 4 H 10 :
5 kmol O 2 100 kg gas 80 kg C 3 H 8 1 kmol C 3 H 8 h 100 kg gas 44.09 kg C 3 H 8 1 kmol C 3 H 8
9.07 kmol O 2 / h
6.5 kmol O 2 100 kg gas 15 kg C 4 H 10 1 kmol C 4 H 10 h 100 kg gas 58.12 kg C 4 H 10 1 kmol C 4 H 10
1.68 kmol O 2 / h
Total: (9.07 + 1.68) kmol O2/h = 10.75 kmol O2/h Air feed rate:
10.75 kmol O 2 1 kmol Air 1.3 kmol air fed h .21 kmol O 2 1 kmol air required
66.5 kmol air / h
The answer does not change for incomplete combustion 4.65
5 L C 6 H 14 0.659 kg C 6 H 14 1000 mol C 6 H 14 L C 6 H 14 86 kg C 6 H 14
38.3 mol C 6 H 14
4 L C 7 H 16 0.684 kg C 7 H 16 1000 mol C 7 H 16 L C 7 H 16 100 kg C 7 H 16
27.36 mol C 7 H 16
C6H14 +19/2 O2 o 6 CO2 + 7 H2O
C6H14 +13/2 O2 o 6 CO + 7 H2O
C7H16 + 11 O2 o 7 CO2 + 8 H2O
C7H16 + 15/2 O2 o 7 CO + 8 H2O
Theoretical oxygen: 38.3 mol C 6 H14
9.5 mol O 2 27.36 mol C 7 H 16 11 mol O 2 mol C 6 H 14 mol C 7 H16
O2 fed: (4000 mol air )(.21 mol O2 / mol air) = 840 mol O2 fed Percent excess air:
840 665 u 100% 665
26.3% excess air
4-58
665 mol O 2 required
4.66
CO
1 O 2 o CO 2 2
H2
1 O2 o H2O 2
130 kmol/h 0.500 kmol N2/kmol x (kmol CO/mol) (0.500–x) (kmol H2/kmol) 20% excess air
Note: Since CO and H 2 each require 0 .5 mol O 2 / mol fuel for complete combustion, we can calculate the air feed rate without determining x CO . We include its calculation for illustrative purposes.
b
A plot of x vs. R on log paper is a straight line through the points R1
bR
99.7, x 2
2
@
x
g
0.05 and
10 . .
b ln R ln a
ln x
R
g
10.0, x1
a Rb
38.3 x
0.288
. g 1303 ln a lnb10 . g 1303 . lnb99.7g 6.00 a expb 6.00g 2.49 u 10 3
b
b
g b
ln 10 . 0.05 ln 99.7 10.0
x
. 2.49 u 10 3 R 1303
moles CO mol
Theoretical O : 175 kmol 0.288 kmol CO 0.5 kmol O 2 2 h kmol kmol CO Air fed: 4.67
a.
Theoretical O 2 :
o CO 2 2H 2 O
o 3CO 2 4H 2 O
17% excess air na (kmol air/h) 0.21 O2 0.79 N2
o 4CO 2 5H 2 O
b g
b g
h
0.0060 100 kmol C 3 H 8 h 207.0 kmol O 2 h
2
100 kmol/h 0.944 CH4 0.0340 C2H6 0.0060 C3H8 0.0050 C4H10
o 2CO 2 3H 2 O
0.944 100 kmol CH 4
kmol O 0.5 kmol O 2 2 43.75 kmol H h 2 1.2 kmol air fed kmol air 250 1 kmol air required h
0.212 kmol H 2 kmol
43.75 kmol O required 1 kmol air 2 h 0.21 kmol O
CH 4 2O 2 7 C2 H 6 O2 2 C 3 H 8 5O 2 13 C 4 H 10 O 2 2
175 kmol h
b g
2 kmol O 2 0.0340 100 kmol C 2 H 6 1 kmol CH 4 h
3.5 kmol O 2 1 kmol C 2 H 6
0.0050 100 kmol C 4 H 10 5 kmol O 2 1 kmol C 3 H 3 h
6.5 kmol O 2 1 kmol C 4 H 10
b g
4-59
4.67 (cont’d) Air feed rate: n f
207.0 kmol O 2
1 kmol air
h
0.21 kmol O 2
b
gb
gb
1.17 kmol air fed kmol air req.
g
b.
na
n f 2 x1 35 . x 2 5x 3 6.5x 4 1 Pxs 100 1 0.21
c.
n f
aR f , ( n f
n a
bRa , (n a
xi
kAi
75.0 kmol / h, R f 550 kmol / h, Ra
¦x
k
i
i
¦A
60) n f 25) n a
1 k
i
i
1153 kmol air h
125 . Rf 22.0 Ra
1
¦A
i
i
Ai
xi
¦A
, i = CH 4 , C 2 H 4 , C 3 H 8 , C 4 H 10
i
i
4.68
Run 1 2 3
Pxs 15% 15% 15%
Rf 62 83 108
A1 248.7 305.3 294.2
A2 19.74 14.57 16.61
A3 6.35 2.56 4.78
A4 1.48 0.70 2.11
Run 1 2 3
nf 77.5 103.8 135.0
x1 0.900 0.945 0.926
x2 0.0715 0.0451 0.0523
x3 0.0230 0.0079 0.0150
x4 0.0054 0.0022 0.0066
na 934 1194 1592
d.
Either of the flowmeters could be in error, the fuel gas analyzer could be in error, the flowmeter calibration formulas might not be linear, or the stack gas analysis could be incorrect.
a.
C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Basis:
100 mol C4H10
nCO2 (mol CO2) nH2O (mol H2O) nC4H10 (mol C4H10) nO2 (mol O2) nN2 (mol N2)
Pxs (% excess air) nair (mol air) 0.21 O2 0.79 N2 D.F. analysis 6 unknowns (n, n1, n2, n3, n4, n5) -3 atomic balances (C, H, O) -1 N2 balance -1 % excess air -1 % conversion 0 D.F.
4-60
Ra 42.4 54.3 72.4
4.68 (cont’d) b.
i) Theoretical oxygen = (100 mol C4H10)(6.5 mol O2/mol C4H10) = 650 mol O2 n air
(650 mol O 2 )(1 mol air / 0.21 mol O 2 )
100% conversion n C4H10 n N2 nCO2 n H2O
0 , nO 2
3095 mol air
0
b0.79gb3095 molg 2445 mol b100 mol C H reactgb4 mol CO mol C H g b100 mol C H reactgb5 mol H O mol C H g 4
4
10
2
10
2
4
4
10
10
U| 73.1% N 400 mol CO V12.0% CO 500 mol H O |W14.9% H O 2
2
2
2
2
ii) 100% conversion nC4H10 = 0 20% excess nair = 1.2(3095) = 3714 mol (780 mol O2, 2934 mol N2) Exit gas: 400 mol CO2
10.1% CO2
500 mol H2O
12.6% H2O
130 mol O2
3.3% O2
2934 mol N2
74.0% N 2
iii) 90% conversion nC4H10 = 10 mol C4H10 (90 mol C4H10 react, 585 mol O2 consumed) 20% excess: nair = 1.2(3095) = 3714 mol (780 mol O2, 2483 mol N2) Exit gas: 10 mol C4H10
0.3% C4H10 9.1% CO2
360 mol CO2 450 mol H2O (v)
4.69
a.
11.4% H2O
195 mol O2
4.9% O2
2934 mol N2
74.3% N 2
C3H8 + 5 O2 o 3 CO2 + 4 H2O
H2 +1/2 O2 o H2O
C3H8 + 7/2 O2 o 3 CO + 4 H2O Basis: 100 mol feed gas 100 mol 0.75 mol C3H8 0.25 mol H2
n1 (mol C3H8) n2 (mol H2) n3 (mol CO2) n4 (mol CO) n5 (mol H2O) n6 (mol O2) n7 (mol N2)
n0 (mol air) 0.21 mol O2/mol 0.79 mol N2/mol Theoretical oxygen:
75 mol C 3 H 8
5 mol O 2 25 mol H 2 0.50 mol O 2 mol C 3 H 8 mol H 2
4.69 (cont’d) 4-61
387.5 mol O 2
Air feed rate: n0
387.5 mol O 2 1 kmol air 125 . kmol air fed h 0.21 kmol O 2 1 kmol air req' d.
90% propane conversion n1
2306.5 mol air
0100 . (75 mol C 3 H 8 ) = 7.5 mol C 3 H 8
(67.5 mol C 3 H 8 reacts) 85% hydrogen conversion n2
0150 . (25 mol C 3 H 8 ) = 3.75 mol H 2
0.95(67.5 mol C 3 H 8 react) 3 mol CO 2 generated mol C 3 H 8 react
95% CO 2 selectivity n3
192.4 mol CO 2 0.05(67.5 mol C 3 H 8 react) 3 mol CO generated mol C 3 H 8 react
5% CO selectivity n3
FG H
H balance: (75 mol C 3 H 8 ) 8
. mol CO 101
IJ K
mol H ( 25 mol H 2 )(2) mol C 3 H 8
(7.5 mol C 3 H 8 )(8) (3.75 mol H 2 )(2) n5 ( mol H 2 O)(2) n5
2912 . mol H 2 O
mol O ) (192.4 mol CO 2 )(2) mol O 2 (101 . mol CO)(1) ( 2912 . mol H 2 O)(1) + 2n6 ( mol O 2 ) n6
1413 . mol O 2
O balance: (0.21 u 2306.5 mol O 2 )(2
N 2 balance: n7
0.79(2306.5) mol N 2
1822 mol N 2
Total moles of exit gas = (7.5 + 3.75 + 192.4 + 10.1 + 291.2 + 141.3 + 1822) mol = 2468 mol CO concentration in exit gas =
b.
101 . mol CO u 10 6 2468 mol
4090 ppm
If more air is fed to the furnace, (i)
more gas must be compressed (pumped), leading to a higher cost (possibly a larger pump, and greater utility costs)
(ii) The heat released by the combustion is absorbed by a greater quantity of gas, and so the product gas temperature decreases and less steam is produced.
4-62
4.70
a.
C5H12 + 8 O2 o 5 CO2 + 6 H2O Basis: 100 moles dry product gas n1 (mol C5H12)
100 mol dry product gas (DPG) 0.0027 mol C5H12/mol DPG 0.053 mol O2/mol DPG 0.091 mol CO2/mol DPG 0.853 mol N2/mol DPG n3 (mol H2O)
Excess air n2 (mol O2) 3.76n2 (mol N2)
3 unknowns (n1, n2, n3) -3 atomic balances (O, C, H) -1 N2 balance -1 D.F. Problem is overspecified b.
N2 balance: 3.76 n2 = 0.8533 (100) n2 = 22.69 mol O2 C balance: 5 n1 = 5(0.0027)(100) + (0.091)(100) n1 = 2.09 mol C5H12 H balance: 12 n1 = 12(0.0027)(100) + 2n3 n3 = 10.92 mol H2O O balance: 2n2 = 100[(0.053)(2) + (0.091)(2)] + n3 45.38 mol O = 39.72 mol O Since the 4th balance does not close, the given data cannot be correct.
c. n1 (mol C5H12) Excess air n2 (mol O2) 3.76n2 (mol N2)
100 mol dry product gas (DPG) 0.00304 mol C5H12/mol DPG 0.059 mol O2/mol DPG 0.102 mol CO2/mol DPG 0.836 mol N2/mol DPG n3 (mol H2O)
N2 balance: 3.76 n2 = 0.836 (100) n2 = 22.2 mol O2 C balance: 5 n1 = 100 (5*0.00304 + 0.102) n1 = 2.34 mol C5H12 H balance: 12 n1 = 12(0.00304)(100) + 2n3 n3 = 12.2 mol H2O O balance: 2n2 = 100[(0.0590)(2) + (0.102)(2)] + n3 44.4 mol O = 44.4 mol O ¸ Fractional conversion of C5H12:
2.344 100 u 0.00304 2.344
0.870 mol react/mol fed
Theoretical O2 required: 2.344 mol C5H12 (8 mol O2/mol C5H12) = 28.75 mol O2 % excess air:
22.23 mol O 2 fed - 18.75 mol O 2 required u 100% 18.6% excess air 18.75 mol O 2 required
4-63
4.71
a.
12 L CH 3 OH 1000 ml 0.792 g mol h L ml 32.04 g
296.6 mol CH 3 OH / h
CH3OH + 3/2 O2 o CO2 +2 H2O, CH3OH + O2 o CO +2 H2O 296.6 mol CH3OH(l)/h
n 2 ( mol dry gas / h) 0.0045 mol CH3OH(v)/mol DG 0.0903 mol CO2/mol DG 0.0181 mol CO/mol DG x (mol O2/mol DG) (1–x) (mol N2/mol DG) n 3 ( mol H 2 O(v) / h)
n1 (mol O 2 / h) 3.76n1 (mol N 2 / h)
4 unknowns (n1 , n 2 , n 3 , x ) – 4 balances (C, H, O, N2) = 0 D.F. b.
Theoretical O2: 296.6 (1.5) = 444.9 mol O2 / h C balance: 296.6 = n 2 (0.0045 + 0.0903 + 0.0181) n 2 = 2627 mol/h H balance: 4 (296.6) = n 2 (4*0.0045) + 2 n 3 n 3 = 569.6 mol H2O / h
O balance : 296.6 2n1 2627[0.0045 2(0.0903) 0.0181 2(0.8871- x)] 569.6 N2 balance: 3.76 n 1 = x ( 2627) Solving simultaneously n1 Fractional conversion: % excess air:
296.6 2627(0.0045) 296.6
0.960 mol CH 3 OH react/mol fed
574.3 444.9 u 100% 29.1% 444.9 569.6 mol H 2 O (2627 569.6) mol
Mole fraction of water:
4.72
574.3 mol O 2 / h, x = 0.822 mol N 2 / mol DG
0.178 mol H 2 O/mol
c.
Fire, CO toxicity. Vent gas to outside, install CO or hydrocarbon detector in room, trigger alarm if concentrations are too high
a.
G.C. Say ns mols fuel gas constitute the sample injected into the G.C. If xCH 4 and xC2 H 6 are the mole fractions of methane and ethane in the fuel, then
b g b n b molg x b mol CH
gb g molgb1 mol C 1 mol CH g
ns mol xC2 H 6 mol C2 H 2 mol 2 mol C 1 mol C 2 H 6
E
s
CH 4
b bmol CH
4
4
xC2 H 6 mol C2 H 6 mol fuel xCH 4
4
mol fuel
g
g
20 85
01176 . mole C 2 H 6 mole CH 4 in fuel gas
4-64
4.72 (cont’d)
b1.134 g H Ogb1 mol 18.02 gg
mole H 2 O 0.50 mol product gas mole product gas Basis: 100 mol product gas. Since we have the most information about the product stream composition, we choose this basis now, and would subsequently scale to the given fuel and air flow rates if it were necessary (which it is not). 2
Condensation measurement:
0126 .
CH 4 2O 2 o CO 2 2H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 100 mol dry gas / h
n1 (mol CH4 ) 0.1176 n1 (mol C2H6) n2 (mol CO2)
0.126 mol H2O / mol 0..874 mol dry gas / mol 0.119 mol CO2 / mol D.G. x (mol N2 / mol) (0.881-x) (mol O2 / mol D.G.)
n3 (mol O2 / h) 376 n3 (mol N2 / h)
Strategy: H balance n ; 1
C balance n 2 ;
b gb
H balance: 4n1 6 01176 . n1
UV W
N 2 balance n3 , x O balance
. gb2g n g b100gb0126
1
5.356 mol CH 4 in fuel
0.1176(5.356) = 0.630 mol C2H6 in fuel
b gb
g
C balance: 5.356 2 0.630 n2
. gn b100gb0.874gb0119
2
3.784 mol CO 2 in fuel
Composition of fuel: 5.356 mol CH 4 , 0.630 mol C 2 H 6 , 3.784 mols CO 2 0.548 CH 4 , 0.064 C 2 H 6 , 0.388 CO 2
b100gb0.874gx b0.881 x g . g b100gb0.874gb2g 0119 . b2gb3.784g 2n b100gb0126
N 2 balance: 3.76n3 O balance:
3
Solve simultaneously: n3 18.86 mols O 2 fed , x 0.813 5.356 mol CH 4 2 mol O 2 0.630 mol C 2 H 6 3.5 mol O 2 Theoretical O 2 : 1 mol CH 4 1 mol CH 4 12.92 mol O 2 required Desired O2 fed:
(5.356 0.630 3.784) mol fuel 7 mol air 0.21 mol O 2 = 14.36 mol O2 1 mol fuel mol air
Desired % excess air:
b.
Actual % excess air:
14.36 12.92 u 100% 11% 12.92
18.86 12.92 u 100% 46% 12.92
Actual molar feed ratio of air to fuel:
(18.86 / 0.21) mol air 9.77 mol feed
4-65
9 :1
4.73
a.
C3H8 +5 O2 o 3 CO2 + 4 H2O, C4H10 + 13/2 O2 o 4 CO2 + 5 H2O Basis 100: mol product gas n1 (mol C3H8) n2 (mol C4H10)
100 mol 0.474 mol H2O/mol x (mol CO2/mol) (0.526–x) (mol O2/mol)
n3 (mol O2) Dry product gas contains 69.4% CO2
x 0.526 x
69.4 x 30.6
0.365 mol CO 2 /mol
3 unknowns (n1, n2, n3) – 3 balances (C, H, O) = 0 D.F. O balance: 2 n3 = 152.6 n3 = 76.3 mol O2 n1 7.1 mol C 3 H 8 C balance : 3 n1 4 n 2 36.5 ½ 65.1% C 3 H 8 , 34.9% C 4 H10 ¾ H balance : 8 n1 10 n 2 94.8¿ n 2 3.8 mol C 4 H10 b.
nc=100 mol (0.365 mol CO2/mol)(1mol C/mol CO2) = 365 mol C nh = 100 mol (0.474 mol H2O/mol)(2mol H/mol H2O)=94.8 mol H 27.8%C, 72.2% H From a: 7.10 mol C 3 H 8
3.80 mol C 4 H10 4 mol C 3 mol C mol C 4 H10 mol C 3 H 8
7.10 mol C 3 H 8 11 mol (C H) 3.80 mol C 4 H10 14 mol (C H) mol C 3 H 8 mol C 4 H10 4.74
Basis: 100 kg fuel oil Moles of C in fuel:
100 kg 0.85 kg C 1 kmol C kg 12.01 kg C
Moles of H in fuel:
100 kg 0.12 kg H 1 kmol H 12.0 kmol H kg 1 kg H
Moles of S in fuel:
100 kg 0.017 kg S 1 kmol S kg 32.064 kg S
1.3 kg non-combustible materials (NC)
4.74 (cont’d) 4-66
7.08 kmol C
0.053 kmol S
u 100% 27.8% C
100 kg fuel oil 7.08 kmol C 12.0 kmol H 0.053 kmol S 1.3 kg NC (s) 20% excess air n1 (kmol O2) 3.76 n1 (kmol N2)
n2 (kmol N2) n3 (kmol O2) n4 (kmol CO2) (8/92) n4 (kmol CO) n5 (kmol SO2) n6 (kmol H2O)
C + O2 o CO2 C + 1/2 O2 o CO 2H + 1/2 O2 o H2O S + O2 o SO2
Theoretical O2: 7.08 kmol C 1 kmol O 2 12 kmol H .5 kmol O 2 0.053 kmol S 1 kmol O 2 1 kmol C 2 kmol H 1 kmol S
10.133 kmol O 2
20 % excess air: n1 = 1.2(10.133) = 12.16 kmol O2 fed O balance: 2 (12.16) = 2 (6.5136) + 0.5664 + 2 (0.053) + 6 + 2 n3 n3 = 2.3102 kmol O2 C balance: 7.08 = n4+8n4/92 n4 = 6.514 mol CO2 8 (6.514)/92 = 0.566 mol CO S balance: n5 = 0.53 kmol SO2 H balance: 12 = 2n6 n6 = 6.00 kmol H2O N2 balance: n2 = 3.76(12.16) = 45.72 kmol N2 Total moles of stack gas = (6.514 + 0.566 + 0.053 + 6.00 + 2.310 + 45.72) kmol = 61.16 kmol 10.7% CO, 0.92% CO, 0.087% SO 2 , 9.8% H 2 O, 3.8% O 2 , 74.8% N 2
4.75
a. Basis: 5000 kg coal/h; 50 kmol air min
3000 kmol air h
5000 kg coal / h 0.75 kg C / kg 0.17 kg H / kg 0.02 kg S / kg 0.06 kg ash / kg
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2 C + 1/2 O2 --> CO
3000 kmol air / h 0.21 kmol O2 / kmol 0.79 kmol N2 / kmol
n1 (kmol O2 / h) n2 (kmol N2 / h) n3 (kmol CO2 / h) 0.1 n3 (kmol CO / h) n4 (kmol SO2 / h) n5 (kmol H2O / h)
mo kg slag / h
Theoretical O 2 : C:
0.75 5000 kg C
b g
1 kmol C
1 kmol O 2
h
12.01 kg C
1 kmol C
4.75 (cont’d)
4-67
312.2 kmol O 2 h
b g
H:
0.17 5000 kg H 1 kmol H 1 kmol H 2 O 1 kmol O 2 h 2 kmol H 2 kmol H 2 O 101 . kg H
S:
0.02 5000 kg S 1 kmol S 1 kmol O 2 = 3.1 kmol O2/h 32.06 kg S 1 kmol S h
210.4 kmol O 2 h
b g
Total = (312.2+210.4 + 3.1) kmol O2/h = 525.7 kmol O 2 h
b
g
O 2 fed = 0.21 3000 Excess air:
630 kmol O 2 h
630 525.7 u 100% 19.8% excess air 525.7
b. Balances: 0.94 0.75 5000 kg C react 1 kmol C C: h 12.01 kg C
b gb gb g n 3
266.8 kmol CO 2 h , 01 . n 3
H:
. gb5000g kg H b017
S:
(from part a)
N2 :
O:
h
n 3 0.1n 3
26.7 kmol CO h
1 kmol H 1 kmol H 2 O n 5 n5 101 . kg H 2 kmol H
b
3.1 kmol O 2 for SO 2 h
g
1 kmol SO 2 1 kmol O 2
420.8 kmol H 2 O h
n 4 n 4
31 . kmol SO 2 h
b0.79gb3000g kmol N h n n 2370 kmol N h b0.21g(3000)b2g 2n 2b266.8g 1b26.68g 2b31. g b1gb420.8g 2
2
2
2
1
n1
136.4 kmol O 2 / h
Stack gas total
3223 kmol h
Mole fractions:
c.
x CO
26.7 3224 8.3 u 10 3 mol CO mol
xSO2
31 . 3224
9.6 u 10 4 mol SO 2 mol
1 SO 2 O 2 o SO 3 2 SO 3 H 2 O o H 2SO 4 3.1 kmol SO 2 1 kmol SO 3 1 kmol H 2SO 4
98.08 kg H 2SO 4
1 kmol SO 2
kmol H 2SO 4
h
1 kmol SO 3
4-68
304 kg H 2SO 4 h
4.76 a.
Basis: 100 g coal as received (c.a.r.). Let a.d.c. denote air-dried coal; v.m. denote volatile matter 100 g c.a. r. 1.147 g a.d.c. 1.207 g c.a. r.
95.03 g a.d.c
95.03 g air - dried coal; 4.97 g H 2 O lost by air drying
. ggH O b1.234 1204 2
1.234 g a.d.c. Total H 2 O 95.03 g a.d.c
4.97 g 2.31 g
2.31 g H 2 O lost in second drying step
7.28 g moisture
0.811g g b v. m. H Og . b1347 2.31 g H O 2
2
1.347 g a.d.c.
95.03 g a.d.c
0.111 g ash 1.175 g a.d.c.
8.98 g ash
b100 7.28 35.50 8.98gg
Fixed carbon
3550 . g volatile matter
48.24 g fixed carbon
7.28 g moisture 7.3% moisture 48.24 g fixed carbon 48.2% fixed carbon 35.50 g volatile matter 35.5% volatile matter 8.98 g ash 9.0% ash 100 g coal as received b. Assume volatile matter is all carbon and hydrogen. C CO 2 o CO 2 :
2H
1 mol O 2 1 mol C
1 mol C 10 3 g 1 mol air 12.01 g C 1 kg 0.21 mol O 2
0.5 mol O 2 1 O2 o H 2O : 2 2 mol H
Air required:
396.5 mol air kg C
1 mol H 10 3 g 1 mol air 1.01 g H 1 kg 0.21 mol O 2
1179 mol air kg H
1000 kg coal 0.482 kg C 396.5 mol air kg coal
1000 kg 0.355 kg v. m.
kg C 6 kg C
396.5 mol air
kg 7 kg v. m. kg C 1000 kg 0.355 kg v. m. 1 kg H 1179 mol air kg
7 kg v. m.
4-69
kg H
3.72 u 105 mol air
4.77
a.
Basis 100 mol dry fuel gas. Assume no solid or liquid products!
n1 (mol C) n2 (mol H) n3 (mol S)
100 mol dry gas C + 02 --> CO2 C + 1/2 O2 --> CO 2H + 1/2 O2 -->H2O S + O2 --> SO2
0.720 mol CO2 / mol 0.0257 mol CO / mol 0.000592 mol SO2 / mol 0.254 mol O2 / mol
n4 (mol O2) (20% excess)
n5 (mol H2O (v))
½ H balance : n 2 2 n 5 ° O balance : 2 n 4 100 [ 2(0.720) 0.0257 2 (0.000592) 2 (0.254)] n 5 ¾ ° 20 % excess O 2 : (1.20) (74.57 0.0592 0.25 n 2 ] n 4 ¿ n2 = 183.6 mol H, n4 = 144.6 mol O2, n5 = 91.8 mol H2O Total moles in feed: 258.4 mol (C+H+S) 28.9% C, 71.1% H, 0.023% S
4.78
Basis: 100 g oil Stack SO 2 , N 2 , O 2, CO 2, H 2O (612.5 ppm SO 2)
x n 3 mol SO 2 (N2 , O2 , CO2 , H 2 O)
0.10 (1 – x ) n 5 mol SO 2 (N2 , O2 , CO2 , H 2 O)
100 g oil 0.87 g C/g 0.10 g H/g 0.03 g S/g n 1 mol O2 3.76 n 1 mol N2 (25% excess)
furnace
n 2 mol N 2 n 3 mol O 2 n 4 mol CO2 n 5 mol SO 2 n 6 mol H 2 O
CO 2 : H 2 O:
b g
0.87 100 g C
(1 – x ) n 5 mol SO 2 (N2 , O2 , CO2 , H 2 O)
scrubber
Alkaline solution
0.90 (1 – x ) n 5 mol SO 2
1 mol C 1 mol CO 2 n4 12.01 g C 1 mol C
b g
0.10 100 g H 1 mol H 1 mol H 2 O n6 2 mol H 101 . gH
4-70
FG 7.244 mol O IJ H consumed K F 2.475 mol O IJ 4.95 mol H OG H consumed K
7.244 mol CO 2
2
2
2
4.78 (cont’d) SO 2 :
b g
0.03 100 g S
1 mol S 1 mol SO 2 n5 32.06 g S 1 mol S
b
0.0936 mol SO 2
g
FG 0.0956 mol O IJ H consumed K 2
125 . 7.244 2.475 0.0936 12.27 mol O 2
25% excess O 2 : n1
b
g
12.27 mol O 2 fed 7.244 2.475 0.0936 mol O 2 consumed
O 2 balance: n3
2.46 mol O 2 N 2 balance: n 2
b
b
3.76 12.27 mol
g
4614 . mol N 2
g
SO 2 in stack SO 2 balance around mixing point :
F H
I K
b gb
g
. 1 x 0.0936 x 0.0936 010 n5
b
0.00936 0.0842 x mol SO 2
g
Total dry gas in stack (Assume no CO2 , O2 , or N 2 is absorbed in the scrubber)
b
7.244 2.46 4614 . 0.00936 0.0842 x
bCO g bO g 2
bN g
2
2
b
g
bSO g
g
b
g
5585 . 0.0842 x mol dry gas
2
612.5 ppm SO 2 dry basis in stack gas 0.00936 0.0842 x 5585 . 0.0842 x
612.5 x 10 . u 10 6
0.295 30% bypassed
Basis: 100 mol stack gas
4.79
n 1 (mol C) n 2 (mol H) n 3 (mol S) n 4 (mol O2 ) 3.76 n 4 (mol O2 )
a.
C balance: n1 H balance: n2
C + O2 o CO2 1 2H + O2 o H 2 O 2 S + O 2 o SO 2
100 mol 0.7566 N 2 0.1024 CO2 0.0827 H 2 O 0.0575 O 2 0.000825 SO 2
. b100gb01024 g 10.24 mol C b100gb0.0827gb2g 16.54 mol H
10.24 mol C 16.54 mol H
0.62
mol C mol H
The C/H mole ratio of CH 4 is 0.25, and that of C2 H 6 is 0.333; no mixture of the two could have a C/H ratio of 0.62, so the fuel could not be the natural gas. b.
b100gb0.000825g 0.0825 mol S b10.24 mol Cgb12.0 g 1 molg 122.88 g CU| 122.88 7.35 g C g H b16.54 mol Hgb1.01 g 1 molg 16.71 g H V 216.65.71 b0.0825 mol Sgb32.07 g 1 molg 2.65 g S |W 142.24 u 100% 1.9% S
S balance: n 3
4-71
No. 4 fuel oil
4.80
a.
Basis: 1 mol CpHqOr 1 mol CpHqOr no (mol S) Xs (kg s/ kg fuel)
C + 02 --> CO2 2H + 1/2 O2 -->H2O S + O2 --> SO2
P (% excess air) n1 (mol O2) 3.76 n1 (mol N2)
n2 (mol CO2) n3 (mol SO2) n4 (mol O2) 3.76 n1 (mol N2) n5 (mol H2O (v))
Hydrocarbon mass: p (mol C) ( 12 g / mol) = 12 p (g C) q (mol H) (1 g / mol) =
q (g H)
(12 p + q + 16 r) g fuel
r (mol O) (16 g / mol) = 16 r (g O) S in feed: no=
(12 p q 16r) g fuel
Theoretical O2:
X s (g S) 1 mol S (1 - X s ) (g fuel) 32.07 g S
X s (12 p q 16 r) (mol S) (1) 32.07(1 - X s )
p (mol C) 1 mol O 2 q (mol H) 0.5 mol O 2 ( r mol O) 1 mol O 2 1 mol C 2 mol H 2 mol O (p 1/4 q 1/2 r) mol O 2 required
% excess n1 = (1 + P/100) (p +1/4 q – ½ r) mol O2 fed
(2)
C balance: n2 = p
(3)
H balance: n5 = q/2
(4)
S balance: n3 = n0
(5)
O balance: r + 2n1 = 2n2 + 2n3 + 2n4 + n5 n4 = ½ (r+2n1-2n2-2n3-n5)
(6)
Given: p = 0.71, q= 1.1, r = 0.003, Xs = 0.02 P = 18% excess air (1) n0 = 0.00616 mol S
(5) n3 = 0.00616 mol SO2
(2) n1 = 1.16 mol O2 fed
(6) n4 = 0.170 mol O2
(3) n2 = 0.71 mol CO2
(4) n5 = 0.55 mol H2O
(3.76*1.16) mol N2 = 4.36 mol N2 Total moles of dry product gas = n2 + n3 + n4 + 3.76 n1=5.246 mol dry product gas Dry basis composition yCO2 = (0.710 mol CO2/ 5.246 mol dry gas) * 100% = 13.5% CO2 yO2 = (0.170 / 5.246) * 100% = 3.2% O2 yN2 = (4.36 / 5.246) * 100% = 83.1% N2 ySO2 = (0.00616 / 5.246) * 106 = 1174 ppm SO2
4-72
CHAPTER FIVE 5.1
Assume volume additivity
m
m0 m = mt
mass of tank at time t
empty tank
a.
A
0.400 0.600 U 0.703 kg L 0.730 kg L
1 U
Av. density (Eq. 5.1-1):
A mass of
A
A
UO
UD
b250 150gkg b10 3g min
14.28 kg min
0.719 kg L
bm
mass flow rate of liquid
1L / min) / min) = m(kg 14.28 kg V(L V 19.9 L min min 0.719 kg U ( kg / L) b. m0
5.2
bg
150 14.28 3 m(t) - mt
107 kg
b
g
void volume of bed: 100 cm 3 2335 . 184 cm3
porosity: 50.5 cm3 void 184 cm3 total bulk density: 600 g 184 cm3
b
50.5 cm 3
0.274 cm3 void cm3 total
3.26 g cm3
g
absolute density: 600 g 184 50.5 cm3
4.49 g cm3
5.3 C 6 H 6 (l ) B (kg / min) m VB = 20.0 L / min (kg / min) m (L / min) V
C 7 H 8 (l )
T (kg / min) m (L / min) V T 2 . m) 2 015 . m = 'V SD 'h S (55 V 0.0594 m3 / min 't 4 60 min 4 't Assume additive volumes -V V V 59.4 20.0 L / min = 39.4 L / min T B
b
m xB
g
0.879 kg 20.0 L 0.866 kg 39.4 L L min L min (0.879 kg / L)(20.0 L / min) 0.34 kg B / kg (517 . kg / min)
B UT V T UB V B m m
5-1
. kg / min 517
g
a.
b.
1 U sl
U| V| W
b
F IF I e jge jhbmgGH 11 N JK GH 11 Pa JK kg
U sl
m s2
m3
kgm s2
N m2
U sl gh
g
1 xc xc check units! Uc Ul
1 kg crystals / kg slurry kg liquid / kg slurry kg slurry / L slurry kg crystals / L crystals kg liquid / L liquid L slurry L crystals L liquid L slurry kg slurry kg slurry kg slurry kg slurry 'P 2775 1415 kg / m3 c. i.) U sl gh 9.8066 0.200 ii.)
1 U sl
xc
xc Uc
b gb g b1- x g x FG 1 1 IJ FG 1 U HU U K HU c
l
F 1 GG 1415 kg / m H
c
3
l
sl
1 . 1000 kg / m3 12
d
I iJJK
1 Ul
F I GG 2.3d10001kg / m i 12. d10001kg / m iJJ H K msl U sl
iv.) mc
x c msl
IJ K
0.316 kg crystals / kg slurry
3
175 kg 1000 L 3 1415 kg / m m3
iii.) Vsl
1238 . L
b0.316 kg crystals / kg slurrygb175 kg slurryg
55.3 kg crystals
55.3 kg CuSO 4 5H 2 O 1 kmol 1 kmol CuSO 4 159.6 kg 249 kg 1 kmol CuSO 4 5H 2 O 1 kmol
v.) mCuSO 4 vi.) ml
c
3
b1 x gm b0.684 kg liquid / kg slurrygb175 kg slurryg c
ml Ul
vii.)
Vl
h(m) Ul(kg/m^3) Uc(kg/m^3) 'P(Pa) xc Usl(kg/m^3)
0.2 1200 2300 2353.58 0 1200.00
sl
120 kg
1000 L m3
b1.2gd1000 kg / m i 3
35.4 kg CuSO 4
120 kg liquid solution
100 L
d. 2411.24 0.05 1229.40
2471.80 0.1 1260.27
2602.52 0.2 1326.92
2747.84 0.3 1401.02
2772.61 0.316 1413.64
2910.35 0.4 1483.87
3093.28 0.5 1577.14
Effect of Slurry Density on Pressure Measurement
0.6 Solids Fraction
5.4
P0 U sl gh1 P0 U sl gh 2 'P = P1 P2 h = h1 h 2 P1 P2
0.5 0.4 0.3 0.2
'P = 2775, U = 0.316
0.1 0 2300.00
2500.00
2700.00
2900.00
Pressure Difference (Pa)
5-2
3100.00
5.4 (cont’d) e.
b
g d
b
b1- x gbkg liquid g, V dm c
l
3
liquid
d
U c kg / m
1 kg
5.7
a.
c
29.0 g
3
0.0064 m air
nRT P
b3.06L - 2.8Lg u 100%
Assume Patm
1013 . bar
b.
PV Ps Vs
n
. gbar b10 1013
nRT n n s RTs
20.0 m3
V
273K 298.2K
b
3.06 L
9.3%
2.8L
nRT n
0.0064 m3 mol
4.5 kg m 3
mol 103 g
100 . mol 0.08206 L atm 373.2 K mol K 10 atm
a.
5.8
1 kg
b. % error =
PV
3
1 atm
1 mol
V=
i
g
3 = 0.08206 m atm 313.2 K 1 kmol RT V kmol K 4.0 atm 103 mol
PV
5.6
l
Assume Patm
U
b
bV V gdm i c
5.5
1 1 xc xc Uc Ul
3
3
g i b1-Ux dgbkgkg/ mliquid i l
U sl
g
x c kg crystals
i
Basis: 1 kg slurry x c kg crystals , Vc m3 crystals
kmol K 28.02 kg N 2 20.0 m3 kmol 25 + 273.2 K 0.08314 m3 bar
g
249 kg N 2
Ts P n s T Ps Vs
. gbar b10 1013 1.013 bar
1 kmol
28.02 kg N 2 kmol 22.415 m STP 3
a.
R=
Ps Vs n sTs
1 atm 22.415 m3 1 kmol 273 K
b.
R=
Ps Vs n sTs
1 atm 760 torr 359.05 ft 3 1 lb - mole 1 atm 492 $ R
8.21 u 10 2
5-3
b g
atm m3 kmol K 555
torr ft 3 lb - mole $ R
249 kg N 2
5.9
P = 1 atm +
10 cm H 2 O
1m 1 atm 2 10 cm 10.333 m H 2 O 3
= 2.0 m T = 25$ C = 298.2 K , V 5 min m = n mol / min MW g / mol
b
a.
g
= m = m
b.
b
101 . atm
0.40 m3 min = 400 L min
g
L 28.02 PV 1.01 atm 400 min MW = Latm RT 0.08206 molK 298.2 K
400
L min
28.02 273 K 1 mol 298.2 K 22.4 L STP
b g
g mol
F mI V dm si uG J H s K Ad m i 3
5.10 Assume ideal gas behavior:
458 g min
458 g min
T2 P1 D12 nR 2 nR T1 P2 D 2
nRT P u 2 2 u1 SD 4
2
. 1013 . gbar b7.50 cmg b180 134 m sec sec 333.2 K b153 . 1013 . gbar b5.00 cmg . 100 . g atm 5 L PV b100 Assume ideal gas behavior: n 0.406 mol u2
5.11
g mol
u1
T2 P1D12 T1P2 D 22
2
60.0 m 300.2 K
2
Latm 0.08206 mol RT K 32.0 g mol Oxygen
MW 13.0 g 0.406 mol
5.12 Assume ideal gas behavior: Say m t
g t
0.009391 mol
b g
std cm3 STP min V
b g
' V liters 273K 763 mm Hg 103 cm3 1L ' t min 296.2K 760 mm Hg
I
std cm3 STP min V
5.0 9.0 12.0
139 268 370
U| |V straight line plot ||I 0.031EV 0.93 W std
b.
b g
0.010 mol N 2 22.4 liters STP 103 cm3 min 1 mole 1L
d
mol of gas in tank
gU|V n 0.009391 mol g |W m 37.0256 g b37.062 37.0256gg 3.9 g mol Helium b b
unknown: MW
std V
mass of tank, n g
m t n g 28.02 g mol m t n g 44.1 g mol
N 2 : 37.289 g CO 2 : 37.440 g
5.13 a.
300 K
i
I = 0.031 224 cm3 / min 0.93 7.9 5-4
224 cm3 / min
925.3
'V 't
/ kmol) g nbkmolgM(kg Vb Lg
b FU I si G J HU K
5.14 Assume ideal gas behavior U kg L
d
3
i V dcm
V2 cm s
1
PM
! RT
12
3
1
V1 P1 M1T2 P2 M 2 T1
12
2
LM N
cm3 758 mm Hg 28.02 g mol 323.2K s 1800 mm Hg 2.02 g mol 298.2K
a.
VH 2
b.
M
0.25M CH 4 0.75M C 3H 8
Vg
cm3 350 s
350
n P V RT
OP Q
12
881 cm3 s
b0.25gb16.05g b0.75gb44.11g
LM b758gb28.02gb323.2g OP N b1800gb37.10gb298.2g Q
37.10 g mol
12
205 cm3 s
5.15 a. Reactor
'h soap
b.
d
2 S 0.012 m 4
i
2
n CO 2
2 PV = S R 'h V 't RT
n CO 2
755 mm Hg 1 atm 11 . u 10-3 m3 / min 1 kmol 3 m atm 760 mm Hg 300 K 1000 mol 0.08206 kmol K
. m 60 s 12 . u 10 3 m3 / min 11 7.4 s min 0.044 mol / min
5.16 air m
10.0 kg / h
n air (kmol / h)
n (kmol / h) y CO (kmol CO 2 / kmol) 2
CO = 20.0 m3 / h V 2 n CO (kmol / h) 2
150 o C, 1.5 bar
Assume ideal gas behavior 10.0 kg 1 kmol n air 0.345 kmol air / h h 29.0 kg air n CO 2
PV RT
y CO 2 u 100%
15 100 kPa 20.0 m3 / h . bar 3 1 bar 423.2 K 8.314 mkmolkPa K
b
0.853 kmol CO 2 / h
0.853 kmol CO 2 / h u 100% 0.853 kmol CO 2 / h + 0.345 kmol air / h
g
5-5
712% .
5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior 3
o
1 (kg / min) m 0.70 kg H 2 O / kg 0.30 kg S / kg
311 m / min, 83 C, 1 atm n 3 (kmol / min) 0.12 kmol H 2 O / kmol 0.88 kmol dry air / kmol
n 2 (kmol air / min) (m 3 / min) V 2 167 o C, - 40 cm H 2 O gauge
4 (kg S / min) m
a.
n 3
1 atm 311 m3 365.2K min
kmol K 10.38 kmol min 0.08206 m3 atm 10.38 kmol 0.12 kmol H 2 O 18.02 kg kmol kmol min
H 2 O balance: 0.70 m1 1 m
32.2 kg min milk
b g
b
S olids balance: 0.30 32.2 kg min
4 m 4 m
b
9.6 kg S min
g
0.88 10.38 kmol min n 2
Dry air balance: n 2
2 V
g
9.13 kmol 0.08206 m3 atm min kmol K
9.13 kmol min air
440K 1033 cm H 2 O 1033 40 cm H 2 O 1 atm
b
g
3
343 m air min u air (m / min) =
air (m3 / s) V A (m2 )
343 m3 1 min min 60 s
S
4 (6
m) 2
0.20 m / s
b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor by the air instead of falling to the conveyor belt.
5.18 SG CO 2
5.19 a.
x CO 2
U CO 2 U air
PM CO2 RT PM air RT
0.75 x air
M CO 2 M air
152 .
1 0.75 0.25
Since air is 21% O 2 , x O 2 b.
44 kg / kmol 29 kg / kmol
mCO 2 = n x CO 2 M CO 2
0.0525 5.25 mole% O 2
(0.25)(0.21)
b
g
2 u 1.5 u 3 m3 0.75 kmol CO 2 44.01 kg CO 2 1 atm 12 kg m 3 atm 298 2 K kmol kmol CO . 0.08206 kmol 2 K
More needs to escape from the cylinder since the room is not sealed.
5-6
5.19 (cont’d) c. With the room closed off all weekend and the valve to the liquid cylinder leaking, if a person entered the room and closed the door, over a period of time the person could die of asphyxiation. Measures that would reduce hazards are: 1. Change the lock so the door can always be opened from the inside without a key. 2. Provide ventilation that keeps air flowing through the room. 3. Install a gas monitor that sets off an alarm once the mole% reaches a certain amount. 4. Install safety valves on the cylinder in case of leaks. 15.7 kg
5.20 n CO 2
1 kmol
0.357 kmol CO 2
44.01 kg
b
Assume ideal gas behavior, negligible temperature change T 19q C 292.2 K
a.
P1V P2 V
b
n1 RT n1 n1 0.357 RT n1 0.357
g
n1 b. Vtank
102kPa 3.27 u 103 kPa
0.0115 kmol air in tank
n1RT P1
0.0115 kmol 292.2 K 8.314 m3 kPa 103 L m3 102kPa kmol K
15700 g CO 2 + 11.5 mol air (29.0 g air / mol) 274 L
Uf c.
P1 P2
g
274 L
58.5 g / L
CO 2 sublimates large volume change due to phase change rapid pressure rise. Sublimation causes temperature drop; afterwards, T gradually rises back to room temperature, increase in T at constant V slow pressure rise.
5.21 At point of entry, P1
b10 ft H Ogb29.9 in. Hg 33.9 ft H Og 28.3 in. Hg 2
2
37.1 in. Hg .
At surface, P2
28.3 in. Hg, V2 bubble volume at entry 1 x solid x solution 0.20 0.80 Mean Slurry Density: 3 U sl U solid U solution (1.2)(1.00 g / cm ) (100 . g / cm 3 ) 0.967
cm 3 U sl g
1.03 g 2.20 lb 5 u 10 4 ton 10 6 cm 3 1 lb 264.17 gal cm 3 1000 g
a.
300 ton gal 40.0 ft 3 (STP) 534.7 o R 29.9 in Hg hr 4.3 u 10 3 ton 1000 gal 492 o R 37.1 in Hg
b.
P2 V2 P1V1
nRT V 2 nRT V1
% change =
4 3S P1 P2 4 3S
e j e j
b2.2 - 2.0g mm u 100 2.0 mm
D2 3 2
37.1 D 32 28.3
D1 3 2
10%
5-7
1.31D13
4.3 u 10 3 ton / gal
2440 ft 3 / hr
!D
D1 2 mm
2
2.2 mm
5.22 Let B = benzene n1 , n 2 , n 3 moles in the container when the sample is collected, after the helium is added, and after the gas is fed to the GC. n inj moles of gas injected n B , n air , n He moles of benzene and air in the container and moles of helium added n BGC , m BGC moles, g of benzene in the GC y B mole fraction of benzene in room air
a.
P1V1
n1 P2 V2
n2 P3 V3
n3
n1RT1 (1 { condition when sample was taken): P1 = 99 kPa, T1
mol K 99 kPa 2 L kPa 101.3 atm 306 K .08206 L atm
0.078 mol = n air n B
n 2 RT2 (2 { condition when charged with He): P2 = 500 kPa, T2
mol K 500 kPa 2 L 101.3 kPa 306 K .08206 L atm atm
mol K 400 kPa 2 L kPa 101.3 atm 296 K .08206 L atm
n BGC u
y B (ppm) =
306K
0.393 mol = n air + n B n He
n 3 RT3 (3 { final condition in lab): P3 = 400 kPa, T3
n inj = n 2 n 3 nB
306K
296K
0.325 mol = (n air n B n He ) n inj
0.068 mol n2 n inj
0.393 mol m BGC (g B) 1 mol 0.068 mol 78.0 g
nB u 106 n1
0.0741 m BGC u 106 0.078
9 am: y B
(0.950 u 106 )(0.656 u 10 6 )
1 pm: y B
(0.950 u 106 )(0.788 u 10 6 )
5 pm: y B
(0.950 u 106 )(0.910 u 10 6 )
0.0741 m BGC
0.950 u 106 m BGC
U| | 0.749 ppm VThe avg. is below the PEL | 0.864 ppm| W 0.623 ppm
b. Helium is used as a carrier gas for the gas chromatograph, and to pressurize the container so gas will flow into the GC sample chamber. Waiting a day allows the gases to mix sufficiently and to reach thermal equilibrium. c. (i) It is very difficult to have a completely evacuated sample cylinder; the sample may be dilute to begin with. (ii) The sample was taken on Monday after 2 days of inactivity at the plant. A reading should be taken on Friday. (iii) Helium used for the carrier gas is less dense than the benzene and air; therefore, the sample injected in the GC may be Herich depending on where the sample was taken from the cylinder. (iv) The benzene may not be uniformly distributed in the laboratory. In some areas the benzene concentration could be well above the PEL.
5-8
b g
4 S 10 m 3 Moles of gas in balloon
5.23 Volume of balloon
b
n kmol a.
b.
4189 m3
g
3
4189 m3
492q R 3 atm 1 kmol 535q R 1 atm 22.4 m3 STP
b g
515.9 kmol
He in balloon:
m
b515.9 kmolg b4.003 kg kmolg
mg
2065 kg 9.807 m 1N 2 s 1 kg m / s2
dP dP
iV iV
gas in balloon air displaced
Fbuoyant
n gas RT n air RT
Fbuoyant
2065 kg He 20,250 N
Pair n gas Pgas
n air
172.0 kmol 29.0 kg 9.807 m 1 N 1 kmol s2 1 kg2m s
Wair displaced
Since balloon is stationary, Wtotal
Fcable
Fcable
Fbuoyant Wtotal
c. When cable is released, Fnet
b
¦F
1
48920 N
dAi = 27200 N a
1 atm 515.9 kmol 172.0 kmol 3 atm
48,920 N
0
b2065 150gkg
9.807 m 1 N s2 1 kgs2m
M tot a
27200 N 1 kg m / s2 2065 + 150 kg N
g
12.3 m s2
d. When mass of displaced air equals mass of balloon helium the balloon stops rising. Need to know how density of air varies with altitude. e. The balloon expands, displacing more air buoyant force increases balloon rises until decrease in air density at higher altitudes compensates for added volume. 5.24 Assume ideal gas behavior, Patm a. b.
1 atm
PN VN 5.7 atm 400 m3 / h 240 ft 3 h Pc 9.5 atm Mass flow rate before diversion: PN VN
240 m3 h
Pc Vc Vc
273 K 5.7 atm 1 kmol 303 K 1 atm 22.4 m3 STP
b g 5-9
44.09 kg kmol
2426
kg C 3 H 6 h
27,20
5.24 (cont’d) Monthly revenue:
b2426 kg hgb24 h daygb30 days monthgb$0.60 kgg
c.
$1,048,000 month
Mass flow rate at Noxious plant after diversion: 240 m3 hr
273 K 2.8 atm 1 kmol 306 K 1 atm 22.4 m3
Propane diverted 5.25 a. PHe PCH 4 PN 2
b2426 1180g kg h
44.09 kg 1180 kg hr kmol
1246 kg h
y He P = 0.35 (2.00 atm) = 0.70 atm y CH 4 P = 0.20 (2.00 atm) = 0.40 atm y N 2 P = 0.45 (2.00 atm) = 0.90 atm
b. Assume 1.00 mole gas 4.004 g 0.35 mol He mol
U| || 16.05 g I F 0.20 mol CH G H mol JK 3.21 g CH V|17.22 g mass fraction CH F 28.02 g IJ 12.61 g N || 0.45 mol N G H mol K W FG H
IJ K
140 . g He
4
2
c.
MW
d.
U gas
4
3.21 g 17.22 g
0186 .
2
g of gas 17.2 g / mol mol P MW m n MW V
4
d i d i b2.00 atmgb17.2 kg / kmolg V RT e0.08206 jb363.2 Kg m 3 atm kmolK
115 . kg / m3
5.26 a. It is safer to release a mixture that is too lean to ignite. If a mixture that is rich is released in the atmosphere, it can diffuse in the air and the C3H8 mole fraction can drop below the UFL, thereby producing a fire hazard. b. fuel-air mixture n 1 ( mol / s) y C 3H 8 0.0403 mol C 3 H 8 / mol n C 3H 8 150 mol C3 H 8 / s
n 3 ( mol / s) 0.0205 mol C 3 H 8 / mol
diluting air n 2 ( mol / s)
n 1
150 mol C3 H 8 mol s 0.0403 mol C 3 H 8
Propane balance: 150 = 0.0205 n 3 n 3
5-10
3722 mol / s
7317 mol / s
5.26 (cont’d) Total mole balance: n 1 n 2
c.
b g
n 3 n 2
7317 3722
n 2
. n 2 13
2 V
4674 mol / s 8.314 m3 Pa 398.2 K mol K 131,000 Pa
3595 mol air / s
4674 mol / s
min
1 V
3722 mol 8.314 m3 Pa 298.2 K s mol K 110000 Pa
y2
150 mol / s n 1 n 2
b
U| |V V || V W
118 m3 / s
2
. 141
1
83.9 m3 / s
m3 diluting air m3 fuel gas
150 mol / s u 100% 18% . 3722 mol / s + 4674 mol / s
g
d. The incoming propane mixture could be higher than 4.03%. If n 2 n 2 min , fluctuations in the air flow rate would lead to temporary explosive
b g
conditions. 5.27
b
gb
Basis: 12 breaths min 500 mL air inhaled breath
g
600 mL inhaled min
o
24 C, 1 atm 6000 mL / min
lungs
n in (mol / min) 0.206 O 2 0.774 N 2 0.020 H 2 O
a.
n in
blood
37 o C, 1 atm n out (mol / min) 0.151 O 2 0.037 CO 2 0.750 N 2 0.062 H 2 O
6000 mL 1L 273K 1 mol 3 min 10 mL 297K 22.4 L STP
b g
b
gb
g
N 2 balance: 0.774 0.246
O 2 transferred to blood:
0.750n out n out
0.246 mol min 0.254 mol exhaled min
. g b mol O b0.246gb0.206g b0.254gb0151
2
g
min 32.0 g mol
0.394 g O 2 min CO 2 transferred from blood:
b0.254gb0.037g bmol CO
2
g
min 44.01 g mol
0.414 g CO 2 min H 2 O transferred from blood:
b0.254gb0.062g b0.246gb0.020g bmol H O ming 18.02 g mol 2
g H 2 O min 0195 .
5-11
5.27 (cont’d) PVin PVout
FG n IJ FG T IJ FG 0.254 mol min IJ FG 310K IJ 1078 H n K H T K H 0.246 mol min K H 297K K . mL exhaled ml inhaled . g H O lost ming b0.394 g O gained ming 0.215 g min b0.414 g CO lost ming b0195
b.
n in RTin n out RTout
Vout Vin
out
out
in
in
2
2
2
STACK
5.28
Ta (K) Ts (K) M s (g/mol) M a (g/mol) Ps (Pa) Pc (Pa)
LL(m) (M)
PM RT
Ideal gas: U
bUgLg
a.
D
b.
Ms
Pa
755 mm Hg 29.0 g mol , Ta
755 mm Hg
u
b g
UgL
Pa M a PM gL a s gL RTa RTs
stack
. gb44.1g b0.02gb32.0g b0.80gb28.0g b018
Ma
D
combust.
LM N
Pa gL M a M a R Ta Ts
310 . g mol , Ts
OP Q
655K ,
294 K , L 53 m
1 atm 53.0 m 9.807 m kmol - K 2 s 0.08206 m3 atm 760 mm Hg
LM 29.0 kg kmol 310. kg kmol OP u FG 1 N IJ 655K N 294K Q H 1 kg m / s K 2
323 N 1033 cm H 2 O m2 1.013 u 105 N m2
3.3 cm H 2 O
b.
b g
MWCCl2O 98.91 g / mol U P MW 98.91 CCl2O 3.41 U air 29.0 RT Phosgene, which is 3.41 times more dense than air, will displace air near the ground. 2 S D in L S 2 Vtube 0.635 cm - 2 0.0559 cm 15.0 cm 3.22 cm3 4 4
5.29 a. U
b g
m CCl O 2
c.
!
n CCl 2 O(l)
b
Vtube U CCl O 2
3.22 cm3
gb
g
1L 1 atm 98.91 g / mol 3 3 Latm 296.2 K 10 cm 0.08206 molK
g mol 3.22 cm3 137 . u 1000 . 3 cm 98.91 g
5-12
0.0446 mol CCl 2 O
0.0131 g
5.29 (cont’d) mol K 1 atm 2200 ft 3 28.317 L 2563 mol air 3 296.2K .08206 L atm ft
PV RT
n air n CCl 2 O
0.0446 17.4 u 10 6 2563
n air
17.4 ppm
The level of phosgene in the room exceeded the safe level by a factor of more than 100. Even if the phosgene were below the safe level, there would be an unsafe level near the floor since phosgene is denser than air, and the concentration would be much higher in the vicinity of the leak. d. Pete’s biggest mistake was working with a highly toxic substance with no supervision or guidance from an experienced safety officer. He also should have been working under a hood and should have worn a gas mask. 5.30 CH 4 2O 2 o CO 2 2 H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 C 3 H 8 5O 2 o 3CO 2 4 H 2 O 1450 m3 / h @ 15o C, 150 kPa n 1 (kmol / h) 0.86 CH 4 , 0.08 C 2 H 6 , 0.06 C 3 H 8 n 2 (kmol air / h) 8% excess, 0.21 O 2 , 0.79 N 2
n 1
1450 m3 h
273.2K 288.2K
b1013. 150gkPa 101.3 kPa
1 kmol 22.4 m3 STP
b g
152 kmol h
FG H
IJ OP K PQ
Theoretical O 2 :
LM F MN GH
IJ K
IJ K
FG H
152 kmol 2 kmol O 2 3.5 kmol O 2 5 kmol O 2 0.08 0.06 0.86 h kmol CH 4 kmol C 2 H 6 kmol C 3 H 8 air Air flow: V
b
g
108 . 349.6 kmol O 2 h
1 kmol Air 0.21 kmol O 2
5-13
b g
22.4 m3 STP kmol
349.6 kmol h O 2
b g
4.0 u 104 m3 STP h
5.31 Calibration formulas
bT dP dV dV
25.0; R T
g
0; R p
F
0; R p
A
0; R A
27g Tbq Cg 0.77R 14.2 g b 0i , d P 20.0, R 6i P bkPag 3.33R d m hi 200R 0i , d V 2.0 u 10 , R 10i V d m hi 4000R 0h , d V 1.0 u 10 , R 25i V 14 , T 35.0, R T g
T
r
gauge
3
F
F
5
A
p
F
A
A
3
F
3
A
(m 3 / h), T, P V F g CH 4 2O 2 o CO 2 2 H 2 O 7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2 C 3 H 8 5O 2 o 3CO 2 4 H 2 O 13 C 4 H 10 O 2 o 4CO 2 5H 2 O 2
n F (kmol / h) x A (mol CH 4 / mol) x B (mol C 2 H 6 / mol) x C (mol C 3 H 8 / mol) x D (mol n - C 4 H 10 / mol) x E (mol i - C 4 H 10 / mol) ( m 3 / h) (STP) V A n A (kmol / h) 0.21 mol O 2 / mol 0.79 mol N 2 / mol
n F
d
F m3 h V
i
273.2K
dP 101.3ikPa
1 kmol
g
bT 273.2gK 101.3 kPa d P 101.3i kmol 0.12031V FG IJ T + 273 b g H h K F
b g
22.4 m3 STP
g
Theoretical O 2 :
dn i
o 2 Th
b
c
Air feed: n A
dn i
o 2 Th
kmol O 2 req.
1 kmol air
h
0.21 kmol O 2
FG P IJ dn i H 100K bkmol air hg d22.4 m bSTPg kmoli 4.762 1
A V
n a
gh
n F 2x A 3.5x B 5x C 6.5 x D x E kmol O 2 req. h
x
b1 P
x
g
100 kmol feed
1 kmol req.
o 2 Th
3
b g
22.4n A m 3 STP h
RT T(C) Rp Pg(kPa) Rf xa xb xc xd xe PX(%) nF nO2, th nA Vf (m3/h) Va (m3/h) Ra 25.0 7.25 0.81 0.08 0.05 0.04 0.02 1450 22506.2 5.63 23.1 32.0 7.5 15 72.2 183.47 1004.74 64.3 5.8 0.58 0.31 0.06 0.05 0.00 1160 29697.8 7.42 7.5 20.0 19.3 23 78.9 226.4 1325.8 52.6 2.45 0.00 0.00 0.65 0.25 0.10 490 22022.3 5.51 46.5 50.0 15.8 33 28.1 155.2 983.1 10.0 1200 30439.2 7.6 21 30.4 3 6 0.02 0.4 0.35 0.1 0.13 15 53.0 248.1 1358.9 13.3 1400 29283.4 7.3 23 31.9 4 7 0.45 0.12 0.23 0.16 0.04 15 63.3 238.7 1307.3 16.7 1800 32721.2 8.2 25 33.5 5 9 0.5 0.3 0.1 0.04 0.06 15 83.4 266.7 1460.8 20.0 10 0.5 0.3 0.1 0.04 0.06 2000 37196.7 9.3 27 35.0 6 15 94.8 303.2 1660.6
5-14
5.32 NO 21 O 2 NO 2 1 mol 0.20 mol NO / mol 0.80 mol air / mol 0.21 O 2 0.79 N 2 P0 380 kPa
R| S| T
a.
n1 (mol NO) n 2 (mol O 2 ) n 3 (mol N 2 ) n 4 (mol NO 2 ) Pf (kPa)
U| V| W
Basis: 1.0 mol feed 90% NO conversion: n1
O 2 balance: n 2
0.80(0.21)
018 . mol NO 0.5 mol O 2 mol NO
N 2 balance: n 3
0.80(0.79)
0.632 mol N 2
n4 y NO yO2
018 . mol NO 1 mol NO 2 018 . mol NO 2 n f 1 mol NO mol NO 0.020 mol NO 0.022 0.91 mol mol 0.086
mol O 2 yN2 mol
Pf V n f RT Pf = P0 V n 0 RT b.
0.020 mol NO NO reacted = 0.18 mol
010 . (0.20)
n f = n0
Pf P0
P0
(1 mol)
0.695
nf n0
mol N 2 y NO 2 mol
380 kPa
360 kPa 380 kPa
FG 0.91 mol IJ H 1 mol K
0.0780 mol O 2
n1 n 2 n 3 n 4
0198 .
0.91 mol
mol NO 2 mol
346 kPa
0.95 mol
n i0 X i[
ni
E
n1 (mol NO)
0.20 [
n 2 ( mol O 2 ) (0.21)(0.80) 0.5[ n 3 (mol N 2 ) ( 0.79)(0.80) n 4 ( mol NO 2 ) [ 1 0.5[
nf
0.95 [
010 .
n1 010 mol O 2 , n 3 0.632 mol N 2 , . mol NO, n 2 0118 . n 4 010 . mol NO 2 y NO 0105 . , y O 2 0124 . , y N 2 0.665, y NO 2 NO conversion = P (atm) =
Kp
b0.20 - n g u 100% 1
3.55 atm
(y NO 2 P) ( y NO P )( y O 2 P)
50%
0.20
360 kPa 101.3 kPa atm
0105 .
(y NO 2 ) 0.5
0.5
( y NO )( y O 2 ) P
0.5
5-15
b
0.105
(0105 . ) 0.124
g b3.55g 0.5
1
0.5
151 . atm 2
5.33 Liquid composition: 49.2 kg M 1 kmol 112.6 kg
0.437 kmol M
29.6 kg D 1 kmol 147.0 kg
0.201 kmol D
0.221 kmol D / kmol
21.2 kg B 1 kmol 78.12 kg
0.271 kmol B
0.298 kmol B / kmol
100 kg liquid
0.481 kmol M / kmol
0.909 kmol a.
Basis: 1 kmol C 6 H 6 fed V1 (m 3 ) @ 40 o C, 120 kPa n1 (kmol) 0.920 HCl 0.080 Cl 2 1 kmol C 6 H 6 (78.12 kg) n 0 (kmol Cl 2 )
n 2 (kmol) 0.298 C 6 H 6 0.481 C 6 H 5Cl 0.221 C 6 H 4 Cl 2
C 6 H 6 Cl 2 o C 6 H 5Cl + HCl
1 kmol C 6 H 6
C balance:
n2
6 kmol C 1 kmol C 6 H 6
n 2 0.298 u 6 0.481 u 6 0.221 u 6
100 . kmol
1 kmol C 6 H 6
H balance:
C 6 H 5Cl Cl 2 o C 6 H 5Cl 2 + HCl
6 kmol H 1 kmol C 6 H 6
b
g
n1 0.920 (1)
n 2 0.298 u 6 0.481 u 5 0.221 u 4 n1 100 . kmol
b.
100 . kmol 1013 . kPa 0.08206 m3 atm 313.2 K kmol K 120 kPa 1 atm
n1RT P
V1
217 . m3 7812 . kg B
V1 mB
( m3 / s) V gas
d2 =
0.278 m3 / kg B
u(m / s) A(m 2 )
u
4V Sd 2 gas d2 = S u 4
B0 ( kg B) 0.278 m3 4m s 1 min 104 cm2 kg B S (10) m 60 s min m2
d(cm)
b g
B0 2.43 m
217 . m3
B0 (cm2 ) 5.90m
1 2
c. Decreased use of chlorinated products, especially solvents. 5-16
5.34 Vb ( m3 / min) @900$ C, 604 mtorr 60% DCS conversion n 1 (mol DCS / min) n 2 (mol N 2 O / min) n b (mol / min) n 3 (mol N 2 / min) n 4 (mol HCl(g) / min)
3.74 SCMM
U| V| W
n a ( mol / min) 0.220 DCS 0.780 N 2 O
SiH 2 Cl 2(g) 2 N 2 O (g) o SiO 2(s) 2 N 2(g) 2 HCl (g) a.
n a
3.74 m3 (STP) 103 mol 167 mol / min min 22.4 m3 (STP)
mol DCS I gFGH 0.220 mol JK b167 mol / ming 14.7 mol DCS / min mol DCS DCS reacted: b0.60gb0.220gb167g 22.04 mol DCS reacted / min min
b
60% conversion: n 1 = 1- 0.60
N 2 O balance: n 2
b g molminN O
0.780 167
N 2 balance: n 3 HCl balance: n 4
n B
b.
n B RT P
p DCS
x DCS P =
p N 2O
22.04 mol DCS 2 mol N 2 O mol DCS min
22.04 mol DCS 2 mol N 2 mol DCS min 22.04 mol DCS 2 mol HCl min mol DCS
n 1 n 2 n 3 n 4
B V
2
8618 . mol N 2 O / min
44.08 mol N 2 / min 44.08 mol HCl / min
189 mol / min
189 mol 62.36 L torr 0.001 m3 1173 K min mol K L 0.604 torr
2.29 u 104 m3 / min
n 1 14.7 mol DCS / min 640 mtorr = 49.8 mtorr P= n B 189 mol / min 86.2 mol N 2 O / min n 640 mtorr = 292 mtorr x N 2O P = 2 P = n B 189 mol / min
r = 3.16 u 10-8 p DCS p N 2 O = r t MW c. h(A) U SiO2
0.65
b
gb g
3.16 u 10-8 49.8 292
0.65
6.3 u 10 5
mol SiO 2 m2 s
6.3 u 10 5 mol SiO 2 60 s 120 min 60.09 g / mol 1010 A 2 6 3 m s min 2.25 u 10 g / m 1 m (Table B.1)
= 1.2 u 10 A 5
The films will be thicker closer to the entrance where the lower conversion yields higher pDCS and p N 2 O values, which in turn yields a higher deposition rate.
5-17
5.35 Basis: 100 kmol dry product gas n1 (kmol C x H y ) m1 (kg C x H y ) kmol dry gasU R|100 |V 0.105 CO S|0.053 O |W T0.842 N
(m 3 ) V 2
n 2 (kmol air) 0.21 O 2 0.79 N 2
2
2
2
o
n 3 (kmol H 2 O)
30 C, 98 kPa
a.
N 2 balance: 0.79n 2
b
O balance: 2 0.21n 2 C balance:
d
g
0.842(100) n 2 = 106.6 kmol air
b
g b
n1 kmol C x H y
i x bkmol Cg dkmol C H i x
H balance: n1y = 2n 3
g
2 0.053 n 3 n 3 100 2 0105 .
n 3 13.17
! n1 y
b
1317 . kmol H 2 O
g
b1g
n1x = 10.5 100 0105 .
y
b 2g
26.34
b g b g yx 2610.34.5 2.51 mol H / mol C fed: 0.21b106.6 kmol air g 22.4 kmol in excess = 5.3 kmol Theoretical O = b22.4 - 5.3g kmol = 17.1 kmol
Divide 2 by 1 O2
O2
2
% excess = b.
V2 m1 =
5.3 kmol O 2 u 100% 17.1 kmol O 2
31% excess air
106.6 kmol N 2 22.4 m3 (STP) 1013 . kPa 303 K kmol 98 kPa 273 K
b
g
b
g
n1x kmol C 12.0 kg n1 y kmol H 101 . kg kmol kmol
V2 2740 m3 air = m1 152.6 kg fuel
18.0
2740 m3
n1x=10.5
! m1
n1y=26.34
152.6 kg
m3 air kg fuel
5.36 3N 2 H 4 o 6xH 2 (1 2 x)N 2 (4 4 x)NH 3 a. 0 d x d 1 b.
nN2H4
50 L 0.82 kg 1 kmol L 32.06 kg
128 . kmol
LM 6x kmol H b1 2xg kmol N b4 4xg kmol NH OP 3 kmol N H N 3 kmol N H 3 kmol N H Q 128 . x + 2.13 kmol . b6x 1 2x 4 4xg 1707 3
n product
2
2
. kmol N 2 H 4 128
2
4
5-18
2
4
3
2
4
5.36 (cont’d) nproduct 2.13 2.30 2.47 2.64 2.81 2.98 3.15 3.32 3.50 3.67 3.84
Vp (L) 15447.92 16685.93 17923.94 19161.95 20399.96 21637.97 22875.98 24113.99 25352.00 26590.01 27828.02
Volume of Product Gas 30000.00
25000.00
20000.00 V (L)
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
15000.00
10000.00
5000.00
0.00 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
c.
Hydrazine is a good propellant because as it decomposes generates a large number of moles and hence a large volume of gas.
5.37 A (g A / h) m
c
h
C A (g A / m 3 )
m3 / h V air
a. (i) Cap left off container of liquid A and it evaporates into room, (ii) valve leak in cylinder with A in it, (iii) pill of liquid A which evaporates into room, (iv) waste containing A poured into sink, A used as cleaning solvent.
A b. m
c. d.
FG kg A IJ H hK
yA
in
mol A mol air
yA 50 u106
d h V air
min
FG kg A IJ H hK C e j V M e j n
A m
out
gA A m3 gA A mol
F I FG GH JK H
3 m C kg A V air A h m3
air
CA
A m PV ; nair kVair RT
!
IJ K
yA
A RT m k Vair M A P
A 90 m . g/ h
A RT m kyA MA P
3
9.0 g / h
d
0.5 50 u 10
6
i
m Pa 8.314 mol 293 K K 3 101.3 u 10 Pa 104.14 g / mol
83 m3 / h
Concentration of styrene could be higher in some areas due to incomplete mixing (high concentrations of A near source); 9.0 g/h may be an underestimate; some individuals might be sensitive to concentrations PEL. e. Increase in the room temperature could increase the volatility of A and hence the rate of air : At higher T, need evaporation from the tank. T in the numerator of expression for V a greater air volume throughput for y to be PEL.
5-19
C3H 6 H 2 C3H 8
5.38 Basis: 2 mol feed gas
235$ C, P2
a. At completion, n p P2 V P1V
35.1 atm
n2
P2 T1 n1 P1 T2
g
2 np
1 mol 508K 32.0 atm 27.3 atm 2 mol 298K
n 2 T2 P1 n1 T1
35.1 atm 298K 2 mol 1.29 mol 32.0 atm 508K
1.29 2 n p n p
b
n p 2(1 n p )
2 1 1 mol
1 mol , n 2
n 2 RT2 P2 n1RT1
b. P2
U| V| W
n p (mol C 3 H 8 ) (1 - n p )(mol C 3 H 6 ) n 2 (1 - n p )(mol H 2 )
1 mol C 3 H 6 1 mol H 2 25$ C, 32 atm
0.71 mol C 3 H 8 produced
1- 0.71
0.29 mol C 3 H 6 unreacted 71% conversion of propylene
P2 (atm) 27.5 28.0 29.5 30.0 31.5 32.0 33.0 33.5 34.0 34.5 35.0 37.0 39.0 40.0
C3H8 prod. 0.99075 0.9724 0.91735 0.899 0.84395 0.8256 0.7889 0.77055 0.7522 0.73385 0.7155 0.6421 0.5687 0.532
c. %conv. 99.075 97.24 91.735 89.9 84.395 82.56 78.89 77.055 75.22 73.385 71.55 64.21 56.87 53.2
Pressure vs Fraction Conversion 120
100
80 % conversion
n2 1.009 1.028 1.083 1.101 1.156 1.174 1.211 1.229 1.248 1.266 1.285 1.358 1.431 1.468
60
%conv.
40
20
0 25.0
27.0
29.0
31.0
33.0 Pressure (atm)
5-20
35.0
37.0
39.0
41.0
5.39
Convert fuel composition to molar basis Basis: 100 g
b b
UV W
g g
97.2 mol % CH 4 5.92 mol CH 4 2.8 mol % C 2 H 6 . mol C 2 H 6 017
95 g CH 4 1 mol 16.04 g 5 g C2 H 6 1 mol 30.07 g
500 m3 / h
n 2 (kmol CO 2 / h) n 3 (kmol H 2 O / h)
n 1 (mol / h) 0.972 CH 4
n 4 (kmol O 2 / h) n 5 (kmol N 2 / h)
0.028 C 2 H 6 40$ C, 1.1 bar
(SCMH) V air 25% excess air
n 1
1 P1V RT1
. bar 500 m3 11 313K h
kmol K 0.08314 m3 bar
7 C 2 H 6 O 2 o 2CO 2 3H 2 O 2
CH 4 2O 2 o CO 2 2 H 2 O Theoretical O 2 =
. kmol h 211
LM N
21.1 kmol 0.972 kmol CH 4 kmol h
0.028 kmol C 2 H 6 kmol
b
125 . 431 . kmol O 2 Air Feed: h
g
3.5 kmol O 2 1 kmol C 2 H 6
1 kmol Air 0.21 kmol O 2
5.40 Basis: 1 m3 gas fed @ 205q C, 1.1 bars Ac 1 m 3 @205$ C, 1.1 bar
2 kmol O 2 1 kmol CH 4
OP Q
431 .
b g
22.4 m3 STP 1 kmol
kmol O 2 h
5700 SCMH
acetone
n 3 (kmol), 10$ C, 40 bar
condenser
n1 (kmol) y1 (kmol Ac / kmol) (1 - y 1 )(kmol air / kmol) p AC 0.100 bar
y 3 (kmol Ac / kmol) (1- y 3 )(kmol air / kmol) p AC 0.379 bar
n 2 (kmol Ac(l))
a.
n1 y1
100 . m3
273K
1 kmol 110 . bars 478K 10132 bars 22.4 m3 STP .
bar 0100 . 1.1 bars
b g
0.0909 kmol Ac kmol , y 3
b
gb
g
Air balance: 0.0277 0.910 Mole balance: 0.0277 Acetone condensed
0.0277 kmol
0.379 bar 40.0 bars
(1 9.47 u 10 3 )n 3 n 3
0.0254 n 2 n 2
0.0254 kmol
0.0023 kmol Ac condensed
0.0023 kmol Ac 58.08 kg Ac 1 kmol Ac
5-21
9.47 u 10 3 kmol Ac kmol
0.133 kg acetone condensed
5.40 (cont’d)
b g
0.0254 kmol 22.4 m3 STP
Product gas volume
b.
. 283K 10132 bars 0.0149 m3 273K 40.0 bars
20.0 m3 effluent 0.0277 kmol feed 0.0909 kmol Ac 58.08 kg Ac 196 kg Ac h h 0.0149 m3 effluent kmol feed kmol Ac
5.41 Basis: 1.00 u 10 6 gal. wastewater day. Neglect evaporation of water. 1.00 u 10 6 gal / day
Effluent gas: 68$ F, 21.3 psia(assume)
n1 (lb - moles H2 O / day) 0.03n1 (lb - moles NH3 / day)
n2 (lb - moles air / day) n3 (lb - moles NH 3 / day)
300 u 10 6 ft 3 air / day
Effluent liquid
n1 (lb - moles H2 O / day) n4 (lb - moles NH3 / day)
$
20 C, 21.3 psia n2 (lb - moles air / day)
a.
Density of wastewater: Assume U
LMn N
62.4 lb m ft 3
OP Q
lb - moles H 2 O 18.02 lb m 0.03n1 lb m NH 3 17.03 lb m 1 ft 3 u day 1 lb - mole day 1 lb - mole 62.4 lb m gal 100 . u 106 day 1
n1 n2
450000 lb - moles H 2 O fed day , 0.03n1
13500 lb - moles NH 3 fed day
. psi 1 lb - mole 300 u 106 ft 3 492 $ R 213 $ day 527.7 R 14.7 psi 359 ft 3 STP
b g
93% stripping: n 3
7.4805 gal 1 ft 3
. u 106 lb - moles air day 113
0.93 u 13500 lb - moles NH 3 fed day 12555 lb - moles NH 3 day
Volumetric flow rate of effluent gas PVout PVin
n out RT Vout n in RT
Vin
n out n in
300 u 106 ft 3
d1.13 u 10
6
i
12555 lb - moles day 6
1.13 u 10 lb - moles day
day
303 u 106 ft 3 day Partial pressure of NH 3
y NH 3 P
12555 lb - moles NH 3 day
d1.129 u 10
0.234 psi
5-22
6
i
12555 lb - moles day
u 213 . psi
5.42 Basis: 2 liters fed / min Cl ads.=
2.0 L soln 1130 g 012 . g NaOH 1 mol 0.23 NaOH ads. 1 mol Cl 2 L g soln 40.0 g mol NaOH 2 mol NaOH 60 min
2 L / min @ 23$ C, 510 mm H 2 O n 1 (mol / min) y (mol Cl 2 / mol) (1 - y)(mol air / mol)
0.013
mol min
n 2 (mol air / mol)
0.013 mol Cl 2 / min Assume Patm
n 1
2L
b g b10.33 0.510g m H O
10.33 m H 2 O Pabs
273K 10.84 m H 2 O
1 mol
b g
0.013 y
0150 .
mol Cl 2 ,? specification is wrong mol
125 L / min @ 25o C, 105 kPa n 1 ( mol / min) y1 (mol H 2 O / mol) 1- y1 ( mol dry gas / mol) 0.235 mol C 2 H 6 / mol DG 0.765 mol C 2 H 4 / mol DG
R|b g S| T
10.84 m H 2 O
0.0864 mol min
min 296K 10.33 m H 2 O 22.4 L STP
Cl balance: 0.0864y
5.43
2
in
(L / min) @ 65o C, 1 atm V 3 n C 2 H 6 (mol C 2 H 6 / min) n C 2 H 4 (mol C 2 H 4 / min) n air (mol air / min) n H 2 O (mol H 2 O / min)
U| V| W
355 L / min air @ 75o C, 115 kPa n 2 (mol / min) y 2 ( mol H 2 O / mol) (1- y 2 )( mol dry air / mol)
a.
Hygrometer Calibration ln y b ln a
b. n 1 n 2 R1
b
ln y1 y 2 R 2 R1
g lnd0.2 10 i
dy
bR ln a
i
4
90 5
ln y1 bR 1
0.08942
bg
ln 10 4 0.08942 5 a
125 L 273K 105 kPa 1 mol min 298K 101 kPa 22.4 L STP
b g
355 L 273K 115 kPa min
ae bR
1 mol
b g
348K 101 kPa 22.4 L STP
86.0 o y1
0.140 , R 2
12.8 o y 2
5-23
6.395 u 105 y 6.395 u 105 e 0.08942R
5.315 mol min wet gas 14.156 mol min wet air
2.00 u 104 mol H 2 O mol
5.43 (cont’d)
C H I b5.315 mol mingFGH b1 0.140g molmolDG IJK FGH 0.235 mol J mol DG K 2
C 2 H 6 balance: n C2 H 6
6
1.07 mol C 2 H 6 min
b5.315gb0.860gb0.765g 3.50 mol C H min Dry air balance: n b14.156gd1 2.00 u 10 i 14.15 mol DA min Water balance: n b5.315gb0.140g b14.156gd1.00 u 10 i 0.746 mol H O min n b1.07 3.50 14.15g mol min 18.72 mol min , n b18.72 0.746g 19.47 mol min 19.47 mol min 22.4 L bSTPg 338K 540 liters min V C 2 H 4 balance: n C2 H 4
2
4
4
air
4
H 2O
2
dry product gas total
3
mol
Dry basis composition:
c.
273K
FG 1.07 IJ u 100% H 18.72 K
5.7% C 2 H 6 , 18.7% C 2 H 4 , 75% dry air
0.746 mol H 2 O u 1 atm 0.03832 atm 19.47 mol
p H 2O
y H 2Ol P
y H 2O
0.03832 R
FG H
1 0.03832 ln 0.08942 6.395 u 10 5
IJ K
71.5
5.44 CaCO 3 o CaO CO 2 n CO 2
1350 m3 273K 1 kmol h 1273K 22.4 m3 STP
b g
12.92 kmol CO 2
12.92 kmol CO 2 h
1 kmol CaCO 3 100.09 kg CaCO 3
h
1 kmol CO 2
1 kmol CaCO 3
1362 kg limestone
0.17 kg clay
h
0.83 kg limestone
279 kg clay h
Weight % Fe 2 O 3 kg Fe 2 O 3 kg clay
b g
279 0.07 u 100% 18% Fe 2 O 3 . 1362 279 12.92 44.1 kg limestone kg clay
b g
kg CO 2 evolved
5-24
1 kg limestone 0.95 kg CaCO 3
1362 kg limestone h
5.45 Basis:
R|864.7 g C b1 mol 12.01 gg 72.0 mol C |116.5 g H b1 mol 1.01 gg 115.3 mol H 1 kg Oil S ||13.5 g S b1 mol 32.06 gg 0.4211 mol S T5.3 g I
C O 2 o CO 2 1 C + O 2 o CO 2 S O 2 o SO 2 1 2H O 2 o H 2 O 2
5.3 g I n 1 (mol CO 2 ) n 2 (mol CO) n 3 (mol H 2 O) n 4 (mol SO 2 ) n 5 (mol O 2 ) n 6 (mol N 2 )
72.0 mol C 115.3 mol H 0.4211 mol S 5.3 g I
n a (mol), 0.21 O 2 , 0.79 N 2 15% excess air 175$ C, 180 mm Hg (gauge)
a.
Theoretical O 2 : (0.95)72.0 mol C 1 mol O 2 (0.05)72.0 mol C 0.5 mol O 2 1 mol C 1 mol C 115.3 mol H 0.25 mol O 0.4211 mol S 1 mol O 2 2 1 mol H 1 mol S Air Fed:
e
1.15 99.45 mol O
2
j
b g
544 mol Air 22.4 liter STP 1 kg oil mol
1 mol Air 0.21 mol O
544 mol Air
n
2
99.45 mol O
2
a
1 m3 448K 760 mm Hg 16.2 m3 air kg oil 3 10 liter 273K 940 mm Hg
b. S balance: n 4 0.4211 mol SO 2 H balance: 115.3 = 2n 3 n 3 57.6 mol H 2 O C balance:
b g
0.95 72.0 = n1 n1 0.05(72.0)
n2
68.4 mol CO 2
3.6 mol CO
429.8 mol N b g O balance: 0.21b544g2 = 57.6 + 3.6 + 2(68.4) + 2b0.4211g + 2n n Total moles bexcluding inertsg wet: 559.8 mols dry: 502.2 mols N 2 balance: 0.79 544 = n 6 n 6
2
5
dry basis:
3.6 mol CO 502.2 mol
wet basis:
3.6 mol CO u 10 6 559.8 mol
7.2 u 10 3
mol CO 0.4211 mol SO 2 , mol 502.2 mol
6400 ppm CO ,
5-25
5
14.8 mol O 2
8.4 u 10 4
0.4211 mol SO 2 u 10 6 559.8 mol
mol SO 2 mol
750 ppm SO 2
bg
5.46 Basis: 50.4 liters C 5 H 12 l min 50.4 L C 5 H 12 ( l ) / min
b
n 1 kmol C 5 H 12 / min
g
heater
(L / min) 15% excess air, V air n 2 kmol air 0.21 O 2 0.79 N 2 336 K, 208.6 kPa (gauge)
a f n b kmol air / ming
n 1 kmol C5 H 12 / min
Combustion chamber
2
n 3 (kmol C 5 H 12 / min) n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min) n 7 (kmol H 2 O / min) (L / min) V gas
Condenser
(L / min), 275 K, 1 atm V liq = 3.175 kg C 5O12 / min m n 3 (kmol C 5O12 / min) n 7 (kmol H 2 O(l ) / min)
C5 H 12 802 o 5CO 2 6H 2 O
a.
n 1
n 3
50.4 L 0.630 kg min
. kg 1 kmol 3175 min 72.15 kg
frac. convert = n 2 air V
L
1 kmol 72.15 kg
0.440 kmol min
0.044 kmol / min
0.440 - 0.044 kmol u 100 90% C5 H12 converted 0.440
b
0.440 kmol C 2 H 12 1.15 8 kmol O 2 min kmol C 5 H 12
b g
19.28 kmol 22.4 L STP min mol
g
1 mol air 0.21 mol O 2
19.28 kmol air min
336K
101 kPa 1000 mol 273K 309.6 kPa kmol
n 4
0.440 kmol C 2 H12 0.15(8 kmol O 2 ) min kmol C 2 H12
n 5
19.28 mol air 0.79 mol N 2 min mol air
n 6
0.90(0.440 kmol C5 H12 ) 5 kmol CO 2 min kmol C5 H 12
gas V
n 4 (kmol O 2 / min) n 5 (kmol N 2 / min) n 6 (kmol CO 2 / min)
173000 L min
0.528 kmol O 2
15.23 kmol N 2 / min
. kmol CO 2 / min 198
0.528 + 15.23 + 1.98 kmol 22.4 L(STP) 548 K 1000 mol min mol 273 K kmol
5-26
7.98 u 105 L / min
5.46 (cont’d) 0.9(0.440 kmol C5 H12 ) 6 kmol H 2 O min kmol C5 H 12
n 7
2.38 kmol H 2 O(l ) / min
Condensate: 0.044 kmol 72.15 kg
V C5 H 12 V H 2O
min
kmol
2.38 kmol 18.02 kg min
L
kmol
5.04 L min
0.630 kg L
42.89 L min
1 kg
Assume volume additivity (liquids are immiscible) liq 5.04 42.89 47.9 L min V b.
bg
C5 H 12 l
C 5 H 12 ( l )
bg
H 2O l
bg
H2O l
5.47 n air (kmol / min), 25$ C, 1 atm 0.21 O 2 0.79 N 2 n 0 (kmol / min)
n 1 (kmol H 2S / min) Furnace
H 2 S + 23 O 2 o SO 2 H 2 O
0.20 kmol H 2 S / mol 0.80 kmol CO 2 / mol
Reactor
n 2 (kmol H 2 S / min)
2H 2 S + SO 2 o 3S(g) 2 H 2 O
10.0 m 3 / min @ 380$ C, 205 kPa n exit (kmol / min) n 3 (kmol N 2 / min) n 4 (kmol H 2 O / min) n 5 (kmol CO 2 / min) n 6 (kmol S / min)
n exit n 1
PV RT
b0.20gn
205 kPa 10.0 m3 / min m3 kPa 653 K 8.314 kmol K 0
/ 3 = 0.0667n 0 ; n 2
0.377 kmol / min
2 n 1 = 0.133n 0
5-27
5.47 (cont’d) 0.0667 n 0 (kmol H 2 S fed) 15 . kmol O 2 1 kmol air 1 kmol H 2 S 0.21 kmol O 2 (min)
Air feed to furnace: n air
0.4764 n 0 kmol air / min 0.4764 n 0 (kmol air) 0.79 kmol N 2 (min) min
Overall N 2 balance: n 3
Overall S balance: n 6
0.200n 0 (kmol H 2S) 1 kmol S (min) 1 kmol H 2S
Overall CO 2 balance: n 5 Overall H balance:
0.3764 n 0 ( kmol N 2 / min)
0.200n 0 (kmol S / min)
0.800n 0 (kmol CO 2 / min)
0.200n 0 (kmol H 2 S) 2 kmol H (min) 1 kmol H 2 S
n 4 kmol H 2 O 2 kmol H min 1 kmol H 2 O
n 4 = 0.200n 0 (kmol H 2 O / min) n exit
b
g
n 0 0.376 0.200 + 0.200 + 0.800 = 0.377 kmol / min n 0 = 0.24 kmol / min
n air = 0.4764(0.24 kmol air / min)
0114 . kmol air / min
5.48 Basis: 100 kg ore fed 82.0 kg FeS 2 (s), 18.0 kg I.
b
gb
n FeS2 fed = 82.0 kg FeS2 1 kmol / 120.0 kg
100 kg ore 06833 . kmol FeS2 18 kg I
g
0.6833 kmol FeS2
Vout m3 (STP)
40% excess air n 1 (kmol) 0.21 O 2 0.79 N 2 V1 m 3 (STP)
n 2 (kmol SO 2 ) n 3 (kmol SO 3 ) n 4 (kmol O 2 ) n5 (kmol N 2 )
m6 (kg FeS2 ) m7 (kg Fe 2 O 3 ) 18 kg I
2 FeS 2(s) 112 O 2(g) o Fe 2 O 3(s) 4SO 2(g) 2 FeS 2(s) 152 O 2(g) o Fe 2 O 3(s) 4SO 3(g) a.
n1
0.6833 kmol FeS2 7.5 kmol O 2 1 kmol air req'd 140 . kmol air fed 2 kmol FeS2 0.21 kmol O 2 kmol air req'd
V1
b17.08 kmolgb22.4 SCM / kmolg
n2
(0.85)(0.40)0.6833 kmol FeS2 4 kmol SO 2 2 kmol FeS2
382 SCM / 100 kg ore
5-28
0.4646 kmol SO 2
17.08 kmol air
5.48 (cont’d) n3
(0.85)(0.60)0.6833 kmol FeS2 4 kmol SO 2 2 kmol FeS2
n4
b0.21 u 17.08gkmol O
n5 Vout
2
fed
0.6970 kmol SO 3
.4646 kmol SO 2 55 . kmol O 2 4 kmol SO 2
.702 kmol SO 3 7.5 kmol O 2 4 kmol SO 3
1633 . kmol O 2
b0.79 u 17.08g kmol N 13.49 kmol N = b0.4646 + 0.6970 + 1.633 + 13.49 g kmol 2
2
22.4 SCM (STP) / kmol
365 SCM / 100 kg ore fed ySO 2
0.4646 kmol SO 2 u 100% 16.285 kmol
2.9%; y SO3
4.3%; y O 2
10.0%; y N 2
82.8%
b.
e j
Product gas, T o C Converter
0.4646 kmol SO 2 SO 3 0.697 kmol 21 OO2 2 SO SO 3 . 2 kmol 1633 13.49 kmol N 2
n SO 2 (kmol) n SO 3 (kmol) n O 2 (kmol) n N 2 (kmol)
Let [ (kmol) = extent of reaction
n SO 2 0.4646 [ n SO 3 0.697 [ 21 [ n O 2 1633 . n N 2 13.49 n = 16.28 - 21 [ K p (T) =
U| |V y || y W
SO 2
1 2
P = 1 atm, T = 600o C, K p n SO 2
1 2
-1
9.53 atm 2 [
kmol fSO2 01847 .
P = 1 atm, T = 400o C, K p n SO 2
c h [h (0.4646 [ )c1633 . (0.697 [ ) 16.28 21 [
P y SO 3 P y SO 2 ( P y O 2 )
O2
0.4646 [ 0.697 [ , y SO 3 1 16.28 - 2 [ 16.28 - 21 [ 1 2[ . 1633 13.49 , y N2 1 16.28 - 2 [ 16.28 - 21 [
1 2
P
- 21
1
K p (T) P 2
0.2799 kmol
. b0.4646 01847 g kmol SO
2
0.4646 kmol SO 2 fed -1
397 atm 2 [
0.00059 kmol fSO 2
1 2
reacted
0.602
0.4587 kmol
0.987
The gases are initially heated in order to get the reaction going at a reasonable rate. Once the reaction approaches equilibrium the gases are cooled to produce a higher equilibrium conversion of SO2.
5-29
5.48 (cont’d) c.
SO 3 leaving converter: (0.6970 + 0.4687) kmol = 1.156 kmol 1.156 kmol SO 3 1 kmol H 2 SO 4 98 kg H 2 SO 4 min kmol 1 kmol SO 3
Sulfur in ore:
0.683 kmol FeS 2 2 kmol S 32.1 kg S kmol FeS2 kmol
113.3 kg H 2SO 4 43.8 kg S
100% conv.of S:
438 . kg S
kg H 2SO 4 kg S
2.59
0.683 kmol FeS 2 2 kmol S 1 kmol H 2SO 4 98 kg kmol FeS2 1 kmol S kmol
133.9 kg H 2SO 4 43.8 kg S
113.3 kg H 2 SO 4
3.06
133.9 kg H 2SO 4
kg H 2SO 4 kg S
The sulfur is not completely converted to H2SO4 because of (i) incomplete oxidation of FeS2 in the roasting furnace, (ii) incomplete conversion of SO2 to SO3 in the converter. 5.49 N 2 O 4 2 NO 2 a.
n0
b.
n1
p NO 2
dP
gauge
i
b2.00 atmgb2.00 Lg b473Kgb0.08206 L atm mol - Kg
1.00 V
RT0 mol NO 2 , n 2
y NO 2 P
mol N 2 O 4
FG n IJ P , p Hn n K 1
1
2
N 2O4
Ideal gas equation of state PV
0.103 mol NO 2
FG n IJ P K n P n bn n g Hn n K bn n gRT n n PV / RT b1g 2
p
1
1
2 1
2
2
2
1
1
2
2
Stoichiometric equation each mole of N 2 O 4 present at equilibrium represents a loss of two moles of NO 2 from that initially present n1 2n 2 Solve (1) and (2) n1
b3g ,
2(PV / RT) 0.103
n2
0.103
b2 g
b4 g
0.103 (PV / RT)
Substitute (3) and (4) in the expression for K p , and replace P with Pgauge 1 t
gauge
t
T(K) 350 335 315 300
Pgauge(atm) 0.272 0.111 -0.097 -0.224
t
nt Kp(atm) 0.088568 5.46915 0.080821 2.131425 0.069861 0.525954 0.063037 0.164006
i
1 where n t
(1/T) ln(Kp) 0.002857 1.699123 0.002985 0.756791 0.003175 -0.64254 0.003333 -1.80785
dP
gauge
i
1 V
RT
nt
i
T
V 2L
2
y = -7367x + 22.747 R2 = 1
1 0 -1 -2 0.0028
0.003
0.0032 1/T
5-30
d
24.37 Pg 1
Variation of Kp with Temperature
ln Kp
b2n 0.103g dP n b0.103 n g 2
Kp
0.0034
5.49 (cont’d) c.
A semilog plot of K p vs.
1 is a straight line. Fitting the line to the exponential law T
yields ln K p
7367 22.747 K p T
7.567 u 109 exp
FG 7367 IJ a 7.567 u 10 H T K b = 7367K
9
atm
10.00 atm
5.50 n 1 (kmol A / h) n 2 (kmol H 2 / h)
5.00 kmol S / h n 4 (kmol A / h) n 5 (kmol H 2 / h)
3n 3 (kmol A / h) n 3 (kmol H 2 / h)
5.00 kmol S / h
n 4 (kmol A / h) n 5 (kmol H 2 / h)
Vrcy (SCMH)
Extent of reaction equations: n i
n i0 Q i[
A + H2 l S A:
n 4
H 2 : n 5 S:
3n 3 [ n [ 3
5.00 = [
! n 4 n 5 n tot
U| V| p 10.0 W
3n 3 5.00 n 3 5.00 4 n 3
A
pH2 pS Kp n 4 n 5 rcy V
pS pA p H2
b
g
5.00 4n 3 10.0 10.0 3n 3 5.00 n 3 5.00
b
gb
g
yA P =
n 4 P n tot
y H2 P = yS P =
n 3 . 0100
n 5 P n tot
3n 3 - 5.00 10.0 4 n 3 10.0 n 3 - 5.00 10.0 4 n 3 10.0
5.00 10.0 4 n 3 10.0 . kmol H 2 / h 1108
3(1108 . ) - 5.00 28.24 kmol A / h 1108 . 5.00 6.08 kmol H 2 / h
b28.24 + 6.08g kmol / h d22.4 m (STP) / kmoli 3
5-31
769 SCMH
5.51 n 4 (kmol CO / h) n 5 (kmol H 2 / h)
Reactor
Separator
100 kmol CO / h
n 1 (kmol CO / h) n 2 (kmol H 2 / h)
H xs (% H 2 excess)
a.
n 6 (kmol M / h)
n 4 (kmol CO / h) n 5 (kmol H 2 / h) n 6 (kmol M / h) T, P
n 3 (kmol H 2 / h) T (K), P (kPa)
Balances on reactor four equations in n 3 , n 4 , n 5 , and n 6 1 kmol CO CO balance n 4 100 kmol CO / h n 6 n 4 100 n 6 1 kmol M 100 kmol CO 1 kmol CO H 2 balance n 3 105 . 210 kmol H 2 / h h 1 kmol M 2 kmol CO n 5 210 kmol H 2 / h n 6 n 5 210 2 n 6 1 kmol M n T n 4 n 5 n 6 100 n 6 210 2 n 6 n 6 310 2 n 6
b g
b1g
b g
b g
b
b
K p T = 500K
g
u 10 4 1390 .
b 2g g b g b3g F 21.225 + 9143.6 7.492lnb500Kg I G exp GH +4.076 u 10500bK500Kg -1.161 u 10 b500Kg JJK -3
2
-8
9.11 u 10 7 kPa -2 Kp
yMP
d
y CO P y H 2 P
i
2
Kp P2
yM
d i
y CO y H 2
b
(1) ( 3) 2
!
n 6 310 2 n 6
g
b100 n g b210 2n g b310 2n g b310 2n g n b310 2 n g b100 n gb210 2n g 6
6
Kp P2
b
9.11 u 10 7 kPa -2 5000 kPa
g
22.775
6
6
2
6
Solving for n 6 n 6
100 n 6
n 5
210 2 n 6
n 1 n 2
rec V
bn
4
n 5
75.7 kmol M / h
n 4
24.3 kmol CO / h 58.6 kmol H 2 / h
1 kmol CO n 6 75.7 kmol CO / h 1 kmol M 2 kmol H 2 n 6 151 kmol H 2 / h 1 kmol M 3
m (STP) g 22.4 kmol
6
2
2
1860 SCMH
5-32
2
6
6
2
5.51 (cont’d) b. P(kPa) 1000 5000 10000 5000 5000 5000 5000 5000 5000
`
T(K) Hxs(%) 500 5 500 5 500 5 400 5 500 5 600 5 500 0 500 5 500 10
Kp(T)E8 9.1E+01 9.1E+01 9.1E+01 3.1E+04 9.1E+01 1.6E+00 9.1E+01 9.1E+01 9.1E+01
KpP^2 0.91 22.78 91.11 7849.77 22.78 0.41 22.78 22.78 22.78
KpP^2- n1(kmol KpcP^2 CO/h) 1.3E-05 25.55 2.3E+01 9.00 4.9E-03 86.72 3.2E-08 98.93 3.4E-03 75.68 -2.9E-04 14.58 9.8E-03 73.35 3.4E-03 75.68 -3.1E-03 77.77
ntot n6(kmol M/h) (kmol/h) KpcE8 25.55 258.90 9.1E-01 9.00 292.00 2.3E-01 86.72 136.56 9.1E+01 98.93 112.15 7.8E+03 75.68 158.64 2.3E+01 14.58 280.84 4.1E-01 73.35 153.30 2.3E+01 75.68 158.64 2.3E+01 77.77 164.45 2.3E+01
n3(kmol n4(kmol n5(kmol H2/h) H2/h) CO/h) 210 74.45 158.90 210 91.00 192.00 210 13.28 36.56 210 1.07 12.15 210 24.32 58.64 210 85.42 180.84 200 26.65 53.30 210 24.32 58.64 220 22.23 64.45 n2(kmol H2/h) 51.10 18.00 173.44 197.85 151.36 29.16 146.70 151.36 155.55
Vrec (SCMH) 5227 6339 1116 296 1858 5964 1791 1858 1942
c. Increase yield by raising pressure, lowering temperature, increasing Hxs. Increasing the pressure raises costs because more compression is needed. d. If the temperature is too low, a low reaction rate may keep the reaction from reaching equilibrium in a reasonable time period. e. Assumed that reaction reached equilibrium, ideal gas behavior, complete condensation of methanol, not steady-state measurement errors. 5.52
CO 2 CO + 21 O 2
1.0 mol CO 2 1.0 mol O 2 1.0 mol N 2 T = 3000 K, P = 5.0 atm
dp
CO p O 2
d
1 A B C 2 1 1 C D E 2 2
1/ 2
i
0.3272 atm p CO 2 1 O 2 12 N 2 NO 2 p NO 0.1222 K2 1/ 2 p N 2 p O2 K1
1/ 2
i
A CO 2 , B CO , C O 2 , D N 2 , E NO n A0
n C0
n D0
1 , n B0
n E0
5-33
0
[ 1 - extent of rxn 1 [ 2 - extent of rxn 2
5.52 (cont’d) 1 [1 [1 1 1 1 [1 [ 2 2 2 1 1 [2 2 [2 1 6 [1 3 [1 2 2
nA nB nC nD nE n tot
U| || y n n 2 b1 [ g b 6 [ g | yy b22[ [b 6 [[ gg b 6 [ g V| || yy 2b 2[ [b 6 g b[6 g [ g || W A B
A
tot
1
1
C
1
D
2
1
y B y1C2 b1 12 1g p yA
CO 2
1
b
2[ 1 2 [ 1 [ 2
12
12
12
1
K2
d
p NO pO2 p N 2
b
i
12
12
1
yE 12 12 yC y D
01222 2 [1 [ 2 .
1
p11 2 1 2
g b2 [ g 12
12
2
1
b2 [
5.53 a.
b
2[ 2 1
[2
g b2 [ g 12
12
01222 .
2
2[ 2
(2)
g b6 [ g
2 1 [1 yB yC
(1)
2
Solve (1) and (2) simultaneously with E-Z Solve [ 1
yA
0.3272
1
12
1
yiP
1
2
g b5g K p 2b1 [ gb6 [ g 0.3272b1 [ gb6 [ g 2.236[ b2 [ [ g p CO p1O22
pi
1
2
E
1
1
0.2574 mol CO 2 mol y D yE 0.0650 mol CO mol 0.3355 mol O 2 mol 1
0.20167, [ 2
012081 . ,
0.3030 mol N 2 mol 0.0390 mol NO mol
n 4 (kmol / h)
0.04 O 2 0.96 N 2
PX = C8 H 10 , TPA = C8 H 6 O, S = Solvent (m 3 / h) @105o C, 5.5 atm V 3 n 3O (kmol O 2 / h) n 3N (kmol N 2 / h) n 3W (kmol H 2 O(v) / h) (m 3 / h) at 25o C, 6.0 atm V 2 n 2 (kmol / h) 0.21 O 2 0.79 N 2
condenser
n 3W (kmol H 2 O(v) / h) (m 3 / h) V 3W
reactor
n 1 (kmol PX / h) ( n 1 n 3p ) kmol PX / h s (kg S / h) m 3 kg S / kg PX
5-34
n 3p ( kmolPX / h) 100 kmol TPA / h s (kg S / h) m
separator
100 mol TPA / s
n 3p (kmol PX / h) s (kg S / h) m
5.53 (cont’d) b. Overall C balance: n 1
c.
FG kmol PX IJ 8 kmol C H h K kmol PX
O 2 consumed =
100 kmol TPA 8 kmol C n 1 h kmol TPA
100 kmol TPA 1.5 kmol O 2 h 1 kmol TPA
150
Overall N 2 balance: 0.79n 2 Overall H 2 O balance: n 3W
2 V 3 V
n 2 RT P
bn
3W V
3W
150 kmol O 2 / h
kmol O 2 + 0.04n 4 h 0.96n 4
Overall O 2 balance: 0.21n 2
g
U| V| W
n 2 n 4
100 kmol TPA 2 kmol H 2 O h 1 kmol TPA
848 kmol 0.08206 m3 atm 298 K h kmol K 6.0 atm
n 4 RT P
100 kmol PX / h
200 kmol H 2 O / h
3450 m3 air / h
b200 + 698g kmol 0.08206 m atm 378 K 3
kmol K 5.5 atm
h
200 kmol H 2 O (l) 18.0 kg 1 m3 h kmol 1000 kg
d
b100 + 11.1g kmol PX h
5065 m3 / h
3.60 m3 H 2 O(l) / h leave condenser
i
n 1 100
d. 90% single pass conversion n 3p = 0.10 n 1 n 3p ====> n 3p recycle m
848 kmol air / h 698 kmol / h
106 kg 4 kg recycle kg PX 1 kmol PX
111 . kmol PX / h
4.71 u 104 kg recycle / h
e. O2 is used to react with the PX. N2 does not react with anything but enters with O2 in the air. The catalyst is used to accelerate the reaction and the solvent is used to disperse the PX. f. The stream can be allowed to settle and spearate into water and PX layers, which may then be separated. 5.54
n 1 (kmol CO / h), n 3 (kmol H 2 / h), 0.10n 2 (kmol H 2 / h)
Separator n 6 (kmol CO / h) n 7 (kmol H 2 / h) n 8 (kmol CO 2 / h) 2 kmol N 2 / h
0.90 n 2 2 kmol N 2 / h
n 1 , n 2 , n 3 2 kmol N 2 / h 0.300 kmol CO / kmol 0.630 kmol H 2 / kmol 0.020 kmol N 2 / kmol 0.050 kmol CO 2 / kmol
Reactor
n 1 (kmol CO / h) n 2 (kmol H 2 / h) n 3 (kmol CO 2 / h) n 4 (kmol M / h) n 5 (kmol H 2 O / h) 2 5-35 kmol N 2 / h
Separator
n 4 (kmol M / h) n 5 (kmol H 2 O / h)
5.54 (cont’d) CO + 2H 2 CH 3OH(M) CO 2 3H 2 CH 3OH + H 2 O
a.
Let [ 1 ( kmol / h)
extent of rxn 1, [ 2 ( kmol / h)
extent of rxn 2
CO: n 1 = 30 - [ 1 H 2 : n 2 = 63 - 2[ 1 3[ 2
U| M: n = [ [ || Py Py H O: n = [ V| dK i P y P Py y , dK i bP y gd P y i d i d id i N : n = 2 || n 100 - 2[ 2[ W n 2[ 2[ g n 84.65 (1) dK i P = n F n I bb[30[[ gbgb100 63 2[ 3[ g G J n H n K FG n IJ FG n IJ . (2) dK i P = FH nn IKFH nn IK [ bb[5 [[ gbgb63100 2[2[ 3[2[g g 1259 GH n JK GH n JK CO 2 : n 3 = 5 - [ 2 2 2
4
1
5
2
2
M
p 1
CO
N2
tot
p 2
H2
2
1
tot
2
1
2
2
1
2
tot
tot
2
CO 2
3
H2
4
p 1
H 2O
2
1
2
M
p 2
2
2
1
4
5
tot
tot
3
2
tot
tot
1
2
2
2
2
1
1
2 2
3
2
1
2
Solve (1) and (2) for [ 1 , [ 2 [ 1 = 25.27 kmol / h [ 2 = 0.0157 kmol / h
n 1
30.0 25.27
4.73 kmol CO / h
9.98% CO
n 2
63.0 2(25.27) 3(0.0157) 12.4 kmol H 2 / h
26.2% H 2
n 3
5.0 0.0157
10.5% CO 2
n 4
25.27 0.0157
n 5
0.0157
n total
4.98 kmol CO 2 / h
0.0157 kmol H 2 O / h
49.4 kmol / h
C balance: n 4 25.3 kmol / h O balance: n 6 2 n 8 n 4 n 5 H balance: 2n 7 b. (n 4 ) process
53.4% M 0.03% H 2 O
UV n 25.4 kmol CO / h 25.44 mol / sW n = 0.02 kmol CO / h
2(0.9 n 2 ) 4 n 4 2 n 5
6
8
123.7 n 7
237 kmol M / h
Scale Factor =
25.3 kmol M / h
237 kmol M / h 25.3 kmol / h
5-36
2
618 . mol H 2 / s
5.54 (cont’d)
237 kmol / h I F 22.4 m (STP) I gFGH 25.3 J kmol JK 18,700 SCMH kmol / h K GH F 237 kmol / h IJ 444 kmol / h Reactor effluent flow rate: b 49.4 kmol / hgG H 25.3 kmol / sK F kmol IJ FG 22.4 m (STP) IJ 9946 SCMH V G 444 H h K H kmol K 3
b
Process feed: 25.4 618 . 0.02 2.0
3
std
. kPa 9950 m 3 (STP) 473.2 K 1013 h 273.2 K 4925 kPa
Vactual c.
=V V n
354 m 3 / h 1000 L 1 kmol 444 kmol / h m 3 1000 mol
354 m 3 / h
0.8 L / mol
(5.2-36)
< 20 L / mol ====> ideal gas approximation is poor V from n using the ideal gas equation of state is likely Most obviously, the calculation of V to lead to error. In addition, the reaction equilibrium expressions are probably strictly valid only for ideal gases, so that every calculated quantity is likely to be in error.
5.55 a.
RTc PV B Bo ZB1 1 B = RT Pc V From Table B.1 for ethane: Tc 305.4 K, Pc 48.2 atm From Table 5.3 -1 Z = 0.098 0.422 0.422 0.333 Bo 0.083 1.6 0.083 1.6 Tr 308.2 K 305.4 K . . 0172 0172 4.2 0139 0.0270 B1 0139 . . 4 .2 Tr 308.2 K 305.4K RTc 0.08206 L atm 305.4 K 0.333 0.098 0.0270 B(T) = B o ZB1 Pc mol K 48.2 atm 01745 L / mol .
b
b
g
e
j
e
j
g
FG H
b
IJ K
g
2 PV mol K 2 - B = 10.0 atm V V V + 0.1745 = 0 RT 308.2K 0.08206 L atm = V
b
gb
1 r 1 - 4 0.395 mol / L 01745 . L / mol
b
g
2 0.395 mol / L
Videal
b.
g
2.343 L / mol, 0.188 L / mol
RT / P 0.08206 u 308.2 / 10.0 2.53, so the second solution is likely to be a mathematical artifact. 10.0 atm 2.343 L / mol PV 0.926 z= Latm RT 0.08206 mol 308.2K K
5-37
c.
5.56
= m
V 1000 L mol 30.0 g 1 kg MW = h 2.343 L mol 1000 g V
PV RT
1
b
RTc B B= Bo ZB1 Pc V
12.8 kg / h
g
b g 513.2 K, P 78.50 atm T bC H g 369.9 K, P 42.0 atm From Table 5.3 -1 Z bCH OH g = 0.559, Z bC H g = 0.152 From Table B.1 Tc CH 3OH c
3
c
8
c
3
Bo (CH 3OH)
0.083
Bo ( C 3 H 8 )
0.422 1.6 Tr
. 0139
B1 (CH 3OH)
0139 .
B1 (C 3 H 8 )
3
0.422 0.083 1.6 Tr
B(CH 3OH) =
0.083 0.083
. 0172 4 .2 Tr
0172 . 4 .2 Tr
RTc Bo ZB1 Pc
e
e
e
513.2K 0.422
j
0.333
j
1.6
369.9K . 0172
373.2K
513.2K 0172 .
373.2K
0.619
1.6
373.2K
373.2K
. 0139 0139 .
b
e
8
0.422
369.9K
j
j
0.516
4 .2
0.0270
4 .2
g
0.08206 L atm 513.2K 0.619 0.559 0.516 mol K 78.5 atm RTc Bo ZB1 B(C 3 H 8 ) = Pc
b
c
b
g
h
b
c
¦¦y y B i
i
B ij B mix
b
L mol
g
0.08206 L atm 369.9 K . 0.0270 0.333 0152 mol K 42.0 atm
B mix
0.4868
j
ij
B ij
j
d
0.5 B ii B jj
g
h
0.2436
L mol
i
g
0.5 0.4868 0.2436 L / mol = -0.3652 L / mol
b0.30gb0.30gb0.4868g 2b0.30gb0.70gb0.3652g b0.70gb0.70gb0.2436g
0.3166 L / mol
IJ K
FG H
2 PV mol K 2 - B = 10.0 atm V V V + 0.3166 = 0 mix RT 373.2K 0.08206 L atm = Solve for V:V V ideal
RT P
b
gb
1 r 1- 4 0.326 mol / L 0.3166 L / mol
b
g
2 0.326 mol / L
0.08206 L atm 373.2 K mol K 10.0 atm
g
2.70 L / mol, 0.359 L / mol
3.06 L / mol V virial
5-38
2.70 L / mol
5.57 a.
15.0 kmol CH 3OH / h 1000 mol 1 m3 135 m3 / h 0.30 kmol CH 3OH / kmol 1 kmol 1000 L
2.70 L / mol
= Vn V
van der Waals equation: P =
d
i
d
RT a2 2 -b V V
i
2 V - b PV 3 PV 2 b = RTV 2 aV + ab Multiply both sides by V
b
g
+ -Pb - RT V aV - ab = 0 PV 3
c3
2
P = 50.0 atm
c2
b-Pb - RTg b50.0 atmgb0.0366 L / molg c0.08206
c1
a = 133 . atm L2 / mol 2
c0
ab = - 133 . atm L2 / mol 2 0.0366 L / mol
ib
d
0.0487
hb223 Kg
201 . L atm / mol
g
3
atm L mol 3
RT P
b. V ideal
Latm molK
0.08206 L atm 223 K mol K 50.0 atm
0.366 L / mol
c. T(K)
P(atm)
223 223 223 223 223
1.0 10.0 50.0 100.0 200.0
c3
c2
1.0 10.0 50.0 100.0 200.0
c1
-18.336 -18.6654 -20.1294 -21.9594 -25.6194
c0
1.33 1.33 1.33 1.33 1.33
-0.0487 -0.0487 -0.0487 -0.0487 -0.0487
V(ideal) V f(V) % error (L/mol) (L/mol) 18.2994 18.2633 0.0000 0.2 1.8299 1.7939 0.0000 2.0 0.3660 0.3313 0.0008 10.5 0.1830 0.1532 -0.0007 19.4 0.0915 0.0835 0.0002 9.6
d. 1 eq. in 1 unknown - use Newton-Raphson.
b1g gdV i
b
Eq. (A.2-13) a Eq. (A.2-14) ad (k +1) Then V
g
b g
3 + -20.1294 V 2 133 50.0V =0 . V-.0487 wg wV
2 40.259 V + 1.33 150V solve
g d
g a
(k) d Guess V (1) V
1 2 3 4
V ideal
(k) V 0.3660 0.33714 0.33137 0.33114
5-39
0.3660 L / mol . (k +1) V 0.33714 0.33137 0.33114 0.33114
converged
b
5.58 C 3 H 8 : TC
369.9 K
5.0 m3 75 kg
Specific Volume
d
42.0 atm 4.26 u 106 Pa
PC
44.09 kg 1 kmol 1 kmol 103 mol
i
Z
0152 .
2.93 u 10 3 m3 mol
Calculate constants 0.42747
a
d8.314 m Pa mol Ki b369.9 Kg 2
3
6
4.26 u 10 Pa
0.08664
b
d8.314 m Pa mol Ki b369.9 Kg 3
4.26 u 106 Pa
b
g
b
015613 0152 0152 m 0.48508 155171 . . . .
e
D
1 0.717 1 298.2 369.9
j
2
g
2
2
0.949 m6 Pa mol 2 6.25 u 10 5 m3 mol
0.717
115 .
SRK Equation:
d8.314 m Pa mol Kib298.2 Kg d2.93 u 10 6.25 u 10 i m mol 2.93 u 10
d
3
P
3
5
3
115 . 0.949 m6 Pa mol 2 3
d
i
i
m3 mol 2.93 u 10 3 6.25 u 10 5 m3 mol
7.40 u 106 Pa 7.30 atm
P Ideal:
RT V
P
TC
3
3
8.46 u 106 Pa 8.35 atm
3
2.93 u 10 m mol (8.35 7.30) atm u 100% 14.4% 7.30 atm
Percent Error: 5.59 CO 2 :
d8.314 m Pa mol Kib298.2 Kg
304.2 K
PC
72.9 atm Z
0.225
151.2 K PC 48.0 atm Z 0.004 35.0 L / 50.0 mol 0.70 L mol P 510 . atm , V
Ar:
TC
Calculate constants (use R
L2 atm L , m 0.826 , b 0.0297 ,D 2 mol mol L2 atm L a 137 . , m 0.479 , b 0.0224 ,D 2 mol mol
CO 2 : a Ar:
bg
f T
0.08206 L atm mol K )
3.65
RT a 1 m 1 T TC V b Vb V
d
i
e
j
2
P=0
Use E-Z Solve. Initial value (ideal gas): L L atm Tideal 510 . atm 0.70 0.08206 mol mol K
b
gFGH
e 1 0.479e1
IJ FG K H
5-40
IJ K
j . j T 1512
1 0.826 1 T 304.2
435.0 K
2
2
b g
E - Z Solve Tmax
5.60 O 2 : TC
CO 2
154.4 K ; PC
455.4 K ,
bT g
431.2 K
max Ar
49.7 atm ; Z
b
g
0.021 ; T 208.2 K 65q C ; P 8.3 atm ;
0.08206 L atm mol K
250 kg h ; R m
SRK constants: a 138 . L2 atm mol 2 ; b
0.0221 L mol ; m 0.517 ; D
0.840
d i dVRT bi V dVaD bi P = 0=====> V = 2.01 L / mol E-Z Solve
f V
SRK equation:
V
250 kg kmol 103 mol 2.01 L 15,700 L h h 32.00 kg 1 kmol mol
W
5.61
e
¦F
PCO2 A - W = 0 where W = mg = 5500 kg 9.81
y
m s2
j
53900 N
PCO 2 A
a.
PCO 2
W
53900 N
A piston
S 4
. mg b015
2
b. SRK equation of state: P = For CO 2 : Tc
304.2, Pc
1 atm 1.013 u 105 N / m2
Da RT -b V V +b V
d i d
. atm 301
i
72.9 atm , Z = 0.225
a = 3.654 m atm / kmol , b = 0.02967 m3 / kmol, m 0.8263, D (25o C) 1016 . 6
2
. ge3.654 e0.08206 jb298.2 Kg b1016 j 301 . atm = eV - 0.02967 j V dV + 0.02967i m 3 atm kmolK
m6 atm kmol 2
m6 kmol 2
m3 kmol
E-Z Solve
= 0.675 m 3 / kmol =====> V
b g 0.030 m Vbafter expansiong 0.030 m b015 . mg b15 . mg 3
V before expansion
3
mCO 2
2
S 4
44.01 kg V 0.0565 m3 MW = 3 0.675 m / kmol kmol V
mCO 2 (initially) =
0.0565 m3
3.68 kg
PV 1 atm 0.030 m3 44.01 kg MW = m 3 atm 298.2 K RT kmol 0.08206 kmol K
mCO 2 (added) = 3.68 - 0.0540 kg = 3.63 kg
5-41
0.0540 kg
5.61 (cont’d) c.
W = 53,900 N
V
n o (kmol) Vo (m3 ) 1 atm, 25o C
! n (kmol) P (atm), 25o C
ho
ho
d(m)
d(m)
Given T, Vo , h, find d Vo Initial: n o Po 1 RT Sd 2 h 3.63 (kg) Final: V = Vo , n = no 4 44 (kg / kmol)
b
=V V n
h
add 3.63 kg CO 2
Vo
g
Vo 0.0825 RT
Sd 2 h 4
Vo 0.0825 RT Da 53,900 W RT 2 P= V +b A piston V - b V Sd / 4
d
Da RT V +b V-b V
i d i in b1g one equation in one unknown. Substitute expression for V
b1g Solve for d .
5.62 a. Using ideal gas assumption: Pg
nRT Patm V
35.3 lb m O 2 1 lb - mole 10.73 ft 3 psia 509.7 o R 14.7 psia = 2400 psig 32.0 lb m lb - mole o R 2.5 ft 3
b. SRK Equation of state: P =
Da RT -b V V +b V
d i d
3 = 2.5 ft 32.0 lb m / lb - mole V 35.3 lb m
For O 2 : Tc
277.9 o R, Pc
2.27
ft psi , b lb - mole 2
ft 3 lb - mole
730.4 psi, Z = 0.021
6
a = 52038 .
i
0.3537
ft 3 , m = 0.518, D 50o F lb - mole
d
i
e10.73 jd509.7 Ri b0.667ge52038. b2400 + 14.7g psi = V - 0.3537 dV + 0.3537i V d i ft 3 psi
o
ft 6 psi
lb-mole 2
o
lb-mole R
ft 3 lb-mole
= 2.139 ft 3 / lb - mole E - Z Solve V 32.0 lb m V 2.5 ft 3 mO 2 MW = 3 2.139 ft / lb - mole lb - mole V
5-42
37.4 lb m
0.667
j
ft 6 lb-mole 2
5.62 (cont’d) Ideal gas gives a conservative estimate. It calls for charging less O2 than the tank can safely hold. c. 1. 2. 3. 4.
Pressure gauge is faulty The room temperature is higher than 50qF Crack or weakness in the tank Tank was not completely evacuated before charging and O2 reacted with something in the tank 5. Faulty scale used to measure O2 6. The tank was mislabeled and did not contain pure oxygen.
5.63 a. SRK Equation of State: P =
Da RT +b V-b V V
d i d
i
d id i = PV dV - bid V + bi RTV dV + bi Dad V - bi f dV i PV RTV dDa - b P - bRTiV - Dab = 0 f dV
i
V -b V +b : multiply both sides of the equation by V
3
2
0
2
b. Problem 5.63-SRK Equation Spreadsheet Species Tc(K) Pc(atm) Z a b m
CO2 304.2 R=0.08206 m^3 atm/kmol K 72.9 0.225 3.653924 m^6 atm/kmol^2 0.029668 m^3/kmol 0.826312
f(V)=B14*E14^3-0.08206*A14*E14^2+($B$7*C14-$B$8^2*B14-$B$8*0.08206*A14)*E14-C14*$B$7*$B$8 T(K) 200 250 300 300 300
P(atm) 6.8 12.3 6.8 21.5 50.0
alpha 1.3370 1.1604 1.0115 1.0115 1.0115
V(ideal) 2.4135 1.6679 3.6203 1.1450 0.4924
V(SRK) 2.1125 1.4727 3.4972 1.0149 0.3392
f(V) 0.0003 0.0001 0.0001 0.0000 0.0001
c. E-Z Solve solves the equation f(V)=0 in one step. Answers identical to VSRK values in part b. d. REAL T, P, TC, PC, W, R, A, B, M, ALP, Y, VP, F, FP INTEGER I CHARACTER A20 GAS DATA R 10.08206/ READ (5, *) GAS WRITE (6, *) GAS 10 READ (5, *) TC, PC, W READ (5, *) T, P IF (T.LT.Q.) STOP
5-43
5.63 (cont’d) R 0.42747 *R*R/PC*TC*TC B 0.08664 *R*TC/PC W 015613 M 0.48508 W 155171 . .
b
ALP
d1. M c1 bT / TCg
0.5hi
2 .
g
VP R T / P DO 20 I 7, 15 V = VP F = R * T/(V – B) – ALP * A/V/(V + B) – P FP = ALP * A * (2. * V + B)/V/V/(V + B) ** 2 – R * T/(V – B) ** 2. VP = V – F/FP IF (ABS(VP – V)/VP.LT.0.0001) GOTO 30 20 CONTINUE WRITE (6, 2) 2 FORMAT ('DID NOT CONVERGE') STOP 30 WRITE (6, 3) T, P, VP 3 FORMAT (F6.1, 'K', 3X, F5.1, 'ATM', 3X, F5.2, 'LITER/MOL') GOTO 10 END $ DATA CARBON 304.2 200.0 250.0 300.0 –1
72.9 6.8 12.3 21.5 0.
DIOXIDE 0.225
RESULTS CARBON DIOXIDE 200.0 K 6.8 ATM 250.0 K 12.3 ATM 300.0 K 6.8 ATM 300.0 K 21.5 ATM 300.0 K 50.0 ATM 5.64 a.
b.
N 2 : TC PC
Tr 126.2 K 33.5 atm Pr
He: TC PC
Tr 5.26 K 2.26 atm Pr
2.11 LITER/MOL 1.47 LITER/MOL 3.50 LITER/MOL 1.01 LITER/MOL 0.34 LITER/MOL
U| 40 MPa 10 atm V z 12. 1178 . | 33.5 atm 1.013 MPa W . |U b200 273.2g b5.26 8g 552 V| z 16. 350 b2.26 8g 34.11 W b40 273.2g 126.2
2.48
Fig. 5.4-4
Fig. 5.4-4
n Newton’s correction
5-44
5.65 a.
d
U kg / m3
i
m (kg) V (m3 )
(MW)P RT
30 kg kmol 9.0 MPa 10 atm 3 m atm 465 K 0.08206 kmolK 1.013 MPa b.
UV W
Tr Pr
. 465 310 15 9.0 4.5 2.0
U
(MW)P zRT
5.66 Moles of CO 2 : TC PC
z
V r
VP C RTC
, Vr 1507 .
Tr1
154.4 K 49.7 atm
Pr1 Pr2
z 2 T2 P1 z1 T1 P2
127 m3 1.61 358 K
5.68 O 2 : TC PC
n1 n 2
b1600 14.7gpsi 72.9 atm
1 atm . 1507 14.7 psi
10.0 ft 3 72.9 atm lb - moleq R 1k 3 2.27 lb - moles 304.2 k 0.7302 ft atm 1.8 q R 0.80 z 0.85
1614.7 psi 10.0 ft 3 lb - moleq R 1 atm 3 0.85 2.27 lb - moles 0.7302 ft atm 14.7 psi
PV znR
Tr2
V2
2.27 lb - moles
44.01 lb m CO 2
P PC
V1
831 . kg m3
100 lb m CO 2 1 lb - mole CO 2
UV W
5.67 O : T 2 C PC
V2
0.84
69.8 kg m3 0.84
304.2 K Pr 72.9 atm
Fig. 5.4-3: Pr T
Fig. 5.4-3
69.8 kg m3
h
|UVz 1 49.7 0.02 |W 358 154.4 2.23 U| Vz 1000 49.7 20.12 |W 298 154.4 1.93
1.00 298 K 1000 atm
154.4 K 49.7 atm
FG H
V P1 P2 RT z1 z 2
IJ K
320 q F
1.00 (Fig. 5.4 - 2)
1
1 atm
779q R
0.80
b
1.61 Fig. 5.4 - 4
2
g
0.246 m3 h
Tr
b27 273.2g 154.4
Pr1
175 49.7
3.52 z1
0.95
Pr2
1.1 49.7
0.02 z 2
1.00
1.94
FG H
(Fig. 5.3-2)
mol K 175 atm 11 . atm 300.2 K 0.08206 L atm 0.95 1.00 10.0 L
5-45
IJ K
74.3 mol O 2
5.69 a.
= V 50.0 mL 44.01 g 4401 V . mL / mol n 5.00 g mol RT 82.06 mL atm 1000 K P= 186 atm mol K 440.1 mL / mol V
b. For CO 2 : Tc 304.2 K, Pc 72.9 atm T 1000 K Tr 3.2873 Tc 304.2 K VP . mL 72.9 atm mol K 4401 c Vr ideal RTc mol 304.2 K 82.06 mL atm Figure 5.4 - 4: Vr ideal P= c.
zRT V
128 . and Tr
. 128
3.29 z = 1.02
. 82.06 mL atm mol 1000 K 102 mol K 440.1 mL
190 atm
a = 3.654 u 10 6 mL2 atm / mol 2 , b = 29.67 mL / mol, m 0.8263, D (1000 K ) P=
. ge j c82.06 hb1000 Kg b01077 b440.1- 29.67g 440.1b440.1 + 29.67 g 2 3.654 u 106 mLmolatm 2
mLatm molK
mL2 mol 2
mL mol
01077 .
198 atm
5.70 a. The tank is being purged in case it is later filled with a gas that could ignite in the presence of O2. b. Enough N2 needs to be added to make x O 2
10 u 10 6 . Since the O2 is so dilute at this
condition, the properties of the gas will be that of N2. . atm, Tr 2.36 Tc 126.2 K, Pc 335 n initial nO2 n O2 n2
n1
PV RT
1 atm 5000 L Latm 0.08206 molK 298.2 K
204.3 mol air
FG 0.21 mol O IJ H mol air K
10 u 10 6 n 2
2
204.3 mol
42.9 mol O 2
4.29 u 10 6 mol
5000 L 116 . u 10 -3 L / mol 6 4.29 u 10 mol mol K 335 . u 10 3 L . atm ideal VPc 116 V r RTc mol 0.08206 L atm 126.2 K not found on compressibility charts = V
Ideal gas: P =
RT V
0.08206 L atm 298.2 K mol K 116 . u 10 3 L / mol
38 . u 10 3
2.1 u 10 4 atm
The pressure required will be higher than 2.1 u 10 4 atm if z t 1, which from Fig. 5.3 - 3 is very likely. n added
ib
d
4.29 u 106 204.3 # 4.29 u 106 mol N 2 0.028 kg N 2 / mol
5-46
g
. u 105 kg N 2 120
5.70 (cont’d) c.
143 . kmol N 2
143 . kmol N 2
n initial 0.204 kmol yO 0.21 kmol O 2 / kmol 2
143 . kmol N 2 y1
N 2 at 700 kPa gauge = 7.91 atm abs. Pr n2
P2 V zRT
y1
y init n init 1.634
y2
y 1 n init 1.634
y init
0.236, Tr
7.91 atm 5000 L Latm 298.2 K 0.99 0.08206 mol K
b0.21g0.204 1634 .
y init
FG n IJ K H 1634 .
Fig 5.4-2
2.36 =======> z = 0.99
1633 . kmol
0.026
2
init
0.0033
FG y IJ FG n IJ n = H y K 4.8 Need at least 5 stages K H 1634 . F n IJ lnG H 1.634 K 5b143 . kmol N gb28.0 kg / kmolg 200 kg N ln
n
yn
143 . kmol N 2 y2
n
init
init
init
Total N 2
2
2
d. Multiple cycles use less N2 and require lower operating pressures. The disadvantage is that it takes longer. 5.71
= MW a. m b.
Tc Pc
PV SPV = MW Cost ($ / h) = mS RT RT
UV W
369.9 K = 665.8o R Tr 0.85 Pr 016 42.0 atm .
= 60.4 m
PV zT
ideal m z
F 44.09 lb / lb - mol I SPV GH 0.7302 JK T m
3
ft atm lb-molo R
60.4
SPV T
Fig. 5.4-2
z = 0.91
ideal 110 . m
Delivering 10% more than they are charging for (undercharging their customer)
5-47
5.72 a.
For N 2 : Tc
After heater: Tr
n =
b. tank =
4.65 ft 3 / min
34,900 gal 133.0 K, Pc
34.5 atm
300 K 2.26 133.0 K 2514.7 psia 1 atm 34.5 atm 14.7 psia
Pr1
U| |V 5.0| |W
Fig. 5.4-3
z = 1.02
mol K 2514.7 psia 150 L 1 atm 1022 mol 1.02 300 K 14.7 psia 0.08206 L atm
After 60h: Tr1 Pr1
b.
0.418 lb - mole / min
b g
Initially: Tr1
n leak
1.02
0.418 lb - mole 28 lb m / lb - mole 60 min 24 h 7 days 2 weeks min h day week 0.81 62.4 lb m / ft 3
For CO: Tc
n2
U| |V z . | 12 |W
. 0.418 lb - mole 10.73 ft 3 psia 609.7 o R 102 min lb - mole o R 600 psia
4668 ft 3
n1
335 . atm
609.7 o R 2.68 227.16o R 600 psia 1 atm Pr . atm 14.7 psia 335
150 SCFM 359 SCF / lb - mole
= zRTn V P
5.73 a.
227.16o R, Pc
126.20 K
300 K 2.26 133.0 K 2258.7 psia 1 atm 34.5 atm 14.7 psia
U| |V 4.5| |W
Fig. 5.4-3
z = 1.02
mol K 2259.7 psia 150 L 1 atm 1.02 300 K 14.7 psia 0.08206 L atm n n2 173 = 1 . mol / h 60 h PV RT
918 mol
200 u 106 mol CO 1 atm 30.7 m3 1000 L Latm mol air 0.08206 mol m3 K 300 K
n2
y 2 n air
y2
t min
n2 n leak
0.25 mol 1.73 mol / h
0.25 mol
014 . h
t min would be greater because the room is not perfectly sealed c. (i) CO may not be evenly dispersed in the room air; (ii) you could walk into a high concentration area; (iii) there may be residual CO left from another tank; (iv) the tank temperature could be higher than the room temperature, and the estimate of gas escaping could be low. 5-48
5.74 CH 4 : Tc
190.7 K , Pc
45.8 atm
C 2 H 6 : Tc
305.4 K , Pc
48.2 atm
C 2 H 4 : Tc
2831 . K , Pc
50.5 atm
b0.20gb190.7g b0.30gb305.4g b0.50gb2831. g 2713. K Pseudocritical pressure: P c b0.20gb458 . g b0.30gb48.2g b0.50gb50.5g 48.9 atm U| b90 273.2gK 134 Reduced temperature: T . |V z 0.71 2713 . K 200 bars 1 atm Reduced pressure: P 4.04 | |W 48.9 atm 1.01325 bars Mean molecular weight of mixture: b0.30gM b0.50gM M b0.20gM b0.20gb16.04g b0.30gb30.07g b0.50gb28.05g Pseudocritical temperature: Tcc c
r
Figure 5.4-3
r
CH 4
C2H 6
C2 H 4
26.25 kg kmol V
znRT P
0.71 10 kg
1 kmol 0.08314 m 3 bar 26.25 kg kmol K
UV W
5.75 N 2 : Tc 126.2 K, PC 33.5 atm Tcc N 2 O: Tc 309.5 K, PC 71.7 atm Pcc M n a.
b g b g
200 bars
b
0.041 m 3 (41 L)
g
0.10 309.5 0.90 126.2 144.5 K 0.10 71.7 0.90 33.5 37.3 atm
b g b g 29.62 5.0 kgb1 kmol 29.62 kgg 0.169 kmol T b24 273.2 g 144.5 2.06
b g
0.10 44.02 0.90 28.02
169 mol
r
V r
mol K
30 L
Pr V
r
T
30 L 37.3 atm mol K 169 mol 144.5 K 0.08206 L atm
U| Vz 0.56| W
0.97 169 mol 297.2 K 0.08206 L atm
P
b.
b90 + 273g K
U|V z 0.56 b from a.g |W
273 37.3 7.32
b
g
518 K 245q C
5-49
g
133 atm 132 atm gauge
1.14 Fig. 5.4 - 3
273 atm 30 L mol K 1.14 169 mol 0.08206 L atm
b
0.97 Fig. 5.4 - 3
b g b g
UV W
5.76 CO: Tc 133.0 K, Pc 34.5 atm Tcc H 2 : Tc 33 K, Pc 12.8 atm Pcc Turbine inlet:
Turbine exit: Tr Pr in Pin V out Pout V
Tr
b150 273.2g 96.2
Pr
2000 psi 1 atm 29.0 atm 14.7 psi
373.2 96.2 3.88 1 29.0 0.03
z in nRTin Vin z out n RTout
b
g g
0.60 133.0 0.40 33 8 0.60 34.5 0.40 12.8 8
b
96.2 K 29.0 atm
U| V o z | 1.01 4.69| W 4.4
Fig. 5.4-1
z=1.0
Vout u
Pout z in Tin Pin z out Tout
15,000
ft 3 14.7 psia 1.01 423.2K min 2000 psia 1.00 373.2
126 ft 3 / min If the ideal gas equation of state were used, the factor 1.01 would instead be 1.00 1% error
UV W
5.77 CO: Tc 133.0 K, Pc 34.5 atm Tcc CO 2 : Tc 304.2 K, Pc 72.9 atm Pcc Initial: Tr Pr Final: Pr
303.2 138.1 2.2 2014.7 524.8 3.8
b g b g
UV o z W
1889.7 524.8 3.6 z1
b
g
0.97 133.0 0.03 304.2 138.1 K 0.97 34.5 0.03 72.9 35.7 atm 524.8 psi
Fig. 5.4-3
1
b g
0.97
0.97
Total moles leaked: n1 n 2
FG P P IJ V b2000 1875gpsi 0.97 H z z K RT 1
2
1
2
30.0 L 1 atm mol K 303 K 14.7 psi 0.08206 L atm
10.6 mol leaked
b g
Moles CO leaked: 0.97 10.6 Total moles in room: Mole% CO in room =
10.3 mol CO
24.2 m3 103 L 273 K 1 mol 3 1m 303 K 22.4 L STP
b g
10.3 mol CO u 100% 10% . CO 973.4 mol
5-50
973.4 mol
CO 2H 2 o CH 3OH
5.78 Basis: 54.5 kmol CH 3OH h n 1 (kmol CO / h) 2n 1 (kmol H 2 / h) 644 K 34.5 MPa
Catalyst Bed
CO, H 2
Condenser
54.5 kmol CH 3OH (l ) / h
a.
54.5 kmol CH 3OH 1 kmol CO react
n 1
h
b g
2n 1
2 218
1 kmol CH 3OH
1 kmol CO fed 0.25 kmol CO react
b
g
436 kmol H 2 h 218 436
CO: Tc
133.0 K
Pc
34.5 atm
H 2 : Tc
33 K
Pc
12.8 atm
218 kmol h CO
654 kmol h (total feed)
Newton’s corrections Tcc Pcc Tr
b g b g 1 b3 34.5g 23 b12.8 8g 1 2 133.0 33 8 3 3
717 . K 25.4 atm
U| V| W
644 71.7 8.98 5.4-4 Fig. o z 1 34.5 MPa 10 atm 13.45 24.5 atm 1.013 MPa
Pr feed V
1.18
1.18 654 kmol 644 K 0.08206 m3 atm 1.013 MPa 120 m3 h h 34.5 MPa kmol K 10 atm Vcat
120 m3 h
1 m3 cat 25,000 m3 / h
0.0048 m3 catalyst (4.8 L)
b. CO, H 2 54.5 kmol CH 3OH (l ) / h
n 4 kmol CO / h 2n 4 kmol H 2 / h
Overall C balance n 4 Fresh feed:
feed V
54.5 mol CO h
54.5 kmol CO h 109.0 kmol H 2 h 163.5 kmol feed gas h
1.18 163.5 kmol 644 K 0.08206 m3 atm 1.013 MPa h 34.5 MPa kmol K 10 atm
5-51
29.9 m3 h
5.79
H 2 : Tc Pc
(33.3 8) K = 41.3 K
1 - butene: Tc
419.6 K
Pc
39.7 atm
(12.8 8) atm = 20.8 atm
. (413 . K) + 0.85(419.6 K) = 362.8 K Tc ' 015 . (20.8 atm) + 0.85(39.7 atm) = 36.9 atm Pc ' 015 = znRT V P
0.86 35 kmol 0.08206 m3 atm 323 K 1 h h kmol K 10 atm 60 min
CH 3 :
Tc
190.7 K Pc
458 . atm
C 2 H 4 : Tc
2831 . K Pc
50.5 atm
C 2 H 6 : Tc
305.4 K Pc
48.2 atm
UV W
Fig. 5.4-2
z = 0.86
133 . m3 / min
d
i FG 100 cmIJ S b150 m / ming H m K
F I FG IJ d i GH JK H K
4 133 . m 3 / min
3 2 m = u m A m 2 = u u Sd d = 4V V Su 4 min min
5.80
Tr ' 0.89 Pr ' 0.27
T=90 o C
. (190.7 K) + 0.60(283.1 K) + 0.25(305.4 K) = 274.8 K ====> Tc ' 015
10.6 cm
U| V| . W 35
. Tr ' 132
P=175 bar
. (458 . atm) + 0.60(50.5 atm) + 0.25(48.2 atm) = 49.2 atm =====> Pr ' Pc ' 015 5.4-3 Fig. o z = 0.67
F I FG IJ d i FG GH JK H K H
3 m = u m A m 2 = 10 m V s s s
n = 5.81
PV zRT
N2:
IJ FG 60 s IJ S b0.02 mg K H min K 4
2
. 0188
kmol K 0188 . m3 / min 175 bar 1 atm 0.67 1.013 bar .08206 m3 atm 363 K Tc
acetonitrile: Tc
126.2 K = 227.16o R Pc
335 . atm
548 K = 986.4 o R
47.7 atm
Pc
Tank 1 (acetonitrile): T1 = 550o F, P1 = 4500 psia Tr1 n 1 =
P1 V1 z 1 RT1
306 atm 0.200 ft 3 0.80 1009.7 o R
P2 V2 z 2 RT2
10.0 atm 2.00 ft 3 1.00 1009.7 o R
5-52
. kmol / min 163
. Pr1 102
Fig. 5.4-3
6.4 z1 = 0.80
lb - mole o R = 0.104 lb - mole 0.7302 ft 3 atm
Tank 2 (N 2 ): T2 = 550 o F, P2 = 10 atm Tr2 n 2 =
m3 min
4.4 , Pr2
Fig. 5.4-3
6.4 z 2 = 1.00
lb - mole o R = 0.027 lb - mole .7302 ft 3 atm
5.81 (cont’d) . I .027 I FG 0104 J 986.4 R + FGH 00131 J 227.16 R = 830 R H 0131 . K . K . I FG 0104 F 0.027 IJ 33.5 atm = 44.8 atm 47.7 atm + G J H 0131 K H 0131 . . K o
Final: Tc ' Pc '
dV i r
P=
=
ideal
o
o
T=550 Fo
o
Tr ' 122 .
' Fig. 5.4-2 VP 2.2 ft 3 44.8 atm lb - mole o R c z = 0.85 = = 1.24 RTc ' 0.131 lb - mole 830 o R 0.7302 ft 3 atm 0.85 0131 . lb - mole .7302 ft 3 atm 1009.7 o R lb - mole o R 2.2 ft 3
znRT V
37.3 atm
5.82 3.48 g Ca H b O c , 26.8o C, 499.9 kPa n c (mol C), n H (mol H), n O (mol O)
1 L @483.4 o C, 1950 kPa n p (mol) 0.387 mol CO 2 / mol 0.258 mol O 2 / mol 0.355 mol H 2 O / mol
n O 2 (mol O 2 ) 26.8o C, 499.9 kPa
a.
d
Volume of sample: 3.42 g 1 cm3 159 . g
i
2.15 cm3
O 2 in Charge:
d
1.000 L 2.15 cm 3 10 3 L km 3 L atm 0.08206 mol K
n O2
i
499.9 kPa
1 atm
300 K
101.3 kPa
0.200 mol O 2
Product
1.000 L 1950 kPa 1 atm L atm 0.08206 756.6 K 101.3 kPa mol K Balances: np
b
g
O: 2 0.200 n O
b
b
g
0.387 0.310
H: n H
2 0.355 0.310
gb
g
0.310 2 0.387 2 0.258 0.355 n O
C: n C
b
g b
0.310 mol product
0.110 mol O in sample
0.120 mol C in sample
g
0.220 mol H in sample
Assume c 1 a 0.120 0.110 1.1 b 0.220 0.110 2 Since a, b, and c must be integers, possible solutions are (a,b,c) = (11,20,10), (22,40,20), etc. b.
b g
b g
MW 12.01a 1.01b 16.0c 12.01 1.1c 1.01 2c 16.0c 300 MW 350 c 10 C11 H 20 O 10
5-53
31.23c
bg
C5 H 10
5.83 Basis: 10 mL C5 H 10 l charged to reactor
15 O 2 o 5CO 2 5H 2 O 2
bg
10 mL C5 H 10 l n1 (mol C5 H 10 )
n 2 (mol air) 0.21 O 2 0.79 N 2 27 o C, 11.2 L, Po (bar)
a.
n1
b
bg
10.0 mL C5 H 10 l
Stoichiometric air: n 2 Po
n 3 (mol CO 2 ) n 4 mol H 2 O(v) n 5 (mol N 2 ) 75.3 bar (gauge), Tad
nRT V
0.745 g 1 mol mL 70.13 g 0.1062 mol C5 H 10
d Ci o
0.1062 mol C5 H 10 7.5 mol O 2
1 mol air
1 mol C 2 H 10
0.21 mol O 2
3.79 mol 0.08314 L bar 300K 11.2 L
g
mol K
3.79 mol air
8.44 bars
(We neglect the C5 H 10 that may be present in the gas phase due to evaporation) Initial gauge pressure 8.44 bar 1 bar b.
7.44 bar
U| || V| || W
0.1062 mol C 5 H 10
5 mol CO 2 0.531 mol CO 2 1 mol C 5 H 10 0.531 mol CO 2 1 mol H 2 O 4.052 mol product gas 0.531 mol H 2 O 1 mol CO 2 0.79 3.79 2.99 mol N 2
n3 n4
b g
n5
CO 2 : y 3 = 0.531 / 4.052 = 0.131 mol CO 2 / mol, Tc = 304.2 K Pc = 72.9 atm H 2 O: y 4 = 0.531 / 4.052 = 0.131 mol H 2 O / mol, N2:
y 5 = 2.99 / 4.052 = 0.738 mol N 2 / mol,
Tc = 647.4 K Pc = 218.3 atm Tc = 126.2 K Pc = 33.5 atm
. (304.2 K) + 0.131(647.4 K) + 0.738(126.2 K) = 217.8 K Tc ' 0131 . (72.9 atm) + 0.131(218.3 atm) + 0.738(33.5 atm) = 62.9 atm Pr ' 121 . Pc ' 0131 ideal V r T=
PV znR
' VP c RTc '
11.2 L 62.9 atm mol K 4.052 mol 217.8 K .08206 L atm
b75.3 1gbars 1.04
112 . L mol K 4.052 mol 0.08314 L bar
5-54
9.7 z | 1.04 (Fig. 5.4 - 3) 2439 K - 273 = 2166o C
CHAPTER SIX 6.1
a.
AB: Heat liquid - -V | constant BC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T 100 o C. CD: Heat vapor - -T increases, V increases .
b. Point B: From Perry’s Handbook, Table 3.8 U H 2 O(l), 80q C 0.9718 g ml , U H 2 O(l), 100q C 10 mL 0.9718 g / mL
bg
H 2 O l , 100q C: V B
0.9584 g / mL
0.9584 g mL
10.14 mL
Point C: H2O (v, 100qC) n
10 mL 0.9718 g
0.5393 mol 18.02 g nRTC 0.5393 mol 0.08206 L atm 373 K 16.5 L 1 atm mol K PC
mL
nRTC VC
PCVC
6.2
1 mol
a. Pfinal
243 mm Hg . Since liquid is still present, the pressure and temperature must lie on the
vapor-liquid equilibrium curve, where by definition the pressure is the vapor pressure of the species at the system temperature. b. Assuming ideal gas behavior for the vapor, m(vapor) m(liquid)
6.3
a.
mol K
(3.000 - 0.010) L (30 + 273.2) K 10 mL
0.08206 L atm
1.489 g mL
m(vapor) + m(liquid) = 19.5 g
x vapor =
4.59 19.48
b. ln p
B
119.39 g
760 mm Hg
mol
0.235 g vapor / g total
7.09808
1238.71 45 217
'H v 1 'H v B R T R
ln( p1* )
1 atm
14.89 g
m total
log 10 p
243 mm Hg
'H v / R T1
b
2.370 p *
d
ln p2* / p1* 1 T2
1 T1
10 2.370
i
b
6-1
g
ln 760 / 118.3
b
1 77 .0 273.2 K
K g b29.54151 273.2gK
ln 118.3
234.5 mm Hg
g
18.49
b
1 29 .5 273.2 K
g
4151K
4.59 g
6.3 (cont’d) ln p (45o C)
b
4151 18.49 p 45 273.2
g
. 234.5 2310 u 100% 234.5 c.
327.7 234.5 u 100% 234.5
b
327.7 mm Hg
39.7% error
g
1 (rect. scale) on semilog paper T 273.2 straight line: slope 7076K , intercept 2167 .
Plot p log scale vs
b
g
ln p mm Hg ' Hv R
b
7076K ' H v
7076 K
8.314 J
g
exp
1 kJ
mol K 10 3 J
LM 7076 2167 O . P N T ( C) 2732. Q o
58.8 kJ mol
ln p* = A/T(K) + B p*(mm Hg) 5 20 40 100 400 760
1/T(K) 0.002834 0.002639 0.002543 0.002410 0.002214 0.002125
ln(p*) p*(fitted) 1.609 5.03 2.996 20.01 3.689 39.26 4.605 101.05 5.991 403.81 6.633 755.13
7 6 5 4 3 2 1 0
1/T
6-2
0.003
0.0028
0.0026
0.0024
y = -7075.9x + 21.666
0.002
o
T( C) 79.7 105.8 120.0 141.8 178.5 197.3
ln(p*)
6.5
7076 2167 . p mm Hg T ( C) 273.2 o
0.0022
6.4
15% . error
FG 118.3 760IJ b45 29.5g 118.3 H 29.5 77 K
p
. mm Hg 2310
o
T( C) p*(fitted) 50 0.80 80 5.12 110 24.55 198 760.00 230 2000.00 Least confidence (Extrapolated)
6.6
a. T(°C)
1/T(K)
42.7 58.9 68.3 77.9 88.6 98.3 105.8
3.17u10-3 3.01u10-3 2.93u10-3 2.85u10-3 2.76u10-3 2.69u10-3 2.64u10-3
p*(mm Hg) =758.9 + hright -hleft 34.9 78.9 122.9 184.9 282.9 404.9 524.9
b. Plot is linear, ln p
'H v B ln p RT
5143.8 K 19.855 T
At the normal boiling point, p 760 mmHg Tb ' H v
8.314 J 5143.8 K
1 kJ
mol K
10 3 J
116q C
42.8 kJ mol
c. Yes — linearity of the ln p vs 1 / T plot over the full range implies validity.
6.7
a.
b
g
ln p a T 273.2 b y
ax b
y
b
ln p ; x 1 T 273.2
g
Perry' s Handbook, Table 3 - 8: 39.5q C , p1 400 mm Hg x1
T1
56.5q C , p2 760 mm Hg x 2
T2
50q C x 3.0941 u 10 x x1 y1 y 2 y1 x 2 x1
T
FG H
y
IJ b K
u 10 3 , y1 31980 . 3
3.0331 u 10 , y 2
5.99146 6.63332
3
g
b
6.39588 p 50q C
g
e 6.39588
599 mm Hg
b. 50q C 122q F 12 psi 760 mm Hg
Cox Chart p
c.
6.8
log p 7.02447
b
. 11610 50 224
2.7872 p 10 2.7872
g
Estimate p 35q C : Assume ln p a
b
g
ln p2 p1 1 T2
1 T1
b ln p1
a T1
625 mm Hg
14.6 psi
b
g
ln 200 50 1 45 273.2
b g
ln 50
1 25 273.2
6577.1 25 273.2
613 mm Hg
a b , interpolate given data. T K
b g
U| ln p b35q Cg 6577.1 25.97 4.630 35 273.2 |V 25.97 | p b35q Cg e 102.5 mm Hg |W
6577.1
4 .630
6-3
6.8 (cont’d) Moles in gas phase: n
150 mL
a.
m 2 S
2 F
g
4
mol
222
2.
8.0 u 10 6.9
b
273 K 102.5 mm Hg 1L 1 mol 3 35 + 273.2 K 760 mm Hg 10 mL 22.4 L STP
b g
Two intensive variable values (e.g., T & P) must be
specified to determine the state of the system. 1209.6 2.5107 p MEK 10 2.5107 324 mm Hg b. log p MEK 6.97421 55 216. Since vapor & liquid are in equilibrium p MEK p MEK 324 mm Hg y MEK
p MEK / P
324 1200 0.27 ! 0115 . The vessel does not constitute an explosion
hazard. 6.10 a. The solvent with the lower flash point is easier to ignite and more dangerous. The solvent with a flash point of 15qC should always be prevented from contacting air at room temperature. The other one should be kept from any heating sources when contacted with air. b. At the LFL, y M 0.06 p M p *M 0.06 u 760 mm Hg = 45.60 mm Hg 1473.11 T 6.85q C Antoine log 10 45.60 = 7.87863 T + 230 c. The flame may heat up the methanol-air mixture, raising its temperature above the flash point. 6.11 a. At the dew point, p ( H 2 O) = p( H 2 O) = 500 u 0.1 = 50 mm Hg T = 38.1q C from Table B.3. b. VH2O
30.0 L
273 K 500 mm Hg 1 mol 0.100 mol H2 O 18.02 g 1 cm3 (50 + 273) K 760 mm Hg 22.4 L (STP) mol mol g
c. (iv) (the gauge pressure)
6-4
134 . cm3
6.12 a.
60 mm Hg b g 755 mm Hg b577 222gmm Hg 400 mm Hg
T1
58.3q C , p1 755 mm Hg 747 52 mm Hg
T2
110q C , p2
a b T K
ln p a b
b g
b
g
b
ln p2 p1 1 T2
ln p1
a T1
g
ln 400 60
1 T1
1 110 273.2
b g
ln 60
46614 .
58.31273.2
4661.4 58.3 273.2
18.156
T=130oC=403.2 K 46614 . 18156 . T ln p 130q C 6.595 p 130q C e 6.595 7314 . mm Hg
ln p
b
b.
g
b
g
Basis: 100 mol feed gas CB denotes chlorobenzene. n1 mol @ 58.3qC, 1atm y1 (mol CB(v)/mol) (sat’d) (1-y1) (mol air/mol)
100 mol @ 130qC, 1atm y0 (mol CB(v)/mol) (sat’d) (1-y0) (mol air/mol)
n2 mol CB (l)
b
g
731 mm Hg 760 mm Hg
pCB 130q C y o
Saturation condition at inlet: y o P
0.962 mol CB mol
b
60 mm Hg 0.0789 mol CB mol g 760 mm Hg Air balance: 100b1 y g n b1 y g n b100gb1 0.962g b1 0.0789g 4.126 mol Total mole balance: 100 n n n 100 4.126 9587 . mol CBbl g pCB 58.3q C y1
Saturation condition at outlet: y1 P o
1
1
1
% condensation:
1
2
2
95.87 mol CB condensed u 100% 0.962 u 100 mol CB feed
b
g
99.7%
c. Assumptions: (1) Raoult’s law holds at initial and final conditions; (2) CB is the only condensable species (no water condenses); (3) Clausius-Clapeyron estimate is accurate at 130qC. 6.13 T
78q F = 25.56q C , Pbar
y H 2O P
b
0.87 p 25.56q C
d i
Dew Point: p Tdp
yp
29.9 in Hg = 759.5 mm Hg , hr
g
Table B.3
y H 2O
b
g
0.0293 759.5
87%
b
0.87 25544 . mm Hg 759.5 mm Hg
22.22 mm Hg
6-5
Table B.3
g
0.0293 mol H 2 O mol air Tdp
23.9q C
6.13 (cont’d) hm ha
hp
0.0293 1 0.0293
0.0302 mol H 2 O mol dry air
0.0293 mol H 2 O 18.02 g H 2 O mol dry air
b
p 2556 . qC
mol H 2 O
hm
g
mol dry air
b
29.0 g dry air
P p 2556 . qC
g
0.0182 g H 2 O g dry air
0.0293 u 100 84% 24.544 759.5 24.544
u 100%
6.14 Basis I : 1 mol humid air @ 70q F (21.1q C), 1 atm, hr hr
50% y H 2 O P Table B.3
Mass of air:
y H 2O
b
0.50 p H 2 O 21.1q C
1 mol
Volume of air:
mol H 2 O mol
0.012
0.988 mol dry air 29.0 g
1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP 1 mol
28.87 g 24.13 L
Density of air
g
0.50 u 18.765 mm Hg 760.0 mm Hg
0.012 mol H 2 O 18.02 g
50%
273.2K
80% y H 2 O P Table B.3
Mass of air:
y H 2O
b
0.80 p H 2 O 21.1q C
1 mol
Density of air
80%
g
0.80 u 18.765 mm Hg 760.0 mm Hg
0.020 mol H 2 O 18.02 g
Volume of air:
0.020
mol H 2 O mol
0.980 mol dry air 29.0 g 1 mol
b g b273.2 21.1gK
1 mol
22.4 L STP 1 mol
28.78 g 24.13 L
273.2K
80% y H 2 O P Table B.3
y H 2O
24.13 L
1193 . g L
Basis III: 1 mol humid air @ 90q F (32.2q C), 1 atm, hr hr
24.13 L
1196 . g L
Basis II: 1 mol humid air @ 70q F (21.1q C), 1 atm, hr hr
28.87 g
b
0.80 p H 2 O 32.2q C
80%
g
0.80 u 36.068 mm Hg 760.0 mm Hg
6-6
0.038
mol H 2 O mol
28.78 g
6.14 (cont’d) Mass of air:
0.038 mol H 2 O 18.02 g
Volume of air: Density
1 mol
0.962 mol dry air 29.0 g 1 mol
b g b273.2 32.2gK
1 mol
28.58 g 25.04 L
22.4 L STP 1 mol
273.2K
28.58 g
25.04 L
1141 . g L
Increase in T increase in V decrease in density Increase in hr more water (MW = 18), less dry air (MW = 29) decrease in m decrease in density Since the density in hot, humid air is lower than in cooler, dryer air, the buoyancy force on the ball must also be lower. Therefore, the statement is wrong. 6.15 a. hr
50% y H 2 O P Table B.3
b
0.50 p H 2 O 90q C
0.50 u 525.76 mm Hg 760.0 mm Hg
y H 2O
Dew Point: y H 2 O p
d i
p Tdp
Degrees of Superheat
b. Basis:
g
b g
0.346 760
0.346 mol H 2 O / mol
262.9 mm Hg
1 m 3 feed gas 10 3 L 273K mol 3 m 363K 22.4 L STP
n1 mol @ 25qC, 1atm y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol air/mol)
0.346 H2O mol /mol 0.654 mol air/mol
n2 mol H2O (l)
b g 23.756 P 760 Dry air balance: 0.346b33.6g n b1 0.0313g n
Saturation Condition: y1
p H* 2 O 25q C 1
g
p * ( 90q C) y H 2O
0.0313 mol H 2 O mol
1
Total mol balance: 33.6 = 12.0 + n2 n2
b
72.7q C
33.6 mol
b g
p 90q C P
Tdp
90 72.7 17.3q C of superheat
33.6 mol @ 90qC, 1atm
c. y H 2 O P
Table B.3
12.0 mol
216 . mol H 2 O condense / m 3
525.76 mmHg 0.346
6-7
1520 mm Hg = 2.00 atm
6.16 T
90q F = 32.2q C , p
29.7 in Hg = 754.4 mm Hg , hr
95%
Basis: 10 gal water condensed/min n condensed
1 ft 3 62.43 lb m 7.4805 gal ft 3
10 gal H 2 O
y2 (lb-mol H2O (v)/lb-mol) (sat’d) (1-y2) (lb-mol DA/lb-mol) 40oF (4.4oC), 754 mm Hg
y1 (lb-mol H2O (v)/lb-mol) (sat’d) (1-y1) (lb-mol DA/lb-mol) hr=95%, 90oF (32.2oC), 29.7 in Hg (754 mm Hg)
Raoult' s law: y 2 P
4.631 lb-moles H2O (l)/min
b
0.95 p 32.2q C
95% hr at inlet: y H 2 O P
y H 2O
4.631 lb - mole / min
n2 (lb - moles / min)
V1 (ft 3 / min) n1 (lb - moles / min)
Table B.3
1 lb - mol 18.02 lb m
b
g
0.95 36.068 mm Hg
g
0.045 lb - mol H 2 O lb - mol
754.4 mm Hg
b
g
Table B.3
p * 4.4q C
y2
6.292 754.4
0.00834 lb - mol H 2 O lb - mol
UV RS W T
n1 125.3 lb - moles / min Mole balance: n1 n 2 4.631 n 2 120.6 lb - moles / min Water balance: 0.045n1 0.00834n 2 4.631 125.3 lb - moles 359 ft 3 (STP) (460 + 90) o R 760 mm Hg Volume in: V = min lb - moles 492 o R 754 mm Hg 5.07 u 10 4 ft 3 / min 6.17 a. Assume no water condenses and that the vapor at 15qC can be treated as an ideal gas. 760 mm Hg
p final
(15 273) K (200 + 273) K
462.7 mm Hg ( p H 2 O ) final
0.20 u 462.7
p * (15q C) = 12.79 mm Hg < p H 2 O . Impossible condensation occurs. ( pair ) final
P
( pair ) initial
p H 2O pair
b. Basis:
Tfinal Tinitial
(0.80 u 760) mm Hg u
370.2 12.79
1 L 273 K
383 mm Hg
mol
473 K 22.4 L (STP)
0.0258 mol
6-8
288 K 473 K
370.2 mm Hg
92.6 mm Hg
6.17 (cont’d) n1 mol @ 15qC, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol dry air/mol)
0.0258 mols @ 200qC, 760 mm Hg 0.20 H2O mol /mol 0.80 mol air/mol
n2 mol H2O (l)
b
p H* 2 O 15q C
Saturation Condition: y1
12.79 mm Hg . mm Hg 3831
P
b
c.
g
g n b1 0.03339g n
Dry air balance: 0.800 0.0258
1
Mass of water condensed =
0.02135 mol
1
Total mole balance: 0.0258 = 0.02135 + n2 n2 0.00445 mol
0.03339 mol H 2 O mol
0.00445 mol
18.02 g mol
0.0802 g
6.18 Basis: 1 mol feed n2 (mol), 15.6°C, 3 atm y 2 (mol H 2 O (v)/mol)(sat'd) (1 – y 2) (mol DA/mol)
3
V1 (m ) 1 mol, 90°C, 1 atm 0.10 mol H 2O (v)/mol 0.90 mol dry air/mol
heat
100°C, 3 atm n2 (mol) 3 V2 (m )
n3 (mol) H 2 O( l ), 15.6°C, 3 atm
b
p H* 2 O 15.6q C
Saturation: y 2
g
Table B.3
P
atm 13.29 mm Hg 0.00583 3 atm 760 mm Hg
y2
b g n b1 0.00583g n H O mol balance: 0.10b1g 0.00583b0.9053g n Dry air balance: 0.90 1
2
2
Fraction H 2 O condensed: hr
y 2 P u 100% p 100q C
b
g
0.9053 mol
2
3
n3
0.0947 mol condensed 0100 . mol fed
b
g u 100%
0.00583 3 atm 1 atm
0.0947 mol
0.947 mol condense mol fed
. 175% n1 mol @ 15qC, 383.1 mm Hg y1 (mol H2O (v)/mol) (sat’d) (1-y1) (mol dry air/mol)
0.0258 mols @ 200qC, 760 mm Hg 0.20 H2O mol /mol 0.80 mol air/mol
n2 mol H2O (l)
6-9
6.18 (cont’d)
b g
V2
0.9053 mol 22.4 L STP mol
V1
1 mol 22.4 L STP mol
V2 V1
b g
363K 1 m 3 2.98 u 10 2 m 3 feed air @ 90q C 3 273K 10 L
9.24 u 10 3 m 3 outlet air 2.98 u 10 2 m 3 feed air
6.19 Liquid H 2 O initially present:
0.0208 1 0.0208
0.310 m 3 outlet air m 3 feed air
25 L 1.00 kg
b
p H* 2 O 25q C
Saturation at outlet: y H 2 O
373K 1 atm 1 m 3 9.24 u 10 3 m 3 outlet air @ 100q C 273K 3 atm 10 3 L
g
1 kmol 18.02 kg
L
bg
1.387 kmol H 2 O l
23.76 mm Hg 15 . u 760 mm Hg
P
0.0208 mol H 2 O mol air
0.0212 mol H 2 O mol dry air
b g
15 L STP 1 mol 0.670 mol dry air min min 22.4 L STP 0.670 mol dry air 0.0212 mol H 2 O 0.0142 mol H 2 O min Evaporation Rate: min mol dry air Flow rate of dry air:
Complete Evaporation:
b g
min 1.387 kmol 10 3 mol 1h 1628 h 0.0142 mol 60 min kmol
b67.8 daysg
7.069 u 10 3 ft 3 7.481 gal S 2 5.288 u 104 gal / day 6.20 a. Daily rate of octane use = 30 (18 8) 3 4 day ft ( SG ) C8 H18
b. 'p
5.288 u 10 4 gal 1 ft 3 0.703 u 62.43 lb m day 7.481 gal ft 3 0.703 3.10 u 105 lb m C 8 H 18 / day
0.703 u 62.43 lb m ft
32.174 ft
3
c. Table B.4: pC* 8 H18 (90 o F)
s
2
1 lb f lb ft 32.174 m2 s
20.74 mm Hg
(18 - 8) ft
14.696 psi
29.921 in Hg 14.696 lb f / in 2
0.40 lb f / in 2
6.21 in Hg
poctane y octane P 760 mm Hg Octane lost to environment = octane vapor contained in the vapor space displaced by liquid during refilling.
Volume:
5.288 u 10 4 gal
1 ft 3 7.481 gal
7069 ft 3
6-10
6.20 (cont’d) (16.0 + 14.7) psi 7069 ft 3 36.77 lb - moles 10.73 ft 3 psi / (lb - mole o R) (90 + 460) o R pC H 0.40 psi Mole fraction of C 8 H 18 : y = 8 18 0.0130 lb - mole C 8 H 18 / lb - mole P (16.0 + 14.7) psi pV RT
Total moles: n
Octane lost
0.0130(36.77) lb - mole 0.479 lb - mole ( 55 lb m
25 kg)
d. A mixture of octane and air could ignite. * * (85o F) = ptol (29.44 o C) = 35.63 mmHg = ptol 6.21 a. Antoine equation ptol
ptol P
Mole fraction of toluene in gas: y yPV RT 0.0469 lb - mole tol
Toluene displaced
35.63 mmHg 760 mmHg
0.0469 lb - mole toluene / lb - mole
yntotal
1 atm
lb - mole
0.7302
1 ft 3
900 gal
3
ft atm o
lb - mole R
(85 460) o R 7.481 gal
92.13 lb m tol lb - mole
1.31 lb m toluene displaced
b. Basis: 1mol 0.0469 mol C7H8(v)/mol 0.9531 mol G/mol
nV (mol) y (mol C7H8(v)/mol) (1-y) (mol G/mol) T(oF), 5 atm
Assume G is noncondensable
nL [mol C7H8 (l)] 90% of C7H8 in feed
90% condensation n L
0.90(0.0469)(1) mol C 7 H 8
Mole balance: 1 nV 0.0422 nV Toluene balance: 0.0469(1) Raoult’s law:
ptol
0.0422 mol C 7 H 8 (l )
0.9578 mol
y (0.9578) 0.0422 y
0.004907 mol C 7 H 8 / mol
* (0.004907)(5 u 760) 18.65 mmHg = ptol (T )
yP
Antoine equation: T
B C( A log 10 p * ) A log 10 p
*
1343.943 219.377(6.95334 log 10 18.65) 6.95334 log 10 18.65
6-11
17.12 o C = 62.8 o F
6.22 a. Molar flow rate: n =
VP RT
100 m 3 h
kmol K 2 atm 3 82.06 u 10 m atm (100 + 273) K -3
6.53 kmol / h
b. Antoine Equation: 1171.530 100 + 224.366 1845 mm Hg
log 10 p *Hex (100q C) = 6.87776 p* p Hex
y Hex P
p *Hex (T )
0.150(2.00) atm 760 mm Hg
* not saturated 228 mm Hg < p Hex
atm
228 mm Hg log 10 228 = 6.87776 -
3.26601
1171.530 T + 224.366
2.35793 T
34.8q C
nV (kmol/h) y (kmol C6H14 (v)/kmol), sat’d (1-y) (kmol N2/kmol) T (oC), 2 atm
c. 6.53 kmol/h 0.15 C6H14 (v) 0.85 N2
nL (kmol C6H14 (l)/h) 80% of C6H14 in feed
n L 0.80(0.15)(6.53 kmol / h) = 0.7836 kmol C 6 H 14 (l ) / h 80% condensation: Mole balance: 6.53 nV 0.7836 nV 5.746 kmol / h Hexane balance: 015 . (6.53) y (5.746) 0.7836 y 0.03409 kmol C 6 H 14 / kmol
(0.03409)(2 u 760 mmHg) = 51.82 mmHg = p *Hex (T ) 1171530 . T 2.52 o C Antoine equation: log 10 5182 . 6.87776 T 224.366 Raoult’s law:
p Hex
yP
6.23 Let H=n-hexane a. n0 ( kmol / min) y0 (kmol H(v)/kmol (1-y0) (kmol N2/kmol) 80oC, 1 atm, 50% rel. sat’n
50% relative saturation at inlet: y o P Table B.4
yo
Condenser
n1 ( kmol / min) 0.05 kmol H(v)/kmol, sat’d 0.95 kmol N2/kmol T (oC), 1 atm 1.50 kmol H(l)/min
0.500 p H* (80 o C)
(0.500)(1068 mmHg) = 0.703 kmolH / kmol 760 mmHg
Saturation at outlet: 0.05 P
p H* (T1 ) p H* (T1 )
6-12
0.05(760 mmHg) = 38 mmHg
6.23 (cont’d) Antoine equation: log 10 38 6.87776
1171530 . T1 T1 224.366
UV RS W T
Mole balance: n 0 n1 150 . n 0 N 2 balance: (1 0.703)n 0 0.95n1 n1 N2 volume: VN 2 b.
3.24 o C
2.18 kmol / min 0.682 kmol / min
(0.95)0.682 kmol 22.4 m 3 (STP) 14.5 SCMM min kmol
Assume no condensation occurs during the compression
2.18 kmol/min 0.703 H(v) 0.297 N2 80oC, 1 atm
Compressor
V0 ( m3 / min) 2.18 kmol/min 0.703 H(v) 0.297 N2 T0 (oC), 10 atm, 50% R.S.
V1 (m 3 / min) 0.682 kmol/min 0.05 H(v), sat’d 0.95 N2 T1 (oC), 10 atm Condenser
1.5 kmol H(l)/min
50% relative saturation at condenser inlet: 0.500 p *H (T0 )
u 10 4 mmHg 0.703(7600 mmHg) p H* (T0 ) 1068 .
Saturation at outlet: 0.050(7600 mmHg) = 380 mmHg = p *H (T1 ) Volume ratio:
V1 V0
n1 RT1 / P n0 RT0 / P
n1 (T1 273.2) n0 (T0 273.2)
Antoine
Antoine
0.682 kmol / min 1321 K u 2.18 kmol / min 460 K
T0
187 o C
T1
48.2q C
m 3 out 0.22 3 m in
c. The cost of cooling to 3.24 o C (installed cost of condenser + utilities and other operating costs) vs. the cost of compressing to 10 atm and cooling at 10 atm. 6.24 a.
Maximum mole fraction of nonane achieved if all the liquid evaporates and none escapes. (SG)nonane n max
Assume T n gas
15 L C 9 H 20 (l ) 0.718 u 1.00 kg L C 9 H 20
kmol 128.25 kg
25o C, P = 1 atm 1 kmol 2 u 10 4 L 273 K 3 298 K 22.4 u 10 L(STP)
6-13
0.818 kmol
0.084 kmol C 9 H 20
6.24 (cont’d) y max
n max n gas
0.084 kmol C 9 H 20 0.818 kmol
010 . kmol C 9 H 20 / kmol (10 mole%)
As the nonane evaporates, the mole fraction will pass through the explosive range (0.8% to 2.9%). The answer is therefore yes . The nonane will not spread uniformly—it will be high near the sump as long as liquid is present (and low far from the sump). There will always be a region where the mixture is explosive at some time during the evaporation. b. ln p *
A B T
T1
258 . o C = 299 K, p1*
5.00 mmHg
T2
66.0 o C = 339 K, p2*
40.0 mmHg
5269 5269 ln( 40.0 / 5.00) A 5269, B = ln(5.00) + 19.23 p * exp(19.23 ) 1 1 T ( K) 299 339 299 At lower explosion limit, y 0.008 kmol C 9 H 20 / kmol p * (T ) yP (0.008)(760 mm Hg) A
= 6.08 mm Hg
Formula for p
T
*
302 K = 29 o C
c. The purpose of purge is to evaporate and carry out the liquid nonane. Using steam rather than air is to make sure an explosive mixture of nonane and oxygen is never present in the tank. Before anyone goes into the tank, a sample of the contents should be drawn and analyzed for nonane. 6.25 Basis: 24 hours of breathing n0 (mol H 2 O) 23°C, 1 atm n1 (mol) @ hr = 10% 0.79 mol N 2/mol y 1 (mol H 2 O/mol) + O2 , CO2
Air inhaled: n1
Lungs
O2
CO2
37°C, 1 atm n2 (mol), saturated 0.75 mol N 2/mol y 2 (mol H 2 O/mol) + O2 , CO2
12 breaths 500 ml 1 liter min breath 10 3 ml
b
273K 1 mol 23 273 K 22.4 liter STP
g
b g
60 min 24 hr 1 hr 1 day
356 mol inhaled day
Inhaled air - -10% r. h.: y1 Inhaled air - -50% r. h.: y1
b
g
b
g
010 . p H 2O 23q C
010 . 2107 . mm Hg
P
760 mm Hg
b
g
b
0.50 p H 2O 23q C
0.50 2107 . mm Hg
P
760 mm Hg
6-14
g
2.77 u 10 3
mol H 2 O mol
139 . u 10 2
mol H 2 O mol
6.25 (cont’d) n2 y 2 n1 y1 (n0 ) 10% rh (n0 ) 50% rh
H 2 O balance: n0
(n1 y1 ) 50% (n1 y1 ) 10%
FG 356 mol IJ L(0.0139 0.00277) mol H O OFG 18.0 g IJ mol PQH 1 mol K H day K MN 2
71 g / day
Although the problem does not call for it, we could also calculate that n2 = 375 mol exhaled/day, y2 = 0.0619, and the rate of weight loss by breathing at 23oC and 50% relative humidity is n0 (18) = (n2y2 - n1y1)18 = 329 g/day. 6.26 a. To increase profits and reduce pollution. b. Assume condensation occurs. A=acetone n 1 mol @ To C, 1 at m y1 mol A(v)/ mol (sat’d) (1-y1 ) mol N2 /mo l
1 mo l @ 90o C, 1 atm 0.20 mol A(v)/ mol 0.80 mol N2 /mo l
n 2 mol A(l)
For cooling water at 20oC
d
log 10 p *A 20 o C
i
Saturation: y1 P
7.02447
d
d
11610 . 20 224
i
p *A 20 o C y1
2.26627 p *A 20 o C 184.6 760
i
184.6 mmHg
0.243 ! 0.2 , so no saturation occurs.
For refrigerant at –35oC
d
log 10 p *A 35o C
i
7.02447
11610 . 35 224
d
0.88161 p *A 35o C
i
7.61 mmHg
7.61 0.0100 d i 760 N mole balance: 1b0.8g n b1 0.01g n 0.808 mol Saturation: y1 P 2
p *A 35o C y1 1
1
Total mole balance: 1 0.808 n2 n2 Percentage acetone recovery: c. d.
0.192 mol
0192 . u 100% 1
96%
Costs of acetone, nitrogen, cooling tower, cooling water and refrigerant The condenser temperature could never be as low as the initial cooling fluid temperature because heat is transferred between the condenser and the surrounding environment. It will lower the percentage acetone recovery.
6-15
6.27
Basis:
12500 L 1 mol 273 K 103000 Pa h 22.4 L(STP) 293 K 101325 Pa
n o (mol/h) @ 35o C, 103 KPa y0 [mol H2O(v)/mol] y1– H2 O(v)/ mol o mol y0 (mol DA/mol) ) mol DA/mo l (1-y o hr=90% h r =90%
Inlet: y o
d
hr p H* 2 O 35o C P
d
p H* 2 O 20 o C
i
i
528.5 mol / h
528.5 (mo l/h) @ 20o C, 103 KPa y0 [mol H2O(v)/mol] y1– H2 O(v)/ mol (sat’d) 1 mol y0 (mol DA/mol) (1-y1 ) mol DA/mo l
[molH2O(l)/h H2O(l)/h] n 2n2mol
0.90 u 42.175 mmHg 101325 Pa 103000 Pa 760 mmHg
0.4913 mol H 2 O / mol
17.535 mmHg 101325 Pa 0.02270 mol H 2 O / mol P 103000 Pa 760 mmHg Dry air balance: 1 0.04913 no 1 0.02270 528.5 no 543.2 mol / h Outlet: y1
b
g b
gb
g
543.2 mol 22.4 L(STP) 308 K 101325 Pa 13500 L / h h mol 273 K 103000 Pa Total balance: 543.2 528.5 n2 n2 14.7 mol / h
Inlet air:
14.7 mol 18.02 g H 2 O 1 kg h 1 mol H 2 O 1000 g
Condensation rate:
6.28
Basis:
0.265 kg / h
10000 ft 3 1 lb - mol 492 o R 29.8 in Hg min 359 ft 3 (STP) 550 o R 29.92 in Hg
n1 lb-mole/min 40oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/mol)
24.82 lb-mole/min 90oF, 29.8 in.Hg y0 [lb-mole H2O(v)/mol 1- y0 (lb-mole DA/mol) hr = 88%
Inlet: y o Outlet: y1
d
i
b
0.88 36.07 mmHg
P
29.8 in Hg
i
p H* 2 O 40 o F P
b
g
1 in Hg 25.4 mmHg
6.274 mmHg 1 in Hg 29.8 in Hg 25.4 mmHg
Dry air balance: 24.82 1 0.0419 Total balance: 24.82
n1 lb-mole/min 65oF, 29.8 in.Hg y1 [lb-mole H2O(v)/lb-mole] 1- y1 (lb-mole DA/lb-mole)
n2 [lb-mole H2O(l)/min]
hr p H* 2O 90 o F
d
24.82 lb - mol / min
0.00829 lb - mol H 2 O / lb - mol
g n b1 0.00829g n 1
23.98 n2 n2
1
23.98 lb - mol / min
0.84 lb - mole / min
6-16
0.0419 lb - mol H 2 O / lb - mol
6.28 (cont’d) 0.84 lb - mol 18.02 lb m 1 ft 3 7.48 gal min lb mol 62.4 lb m 1 ft 3
Condensation rate:
Air delivered @ 65oF:
181 . gal / min
23.98 lb - mol 359 ft 3 (STP) 525o R 29.92 in Hg min 1 lb mol 492 o R 29.8 in Hg
9223 ft 3 / min
6.29 Basis: 100 mol product gas
no mol, 32oC, 1 atm yo mol H2O(v)/mol (1-yo) mol DA/mol hr=70%
100 mol, T1, 1 atm
100 mol, 25oC,1 atm
y1 mol H2O(v)/mol, (sat’d) (1-y1) mol DA/mol
y1 mol H2O(v)/mol, (1-y1) mol DA/mol hr=55%
(mol HH22O(l)) nn22lb-mol O(l)/min
Outlet: y1
d
hr p H* 2 O 25o C
i
b
g
0.55 23.756
P
b g 13.07 p bT g T 15.3 C d32 Ci 0.70b35.663g 0.0328 mol H O / mol
hr p H* 2O
b
g
Total balance: 1016 . n2
Ratio:
b
2
b
g
100 1 0.0172 no
100.0 n2
16 . mol 18.02 g 1 kg 1 mol 1000 g
g
1016 . mol
. mol (i.e. removed) 16
0.0288 kg H 2 O
100 1 0.0172 mol 29.0 g 1 kg 1 mol 1000 g
0.0288 2.85
1
760
Dry air balance: no 1 0.0328
kg dry air:
1
o
P
kg H 2 O removed :
o
* H 2O
Saturation at T1 : 0.0172 760 Inlet: y o
0.0172 mol H 2 O / mol
760
2.85 kg dry air
0.0101 kg H 2 O removed / kg dry air
6-17
6.30 a.
Room air T
22q C , P 1 atm , hr
b
Second sample T
g
9251 mm Hg 839 mm Hg
ae bH , y1
ln y
bH ln a y
b
0.01044g . g lnb01103
H 2 H1
y Basis:
b
0.01044, H1
g b gb g expb0.054827 H g
ln 0.01044 0.054827 5
7.937 u 10 3
1 m 3 delivered air
b
n o mol, 35o C, 1 at m yo mol H2 O(v)/ mol (1-yo ) mol DA/mo l H=30
01103 . mol H 2 O mol 5 , y2
01103 . , H2
48
0.054827
48 5
ln a ln y1 bH1
b.
50q C , P 839 mm Hg , saturated:
b
ln y 2 y1
0.01044 mol H 2 O mol
760 mm Hg
P H 2 O 50q C y 2
y2 P
b
b0.40g19.827 mm Hg
g
0.40 P H 2 O 22q C y1
y1 P
40% :
b
g
4.8362 a exp 4.8362
273K 1 k mol 22 273 K 22.4m 3 STP
g
b g
10 3 mol 1 kmol
7.937 u 10 3
4131 . mol air delivered
41.31 mol, T, 1 at m
41.4 mol, 22o C,1 at m
0.0104 mo l H2 O(v)/mo l, (sat’d) 0.09896 mo l DA/ mol
0.0104 mo l H2 O(v)/mo l 0.09896 mo l DA/ mol
n 1 mol H2 O(l)
Saturation condition prior to reheat stage: y H 2O P
bg
b g b0.01044gb760 mm Hgg
PH*2 O T PH 2 O T T
7.93 mm Hg Table 8.3
7.8q C
bg
Part a
30 y 0
Humidity of outside air: H
b
Overall dry air balance: n0 1 V0 Overall water balance: n0 y 0
n2
0.0411 mol H 2 O mol
. gb0.9896g b4131 b g b1 0.0411g 42.63 mol b4131 . gb0.0104g n b42.63gb0.0411g b4131 . gb0.0104g
g
4131 . 0.9896 n0
2
132 . mol H 2 O condensed Mass of condensed water
132 . mol H 2 O 18.02 g H 2 O 1 kg 1 mol H 2 O 10 3 g 0.024 kg H 2 O condensed m 3 air delivered
6-18
6.31 a.
Basis: n 0 mol feed gas . S solvent , G
solvent - free gas n1 (mol) @ Tf (qC), P4 (mm Hg) y1 [mol S(v)/mol] (sat’d) (1–y1) (mol G/mol)
n0 (mol) @ T0 (qC), P0 (mm Hg) y0 (mol S/mol) (1-y0) (mol G/mol) Td0 (qC) (dew point)
n2 (mol S (l))
p Tdo y o
n1 n2 n1
n0 y 0 f n 0 n2
(1)
d i
p T f
d i
Fractional condensation of S = f n2
b gb g
Po
p T f y1
Saturation condition at outlet: y1 Pf
Total mole balance: n 0
b g
p Tdo
b g
Inlet dew point = T0 y o Po
(2)
Pf (1) o n2
bg
Eq. 3 for n1
n1
b gP n fp bT g n0 fp T0
n 0
0
0
do
Po
n1 y1 n2
S balance: n0 y 0
(1) - (4)
b g OPFG p dT iIJ n fp bT g P PQGH P JK p L fp eTdo j OP p dT i M1 MM Po PP b1 f g p bT g LM1 fp bT g OP p dT i P N Q P MN P PQ P p T b1 f g eP do j b g LMn P MN
n 0 p Tdo
0
n 0 fp Tdo
o
f
o
0
f
do
o
f
do
f
do
o
o
f
f
o
b.
Condensation of ethylbenzene fromnitrogen Antoine constants for ethylbenzene A= 6.9572 B= 1424.3 C= 213.21 Run T0 P0 Td0 f 1 2 3 4
50 50 50 50
765 765 765 765
40 40 40 40
0.95 0.95 0.95 0.95
Tf
p* (Td0) p*(Tf)
45 40 35 20
21.492 21.492 21.492 21.492
6-19
27.62 21.49 16.56 7.08
Pf 19137 14892 11472 4904
Crefr Ccomp Ctot 2675 4700 8075 26300
107014 109689 83327 88027 64244 72319 27595 53895
(3) (4)
6.31 (cont’d) When Tf decreases, Pf decreases. Decreasing temperature and increasing pressure both to c. increase the fractional condensation. When you decrease Tf, less compression is required to achieve a specified fractional condensation. d.
6.32 a.
A lower Tf requires more refrigeration and therefore a greater refrigeration cost (Crefr). However, since less compression is required at the lower temperature, Ccomp is lower at the lower temperature. Similarly, running at a higher Tf lowers the refrigeration cost but raises the compression cost. The sum of the two costs is a minimum at an intermediate temperature.
Basis : 120 m 3 min feed @ 1000 o C(1273K), 35 atm . Use Kay’s rule. Cmpd. H2 CO CO 2 CH 4 Tcc Pcc Tr Pr V
b g P batmg bT g
Tc K 33.2
c corr .
c
12.8
133.0 304.2 190.7
34.5 72.9 458 .
413 .
bP g
c corr
bApply Newton' s corrections for H g 2
20.8
b g b g b g b g 133.4 K 0.40b20.8g 0.35b34.5g 0.20b72.9g 0.05b458 . g 37.3 atm 1273 K 133.4 K 9.54 U Generalized compressibility charts (Fig. 5.3 - 2) V z 102 35.0 atm 37.3 atm 0.94 W . ¦yT ¦y P
i ci
i
ci
8.314 N m
1.02 35 atm
120 m 3 min
0.40 413 . 0.35 133.0 0.20 304.2 0.05 190.7
1273 K
mol K
mol 1 kmol 3.04 u 10 m 3 10 3 mol 3
1 atm 101325 N m
3
3.04 u 10 3 m 3 mol
39.5 kmol min
n1 (kmol/min), 261 K, 35atm yNaOH sat’d yH2 yCH4 (2% of feed) yCO
1.2(39.5) kmol/min MeOH(l) 39.5 kmol/min, 283K, 35 atm 0.40 mol H2/mol 0.35 mol CO/mol 0.20 mol CO2/mol 0.05 mol CH4/mol
n2 (kmol/min), liquid yMeOH yCO2 yCH4 (98% of feed)
6-20
6.32 (cont’d)
b
p MeOH 261K
Saturation at Outlet: y McOH
b.
n MeOH
n MeOH n + H 2 nCH 4
A input
10
P 4.97 u 10
y McOH
g
b
7 .878631473.11 12 2300
b
g
mm Hg
g
35 atm 760 mm Hg atm 4
mol MeOH mol n MeOH
A 0.02 of input
n MeOH 39.5 0.40 0.02 0.05 0.35
nCO
A input
E
n MeOH
0.0148 kmol min MeOH in gas
The gas may be used as a fuel. CO2 has no fuel value, so that the cost of the added energy required to pump it would be wasted.
6.33 n0 (kmol/min wet air) @ 28qC, 760 mmHg
n1 (kmol/min wet air) @ 80qC, 770
y1 (mol H2O/mol) (1-y1) (mol dry air/mol) 50% rel. sat.
y2 (mol H2O/mol) (1-y2) (mol dry air/mol) Tdew point = 40.0oC
1500 kg/min wet pulp
m 1 (kg/min wet pulp)
0.75 /(1 + 0.75) kg H2O/kg 1/1.75 kg dry pulp/kg
0.0015 kg H2O/kg 0.9985 kg dry pulp/kg
Dry pulp balance: 1500 u
1 1 0.75
50% rel. sat’n at inlet: y1 P
m 1 (1 0.0015) m 1
0.50 p H* 2 O (20 o C) y1
858 kg / min
0.50(28.349 mm Hg) / (760 mm Hg)
= 0.0187 mol H 2 O / mol o
40 C dew point at outlet: y 2 P
p H* 2 O (40 o C)
y2
(55.324 mm Hg) / ( 770 mm Hg) = 0.0718 mol H 2 O / mol
Mass balance on dry air: n 0 (1 0.0187) n1 (1 0.0718)
(1)
Mass balance on water: n 0 ( 0.0187 )(18.0 kg / kmol ) 1500( 0.75 / 1.75) Solve (1) and (2) n 0
622.8 kmol / min, n1
n 1 ( 0.0718)(18) 858( 0.0015) ( 2 )
658.4 kmol / min
Mass of water removed from pulp: [1500(0.75/1.75)–858(.0015)]kg H2O = 642 kg / min Air feed rate: V0
622.8 kmol 22.4 m 3 (STP) (273 + 28) K 1.538 u 10 4 m 3 / min kmol 273 K min
6-21
6.34
Basis: 500 lb m hr dried leather (L) n1 (lb - moles / h)@130o F, 1 atm n0 (lb - moles dry air / h)@140o F, 1 atm
y1 (lb - moles H2 O / lb - mole) (1- y1 )(lb - moles dry air / lb - mole)
0 (lb m / h) m
500 lb m / h 0.06 lb m H2 O(l) / lb m 0.94 lb m L / lb
0.61 lb m H2 O(l) / lb m 0.39 lb m L / lb m
b0.94gb500g m
Dry leather balance: 0.39m0 Humidity of outlet air: y1 P
0
b
1205 lb m wet leather hr
g
0.50 p H 2 O 130q F y1
b gb
H 2 O balance: 0.61 1205 lb m hr n1
g
a
fb
0.06 500 lb m hr
E
b1 0.0756g(517.5) lb - moles hr 478.4 lb - moles 359 ft bSTPg b140 460gq R 3
6.35 a.
hr
0.0756
mol H 2 O mol
g
0.0756n1 lb - moles H 2 O
18.02 lb m
hr
1 lb - mole
517.5 lb - moles hr
Dry air balance: n0
Vinlet
0.50(115 mm Hg) 760 mmHg
492q R
1 lb - mole
478.4 lb - moles hr
2.09 u 10 5 ft 3 hr
Basis: 1 kg dry solids n 1 (kmol)N 2, 85°C
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N 2 /mol) 70% rel. sat.
dryer 1.00 kg solids 0.78 kg Hex
n 4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80q C
b0.78 0.05gkg Antoine eq.
B
70% rel. sat.: y 2
condenser
b
0.70 p hex 80q C
n 3 (kmol) 28°C, 5.0 atm y 3 (mol Hex/mol) sat'd (1 – y 3) (mols N 2 /mol)
kmol 8.47 u 10 3 kmol Hex 86.17 kg
g b0.70g10
P
b
6.87776 1171.530 80 224 .366
760
6-22
g 0.984 mol Hex mol
6.35 (cont’d) n2
8.47 u 10 3 kmol Hex
1 kmol 0.984 kmol Hex
b1 0.984g0.0086
N 2 balance on dryer: n1
Antoine Eq.
B
b
p hex 28q C
Saturation at outlet: y 3
g
10
0.0086 kmol u 10 4 kmol 1376 .
b
6.877761171.580 28 224 .366
5 760
b
g
. u 10 4 kmol 144
n3 1 0.0452 n3
Mole balance on condenser: 0.0086 144 . u 10 4 n4 n4 Fractional hexane recovery:
0.0452 mol Hex mol
b g
P
Overall N 2 balance: 1.376 u 10 -4
g
0.0085 kmol cond. 86.17 kg 0.78 kg feed
kmol
0.0085 kmol 0.939 kg cond. kg feed
b. Basis: 1 kg dry solids
0.9n
heater
3
0.9n 3 (kmol) @ 28°C, 5.0 atm
y3 (1 – y3)
y 3 (mol Hex/mol) sat'd (1 – y 3) (mol N 2/mol)
n 2 (kmol) 80°C, 1 atm y 2 (mol Hex/mol) (1 – y 2) (mols N2 /mol) 70% rel. sat.
n 1 (kmol)N 2 85°C dryer 1.00 kg solids 0.78 kg Hex
condenser
0.1n3
n4 (kmol) Hex(l)
0.05 kg Hex 1.00 kg solids
Mol Hex in gas at 80q C: .8.47 u 10 3 0.9n3 (0.0452) N2 balance on dryer: n1 0.9n3 (1 0.0452) Overall N2 balance: n1
n3 (kmol) y3 (1 – y3)
n2 (0.984)
n2 (1 0.984)
( 2)
0.3n3 (1 0.0452)
R|n Equations (1) to (3) Sn |Tn
2 3
u 10 5 kmol 1515 .
Saved fraction of nitrogen =
(3)
5
u 10 kmol 1447 . 0.0086 kmol
1
(1)
u 10 5 1.376 u 10 -4 1447 . u 100% 1.376 u 10 4
90%
Introducing the recycle leads to added costs for pumping (compression) and heating.
6-23
6.36 b. m 1 (lbm/h)
300 lbm/h wet product 0.2 . 0167 lb m T(l) / lb m 1 0.2 0.833 lb m D / lb m
0.02 / (102 . ) 0.0196 lb m T(l) 0.9804 lb m D / lb m
Dryer
n3 (lb-mole/h) @ 200
n1 (lb-mole/h)
O
F,
y3 (lb-mole T/lb-mole) y1 (lb-mole T(v)/lb-mole) (1-y3)( lb-mole N2/lb-mole) (1–y1) (lb-mole N2/lbmole) T=toluene Heater 70% r.s.,150oF, 1.2 atm D=dry solids n3 (lb-mole/h)
y3 (lb-mole T(v)/lb-mole) (1-y3) (lb-mole N2/lb-mole)
Condenser
Eq.@ 90OF, 1atm
n2 ( lb-mole T(l)/h
Strategy: Overall balance m 1 & n 2 ; Relative saturationy1;, Gas and liquid equilibriumy3 Balance over the condenser n1 & n 3
UV RS W T
m 1 u 0.0196 n 2 u 92.13 Toluene Balance: 300 u 0167 . m 1 255 lb m / h m 1 u 0.9804 Dry Solids Balance: 300 u 0833 . n 2 0.488 lb - mole / h 70% relative saturation of dryer outlet gas: pC* 7 H8 y1 P
O
O
(150 F = 6556 . C) = 10
(6.95334-
0.70 pC* 7 H8 (150 O F) y1
1343.943 ) 65.56 219 .37
0.70 pC*7 H8 P
172.47 mmHg
(0.70)(172.47) . u 760 12
lb - mole T(v) / lb - mole . 01324
Saturation at condenser outlet: pC* 7 H8 y3
O
O
(90 F = 32.22 C) = 10 pC* 7 H8 P
40.90 760
(6.95334-
1343.943 ) 32 .22 219 .37
40.90 mmHg
0.0538 mol T(v) / mol
UV RS W T
Condenser Toluene Balance: n1 u 01324 0.488 n 3 u 0.0538 . n1 Condenser N 2 Balance: n1 u (1 01324 . ) n 3 u (1 0.0538) n 3
6-24
5875 lb - mole / h . 5.387 lb - mole / h
6.36 (cont’d) Circulation rate of dry nitrogen = 5.875 u (1 - 0.1324) =
5.097 lb - mole
lb - mole
h
28.02 lb m
lb m / h 0182 .
Vinlet
6.37
b g
5.387 lb - moles 359 ft 3 STP hr 1 lb - mole
C 6 H 14
Basis: 100 mol C 6 H 14
(200 460)q R 492q R
2590 ft 3 h
19 O 2 o 6CO 2 7H 2 O 2 n1 (mol) dry gas, 1 atm 0.821 mol N 2 /mol D.G. 0.069 mol CO2 /mol D.G. 0.021 mol CO/mol D.G. 0.086 mol O 2 /mol D.G. 0.00265 mol C 6 H 14/mol D.G. n2 (mol H 2 O)
100 mol C 6 H 14 n0 (mol) air 0.21 mol O 2 /mol 0.79 mol N 2 /mol
b g n LMM0b.069g 0b.021g 6b0b.00265g gOPP n 5666 mol dry gas Q N 100 0.00265b5666g mol reacted u 100% 85.0% Conversion: C balance: 6 100
1
CO 2
1
CO
C 6 H 14
100 mol fed
N 2 balance: 0.79n0 Theoretical air:
Excess air:
b g
0.821 5666 n0
100 mol C 2 H 14
5888 4524 u 100% 4524
b g
H balance: 14 100 Dew point: y H 2 O
5888 mol air
19 mol O 2
1 mol air
2 mol C 2 H 14
0.21 mol O 2
30.2% excess air
b gb
g
2n2 5666 14 0.00265 n2 595 595 + 5666
4524 mol air
d i
p Tdp
760 mm Hg
6-25
595 mol H 2 O
d i
p Tdp
Table B.3
72.2 mm Hg Tdp
451 . qC
6.38 Basis: 1 mol outlet gas/min n0 ( mol / min) y 0 ( mol CH 4 / mol) (1 y0 ( mol C 2 H 6 / mol)
1 mol / min @ 573K, 105 kPa y1 (mol CO 2 / mol) y 2 (mol H 2 O / mol) (1 y1 y 2 ) mol N 2 / mol
n1 (mol O 2 / min) 3.76n1 (mol N 2 / min)
CH 4 2O 2 o CO 2 2H 2 O pCO 2
80 mmHg y1
80 mmHg 101325 Pa 105000 Pa 760 mmHg
b
100% O2 conversion : 2no yo 7 no 1 yo
b
C balance: no yo 2no 1 yo
. n1 N2 balance: 376
C2 H6
2
g
g
7 O 2 o 2CO 2 3H 2 O 2 0.1016 mol CO 2 / mol
(1)
n1
01016 .
(2)
1 y1 y2
b
H balance: 4no yo 6no 1 yo
g
(3) 2 y2
(4)
R|n 0.0770 mol | y 0.6924 mol CH / mol Solve equations 1 to 4 S mol O . ||n 01912 mol H O / mol . Ty 01793 Dew point: 01793 . b105000g Pa 760 mmHg 1412. mmHg T p dT i o
4
o
* H2 O
1
2
2
2
dp
dp
101325 Pa
b
6.39 Basis: 100 mol dry stack gas n P (mol C 3 H 8) n B (mol C 4H10 ) n out (mol) 0.21 O2 0.79 N2
P = 780 mm Hg Stack gas: Tdp = 46.5°C 100 mol dry gas 0.000527 mol C 3 H 8/mol 0.000527 mol C 4H 10/mol 0.0148 mol CO/mol 0.0712 mol CO 2/mol + O 2, N 2
nw (mol H2O)
C 3 H 8 5O 2 o 3CO 2 4H 2 O
C 4 H 10
6-26
g
58.8 o C Table B.3
13 O 2 o 4CO 2 5H 2 O 2
Dew point But y w
b
46.5q C y w P
nw 100 n w
g
77.6 mm Hg 780 mm Hg
p w 46.5q C y w
0.0995 n w
0.0995
mol H 2 O mol
11.0 mol H 2 O
b100g b0.000527gb3g b0.000527gb4g 0.0148 0.0712 3n 4n 8.969 b1g H balance: 8n 10n b100g b0.000527gb8g b0.000527gb10g b110 . gb2g 8n 10n 23149 . b2 g R| 56% C H . mol C H U Rn 1454 S Solve b1g & b2g simultaneously: S V . mol C H W |T 44% C H Tn 1152 bAnswers may vary r 2% due to loss of precisiong C balance: 3n p 4n B p
B
p
B
p
B
p
3
8
3
8
B
4
10
4
10
6.40 a. L1 (lb - mole C 10 H 22 / h)
L 2 (lb - mole / h) x 2 (lb - mole C 3 H 8 / lb - mole) 1 x 2 (lb - mole C 10 H 22 / lb - mole)
G 1 (lb - mole / h) y1 (lb - mole C 3 H 8 / lb - mole) 1 y1 (lb - mole N 2 / lb - mole) Basis: G 2
G 2 = 1 lb - mole / h 0.07 (lb - mole C 3 H 8 / lb - mole) 0.93 (lb - mole N 2 / lb - mole)
1 lb - mole h feed gas
b gb g
N 2 balance: 1 0.93
b
g
98.5% propane absorption G 1 y1
b1g & b2g G
1
b
g 0.93 b1 0.985gb1gb0.07g G y
G 1 1 y1 G 1 1 y1
1 1
. u 10 3 105
b1g b2g
u 10 3 mol C 3 H 8 mol 1128 .
0.93105 lb - mol h , y1
Assume G 2 L 2 streams are in equilibrium
From Cox Chart (Figure 6.1-4), p *C3 H8 (80 o F ) 160 lb / in 2
b
g
Raoult' s law: x 2 p C3H 8 80q F
b gb g
Propane balance: 0.07 1 Decane balance: L1
0.07 p x 2
b0.07gb10. atmg
G 1 y1 L 2 x 2 L 2
10.89 atm
b
gd h b d L / G h 1
2 min
0.006428
gd
mol H 2 O mol
u 10 3 0.07 0.93105 1128 .
gb
g
10.66 lb - mole h
10.7 mol liquid feed / mol gas feed
6-27
i
0.006428
10.726 lb - mole h 1 x 2 L 2 1 0.006428 10.726
b
10.89 atm
6.40 (cont’d) b. The flow rate of propane in the exiting liquid must be the same as in Part (a) [same feed rate and fractional absorption], or n C3H 8
10.726 lb - mole 0.006428 lb - mole C 3 H 3 h
lb - mole
0.06895 lb - mol C 3 H 8 h
The decane flow rate is 1.2 x 10.66 = 12.8 lb-moles C10H22/h x2
b
0.06895 lb - mole C 3 H 8 h 0.06895 + 12.8 lb - moles h
g
0.00536 lb - mole C 3 H 8 / lb - mole
c. Increasing the liquid/gas feed ratio from the minimum value decreases the size (and hence the cost) of the column, but increases the raw material (decane) and pumping costs. All three costs would have to be determined as a function of the feed ratio. 6.41 a. Basis: 100 mol/s liquid feed stream
Let B n - butane , HC = other hydrocarbons
o
100 mol/s @ 30 C, 1 atm
n4 (mol/s) @ 30qC, 1 atm
xB =12.5 mol B/s 87.5 mol other hydrocarbon/s
y4 (mol B/mol) (1-y4) (mol N2/mol)
88.125 mol/s
n3 (mol N2/s)
0.625 mol B/s (5% of B fed) 87.5 mol HC/s
p *B (30 o C) # 41 lb / in 2 Raoult' s law: y 4 P
2120 mm Hg (from Figure 6.1-4)
b
95% n-butane stripped: n 4 0.3487 Total mole balance: 100 n 3 mol gas fed mol liquid fed
b. If y 4
0.8 u 0.3487
x B p B* (30 o C) 0125 . u 2120 760 P 12.5 0.95 n 4 34.06 mol / s
x B p *B (30 o C) y 4
0.3487
g b gb g
34.08 88125 . n 3
22.20 mol / s 100 mol / s
22.20 mol / s
0.2220 mol gas fed / mol liquid fed
0.2790 , following the same steps as in Part (a),
b
g b gb g
12.5 0.95 n 4 42.56 mol / s 95% n-butane is stripped: n 4 0.2790 Total mole balance: 100 n 3 42.56 88.125 n 3 30.68 mol / s mol gas fed 30.68 mol / s 0.3068 mol gas fed / mol liquid fed mol liquid fed 100 mol / s c. Increasing the nitrogen feed rate can cause an increase of nitrogen consumption, but more butane can be recovered. Therefore, costs of nitrogen (including compression) and butane are needed to determine the most cost-effective value of the gas/liquid feed rate.
6-28
6.42 Basis: 100 mol NH 3 Preheated air 100 mol NH 3 780 kPa sat'd
N2 O2
converter
n3 n4 n5 n6
n 1 (mol) O 2 3.76 n 1 (mol) N 2 n 2 (mol) H 2 O 1 atm, 30°C h r = 0.5
a. i)
b g
P Tsat
NH 3 feed: P
Antoine: log 10 6150
b
g
820 kPa
55 wt% HNO 3 (aq ) n 8 (mol HNO 3 ) n 9 (mol H 2O)
n7 (mol H 2O)
6150 mm Hg 8.09 atm
b
g
7.55466 1002.711 Tsat 247.885 Tsat
Table B.1
VNH 3
absorber
(mol NO) (mol N 2) (mol O 2 ) (mol H 2O)
Pc Tc
18.4q C 291.6 K
UV W
. atm Pr 8.09 / 1113 . 1113 0.073 z . K Tr 2916 . / 4055 . 0.72 4055
b
0.92 100 mol
g
8.314 Pa 2916 . K mol - K 820 u 10 3 Pa
0.92
(Fig. 5.3-1)
0.272 m 3 NH 3
Air feed: NH 3 2O 2 o HNO 3 H 2 O n1
100 mol NH 3
2 mol O 2 mol NH 3
b
hr p * 30q C
Water in Air: y H 2 O
p 0.02094
200 mol O 2
g
. 0.500 u 31824 0.02094 760 n2 n2 20.36 mol H 2 O n2 4.76 200
A
( 4 .76 mol air mol O 2 )
Vair
b g
b g
4.76 200 20.36 mol 22.4 L STP 1 mol
303K
1 m3
273K 10 3 L
24.2 m 3 air
ii) Reactions: 4 NH 3 5O 2 o 4 NO 6H 2 O , 4 NH 3 3O 2 o 2 N 2 6H 2 O Balances on converter NO: n3
97 mol NH 3
4 mol NO 4 mol NH 3
6-29
97 mol NO
6.42 (cont’d)
b g
N 2 : n4
3.76 2.00 mol +
O 2 : n5
200 mol
3 mol NH 3
2 mol N 2 4 mol NH 3 5 mol O 2 4 mol NH 3
97 mol NH 3
. mol N 2 7535
3 mol O 2 76.5 mol O 2 4 mol NH 3 100 mol NH 3 6 mol H 2 O H 2 O: n6 20.36 mol + 170.4 mol H 2 O 4 mol NH 3 n total (97 7535 . 76.5 170.4) mol = 1097 mol converter effluent
3 mol NH 3
8.8% NO, 68.7% N 2 , 7.0% O 2 , 15.5% H 2 O iii) Reaction: 4 NO 3O 2 2 H 2 O o 4 HNO 3 HNO 3 bal. in absorber: n8
H 2 O in product: n9
97 mol NO react 4 mol HNO 3 4 mol NO
97 mol HNO 3
63.02 g HNO 3 mol
277.56 mol H 2 O
b
gb g
H balance on absorber: 170.4 2 2n7
b.
b
1 mol H 2 O 45 g H 2 O 55 g HNO 3 18.02 g H 2 O
gb gb
97 277.6 2 mol H
n7
155.7 mol H 2 O added
VH 2O
155.7 mol H 2 O 18.02 g H 2 O 1 cm 3 1 m3 1 mol 1 g 10 6 cm 3
V NH 3 Vair VH 2O
g bg
2.81 u 10 3 m 3 H 2 O l
97 mol HNO 3
63.02 g HNO 3 277.6 mol H 2 O 18.02 g HNO 3 mol mol 11115 g 11115 . kg
M acid in old basis
Scale factor
97 mol HNO 3
b1000 metric tonsgb1000 kg metric tong
8.997 u 10 4
11.115 kg
d8.997 u 10 id0.272 m NH i 2.45 u 10 m NH d8.997 u 10 id24.2 m airi 2.18 u 10 m air d8.997 u 10 id2.81 u 10 m H Oi 253 m H Oblg 4
3
4
3
3
4
4
6
3
3
2
6-30
3
3
3
3
2
6.43 a.
Basis: 100 mol feed gas 100 mol 0.10 mol NH3 /mol 0.90 mol G/mol
G = NH3 -free gas Absorber
n 1 (mol H2 O( l))
n 2 (mol) in equilibrium y A (mol NH 3 /mol) at 10°C(50°F) (mol H O/mol) yW 2 and 1 atm (mol G/mol) yG (mol) n3 x A (mol NH 3 /mol) (1 – x A) (mol H 2O/mol)
Composition of liquid effluent . Basis: 100 g solution Perry, Table 2.32, p. 2-99: T = 10oC (50oF), U = 0.9534 g/mL 0.120 g NH3/g solution
12.0 g NH 3 88.0 g H 2 O = 0.706 mol NH 3 , = 4.89 mol NH 3 (17.0 g / 1 mol) (18.0 g / 1 mol)
12.6 mole% NH 3 ( aq), 87.4 mole% H 2 O(l)
Composition of gas effluent T
o
50 F, x A
o . 0126
yA yW yG
p H 2O p total
. / 14.7 0.0823 mol NH 3 mol 121 . / 14.7 0.0105 mol H 2 O mol 0155 1 y A yW 0.907 mol G mol
b gb g n y n b100gb0.90g b0.907g 99.2 mol absorbed b100gb0.10g b99.2gb0.0823g . mol NH 184
G balance: 100 0.90 NH 3
g U| . 0155 psiabTable 2 - 21gV |W 14.7 psia b
. psia Table 2 - 23 121
p NH 3
Perry
% absorption
2
G
2
in
3
out
1.84 mol absorbed u 100% 18.4% 100 010 . mol fed
b gb g
b. If the slip stream or densitometer temperature were higher than the temperature in the contactor, dissolved ammonia would come out of solution and the calculated solution composition would be in error. 6.44 a. 15% oleum: Basis - 100kg 15 kg SO 3
85 kg H 2 SO 4
1 kmol H 2 SO 4 98.08 kg H 2 SO 4
84.4% SO 3
6-31
1 kmol SO 3 1 kmol H 2 SO 4
80.07 kg SO 3 1 kmol SO 3
84.4 kg
6.44 (cont’d) b.
Basis 1 kg liquid feed n o (mol), 40o C, 1.2 at m
n1 (mol), 40o C, 1.2 at m
0.90 mol SO3 /mol 0.10 mol G/ mol
y1 mol SO3 /mo l (1-y1 ) mol G/ mol Equilib riu m @ 40o C
1 kg 98% H2 SO4
m1 (kg) 15% oleu m
0.98 kg SO3 0.02 kg H2 O
0.15 kg SO3 /kg 0.85 kg H2 SO4 /kg
b
pSO3 40q C, 84.4%
g
115 . 760
151 . u 10 3 mol SO 3 mol
i)
y1
ii)
0.02 kg H 2 O 2.02 kg H 2.02 kg H 98.08 kg H 2 SO 4 18.02 kg H 2 O 0.85 m1 H 2 SO 4 2.02 kg H m1 128 . kg 98.08 kg H 2 SO 4 But since the feed solution has a mass of 1 kg, 0.28 kg SO 3 10 3 g 1 mol SO 3 absorbed 128 . 10 . kg 350 . mol kg 80.07 g 3.5 mol n0 n1 . u 10 3 n1 G balance: 0.10n0 1 151 H balance:
P 0.98 kg H 2 SO 4
b
g
d
i
E
n0 n1 V
389 . mol 0.39 mol
b g
3.89 mol 22.4 L STP 1 kg liquid feed mol
313K 1 atm 1 m 3 273K 1.2 atm 10 3 L
8.33 u 10 -2 m 3 kg liquid feed 6.45 a. Raoult’s law can be used for water and Henry’s law for nitrogen. b. Raoult’s law can be used for each component of the mixture, but Henry’s law is not valid here. c. Raoult’s law can be used for water, and Henry’s law can be used for CO2.
b g p b100q Cg
6.46 p B 100q C
T
b gh 1350.5 mm Hg c c6.95334 1343.943 b100 219.377gh 556.3 mm Hg 0.40b1350.5g 0.0711 mol Benzene mol P x p y 10b760g 0.60b556.3g 0.0439 mol Toluene mol y 10b760g
10
6.90565 1211.033 100 220.790 10
Raoult' s Law: y B
B
B
B
T
y N2
1 0.0711 0.0439
6-32
0.885 mol N 2 mol
6.47 N 2 - Henry' s law: Perry' s Chemical Engineers' Handbook, Page. 2 - 127, Table 2 - 138
b
H N 2 80q C
g
12.6 u 10 4 atm mole fraction
b0.003gd12.6 u 10 i 378 atm . mm Hg 1 atm 3551 H O - Raoult' s law: p b80q Cg 760 mm Hg p d x id p i b0.997gb0.467g 0.466 atm pN2
4
xN2 H N2
H 2O
2
H 2O
H 2O
H 2O
Total pressure: P
p N 2 p H 2O
Mole fractions: y H 2 O y N2
378 0.466 378.5 atm
p H 2O P 1 y H 2O
b
6.48 H 2 O - Raoult' s law: p H 2 O 70q C p H 2O Methane Henry' s law: p m Total pressure: P
0 / 466 / 378.5 123 . u 10 3 mol H 2 O mol gas 0.999 mol N 2 mol gas
g
233.7 mm Hg
1 atm
0.3075 atm
760 mm Hg
x H 2 O p H 2 O
b1 x gb0.3075g m
xm Hm
p m p H 2O
x m 6.66 u 10 4 (1 x m )(0.3075) 10
146 . u 10 4 mol CH 4 / mol
xm 6.49 a.
0.467 atm
Moles of water: n H 2 O
1000 cm 3
1g mol 55.49 mol 3 cm 18.02 g
Moles of nitrogen: nN2
(1 - 0.334) u 14.1 cm 3 (STP)
1 mol 1L 22.4 L (STP) 1000 cm 3
4.192 u 10 4 mol
Moles of oxygen: n O2
(0.334) 14.1 cm 3 (STP)
mol
L
22.4 L (STP) 1000 cm
3
2.102 u 10 4 mol
Mole fractions of dissolved gases: nN2 4.192 u 10 4 x N2 n H 2 O n N 2 nO 2 55.49 4.192 u 10 4 2.102 u 10 4 7.554 u 10 6 mol N 2 / mol x O2
nO 2 n H 2 O n N 2 nO 2
2.102 u 10 4 55.49 4.192 u 10 4 2.102 u 10 4
3.788 u 10 6 mol O 2 / mol
6-33
6.49 (cont’d) Henry' s law p N2
Nitrogen: H N 2
x N2 pO2
Oxygen: H O 2
x O2
0.79 1 7.554 u 10 6
u 10 5 atm / mole fraction . 1046
0.21 1 3.788 u 10 6
u 10 4 atm / mole fraction . 5544
b. Mass of oxygen dissolved in 1 liter of blood: 2.102 u 10 -4 mol 32.0 g 6.726 u 10 3 g mol 0.4 g O 2 1 L blood Mass flow rate of blood: m blood 59 L blood / min min 6.72 u 10 -3 g O 2 c. Assumptions: (1) The solubility of oxygen in blood is the same as it is in pure water (in fact, it is much greater); (2) The temperature of blood is 36.9qC. m O2
6.50 a.
bg
Basis: 1 cm 3 H 2 O l (SG)
1.0
H2O o
( SC) CO2
1 g H 2 O 1 mol 18.0 g
0.0555 mol H 2 O
b g
0.0901 cm 3 STP CO 2
0.0901
o
1 mol 22,400 cm 3 STP
b g
d4.022 u 10 i mol CO 7.246 u 10 mol CO d0.0555 + 4.022 u 10 i mol 1 atm H b20q Cg 13800 atm mole fraction 7.246 u 10 6
b.
4.022 u 10 6 mol CO 2
pCO 2
1 atm x CO2
pCO 2
x CO 2 H CO2
2
5
6
CO 2
2
mol
5
b g
For simplicity, assume n total | n H 2 O mol x CO2 nCO 2
pCO 2 H 12 oz
b35. atmg b13800 atm mole fractiong
2.536 u 10 4 mol CO 2 mol
1L 10 3 g H 2 O 1 mol H 2 O 2.536 u 10 4 mol CO 2 33.8 oz 1L 18.0 g H 2 O 1 mol H 2 O
44.0 g CO 2 1 mol CO 2
0.220 g CO 2
c. V
0.220 g CO 2
b g b273 37gK
1 mol CO 2
22.4 L STP
44.0 g CO 2
1 mol
6-34
273K
0127 . L 127 cm 3
6.51 a.
– SO2 is hazardous and should not be released directly into the atmosphere, especially if the analyzer is inside. – From Henry’s law, the partial pressure of SO2 increases with the mole fraction of SO2 in the liquid, which increases with time. If the water were never replaced, the gas leaving the bubbler would contain 1000 ppm SO2 (nothing would be absorbed), and the mole fraction of SO2 in the liquid would have the value corresponding to 1000 ppm SO2 in the gas phase.
b
g
b
b. Calculate x mol SO 2 mol in terms of r g SO 2 100 g H 2 O
b b
Basis: 100 g H 2 O 1 mol 18.02 g r (g SO 2 ) 1 mol 64.07 g
FG H
g g
555 . mol H 2 O 0.01561r (mol SO 2 )
IJ K
mol SO 2 0.01561r mol 555 . 0.01561r From this relation and the given data, pSO 2 xSO 2
g
0 mmHg xSO2
0 mol SO 2 mol 1.4 x 10–3 2.8 x 10–3 4.2 x 10–3 5.6 x 10–3
42 85 129 176
A plot of pSO 2 vs. xSO 2 is a straight line. Fitting the line using the method of least squares (Appendix A.1) yields
dp
HSO 2 xSO2
SO 2
c.
100 ppm SO 2 ySO2 pSO2
ySO2 P
,
H SO 2
u 104 . 3136
mm Hg mole fraction
100 mol SO 2 106 mols gas
d10. u 10 ib760 mm Hgg
Henry's law xSO 2
Since xSO 2
i
4
pSO 2 H SO 2
0.0760 mm Hg
0.0760 mm Hg 3.136 u 104 mm Hg mole fraction
2.40 u 106 mol SO 2 mol is so low, we may assume for simplicity that Vfinal | Vinitial
nfinal | ninitial
bg
140 L 103 g H 2 O l 1L
1 mol 18 g
7.78 u 103 moles
7.78 u 103 mol solution 2.40 u 10 6 mol SO 2 1 mol solution 0.0187 mol SO 2 dissolved 134 . u 10 4 mol SO 2 L 140 L
nSO 2
140 L , and
0.0187 mol SO 2 dissolved
d. Agitate/recirculate the scrubbing solution, change it more frequently. Add a base to the solution to react with the absorbed SO2.
6-35
6.52 Raoult’s law + Antoine equation (S = styrene, T = toluene): 0.650(150 mm Hg)
yS P
x S pS xS
yT P
xT pT xT
10
10
b
6.92409 1420.0 T 206
0.65(150)
b
6.92409 1420.0 85.9 206
1 xS
xT
b g P b85q Cg
6.53 PB 85q C
T
b
6.95334 1343.943 T 219 .377
1 0.852
g
g
0.35(150)
b g 10 10 85.9q C (Determine using E - Z Solve or a spreadsheeet)
T
10
g
0.350(150 mm Hg)
0.65(150)
1 x S xT
xS
b
6.92409 1420.0 T 206
6.95334 1343.943 T 219 .377
0.852 mol styrene / mol
g
0148 . mol toluene / mol
b g 6.95334 1343.943 b85 219 .377 g 10 10
6.905651211.033 85 220.790
Raoult's Law: y B P
x B PB
. mm Hg 8817 . mm Hg 3451
b g b g b g b g
0.35 8817 . 0.0406 mol Benzene mol 10 760 0.65 3451 . 0.0295 mol Toluene mol 10 760 1 0.0406 0.0295 0.930 mol N 2 mol
yB yT yN 2
6.54 a. From the Cox chart, at 77q F, p*P
140 psig, p*nB
35 psig, p*iB
51 psig
Total pressure P = x p p *p + x nB p *nB + x iB p *iB 0.50(140) 0.30(35) 0.20(51) 150.8 psig P 200 psig, so the container is technically safe. b. From the Cox chart, at 140q F, pP*
* 300 psig, pnB
* 90 psig, piB
120 psig
Total pressure P = 0.50(300) 0.30(90) 0.20(120) # 200 psig The temperature in a room will never reach 140oF unless a fire breaks out, so the container is adequate.
b g P b120q Cg
6.55 a. Antoine: Pnp 120q C
ip
10 10
b
6.852211064.630 120 232 .000
b
6.78967 1020.012 120 233.097
g
g
6725 mm Hg 7960 mm Hg
When the first bubble of vapor forms, xnp
0.500 mol n - C5 H 12 (l) / mol
* Total pressure: P = xnp pnp + xip pip*
xnp
0.500 mol i - C5 H 12 (l) / mol
0.50(6725) 0.50(7960)
6-36
7342 mm Hg
6.55 (cont’d) * xnp pnp
ynp
0.500(6725) 7342
P
1 ynp
yip
0.458 mol n - C5 H 12 (v) / mol
1 0.458 0.542 mol i - C5 H 12 (v) / mol
When the last drop of liquid evaporates, 0.500 mol n - C5 H 12 (v) / mol
ynp
ynp P * pnp (120o C)
xnp xip
yip
yip P * pip (120o C)
0.5 mol i - C5 H 12 (v) / mol
0.500 P 0.500 P 6725 7960
1 P
7291 mm Hg
0.5 * 7291 mm Hg 0.542 mol n - C5 H 12 (l) / mol 6725 mm Hg 1 xnp 1 0.542 0.458 mol i - C5 H 12 (l) / mol
xnp xip
b. When the first drop of liquid forms, 0.500 mol n - C5 H 12 (v) / mol
ynp
yip
0.500 mol i - C12 H 12 (v) / mol
P = (1200 + 760) = 1960 mm Hg 0.500 P 0.500 P * * pnp (Tdp ) pip (Tdp )
xnp xip Tdp
10
b
6.852211064 .63 63.1 232 .000
10
Pip
10
xnp
0.5 * 1960 mm Hg * (631 . o C) pnp
b
g
6.78967 1020.012 63.1 233.097
1 xnp
980 10
6.78967 1020.012 /( Tdp 233.097)
631 . oC
Pnp
xip
980 6.852211064 .63/( Tdp 232 )
1756 mm Hg
g
2218 mm Hg
0.558 mol n - C5 H12 / mol
1 0.558 0.442 mol i - C5 H 12 / mol
When the last bubble of vapor condenses, xnp
0.500 mol n - C5 H 12 (l) / mol
xip
0.500 mol i - C5 H12 (l) / mol
* Total pressure: P = xnp pnp + xip pip*
1960 (0.5)10 T 62.6q C ynp
yip
* (62.6o C) xnp pnp
1 ynp
b
6.852211064 .63 T 232 .000
P 1 0.442
g (0.5)106.78967 1020.012 /( T
bp 233.097)
0.5(1731) 0.442 mol n - C5 H 12 (v) / mol 1960 0.558 mol i - C5 H 12 (v) / mol
6-37
1
6.56 B = benzene, T = toluene nv ( mol / min) at 80o C, 3 atm 10 L(STP)/min nN 2 ( mol / min)
xB [mol B(l)/mol]
yN2 (mol N2/mol) yB [mol B(v)/mol]
xT [mol T(l)/mol]
yT [mol T(v)/mol]
10.0 L(STP) / min = 0.4464 mol N 2 / min 22.4 L(STP) / mol
nN 2
b g p b80q Cg
b g 6.95334 1343.943 b80 219 .377 g 10
Antoine: p B 80q C
10
T
6.905651211.033 80 220.079
757.7 mm Hg . mm Hg 2912
a. Initially, xB = 0.500, xT = 0.500. Raoult's law:
b
g
mol B(v) mol 0166 .
b
g
0.0639 mol T(v) mol
yB
x B p B (80o C) P
0.500 757.7 3(760)
yT
xT pT (80o C) P
0.500 2912 . 3(760)
0.0639) n v N 2 balance: 0.4464 mol N 2 / min = nv (1 0166 . mol BI FG 0.5797 mol IJ FG 0166 . J H K H min mol K FG 0.5797 mol IJ FG 0.0639 mol BIJ H min K H mol K
nB0 nT0
0.5797 mol / min
mol B(v) min
0.0962 0.0370
mol T(v) min
b. Since benzene is evaporating more rapidly than toluene, xB decreases with time and xT (= 1–xB) increases. c. Since xB decreases, yB (= xBpB*/P) also decreases. Since xT increases, yT (= xTpT*/P) also increases. 6.57 a. P
xhex phex
dT i
760 mm Hg
bp
xhep phep
0.500 10
dT i
, yi
bp
xi
yi P pi
dT i dp
¦x i
i
P
6.87776 1171.53/( Tbp 224 .366)
E-Z Solve or Goal Seek Tbp
b.
d i
xi pi Tbp
P
¦
pi
0.500 10
80.5q C yhex yi
i
dT i
1
dp
6-38
, Antoine equation for Pr 6.90240 1268.115/( Tbp 216.9 )
0.713, yhep
0.287
6.57 (cont’d) 760 mmHg
LM N10
0.30 6.87776 1171.53/( Tdp 224 .366)
E-Z Solve or Goal Seek Tdp
6.58
a.
f (T )
P
N
¦
xi pi* (T )
OP Q
0.30 10
6.90240 1268.115/( Tdp 216.9 )
711 . q C xhex
0 T , where
0.279, xhep
1
0.721
FG A B IJ T C K 10H i
pi* (T )
i
i
i 1
yi (i 1,2,, N )
xi pi* (T ) P
b. Calculation of Bubble Points A B Benzene 6.90565 1211.033 Ethylbenzene 6.95719 1424.255 Toluene 6.95334 1343.943
C 220.79 213.206 219.377
P(mmHg)= 760 xB 0.226 0.443 0.226 When x B When x EB When xT
xEB 0.443 0.226 0.226
Tbp(oC) 108.09 96.47 104.48
xT 0.331 0.331 0.548
b
g
dT i
1 pure benzene , Tbp
b
g
bp C H 6 6
1 pure ethylbenzene , Tbp
b
g
1 pure toluene , Tbp
. oC 801
dT i
dT i
bp C H 7 8
pEB pT f(T) pB 378.0 148.2 233.9 -0.086 543.1 51.6 165.2 0.11 344.0 67.3 348.6 0.07
bp C H 8 10
136.2 o C Tbp , EB ! Tbp ,T ! Tbp , B
110.6o C
Mixture 1 contains more ethylbenzene (higher boiling point) and less benzene (lower bp) than Mixture 2, and so (Tbp)1 > (Tbp)2 . Mixture 3 contains more toluene (lower bp) and less ethylbenzene (higher bp) than Mixture 1, and so (Tbp)3 < (Tbp)1. Mixture 3 contains more toluene (higher bp) and less benzene (lower bp) than Mixture 2, and so (Tbp)3 > (Tbp)2
6-39
6.59
a. Basis: 150.0 L/s vapor mixture
n 1 (mol/s) @ T(oC), 1100 mm Hg 0.600 mol B(v)/mol 0.400 mol H(v)/mol
n 0 (mol/s) @ 120qC, 0.500 mol B(v)/mol 0.500 mol H(v)/mol
Gibbs phase rule: F = 2 + m - S = 2 + 2 - 2 = 2
n 2 (mol/s) x2 [mol B(l)/mol] (1- x2) [mol H(l)/mol]
Since the composition of the vapor and the pressure are given, the information is enough. Equations needed: Mole balances on butane and hexane, Antoine equation and Raoult’s law for butane and hexane 150.0 L 393 K
b. Molar flow rate of feed: n 0
s
mol
273 K 22.4 L (STP)
9.640 mol / s
Raoult' s law for butane: 0.600(1100) = x 2 10 6.82485 943.453/( T 239.711) Raoult's law for hexane: 0.400(1100) = (1- x 2 ) 106.877761171.530 /( T 224.366) (1) Mole balance on butane: 1(0.5) = n 1 0.6 n 2 x 2 (2) Mole balance on hexane: 1(0.5) = n1 0.4 n 2 (1 x 2 ) c. From Raoult's law, 1 =
1100(0.4) 1100(0.6) . 1171530 943.453 ) ) 10 **(6.87776 10 **(6.82485 T 239.711 T 224.366
T = 57.0 q C x2
1100(0.6) 10
6.82485 943.453/( 57 .0 239 .711)
015 . mol butane / mol
Solving (1) and (2) simultaneously n1 d.
6.60
0.778 mol C 4 H 10 / s; n 2
0.222 molC 6 H 14 / s
Assumptions: (1) Antoine equation is accurate for the calculation of vapor pressure; (2) Raoult’s law is accurate; (3) Ideal gas law is valid.
P = n-pentane, H = n-hexane 170.0 kmol/h, T1a (oC), 1 atm
85.0 kmol/h, T1b (oC), 1 0.98 mol P(l)/mol 0.02 mol H(l)/mol
n 0 (kmol/h) 0.45 kmol P(l)/kmol 0.45 kmol H(l)/kmol
n 2 (kmol/h) (l), x2 (kmol B(l)/kmol) (1- x2) (kmol H(l)/kmol)
6-40
6.60 (cont’d) a. Molar flow rate of feed: n 0 (0.45)(0.95) 85(0.98) n 0 Total mole balance : 195 85.0 n 2 n 2
195 kmol / h
110 kmol / h
Pentane balance: 195( 0.45) 85.0(0.98) 110 x 2 x 2
0.0405 mol P / mol
b. Dew point of column overhead vapor effluent: Eq. 6.4 - 7, Antoine equation 0.02(760) 0.98(760) 6.877761171.530/( T 224.366) 6.852211064.63/ ( T1a 232.000) 1a 10 10
1 T1a
37.3o C
Flow rate of column overhead vapor effluent. Assuming ideal gas behavior, Vvapor
170 kmol 0.08206 m 3 atm (273.2 + 37.3) K h kmol K 1 atm
4330 m 3 / h
Flow rate of liquid distillate product. Table B.1 U P = 0.621 g / mL, U H = 0.659 g / mL Vdistillate
0.98(85) kmol P 72.15 kg P L h kmol P 0.621 kg P +
0.02(85) kmol H 86.17 kg H L h kmol H 0.659 kg H
9.9 u 10 3 L / h
c. Reboiler temperature. 0.04 10 6.852211064.63/ ( T2 232.000) 0.96 10 6.877761171.530 /( T2 224 .366)
760 T2 = 66.6q C
Boilup composition. y2
x 2 p P* (66.6 o C) P (1 - y 2 )
0.04 10 6.852211064.63/ ( 66.6232 ) 760
0.102 mol P(v) / mol
0.898 mol H(v) / mol
d. Minimum pipe diameter
F I GH JK
m3 V s
Dmin
u max
FG mIJ u SD H sK 4
4Vvapor
S u max
2 min
(m2 )
4 4330 m 3 / h 1 h S 10 m / s 3600 s
0.39 m (39 cm)
Assumptions: Ideal gas behavior, validity of Raoult’s law and the Antoine equation, constant temperature and pressure in the pipe connecting the column and the condenser, column operates at steady state.
6-41
6.61 a.
Condenser F (mol) x 0 (mol butane/mol)
V (mol) 0.96 mol butane/mol R (mol) x 1 (mol butane/mol)
T P
Partial condenser: 40q C is the dew point of a 96% C 4 H 10 4% C 5 H 12 vapor mixture at P Pmin Total condenser: 40q C is the bubble point of a 96% C 4 H 10 - 4% C 5 H 12 liquid mixture at P Pmin yi P 1 xi Pmin Dew Point: 1
pi 40q C yi pi 40q C
¦
¦
b
(Raoult's Law)
g
4
10
i
5
12
P
2595.63 mm Hg partial condenser
¦y P ¦x p i
75 kmol / h , R V
i
Total balance: F
15 . R
75 u 15 . kmol / h 112.5 kmol / h
b b
g
Butane balance: 187.5x 0 yi xi
Raoult' s law:
d
i
*
o
p B (85 C)
1
0
yA xA yB xB
p A P
p A
p B P
p B
110.76 mm Hg
FG 6.95719 1424.255 IJ 85 213.206 K
p *EB (85o C) 10 H
187.5 kmol / h
UV P 2596 mm Hg gb gW x 0.8803 mol butane mol 112.5b0.8803g 0.96b75g x 0.9122 mol butane mol reflux
pi D AB P
FG 6.92409 1420.0 IJ 85 206 K
10 H
F
g
Equilibrium: 0.96 P x1 2830.70 Raoult' s law 0.04 P 1 x1 867.22
b
0.96 kmol butane kmol
75 112.5 187.5 kmol / h
Total balance as in b. R 112.5 kmol / h
b. p S* 85o C
g
i
Feed and product stream compositions are identical: y
6.62 a.
867.22 mmHg
b40q Cg 0.96b2830.70g 0.04b867.22g 2752.16 mm Hg b total condenser g
Bubble Point: P
c.
2830.70 mmHg
b
1 0.96 2830.70 0.04 867.22
Pmin
b. V
pi
g
FG 6.82485 943.453 IJ 40 239 .711 K 10 H FG 6.85221 1064.63 IJ 40 232 .00 K 10 H
bC H g Antoine Eq. for p bC H g Antoine Eq. for
b
¦
FG 6.90565 1211.033 IJ 85 220.790 K 10 H
15174 . mm Hg 88167 . mm Hg
6-42
D AB
6.62 (cont’d) p S*
D S,EB
110.76 15174 .
p *EB
p B*
0.730 , D B,EB
* p EB
88167 . 15174 .
5810 .
Styrene ethylbenzene is the more difficult pair to separate by distillation because D S,EB is closer to 1 than is D B,EB . c.
yi xi yj xj
D ij
d. D B , EB
D
yi xi yi (1 yi ) 1 xi
b
ij
g
x B D B , EB
yB 5810 .
1 (D B , EB 1) x B
d
D ij xi
i
1 D ij 1 xi
581 . xB , P 1 4.81x B
x B p *B (1 x B ) p *EB
bg bg
xB yB P 6.63 a.
y j 1 yi x j 1 xi
0.0 0.2 0.4 0.6 0.8 10 . mol B l mol 0.0 0.592 0.795 0.897 0.959 10 . mol B v mol 152 298 444 5900 736 882 mmHg
Since benzene is more volatile, the fraction of benzene will increase moving up the column. For ideal stages, the temperature of each stage corresponds to the bubble point temperature of the liquid. Since the fraction of benzene (the more volatile species) increases moving up the column, the temperature will decrease moving up the column.
b. Stage 1: n l y0
150 mol / h, n v
0.55 mol B mol 0.45 mol S mol ;
0.65 mol B mol 0.35 mol S mol
¦ x p bT g
Bubble point T : P P1
200 mol / h ; x1
i
(0.400 u 760) mmHg E-Z Solve
o T1 y1
i
b0.55g10
6.905651211.033/ ( T 220.79 )
b g
0.45 10 6.92409 1420 /( T 206 )
67.6 o C
bg
x1 p B T
P B balance: y 0 n v x 2 n l
b g
0.55 508
0.920 mol B mol 0.080 mol S mol 0.400 u 760 y1 n v x1n l x 2 0.910 mol B mol 0.090 mol S mol
Stage 2: Solve o T2 (0.400 u 760) mmHg 0.910 p B* (T2 ) 0.090 p S* (T2 ) E-Z
b
55.3o C
g
0.910 3310 .
0.991 mol B mol 0.009 mol S mol 760 u 0.400 B balance: y1 n v x 3 n l y 2 n v x 2 n l x 3 | 1 mol B mol | 0 mol S mol y2
c. In this process, the styrene content is less than 5% in two stages. In general, the calculation of part b would be repeated until (1–yn) is less than the specified fraction. 6-43
6.64 Basis: 100 mol/s gas feed. H=hexane. 200 mol oil/s
n l (mol/s) x +i 1 (mol H/mol)
n F (mol/s) y F (mol H/mol) 1 – y F (mol N 2/mol)
100 mol/s 0.05 mol H/mol 0.95 mol N /mol 2
n 2 (mol/s) x 2 (mol H/mol) 1 – x 2 (mol Oil/mol)
Stage i n l (mol/s) x i (mol H/mol)
b g b g b gb g
y F nF
U|V n |W y
Mole Balance: 100 200 95.025 n2 n2
b g
Hexane Balance: 0.05 100 n L
b
g
1 200 205 n L 2 Antoine
y1
b.
b
B
g
x1 p H 50q C / P
n v (mol/s) y i– 1 (mol H/mol)
99.5% of H in feed.
N 2 balance: 0.95 100 1 y F nF 99.5% absorption: 0.05 100 0.005
a.
n v (mol/s) y i (mol H/mol)
b
F
95.025 mol s
F
2.63 u 10 4 mol H(v) mol
205 mol s 0.0243 mol H(l) mol g b g 1 b100 95.025g n 97.52 mol s 2
2.63 u 10 4 95.025 x1 204.99 x1 202.48 mol s , n G
b
G
g
0.0243 403.73 / 760 0.0129 mol H(v) mol
H balance on 1st Stage: y 0 n v x 2 n l
y1 n v x1 n l x 2
0.00643 mol H(l) mol
c. The given formulas follow from Raoult’s law and a hexane balance on Stage i. d. Hexane Absorption P= y0= nGf= A=
760 0.05 95.025 6.87776
T 30
p*(T) 187.1
i 0 1 2 3
x(i) 2.43E-02 3.07E-03 5.57E-04
PR= x1= nL1= B=
y(i) 5.00E-02 5.98E-03 7.56E-04 1.37E-04
1 0.0243 204.98 1172
ye= 2.63E-04 nG= 97.52 nL= C= 224.366
T 50
p*(T) 405.3281
i 0 1 2 3 4 5
x(i) 2.43E-02 6.42E-03 1.84E-03 6.63E-04 3.60E-04
6-44
y(i) 5.00E-02 1.29E-02 3.43E-03 9.81E-04 3.53E-04 1.92E-04
202.48
T 70
p*(T) 790.5304
i 0 1 2 3 4 5 ... 21
x(i)
y(i) 5.00E-02 2.43E-02 2.52E-02 1.23E-02 1.28E-02 6.38E-03 6.63E-03 3.38E-03 3.52E-03 1.89E-03 1.96E-03 ... ... 3.80E-04 3.96E-04
6.64 (cont’d) e. If the column is long enough, the liquid flowing down eventually approaches equilibrium with the entering gas. At 70oC, the mole fraction of hexane in the exiting liquid in equilibrium with the mole fraction in the entering gas is 3.80x10–4 mol H/mol, which is insufficient to bring the total hexane absorption to the desired level. To reach that level at 70oC, either the liquid feed rate must be increased or the pressure must be raised to a value for which the final mole fraction of hexane in the vapor is 2.63x10–4 or less. The solution is Pmin 951 mm Hg. 0.30 at T
6.65 a. Intersection of vapor curve with y B b. T
100q C x B
b
104q C 13% B(l), 87%T(l)
g
0.46 mol B mol liquid
b
g
0.727 mol nV 0.273 mol nL
mol vapor mol liquid
0.24 mol B mol liquid , y B n V (mol vapor) 0.46 mol B(v)/mol n L (mol liquid) 0.24 mol B(l)/mol
Basis: 1 mol 0.30 mol B(v)/mol
Balances
UV W
nL Total moles: 1 nV n L nV B: 0.30 0.46nV 0.24n L c. Intersection of liquid curve with x B
0.3 at T
6.66 a.
P
798 mm Hg, y B
0.50 mol B(v) mol
b.
P
690 mm Hg, y B
015 . mol B(l) mol
c.
P
750 mm Hg, y B
0.24 mol B(v) mol , x B
0.375
98q C 50% B(v), 50%T(v)
0.43 mol B(l) mol
nV (mol) 0.43 mol B/mol nL (mol) 0.24 mol B/mol
3 mol B 7 mol T
UV W
nV Mole bal.: 10 nV n L nL B bal.: 3 0.43nV 0.24n L
316 . mol 6.84 mol
nv nl
0.46
mol vapor mol liquid
Answers may vary due to difficulty of reading chart. d. i)
P 1000 mm Hg all liquid . Assume volume additivity of mixture components. V
3 mol B 78.11 g B 10 3 L 7 mol T 92.13 g T 10 3 L 10 . L mol B 0.879 g B mol T 0.866 g T
ii) 750 mmHg. Assume liquid volume negligible
6-45
6.66 (cont’d) 3.16 mol vapor
V
0.08206 L atm
373 K
760 mm Hg
mol K 750 mm Hg
1 atm
0.6 L
97.4 L
(Liquid volume is about 0.6 L) iii) 600 mm Hg v
10 mol vapor 0.08206 L atm
373K
760 mm Hg
mol K 600 mm Hg
1 atm
388 L
6.67 a. M = methanol n V (mol) y (mol M (v)mol) n L (mol) x (mol M (l)/mol)
n f (mol) x F (mol M (l)/mol)
UV W
n f nV n L Mole balance: x F nV x F n L MeOH balance: x F n f ynV xn L xF b. Tmin
0.4, x
0.23, y
75o C, f
0.62 f
0 , Tmax
0.4 0.23 0.62 0.23
87 o C, f
ynV xn L f
0.436
1
6.68 a.
Txy diagram (P=1 atm) 80 75 T(oC)
70
Vapor
65 liquid
60 55 50 0
0.2
0.4
0.6
Mole fraction of Acetone
b.
xA
0.47; y A
0.66
6-46
0.8
1
nV nL
xF x yx
6.68 (cont’d) c. (i) x A
0.34; y A
0.55
UV W
(ii) Mole bal.: 1 nV n L nV B bal.: 0.50 0.55nV 0.34n L (iii) U A( l ) Ul MA
0.791 g / cm 3 , U E(l)
0.789 g / cm 3 0.790 g / cm 3
46.07 g / mol
b0.34gb58.08g b1 0.34gb46.07g
Ml
0.238 mol liquid
76.2 mole% vapor
b0.34gb0.791g b1 0.34gb0.789g
58.08 g / mol, M E
0.762 mol vapor, n L
5015 . g / mol
Basis: 1 mol liquid (0.762 mol vapor / 0.238 mol liquid) = 3.2 mol vapor (1 mol)(5015 . g / mol) Liquid volume: Vl 63.48 cm 3 3 (0.790 g / cm ) Vapor volume: 3.2 mol 22400 cm 3 (STP) (65 + 273)K 88,747 cm 3 mol 273K 88,747 u 100% 99.9 volume% vapor Volume percent of vapor 88747 63.48 Vv =
d. For a basis of 1 mol fed, guess T, calculate nV as above; if nV z 0.20, pick new T. T 65 qC 64.5 qC e.
xA 0.34 0.36
Raoult' s law: yi P = xi pi* P 760 0.5 u 10
7.02447 1161/( Tbp 224 )
yA 0.55 0.56
fV 0.333 0.200
x A p *A x E p E* 0.5 u 10
8.04494 1554.3/ ( Tbp 222.65)
Tbp
66.25o C
0.5 u 10 7.02447 1161/( 66.25 224) 0.696 mol acetone / mol 760 'Tbp . 66.25 618 u 100% 7.20% error in Tbp The actual Tbp 618 . oC Tbp (real) 618 . y
xp *A P
yA
0.674
'y A y A (real)
0.696 0.674 u 100% 0.696
error in y A 316% .
Acetone and ethanol are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for acetone mole fractions that are not very close to 1.
6-47
6.69 a. B = benzene, C = chloroform. At 1 atm, (Tbp)B = 80.1oC, (Tbp)C = 61.0oC The Txy diagram should look like Fig. 6.4-1, with the curves converging at 80.1oC when xC = 0 and at 61.0oC when xC = 1. (See solution to part c.) b. Txy Diagram for an Ideal Binary Solution A B C 6.90328 1163.03 227.4 Chloroform 6.90565 1211.033 220.79 Benzene 760 P(mmHg)= x 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1
T 80.10 78.92 77.77 76.66 75.58 74.53 73.51 72.52 71.56 70.62 69.71 68.82 67.95 67.11 66.28 65.48 64.69 63.93 63.18 62.45 61.73
y 0 0.084 0.163 0.236 0.305 0.370 0.431 0.488 0.542 0.593 0.641 0.686 0.729 0.770 0.808 0.844 0.879 0.911 0.942 0.972 1
p1 0 63.90 123.65 179.63 232.10 281.34 327.61 371.15 412.18 450.78 487.27 521.68 554.15 585.00 614.02 641.70 667.76 692.72 716.27 738.72 760
p2 760 696.13 636.28 580.34 527.86 478.59 432.30 388.79 347.85 309.20 272.79 238.38 205.83 175.10 145.94 118.36 92.17 67.35 43.75 21.33 0
p1+p2 760 760.03 759.93 759.97 759.96 759.93 759.91 759.94 760.03 759.99 760.07 760.06 759.98 760.10 759.96 760.06 759.93 760.07 760.03 760.05 760
T xy diagram (P =1 atm ) 85
75
V apor
o
T( C)
80
70 Liquid 65 60 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
M ole fraction of chloroform
6-48
0.8
0.9
1
6.69 (cont’d) d. Txy diagram (P=1 atm) 85
T(oC)
80
yc
xc
75 70 x
65
y
60 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
Mole fraction of choloroform
'T Tactual Raoult’s law: Tbp = 71o C, y = 0.58 'y y actual
71 75.3 u 100% 5.7% error in Tbp 75.3 0.58 0.60 u 100% 3.33% error in y 0.60
Benzene and chloroform are not structurally similar compounds (as are, for example, pentane and hexane or benzene and toluene). There is consequently no reason to expect Raoult’s law to be valid for chloroform mole fractions that are not very close to 1. 6.70 P | 1 atm 760 mm Hg 760 0.40 u 10
d i b
g d i
x m pm* Tbp 1 x m p *P Tbp
7 .878631411/( Tbp 230 )
0.60 u 10
7 .74416 1437 .686/( Tbp 198.463)
Solve E-Z o T
79.9 o C
We assume (1) the validity of Antoine’s equation and Raoult’s law, (ii) that pressure head and surface tension effects on the boiling point are negligible. The liquid temperature will rise until it reaches 79.9 qC, where boiling will commence. The escaping vapor will be richer in methanol and thus the liquid composition will become richer in ethanol. The increasing fraction of the less volatile component in the residual liquid will cause the boiling temperature to rise.
6-49
6.71 Basis: 1000 kg/h product nH4 (mol H 2 /h) E = C2 H5 OH ( M = 46.05) A = CH 3 CHO ( M = 44.05)
scrubber n3 (mol/h) y A3 (mol A/mol), sat'd y E3 (mol E/mol), sat'd y H3 (mol H 2 /h) vapor, –40°C
P = 760 mm Hg Fresh feed n0 (mol E/h) nA1 (mol A/h) nE1 (mol E/h) 280°C
reactor
nA2 (mol A/h) nE2 (mol E/h) nH2 (mol H 2/h)
condenser
nC (mol/h) 0.550 A 0.450 E liquid, –40°C
Scrubbed Hydrocarbons nA4 (mol A/h) nE4 (mol E/h)
still
Product 1000 kg/h np (mol/h) 0.97 A 0.03 E
nr (mol/h) 0.05 A 0.95 E
Strategy
Calculate molar flow rate of product n p from mass flow rate and composition
x
Calculate y A3 and y E3 from Raoult’s law: y H3 the still involve fewest unknowns (n c and n r )
x x
a.
d i
x
1 y A3 y E3 . Balances about
UV W
Total mole balance about still n c , n r A balance about still A, E and H 2 balances about scrubber n A4 , n E4 , and n H4 in terms of n 3 . Overall atomic balances on C, H, and O now involve only 2 unknowns ( n 0 , n 3 )
UV W
x
Overall C balance n 0 , n 3 Overall H balance
x x x x
A balance about fresh feed-recycle mixing point n A1 E balance about fresh feed-recycle mixing point n E1 A, E, H 2 balances about condenser n A2 , n E2 , n H2 All desired quantities may now be calculated from known molar flow rates.
Molar flow rate of product M
0.97 M A 0.03 M E
b0.97gb44.05g b0.03gb46.05g
n p
1000 kg 1 kmol h 44.11 kg
22.67 kmol h
b g p b 40q Cg 0.550 p b 40 q Cg
Table B.4 (Antoine)
pA* 40q C * E
Raoult’s law y A3
* A
P
44.11 g mol
38.9 mm Hg 0.343 mm Hg 0.550(38.9) 760
6-50
0.02814 kmol A / kmol
6.71 (cont’d) y E3
b
0.450 p E* 40 q C
g
0.450( 0.343) 2.03 u 10 4 kmol E kmol 760 0.9716 kmol H 2 kmol
P 1 y A3 y E3
y H3
UV W
n r Mole balance about still: n c n p n r n c 22.67 n r A balance about still: 0.550n c 0.97(22.67) 0.05n r n c
29.5 kmol / h recycle 52.1 kmol / h
A balance about scrubber: n A4
n 3 y A3
0.02815n 3
(1)
E balance about scrubber: n E4
n 3 y E3
2.03 u 10 4 n 3
(2)
0.9716n 3
(3)
H 2 balance about scrubber: n H4
n 3 y H3
Overall C balance: n 0 (mol E) 2 mol C h
bn gb2g bn gb2g d0.97n ib2g d0.03n ib2g A4
1 mol E
n 0
E4
p
p
n A 4 n E 4 22.67
(4)
Overall H balance: 6n 0
b gb g b gb g
2n H4 4n A4 6n E4 n p 0.97 4 0.03 6
(5)
Solve (1)–(5) simultaneously (E-Z Solve): n 0
23.3 kmol E / h (fresh feed), n H 4
22.6 kmol H 2 / h (in off - gas)
n 3 = 23.3 kmol / h, n A 4 = 0.66 kmol A / h, n E 4 = 0.0047 kmol E / h
A balance about feed mixing point: n A1
0.05n r
E balance about feed mixing point: n E1
n 0 0.95n r
E balance about condenser: n E2
1.47 kmol A h
n 3 y E3 0.450n c
51.33 kmol E h 23.47 kmol E h
Ideal gas equation of state: . 5133 . g kmol 22.4 m bSTP g b273 + 280gK b147 3
Vreactor feed b. Overall conversion
h n 0 0.03n p
Single-pass conversion
1 kmol u 100%
n 0 n E1 n E2 u 100% n E1
b gb
23.33 0.03 22.67
n E4
0.0047 kmol E h
6-51
2396 m 3 h
g u 100%
23.33 513 . 235 . u 100% 51.3
Feed rate of A to scrubber: n A4 = 0.66 kmol A / h Feed rate of E to scrubber:
273K
54%
97%
6.72 a. G = dry natural gas, W = water n 3 (lb - mole G / d) n 4 (lb - mole W / d) 10 lb m W / 10 6 SCF gas 90 o F, 500 psia
Absorber n 7 (lb - mole W / d)
FG lb - mole TEG IJ H d K F lb - mole W IJ n G H d K
4.0 u 10 6 SCF / d 4 u 80 = 320 lb m W / d n1 (lb - mole G / d) n 2 [lb - mole W(v) / d]
n 5
Distillation Column
6
FG lb- mole TEGIJ H d K F lb - mole WIJ n G H d K n5 8
Overall system D. F. analysis:
5 unknowns (n1 , n 2 , n 3 , n 4 , n 7 ) 2 feed specifications (total flow rate, flow rate of water) 1 water content of dried gas 2 balances (W, G) 0 D. F.
320 lb m W 1 lb - mole
Water feed rate: n 2
d
18.0 lb m
17.78 lb - moles W / d
Dry gas feed rate: 4.0 u 106 SCF 1 lb - mole lb - moles W . u 104 lb - moles G / d 17.78 1112 n1 d d 359 SCF Overall G balance: n1
n 3 n 3
. 1112 u 10 4 lb - moles G / d
Flow rate of water in dried gas: n 4
(n 3 n 4 ) lb - moles d n3 1.112 u10 4
o n 4
359 SCF gas 10 lb m W 1 lb - mole W lb - mole
10 6 SCF
18.0 lb m
2.218 lb - mole W(l) / d
Overall W balance: n 7
(17.78 2.218) lb - moles W
18.0 lb m
d
1 lb - mole
6-52
280
lb m W u d
F 1 ft I GH 62.4 lb JK 3
m
4.5
ft 3 W d
6.72 (cont’d) b. Mole fraction of water in dried gas = n 4 n 3 n 4
yw
2.218 lb - moles W / d (2.218 + 1.112 u 10 4 ) lb - moles / d
. u 10 4 199
lb - moles W(v) lb - mole
Henry’s law: ywP = Hwxw ( x w ) max
(199 . u 10 4 )(500 psia)(1 atm / 14.7 psia) 0.398 atm / mole fraction
0.0170
lb - mole dissolved W lb - mole solution
c. Solvent/solute mole ratio 37 lb m TEG 1 lb - mole TEG 18.0 lb m W lb - mole TEG 4.434 lb m W 150.2 lb m TEG 1 lb m W lb - mole W absorbed
n5 n2 n4
n5 4.434(17.78 2.22) 69.0 lb - moles TEG / d xw = 0.80(0.0170) = 0.0136
n6 lb - mole W 5 69.0 . lb - mole W/ d = n o n6 0951 lb - mole n5 n6
Solvent stream entering absorber m
0.951 lb - moles W 18.0 lb m 69.0 lb - moles TEG d lb - mole d
150.2 lb m lb - mole
= 1.04 u 104 lb m / d W balance on absorber . 095 . 222 . ) lb - moles W/ d = 16.51 lb - moles W/ d n8 (1778 1651 . lb - moles W / d xw 019 . lb - mole W / lb - mole (16.51 + 69.9) lb - moles / d
c. The distillation column recovers the solvent for subsequent re-use in the absorber. 6.73 Basis: Given feed rates G2 G1 G3 100 mol/h 200 mol air/h n1 (mol/h) 0.999 H 2 0.96 H2 0.04 H2 S, sat'd 0.001 H 2S 1.8 atm absorber stripper 40°C L2 0°C L1 (mol/h) (mol/h) n3 n4 0.002 H 2S x 3 (mol H 2 S/mol) (1 – x 3) (mol solvent/mol) 0.998 solvent 0°C heater
6-53
G4 200 mol air/h n2 mol H 2S/mol 0.40°C, 1 at m
n3 (mol/h) x 3 (mol H 2S/mol) (1 – x 3) (mol solvent/mol) 40°C
6.73 (cont’d) Equilibrium condition: At G1, p H 2S p H 2S
x3
b0.04gb18. atmg
0.072 atm 27 atm mol fraction
H H 2S
0.072 atm
2.67 u 10 3 mole H 2 S mole
Strategy: Overall H 2 and H 2 S balances n1 , n 2 n 2 air flow rate volumetric flow rate at G4 H 2 S and solvent balances around absorber n 3 , n 4 0.998n 4 solvent flow rate
b100gb0.96g 0.999n n Overall H S balance: b100gb0.04g 0.001n n Overall H 2 balance:
1
2
961 . mol h
1
1
n1 96.1
n 2
2
3.90 mol H 2 S h
Volumetric flow rate at stripper outlet VG4
b200 + 3.90gmol h
b g b273 40gK
22.4 liters STP 1 mol
273 K
5240 L hr
H 2 S and solvent balances around absorber:
b100gb0.04g 0.002n 0.001n 0.998n n d1 2.67 u 10 i 4
4
1
3
3
Solvent flow rate
0.998n 4
n 3 x 3 n 4
1335 . n 3 1952
5820 mol solvent h
6.74 Basis: 100 g H 2 O Sat'd solution @ 60°C 100 g H 2 O 16.4 g NaHCO 3
Sat'd solution @ 30°C 100 g H 2 O 11.1 g NaHCO3 ms (g NaHCO3 ( s))
bg
NaHCO 3 balance 16.4 111 . ms ms
% crystallization
5.3 g NaHCO 3 s
5.3 g crystallized u 100% 16.4 g fed
32.3%
6.75 Basis: 1275 kg/h feed solution m1 (kg H2 O(v )/h)
875 kg/h x 0 (kg KOH/kg) (1 – x0) (kg H 2O/kg)
Sat'd solution 10°C m2 (kg H2 O(1)/h) 1.03 m2 (kg KOH/h)
m3 (kg KOH-2H 2O( s)/h) 60% of KOH in feed
6-54
|UV n W|
3
| n 4
5830 mol h
6.75 (cont’d) Analysis of feed: 2KOH H 2 SO 4 o K 2 SO 4 2H 2 O x0
bg
22.4 mL H 2 SO 4 l 1L 0.85 mol H 2 SO 4 3 5 g feed soln 10 mL L 0.427 g KOH g feed
b
gb g
60% recovery: 875 0.427 0.60 h
b
737.2 kg KOH 2H 2 O h 288.3 kg H 2 O h
56.11 kg KOH
b g
KOH balance: 0.503 1275
56.11 g KOH 1 mol KOH
224.2 g KOH h
448.9 kg KOH 92.15 kg KOH 2H 2 O
m3
2 mol KOH 1 mol H 2 SO 4
448.9 0.97m2 m2
b
g
198.4 kg h
g
Total mass balance: 875 737.2 197 . 198.4 m1 m1
147.0 kg H 2 O h evaporated
6.76 a. 45 R 0 30 g A dissolved CA 0 0.200 0.300 mL solution Plot CA vs. R CA = R / 150
CA
b. Mass of solution:
500 mol 1.10 g
550 g (160 g A, 390 g S) ml The initial solution is saturated at 10.2 qC. 160 g A 0.410 g A g S 410 . g A 100 g S @ 10.2q C Solubility @ 10.2 qC 390 g S 17.5 150 g A 1 mL soln At 0qC, R 17.5 CA 0.106 g A g soln mL soln 110 . g soln Thus 1 g of solution saturated at 0qC contains 0.106 g A & 0.894 g S. 0106 . gA Solubility @ 0qC 0118 . g A g S 118 . g A 100 g S @ 0q C 0.894 g S 390 g S 11.8 g A Mass of solid A: 160 g A 114 g A s 100 g S
bg
c.
g A remaining in soln
g A initial
0.5 u 390 g S 11.8 g A 160 114 g A 100 g S
b
g
bg
23.0 g A s
6.77 a. Table 6.5-1 shows that at 50oF (10.0oF), the salt that crystallizes is MgSO 4 7 H 2 O , which contains 48.8 wt% MgSO4. b. Basis: 1000 kg crystals/h. m 0 (g/h) sat’d solution @ 130oF
m 1 (g/h) sat’d solution @ 50oF 0.23 g MgSO4/g 0.77 g H2O/g
0.35 g MgSO4/g 0.65 g H2O/g
1000 kg MgSO4∙7H2O(s)/h
6-55
6.77 (cont’d) m 0 Mass balance: m 0 m 1 1000 kg / h MgSO 4 balance: 0.35m0 0.23m 1 0.488(1000) kg MgSO 4 / h m 1 The crystals would yield 0.488 u 1000 kg / h = 488 6.78
2150 kg feed / h 1150 kg soln / h
kg anhydrous MgSO 4 h
Basis: 1 lbm feed solution. Figure 6.5-1 a saturated KNO3 solution at 25oC contains 40 g KNO3/100 g H2O x KNO 3
40 g KNO 3 (40 + 100) g solution
0.286 g KNO 3 / g = 0.286 lb m KNO 3 / lb m x
1 lbm solution @ 80oC 0.50 lbm KNO3/lbm 0.50 lbm H2O/lbm
Mass balance: 1 lb m m1 m2 KNO3 balance: 0.50 lb m KNO3
m1(lbm) sat’d solution @ 25oC 0.286 lbm KNO3/lbm soln 0.714 lbm H2O/lbm soln m2 [lbm KNO3(s)]
0.286m1 m2
m1 = 0.700 lb m solution / lb m feed m2 0.300 lb m crystals / lb m feed
0.300 lb m crystals / lb m feed = 0.429 lb m crystals / lb m solution 0.700 lb m solution / lb m feed
Solid / liquid mass ratio =
6.79 a. Basis: 1000 kg NaCl(s)/h. Figure 6.5-1 a saturated NaCl solution at 80oC contains 39 g NaCl/100 g H2O x NaCl
39 g NaCl (39 + 100) g solution
0.281 g NaCl / g = 0.281 kg NaCl / kg m 2 [kg H 2 O(v) / h] m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
m 0 (kg/h) solution 0.100 kg NaCl/kg 0.900 kg H2O/kg
0 m 1 m 2 Mass balance: m 1 m 2 NaCl balance: 0.100 kg NaCl 0281 . m Solid / liquid mass ratio =
1 = 0.700 lb m solution / lb m feed m 2 0.300 lb m crystals / lbm feed m
0.300 lb m crystals / lbm feed = 0.429 lb m crystals / lb m solution 0.700 lbm solution / lbm feed
The minimum feed rate would be that for which all of the water in the feed evaporates to produce solid NaCl at the specified rate. In this case
6-56
6.79 (cont’d)
0 ) min 0100 . (m
1000 kg NaCl / h (m 0 ) min
Evaporation rate: m 2
10,000 kg / min
9000 kg H 2 O / h
Exit solution flow rate: m 1
0
b.
m 2 [kg H 2 O(v) / h] m 1 (kg/h) sat’d solution @ 80oC 0.281 kg NaCl/kg soln 0.719 kg H2O/kg soln 1000 kg NaCl(s)/h
m 0 (kg/h) solution 0.100 kg NaCl/kg 0.900 kg H2O/kg
40% solids content in slurry 1000
0 NaCl balance: 0.100m 0 Mass balance: m
kg NaCl = 0.400( m 1 ) max ( m 1 ) max h
0 0281 . (2500) m
2 m 2 2500 m
2500
kg h
7025 kg / h
4525 kg H2 O evaporate / h
6.80 Basis: 1000 kg K 2 Cr2 O 7 (s) h . Let K = K 2 Cr2 O 7 , A = dry air, S = solution, W = water. Composition of saturated solution:
0.20 kg K 0.20 kg K 01667 . kg K kg soln kg W 1+ 0.20 kg soln
b
g
n2 (mol / h) y2 (mol W(v) / mol)
m e [kg W(v) / h)
(1 y2 )(mol A/ mol) 90o C, 1 atm, Tdp 392 . oC
1 (kg / h) m f (kg / h) m
f m r (kg / h) m
CRYSTALLIZERCENTRIFUGE
0.210 kg K/ kg
0.90 kg K(s) / kg 0.10 kg soln / kg
DRYER
1000 kg K(s) / h
0.1667 kg K / kg
0.790 kg W(l) / kg
0.8333 kg W/ kg na (mol A / h)
m r (kg recycle / h) 0.1667 kg K / kg 0.8333 kg W / kg
Dryer outlet gas: y 2 P
b
g
* 39.2q C y 2 pW
f Overall K balance: 0.210m
53.01 mm Hg 760 mm Hg
1000 kg K h m f
6-57
0.0698 mol W mol
4760 kg h feed solution
6.80 (cont’d)
b
gb
1 01667 . 010 . m 1 K balance on dryer: 0.90m
g
1000 kg h m 1 1090 kg h
Mass balance around crystallizer-centrifuge
f m r m
1 m r me m e m
95% solution recycled m r
4760 1090 3670 kg h water evaporated
b0.10 u 1090g kg
h not recycled
95 kg recycled 5 kg not recycled
2070 kg h recycled Water balance on dryer
. gb1090g kg W h b0.8333gb010 18.01 u 10
3
kg mol
0.0698n 2 n 2
7.225 u 104 mol h
Dry air balance on dryer
na
b1 0.0698g7.225 u 10
4
b g
mol 22.4 L STP h 1 mol
6-58
b g
151 . u 10 6 L STP h
6.81. Basis : 100 kg liquid feed. Assume Patm=1 atm n 2w (kmol H 2 O )(sat' d)
100 kg Feed 0.07 kg Na 2 CO 3 / kg 0.93 kg H 2 O / kg
n 2c (kmol CO 2 )
Reactor Reactor
n 2a (kmol Air)
e
70 o C, 3 atm(absolute)
n1 (kmol)
Filtrate
0.70 kmol CO 2 / kmol m3 ( kg NaHCO 3 (s))
0.30 kmol Air / kmol
Filter
R|m (kg solution) U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 4
3
2
m5 (kg) 0.024 kg NaHCO 3 / kg 0.976 kg H 2 O / kg Filter cake m6 (kg) 0.86 kg NaHCO 3 (s) / kg
R|0.14 kg solution U| S|0.024 kg NaHCO / kgV| T0.976 kg H O / kg W 3
2
Degree of freedom analysis: Reactor 6 unknowns (n1, n2, y2w, y2c, m3, m4) –4 atomic species balances (Na, C, O, H) –1 air balance –1 (Raoult's law for water) 0 DF
Filter 2 unknowns –2 balances 0 DF
Na balance on reactor 100 kg 0.07 kg Na 2 CO3 46 kg Na (m3 0.024m4 ) kg NaHCO3 23 kg Na kg 106 kg Na 2 CO3 84 kg NaHCO3 3.038 0.2738(m3 0.024m4 ) (1) Air balance: 0.300 n1
n2 a
( 2)
C balance on reactor : n1 (kmol) 0.700 kmol CO 2 kmol
12 kg C
100 kg 0.07 kg Na 2 CO 3
12 kg C
1 kmol CO 2 kg 106 kg Na 2 CO 3 12 (n2c )(12) (m3 0.024m2 )( ) 8.40n1 0.7924 12n2c 01429 . (m3 0.024m4 ) (3) 84 H balance : 1 2 2 ) (n2 w )(2) ( m3 0.024m4 )( ) 0.976m4 ( ) 18 84 18 ( 4) 10.33 2n2 w 0.01190(m3 0.024m4 ) 0.1084m4 (100)(0.93)(
6-59
6.81(cont'd) O balance (not counting O in the air): 48 16 n1 (0.700)(932) 100 (0.07)( ) 100 (0.93)( ) 106 18 48 16 (n2 w )(16) n2 c ( 32) (m3 0.024m4 )( ) 0.976m4 ( ) 84 18 22.4n1 85.84 16n2 w 32n2 c 0.5714(m3 0.024m4 ) 0.8676m4
(5)
Raoult's Law : yw P
p w* (70 o C)
n2 w
01025 . (n2 w
n2 w n2 w n 2 c n 2 a n2 c n2 a )
233.7 mm Hg (3 * 760) mm Hg (6)
Solve (1)-(6) simultaneously with E-Z solve (need a good set of starting values to converge). n1 0.8086 kmol, n2a 0.2426 kmol air, n2c 0.500 kmol CO 2 , n2w 0.0848 kmol H 2 O(v), m3 8.874 kg NaHCO 3 (s), m4 92.50 kg solution
NaHCO3 balance on filter: m3 0.024m4
0.024m5 m6 [0.86 (014 . )(0.024)]
m3 8.874
11.09
0.024m 5 0.8634m 6
(7)
m4 92 .50
Mass Balance on filter: 8.874 92.50 1014 . Solve (7) & (8) Scale factor
m5 m6
91.09 kg filtrate 10.31 kg filter cake
500 kg / h 8.867 kg
m5 m6
(8)
(0.86)(10.31) 8.867 kg NaHCO 3 (s)
56.39 h 1
(a) Gas stream leaving reactor
R| |S || T
46.7kmol / h n 2w (0.0848)(56.39) 4.78 kmol H 2 O(v) / h 0.102 kmol H 2 O(v) / kmol n 2c (0.500)(56.39) 28.2 kmol O 2 / h 0.604 kmol CO 2 / kmol n 2a (0.2426)(56.39) 13.7 kmol air / h 0.293 kmol Air / kmol
U| V| W
V2
n 2 RT P
(b) Gas feed rate: V1
(46.7 kmol / h)(0.08206 3 atm
m 3 atm )(343 K) kmol K
438 m 3 / h
56.39 u 0.8086 kmol 22.4 m 3 (STP) 1h 17.0 SCMM h kmol 60 min
6-60
6.81(cont'd) (c) Liquid feed: (100)(56.39) 5640 kg / h To calculate V , we would need to know the density of a 7 wt% aqueous Na2CO3 solution. (d) If T dropped in the filter, more solid NaHCO3 would be recovered and the residual solution would contain less than 2.4% NaHCO3. (e) Henry's law
Benefit: Higher pressure greater pCO2
higher concentration of CO 2 in solution
higher rate of reaction smaller reactor needed to get the same conversion lower cost Penalty: Higher pressure greater cost of compressing the gas (purchase cost of compressor, power consumption) 6.82 600 lb m / h
Dissolution Dissolution 0.90 MgSO4 7H 2 O Dissolution Tank Tank Tank 0.10 I 1 (lb m H 2 O / h) m
Filter I
R|m (lb soln / h) U S|0.32 kg MgSO / kg T0.68 kg H O / kg W 2
4
4
6000 lb m I / h
6 (lb m / h) m
o
2
3 (lb m so ln/ h) m 0.32 MgSO 4
110 F
0.23 lb m MgSO 4 / lb m
R|300 lb soln / h|U S|0.32 MgSO V| T0.68 H O W m
m
2
6000 lb m I / h
0.68 H 2 O
0.77 lb m H 2 O / lb m 4 (lb m MgSO 4 7 H 2 O / h m
Filter II
R|m (lb |S0.23 lb ||0.77 lb T 5
m
soln) MgSO 4 / lb m
m
H 2 O / lb m
m
U| |V || W
Crystallizer
4 (lb m MgSO 4 7 H 2 O) m
R|0.05m S|0.23 lb T0.77 lb
(lb m soln)
4 m
MgSO 4 / lb m
m
H 2 O / lb m
U| V| W
a. Heating the solution dissolves all MgSO4; filtering removes I, and cooling recrystallizes MgSO4 enabling subsequent recovery. (b) Strategy: Do D.F analysis.
6-61
6.82(cont'd)
UV W
Overall mass balance m 1 , m 4 Overall MgSO 4 balance
( MW) MgSO4
UV W
Diss. tank overall mass balance m 2 , m 6 Diss. tank MgSO 4 balance
(24.31 32.06 64.00) 120.37, ( MW) MgSO4 7H2O
(12037 . 7 * 18.01) 246.44
Overall MgSO4 balance: 60,000 lb m 0.90 lb m MgSO 4 7H 2 O 120.37 lb m MgSO 4 h lb m 246.44 lb m MgSO 4 7H 2 O (300 lb m / h)(0.32 lb m MgSO 4 / lb m ) m 4 (120.37 / 246.44) 0.05m 4 (0.23) m 4
5.257 x10 4 lb m crystals / h m 4 5.257 x104 lb m / h
Overall mass balance: 60,000 m 1
6300 105 . m 4
m 1
1494 lb m H 2 O / h
c. Diss. tank overall mass balance: Diss. tank MgSO 4 balance:
60,000 m 1 m 6 m 2 6000 54,000(120.37 / 246.44) 0.23m 6
m 2
1512 . x10 5 lb m / h
m 6
9.575x10 4 lb m / h recycle
Recycle/fresh feed ratio
9.575x10 4 lb m / h 1494 lb m / h
0.32m 2
UV W
64 lb m recycle / lb m fresh feed
6.83 a. n 1 (kmol CO 2 / h) Cryst Filter
1000 kg H 2SO 4 / h (10 wt%) 1000 kg HNO 3 / h w (kg H 2 O / h) m
2 (kg CaSO4 / h) m 3 (kg Ca(NO3 )2 / h) m
Filter cake 5 (kg / h) m 0.96 kg CaSO 4 (s) / kg
4 (kg H2O/ h) m
0.04 kg soln / kg
0 (kg CaCO 3 / h) m 0 (kg solution / h) 2m
0 (kg CaCO 3 / h) m
8 (kg soln / h) m
0 (kg solution / h) 2m
R| X (kg CaSO / kg) U| 500 X (kg H O / kg) Solution composition: S |T(1 501X )(kg Ca(NO ) / kg)V|W a
4
a
a
6-62
2
3 2
6.83 (cont’d) b. Acid is corrosive to pipes and other equipment in waste water treatment plant. c. Acid feed:
1000 kg H 2 SO 4 / h (2000 m w ) kg / h
0.10 m w
8000 kg H 2 O / h
Overall S balance: 1000 kg H 2 SO 4 h
32 kg S
5 (kg / h) (0.96 0.04 X a ) (kg CaSO 4 ) m
98 kg H 2 SO4 m8 (kg / h) X a (kg CaSO 4 ) kg
32 kg S
kg
136 kg CaSO 4
32 kg S 136 kg CaSO 4
5 (0.96 0.04 X a ) 0.2353m 8 Xa 3265 . 0.2353m
(1)
Overall N balance: 1000 kg HNO 3
14 kg N
h
63 kg HNO 3
0.04m 5 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) kg
8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) m kg
28 kg N 164 kg Ca(NO 3 ) 2
28 kg N 164 kg Ca(NO 3 ) 2
222.2 0.00683m 5 (1 501X a ) 0171 . m 8 (1 501X a )
(2)
Overall Ca balance: 5 (kg / h) (0.96 0.04X a ) (kg CaSO 4 ) 40 kg Ca m 40 kg Ca 100 kg CaCO 3 kg 136 kg CaSO 4 5 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) 0.04m 40 kg Ca kg 164 kg Ca(NO 3 ) 2 8 (kg / h) X a (kg CaSO 4 ) m 40 kg Ca kg 136 kg CaSO 4 8 (kg / h) (1 501X a ) (kg Ca(NO 3 ) 2 ) m 40 kg Ca kg 164 kg Ca(NO 3 ) 2
0 (kg / h) m
0.40m 0
0.294m 5 (0.96 0.04 X a ) 0.00976m 5 (1 501 X a ) 0.294m 8 X a 0.244m 8 (1 501 X a )
(3)
Overall C balance : 0 (kg / h) m 0.01m 0
12 kg C 100 kg CaCO 3 n1
n 1 (kmol CO 2 / h)
(4)
6-63
1 kmol C 1 kmol CO 2
12 kg C 1 kmol C
6.83 (cont’d) Overall H balance : 1000 (kg H 2SO4 )
2 kg H
h
98 kg H 2 SO4
1000 kg HNO3
1 kg H
h
w (kg / h) m
2 kg H
63 kg HNO3 18 kg H 2 O 5 (kg / h) 500 X a (kg H 2 O) 2 kg H 8 (kg / h) 500 X a (kg H 2 O) 2 kg H m 0.04m kg 18 kg H 2 O kg 18 kg H 2 O 5 X a 5556 8 Xa 92517 2.22m . . m
(5)
Solve eqns. (1)-(5) simultaneously, using E-Z Solve. m 0
1812.5 kg CaCO 3 (s) / h,
m 5
n1
18.1 kmol CO 2 / h(v),
Xa
Recycle stream
2 * m 0
1428.1 kg / h,
m 8
9584.9 kg soln / h,
0.00173 kg CaSO 4 / kg
3625 kg soln / h
CaSO . R| 0.00173(kg CaSO / kg) U| R||0173% S| 500 * 0.00173(kg H O / kg) V| S|86.5% H O T(1 501* 0.00173)(kg Ca(NO ) / kg)W |T13.3% Ca(NO ) 4
4 2
2
3 2
d.
3 2
From Table B.1, for CO2:
Pc 72.9 atm 304.2 K , T (40 273.2) K Tr 103 . , Tc 304.2
U| |V || W
Tc
Pr
30 atm 72.9 atm
0.411
From generalized compressibility chart (Fig. 5.4-2): z
0.86 V
0.86 0.08206 L atm 313.2 K L 0.737 mol K 30 atm mol CO 2
zRT P
Volumetric flow rate of CO2: V
e.
n1 * V
18.1 kmol CO 2
0.737 L
1000 mol
h
mol CO 2
1 kmol
1.33x10 4 L / h
Solution saturated with Ca(NO3)2:
1 501X a (kg Ca(NO 3 ) 2 / kg) 500Xa (kg H 2 O / kg)
1.526 X a
0.00079 kg CaSO 4 / kg
Let m 1 (kg HNO3/h) = feed rate of nitric acid corresponding to saturation without crystallization.
6-64
6.83 (cont’d) Overall S balance: 5 (kg / h) (0.96 (0.04)(0.00079)) (kg CaSO4 ) m
1000 kg H 2SO4
32 kg S
h
98 kg H2SO4
kg 8 (kg / h) 0.00079 (kg CaSO4 ) m kg
32 kg S 136 kg CaSO4
32 kg S 136 kg CaSO4
5 0.000186m 8 326.5 0.226m
(1')
Overall N balance: 1 (kg HNO3 ) m
14 kg N
h
63kg HNO3
5 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 ) 2 ) 0.04m kg
28 kg N 164 kg Ca(NO3 ) 2
8 (kg / h) (1 (501)(0.00079)) (kg Ca(NO3 )2 ) m kg
28 kg N 164 kg Ca(NO3 ) 2
1 000413 5 0103 8 0222 m . m . . m
(2')
Overall H balance: 1 kg HNO 3 m 1000 (kg H 2 SO 4 ) 2 kg H 1 kg H h 98 kg H 2 SO 4 h 63 kg HNO 3 5 (kg / h) 500(0.00079) (kg H 2 O) 8000 (kg / h) 2 kg H 0.04m 2 kg H 18 kg H 2 O kg 18 kg H 2 O 8 (kg / h) 500(0.00079) (kg H 2 O) m 2 kg H kg 18 kg H 2 O 1 0.00175m 5 0.0439m 8 909.30 0.0159m Solve eqns (1')-(3') simultaneously using E-Z solve: m 1
1155 . x10 4 kg / h;
5 m
x10 3 kg / h; 1424 .
Maximum ratio of nitric acid to sulfuric acid in the feed . x10 4 kg / h 1155 1000 kg / h
. kg HNO 3 / kg H 2 SO 4 115
6-65
8 m
2.484 x10 4 kg / h
(3' )
6.84 Moles of diphenyl (DP): Moles of benzene (B):
0.363 619 . 0.363
x DP
bg
bg
p B* T 'Tm
(1 x DP ) p B* T
'Tbp
'Tbp
o
m
55 . 3.6 19 . qC
o
801 82.0 q C . 185 .
RTbp2 'H mp
Tbp
2
30,765
0.0q C, 'Tfp
114.0 mm Hg
2
v
6.85 Tfp
g
g b0.0554g 3.6 K = 3.6 C T 9837 8.314b273.2 801 .g . K = 1.85 C b0.0554g 185
RTb02 x DP 'H Tb
b
0.945 120.67 mm Hg
b
U| |V 619 . mol | |W
0.0544 mol DP mol
8.314 273.2 55 .
2 RTm0 x DP 'H m
'Tbp
56.0 g 0.363 mol 154.2 g mol 550.0 ml 0.879 g 1 mol ml 78.11 g
4.6q C
xu
b
g b0.0445g
8.314 273.2 100.0 40,656
2
13 . qC
100.0 13 . 101.3 q C
30 . q C xu
'Tbp 'H mp RTbp2
b
g 8.314b373.2g 3.0 40,656
2
mol urea mol 0105 .
Initial mass of urea:
1000 g soln
1 mol soln 0.0445 mol U 60.06 g U 60.06 0.0445 18.02 0.9555 g soln mol soln mol U
b
g
b
g
134 g urea
Moles of H 2 O:
b1000 134g g
Final moles of urea:
Added Urea:
1 mol 18.02 g
m m 48.0
5.63 mol 60.06 g mol
48.0 mol H 2 O
m 5.63 mol urea 0105 . 134 g
204 g urea
6-66
b0.5150 gg b1101. g molg b0.5150 gg b1101. g molg b100.0 gg b94.10 g molg
6.86 x aI
RTm20 'TmI xs 'TmII 'H m
'Tm
x sII
b1 0.00523g mol solvent
b g
b
0.49q C 0.41q C
'H vI B , ln p s* Tbs RTb 0
b g
6380 J mol
b g
ln Ps* Tb 0 ln P0* Tbs * s
FG H
* 0
bs
6.88 m 1 (g styrene) 90 g ethylbenezene 100 g EG 90 g ethylbenzene 30 g styrene
Styrene balance: m1 m2
m 2 (g styrene) 100 g EG
30 g styrene
m2 100 m2
6.38 kJ / mol
'H v 1 'H v Tbs Tb 0 1 # R Tb 0 Tbs R Tb20
b0
Equilibrium relation:
mol solute mol solution
'H vII B RTbs
IJ K p bT g b1 x g p bT g lnb1 x g | x
b g
0.00523
2
Assume 'H vI # 'H vII ; T0 Ts # T02
b. Raoult’s Law:
0.00438
'TmI
g b0.00523g
0.49
'TmII
x sI
8.314 273.2 5.00
RTm20 xs 'Tm
ln p s* Tb 0
x sII
94.10 g solvent 0.4460 g solute 8350 . g solute mol 1 mol solvent 95.60 g solvent
0.00523 mol solute
'H m 6.87 a.
x sI
0.00438 mol solute mol
. 019
FG m IJ H 90 m K 1
1
solve simultaneously
m1
25.6 g styrene in ethylbenzene phase
m2
4.4 g styrene in ethylene glycol phase
6-67
'H v 'Tb RTb20
'Tb
RTb20 x 'H v
6.89 Basis: 100 kg/h.
A=oleic acid; C=condensed oil; P=propane
100 kg / h 0.05 kg A / kg 0.95 kg C / kg
95.0 kg C / h 2 kg A / h m 3 kg A / h m 1 kg P / h m
1 kg P / h m
a. 90% extraction: m 3
(0.09)(0.05)(100 kg / h) = 4.5 kg A / h
Balance on oleic acid: (0.05)(100) Equilibrium condition:
015 .
m 2 4.5 kg A / h m 2
0.5 / (n1 0.5) n1 4.5 / (4.5 95)
0.5 kg A / h
73.2 kg P / h
b. Operating pressure must be above the vapor pressure of propane at T=85oC=185oF * 500 psi 34 atm Figure 6.1-4 p propane c. Other less volatile hydrocarbons cost more and/or impose greater health or environmental hazards. 6.90 a. Benzene is the solvent of choice. It holds a greater amount of acetic acid for a given mass fraction of acetic acid in water. Basis: 100 kg feed.
A=Acetic acid, W=H2O, H=Hexane, B=Benzene
100 (kg)
m1 (kg)
0.30 kg A / kg
0.10 kg A / kg
0.70 kg W / kg
0.90 kg W / kg m2 (kg A)
m H (kg H)
m H (kg H) or m B (kg B)
or m B (kg B)
Balance on W:
100 * 0.70 m1 * 0.90 m1
Balance on A:
100 * 0.30 m2 77.8 * 0.10 m2
77.8 kg 22.2 kg
Equilibrium for H: KH
m2 / (m2 m H ) xA
22.2 / (22.2 m H ) 0.10
0.017 m H
22.2 / (22.2 mB ) 0.10
0.098 m B
1.30 x10 4 kg H
Equilibrium for B: KB
m2 / (m2 m B ) xA
2.20 x10 3 kg B
(b) Other factors in picking solvent include cost, solvent volatility, and health, safety, and environmental considerations.
6-68
6.91
a. Basis: 100 g feed 40 g acetone, 60 g H 2 O. A = acetone, H = n - C 6 H 14 , W = water 40 g A 60 g W
e 1 (g A) 60 g W
25°C
100 g H r 1 (g A)
100 g H
25°C
75 g H
b
xA in H phase / xA in W phase
0.343 x mass fraction
Balance on A stage 1: Equilibrium condition stage 1: Balance on A stage 2: Equilibrium condition stage 2:
75 g H r 2 (g A)
g
40 e1 r1 r1 100 r1 e1
b g b60 e g
U| e 0.343V| r W U| r 0.343V| e W
1
1
1
27.8 e2 r2 r2 75 r2 e2
b g b60 e g
2
2
20.6 g A remaining u 100% 40 g A fed
% acetone not extracted
e 2 (g A) 60 g W
2
27.8 g acetone 12.2 g acetone 7.2 g acetone 20.6 g acetone
515% .
b. 40 g A 60 g W
e1 g A 60 g W
r1 g A 175 g H
175 g H
Balance on A stage 1: Equilibrium condition stage 1: % acetone not extracted c.
U| r b g 0.343V e |W b60 e g
40.0 e1 r1 r1 175 r1 e1
1
1
22.2 g A remaining u 100% 40 g A fed
40 g A 60 g W
20.6 / (m 20.6) 19.4 / (60 19.4)
17.8 g acetone 22.2g acetone
555% .
19.4 g A 60 g W 20.6 g A m (g H)
m (g H)
Equilibrium condition:
1
0.343 m
225 g hexane
d. Define a function F=(value of recovered acetone over process lifetime)-(cost of hexane over process lifetime) – (cost of an equilibrium stage x number of stages). The most costeffective process is the one for which F is the highest.
6-69
6.92
a. P--penicillin; Ac--acid solution; BA--butyl acetate; Alk--alkaline solution
Broth Mixing tank
100 kg 0.015 P 0.985 Ac
m1 (kg BA) Extraction Unit I
Acid
D.F. analysis: Extraction Unit I 3 unknown (m1, m2p, m3p) –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
m3P (kg P) 98.5 (kg Ac) pH=2.1 m4 (kg Alk)
Extraction II m6P (kg P) m1 (kg BA)
m5P (kg P) m4 (kg Alk) pH=5.8
Extraction Unit II (consider m1, m3p) 3 unknowns –1 balance (P) –1 distribution coefficient –1 (90% transfer) 0 DF
b. In Unit I, 90% transfer m3 P 0.90(15 . ) 135 . kg P P balance: 15 . m2 P 1.35 m2 P 015 . kg P . / (135 . m1 ) 135 m1 34.16 kg BA pH=2.1 K 25.0 . / (015 . 98.5) 015 In Unit II, 90% transfer: m5 P 0.90(m3 P ) 1.215 kg P m3 P 1215 P balance: . m6 P m6 P 0.135 kg P m6 P / (m6 P 34.16) pH=5.8 K 0.10 m4 29.65 kg Alk 1.215 / (1.215 m4 ) m1 34.16 kg BA 0.3416 kg butyl acetate / kg acidified broth 100 100 kg broth m4 29.65 kg Alk 0.2965 kg alkaline solution / kg acidified broth 100 100 kg broth Mass fraction of P in the product solution: m5 P 1215 . P xP m4 m5 P (29.65 + 1.215) kg
0.394 kg P / kg
c. (i). The first transfer (low pH) separates most of the P from the other broth constituents, which are not soluble in butyl acetate. The second transfer (high pH) moves the penicillin back into an aqueous phase without the broth impurities. (ii). Low pH favors transfer to the organic phase, and high pH favors transfer back to the aqueous phase. (iii).The penicillin always moves from the raffinate solvent to the extract solvent.
6-70
6.93
W = water, A = acetone, M = methyl isobutyl ketone xW xA xM
0.20 0.33 0.47
U| V| W
Figure 6.6-1
Phase 1: x W
0.07, x A
0.35, x M
0.58
Phase 2: x W
0.71, x A
0.25, x M
0.04
Basis: 1.2 kg of original mixture, m1=total mass in phase 1, m2=total mass in phase 2. H 2 O Balance: Acetone balance:
R| S| T
m1 0.95 kg in MIBK - rich phase 12 . * 0.20 0.07m1 0.71m2 m2 0.24 kg in water - rich phase 12 . * 0.33 0.35m1 0.25m2
6.94 Basis: Given feeds: A = acetone, W = H2O, M=MIBK Overall system composition:
U| b g V 3500 g b20 wt% A, 80 wt% M g 700 g A, 2800 g M |W 2200 g A U | 3500 g WV 25.9% A, 41.2% W, 32.9% M 2800 g M |W
5000 g 30 wt% A, 70 wt% W 1500 g A, 3500 g W
Phase 1: 31% A, 63% M, 6% W
Fig. 6.6-1
Phase 2: 21% A, 3% M, 76% W
Let m1=total mass in phase 1, m2=total mass in phase 2. H 2 O Balance: Acetone balance: 6.95
R| S| T
m1 4200 g in MIBK - rich phase 3500 0.06m1 0.76m2 m2 4270 g in water - rich phase 2200 0.31m1 0.21m2
A=acetone, W = H2O, M=MIBK 41.0 lb m / h
32 lb m / h
x A,1 , x W,1 , 0.70
x AF (lb m A / lb m ) x WF (lb m W / lb m )
2 lb m / h m x A ,2 , x W ,2 , x M ,2
1 (lb m M / h) m
Figure 6.6-1 Phase 1: x M Phase 2: x w ,2 Overall mass balance: MIBK balance:
0.700 x w ,1
0.05; x A,1
0.25 ;
0.81; x A ,2
0.81; x M ,2
0.03
UV W
m 1 32.0 lb m / h m 1 410 . lb m h m 2 m 1 410 m 2 . * 0.7 m 2 * 0.03
6-71
281 . lb m MIBK / h 19.1 lb m h
6.96 a. Basis: 100 kg; A=acetone, W=water, M=MIBK System 1: x a,org = 0.375 mol A, x m,org = 0.550 mol M, x w,org = 0.075 mol W x a,aq = 0.275 mol A, x m,aq = 0.050 mol M, x w,aq = 0.675 mol W
UV W
maq,1 maq ,1 morg ,1 100 Mass balance: Acetone balance: maq ,1 * 0.275 morg ,1 * 0.375 33.33 morg ,1
417 . kg 58.3 kg
System 2: x a,org = 0.100 mol A, x m,org = 0.870 mol M, x w,org = 0.030 mol W x a,aq = 0.055 mol A, x m,aq = 0.020 mol M, x w,aq
= 0.925 mol W
UV W
m aq,2 maq ,2 morg ,2 100 Mass balance: Acetone balance: maq ,2 * 0.055 morg ,2 * 0100 . 9 morg,2 b. K a ,1
x a ,org ,1 x a ,aq ,1
0.375 136 . ; 0.275
K a ,2
x a ,org ,2 x a ,aq ,2
22.2 kg 77.8 kg
0100 . 182 . 0.055
High Ka to extract acetone from water into MIBK; low Ka to extract acetone from MIBK into water. c.
E
x a ,org / x w ,org aw,1
0.375 / 0.075 12.3; E aw,2 0.275 / 0.675
x a ,aq / x w ,aq
If water and MIBK were immiscible, x w ,org d.
0 E
aw
0100 . / 0.040 0.055 / 0.920
418 .
of
Organic phase= extract phase; aqueous phase= raffinate phase
E a ,w
( x a / x w ) org
( x a ) org / ( x a ) aq
( x a / x w ) aq
( x w ) org / ( x w ) aq
Ka Kw
When it is critically important for the raffinate to be as pure (acetone-free) as possible. 6.97 Basis: Given feed rates: A = acetone, W = water, M=MIBK
e 1
e 2
(kg / h)
x1A (kg A / kg)
x 2W (kg W / kg)
x1W (kg W / kg)
x 2M (kg M / kg)
x1M (kg M / kg) 200 kg / h 0.30 kg A / kg 0.70 kg M / kg
r1 Stage I
(kg / h)
x 2A (kg A / kg)
(kg / h)
y1A (kg A / kg) y1W (kg W / kg) y1M (kg M / kg)
300 kg W / h
6-72
Stage II Stage IIS 300 kg W / h
r2
(kg / h)
y 2A (kg A / kg) y 2W (kg W / kg) y 2M (kg M / kg)
6.97(cont'd) Overall composition of feed to Stage 1:
U| V| W
b200gb0.30g
60 kg A h 500 kg h 200 60 140 kg M h 12% A, 28% M, 60% W 300 kg W h
Figure 6.6-1
Extract: x1A Raffinate: y1A
Mass balance Acetone balance:
0.095, x1W
0.880, x1M
0.15, y1W
0.025
0.035, y1M
R| S| T
e1 500 e1 r1 r1 60 0.095e1 015 . r1
0.815
273 kg / h 227 kg / h
Overall composition of feed to Stage 2:
U| 527 kg h . g 34 kg A h b227gb015 b227gb0.815g 185 kg M h V 6.5% A, 35.1% MIBK, 58.4% W b227gb0.035g 300 308 kg W h|W Figure 6.6-1
Extract: x 2 A Raffinate: y 2 A
Mass balance: Acetone balance:
0.04, x 2 W
0.94, x 2 M
0.085, y 2 W
0.025, y 2 M
R| S| T
e2 527 e2 r2 r2 34 0.04e2 0.085r2
Acetone removed: [60 (0.085)(287)] kg A removed / h 60 kg A / h in feed
0.02 0.89
240 kg / h 287 kg / h
0.59 kg acetone removed / kg fed
Combined extract: Overall flow rate = e1 e2
273 240 513 kg / h
( x1 A e1 x 2 A e2 ) kg A
0.095 * 273 0.04 * 240 513
0.069 kg A / kg
Water:
( x1w e1 x 2 w e2 ) kg W e1 e2
0.88 * 273 0.94 * 240 513
0.908 kg W / kg
MIBK:
( x1 M e1 x 2 M e2 ) kg M (e1 e2 ) kg
Acetone:
0.025 * 273 0.02 * 240 513
6-73
0.023 kg M / kg
6.98. a. 1.50 L / min 25o C, 1atm, rh = 25% n0 (mol / min)
M (g gel) Ma (g H2O)
y0 (mol H2 O / mol) (1- y0 ) (mol dry air / mol) n 0
PV RT
(1 atm)(1.50 L / min) (0.08206 L atm / mol K)(298 K)
r.h.=25%
pH2O
0.25
pH* 2O (25o C)
Silica gel saturation condition: Water feed rate: m H
X*
12.5
p H 2O p H* 2 O
0.25 p H* 2 O (25o C )
y0
0.06134 mol / min
p
12.5 * 0.25 3125 .
0.25(23.756 mm Hg) 760 mm Hg
0.06134 mol 0.00781 mol H 2 O 18.01 g H 2 O 2O
min
mol
mol H 2 O
g H 2 O ads 100 g silica gel
0.00781
mol H 2 O mol
0.00863 g H 2 O / min
Adsorption in 2 hours (0.00863 g H 2 O / min)(120min) 1.035 g H 2 O Saturation condition:
1.035 g H 2 O M (g silica gel)
3.125 g H 2 O M 100 g silica gel
33.1 g silica gel
Assume that all entering water vapor is adsorbed throughout the 2 hours and that P and T are constant. b. Humid air is dehumidified by being passed through a column of silica gel, which absorbs a significant fraction of the water in the entering air and relatively little oxygen and nitrogen. The capacity of the gel to absorb water, while large, is not infinite, and eventually the gel reaches its capacity. If air were still fed to the column past this point, no further dehumidification would take place. To keep this situation from occurring, the gel is replaced at or (preferably) before the time when it becomes saturated. 6.99 a.
Let c = CCl4 Relative saturation
0.30
pc * pc (34 o C)
pc
0.30 * (169 mm Hg) 50.7 mm Hg
b. Initial moles of gas in tank: n0
P0V0 RT0
1 atm 50.0 L 1985 . mol 0.08206 L atm / mol K 307 K
Initial moles of CCl4 in tank: nc0
y c 0 n0
pc 0 n0 P0
50.7 mm Hg u 1.985 mol 760 mm Hg
6-74
0.1324 mol CCl 4
6.99 (cont’d) 50% CCl4 adsorbed nc
0.500nc 0 n0 nads
Total moles in tank: n tot
0662 mol CCl 4 (= nads) (1.985 0.0662) mol = 1.919 mol
Pressure in tank. Assume T = T0 and V = V0. n tot RT0 V0
P
nc n tot
yC
. )(0.08206)(307) FG (1919 I F 760 mm Hg IJ atmJ G H K H atm K 50.0
0.0662 mol CCl 4 1919 . mol
pc
nc nc 1853 . n tot
pc
mol CCl 4 mol
0.0345(760 mm Hg) = 26.2 mm Hg n0 nc 0
c. Moles of air in tank: na
yc
0.0345
0.001
nc nair
(1.985 0.1324) mol air = 1.853 mol air
mol CCl 4 nc mol
u 10 3 mol CCl 4 1854 .
mol 1854 .
LM n RT OP = 1854 u 10 mol . 50.0 L N V Q
yc P
735 mm Hg
0.001
3
0
tot
0
0.08206 L atm 307 K 760 mm mol K 1 atm
0.710 mm Hg X*
FG g CCl IJ H g carbon K 4
0.0762 pc X* 1 0.096 pc
Mass of CCl4 adsorbed mads
(nc 0 nc )( MW ) c
0.0762( 0.710) 1 0.096(0.710)
0.001854) mol CCl 4 (01324 .
20.3 mol CCl 4 adsorbed 20.3 g CCl 4 ads Mass of carbon required: mc g CCl 4 ads 0.0506 g carbon a.
X*
E ln X * K F p NO 2
ln(PNO2)
6.100
0.0506
0
E ln K F E ln p NO 2
1
2 ln(X*)
6-75
153.85 g 1 mol CCl 4
400 g carbon
y = 1.406x - 1.965
2 1.5 1 0.5 0 -0.5 -1 -1.5
g CCl 4 adsorbed g carbon
3
6.100 (cont’d) ln X *
X* . ln p NO2 1965 . 1406
.406 e 1.965 p 1NO 2
.406 p 1NO . 0140 2
0.140 (kg NO 2 / 100 kg gel)(mm Hg) 1.406 ; E
KF
1406 .
S * (0.05m) 2 (1m) 10 3 L 0.75 kg gel 5.89 kg gel 1m 3 L Maximum NO2 adsorbed :
b. Mass of silica gel : mg
p NO 2 in feed mads
0.010(760 mm Hg)
0.140(7.60) 1.406 kg NO 2 100 kg gel
7.60 mm Hg 5.89 kg gel
0.143 kg NO 2
Average molecular weight of feed : MW
0.01( MW ) NO2 0.99( MW ) air
(0.01)(46.01) (0.99)(29.0)
29.17
kg kmol
Mass feed rate of NO2: m
8.00 kg
1 kmol
0.01 kmol NO 2
46.01 kg NO 2
h
29.17 kg
kmol
kmol NO 2
Breakthrough time:
tb
0.143 kg NO 2 0.126 kg NO 2 / h
1.13 h
0.126
kg NO 2 h
68 min
c. The first column would start at time 0 and finish at 1.13 h, and would not be available for another run until (1.13+1.50) = 2.63 h. The second column could start at 1.13 h and finish at 2.26 h. Since the first column would still be in the regeneration stage, a third column would be needed to start at 2.26 h. It would run until 3.39 h, at which time the first column would be available for another run. The first few cycles are shown below on a Gantt chart. Run
Regenerate
Column 1 0
1.13
2.63
3.39
4.52
6.02
Column 2 1.13
2.26
3.76
4.52
5.65
Column 3 2.26
6-76
3.39
4.89
5.65
6.78
Let S=sucrose, I=trace impurities, A=activated carbon
Add mA (kg A)
mS (kg S)
mS (kg S) mI (kg I)
m I0 (kg I) R0 (color units / kg S)
R (color units / kg S)
Come to equilibrium
V (L)
V (L) mA (kg A)
mIA (kg I adsorbed)
Assume
x
no sucrose is adsorbed
solution volume (V) is not affected by addition of the carbon m a. R(color units/kg S) kCi (kg I / L) = k I (1) V x
'R = k (Ci 0 Ci )
mIA mI 0 mI
k ( mI 0 m I ) V
'R
kmIA V
'R x100% R0
km IA / V m x100 100 IA kmI 0 / V mI 0 m IA Equilibrium adsorption ratio: X i* mA Normalized percentage color removal:
% removal of color
X
% removal ( 3) 100 m IA / m I 0 = m A / mS m A / mS m X = 100X *i S X i* mI 0
Freundlich isotherm X i* X=
100mS K F E
RE
9.500 9.000
mI 0 X 100mS
E;
8.500 8.000 0.000 1.000 2.000 3.000
6-77
(3) (4)
(5) R KF ( )E k
K F' R E
y = 0.4504x + 8.0718
ln R
(2)
m IA mS m A mI 0
mI 0 X 100mS
(1),( 5)
K F Ci E mI 0 k
100
A plot of ln X vs. ln R should be linear: slope
ln v
6.101
intercept = lnK 'F
6.101 (cont’d) ln X
0.4504 ln p NO2 8.0718 X
K F'
3203, E
X
3203R 0.4504
0.4504
b. 100 kg 48% sucrose solution m S 95% reduction in color
e 8.0718 R 0.4504
480 kg
R = 0.025(20.0) = 0.50 color units / kg sucrose
K F' R E
3203(0.50) 0.4504 2344 % color reduction 97.5 2344 = mA m A / mS m A / 480
6-78
20.0 kg carbon
CHAPTER SEVEN
7.1
0.80 L 35 . u 10 4 kJ 0.30 kJ work 1h 1 kW 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s 2.33 kW 10 3 W 1.341 u 10 3 hp 312 . hp 3.1 hp 1 kW 1W
7.2
All kinetic energy dissipated by friction (a) E k
mu 2 2 5500 lbm 552 miles 2 2 h2 715 Btu
52802 ft 2 12 mile 2
12 h 2 36002 s 2
1 lbf 9.486 u 10 4 B 32.174 lbm ft / s2 0.7376 ft lb f
(b) 3 u 10 8 brakings 715 Btu 1 day 1h 1W 1 MW 4 2 day braking 24 h 3600 s 9.486 u 10 Btu / s 10 6 W
7.3
(a) Emissions: 1000 sacks Paper Plastic
2000 sacks
1000 sacks
(0.0045 0.0146) oz
(724 905) Btu sack
Plastic
2000 sacks
1 lb m
sack 16 oz sack
Energy: Paper
(0.0510 0.0516) oz
(185 464) Btu sack
2617 MW
6.41 lb m
1 lb m 16 oz
2.39 lb m
1.63 u 10 6 Btu 1.30 u 10 6 Btu
(b) For paper (double for plastic)
Materials for 400 sacks
Raw Materials Acquisition and Production
Sack Production and Use
7-1
1000 sacks
Disposal
400 sacks
7.3 (cont’d) Emissions: Paper
400 sacks
Plastic
800 sacks
1 lb m 1000 sacks 0.0510 oz sack 16 oz
1 lb m 0.0516 oz 4.5 lb m sack 16 oz 30% reduction
1 lb m 2000 sacks 0.0045 oz sack 16 oz
1 lb m 0.0146 oz 2.05 lb m sack 16 oz 14% reduction
Energy: Paper
400 sacks
Plastic
(c) .
724 Btu sack
800 sacks
3 u 10 8 persons
185 Btu sack
905 Btu
1000 sacks
sack
119 . u 10 6 Btu; 27% reduction
464 Btu
2000 sacks
sack
1 sack 1 day 1h person - day 24 h 3600 s
1.08 u 10 6 Btu; 17% reduction
649 Btu 1J 1 MW -4 1 sack 9.486 u 10 Btu 10 6 J / s
2,375 MW . (2,375 MW) = 404 MW Savings for recycling: 017 (d) Cost, toxicity, biodegradability, depletion of nonrenewable resources. 7.4
(a) Mass flow rate: m
3.00 gal 1 ft 3 (0.792)(62.43) lb m min 7.4805 gal 1 ft 3
Stream velocity: u
1 3.00 gal 1728 in 3 2 min 7.4805 gal 3 0.5 in 2
Kinetic energy: E k
mu 2 2
b g
0.330 lb m s
d7.70 u 10
3
. g ft b1225 2
2
s
2
3
f
(b) Heat losses in electrical circuits, friction in pump bearings.
7-2
1 ft 1 min . ft s 1225 12 in 60 s
ft lb f 1 1 lb f 7.70 u 103 2 s 2 32.174 lb m ft / s
. u 10 hp I iFGH 01341 .7376 ft lb / sJK
ft lb f / s
1 min 0.330 lb m s 60 s
140 . u 105 hp
7.5
(a) Mass flow rate:
b
g
42.0 m S 0.07 m
m
s 2 mu 2
E k
2
10 3 L 273 K 130 kPa 1 mol 3 573 K 101.3 kPa 22.4 L STP 1m
b g
4
42.0 2 m2 1N 1J 2 2 s 1 kg m / s N m
127.9 g 1 kg 2 s 1000 g
(b)
b g
127.9 g 1 mol 273 K 101.3 kPa 22.4 L STP s 29 g 573 K 130 kPa 1 mol
29 g mol
127.9 g s
113 J s
1 m3 4 3 10 L S (0.07)2 m2
49.32 m s
127.9 g 1 kg 49.32 2 m2 1N 1J 1558 . J/s 2 2 s 1000 g s 1 kg m / s2 N m 'E k = E k (400 $ C) - E k (300 $ C) = (155.8 - 113) J / s = 42.8 J / s 43 J / s 2 mu 2
E k
(c) Some of the heat added goes to raise T (and hence U) of the air 7.6
(a)
'E p
mg'z
1 gal
mu 2 'E p 2
(b) E k
. 1 lbf 1 ft 3 62.43 lb m 32174 ft 10 ft 834 . ft lb f 3 2 7.4805 gal 1 ft s 32.174 lbm ft / s2
b g
mg 'z u
b g
2 g 'z
12
LM2FG 32.174 ft IJ b10 ftgOP NH sK Q 2
12
25.4
ft s
(c) False 7.7 (a)
'E k positive When the pressure decreases, the volumetric flow rate increases, and hence the velocity increases. 'E negative The gas exits at a level below the entrance level. p
b g
2
5 m S 1.5 cm 2 (b) m s
1 m3 10 4 cm 2
273 K 10 bars 1 kmol 303 K 101325 . bars 22.4 m 3 STP
b g
16.0 kg CH 4 1 kmol
0.0225 kg s
d h P d AV h
Pout AVout in
in
V nRT out nRT Vin uout
7.8
'z 'E p mg
uin
Pin Pout
uout (m / s) A(m2) Pin Pout uin (m / s) A(m2) bar b g 109 bar
5ms
Pin Pout
5555 ms .
105 m3 103 L 1 kg H2 O 981 1N 1J 2.778 u 107 kW h . m 75 m h 1 m3 1L s2 1 kg m/ s2 1 N m 1J
204 . u 104 kW h h
The maximum energy to be gained equals the potential energy lost by the water, or 2.04 u 10 4 kW h 24 h 7 days h 1 day 1 week
7-3
3.43 u 10 6 kW h week (more than sufficient)
7.9
(b) Q W
'U 'E k 'E p
b g 0 b no height changeg
'E k
0 system is stationary
'E p Q W
'U , Q 0, W ! 0
(c) Q W
'U 'E k 'E p
b
g
b
g
Q 0 adiabatic , W 0 no moving parts or generated currents 'E k 0 system is stationary 'E p 0 no height change 'U (d). Q W
b b
g
g
0 'U 'E k 'E p
b
g
W 0 no moving parts or generated currents 'E k 0 system is stationary 'E p 0 no height change 'U , Q 0 Even though the system is isothermal, the occurrence of a chemical reaction assures that 'U z 0 in a non-adiabatic reactor. If the temperature went up in the adiabatic reactor, heat must be transferred from the system to keep T constant, hence Q 0 .
Q
b b
g
g
7.10 4.00 L, 30 °C, 5.00 bar V (L), T (°C), 8.00 bar (a). Closed system:
'U 'E k 'E p
RS'E |T'E
'U (b)
Constant T 'U 0 Q W
(c) Adiabatic Q
bg
S 3
0 'U
Q W
b b
0 initial / final states stationary 0 by assumption
k p
g
g
Q W
7.65 L bar
W
8.314 J 0.08314 L bar
765 J
transferred from gas to surroundings
7.65 L bar > 0, Tfinal ! 30q C
1 m2 2.83 u 10 3 m 2 4 2 10 cm (a) Downward force on piston:
7.11 A
Fd
cm 2
Patm A mpiston+weight g 1 atm 1.01325 u 105 N / m2 2.83 u 10 3 m2 atm
7-4
24.50 kg 9.81 m s
2
1N 1 kg m / s2
527 N
7.11 (cont’d) Upward force on piston: Fu
APgas
d2.83 u 10
3
i d
m 2 Pg N m 2
i
Equilibrium condition: Fu
Fd 2.83 u 10 3 m2 P0
V0
303 K 1.01325 u 105 Pa 0.08206 L atm nRT 1.40 g N 2 1 mol N 2 0.677 L P0 28.02 g 1.86 u 105 Pa 1 atm mol K
(b) For any step, 'U 'E k 'E p Step 1: Q | 0 'U Step 2: 'U that V
186 . u 10 5 N m 2
527 P0
Q W 'U 'E k 0 'E p 0
186 . u 105 Pa
Q W
W
Q W As the gas temperature changes, the pressure remains constant, so
nRT Pg must vary. This implies that the piston moves, so that W is not zero. Tfinal 'U
Overall: Tinitial
0 Q W
0
In step 1, the gas expands W ! 0 'U 0 T decreases (c) Downward force Fd
u 10 id2.83 u 10 i b4.50gb9.81gb1g . gd101325 . b100 3
5
331 N (units
as in Part (a)) Final gas pressure Pf Since T0
Tf
F A
30q C , Pf V f
Distance traversed by piston W
331 N 2.83 u 10 3 m 2 P0V0 V f 'V A
. mg b331 Ngb0142
Fd
116 . u 10 5 N m 2 V0
P0 Pf
b1.08 0.677g L
. u 10 b0.677 Lg 186 116 . u 10
5 5
Pa Pa
1 m3 10 3 L 2.83 u 10 3 m2
108 . L m 0142 .
47 N m 47 J
Since work is done by the gas on its surroundings, W
47 J Q Q W 0
47 J
(heat transferred to gas) 7.12 V H
32.00 g 4.684 cm3 mol U PV
103 L
0.1499 L mol 106 cm3 41.64 atm 0.1499 L 8.314 J / (mol K) 1706 J mol mol 0.08206 L atm / (mol K) g
7-5
2338 J mol
d
i
7.13 (a) Ref state U (b) 'U
0 liquid Bromine @ 300 K, 0.310 bar
U final U initial
0.000 28.24
d i
' H
'U ' PV
'U P'V (Pressure Constant)
0.310 bar ' H 2824 . kJ mol
79.94g L . b00516
8.314 J 1 kJ 307 . kJ mol mol 0.08341 L bar 103 J
. kJ 154 kJ b5.00 molgb30.7 kJ / molg 15358 U independent of P U b300 K, 0.205 bar g U b300 K, 0.310 bar g 28.24 kJ mol U d340 K, P i U b340 K, 1.33 bar g 29.62 kJ mol n' H
'H
(c)
28.24 kJ mol
f
'U U final U initial
E
'U
29.62 28.24 1380 kJ mol .
= P' V' V' = PV / P' V changes with pressure. At constant temperature PV (T = 300K, P = 0.205 bar) = (0.310 bar)(79.94 L / mol) 120.88 L / mol V' 0.205 bar 5.00 L 1 mol n 0.0414 mol 120.88 L 'U n'U 0.0414 mol 138 . kJ / mol 0.0571 kJ
b
gb
'U 'E k 'E p 0
Q W Q
0
g
0.0571 kJ
0
(d) Some heat is lost to the surroundings; the energy needed to heat the wall of the container is being neglected; internal energy is not completely independent of pressure. 7.14 (a) By definition H U PV ; ideal gas PV
RT H
U RT
b g U bT g H bT , Pg U bT g RT H bT g independent of P
U T , P
(b) ' H 'H
'U R'T
n' H
3500
b2.5 molgb3599 cal / molg
7.15 'U 'E k 'E p
Q Ws ' Ek 'Ep Ws
'U
cal 1.987 cal 50 K mol K mol
Q W 'U
b b
3599 cal mol
8998 cal 9.0 u 10 3 cal
g
0 no change in m and u 0 no elevation change P'V since energy is transferred from the system to the surroundings
g
b
Q P'V Q
'U P'V
7-6
g
' (U PV )
'H
b b
7.16. (a) ' E k 'Ep
g
0 u1 u 2 0 0 no elevation change
g
'P 0 (the pressure is constant since restraining force is constant, and area is constrant) Ws P'V the only work done is expansion work H 34980 355 . T (J / mol), V1 = 785 cm3, T1 = 400 K, P = 125 kPa, Q = 83.8 J 125 u 103 Pa 785 cm3 1 m3 PV n= 0.0295 mol RT 8.314 m3 Pa / mol K 400 K 106 cm3 -H ) = 0.0295 mol 34980 + 35.5T - 34980 - 35.5(400K) (J / mol) Q = 'H = n(H
b
.
g
2
1
2
83.8 J = 0.0295 35.5T2 - 35.5(400) T2
480 K
0.0295 mol 8.314 m3 Pa 106 cm3 480 K 941 cm3 mol K 125 u 105 Pa 1 m3 125 u 105 N (941 - 785)cm3 1 m3 P'V 19.5 J m2 106 cm3 'U P'V 'U Q 'PV 838 . J 19.5 J 64.3 J
nRT P
i) V ii ) W iii ) Q (b) 'Ep
0
7.17 (a) "The gas temperature remains constant while the circuit is open." (If heat losses could occur, the temperature would drop during these periods.) (b) 'U ' E p ' E R Q 't W 't 'Ep Q
0, ' E k 0.90 u 1.4 W
0, W 1 J s 1W
0, U ( t
0)
0
1.26 J s
U (J ) 126 . t Moles in tank: n U
U n
PV RT
126 . t (J) 0.0859 mol
1 atm
b
2.10 L 25 273 K
g
1 mol K 0.08206 L atm
0.0859 mol
14.67t
Thermocouple calibration: T
aE b
b g
T qC T 0 , E 0.249 T 100 , E 5.27
b g
181 . E mV 4.51
U 14.67t 0 440 880 1320 T 181 . E 4.51 25 45 65 85 (c) To keep the temperature uniform throughout the chamber. (d) Power losses in electrical lines, heat absorbed by chamber walls. (e) In a closed container, the pressure will increase with increasing temperature. However, at the low pressures of the experiment, the gas is probably close to ideal U f T only.
bg
Ideality could be tested by repeating experiment at several initial pressures same results.
7-7
7.18 (b) 'H 'E k 'E p
Q W s (The system is the liquid stream.)
c c
h
' E k 0 no change in m and u ' E p 0 no elevation change Ws 0 no moving parts or generated currents
c
' H
h
h
Q , Q ! 0
(c) 'H 'E k 'E p
Q W s (The system is the water)
c
h
' H 0 T and P ~ constant ' E k 0 no change in m and u Q 0 no ' T between system and surroundings
c
'E p
c
b
g
Q W s (The system is the oil)
c
' E k 0 no velocity change
h
Q W s Q 0 (friction loss); W s 0 (pump work).
(e) 'H 'E k 'E p
Q W s (The system is the reaction mixture)
c
h
' E k ' E p 0 given ' Ws 0 no moving parts or generated current
c
' H
h
W s , W s ! 0 for water system
(d) 'H 'E k 'E p 'H ' E p
h
h
Q , Q pos. or neg. depends on reaction
7.19 (a) molar flow:
1 mol 125 . m3 273 K 122 kPa 423 K 101.3 kPa 22.4 L STP min
b g
' H ' E k ' E p
103 L 43.4 mol min 1 m3
Q W s
c
h
' E k ' E p 0 given Ws 0 no moving parts
c
Q
' H
n' H
h
43.37 mol 1 min min
60s
3640 J
kW
mol 10 3 J / s
2.63 kW
(b) More information would be needed. The change in kinetic energy would depend on the cross-sectional area of the inlet and outlet pipes, hence the internal diameter of the inlet and outlet pipes would be needed to answer this question.
7-8
b g
7.20 (a) H 104 . T q C 25
H in kJ kg P=110 kPa
H out 1.04 34.0 25 9.36 kJ kg H in 1.04 30.0 25 5.20 kJ kg ' H 9.36 5.20 4.16 kJ kg ' H ' E k ' E p
n (mol/s) N2
34 oC
Q W s
c
Q =1.25 kW
h h
' E k ' E p 0 assumed Ws 0 no moving parts
c
' H n' H
Q
n
Q ' H
1.25 kW
kg 1 kJ / s 103 g 1 mol 10.7 mol s 4.16 kJ kW 1 kg 28.02 g
b g
10.7 mol 22.4 L STP V = s mol
303 K 1013 . kPa 273 K 110 kPa
2455 . L / s 246 L s
(b) Some heat is lost to the surroundings, some heat is needed to heat the coil, enthalpy is assumed to depend linearly on temperature and to be independent of pressure, errors in measured temperature and in wattmeter reading.
7.21 (a) H
H 2 H 1 T2 T1 H1 aT1
aT b a b
H
0 Tref
130.2 5.2
b g
Table B.1 S . G.
b
U kJ kg
g
b
U kJ kg
b gb g
b
g
b g
5.2T q C 130.2
25q C
bg
C 6 H 14 l
H PV
U| V| W
129.8 258 . 5.2 H kJ kg 50 30 258 . 5.2 30 130.2
0.659 V
1 m3 152 . u 10 3 m 3 kg 659 kg
b5.2T 130.2gkJ / kg
1J 1 kJ 1 atm 1.0132 u 105 N / m2 1.52 u 10 3 m3 1 atm 1 kg 1 N m 103 J
g
5.2T 130.4
(b) Energy balance: Q 'Ek , 'E p , W 0
'U
20 kg [(5.2 u 20 - 130.4) - (5.2 u 80 - 130.4)] kJ 1 kg
Average rate of heat removal
6240 kJ 1 min 5 min
60 s
7-9
20.8 kW
6240 kJ
7.22
m (kg/s) 260°C, 7 bars H = 2974 kJ/kg u0 = 0
m (kg/s) 200°C, 4 bars H = 2860 kJ/kg u (m/s)
'H 'E k 'E p 'E p
Q
Q Ws
Ws
0
2 mu H out H in 'E k 'H m 2
d
d
u 2 2 H in H out
i
b
i
g
(2) 2974 2860 kJ 103 N m 1 kg m / s2 kg 1 kJ 1N
7.23 (a) 5 L/min
2.28 u 105
5 L/min 100 mm Hg (gauge)
0 mm Hg (gauge)
Qout
Qin
Since there is only one inlet stream and one outlet stream, and m in Eq. (7.4-12) may be written m 'z Q W s m 'U m ' PV ' u 2 mg 2 'U 0 given
d i
'u
Pout Pin mV
Heat input: Q in Efficiency:
V' P Q in
f
V'P
0 assume for incompressible fluid
2
'z
0
W s
0 all energy other than flow work included in heat terms
Q
(b) Flow work: V'P
m out { m ,
d i
a
m 'PV
V'P
m2 u 477 m / s s2
Q in Q out
Q in Q out 5L min
b100 0gmm Hg
5 ml O 2 min
20.2 kJ 1 ml O 2
66.7 J min 1.01 u 105 J min
1 atm 8.314 J 66.7 J min 760 mm Hg 0.08206 liter atm
10 3 J 101 . u 105 J min 1 kJ u 100 0.066%
7-10
Q W s ; 'E k , 'E p , W s
7.24 (a) 'H 'E k 'E p
b b
g
0 'H
Q
H 400q C, 1 atm 3278 kJ kg (Table B.6) H 100q C, sat' d 1 atm 2676 kJ kg (Table B.5)
g
100 kg H 2 O(v) / s
100 kg H 2 O(v) / s
o
400 o C, 1 atm
100 C, saturated Q (kW)
100 kg Q s
b3278 2676.0gkJ
10 3 J kg 1 kJ
(b) 'U 'E k 'E p
b
U 100q C, 1 atm
Q
g
Q W ; 'E k , 'E p , W 0 'U Q 2507 kJ kg U 400q C, 1 atm 2968 kJ kg (Tables B.5 & B.7)
b
g
b
m'U
'U
6.02 u 10 7 J s
g
d
100 kg 2968 2507 kJ kg 10 3 J kJ
i
4.61 u 10 7 J
The difference is the net energy needed to move the fluid through the system (flow work).
c b g h 83.9 kJ kg (Table B.5) H bsteam, 20 bars, sat' d g 2797.2 kJ kg (Table B.6)
7.25 H H 2 O l , 20q C
m [kg H 2 O(l) / h]
m [kg H 2 O(v) / h]
o
20 C
20 bar (sat' d) Q = 0.65(813 kW)
(a) 'H 'E k 'E p 'H
m (b) V
Q ' H
Q W s ; 'E k , 'E p , W s
528 kW
0 'H
Q
m 'H
528 kW
b
kg 1 kJ / s 3600 s 2797.2 83.9 kJ 1 kW 1 h
g
b701 kg hg d0.0995 m kgi A 3
701 kg h
69.7 m 3 h sat' d steam @ 20 bar
Table B.6
(c) V
nRT P
701 kg / h 18.02 g / mol
103 g / kg 485.4 K 0.08314 L bar mol K
20 bar
1 m3 103 L
78.5 m3 / h
The calculation in (b) is more accurate because the steam tables account for the effect of pressure on specific enthalpy (nonideal gas behavior). (d) Most energy released goes to raise the temperature of the combustion products, some is transferred to the boiler tubes and walls, and some is lost to the surroundings.
7-11
c bg
7.26 H H 2 O l , 24q C, 10 bar
h
100.6 kJ kg (Table B.5 for saturated liquid at 24oC; assume H
independent of P).
b
g
2776.2 kJ kg (Table B.6) ' H
H 10 bar, sat' d steam
[kg H2 O(v) / h] m
2776.2 100.6 2675.6 kJ kg
[kg H2 O(v) / h] m 15,000 m3 / h @10 bar (sat'd)
o
24 C, 10 bar Q (kW) 15000 m 3 m h
kg 01943 m3 .
7.72 u 10 4 kg h
A
b Table 8.6 g
d
Energy balance 'E p , W s
'E k
'E k
E kfinal E kinitial f mu
2
i
0 : 'H 'E k
E kinitial |0
7.72 u 10 4 kg
'E k
E kfinal
d15,000 m hi 3
2
2
015 . 2 S 4 m2
h
2
Q
1
1
h3
1J
2 3600 3 s 3
A
1 kg m 2 / s 2
A S D2 4 5.96 u 10 5 J / s 1h 7.72 u 10 4 kg 2675.6 kJ 5.96 u 10 5 J 1 kJ h kg 3600 s s 10 3 J
Q m 'H 'E k
57973 kJ s 5.80 u 10 4 kW 7.27 (a)
228 g/min 25oC
228 g/min T(oC) Q ( kW)
Energy balance: Q 'E x , 'E p , Ws 0
=0
b g
'H Q W
b g 0.263Q bW g T bq C g 25 H bJ gg 0.263QbW g 0 (b) H bbT 25g H bJ gg 3.34 T bq Cg 25
228 g 1 min ( H out H in ) J min 60 s g
H out J g
26.4 27.8 29.0 32.4 4.47 9.28 13.4
Fit to data by least squares (App. A.1)
b
7-12
¦ H i
i
24.8
bT 25g ¦ bT 25g i
i
i
2
3.34
7.27 (cont’d) (c) Q
' H
b
g
350 kg 10 3 g 1 min 3.34 40 20 J min kg 60 s g
kW s 390 kW heat input to liquid 10 3 J
(d) Heat is absorbed by the pipe, lost through the insulation, lost in the electrical leads. 7.28 m w [ kg H 2 O(v) / min] 3 bar, sat' d
m w [ kg H 2 O(l) / min] 27 o C
Q ( kW)
m e [ kg C 2 H 6 / min] 16 o C, 2.5 bar
(a) C H mass flow: m 2 6 e
H ei
m e [ kg C 2 H 6 / min] 93 o C, 2.5 bar
795 m 3 10 3 L 2.50 bar min m 3 289 K 2.487 u 103 kg min
1 K - mol 30.01 g 1 kg 0.08314 L - bar mol 1000 g
941 kJ kg , H ef
1073 kJ kg Energy Balance on C 2 H 6 : 'E p , W s 0, 'E k # 0 Q
LMb N
g OPQ
kJ kg Q 2.487 u 103 1073 941 min kg
b bliquid, 27q Cg
(b) H s1 3.00 bar, sat' d vapor H s2
g
'H
2.487 u 103 kJ 1 min 5.47 u 103 kW min 60 s
2724.7 kJ kg (Table B.6)
1131 . kJ kg (Table B.5)
Assume that heat losses to the surroundings are negligible, so that the heat given up by the
d
condensing steam equals the heat transferred to the ethane 5.47 u 10 3 kW Energy balance on H 2 O: Q m Vs
Q H s2 H s1
'H
5.47 u 10 3 kJ s
b2.09 kg / sg d0.606 m kgi A 3
d
m H s2 H s1
i
kg . 2724.7 kJ 1131
b
g
i
2.09 kg s steam
1.27 m 3 s
Table B.6
Too low. Extra flow would make up for the heat losses to surroundings. (c) Countercurrent flow Cocurrent (as depicted on the flowchart) would not work, since it would require heat flow from the ethane to the steam over some portion of the exchanger. (Observe the two outlet temperatures)
7-13
7.29
250 kg H2 O(v )/min 40 bar, 500°C H 1 (kJ/kg)
W
b b
s
Heat exchanger
250 kg/min 5 bar, T 2 (°C), H 2 (kJ/kg)
Turbine
=1500 kW
250 kg/min 5 bar, 500°C H3 (kJ/kg)
Q(kW)
g
H 2 O v , 40 bar, 500q C : H 1 3445 kJ kg (Table B.7) H 2 O v , 5 bar, 500q C : H 3 3484 kJ kg (Table B.7)
g
(a) Energy balance on turbine: 'E p 'H
d
W s m H 2 H 1
i
3445 kJ 1500 kJ s kg H
0, Q
0, 'E k # 0
W s H 2
H 1 W s m
min 60 s 3085 kJ kg 250 kg 1 min
3085 kJ kg and P 5 bars T = 310q C (Table B.7)
(b) Energy balance on heat exchanger: 'E p
d
H 3 H 2 Q 'H m
i
250 kg
d
Q W s m s H 3 H 1
Q
'H 'W s
1 min
1 kW
kg
60 s
1 kJ / s
1663 kW
0, 'E k # 0
i
Q W s
b3484 3445gkJ
250 kg min
0, 'E k # 0
b3484 3085gkJ
min
(c) Overall energy balance: 'E p 'H
0, W s
kg
1 min 1 kW 1500 kJ 1 kW s 1 kJ / s 60 s 1 kJ / s
1663 kW
b g H Obv , 5 bar, 310q Cg: V
(d) H 2 O v , 40 bar, 500q C : V1 2
2
0.0864 m 3 kg (Table B.7) 0.5318 m 3 kg (Table B.7)
u1
250 kg 1 min 0.0864 m 3 min 60 s kg
u2
250 kg min 0.5318 m 3 min 60 s kg
'E k
m 2 u2 u12 2
1 0.5 S 4 m 2 1 0.5 S 4 m 2 2
250 kg 1 1 min
min 2 0.26 kW
