Before the first law of thermodynamics can be applied to systems, let's review the
.... When we apply a methodical solution procedure, thermodynamics problems.
Chapter 4 The First Law of Thermodynamics The first law of thermodynamics is an expression of the conservation of energy principle. Energy can cross the boundaries of a closed system in the form of heat or work. Energy transfer across a system boundary due solely to the temperature difference between a system and its surroundings is called heat. Work energy can be thought of as the energy expended to lift a weight. Closed System First Law A closed system moving relative to a reference plane is shown below where z is the elevation of the center of mass above the reference plane and G V is the velocity of the center of mass.
Heat
Closed System
G V
CM
z
Work
Reference Plane, z = 0 For the closed system shown above, the conservation of energy principle or the first law of thermodynamics is expressed as
FG Total energy IJ − FG Total energy IJ = FG The change in total IJ H entering the systemK H leaving the systemK H energy of the systemK or
Ein − Eout = ∆E system Chapter 4 -1
According to classical thermodynamics, we consider the energy added to be net heat transfer to the closed system and the energy leaving the closed system to be net work done by the closed system. So
Qnet − Wnet = ∆E system Normally the stored energy, or total energy, of a system is expressed as the sum of three separate energies. The total energy of the system, Esystem, is given as
E = Internal energy + Kinetic energy + Potential energy E = U + KE + PE Recall that U is the sum of the energy contained within the molecules of the system other than the kinetic and potential energies of the system as a whole and is called the internal energy. The internal energy U is dependent on the state of the system and the mass of the system. For a system moving relative to a reference plane, the kinetic energy KE and the potential energy PE are given by G V G V =0
KE = ∫ PE = ∫
z
z =0
G G G mV 2 mV dV = 2
mg dz = mgz
The change in stored energy for the system is
∆E = ∆U + ∆KE + ∆PE
Chapter 4 -2
Now the conservation of energy principle, or the first law of thermodynamics for closed systems, is written as
Qnet − Wnet = ∆U + ∆KE + ∆PE If the system does not move with a velocity and has no change in elevation, the conservation of energy equation reduces to
Qnet − Wnet = ∆U We will find that this is the most commonly used form of the fist law. Closed System First Law for a Cycle Since a thermodynamic cycle is composed of processes that cause the working fluid to undergo a series of state changes through a series of processes such that the final and initial states are identical, the change in internal energy of the working fluid is zero for whole numbers of cycles. The first law for a closed system operating in a thermodynamic cycle becomes 0
Qnet − Wnet = ∆U cycle Qnet = Wnet
Chapter 4 -3
Example 4-1 Complete the table given below for a closed system under going a cycle. Process 1-2 2-3 3-1 Cycle
Qnet kJ +5 +20
Wnet kJ
U2 – U1 kJ -5
+10 -5
(Answer to above problem) Row 1: +10, Row 2: +10, Row 3: -10, -5 Row 4: +15, +15, 0
Example 4-2 Find the required heat transfer to the water in Example 3-5. Review the solution procedure of Example 3-5 and then apply the first law to the process. Conservation of Energy:
Ein − Eout = ∆E Qnet ,14 − Wnet ,14 = ∆U14
Chapter 4 -4
In Example 3-5 we found that
. kJ Wnet ,14 = 12 The heat transfer is obtained from the first law as
Qnet ,14 = Wnet ,14 + ∆U14 where
∆U 14 = U 4 − U 1 = m( u4 − u1 )
At state 1, T1 = 100°C, v1 = 0.835 m3/kg and vf < v1 < vg at T1. The quality at state 1 is
v1 = v f + x1v fg x1 =
v1 − v f v fg
=
0.835 − 0.001044 = 0.499 1.6729 − 0.001044
u1 = u f + x1u fg = 418.94 + (0.499)(2087.6) kJ = 1460.26 kg Because state 4 is a saturated vapor state and v4 = 0.841 m3/kg, interpolating in either the saturation pressure table or saturation temperature table at v4 = vg gives
Chapter 4 -5
kJ u4 = 253148 . kg So
∆U14 = m(u4 − u1 ) = (1 kg )(2531.48 − 1460.26)
kJ kg
= 1071.22 kJ The heat transfer is
Qnet ,14 = Wnet ,14 + ∆U 14 = 1.2 kJ + 1071.22 kJ = 1072.42 kJ Heat in the amount of 1072.42 kJ is added to the water.
Review:
Specific Heats and Changes in Internal Energy and Enthalpy for Ideal Gases
Before the first law of thermodynamics can be applied to systems, let’s review the ways to calculate the change in internal energy and enthalpy of the substance enclosed by the system boundary. For real substances like water, the property tables are used to find the internal energy and enthalpy change. For ideal gases we learned in Chapter 2 that the internal energy and enthalpy are found by knowing the specific heats.
Chapter 4 -6
du ∂u CV = = dT ∂T v
dh ∂h and CP = = dT ∂T P
ideal gas
ideal gas
The change in internal energy and enthalpy of ideal gases can be expressed as
z z
2
∆u = u2 − u1 = CV (T )dT = CV ,ave (T2 − T1 ) 1
2
∆h = h2 − h1 = CP (T )dT = CP ,ave (T2 − T1 ) 1
where CV,ave and CP,ave are average or constant values of the specific heats over the temperature range. We will drop the ave subscript shortly. 2a
P
2b 2c
1
T2 T1 V P-V DIAGRAM FOR SEVERAL PROCESSES FOR AN IDEAL GAS.
In the above figure an ideal gas undergoes three different process between the same two temperatures. Process 1-2a: Constant volume Process 1-2b: P = a + bV, a linear relationship Process 1-2c: Constant pressure These ideal gas processes have the same change in internal energy and enthalpy because the processes occur between the same temperature limits.
Chapter 4 -7
z z
2
∆ua = ∆ub = ∆uc = CV (T ) dT 1
2
∆ha = ∆hb = ∆hc = CP (T ) dT 1
To find ∆u and ∆h we often use average, or constant, values of the specific heats. Some ways to determine these values are as follows: 1. The best average value (the one that gives the exact results)
z
See Table A-2(c) for variable specific data.
Cv ,ave =
2
1
CV ( T )dT T2 − T1
z
2
C P ( T ) dT
and
C P ,ave =
and
C P ,ave =
and
C P ,ave = C P (Tave )
1
T2 − T1
2. Good average values are
Cv ,ave = and
CV ( T2 ) + CV ( T1 ) 2 Cv ,ave = CV ( Tave )
C P ( T2 ) + C P ( T1 ) 2
where Tave =
T2 + T1 2
3. Sometimes adequate (and most often used) values are the ones evaluated at 300 K and are given in Table A-2(a).
Cv ,ave = CV (300 K )
and
C P ,ave = CP (300 K )
Chapter 4 -8
Let's take a second look at the definition of ∆u and ∆h for ideal gases. Just consider the enthalpy for now.
z
2
∆h = h2 − h1 = CP (T )dT 1
Let's perform the integral relative to a reference state where h = href at T = Tref.
∆h = h2 − h1 = or
∆h = h2 − h1 =
z z
Tref
T1 T2
Tref
CP (T ′)dT ′ + CP (T ′)dT ′ −
z z
T2
Tref
T1
Tref
CP (T ′)dT ′ CP (T ′)dT ′
= (h2 − href ) − (h1 − href ) At any temperature, we can calculate the enthalpy relative to the reference state as
h − href = or h = href +
z z
T
Tref
T
Tref
CP (T ′)dT ′
CP (T ′)dT ′
This result is the basis of Table A-17 for air and Tables A-18 through A-25 for other gases. A similar result is found for the change in internal energy.
Chapter 4 -9
The Systematic Thermodynamics Solution Procedure When we apply a methodical solution procedure, thermodynamics problems are relatively easy to solve. Each thermodynamics problem is approached the same way as shown in the following, which is a modification of the procedure given in the text: Thermodynamics Solution Method 1. Sketch the system and show energy interactions across the boundaries. 2. Determine the property relation. Is the working substance an ideal gas or a real substance? Begin to set up and fill in a property table. 3. Determine the process and sketch the process diagram. Continue to fill in the property table. 4. Apply conservation of mass and conservation of energy principles. 5. Bring in other information from the problem statement, called physical constraints, such as the volume doubles or the pressure is halved during the process. 6. Develop enough equations for the unknowns and solve.
Chapter 4 -10
Example 4-3 A tank contains nitrogen at 27°C. The temperature rises to 127°C by heat transfer to the system. Find the heat transfer and the ratio of the final pressure to the initial pressure. System: Nitrogen in the tank. System boundary
2
P T 1= 27°C
T2=127°C 1
NITROGEN GAS
V
P-V diagram for a constant volume process
Property Relation: Nitrogen is an ideal gas. The ideal gas property relations apply. Let’s assume constant specific heats. (You are encouraged to rework this problem using variable specific heat data.) Process: Tanks are rigid vessels; therefore, the process is constant volume. Conservation of Mass:
m2 = m1
Using the combined ideal gas equation of state,
PV PV 2 2 = 1 1 T2 T1 Since R is the particular gas constant, and the process is constant volume,
Chapter 4 -11
V2 = V1 P2 T2 (127 + 273) K = = = 1333 . P1 T1 (27 + 273) K Conservation of Energy: The first law closed system is
Ein − Eout = ∆E Qnet − Wnet = ∆U For nitrogen undergoing a constant volume process (dV = 0), the net work is (Wother = 0)
z
2
Wnet ,12 = 0 + Wb ,12 = PdV = 0 1
Using the ideal gas relations with Wnet = 0, the first law becomes (constant specific heats)
z
2
Qnet − 0 = ∆U = m CV dT = mCV (T2 − T1 ) 1
The heat transfer per unit mass is
Chapter 4 -12
qnet
Qnet = = CV (T2 − T1 ) m kJ = 0.743 127 − 27 K kg ⋅ K kJ = 74.3 kg
b
g
Example 4-4 Air is expanded isothermally at 100°C from 0.4 MPa to 0.1 MPa. Find the ratio of the final to the initial volume, the heat transfer, and work. System: Air contained in a piston-cylinder device, a closed system Process: Constant temperature System boundary
2
P 1
AIR
WB
T = const.
V
P-V diagram for T= constant
Property Relation: Assume air is an ideal gas and use the ideal gas property relations with constant specific heats.
Chapter 4 -13
PV = mRT ∆u = CV (T2 − T1 ) Conservation of Energy:
Ein − Eout = ∆E Qnet − Wnet = ∆U The system mass is constant but is not given and cannot be calculated; therefore, let’s find the work and heat transfer per unit mass. Work Calculation:
b g
Wnet ,12 = Wnet
other ,12
2
Wb ,12 = ∫ P dV = ∫ 1
2
1
+ Wb ,12 mRT dV V
V2 =mRT ln V1 Conservation of Mass: For an ideal gas in a closed system (mass = constant), we have
m1 = m2 PV PV 1 1 = 2 2 RT1 RT2 Since the R's cancel and T2 = T1 Chapter 4 -14
V2 P1 0.4 MPa = = =4 V1 P2 01 . MPa Then the work expression per unit mass becomes
wb ,12
FG IJ H K
Wb ,12 V2 = = RT ln m V1
FG H
wb,12 = 0.287 = 148.4
IJ K
bg
kJ (100 + 273) K ln 4 kg − K
kJ kg
The net work per unit mass is
wnet ,12 = 0 + wb ,12
kJ = 148.4 kg
Now to continue with the conservation of energy to find the heat transfer. Since T2 = T1 = constant,
∆U 12 = m∆u12 = mCV (T2 − T1 ) = 0
So the heat transfer per unit mass is
Chapter 4 -15
qnet qnet − wnet qnet
Qnet = m = ∆u = 0 = wnet kJ = 148.4 kg
The heat transferred to the air during an isothermal expansion process equals the work done.
Examples Using Variable Specific Heats Review the solutions in Chapter 4 to the ideal gas examples where the variable specific heat data are used to determine the changes in internal energy and enthalpy.
Extra Problem for You to Try: An ideal gas, contained in a piston-cylinder device, undergoes a polytropic process in which the polytropic exponent n is equal to k, the ratio of specific heats. Show that this process is adiabatic. When we get to Chapter 6 you will find that this is an important ideal gas process.
Chapter 4 -16
Example 4-5 Incompressible Liquid A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage is 35°C, and you always drink your beverage at 10°C. a. How much heat energy must be removed from your two liters of beverage? b. You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required? Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles?
System: The liquid in the constant volume, closed system container
System
boundary My beverage
Qout The heat removed
Property Relation: Incompressible liquid relations, let’s assume that the beverage is mostly water and takes on the properties of liquid water. The specific volume is 0.001 m3/kg, C = 4.18 kJ/kg⋅K. Process: Constant volume
V2 = V1 Chapter 4 -17
Conservation of Mass:
m2 = m1 = m V m= = v
FG H
IJ K
m3 2L = 2 kg 3 m 1000 L 0.001 kg
Conservation of Energy: The first law closed system is
Ein − Eout = ∆E Since the container is constant volume and there is no “other” work done on the container during the cooling process, we have
b g
Wnet = Wnet
other
+ Wb = 0
The only energy crossing the boundary is the heat transfer leaving the container. Assuming the container to be stationary, the conservation of energy becomes
− Eout = ∆E −Qout = ∆U = mC∆T
Chapter 4 -18
−Qout −Qout
kJ = (2 kg )(4.18 )(10 − 35) K kg ⋅ K = −209.2 kJ
Qout = 209.2 kJ The heat transfer rate to cool the 10 bottles in one-half hour is
Q out =
(10 bottles)(209.2
kJ ) bottle
0.5 hr
F I FG 1hr IJ G kW J H 3600s K GG kJ JJ HsK
= 1162 . kW
$0.085 Cost = (1162 kW )(0.5 hr ) . kW − hr = $0.05
Chapter 4 -19
Conservation of Energy for Control volumes The conservation of mass and the conservation of energy principles for open systems or control volumes apply to systems having mass crossing the system boundary or control surface. In addition to the heat transfer and work crossing the system boundaries, mass carries energy with it as it crosses the system boundaries. Thus, the mass and energy content of the open system may change when mass enters or leaves the control volume.
Wnet G VCM
G m i , Vi Zi
Q net
ZCM
Control surface
Ze
G m e , Ve
Reference plane
Typical control volume or open system Thermodynamic processes involving control volumes can be considered in two groups: steady-flow processes and unsteady-flow processes. During a steady-flow process, the fluid flows through the control volume steadily, experiencing no change with time at a fixed position. Let’s review the concepts of mass flow rate and energy transport by mass. Mass Flow Rate Mass flow through a cross-sectional area per unit time is called the mass flow rate m . If the fluid density and velocity are constant over the flow cross-sectional area, the mass flow rate is
K K V A m = ρVav A = av v
where ρ is the density, kg/m3 ( = 1/v), A is the cross-sectional area, m2; and G Vav is the average fluid velocity normal to the area, m/s.
Chapter 4 -20
The fluid volume flowing through a cross-section per unit time is called the volume flow rate V .
G V = VA
( m 3 / s)
The mass and volume flow rate are related by
V m = ρV = v
( kg / s)
Conservation of Mass for General Control Volume The conservation of mass principle for the open system or control volume is expressed as
LMSumof rate OP LMSum of rate OP LMTime rate changeOP MMof mass flowing PP − MMof mass flowing PP = MMof massinside PP Ninto control volumeQ Nfromcontrol volumeQ Ncontrol volume Q or
∑ m − ∑ m in
out
= ∆m system
( kg / s)
Conservation of Energy for General Control Volume The conservation of energy principle for the control volume or open system has the same word definition as the first law for the closed system. Expressing the energy transfers on a rate basis, the control volume first law is
Chapter 4 -21
LMSumof rate OP LMSum of rate OP LMTime rate changeOP MMof energy flowing PP − MMof energy flowing PP = MMof energy inside PP Ninto control volumeQ Nfromcontrol volumeQ Ncontrol volume Q or
E in − E out
Rate of net energy transfer by heat, work, and mass
=
∆E system
( kW )
Rate change in internal, kinetic, potential, etc., energies
Considering that energy flows into and from the control volume with the mass, energy enters because net heat is transferred to the control volume, and energy leaves because the control volume does net work on its surroundings, the open system, or control volume, the first law becomes
dE Q net + ∑ m iθ i − Wnet − ∑ m eθ e = CV dt for each inlet for each exit
4
:
( kW )
where θ is the energy per unit mass flowing into or from the control volume. The energy per unit mass, θ, flowing across the control surface that defines the control volume is composed of four terms: the internal energy, the kinetic energy, the potential energy, and the flow work. The total energy carried by a unit of mass as it crosses the control surface is
G2 V θ = u + Pv + + gz 2 G2 V = h+ + gz 2 Here we have used the definition of enthalpy, h = u + Pv.
Chapter 4 -22
E in − E out = ∆E CV
FG H
IJ K
FG H
IJ K
G2 G2 V V Q net + ∑ m i hi + i + gzi − Wnet − ∑ m e he + e + gze = ∆E CV 2 2
for each inlet
for each exit
Where the time rate change of the energy of the control volume has been written as ∆E CV . Steady-State, Steady-Flow Processes Most energy conversion devices operate steadily over long periods of time. The rates of heat transfer and work crossing the control surface are constant with time. The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions the mass and energy content of the control volume are constant with time.
dmCV = ∆m CV = 0 dt dECV = ∆E CV = 0 dt Steady-state, Steady-Flow Conservation of Mass: Since the mass of the control volume is constant with time during the steady-state, steady-flow process, the conservation of mass principle becomes
LMSumof rate OP LMSum of rate OP MMof mass flowing PP = MMof mass flowing PP Ninto control volumeQ Nfromcontrol volumeQ or
∑ m
in
= ∑ m out Chapter 4 -23
( kg / s)
Steady-state, steady-flow conservation of energy Since the energy of the control volume is constant with time during the steady-state, steady-flow process, the conservation of energy principle becomes
LMSumof rate OP LMSum of rate OP MMof energy flowing PP = MMof energy flowing PP Ninto control volumeQ Nfromcontrol volumeQ or
E in − E out
0
∆E system
=
Rate of net energy transfer by heat, work, and mass
( kW )
Rate change in internal, kinetic, potential, etc., energies
or
1 E in
: E out
=
Rate of net energy transfer by heat, work, and mass into the system
Rate of energy transfer by heat, work, and mass from the system
Considering that energy flows into and from the control volume with the mass, energy enters because heat is transferred to the control volume, and energy leaves because the control volume does work on its surroundings, the steady-state, steady-flow first law becomes
Q in + Win
G I F V + ∑ m G h + + gz J = Q H 2 K 2
i
i
i
i
out
+ Wout
for each inlet
G I F V + ∑ m G h + + gz J H 2 K e
e
2 e
for each exit
Chapter 4 -24
e
Most often we write this result as
Q net − Wnet
G G I I F F V V = ∑ m G h + + gz J − ∑ m G h + + gz J H 2 K H 2 K e
e
2 e
2
e
i
for each exit
i
i
i
for each inlet
where
Q net = ∑ Q in − ∑ Q out
Wnet = ∑ Wout − ∑ Win
Steady-state, steady-flow for one entrance and one exit A number of thermodynamic devices such as pumps, fans, compressors, turbines, nozzles, diffusers, and heaters operate with one entrance and one exit. The steady-state, steady-flow conservation of mass and first law of thermodynamics for these systems reduce to
m 1 = m 2 ( kg / s) 1 G 1 G V1 A1 = V2 A2 v1 v2 G2 G2 V − V1 Q − W = m h2 − h1 + 2 + g ( z2 − z1 ) 2
LM N
OP Q
( kW )
where the entrance is state 1 and the exit is state 2 and m is the mass flow rate through the device. When can we neglect the kinetic and potential energy terms in the first law? Consider the kinetic and potential energies per unit mass.
Chapter 4 -25
G2 V ke = 2 G m kJ ( 45m / s) 2 1kJ / kg For V = 45 = ke = 1 s kg 2 1000m2 / s 2 G m kJ (140m / s) 2 1kJ / kg = 10 V = 140 ke = 2 2 s kg 2 1000m / s
pe = gz For z = 100m
m kJ 1kJ / kg = m . 100 0 98 s2 kg 1000m2 / s 2 m kJ 1kJ / kg = pe = 9.8 2 1000m . 9 8 s kg 1000m2 / s 2
pe = 9.8
z = 1000m
When compared to the enthalpy of steam (h ≅ 2000 to 3000 kJ/kg) and the enthalpy of air (h ≅ 200 to 6000 kJ/kg), the kinetic and potential energies are often neglected. When the kinetic and potential energies can be neglected, the conservation of energy equation becomes
Q − W = m (h2 − h1 )
( kW )
We often write this last result per unit mass flow as
q − w = (h2 − h1 ) where q =
Q W and w = . m m
Chapter 4 -26
( kJ / kg )
Some Steady-Flow Engineering Devices Below are some engineering devices that operate essentially as steady-state, steady-flow control volumes.
Nozzles and Diffusers
G V1
G G V2 >> V1
G G V2
