Food Process Engineering An introduction by Alan

Food Process Engineering An introduction by Alan Friis & Jørgen Risum

BioCentrum-DTU Septemner 2002 First edition (0.5 version)

Table of contents Introduction ........................................................................................................ 1 Food Process Design and Quality................................................................... 1 A strategy for Process Design of a Food Production Line.............................. 1 About the teaching material........................................................................... 2 Flow sheets ........................................................................................................ 4 Process description......................................................................................... 4 Flow sheets..................................................................................................... 4 Mass and energy balances................................................................................. 6 Mass Balances ................................................................................................ 6 Material balance with accumulation ........................................................ 10 Energy balances ........................................................................................... 11 Continues systems.................................................................................... 11 Composition of food products .......................................................................... 16 Physical properties of food products................................................................ 18 Density .......................................................................................................... 18 Specific heat capacity................................................................................... 19 Heat conductivity.......................................................................................... 20 Thermal diffusivity ........................................................................................ 22 Viscosity (Newtonian liquids) ...................................................................... 22 The influence of temperature on viscosity................................................ 23 Other characterizing parameters ................................................................. 23 Rheology of liquid foods ............................................................................... 24 Non-Newtonian fluids ............................................................................... 25 Time dependency...................................................................................... 25 Time independent liquids ......................................................................... 26 Shear-thinning fluids................................................................................. 28 Shear-thickening fluids ............................................................................. 31 Fluids exhibiting an apparent yield stress ............................................... 31 The role of elasticity and time effects ....................................................... 32 Temperature dependency......................................................................... 32 Rheology solid and semi-solid food products............................................... 32 Fluid mechanics ............................................................................................... 33 Newtonian fluids .......................................................................................... 33 Laminar flow ............................................................................................. 35 Turbulent flow ........................................................................................... 37 Non-Newtonian fluids................................................................................... 38 Laminar flow ............................................................................................. 40 Turbulent flow ........................................................................................... 43 Flow through pipe fittings............................................................................ 43 Heat transfer..................................................................................................... 45 Steady state conduction heat transfer ......................................................... 46 Steady state convection heat transfer.......................................................... 49 Natural convection .................................................................................... 51 Forced convection ..................................................................................... 52 Unsteady state heat transfer ........................................................................ 56

Cooking and condensation ........................................................................... 56 Kondenserende vanddamp ....................................................................... 56 Kogning ..................................................................................................... 56 Heat transfer by radiation ............................................................................ 57 Black body radiation ................................................................................. 57 Heating and cooling ......................................................................................... 58 Effects on processing ....................................................................................... 59 References ........................................................................................................ 60 List of symbols ................................................................................................. 61 Appendix 1. Usefull calculations and conversions .......................................... 63 Appendix 2. Units conversions ........................................................................ 64 Appendix 3. Density equations........................................................................ 65 Appendix 4. Heat capacity equations.............................................................. 66 Appendix 5. Heat conductivity equations ....................................................... 67 Beregninger af ikke homogene systemer ..................................................... 67 Appendix 6. Measuring viscosity and flow curves .......................................... 69 Appendix 7. Moody diagram............................................................................ 70 Appendix 8. Data for water, air + foods .......................................................... 71 Appendix 9. Equations for natural convection ................................................ 72 Appendix 10. Equations for forced convection ................................................ 73 Free falling droplets .................................................................................. 73 Drying ........................................................................................................ 73 Agitated tanks........................................................................................... 73

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Introduction Producing foods of high quality is an obvious objective for food engineers. Hence a methodology for objective determination of the relationships between the characteristics of food products and the involved manufacturing processes (including storage and transportation) is needed. The term ‘characteristic’ pertaining to food products is much more operational than the term ‘quality’ when dealing with process design and validation. Characteristics can be determined objectively (quantitatively or qualitatively), however, definition of quality is subjective. Two persons who like a specific product might like it for different reasons. Hence each individual define his or her own preferred characteristic to determine the quality of the product.

Food Process Design and Quality Some definitions relating to quality and characteristics of foods is given below: §

Food companies producing high quality foods will first assure safety (microbiological, chemical and physical) of their products and next focus on uniform properties and characteristics over time. This, however, does not define the quality level. The quality of a product must comply with generally understood perceptions amongst consumers in a specific area. Hence we find that quality conveniently can be defined as what the consumers wants or what the consumers are made to believe they want and are willing to pay for.

§

In the consumer opinion quality is linked to aspects like nutritional content, physical appearance, consistency, structure, taste, safety issues, convenience, ecology, minimal, careful production, etc. For food producers the perceived quality must be translated into measurable characteristics to be able to design products, processes and control systems in a proper manner.

§

In the field of food process engineering we focus on characteristics of food products. We appreciate the total quality of a food product pertains to a specific situation and as such is seen in a subjective light.

A strategy for Process Design of a Food Production Line The toolbox needed when designing food process lines and determining sensible operating conditions origins from the general field of process engineering. The necessary calculations tools are presented in this book in a way, which pertains specifically to food production. It can be beneficiary to begin whit a procedure of designing processes in order relate to the application of the theoretical subjects. Such a strategy is presented below: 1. Deliverables. Develop a product specification data sheet and assign preliminary limits to measurable characteristics

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2. Raw Materials. Identifying raw material (quality parameter) requirements. Identify the key transformations and processes operations needed to produce the product from these materials. Some interactions between the product and the process operations may be identified in this step as well 3. Flow Sheets. Draw a flowsheet to illustrate the flow of different material streams through the process. Heating and cooling media streams should be included on the flowsheet. Remember that different flow sheets are produced to serve different purposes. While we rarely put all information in one sheet we need one concerning flow of material through the process and one concerning temperature and energy as a minimum 4. Mass Balances. Choose a basis and calculate preliminary mass balance estimates for all streams. These calculations can be refined later as the design process proceeds 5. Energy Balances. Energy balances determine the requirements for heating, cooling and mechanical power. Initial energy balance calculations are performed operation by operation. However, reuse or regeneration of energy must be considered in later to design an economically feasible process 6. Design Refinements. Refine the flow sheet if needed. The level of detail should at this point show each unit operation in separate steps. Refine mass and energy balances if needed 7. Process Selection based on Product-Process Interactions. Identify key interactions between product and process. Evaluate potential processes to produce the desired product. More processes may be suitable in theory; this stage ends with the choice of a suitable process made on a qualified basis 8. Equipment Sizing and Selection. For each part of the process, estimate the size of equipment you require. Calculate pumping, heating and cooling requirements 9. Capital Costs. Estimate the purchase costs (first costs) of your process line 10. Operating Costs. Estimate how much it costs each process to run (second costs) 11. Solution Optimization. Repeat steps 6 to 10 until a desired outcome is achieved 12. Communicate the result. Present the result and confirm the next steps. The procedure listed describes an intuitive way to attack the design of a production line and will be the basis we often refer to throughout the food process engineering curriculum.

About the teaching material The text is divided in chapters pertaining to Flow Sheets (of processes), Mass and Energy Balances, Composition of Food Products, Physical Properties of Food (including Rheology), Fluid Mechanics, Heat Transfer, Heating and Cooling and Effects of Processing. -2-

The text aims at explaining the theoretical information needed to perform process engineering calculations in a practical applicable way. The objective is to enable the student to be able to validate and design food process operations. The examples given in the text is designed to support this objective. When doing calculus it is sensible to bring all quantities to SI units before computation is carried out (do always check that the dimensions of equations are correct). In this text SI units are stated for all of the applied physical parameters.

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Flow sheets Process description Describing a food production process and the plant in which it takes place is a somewhat complicated task. Several questions are relevant to ask before starting the task. o What is the purpose of the description? o Who is going to use it? o In how much detail do you need to describe it? The purpose of the description may be to: o inform a government office about your production in relation to an application for increasing or altering the production. o inform the board of directors about a possible enlargement of the facilities. o give an entrepreneur information in relation to a change in a production line. o perform a critical analysis of the production facility with respect to wastewater and energy consumption. The four examples require different kinds of information and the level of detail will be different. Regardless of purpose the first thing to do is to present the overall plan of production. This is best done with a flow sheet or block diagram. However, this may be very simple at first. Basically all we need is a recipe for the product as is used in normal households and a production procedure, but with the exception that the apparatus and amount of materials are much different.

Flow sheets A flow sheet is but a graphical representation of a process. The purpose is to clarify different streams to and from the processes in the production plant. The simplest flow sheet is perhaps: Raw materials

Product

Processing

However, it is not very informative. It indicates the need for raw materials to make the product. Further specifications will enlighten us somewhat more: Raw milk

Separation

skim milk

Culture Fermentation

Cream

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Packing of ’buttermilk’

It should be recognised that the more detailed the flow sheet becomes and the more information we put in we risk loosing the comprehension of the process. Therefore a good principle is KISS:

Keep It Simple Stupid The information needed must be presented, but be critical, only include what is needed. We will give you an example of a superfluous description. Example 1. A first flow sheet The task is to produce a (continental) breakfast at 7 o’clock in the morning Menu:

Orange juice Coffee Buttered toast and marmelade Yoghurt Breakfast cereals and milk.

What do we need to know? o

How to produce orange juice from oranges? No!

o

How to brew coffee? Yes! Because it this case we have a process. Water 8°C, 800 ml

Boil water 100°C

40 g coarse milled roasted coffee beans

Extraction, 5 min

coffee-grounds (waste)

650 ml 80°C coffee Similarly we need to define how the bread is toasted, but not how to produce the bread (if we are buying it). Again we do not need to know how to milk a cow.

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Mass and energy balances Mass Balances The mass balance helps us to make bookkeeping concerning streams passing through a production line much easier. Introduction example The company Panada Aps are producing breaded products like fish sticks. You are going to introduce a new production of Turkey Cordon Bleu. You are planning a series of analysis (1000) for the content of glucose. You are going to use the method applying Glucoseoxidase. You are employed on a poultry processing plant and are given the task of informing the environmental authorities of the environmental impact caused by the production. What is the common denominator of these three tasks? Comments: They are complex and involve numerous different sub-processes. The solution requires knowledge of the amount of different materials being used or produced and the processes involved. There may even be a necessary sequence of processes in time. How would you attack the problems? Draw a flow sheet to illustrate the overall process. Include the needed information. The first example requires at least the recipe for the product. The second example requires knowledge about amounts of chemicals, temperatures and times needed and environmental conditions for disposal of waste. The third example implies very complicated overviews as it concerns streams of different components (live chicken, water, feathers, blood, offal, waste, waste water and finished product) as well as energy consumption for scalding, chilling, freezing etc (electric power, gas, oil etc.). When we are making an accounting for such tasks we use the laws of conservation of mass and constant energy. For a process in steady-state we have: IN = OUT What comes in must come out. If the process is unsteady (or transient) we may observe an accumulation of some materials: IN = OUT + ACCUMULATED or when a component is produced or consumed: IN + PRODUCED = OUT -6-

In generic terms we get: IN + PRODUCED = OUT + ACCUMULATED Correspondingly energy cannot just disappear. It may be converted to other kinds of energy (mechanical, thermal etc.) as we ignore the more exotic such as nuclear and quantum physical phenomena. These observations are valid for absolute measures (mass) as well as for streams (mass per unit time etc.) The generic procedure for establishing a mass balance goes: 1. Choose a basis (eg. 100 kg of raw material or product) 2. Draw the flow sheet with all in-and out-going streams. 3. Identify the streams of known size. 4. Mark the streams of unknown amounts 5. Identify the relationships between different streams 6. Construct the equations describing the relationships 7. If different units are used, transform to one and only one unit. 8. Identify that the established relations are independent. 9. Determine the number of variables (nv) 10. Determine the number of independent equations (ne) 11. Determine the degrees of freedom (df = nv-ne) in the system 12. If df = 0 solve the system of equations 13. Calculate the numeric value of necessary variables 14. Validate the solution. It must to be reasonable. Ad. 1 The chosen basis may be a stream (kg per time) Ad. 2 It is very important to define the boundaries of the system clearly. Ad. 5 & 6 State equations and inequalities relating the variables to one another. In the ideal case it should give us a system of linear equations. Ad. 12 If df = 0 there is only one solution to the set of equations. If df < 0 the system is over-determined and we have to evaluate the relations established. Some of the information may be redundant or more equations are linear combinations of each other. If df > 0 we may find a solution by assigning definite values to some of the variables, i.e. we are seeking a feasible combination of outcomes. This situation is sometimes preferred to the situation with only one solution. However, we might want to fix some variables. Ad. 14 The validation is necessary as negative masses are not allowed, but may occur in the solution. In such a situation the system of equations and constraints has to be revised. Energy balances are established in exactly the same way, including heat capacities etc. and a lot of problems require a simultaneous solution of both.

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Please note that the set of equations and constraints will often be so complex that a manual solution becomes difficult. In spreadsheet programs like Excel and QuatroPro the function Solver may be used to solve such problems. Example 2. Mass balance – salted butter Butter produced continuously is unsalted. To produce normal salted Danish butter salt is added as brine. Determine the amount of brine that has to be added and the allowable water content of the unsalted butter. Data: Brine 60 g salt per 100 g solution, salted butter: 15 g water per 100 g butter, 1,4 g salt per 100 g butter. Solution: The basis is selected as 100 kg salted butter. Indices: b: brine, f: fat, s: salt, sb: salted butter, ub: unsalted butter, w: water Assumptions: Fat is only part of the unsalted butter and salt is only part of brine. Constrains: All mass fractions must be > 0. The sum of mass ratios must be unity (here for salted butter): (a) xsb,w + xsb,s + x sb,f = 1 and for brine : and for unsalted butter :

(a’) (a’’)

xb,w+xb,s = 1 xub,w + x ub,f = 1

Total mass balance: (b) Balances for components:

Mb+ Mub = Msb

Salt: Fat:

(c) (d)

Mb*xb,s = Msb* xsb,s Mub,f *xub,f = Msb * x sb,f

Water:

(e)

Mb*xb,w + Mub*xub,w = Msb*xsb,w

Above the known quantities are marked with bold face. (a) =>

xsb,f = 1 – 0.155 – 0.014 = 0.831

(a’) => (c) =>

xb,w = 1 – 0.60 = 0.40 Mb= (100*0.014)/0.60 = 2.333 kg brine

(c) in (b) => Mub = 100 – 2,3333 = 97.667 kg unsalted butter in (d) => xub,f = (100*0.831)/97.667 = 0.851 g fat per g unsalted butter and (a’’) => xub,w = 1 – 0.851 = 0.149 g water per g unsalted butter Per 100 kg salted butter we mix 97.67kg unsalted butter with 2.33kg 60% brine. The composition of unsalted butter (per 100g) must be 0.851g fat and 0.149g water. Comment: This example shows the general principle. It may be solved directly by identifying the simple information and combining. The problem is a little peculiar as it puts strong limitations on the composition of unsalted butter. The natural choice would be to adjust the brine to suit the composition of the unsalted butter. Actually it is the same equations, but other unknowns. Note that equation (e) was never used. It is a linear combination of (c) and (d) combined with the (a)’s.

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Example 3. Mass balance – raw milk separation Data: Raw milk is to be separated in skimmed milk and cream in a centrifuge. The problem is to determine how much cream and skimmed milk and of what composition is produced under given demands on the composition of the cream. Solution: The basis is chosen to be 100kg raw milk. Indices: c: cream, f: fat, l: lactose (carbohydrates), p: protein, nf: non fat dry matter, rm: raw milk, s: salts, sm: skimmed milk, w: water. Assumptions: All components are water soluble except the fat, ie. that the cream may be considered as a system composed of skimmed milk and fat. The composition of the aqueous phase is constant. Constraints: All sums of mass fractions must equal unity and all mass fractions must be zero or positive.

Equations: Definition of non fat dry matter: (a)

xnf = x p + x l + x s

Sum of mass fractions:

(b)

xrm,nf + x rm, f + x rm,w = 1

(c)

xc,nf + x c,f + x c,w

(d)

xsm,nf + x sm,f + x sm,w = 1

(e)

Mrm = Msm + Mc

Fat:

(f)

Mrm * x rm,f = Msm * xsm,f + Mc * x c,f

Non fat dry matter:

(g)

Mrm * x rm,nf = Msm * x sm,nf + Mc * x c,nf

Water :

(h)

Mrm * x rm,w = Msm * xsm,w + Mc * x c,w

Total mass balance:

=1

Balances for components:

Constant concentration in aqueous phase : (i)

x rm ,nf

x x = sm ,nf = c ,nf x rmw x smw x c ,w , ,

A common assumption in many textbooks is to consider the skimmed milk to be totally fat free. This gives us a convenient system of linear equations, but as we do know that skimmed milk is not void of fat it is a gross error (and the system is not generic any more). Introducing the assumption of constant composition of the aqueous phase complicates the system. It is no longer a set of linear equations. Solutions must be found by iterative methods. Fortunately this is a simple task in spreadsheet programs like Excel and Quatro Pro using the Solver option. But care must taken in validating the results. Below is shown an example of a solution.

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Cream separation Composition Raw milk Lactose Protein Salts Non fat dm Fat Water Mass NF const

Skim milk

0.077 0.045 0.878 100 0.0877

Target

0.080 0.003 0.916 87.98 0.0877

Cream

0.052 0.350 0.598 12.02 0.0877

100

Constrains Mass fractions = 1, SM Mass fractions = 1, C Mass fractions >= 0, SM, nf Mass fractions >= 0, SM, fat Mass fractions >= 0, SM, water Mass fractions >= 0, C, nf Mass fractions >= 0, C, fat Mass fractions >= 0, C, water Mass fractions = 0.080 1 and lastly the case where n = 1 and m = m here the power-law model is identical to Newton's law of viscosity. The power-law model and other models presented later here are merely convenient engineering representations of certain rheological behaviour of liquids over a specific range of shear rates. These models are therefore not suitable for extrapolation outside the range investigated experimentally.

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Shear-thinning fluids The shear-thinning flow behaviour is by far the most important as most fluid foods exhibit such rheological characteristic, where as, only very few are shearthickening. Examples of shear thinning food products are banana purée, apple purée, mayonnaise, orange juice concentrates and starch suspensions. The general shape of a flow curve for a shear-thinning fluid is shown in Figure 3. The curve can be separated in three parts with the shear-tinning region in the middle and two Newtonian plateaus - one in each end.

Figure 3. Typical shear-thinning behaviour (Fryer et al., 1997).

Even though the shear-thinning regime dominates the flow curve for these fluids the regions with constant viscosity at low respectively high shear rates must be recognised. They are called the lower and the upper Newtonian region respectively. The levels refer to the magnitude of the shear rate and not the viscosity. These regions can be more or less pronounced and might some times lie outside the range of practically measurable shear rates. Far to the left (low shear rates) on the curve in Figure 3 the internal molecular forces dominate the structure and thereby the flow properties of the liquid; i.e. we have a random ordering and therefore a constant viscosity. As we move from left to right hydrodynamic forces from the process enforces increased structuring and alignment of molecules and this causes the viscosity to drop. In the end when the fully aligned structure is reached the viscosity again becomes constant. Beyond the upper Newtonian region increased shear rates can cause the structure to breakdown. This is for example intended in mixing processes. The power-law model can with its two parameters (n and m) only fit the shearthinning part of the flow curve shown in figure 3. The model is thus a relatively simplistic way to describe shear-thinning behaviour. Additionally a graphical interpretation of n can be found when plotting the flow curve in a double logarithmic plot, n is namely the slope of the shear-thinning region. This leads to the appreciation that n denotes the degree of shear-thinning behaviour and in praxis values of n between 0.2 and 1 is found for liquid foods. The consistency index m is a measure of the viscosity level. Examples of shear-thinning behaviour are given in Table 5. - 28 -

Table 5. Power-law parameters for selected liquid foods (Fryer et al, 1997) Food

T (°C)

n (-)

m (Pa sn)

Range of (s-1)

Banana purée

22

0.28

107.3

28 – 200

Apple sauce

26

0.45

7.32

0.78 – 1260

Mayonaise

25

0.60

4.2

40 – 1100

The values of n and m shown above are determined by fitting the power law model to experimental data sets as described in the example below. Example 10. Determining the parameters of the power law model Data: A traditional fat rich mayonnaise has been subjected to repeated measurements of the flow curve. In the figure below data from three repeats is plotted along with lines representing a power law fit for each data set.

Shear dependent viscosity of mayonaise

Viscosity [Pa s]

10 Data set 1 Model 1 Data set 2 Data set 3 Model 2 Model 3 1

0,1 10

100

1000

Shear rate [s-1] Solution: The fitting parameters for the power law models are given below: Data set 1

Data set 2

Data set 3

n

0.42

0.43

0.46

m

13,1

15.2

10.1

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Comments: We see that the three repeated measurements coincide pretty well and only small deviations are found in the model fit. In the case presented we could easily have pooled the data and made one model. We have seen the inability of the power-law model to describe the full flow curve shown in Figure 3. We therefore turn to models with more fitting parameters in realisation of that four fitting parameters are needed to describe the plateaus at each end of the curve. In Figure 4, the basic shapes of the shear thinning flow curve is expressed by the power-law, Sisko and Cross models respectively. We see that a three-parameter model like the Sisko model (not stated here) can fit the upper Newtonian plateau. The last model presented in the figure is the often applied Cross model which has the four parameters needed.

Figure 4. Typical viscosity curves from rheological models (Barnes et al., 1989).

The Cross model which is given by equation (19):

1 h - h¥ = n h 0 - h¥ 1 + ( K1g& )

(

1

)

(19)

,

here h0 and h¥ is the Newtonian viscosity at low and high plateau respectively. K1 is a constant with the dimension [s] and n1 is a dimensionless constant equivalent to the power-law exponent. It is seen that the Cross model holds fitting capabilities similar to the power-law model. A popular variant to the Cross model is the Carreau model, which has a slightly different implementation of the fitting parameters.

h - h¥ = h 0 - h¥

1

(20)

(1 + (K2g& ) ) 2

n2 / 2

The constants K2 and n2 are similar to K1 and n1 in the Cross model. - 30 -

When selecting a proper model to describe non-Newtonian flow behaviour it is important to identify the specific application needs. The power-law model can be perfectly appropriate when comparing different fluids, studying a process, which operate at shear rates in the shear-thinning region or for use in simplified fluid mechanical calculations. Where as the Cross and Carreau models gives much more detailed information and can be required especially in order to describe the viscosity levels at low or high shear rates. A drawback in using the more detailed models is that the increased number of fitting parameters (from 2 to 4) requires more experimental data points in order to fit a reliable model. It is difficult to give guideline regarding how many data points one needs to fit a given model, however, a good rule of thumb is that the experimental data points must describe the shape of the part of flow curve one wishes to model very well. If lacking data points an alternative approach can be to fix some of the fitting parameters before the data fit is performed. The viscosities for the plateau’s h0 and/or h¥ can either be measured or assigned assumed values in order to reduce the number of fitting parameters in the Cross and Carreau models. An other approach often taken is to assign n1 in the Cross model the value of 0.66. This might seem as a major restraint on the fit, however, this is found yield good fits for many liquid foods. Choosing the right model depends on the application. Always apply the model, which describes the part of the flow curve in interest. Even though we have realised that the power law model does not describe the full flow curve this model is sufficient to compare the flow characteristics of non-Newtonian fluids. Shear-thickening fluids Shear-thinning behaviour is rarely found for liquid foods; however, one example is highly concentrated starch suspensions. Fluids exhibiting an apparent yield stress In appreciation of the fact that some fluid foods at rest exhibit an extremely high resistance to deformation and the fact that the shear stress in such cases must exceed a certain value, the apparent yield stress t y (Pa), before notable flow occurs this phenomenon needs to be addressed. It is an apparent yield stress we observe when we try to get tomato ketchup to flow out of a bottle. Everybody has experienced that a high force or even violence needs to be applied in order to get the flow started but once it suddenly comes we often get more than we intended. This is an excellent example of a process where we need to apply a force above the apparent yield stress before the fluid flows, however, we easily observe that a much lower shear stress is needed to spread the ketchup on a plate. This is why an apparent yield stress only applies to a specific process and why reported yields stresses are not universal figures. Some scientists argue whether the phenomena yield stress really exists or if the flow in the initial phase is just too small to be measured with the rheological instruments available today and yet some claims that yield stress is a proven fact. No matter who is right it is of practical relevance to operate with the term - 31 -

apparent yield stress for some materials (like the ketchup and tooth paste) in certain processes. Two common models, which are useful when describing fluids exhibiting an apparent yield stress, are presented here. The Bingham plastic fluid has an apparent yield stress t y followed by a Newtonian flow region. The Bingham model is given by equation (21):

t = t y + m × g&

(21)

Examples of Bingham fluids are: toothpaste and tomato products. For fluids where an apparent yield stress is combined with shear-thinning behaviour (i.e. yoghurt) the Hershel-Bulkley model given by equation (22) can be applied:

t = t y + m × ( g& ) n

(22)

Examples of food products obeying the Hershel-Bulkley model are: minced fish paste, raisin paste and some yoghurt types. The role of elasticity and time effects As stated earlier many liquid foods possesses both viscous and elastic behavior as they to different degrees are what we call viscoelastic. This might be a bit difficult to comprehend, however, if we consider a fruit gel (like jam or marmalade) we can by compressing it slightly with a finger observe an elastic effect. The same product can be brought to flow in a tube given that an appropriate shear stress is applied. In most steady state processing applications we choose to neglect the elastic component since the elastic effects die out after a start-up phase in the processing. However, there are processes like extrusion of liquorices where contraction and pressure build-up and other force induced forming processes where elastic components play an important role when selecting the proper process parameters. Such processes are, however, not included here. Temperature dependency In most cases the temperature dependency of the flow curves must be incorporated in the shear viscosity model by applying an additional fit of the model parameters to their variation with temperature. See under Newtonian fluids for reference to a model for temperature dependency (page 23).

Rheology solid and semi-solid food products This part is unfortunately still under construction

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Fluid mechanics Transport of matter is an essential operation in manufacturing plants. When it comes to fluids they are pumped, flow in pipes, are mixed and filled into containers etc. These are all operations, which are based on the flow behaviour of the fluids. The actual flow conditions found in processing equipment are determined where fluid mechanics is combined with the product rheology. Fluid mechanics of Newtonian and Non-Newtonian liquids is therefore a basic necessity to understand in order to work qualified with operation and design of processing equipment. The basics of fluid mechanics are introduced for Newtonian fluids and extended to cover the most relevant Non-Newtonian fluids as well.

Newtonian fluids The nature of the flow pattern depends greatly on the rate of the flow. At low rates fluids flow in an orderly manner, which is highly reproducible. The flow is said to be laminar and ideally no exchange of fluid takes place in the height of the tube. This type of flow can be pictured as parallel layers, which flows smoothly side-byside along the length of the tube and for flow in a tube with a circular crosssection the velocity profile is parabolic which can be seen to the left in Figure 5. In laminar flow the viscous forces dominates this means that the fluid viscosity is of higher importance than the hydrodynamics of the flow it self. At higher flow rates the flow becomes turbulent. The flow pattern is now formed randomly, there is an increased exchange of fluid in the height of the tube and the flow is no longer reproducible but fluctuates continuously. The turbulent velocity profile is shown to the right in Figure 5. The hydrodynamic forces are now dominant and hence the fluid properties less important for the nature of the flow. Between the true laminar and fully developed turbulent flow regimes a transition takes place, in this zone it is difficult to determine the nature of the flow.

Figure 5. Laminar and turbulent velocity profiles for pipe flow. Here vav. is the average velocity (Leniger & Beverloo, 1975).

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The parabolic velocity profile for the laminar can be derived from exact mathematics. However, there is no such exact mathematical equation for the turbulent velocity profile since random movements dominate the flow. By examining the two profiles we can easily see that the turbulent velocity profile is more flat nosed than the laminar and that the difference between the maximum and the average velocity is much less for turbulent flow compared to laminar. When observing the velocity profiles we see that both touch the wall of the tube. This means that in this position where the lowest velocity is observed the fluid velocity matches that of the tube wall. Thus the velocity is zero here. This is the generally accepted boundary condition is that there is no slip between fluid and any hard surface (e.g. a tube wall). In reality this is of course an assumption but it yields realistic results in must situations where the slip is negligible. However, there are situations where lubrication of the surface takes place. In such cases the flow characteristics must be determined by experiments. The Reynolds number (Re) is a dimensionless number, which yields the ratio of inertia to viscous forces and thus determine the nature of the flow. For low values of the Reynolds number the flow is laminar and for higher values first transition takes place and in the end fully developed turbulent flow develops. The limits for the Reynolds number determining the flow type depends on the cross section profile of the geometry studied. In a cylindrical pipe laminar flow is found for values of Re below 2100 (different sources reports the limit to be anywhere in the interval of 2000 to 2300). This value is called the critical Reynolds number (Rec) and is highly geometry depended. The flow is considered fully turbulent when the Reynolds number exceeds 4000. Hence transition flow is found between these two values. The Reynolds number for Newtonian fluids is defined by:

Re =

v ×d× r m

(23)

where v (m/s) is the average velocity of the flow, d (m) is a characteristic distance (for a pipe the diameter), r is the density of the fluid and m the Newtonian viscosity. Alternatively the expression for the Reynolds can be rewritten to depend on the volume flow (V& ) instead of the average velocity.

Re =

4V& d ×r 4 V& r = p d ×m p× d 2 m

(24)

The expression for the volume flow is found in appendix 1. Example 11 – Calculation of Reynolds number Data: We calculate the Reynolds number for water flowing in a 2” (0.0508m) pipe at an average velocity of 2m/s at 25 °C. Physical properties for water at 25°C: ρ = 1000kg/m3 and µ = 9⋅10-4 N⋅s/m3. Hence we get: V& ≈ 0.00405m3/h - 34 -

Solution: Re =

4 × 0.00405 × 1000

-4

3.14 × 0.0508 × 9.0 × 10

» 1.13 × 105

It is seen that the flow is very much in the turbulent flow regime. Comments: Had the fluid been a grape juice the viscosity would be 0.11 Pa⋅s (taken from Table 4), hence yielding a Reynolds number of 923, in which case the flow is laminar. We have assumed that the density does not change, which can seem a bit careless and inaccurate at first. However, applying the right value will only change the Reynolds number slightly and does by no means change the result that the flow is laminar. By carrying out few calculations we can conclude that the density only plays a minor role in determining the flow characteristics. Looking at the expression for the Reynolds number given in equation (24) the significant parameters influencing the magnitude are identified as the flow rate, the pipe diameter and the viscosity. In creasing the flow rate leads to an increase in the Reynolds number, where as an increase in either the diameter of the pipe or the viscosity would decrease the magnitude. The nature of the flow has great implications for the process performance such as heat transfer, residence time and mixing. Turbulent flow will for example yield better heat transfer performance, however, the pressure needed to drive the liquid trough the process plant will increase at the same time and might limit the practical possibility to go to turbulent flow if for example the viscosity is very high. This is indeed the case for many liquid food products as they have a relatively high viscosity and thus yielding moderate Reynolds numbers. This is in contrast to classical chemical processes where high Reynolds numbers are found for flows involving gasses and low viscosity fluids. Laminar flow The parabolic velocity profile of laminar flow can described by equation (25):

æ DP ÷ö é 2 v ( r ) = çç ÷ R - r 2 ùú çè 4 × L ×m ÷÷ø êë û

(25)

where v(r) [m/s] is the velocity in position r [m] in the tube, ∆P [Pa] is the pressure gradient over the tube, L [m] the length of the tube, R [m] the radius of the tube and r [m] the actual radial position in the tube. In engineering calculations the exact shape of the velocity profiles is often not as important as the relation between key characteristics of the flow and selected characteristic parameters. The magnitude of matter flowing is specified by: a volumetric flow rate V& [m3/s], a & [kg/s] or an average velocity v [m/s]. The rate of the flow is of mass flow rate m interest as it is related to the pressure gradient, which can be looked upon as the pump energy requirement needed to keep the fluid flowing. - 35 -

Equation (26) yields the relation between the average velocity and the pressure gradient:

d 2 DP DP 32m × v v= Û = 32m L L d2

(26)

The relationship between the average velocity and the maximum velocity given by equation (27) is relevant when evaluating the effect and choosing process parameters of continuous processes (please refer to appendix 1 for more relevant relationships):

v max = 2 × v

(27)

Example 12. Calculating pressure drop in laminar flow Data: If we look back at the previous example with the grape juice where the flow were found to be laminar it is now possible to calculate the pressure needed to drive the product through a pipe. Solution: As we do not know the length of interest we choose to calculate the pressure drop for pr. meter pipe.

DP / L =

32 × 0.11× 2 0.05082

» 2728Pa / m » 0,027 bar / m

Comments: Which is a reasonable magnitude for a pressure drop. Had the process on the other hand taken place in a 1” pipe the pressure drop pr. meter pipe rises to 0,11bar (Re=1847), which can give cause to considerations as the pressure needs to be build up by a pump. A practical limit in the pressure is considered in the range of 4 to 7bar (or in some cases up to 10bar) depending on the product. Assuming the pressure limit is 7bar yields a maximum length between two pumps of 64m for the 1” pipe, where as it is 257m for the 2” pipe (4bar limit yields a maximum length of 147m). In this case the smaller pipe would most certainly not be applied. The shear stress relation is given by equation (13) and is proportional to the velocity gradient. For the parabolic velocity profile we know that the first derivative is a straight line and furthermore that in the top point this value is zero. We can now intuitively deduct that the shear stress has its maximum at the wall of the tube and that it from this point decreases linearly to zero in the centre of the tube. We will later appreciate the importance of this when dealing with nonNewtonian fluid mechanics. The wall shear stress tw can be expressed by:

tw =

16 ×m× v R

(28)

- 36 -

Turbulent flow No exact mathematical analysis exists for turbulent flow, though some semitheoretical expressions have been suggested. A suitable empirical formulation has shown that the velocity at any position in the cross-section of a tube will be proportional to the 1/7 power of the distance from the wall. The velocity profile can now be expressed using the notation applied in equation (25) for the laminar flow (Coulson & Richardson, 1996). 1/7 æ v( r) rö = 1.22 × çç1 - ÷÷÷ çè v Rø

(29)

The relation between the maximum and the average velocity is found in the centre of the pipe where r equals 0:

v max = 1.22 × v

(30)

The relation between the average velocity and the pressure gradient can be expressed by the Fanning friction equation, which can be derived from equation (26) by introducing the expression for the Reynolds number (Fryer et al., 1997 & Toledo, 1991):

DP 2 × f × v 2 × r = L d

(31)

In equation (31) f is the friction factor, which for laminar flow is 16/Re and for turbulent flow in a smooth pipe f can be computed from equations (32) and (33):

f = 0.048Re-0.20 104 < Re < 106

(32)

f = 0.193Re-0.35 3 × 103 < Re < 104

(33)

Equation (33) can even though the lower limit in the Reynolds number is 3000 it reasonable to assume that this equation is valid down to a Reynolds number of 2100. We see from equations (32) and (33) that the influence of an increase in the Reynolds number on the pressure loss becomes less important the higher the magnitude of the Reynolds number. The value for the friction factor can also be determined from the Moody diagram (see appendix 7), which also holds information for rough pipes. Example 13. Calculating pressure drop in turbulent flow Data: In Example 11 we determined a flow of water to turbulent. Pipe diameter 2” (0.0508m), average velocity: 2m/s, temperature 25 °C, ρ = 1000kg/m3 and µ = 9⋅10-4 N⋅s/m3, V& ≈ 0.00405m3/h and Re = 1.13*105

- 37 -

Solution: Given the magnitude of the Reynolds number the friction factor is 4 -0.20

computed from equation (32), hence f = 0.048(1.13 × 10 )

= 0,00468 .

The pressure drop pr. meter pipe is then determined by insertion in equation (31):

DP 2 × 0,00468 × 22 ×1000 = » 738 Pa / m »= 0,0074 bar / m L 0,0508 Comments: This pressure drop is according to Example 12 very much reasonable. Had the average flow rate instead been 3m/s we would get: V& ≈ 0.0061m3/h, Re = 1.7*105, f =0,04317

DP 2 × 0,0043 × 32 × 1000 = » 1530 Pa / m »= 0,015 bar / m L 0,0508 This new value is twice as high as the previous but still of an applicable magnitude. From the example we find that the magnitude of the friction factor is not much affected by the change in the Reynolds number caused rise in the average velocity as would be the case for any change in the Reynolds number (e.g. as the result of altering the viscosity). The main factor affecting the pressure drop is in this case the average velocity it self. A similar effect could have been seen if the pipe diameter was reduced by a factor of two. The wall shear stress still assumes it maximum value at the wall of the tube and the symmetry still causes the value to be zero in the centre of the tube even though the decay is no longer linear. The wall shear stress can be calculated from the general equation given below.

tw =

d DP 4 L

(34)

Non-Newtonian fluids When applying a model of non-Newtonian behaviour in fluid mechanics a relation between the fluid behaviour and the constitutive equation of the flow need to be determined. Since the velocity profile depends on the viscosity it is of utmost importance to describe the viscosity change with the shear rate, which in some cases be as much as 2 to 4 decades. Such a change cannot be ignored when calculating velocity profiles and pressure drops in process equipment. One of the earliest and most often implemented models is that of the generalised Newtonian fluid. The generalised Newtonian fluid is a minor modification of the Newtonian fluid where the viscosity is allowed to be a function of the shear rate. The practical application of non-Newtonian models in fluid mechanics is handled by implementing the power-law model from equation (18) into the governing equations for fluid flow (Bird et al., 1987). The mathematics, the theoretical consideration and assumptions of applying the generalised Newtonian fluid are not covered here. The reader must refer to Bird et al., 1987 for this information. - 38 -

Analogous to Newtonian fluid mechanics we can compute the Reynolds number. A rewritten equation for the Reynolds number incorporating the exponent n and consistency index m from the power-law model is given by equation (35).

n ö÷ v (2- n) × d n ×r ÷ èçç 3 n + 1ø÷ m

3- n çæ

Re = 2

(35)

where v [m/s] is the average velocity of the flow, d [m] a characteristic length of the geometry being considered, ρ [kg/m3] the density. Example 14. Calculating Reynolds numbers for non-Newtonian fluids Data: Utilising the product data from Table 4 on page 29 we can calculate the Reynolds number. Banana pure (n=0.28 and m=107.3), pipe diameter 1”, average velocity: 1,1m/s, ρ = 1000kg/m3 and V& ≈ 2.0m3/h. Solution: According to equation (35) the Reynolds number yields:

0.28 ö÷ 1.1 (2-0.28) × 0.02540.28 × 1000 » 3.9 ÷ èçç 3 × 0.28 + 1ø÷ 107.3

3-0.28 æç

Re = 2

Comments: Such low Reynolds numbers are no uncommon when dealing with nonNewtonian fluids exhibiting a high viscosity level and as is seen from the table in Example 16 all cases considered proves well within the laminar flow regime. We can also deduct an equation for determination of the critical Reynolds number where shear-thinning behaviour is considered.

Rec = 2100

(4 n + 2)(5 n + 3) 2 3(3 n + 1)

(36)

Following equation (36) it can be observed that Rec is higher than 2100 for n < 1 which denotes shear thinning behaviour. We find that the limit between laminar and turbulent flow shifts to higher Reynolds numbers as n decreases, where as it shifts to lower values for n>1. The relationship between Rec and n is plotted in Figure 6. Example 15. Critical values of the Reynolds number for power law fluids It is very important not to confuse this number with the Reynolds number. Data: As stated above the critical Reynolds number designates the limit between the flow being laminar or turbulent. Solution: For a fluid with an n of 0.5 a Rec of 2463 is found which means that the flow stay laminar for higher flow rates compared to the Newtonian case, where the critical values is 2100. For a dilatant fluid where n is equal to 2 the critical limit drops to 1857. - 39 -

Comments: We see that laminar flow can be found for higher values of the Reynolds number for shear-thinning fluids where as the transition to turbulent flow occurs earlier for the shear-thickening fluids.

3500

n1

1500 0

1

2

3

4

5

Power law index (n)

Figure 6. Critical Reynolds numbers Rec as a function of the power law index n.

Laminar flow For laminar flow of a Newtonian fluid in a tube we found that the velocity profile takes the form of a parabola. That result was deducted from the constitutive equations for the flow. By incorporating the power-law model for the viscosity in the same set of equations we can derive an analytical mathematical formulation for the velocity profile for a power-law fluid in a pipe. The equation describing the velocity profile is given by equation (37).

æ n ÷öæ DP ÷ö1/ n é 1+1/ n çç v ( r ) = çç R - r 1+1/ n ùú ÷ ÷ ê ÷ ÷ çè n + 1øèç 2 × L × m ø ë û

(37)

where v(r) [m/s] is the velocity in position r, DP [Pa] is the pressure gradient over the tube, L [m] the length, R [m] the radius of the tube and r [m] the actual radial position in the tube. We see that for the case of a Newtonian fluid where n = 1 and m = m equation (37) reduces to the parabolic form shown in equation (25).

The expression for the relationship between the average velocity and the pressure drop is given by:

æ DP 1 ÷ö1/ n n v = çç d( n+1)/ n çè L 4 m ÷÷ø 6n+ 2

- 40 -

(38)

Example 16. Calculating pressure drop for non-Newtonian fluids Data: Banana pure (n=0.28 and m=107.3), pipe diameter 1”, average velocity: 1.1m/s, ρ = 1000kg/m3 (as in the example Example 14). Solution: Equation (38) needs to be rearranged before being able to calculate the pressure drop. 0.28

1.1 c

æ 0.28 ö÷0.28 1 DP çç 0.0254 (0.28+1) = ÷÷ ç L 4 × 107.3 è6 × 0.28 + 2 ø

1.10.28 × 4 × 107.3 DP = » 1.0 × 10-5 Pa / m = 1.0 bar / m 0.28 L æ 0.28 ö÷ çç 0.0254(0.28+1) çè 6 × 0.28 + 2 ø÷÷ This is a high pressure drop pr. meter as we can only pump the product up to 7 meter, which will be considered to short in most processing applications. Utilising the data for all three products presented in Table 5 on page 29 a number of examples are given to the relation between the flow rate, the Reynolds number and the pressure drop. Process characteristics for selected power law fluids Product

n

v

m n

Banana pure Apple sauce Mayonaise

d

V& 3

Re

∆P / L

[Pa⋅s ]

[m/s]

[m]

[m /h]

0.28

107.3

1.1

0.0254

2.0

3.9

1.00

0.28

107.3

0.125

0.0762

2.1

0.1

0.13

0.45

7.32

1.4

0.0508

10.2

68

0.07

0.45

7.32

2.8

0.0508

20.4

198

0.10

0.6

4.2

1.0

0.0254

1.8

30

0.23

0.6

4.2

0.445

0.0381

1.8

12

0.07

[bar/m]

Comments: All examples apply realistic volume flow for industrial applications; hence suitable values of the average velocity and pipe diameter are selected in order to achieve pressure drops with in a realistic range. By normalising equation (37) using the average velocity v given in equation (38) we can eliminate all dependencies but n and the radial position in the tube. This allows us to compare the general shape of velocity profiles for different power-law fluids. The normalised velocity profiles are calculated from equation (39):

v ( r ) æç 3n + 1÷ö é 1+1/ n =ç - r1+1/ n ùú ÷÷ ê R ç ë û è n +1 ø v - 41 -

(39)

In Figure 7 velocity profiles for shear-thinning as well as shear-thickening fluids are compared to the parabolic profile of a Newtonian fluid. We see that the velocity profiles for the shear-thinning fluids have a more flat nosed profile similar to that seen for turbulent flow. For the shear-thickening fluids, however, the velocity profile assumes a more triangular shape. The most obvious result is that the relation between the average and the maximum velocity increases for shear-thickening fluids where it decreases for shear-thinning fluid.

Figure 7. Velocity for power-law fluids for different values of the power-law exponent n (Holdsworth, 1992).

The velocity profile for n = 0 is identical to that of plug flow. Plug flow is an ideal case, which never is practically obtainable in the whole cross-section of the tube. For the power-law fluid the relation between the maximum and the average velocity is by the following equation:

v max =

3n+1 ×v n+1

(40)

If we consider flow of a Bingham fluid described by equation (21) we can show that at a radius less that ry the fluid stress does not exceed the apparent yield stress value and material in the centre of the tube moves as a plug as can be seen in Figure 8. The value of ry can be determined from equation (41).

ry =

4t y

(41)

DP / L

- 42 -

Figure 8. Velocity profile in a tube for a Bingham plastic fluid (Fryer et al., 1997).

Turbulent flow When it comes to turbulent flow there are strong indications that non-Newtonian fluids show flow behaviour similar to that of Newtonian fluids. We have earlier shown that the fluid properties are less important to the nature of the flow in the turbulent region. On top of this it has been reported that if we assume a nonNewtonian fluid to be Newtonian in turbulent flow we slightly over-predict the pressure drop. It can therefore be concluded that there is a safety factor build in when treating turbulent flow of non-Newtonian fluids as if it was a Newtonian fluid. Hence we choose to calculate the Reynolds number using equation (35) and after worth determine the friction factor using either equation (32) or equation (33). The pressure drop is now determined by insertion in equation (31).

Flow through pipe fittings Often we need to calculate the pressure drop for other equipment parts than straight pipes. This is performed in a way that the equivalent length of straight pipe is computed using the dimensionless ratio of L’/D. Below L’/D is listed for some often used fittings. Table 6. Equivalent L’/D ratios for fittings. (Adapted from Toledo, 1991) Fitting type

L’/D

90° elbow

35

45° elbow

15

Tee (used as a coupling), flow straight through

20

Tee (used as an elbow)

60

Gate valve, fully open (butterfly ventil)

10

Globe valve, fully open (kugle ventil)

290

Diaphram valve, fully open (membrane ventil)

105

Couplings and unions

neglible

- 43 -

The length L’ is calculated based on a known diameter and subsequently added to the total length of straight pipe in the system. Example 17. Determining the pressure drop in fittings 1 Data: We have a 90° elbow of a dimension of 1” we get a value of L’/D of 35. Solution: L’ yields:

L ' = D × 35 m = 0.0254 × 35 m = 0.9 m This length is to be added to the total length of straight pipes and other fittings and subsequently the pressure prop must be computed using the relevant equation. Comments: We se that an elbow is almost equivalent to a whole metre of tube Example 18. Determining the pressure drop in fittings 2 Data: An other example is the Globe valve, where L’/D is 290 Solution: In this case L’ is as much as:

L ' = D × 290 m = 0.0254 × 35 m = 7.4 m Comments: We expect that there is a significant pressure drop over a valve and here the maximum values is given.

- 44 -

Heat transfer Processes involving heat transfer are highly important to food processing as processes involving heating and cooling is the most common in food processing plants. All processed foods shall be subjected to pasteurisation in order to prevent the presence of pathogenic microorganisms and to reduce the risk of microbial or enzymatic degradation of foods during transport and storage. We also apply heat or cold to introduce changes to the products in order to achieve desired characteristics of a food product e.g. flavour, texture, structure, colour etc. Examples of processes were heating and cooling are involved are: pasteurisation, sterilisation, cooking, dehydration, drying, freezing and storage. A few examples of the physics involved in such processes are given as an introduction. Pasteurisation and sterilisation processes apply fast heating and cooling, which implies that very high rates of transport of heat energy in and out of a product are of interest. In the case of a cool storage room or tempered storage tank insulation is applied to minimise the energy loss to the surroundings. Hence the interest is how low energy transfer rates can be achieved. Transferring heat from one media to another is called heat transfer and it is comprehensive to separate steady state and unsteady state heat transfer. In steady state heat transfer the temperature at a specific location in the equipment is constant over time, hence the process itself is considered to be in steady state. Unsteady state heat transfer involves changing temperatures over time in the equipment and the product, hence yielding an unsteady process. Example 19. Steady state heat transfer 1 For a freeze storage room the temperature inside the room and in the environs can be considered constant. In order to maintain constant temperature inside the freezer we must remove energy constantly. Hence we have a steady state heat transfer process with a transport of energy even though the temperatures are constant. Example 20. Steady state heat transfer 2 In a continuous heating operation product flow through a heat exchanger at a constant rate and both inlet and outlet temperatures are constant. Thus we need to transfer energy to the process at a constant rate. The process involves a continuous flow of product but the temperature in a specific location is constant. Hence we have a steady state heat transfer process. Example 21. Unsteady state heat transfer A piece of meat is put in cooking brine. The temperature of the meat is constant throughout the piece before cooking. When the cooking starts the temperature changes gradually over time. We can observe this by cutting the meat and looking - 45 -

at the colour when the cooking is stopped. If we wait long enough the temperature will be the same all through the piece but at shorter times we might find raw meat near the centre while the outer layers are cooked. This process involves a constant cooking temperature and changing temperature in the meat over time. Hence the heat transfer is unsteady. Transmission at the surface in the freeze storage, at the surfaces in the heat exchanger and at the surface of the meat is controlled by convection whilst the transmission of energy through walls and inside the meat is due to conduction. These to cases of heat transfer will be dealt with separately.

Steady state conduction heat transfer Transport of thermal energy in solid matter can be expressed by Fouriers equation:

dQ & dT = Q = -k × A [J/s] dt dx

(42)

Where Q& [J/s] or [W] is the rate of energy transferred, k [W/(m×K)] is the thermal conductivity, A [m2] is the area over which the energy is transferred, dT [K] is the driving force and dT / dx [K/m] is the temperature gradient. Example 22. Steady state conduction heat transfer Data: An illustration is the case where a stick connects two reservoirs of different temperature T1 and T2 ,where T1 > T2 . The cross section area of the stick is A [m2] and the length is L [m]. The two reservoirs hold a constant temperature (or has a sufficiently high heat capacity that the temperature does not change during the process). Solution: In this case the driving force is (T1 - T2) [K] and the temperature gradient (T1 - T2)/L [K/m]. This makes it possible to determine the temperature at any position in the stick since we know the temperature at both ends and the distance to the point in question. Comments: However simple this example demonstrates the driving temperature gradient in pratice. In the steady state situation equation (42) can be written as:

T - T2 DT Q& = -k × A × 1 =- k × A × [J/s] L L

(43)

The energy flux [W/m2] is often an interesting characteristic:

Q& k = - × (T1 - T2 ) [W/m2 ] A L

(44)

- 46 -

We see that the energy flux is proportional to the temperature difference and to (k/L). Hence this term yields valuable information regarding the characteristics of such transport processes. We make an analogy to Ohms law (from electricity) and get that the energy flux is equivalent to the current (I) and that the temperature difference the voltage (U). Ohms law is written as: U = R × I or I = U / R . Hence (k/L) must be equivalent to the resistance R, which can be described as the resistance towards transport of energy in solid material. We therefore call (k/L) the thermal resistance. Equation (43) can now be written as:

A ×DT Q& = [J/s] R

(45)

If we have a number of layers in close contact with each other we can describe the total resistance as the accumulated sum of the individual resistances. This is again analogous to Ohms law where summations of resistors in series also are applied. Assuming that the cross section area is constant for the layers we get:

xj x x x R = å R j = R1 + R2 + ... + Ri = 1 + 2 + ... + i = å k1 k2 ki kj

(46)

The assumption is of cause that the layers are in immediate contact and that the resistance between the layers therefore can be neglected. Furthermore the temperature of the surfaces is maintained constant (in reality the outer surfaces). If there is a resistance in the transition between layers (and not infinitely good heat conductivity) this has to be expressed by an additional resistance. This is often the case where heat is transferred between fluids or gasses in motion and solid material. Due to friction a layer of the fluid near the solid material will not be in motion or move very slowly. Thus creating an isolating film towards the solid material. This is in reality convection heat transfer, which will be covered in detail later. At this point we merely extend to resistance analogy to cover transfer processes in general. A wall where the surfaces are in contact with a warm and a cold fluid respectively (like in a heat exchanger) the total resistance is expressed by an expression similar to that in equation (46).

Rtotal = Rfilm,cold + Rwall + Rfilm,warm

(47)

The resistance in the thin film layers can be described by the following equation:

x Rfilm = film k film

(48)

- 47 -

The heat conductivity k film of the fluids can be determined, however, the film thickness x film is not easily estimated. Hence the term film heat transfer coefficient h is introduced as an expression of this resistance:

k h = fluid [W/(m 2 × K)] x film

and

Rfilm =

1 x = film h k fluid

(49)

The result from equation (49) is inserted in equation (47), which yields an expression for the total resistance in the transfer of heat between two fluids through a solid material.

Rtotal =

1 hcold

+ Rwall +

1 hwarm

=

1 hcold

+å

xj kj

+

1 hwarm

(50)

It is convenient to introduce the overall heat transfer coefficient U, which is the inverse of the total resistance ( U = 1/ Rtotal ). Hence equation (45) can be rewritten to the more appropriate form:

Q& = U × A ×DT [J/s]

(51)

Hence the calculation of the resistances can be written in a general way as:

xj 1 1 1 = +å + U h1 k j h2

(52)

The reason the reciprocal values are applied is psychological since the larger the magnitude of U the more energy can be transferred. The same is of cause the case concerning the film heat transfer coefficients. The remainder of this chapter concentrates to a great extend on determination of film heat transfer coefficients and over all heat transfer coefficients, since these numbers determine the efficiency of most of the processes where Q& are of interest to us. Example 23. Validation of energy transfer in a freezer Determine the maximum amount of energy (heat) needed to be removed from a domestic freezer on a summer day. Data: temperature of the environs Tenv=25°C, temperature in the freezer Tfreeze=18°C, the thickness of the insulation layer (polyurethane foam) is 5cm and kfoam=0,033W/(m×K). The outer dimensions of the freezer are: H=80cm, D=60cm and L=120cm. Solution: Since we are interested in the maximum effect we only need to know the resistance towards heat transfer in the insulation layer. We set the resistance to zero on both inner and outer surface. Hence the film heat transfer coefficients are infinite.

- 48 -

The resistance towards heat transfer is calculated from the reduced form of equation (50):

0.05 m x Rfoam = foam = = 1.515( m × K ) /W k foam 0.033W /( m × K ) Hence the maximal energy flux yields by insertion in equation (45):

Q& DT 25 - (-18) = = = 28.4W / m2 A R 1.515 2

The outer area of the freezer: A = 2((0.8 × 0.6) + (0.8 × 1.2) + (0.6 × 1.2)) = 4.32 m The total amount of energy then yields:

Q& Q& = × A = 28.4 × 4.32W = 122.7W A Comments: At first it might not be all that obvious that we can neglect the resistance towards heat transfer from the wall to either side. We consider what would change if we include these resistance contributions. R would increase and Q& / A would decrease. Hence Q& would decrease as well not yielding the maximum energy requirement anymore. It is true though that in most practical situations the amount of energy we need to remove from the freezer is lower that the maximum value.

Steady state convection heat transfer Several cases of convection are of interest to us in food process engineering calculations. Natural convection occurs when there are differences in density in the bulk of a fluid. This can be exemplified by a hot fluid being in contact with a cold wall can cause such differences. The fluid near the wall will be cooled and thus the density increases. This will cause the cold fluid to move downwards near the wall and hence a natural convection flow occurs. Forced convection is found where fluid or air is transported past a surface (i.e. by a pump or a ventilator). This type of convection is dominant in flow in closed processing equipment such as heat exchangers or in ventilation systems. The third case is convection due to cooking or condensation. The magnitude of energy transported from a solid surface to a fluid (or vice versa) depends on the temperature difference and the transport velocity (the degree of convection). The heat transfer conditions in convection cannot be calculated directly, however many experimental studies have produced numerous suitable empirical relations. Having many empirical expressions it is especially important to know the background and relevant assumptions for these before using them.

- 49 -

The task is to determine the film heat transfer coefficient h for this purpose a dimensionless relation between heat transfer due to convection and conduction has been introduced. This number is the Nusselt number (Nu), which relates to h through:

Nu = h ×

d

(53)

k fluid

In the following a relevant selection of the empirical expressions for the Nusselt number is presented. The equations determining the Nusselt number are based on other dimensionless numbers. For obvious reasons the Reynolds number (introduced on page 34) is important for forced convection and the Prandlt number (introduced on page 23) yields information on the fluids applied. The additional dimensionless numbers applied in the Nusselt number expressions are given below. Grashofs number Gr describes the relation between the buoyancy forces and viscous forces. In other words this is the relation between gravity forces and density differences due to temperature differences. Hence the Grashof number is only of significance for natural convection.

Gr =

d3 × g × b × r 2D T

(54)

m2

where d is a charteristic dimension [m], g is the acceleration of gravity [m/s2], b is

the volumetric expansion coefficient, DT is the temperature difference between wall and the bulk fluid and m is the viscosity. The physical parameters are evaluated at the film temperature Tf = (Twall - Tbulk ) / 2 .

Graetz number Gz is a convenient short form of the Reynolds number, the Prandlt number and the diameter to length ratio:

pæ dö Gz = ççRe× Pr× ÷÷÷ 4 çè lø

(55)

Also the Rayleigh number Ra is combining two other dimensionless numbers for the purpose of conveniences:

d3 × g × b × r 2DT Ra=Gr × Pr = m× k

(56)

The equations for the Nusselt number take the general form of:

æ æ m öö ædö Nu = f çççRe,Pr,ççç ÷÷÷,Gr,çç b ÷÷÷÷÷÷ çè m ø÷÷ø÷ è Lø çè w

(57)

where d /L is the diameter to length ratio which appears in laminar flow, m b the bulk viscosity and m w the viscosity at the wall. The physical properties of the fluid - 50 -

are determined using the average temperature of the fluid unless otherwise indicated (e.g. for m w which is the evaluated at the wall temperature). Natural convection For natural convection to air, water or other fluids many relations for the Nusselt number exists covering both highly specialised and more general cases. Most equations are on the form, where the parameters a and b can be found in Table 7:

Nu = a × Ra b

(58)

however, some equations are more elaborate. Table 7. Relations for Nu valid in natural convection Geometry

Characteristic dimension d in Gr

Limits for Ra=Gr×Pr

a

b

Type of flow

Vertical plate

Height > 1m

< 104

1.36

0.2

Laminar

Vertical cylinder

As vertical plate if: D>(35*height)/Gr0,25

104