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CHAPTER 5
HEAT TRANSFER THEORY
Heat transfer is an operation that occurs repeatedly in the food industry. Whether it is called cooking, baking, drying, sterilizing or freezing, heat transfer is part of the processing of almost every food. An understanding of the principles that govern heat transfer is essential to an understanding of food processing. Heat transfer is a dynamic process in which heat is transferred spontaneously from one body to another cooler body. The rate of heat transfer depends upon the differences in temperature between the bodies, the greater the difference in temperature, the greater the rate of heat transfer. Temperature difference between the source of heat and the receiver of heat is therefore the driving force in heat transfer. An increase in the temperature difference increases the driving force and therefore increases the rate of heat transfer. The heat passing from one body to another travels through some medium which in general offers resistance to the heat flow. Both these factors, the temperature difference and the resistance to heat flow, affect the rate of heat transfer. As with other rate processes, these factors are connected by the general equation: rate of transfer = driving force / resistance For heat transfer: rate of heat transfer = temperature difference/ heat flow resistance of medium During processing, temperatures may change and therefore the rate of heat transfer will change. This is called unsteady-state heat transfer , in contrast to steady-state heat transfer when the temperatures do not change. An example of unsteady-state heat transfer is the heating and cooling of cans in a retort to sterilize the contents. Unsteady-state heat transfer is more complex since an additional variable, time, enters into the rate equations. Heat can be transferred in three ways: by conduction, by radiation and by convection. In conduction, the molecular energy is directly exchanged, from the hotter to the cooler regions, the molecules with greater energy communicating some of this energy to neighbouring molecules with less energy. An example of conduction is the heat transfer through the solid walls of a refrigerated store. Radiation is the transfer of heat energy by electromagnetic waves, which transfer heat from one body to another, in the same way as electromagnetic light waves transfer light energy. An example of radiant heat transfer is when a foodstuff is passed below a bank of electric resistance heaters that are red-hot.
Convection is the transfer of heat by the movement of groups of molecules in a fluid. The groups of molecules may be moved by either density changes or by forced motion of the fluid. An example of convection heating is cooking in a jacketed pan: without a stirrer, density changes cause heat transfer by natural convection; with a stirrer, the convection is forced. In general, heat is transferred in solids by conduction, in fluids by conduction and convection. Heat transfer by radiation occurs through open space, can often be neglected, and is most significant when temperature differences are substantial. In practice, the three types of heat transfer may occur together. For calculations it is often best to consider the mechanisms separately, and then to combine them where necessary. HEAT CONDUCTION In the case of heat conduction, the equation, rate = driving force/resistance, can be applied directly. The driving force is the temperature difference per unit length of heat-transfer path, also known as the temperature gradient. Instead of resistance to heat flow, its reciprocal called the conductance is used. This changes the form of the general equation to: rate of heat transfer = driving force x conductance, that is: d Q/ d t = k A d T/ d x (5.1) where d Q/ d t is the rate of heat transfer, the quantity of heat energy transferred per unit of time, A is the area of cross-section of the heat flow path, d T/ d x is the temperature gradient, that is the rate of change of temperature per unit length of path, and k is the thermal conductivity of the medium. Notice the distinction between thermal conductance, which relates to the actual thickness of a given material ( k / x ) and thermal conductivity, which relates only to unit thickness. The units of k, the thermal conductivity, can be found from eqn. (5.1) by transposing the terms k = d Q/ d t x 1/A x 1/(d T /d x ) = Js-^1 x m-^2 x l/(oC m-^1 ) = Jm-^1 s-1 oC-^1 Equation (5.1) is known as the Fourier equation for heat conduction. Note: Heat flows from a hotter to a colder body that is in the direction of the negative temperature gradient. Thus a minus sign should appear in the Fourier equation. However, in simple problems the direction of heat flow is obvious and the minus sign is considered to be confusing rather than helpful, so it has not been used.
Figure 5.1. Heat conduction through a slab EXAMPLE 5.1. Rate of heat transfer in cork A cork slab 10cm thick has one face at – 12 oC and the other face at 21oC. If the mean thermal conductivity of cork in this temperature range is 0.042 Jm-^1 s-1 oC-^1 , what is the rate of heat transfer through 1 m^2 of wall? T 1 = 21oC T 2 = - 12 oC T = 33oC A = 1m^2 k = 0.042 Jm-^1 s-1 oC-^1 x = 0.l m q = 0.042 x 1 x 33
= 13.9 Js-^1 Heat Conductances In tables of properties of insulating materials, heat conductances are sometimes used instead of thermal conductivities. The heat conductance is the quantity of heat that will pass in unit time, through unit area of a specified thickness of material, under unit temperature difference. For a thickness x of material with a thermal conductivity of k in Jm-^1 s-1 oC-^1 , the conductance is k/x = C and the units of conductance are Jm-^2 s-1 oC-^1.
Heat conductance = C = k/x. Heat Conductances in Series Frequently in heat conduction, heat passes through several consecutive layers of different materials. For example, in a cold store wall, heat might pass through brick, plaster, wood and cork. In this case, eqn. (5.2) can be applied to each layer. This is illustrated in Fig. 5.2. Figure 5.2 Heat conductances in series In the steady state, the same quantity of heat per unit time must pass through each layer. q = A 1 T 1 k 1 /x 1 = A 2 T 2 k 2 /x 2 = A 3 T 3 k 3 /x 3 = …….. If the areas are the same, A 1 = A 2 = A 3 = ….. = A q = A T 1 k 1 /x 1 = A T 2 k 2 /x 2 = A T 3 k 3 /x 3 = ……..
A = 1m^2 q = UA T = 0.38 x 1 x 36 = 13.7Js-^1 Further, q = A 3 T 3 k 3 / x 3 and for the cork wall A 3 = 1 m^2 , x 3 /k 3 = 2.33 and q = 13.7Js-^1 Therefore 13.7 = 1 x T 3 x 1/2.33 from rearranging eqn. (5.2) T 3 = 32 oC. But T 3 is the difference between the temperature of the cork/concrete surface and the temperature of the cork surface inside the cold store. Therefore T c - (-18) = where T c is the temperature at the cork/concrete surface and so T c = 14oC. If T 1 is the difference between the temperature of the brick/concrete surface T b and the temperature of the external air Then 13.7 = 1 x T 1 x 1/0.16 = 6.25 T 1 Therefore 18 - T b = T 1 = 13.7/6.25 = 2. and so T b = 15.8oC Working it through shows approximate interface temperatures: air/brick 18oC, brick/concrete 16 oC, concrete/cork 14oC, and cork/air - 18 oC. This shows that almost all of the temperature difference occurs across the insulation (cork): the actual temperatures can be significant especially if they lie below the temperature at which the atmospheric air condenses, or freezes. Heat Conductances in Parallel Heat conductances in parallel have a sandwich construction at right angles to the direction of the heat transfer, but with heat conductances in parallel, the material surfaces are parallel to the direction of heat transfer and to each other. The heat is therefore passing through each material at the same time, instead of through one material and then the next. This is illustrated in Fig. 5.3.
Figure 5.3. Heat conductances in parallel An example is the insulated wall of a refrigerator or an oven, in which the walls are held together by bolts. The bolts are in parallel with the direction of the heat transfer through the wall: they carry most of the heat transferred and thus account for most of the losses. EXAMPLE 5.3. Heat transfer in walls of a bakery oven The wall of a bakery oven is built of insulating brick 10 cm thick and of thermal conductivity 0.22 Jm-^1 s-1oC-^1. Steel reinforcing members penetrate the brick, and their total area of cross- section represents 1% of the inside wall area of the oven. If the thermal conductivity of the steel is 45 Jm-^1 s-1oC-^1 , calculate (a) the relative proportions of the total heat transferred through the wall by the brick and by the steel and (b) the heat loss for each m^2 of oven wall if the inner side of the wall is at 230oC and the outer side is at 25oC. Applying eqn. (5.1) q = A T k / x , we know that T is the same for the bricks and for the steel. Also x, the thickness, is the same. (a) Consider the loss through an area of 1 m^2 of wall (0.99m^2 of brick, and 0.01 m^2 of steel) For brick q b = A b T k b/ x = 0.99(230 - 25)0.
= 446Js-^1
thickness x 1 ,...... The coefficient h s is also known as the convection heat-transfer coefficient and values for it will be discussed in detail under the heading of convection. It is useful at this point, however, to appreciate the magnitude of h s under various common conditions and these are shown in Table 5.1. TABLE 5. APPROXIMATE RANGE OF SURFACE HEAT-TRANSFER COEFFICIENTS h Jm-^2 s-1 oC-^1 Boiling liquids 2400 - 24, Condensing liquids 1800 - 18, Still air 6 Moving air (3ms-^1 ) 30 Liquids flowing through pipes 1200 - 6000 EXAMPLE 5.4. Heat transfer in jacketed pan Sugar solution is being heated in a jacketed pan made from stainless steel, 1.6 mm thick. Heat is supplied by condensing steam at 200kPa gauge in the jacket. The surface transfer coefficients are, for condensing steam and for the sugar solution, 12,000 and 3000 Jm-^2 s-1oC-^1 respectively, and the thermal conductivity of stainless steel is 21 Jm-^1 s-1oC-^1. Calculate the quantity of steam being condensed per minute if the transfer surface is 1.4 m^2 and the temperature of the sugar solution is 83oC. From steam tables, Appendix 8, the saturation temperature of steam at 200 kPa gauge(300kPa absolute) =134oC and the latent heat = 2164kJkg-^1. For stainless steel x/k = 0.0016/21 = 7.6 x 10-^5 T = (condensing temperature of steam) - (temperature of sugar solution) = 134 - 83 = 51oC_._ From eqn. (5.5) 1/U = 1/12,000 + 7.6 x 10-^5 + 1/ U = 2032 Jm-^2 s-1oC-^1 and since A = 1.4m^2
q = UA T
= 2032 x 1.4 x 51 = 1.45 x 10^5 Js-^1 Therefore steam required = heat transferred per sec /latent heat from steam = 1.45 x 10^5 / (2.164 x 10^6 ) kgs-^1
= 0.067kgs-^1 = 4 kg min-^1 UNSTEADY-STATE HEAT TRANSFER In food-process engineering, heat transfer is very often in the unsteady state, in which temperatures are changing and materials are warming or cooling. Unfortunately, study of heat flow under these conditions is complicated. In fact, it is the subject for study in a substantial branch of applied mathematics, involving finding solutions for the Fourier equation written in terms of partial differentials in three dimensions. There are some cases that can be simplified and handled by elementary methods, and also charts have been prepared which can be used to obtain numerical solutions under some conditions of practical importance. A simple case of unsteady-state heat transfer arises from the heating or cooling of solid bodies made from good thermal conductors, for example a long cylinder, such as a meat sausage or a metal bar, being cooled in air. The rate at which heat is being transferred to the air from the surface of the cylinder is given by eqn. (5.4) q = d Q/ d t = h s A ( T s - T (^) a) where T a is the air temperature and T s is the surface temperature. Now, the heat being lost from the surface must be transferred to the surface from the interior of the cylinder by conduction. This heat transfer from the interior to the surface is difficult to determine but as an approximation, we can consider that all the heat is being transferred from the centre of the cylinder. In this instance, we evaluate the temperature drop required to produce the same rate of heat flow from the centre to the surface as passes from the surface to the air. This requires a greater temperature drop than the actual case in which much of the heat has in fact a shorter path. Assuming that all the heat flows from the centre of the cylinder to the outside, we can write the conduction equation d Q /d t = ( k/L ) A ( T c – T s ) where T c is the temperature at the centre of the cylinder, k is the thermal conductivity of the material of the cylinder and L is the radius of the cylinder. Equating these rates: h s A ( T s – T (^) a) = ( k/L ) A ( T c – T s ) h s( T s – T (^) a) = ( k/L )( T c – T s ) and so h s L/k = ( T c – T s )/ ( T s – T (^) a )
For this case, the temperatures for any desired interval can be calculated, if the surface transfer coefficient and the other physical factors are known. This gives a reasonable approximation so long as (Bi) is less than about 0.2. Where (Bi) is greater than 0.2 the centre of the solid will cool more slowly than this equation suggests. The equation is not restricted to cylinders, it applies to solids of any shape so long as the restriction in (Bi), calculated for the smallest half-dimension, is obeyed. Charts have been prepared which give the temperature relationships for solids of simple shapes under more general conditions of unsteady-state conduction. These charts have been calculated from solutions of the conduction equation and they are plotted in terms of dimensionless groups so that their application is more general. The form of the solution is: f {( T - T 0 ) / ( T i - T 0 )} = F {( kt/c L^2 )( h s L/k )} (5.7) where f and F indicate functions of the terms following, T i is the initial temperature of the solid, T 0 is the temperature of the cooling or heating medium, T is the temperature of the solid at time t, ( kt/c L^2 ) is called the Fourier number (Fo) (this includes the factor k/c the thermal conductance divided by the volumetric heat capacity, which is called the thermal diffusivity) and (h s L/k) is the Biot number. A mathematical outcome that is very useful in these calculations connects results for two- and three-dimensional situations with results from one-dimensional situations. This states that the two- and three-dimensional values called F(x,y) and F(x,y,z) can be obtained from the individual one-dimensional results if these are F(x), F(y) and F(z), by simple multiplication: F(x,y ) = F ( x ) F ( y ) and F(x,y,z) = F(x)F(y)F(z) Using the above result, the solution for the cooling or heating of a brick is obtained from the product of three slab solutions. The solution for a cylinder of finite length, such as a can, is obtained from the product of the solution for an infinite cylinder, accounting for the sides of the can, and the solution for a slab, accounting for the ends of the can. Charts giving rates of unsteady-state heat transfer to the centre of a slab, a cylinder, or a sphere, are given in Fig.5.4. On one axis is plotted the fractional unaccomplished temperature change, ( T - T 0 ) / ( T i - T 0 ). On the other axis is the Fourier number ( kt/c L^2 ), which may be thought of in this connection as a time coordinate. The various curves are for different values of the reciprocal of the Biot number, k / hr for spheres, k / hL for slabs. More detailed charts, giving surface and mean temperatures in addition to centre temperatures, may be found in McAdams (1954) , Fishenden and Saunders (1950) and Perry (1997).
Figure 5.4. Transient heat conduction Temperatures at the centre of a sphere, slab and cylinder adapted from Henderson and Perry, Agricultural Process Engineering , 1955 EXAMPLE 5.5. Heat transfer in cooking sausages A process is under consideration in which large cylindrical meat sausages are to be processed in an autoclave. The sausage may be taken as thermally equivalent to a cylinder 30cm long and 10cm in diameter. If the sausages are initially at a temperature of 21oC and the temperature in the autoclave is maintained at 116oC, estimate the temperature of the sausage at its centre 2h after it has been placed in the autoclave. Assume that the thermal conductivity of the sausage is 0.48 Jm-^1 s-1oC -^1 , that its specific gravity is 1.07, and its specific heat is 3350 Jkg-1oC-^1. The surface heat-transfer coefficient in the autoclave to the surface of the sausage is l200Jm-^2 s-1 oC-^1. This problem can be solved by combining the unsteady-state solutions for a cylinder with those for a slab, working from Fig. 5.4.
The calculation of radiant heat transfer rates, in detail, is beyond the scope of this book and for most food processing operations a simplified treatment is sufficient to estimate radiant heat effects. Radiation can be significant with small temperature differences as, for example, in freeze drying and in cold stores, but it is generally more important where the temperature differences are greater. Under these circumstances, it is often the most significant mode of heat transfer, for example in bakers' ovens and in radiant dryers. The basic formula for radiant-heat transfer is the Stefan-Boltzmann Law q = A T^4 (5.8) where T is the absolute temperature (measured from the absolute zero of temperature at – 273 oC, and indicted in Bold type) in degrees Kelvin (K) in the SI system, and (sigma) is the Stefan- Boltzmann constant = 5.73 x 10-^8 J m-^2 s-^1 K-^4. The absolute temperatures are calculated by the formula K = (oC + 273). This law gives the radiation emitted by a perfect radiator (a black body, as this is called, though it could be a red-hot wire in actuality). A black body gives the maximum amount of emitted radiation possible at its particular temperature. Real surfaces at a temperature T do not emit as much energy as predicted by eqn. (5.8) , but it has been found that many emit a constant fraction of it. For these real bodies, including foods and equipment surfaces, that emit a constant fraction of the radiation from a black body, the equation can be rewritten q = A T^4 (5.9) where (epsilon) is called the emissivity of the particular body and is a number between 0 and 1. Bodies obeying this equation are called grey bodies. Emissivities vary with the temperature T and with the wavelength of the radiation emitted. For many purposes, it is sufficient to assume that for: *dull black surfaces (lamp-black or burnt toast, for example), emissivity is approximately 1; *surfaces such as paper/painted metal/wood and including most foods, emissivities are about 0.9; *rough unpolished metal surfaces, emissivities vary from 0.7 to 0.25; *polished metal surfaces, emissivities are about or below 0.05. These values apply at the low and moderate temperatures, which are those encountered in food processing. Just as a black body emits radiation, it also absorbs it and according to the same law, eqn. (5.8). Again grey bodies absorb a fraction of the quantity that a black body would absorb, corresponding this time to their absorptivity (alpha). For grey bodies it can be shown that = . The fraction of the incident radiation that is not absorbed is reflected, and thus, there is a further term used, the reflectivity, which is equal to (1 – ).
Radiation between Two Bodies The radiant energy transferred between two surfaces depends upon their temperatures, the geometric arrangement, and their emissivities. For two parallel surfaces, facing each other and neglecting edge effects, each must intercept the total energy emitted by the other, either absorbing or reflecting it. In this case, the net heat transferred from the hotter to the cooler surface is given by: q = AC ( T 1 4 - T 2^4 ) (5.10) where 1/ C = 1/ 1 + 1/ 2 - (^1) , 1 is the emissivity of the surface at temperature T 1 and 2 is the emissivity of the surface at temperature T 2_._ Radiation to a Small Body from its Surroundings In the case of a relatively small body in surroundings that are at a uniform temperature, the net heat exchange is given by the equation q = A ( T 1 4 - T 24 ) (5.11) where is the emissivity of the body, T 1 is the absolute temperature of the body and T 2 is the absolute temperature of the surroundings. For many practical purposes in food process engineering, eqn. (5.11) covers the situation; for example for a loaf in an oven receiving radiation from the walls around it, or a meat carcass radiating heat to the walls of a freezing chamber. In order to be able to compare the various forms of heat transfer, it is necessary to see whether an equation can be written for radiant-heat transfer similar to the general heat transfer eqn. (5.3). This means that for radiant-heat transfer: q = h r A ( T 1 - T 2 ) = h r A ( T 1 - T 2 ) = h r A T (5.12) where h r is the radiation heat transfer coefficient, T 1 is the temperature of the body and T 2 is the temperature of the surroundings. The T would normally be the absolute temperature for the radiation, but the absolute temperature difference is equal to the Celsius temperature difference, because 273 is added and subtracted and so ( T 1 - T 2 ) = ( T 1 - T 2 ) = T. Equating eqn. (5.11) and eqn. (5.12) q = h r A( T 1 - T 2 ) = A ( T 14 - T 24 ) Therefore hr = ( T 1 4 - T 24 )/ ( T 1 - T 2 )
= ( T 1 + T 2 ) ( T 12 + T 2 2 )
CONVECTION HEAT TRANSFER
Convection heat transfer is the transfer of energy by the mass movement of groups of molecules. It is restricted to liquids and gases, as mass molecular movement does not occur at an appreciable speed in solids. It cannot be mathematically predicted as easily as can transfer by conduction or radiation and so its study is largely based on experimental results rather than on theory. The most satisfactory convection heat transfer formulae are relationships between dimensionless groups of physical quantities. Furthermore, since the laws of molecular transport govern both heat flow and viscosity, convection heat transfer and fluid friction are closely related to each other. Convection coefficients will be studied under two sections, firstly, natural convection in which movements occur due to density differences on heating or cooling; and secondly, forced convection, in which an external source of energy is applied to create movement. In many practical cases, both mechanisms occur together. Natural Convection Heat transfer by natural convection occurs when a fluid is in contact with a surface hotter or colder than itself. As the fluid is heated or cooled it changes its density. This difference in density causes movement in the fluid that has been heated or cooled and causes the heat transfer to continue. There are many examples of natural convection in the food industry. Convection is significant when hot surfaces, such as retorts which may be vertical or horizontal cylinders, are exposed with or without insulation to colder ambient air. It occurs when food is placed inside a chiller or freezer store in which circulation is not assisted by fans. Convection is important when material is placed in ovens without fans and afterwards when the cooked material is removed to cool in air. It has been found that natural convection rates depend upon the physical constants of the fluid, density , viscosity , thermal conductivity k, specific heat at constant pressure cp and coefficient of thermal expansion (beta) which for gases = l/ T by Charles' Law. Other factors that also affect convection-heat transfer are, some linear dimension of the system, diameter D or length L, a temperature difference term, T , and the gravitational acceleration g since it is density differences acted upon by gravity that create circulation. Heat transfer rates are expressed in terms of a convection heat transfer coefficient ( hc ) , which is part of the general surface coefficient hs, in eqn. (5.5). Experimentally, if has been shown that convection heat transfer can be described in terms of these factors grouped in dimensionless numbers which are known by the names of eminent workers in this field:
Nusselt number (Nu) = ( hcD/k ) Prandtl number (Pr) = ( cp /k ) Grashof number (Gr) = ( D^3 ^2 g T /^2 ) and in some cases a length ratio ( L /D ). If we assume that these ratios can be related by a simple power function we can then write the most general equation for natural convection: (Nu) = K (Pr)k(Gr ) m( L/D )n^ (5.14) Experimental work has evaluated K, k, m, n, under various conditions. For a discussion, see McAdams (1954). Once K, k, m, n are known for a particular case, together with the appropriate physical characteristics of the fluid, the Nusselt number can be calculated. From the Nusselt number we can find hc and so determine the rate of convection heat transfer by applying eqn. (5.5). In natural convection equations, the values of the physical constants of the fluid are taken at the mean temperature between the surface and the bulk fluid. The Nusselt and Biot numbers look similar: they differ in that for Nusselt, k and h both refer to the fluid, for Biot k is in the solid and h is in the fluid. Natural Convection Equations These are related to a characteristic dimension of the body (food material for example) being considered, and typically this is a length for rectangular bodies and a diameter for spherical/cylindrical ones (1) Natural convection about vertical cylinders and planes, such as vertical retorts and oven walls (Nu) = 0.53(Pr.Gr)0.25^ for 10^4 < (Pr.Gr) < 10^9 (5.15) (Nu) = 0.12(Pr.Gr)0.33^ for 10^9 < (Pr.Gr) < 10^12 (5.16) For air, these equations can be approximated respectively by: hc = 1.3( T/L )0.25^ (5.17) hc = 1.8( T )0.25^ (5.18) Equations (5.17) and (5.18) are dimensional equations and are in standard units [ T in oC, L (or D ) in metres and hc in Jm-^2 s-1oC-^1 ]. The characteristic dimension to be used in the calculation of (Nu) and (Gr) in these equations is the height of the plane or cylinder.
