CHAPTER
5
where dQ/dt is the rate of heat transfer, the quantity of heat energy transferred per unit of time, A is the area of crosssection of the heat flow path, dT/dx 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
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. On the basis of eqn. (5.1) thermal conductivities of materials can be measured. Thermal conductivity does change slightly with temperature, but in many applications it can be regarded as a constant for a given material. Thermal conductivities are given in Appendices 3, 4, 5, 6, which give physical properties of many materials used in the food industry. In general, metals have a high thermal conductivity, in the region 50400 J m^{1} s^{1} °C^{1}. Most foodstuffs contain a high proportion of water and as the thermal conductivity of water is about 0.7 J m^{1} s^{1}°C^{1} above 0°C, thermal conductivities of foods are in the range 0.6  0.7 J m^{1} s^{1}°C^{1}. Ice has a substantially higher thermal conductivity than water, about 2.3 J m^{1} s^{1}°C^{1}. The thermal conductivity of frozen foods is, therefore, higher than foods at normal temperatures. Most
dense nonmetallic materials have thermal conductivities of 0.52 J m^{1}
s^{1}°C^{1}. Insulating materials, such as those
used in walls of cold stores, approximate closely to the conductivity
of gases as they are made from nonmetallic materials enclosing small
bubbles of gas or air. The conductivity of air is 0.024 J m^{1}
s^{1} °C^{1} at 0°C, and insulating materials
such as foamed plastics, cork and expanded rubber are in the range 0.03
0.06 J m^{1} s^{1} °C^{1}. Some of the
new foamed plastic insulating materials have thermal conductivities as
low as 0.026 J m^{1} s^{1} °C^{1}. When using published tables of data, the units should be carefully checked. Mixed units, convenient for particular applications, are sometimes used and they may need to be converted. If a slab of material, as shown in Fig. 5.1, has two faces at different temperatures T_{1} and T_{2} heat will flow from the face at the higher temperature T_{1} to the other face at the lower temperature T_{2}. The rate of heat transfer is given by Fourier's equation:
but dT/dx, the rate of change of temperature per unit length of path, is given by (T_{1}  T_{2})/x where x is the thickness of the slab,
This may be regarded as the basic equation for simple heat conduction. It can be used to calculate the rate of heat transfer through a uniform wall if the temperature difference across it and the thermal conductivity of the wall material are known.
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 J m^{1} s^{1} °C^{1}, the
conductance is k/x = C and the units of conductance
are
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. In the steady state, the same quantity of heat per unit time must pass through each layer.
q = ADT_{1}k_{1}/x_{1} = ADT_{2}k_{2}/x_{2} = ADT_{3}k_{3}/x_{3} = …….. So ADT_{1} = q(x_{1}/k_{1}) and ADT_{2} = q(x_{2}/k_{2}) and ADT_{3} = q(x_{3}/k_{3}).…..
The sum of the temperature differences over each layer is equal to the difference in temperature of the two outside surfaces of the complete system, i.e.
DT_{1}
+ DT_{2}
+ DT_{3}
+ … = DT
where U = the overall conductance for the combined layers, in J m^{2} s^{1} °C^{1}
This is of the same form as eqn (5.2) but extended to cover the composite slab. U is called the overall heattransfer coefficient, as it can also include combinations involving the other methods of heat transfer – convection and radiation.
Further, q = A_{3}DT_{3}k_{3}/x_{3} and for the cork wall A_{3} = 1 m^{2}, x_{3}/k_{3} = 2.33 and q = 13.7 J s^{1}
But DT_{3} is the difference between the temperature of the cork/concrete surface T_{c }and the temperature of the cork surface inside the cold store.
where T_{c}
is the temperature at the cork/concrete surface If DT_{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 DT_{1} x 1/ 0.16 = 6.25 DT_{1} Therefore 18  T_{b} = DT1 = 13.7/6.25 = 2.2 so T_{b} = 15.8 °C Working it through shows approximate boundary temperatures: air/brick 18°C,brick/concrete 16°C, concrete/cork 14°C, cork/air 18°C This shows that almost all of the temperature difference occurs across the insulation (cork): and the actual intermediate temperatures can be significant especially if they lie below the temperature at which the atmospheric air condenses, or freezes. 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..
Applying eqn. (5.1) q = ADTk/x, we know that DT 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 (0.99 m^{2} of
brick, and 0.01 m^{2} of steel)
For steel q_{s} = A_{s}DT k_{s}/x
Therefore q_{b} /q_{s} = 0.48 (b) Total heat loss
q = (q_{b} + q_{s} ) per m^{2}
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