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Thermal Conductivity
Conduction through a Slab
Heat Conductances
Heat Conductances in Series
Heat Conductances in Parallel

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:

                          dQ/dt = kA dT/dx                                (5.1)

where dQ/dt 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, 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

k = dQ/dt x 1/A x 1/(dT/dx)

   = J s-1 x m-2 x 1/(°C m-1)
   = J m-1 s-1 °C-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.

Thermal Conductivity

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 50-400 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 non-metallic materials have thermal conductivities of 0.5-2 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 non-metallic 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.

Conduction through a Slab

If a slab of material, as shown in Fig. 5.1, has two faces at different temperatures T1 and T2 heat will flow from the face at the higher temperature T1 to the other face at the lower temperature T2.

Fig. 5.1. Heat conduction through a slab
Figure 5.1. Heat conduction through a slab

The rate of heat transfer is given by Fourier's equation:

    dQ/dt = kA DT/Dx = kA dT/dx

Under steady temperature conditions dQ/dt = constant, which may be called q:

and so q = kA dT/dx

but dT/dx, the rate of change of temperature per unit length of path, is given by (T1 - T2)/x where x is the thickness of the slab,

      so q = kA(T1 - T2)/x

       or q = kA DT/x = (k/x) A DT                                                                             (5.2)

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.

EXAMPLE 5.1. Rate of heat transfer in cork
A cork slab 10 cm thick has one face at -12°C and the other face at 21°C. If the mean thermal conductivity of cork in this temperature range is 0.042 J m-1 s-1 °C-1, what is the rate of heat transfer through 1 m2 of wall?

T1 = 21°C T2 = -12°C DT = 33°C
   A = 1 m2  k = 0.042 J m-1 s-1 °C-1    x = 0.1 m

q =
  x 1 x 33
   = 13.9 J s-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 J m-1 s-1 °C-1, the conductance is k/x = C and the units of conductance are
J m-2 s-1 °C-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
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 = A1DT1k1/x1 = A2DT2k2/x2 = A3DT3k3/x3 = ……..

If the areas are the same,
                                        A1 = A2 = A3 = ….. = A

          q = ADT1k1/x1 = ADT2k2/x2 = ADT3k3/x3 = ……..

       So ADT1 = q(x1/k1) and ADT2 = q(x2/k2) and ADT3 = q(x3/k3).…..

   ADT1 + ADT2 + ADT3 + … = q(x1/k1) + q(x2/k2) +q(x3/k3) + …

      A(DT1 + DT2 + DT3 + ..) = q(x1/k1 + x2/k2 +x3/k3 + …)

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.

                 DT1 + DT2 + DT3 + … = DT
and since k1/x1 is equal to the conductance of the material in the first layer, C1, and k2/x2 is equal to the conductance of the material in the second layer C2,

    x1/k1 + x2/k2 + x3/k3 + ... = 1/C1 + 1/C2 + 1/C3 ……
                                          = 1/U

where U = the overall conductance for the combined layers, in J m-2 s-1 °C-1

Therefore                       ADT = q(1/U)

                              And so q = UADT                                                                     (5.3)

This is of the same form as eqn (5.2) but extended to cover the composite slab. U is called the overall heat-transfer coefficient, as it can also include combinations involving the other methods of heat transfer – convection and radiation.

EXAMPLE 5.2. Heat transfer in cold store wall of brick, concrete and cork
A cold store has a wall comprising 11 cm of brick on the outside, then 7.5 cm of concrete and then 10 cm of cork. The mean temperature within the store is maintained at -18°C and the mean temperature of the outside surface of the wall is 18°C.
Calculate the rate of heat transfer through the wall. The appropriate thermal conductivities are for brick, concrete and cork, respectively 0.69, 0.76 and 0.043 J m-1 s-1 °C-1.
Determine also the temperature at the interfaces between the concrete and cork layers, and the brick and concrete layers.

For brick   x1/k1    = 0.11/0.69 = 0.16.
For concrete x2/k2 = 0.075/0.76 = 0.10.
For cork   x3/k3     = 0.10/0.043 = 2.33
                But 1/U = x1/k1 + x2/k2 + x3/k3
                            = 0.16 + 0.10 + 2.33
                            = 2.59

Therefore            U = 0.38 J m-2 s-1°C-1
DT = 18 - (-18) = 36°C,
                         A = 1 m2

                         q = UADT
                            = 0.38 x 1 x 36
                           = 13.7 J s-1

          Further,            q = A3DT3k3/x3

 and for the cork wall A3 = 1 m2, x3/k3 = 2.33 and q = 13.7 J s-1

Therefore    13.7    = 1 x DT3 x 1/2.33          from rearranging eqn. (5.2)
DT3 = 32°C.

But DT3 is the difference between the temperature of the cork/concrete surface Tc and the temperature of the cork surface inside the cold store.

Therefore Tc - (-18) = 32

where Tc is the temperature at the cork/concrete surface

                     and so Tc = 14°C.

If DT1 is the difference between the temperature of the brick/concrete surface, Tb, and the temperature of the external air.

Then            13.7 = 1 x DT1 x 1/ 0.16 = 6.25 DT1

Therefore  18 - Tb = DT1 = 13.7/6.25 = 2.2

so                  Tb = 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

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
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 thermal conductivity 0.22 J m-1 s-1 °C-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 J m-1 s-1 °C-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 m2 of oven wall if the inner side of the wall is at 230°C and the outer side is at 25°C.

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 m2 of wall (0.99 m2 of brick, and 0.01 m2 of steel)
For brick qb = Ab
DT kb/x

0.99(230 - 25)0.22
= 446 J s-1

For steel qs = AsDT ks/x

0.01(230 - 25)45
= 923 J s-1

Therefore qb /qs = 0.48

(b) Total heat loss

     q = (qb + qs ) per m2 of wall
        = 446 + 923
        = 1369 J s-1

Therefore percentage of heat carried by steel
        = (923/1369) x 100
        = 67%

Heat-Transfer Theory > SURFACE-HEAT TRANSFER

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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology