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Radiation between Two Bodies
Radiation to a Small Body from its Surroundings

Radiation heat transfer is the transfer of heat energy by electromagnetic radiation. Radiation operates independently of the medium through which it occurs and depends upon the relative temperatures, geometric arrangements and surface structures of the materials that are emitting or absorbing heat.

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 sT 4                                                                          (5.8)                                   

where T is the absolute temperature (measured from the absolute zero of temperature at -273°C, and indicated in Bold type) in degrees Kelvin (K) in the SI system, and s (sigma) is the Stefan-Boltzmann constant = 5.73 x 10-8 J m-2 s-1K-4 The absolute temperatures are calculated by the formula K = (°C + 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 = eA sT 4                                                                        (5.9)                                

where e (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 un-polished 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
a (alpha). For grey bodies it can be shown that a = e. 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 – a ).

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 = ACs (T14- T24 )                                                                                 (5.10)

where 1/C = 1/e1 + 1/e2 - 1, e1 is the emissivity of the surface at temperature T1 and e2 is the emissivity of the surface at temperature T2.

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 = Aes(T14- T24 )                                                                                  (5.11)

where e is the emissivity of the body, T1 is the absolute temperature of the body and T2 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 = hrA(T1 - T2) = hrA DT                                                                       (5.12)

where hr is the radiation heat-transfer coefficient, T1 is the temperature of the body and T2 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 (T1 - T2) = (T1 - T2) = DT

Equating eqn. (5.11) and eqn. (5.12)

                      q = hrA(T1 - T2) = Aes(T14- T24

      Therefore hr =
es(T14- T24 )/ (T1 - T2)

                               = es(T1 + T2 ) (T12 + T22)mmmmmm

If   Tm = (T1 + T2)/2, we can write T1 + e = Tm and T2 - e = Tm

                          2e  = T1 - T2
                 (T1 + T2) = 2 Tm

and then
              (T12 + T22) = Tm2 - 2eTm + e2 + Tm2 +2eTm +e2

                               = 2Tm2 + 2e2

                               = 2Tm2 + (T1 - T2)2/2

Therefore              hr = es(2Tm)[2Tm2 + (T1 - T2)2/2]

Now, if (T1 - T2) « T1 or T2, that is if the difference between the temperatures is small compared with the numerical values of the absolute temperatures, we can write:

                    hr    es 4Tm3
and so
                     q   = hrA
                          = (
e x 5.73 x 10-8 x 4 xTm3 ) x A DT
                          = 0.23
e (Tm/100)3A DT                                                                    (5.13)

EXAMPLE 5.6.Radiation heat transfer to loaf of bread in an oven
Calculate the net heat transfer by radiation to a loaf of bread in an oven at a uniform temperature of 177°C, if the emissivity of the surface of the loaf is 0.85, using eqn. (5.11). Compare this result with that obtained by using eqn. (5.13). The total surface area and temperature of the loaf are respectively 0.0645 m2 and 100°C.

                            q = Aes(T14- T24

                               = 0.0645 x 0.85 x 5.73 x 10-8 (4504- 3734)
                               = 68.0 J s-1.

By eqn. (5.13)

                            q = 0.23e (Tm/100)3A DT 

                               = 0.23 x 0.85(411/100)3 x 0.0645 x 77
                               = 67.4 J s-1.

Notice that even with quite a large temperature difference, eqn. (5.13) gives a close approximation to the result obtained using eqn. (5.11).


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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