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
where
2e = T1 - T2
also
(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 DT
= (e x
5.73 x 10-8 x 4 xTm3
) x A DT
= 0.23e
(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.
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Radiation between Two Bodies