CHEMICAL ENGINEERING

HEAT TRANSFER 165

Equating (i) and (ii):

1. 94 ð 10 ^6 T^4  5. 10 ð 10 ^4 T^3 C 1. 20 ð 105 T^0.^5 D 3. 42 ð 104

or: 5. 67 ð 10 ^11 T^4  1. 49 ð 10 ^8 T^3 C 3. 51 T^0.^5 D 1

Solving by trial and error, the equilibrium temperature of the aluminium is:

TD438 K.

SubstitutingTD438 K in (i), the energy absorbed and emitted is then 2847 W which
represents an increase of some 375 per centcompared with the value for the concrete

alone.

PROBLEM 9.33

A rectangular iron ingot 15 cmð15 cmð30 cm is supported at the centre of a reheating
furnace. The furnace has walls of silica-brick at 1400 K, and the initial temperature of
the ingot is 290 K. How long will it take to heat the ingot to 600 K?
It may be assumed that the furnace is large compared with the ingot, and that the ingot
is always at uniform temperature throughout its volume. Convection effects are negligible.
The total emissivity of the oxidised iron surface is 0.78 and both emissivity and
absorptivity may be assumed independent of the surface temperature. (Density of ironD
7 .2Mg/m^3. Specific heat capacity of ironD 0 .50 kJ/kg K.)

Solution

As there are no temperature gradients within the ingot, the rate of heating is dependent on
the rate of radiative heat transfer to the surface. In addition, since the dimensions of the
ingot are much smaller than those of the surrounding surfaces, the ingot may be treated
as a black body.

Volume of ingotD
15 ð 15 ð 30 D6750 cm^3 or 0.00675 m^3.
Mass of ingotD
7. 2 ð 103 ð 0. 00675 D 48 .6kg.
For an ingot temperature ofTK, the increase in enthalpyDd
mCpT/dtormCpdT/dt
wheretis the time andCpthe specific heat of the ingot.

The heat received by radiationDA9a

T^4 fT^4 where the area,AD 

4 ð 30 ð 15 C

2 ð 15 ð 15 D2250 cm^2 or 0.225 m^2.
The absorptivityawill be taken as the emissivityD 0. 78

and the furnace temperature,TfD1400 K.

Thus: mCpdT/dtDA9a
T^4 fT^4

or:

∫t

0

dtD

mCp

aA9

∫ 600

290

dT



T^4 fT^4