CHEMICAL ENGINEERING

HEAT TRANSFER 163

The radiating source is the furnace walls atT 1 D1500 K and for a black surface,
e 1 D 1 .0.

The heat sink is the refractory atT 2 D1420 K, at whiche 2 D 0 .43.

PuttingA 1 DA 2 in equation 9.150:qD e 1 e 2 9
T^41 T^42 /
e 1 Ce 2 e 1 e 2

D 

1. 0 ð 0. 43 ð 5. 67 ð 10 ^8

15004  14204 /

 1. 0 C 0. 43  

0. 43 ð 1. 0

D 2. 44 ð 10 ^8

9. 97 ð 1011 / 1. 0 D 2. 43 ð 104 W/m^2 or 24.3kW/m^2

PROBLEM 9.30

The total emissivity of clean chromium as a function of surface temperature,TK, is
given approximately by:eD 0. 38
1  263 /T.
Obtain an expression for the absorptivity of solar radiation as a function of surface
temperature, and calculate the values of the absorptivity and emissivity at 300, 400 and
1000 K.
Assume that the sun behaves as a black body at 5500 K.

Solution

It may be assumed that the absorptivity of the chromium at temperatureT 1 is the emissivity
of the chromium at the geometric mean ofT 1 and the assumed temperature of the sun,
T 2 whereT 2 D5500 K.

Since: eD 0. 38
1 
263 /T i

then, taking the geometric mean temperature as
5500 T 10.^5 :

aD 0. 38 f 1 [263/

 5500 T 10.^5 ]g ii

For the given values of T 1 ,valuesofeandaare now calculated from (i) and (ii)
respectively to give the following data:

T 1 

T 1 T 20.^5 ea

300 1285 0.047 0.300

400 1483 0.130 0.312

1000 2345 0.280 0.337

PROBLEM 9.31

Repeat Problem 9.30 for the case of aluminium, assuming the emissivity to be 1.25 times
that for chromium.

Solution

In this case:

eD 

1. 25 ð 0. 38

1  

263 /T 1 D 0 .475[ 

1  

263 /T 1 ] i

and: aD 0. 475  1. 66 T 10.^5 ii