CHEMICAL ENGINEERING

HEAT TRANSFER 133

thermal conductivity 0.3 W/m K is added, what temperatures will its surfaces attain
assuming the inner surface of the furnace to remain at 1400 K? The coefficient of heat
transfer from the outer surface of the insulation to the surroundings, which are at 290 K,
may be taken as 4.2, 5.0, 6.1, and 7.1W/m^2 K for surface temperatures of 370, 420, 470,
and 520 K respectively. What will be the reduction in heat loss?

Solution

Heat flow through the refractory, QDkA
T 1 T 2 /x (equation 9.12)

Thus for unit area, QD

1. 5 ð 1. 0
   1400 T 2 /
   150 ð 10 ^3

D 

14 , 000  10 T 2 W/m^2 (i)

whereT 2 is the temperature at the refractory – insulation interface.
Similarly, the heat flow through the insulation is:

QD 

0. 3 ð 1. 0 

T 2 T 3 /

 25 ð 10 ^3 D 

12 T 2  12 T 3 W/m^2 (ii)

The flow of heat from the insulation surface atT 3 K to the surroundings at 290 K, is:

QDhA

T 3  290 or

T 3  290 hW/m^2 (iii)

wherehis the coefficient of heat transfer from the outer surface.
The solution is now made by trial and error. A value ofT 3 is selected andhobtained
by interpolation of the given data. This is substituted in equation (iii) to giveQ.T 2 is then
obtained from equation (ii) and a second value ofQis then obtained from equation (i).
ThecorrectvalueofT 3 is then given when these two values ofQcoincide. The working
is as follows and the results are plotted in Fig. 9c.

Q = 14000 − 10 T 2

Q = h (T 3 − 290)

4050 W/m^2

T 3 (K)

662 K

300 350 400 450 500 550 600 650 700 750

10000

8000

4000

2000

0

6000

Q

(W/m

2 )

Figure 9c.