CHEMICAL ENGINEERING

134 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

T 3 hQDh

T 3  290 T 2 DT 3 CQ/ 12 QD 

14 , 000  10 T 2

(K) (W/m^2 K) (W/m^2 )(K) (W/m^2 )

300 3.2 32 302.7 10,973
350 3.9 234 369.5 10,305
400 4.7 517 443.1 9569
450 5.6 896 524.7 8753
500 6.5 1355 612.9 7871
550 7.8 2028 719.0 6810
600 9.1 2821 835.1 5649
650 10.4 3744 962.0 4380
700 11.5 4715 1092.9 3071
750 12.7 5842 1236.8 1632

A balance is obtained whenT 3 D662 K, at whichQD4050 W/m^2.

In equation (i): 4050 D
14 , 000  10 T 2 orT 2 D995 K

Thus the temperatures at the inner and outer surfaces of the insulation are

995 K and 662 K respectively

With no insulation,QD 

1. 5 ð 1. 0

1400  540 /

 150 ð 10 ^3 D8600 W/m^2

and hence the reduction in heat loss is
8600  4050 D4550 W/m^2

or:
4540 ð 100 / 8600 D 52 .9%

PROBLEM 9.8

A pipe of outer diameter 50 mm, maintained at 1100 K, is covered with 50 mm of insu-
lation of thermal conductivity 0.17 W/m K. Would it be feasible to use a magnesia
insulation, which will not stand temperatures above 615 K and has a thermal conduc-
tivity of 0.09 W/m K, for an additional layer thick enough to reduce the outer surface
temperature to 370 K in surroundings at 280 K? Take the surface coefficient of heat
transfer by radiation and convection as 10 W/m^2 K.

Solution

For convection to the surroundings

QDhA 3 

T 3 T 4 W/m

whereA 3 is area for heat transfer per unit length of pipe, m^2 /m.

The radius of the pipe,r 1 D 

50 / 2 D25 mm or 0.025 m.

The radius of the insulation,r 2 D 

25 C 50 D75 mm or 0.075 m.