Page 524 - Bird R.B. Transport phenomena

504 Chapter 16 Energy Transport by Radiation

Fig. 16.5-3. Radiation shield.

rh

т = т 2 т = т 3
e = a = e 2 e = a = e 3

SOLUTION The radiation rate between planes 1 and 2 is given by
4
- T ) Aa{T\ - T\)
2
Ql2 = (16.5-11)
1 - е , l - e 2 1 1
since both planes have the same area A and the view factor is unity. Similarly the heat trans-
fer between planes 2 and 3 is
A(T(T\ - 71) Aa(T\ - T\)
1 -e-> l - e 3 (16.5-12)

These last two equations may be combined to eliminate the temperature of the radiation
shield, T , giving
к (16.5-13)
2

Then, since Q = Q = Q , we get
1 2 2 3 13
(16.5-14)
1 + 1 - 1 + 1 + 1 - 1
e e 2 4 \e 2 e 3
Finally, the ratio of radiant energy transfer with a shield to that without one is

(Ql3)with
(16.5-15)
(Ql3/without
1
1 + - 1 + 1 + 1 - 1
* ^ J V e
EXAMPLE 16.5-2 Predict the total rate of heat loss, by radiation and free convection, from a unit length of hori-
zontal pipe covered with asbestos. The outside diameter of the insulation is 6 in. The outer
Radiation and Free- surface of the insulation is at 560°R, and the walls and air in the room are at 540°R.
Convection Heat Losses
from a Horizontal Pipe SOLUTION
Let the outer surface of the insulation be surface 1 and the walls of the room be surface 2.
Then Eq. 16.15-3 gives
4
Q 12 - aA,F (e T\ - aj ) (16.5-16)
u
A
2