CHEMICAL ENGINEERING

HEAT TRANSFER 137

The mass flow,GD 0 .5 kg/s.

∴

hið 19. 0 ð 10 ^3 / 0. 14
87. 0 ð 10 ^3 / 154 ð 10 ^30.^14

D 2. 01



0. 5 ð 2. 1 ð 103 /

 0. 14 ð 1. 50.^33

or:
0. 13 hið 0. 923 D
2. 01 ð 16. 6 andhiD266 W/m^2 K

This is sufficiently close to the assumed value and hence 14 tubeswould be specified.

PROBLEM 9.10

A metal pipe of 12 mm outer diameter is maintained at 420 K. Calculate the rate of heat
loss per metre run in surroundings uniformly at 290 K, (a) when the pipe is covered with
12 mm thickness of a material of thermal conductivity 0.35 W/mK and surface emissivity
0.95, and (b) when the thickness of the covering material is reduced to 6 mm, but the
outer surface is treated so as to reduce its emissivity to 0.10. The coefficients of radiation
from a perfectly black surface in surroundings at 290 K are 6.25, 8.18, and 10.68 W/m^2
K at 310 K, 370 K, and 420 K respectively. The coefficients of convection may be taken
as 1. 22
/d^0.^25 W/m^2 K, where(K) is the temperature difference between the surface
and the surrounding air andd(m) is the outer diameter.

Solution

Case (a)

Assuming that the heat loss isqW/m and the surface temperature isTK, for conduction
through the insulation, from equation 9.12,qDkAm
420 T/x
The mean diameter is 18 mm or 0.018 m, and hence:

AmD

$ð 0. 018 D 0 .0566 m^2 /mxD 0 .012 m

∴ qD
0. 35 ð 0. 0566
420 T/ 0. 012 D
693. 3  1. 67 TW/m (i)

For convection and radiation from the surface, from equation 9.119:

qD

hrChcA 2 

T 290 W/m

wherehris the film coefficient equivalent to the radiation andhcthe coefficient due to
convection given by:

hcD 1 .22[

T 290 /d]^0.^25 wheredD36 mm or 0.036 m

∴ hcD 2. 80
T 2900.^25 W/m^2 K

Ifhbis the coefficient equivalent to radiation from a black body,hrD 0. 95 hbW/m^2 K
The outer diameter is 0.036 m and hence:
A 2 D
$ð 0. 036 ð 1. 0 D 0 .1131 m^2 /m

∴ qD[0. 95 hbC 2. 80
T 2900.^25 ]0. 1131
T 290

D 0. 1074 hb

T 290 C 0. 317 

T 2901.^25 W/m (ii)