CHEMICAL ENGINEERING

162 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

The coefficient for condensing steam including any scale will be taken as 8500 W/m^2
K,kwas 45 W/m K, andklas 0.073 W/m K.
The surface temperature of the lagging will be assumed to be 314 K and
hrChc to
be 10 W/m^2 K.
The thermal resistances are therefore:

1 /hi$dD 1 /

 8500 ð$ð 0. 150 D 0 .00025 mK/W



xw/kw$dw D 0. 009 /

 45 $ð 0. 159 D 0 .00040 mK/W



xl/kl$dm D 0. 050 /

 0. 073 $ð 0. 215 D 1 .0130 mK/W



1 /

hrðhcds D 1 /

 10 ð 0. 268 D 0 .1190 mK/W

Neglecting the first two terms, the total thermal resistanceD 1 .132 mK/W.

From equation 9.261, heat lost per unit lengthD 

444  294 / 1. 132 D 132 .5W/m.

The surface temperature of the lagging is given by:

T

lagging/TD 

1. 013 / 1. 132 D 0. 895

and: T
lagging D 0. 895
444  294 D134 deg K

Therefore the surface temperatureD
444  134 D310 K which approximates to the
assumed value.
Assuming an emissivity of 0.9:

hrD 

5. 67 ð 10 ^8 ð 0. 9

3104  2944 /

 310  294 D 3 .81 W/m^2 K.

For natural convection:hcD 1. 37
T/ds 0.^25 D 1 .37[
310  294 / 0 .268]^0.^25
D 3 .81 W/m^2 K.

∴
hrChc D 9 .45 W/m^2 K

which again agrees with the assumed value.
In practice forced convection currents are usually present and the heat loss would
probably be higher than this value.
For an unlagged pipe andTD150 K,
hrChc would be about 20 W/m^2 Kandthe
heat loss,Q/lD
hrChc$d 0 TD
20 $ð 0. 168 ð 150 D1584 W/m.
Thus the heat loss has been reduced by about 90% by the addition of 50 mm of lagging.

PROBLEM 9.29

A refractory material with an emissivity of 0.40 at 1500 K and 0.43 at 1420 K is at a
temperature of 1420 K and is exposed to black furnace walls at a temperature of 1500 K.
What is the rate of gain of heat by radiation per unit area?

Solution

In the absence of further data, the system will be considered as two parallel plates.