CHEMICAL ENGINEERING

HEAT TRANSFER 201

In this case the outside temperature of the lagging will be
288  19. 8 D 268 .2Kand
the temperature drop through the lagging will be
268. 2  90 D 178 .2degK.
Thus, the heat flow through the lagging is:

1030 D

kA/xTlaggingD 

0. 07 ð 10. 41 /x 178. 2

from which the thickness of the lagging,xD 0 .126 m or 126 mm

This calculation does not take into account the increase in the surface area at the lagging
surface since it was assumed to be that of the tank, 10.41 m^2. In practice, it will be larger
than this and, if this is taken into account, the reasoning is as follows:
Radius of the tankD

1. 82 / 2 D 0 .91 m
   ∴For a lagging thickness ofxm, the new radius is


2. 91 Cxm and the surface area is:

4 $

 0. 91 Cx^2 m^2

∴convective heat gainD 5. 0 ð 4 $
0. 91 Cx^2
288 TD1030 W
and the outside temperature of the lagging:

TD 288  16. 39 /

 0. 91 Cx^2 K(i)

The heat flow through the lagging (taking an arithmetic mean area and neglecting the
curvature),
1030 D
0. 07 /x 4 $
0. 91 Cx/ 22
T 90

Substituting forTfrom (i) into (ii):

1030 D 

0. 07 /x 4 $

 0. 91 Cx/ 22

198  16. 39 /

 0. 91 Cx^2

Solving by trial and error:xD 0 .151 m or 151 mm

PROBLEM 9.70

Water at 293 K is heated by passing it through a 6.1 m coil of 25 mm internal diameter
pipe. The thermal conductivity of the pipe wall is 20 W/m K and the wall thickness is
3.2 mm. The coil is heated by condensing steam at 373 K for which the film coefficient
is 8 kW/m^2 K. When the water velocity in the pipe is 1 m/s, its outlet temperature is
309 K. What will the outlet temperature be if the velocity is increased to 1.3 m/s, if the
coefficient of heat transfer to the water in the tube is proportional to the velocity raised
to the 0.8 power?

Solution

The surface area of the coilD$dolD$

25 C 2 ð 3. 2 / 10006. 1 D 0 .602 m^2

i) When the water velocity is 1 m/s:

Area for flowD$d^2 i/ 4 D$

 25 / 10002 / 4 D 0 .00049 m^2

Vo l u m e fl o wD 

0. 00049 ð 1. 0 D 0 .00049 m^3 /s