CHEMICAL ENGINEERING

HEAT TRANSFER 213

The maximum temperature which can be attained is obtained by putting dT/dtD0in(i)
which gives:
TmaxD 385 .5K.

PROBLEM 9.81

A pipe, 50 mm outside diameter, is carrying steam at 413 K and the coefficient of heat
transfer from its outer surface to the surroundings at 288 K is 10 W/m^2 K. What is the
heat loss per unit length?
It is desired to add lagging of thermal conductivity 0.03 W/m K as a thick layer to
the outside of the pipe in order to cut heat losses by 90%. If the heat transfer from the
outside surface of the lagging is 5 W/m^2 K, what thickness of lagging is required?

Solution

Outside area of pipe:$dlD$
50 / 1000 ð 1. 0 D 0 .157 m^2 /m. Assuming the pipe wall
is at the temperature of the steam, that is the resistance of the wall is negligible, then the
heat loss is:

QDhA

TwTa D 

10 ð 0. 157

413  288 D196 W/m

With the addition of lagging of thickness,xmm, the required heat flow is 19.6 W/m
The diameter of the laggingD
50 C 2 x/1000 m

and the surface area of the laggingD[$
50 C 2 x/1000]ð 1. 0
D
0. 157 C 0. 00628 xm^2 /m.

The heat transferred to the surroundings,

19. 6 D 5

0. 157 C 0. 00628 x

T 2  288 W

and: T 2 D
49. 14 C 1. 809 x/
0. 157 C 0. 00628 xK(i)

For conduction through the lagging:

QDk

 2 $rml

T 1 T 2 /

r 2 r 1 W (equation 9.22)

where QD 19 .6W/m,kD 0 .03 W/mK, lD 1 .0 m and the steam temperature,T 1 D
413 K.

r 1 D25 mm or 0.025 m,

r 2 D 

25 Cxmm or 

25 Cx/ 1000 D 

0. 025 C 0. 001 xm

∴ rmD
0. 025 C 0. 001 x 0. 025 /ln[
0. 025 C 0. 001 x/ 0 .025]

D 0. 001 x/ln 

1 C 0. 04 xm

Thus, in equation 9.22:

19. 6 D 0. 03

2 $

 0. 001 x/ln 

1 C 0. 04 xð 1. 0

413 T 2 /

 0. 025 C 0. 001 x 0. 025

∴ 103 , 981 D[x/ln
1 C 0. 04 x]
413 T 2 / 0. 001 x