CHEMICAL ENGINEERING

198 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Mass flow of waterD 

1000 ð 0. 00112 D 1 .12 kg/s

Heat removed by waterD 1. 12 ð 4. 187

333  293 D 187 .6kW

Surface area of tube, based on inside diameterD

$ð 0. 025 ð 10 D 0 .785 m^2

Vapour temperature = 353 K ∴T 1 D 

353  293 D60 deg K

T 2 D 

353  333 D20 deg K

and from equation 9.9,TmD
60  20 /ln
60 / 20 D 36 .4degK

From equation 9.1, 187. 6 D
Uð 0. 785 ð 36. 4

and the overall coefficient based on the inside diameter is:UD 6 .57 kW/m^2 K.

In equation 9.201, neglecting the wall and scale resistances:

1 /UD 1 /hoC 1 /hi



1 / 6. 57 D 

1 / 15. 0 C 1 /hiandhiD 11 .68 kW/m^2 K

If the latent heat of condensation is 800 kJ/kg, then assuming the vapour enters and the
condensate leaves at the boiling point:

rate of condensationD 

187. 6 / 800 D 0 .235 kg/s

PROBLEM 9.67

A chemical reactor, 1 m in diameter and 5 m long, operates at a temperature of 1073 K.
It is covered with a 500 mm thickness of lagging of thermal conductivity 0.1 W/m K.
The heat loss from the cylindrical surface to the surroundings is 3.5 kW. What is the heat
transfer coefficient from the surface of the lagging to the surroundings at a temperature
of 293 K? How would the heat loss be altered if the coefficient were halved?

Solution

From equation 9.20, the heat flow at any radiusris given by:

QDk

 2 $C 1 dT/drW

∴ dr/rD  2 $kl/QdT

Integrating between the limitsr 1 andr 2 at which the temperatures areT 1 andT 2 respec-
tively:
∫r 2

r 1

dr/rD  2 $kl/Q

∫T 2

T 1

dT

∴ ln
r 2 /r 1 D  2 $kl/Q
T 1 T 2

In this case: r 1 D
1 / 2 D 0 .50 m, r 2 D
0. 50 C 500 / 1000 D 1 .0m,lD5m,kD
0 .1W/mK,QD3500 W andT 1 D1073 K.

∴ ln

1. 0 / 0. 50 D
   2 $ð 0. 1 ð 5 / 3500
   1073 T 2

and: 0. 693 D 0. 00090
1073 T 2 andT 2 D301 K