CHEMICAL ENGINEERING

HEAT TRANSFER 207

temperature is 293 K. When the outside and initial temperatures are only 278 K, it takes
2700 s to heat the contents to 373 K. The area of the steam coil is 2.5m^2 and of the
external surface is 40 m^2. If the overall heat transfer coefficient from the coil to the liquid
in the vessel is 400 W/m^2 K, show that the overall coefficient for transfer from the vessel
to the surroundings is about 5 W/m^2 K.

Solution

Using the argument in Section 9.8.3, the net rate of heating is given by:

mCpdT/dtDUcAc

TsTUoAo

TTo

whereUcandUoare the overall coefficients from the coil and the outside of the vessel
respectively,AcandAoare the areas of the coil and the outside of the vessel andTs,T
andToare the temperature of the steam, the contents and the surroundings respectively.
WritingUcAcDaandUoAoDb:

mCpdT/dtDa

 393 Tb

TTa D 393 aCbTa

aCbT

Integrating:tD
mCp/
aCb

[

ln 

1 /[393aCbTa

aCbT]

] 373

Tas

D

mCp/

aCbln[

a

 393 Ta/

 20 a 373 bCbTa ]s

WhenTaD293 K: 1800 D
mCp/
aCbln
100 a/
20 a 80 b

WhenTaD278 K: 2700 D
mCp/
aCbln[
115 a/
20 a 95 b] (ii)

Dividing (i) by (ii): 0. 667 Dln[5a/
a 4 b]/ln[23a/
4 a 19 b] (iii)

ButUcD400 W/m^2 K,AcD 2 .5m^2 and hence:

aDUcAcD1000 W/K or 1 kW/K.

Substituting in (iii): 1. 5 Dln[23/
4  19 b]/ln[5/
1  4 b]
Solving by trial and error:bD 0 .2 kW/K or 200 W/K.

∴
Uoð 40 D200 andUoD
200 / 40 D5W/m^2 K.

PROBLEM 9.76

Steam at 403 K is supplied through a pipe of 25 mm outside diameter. Calculate the
heat loss per metre to surroundings at 293 K, on the assumption that there is a negligible
drop in temperature through the wall of the pipe. The heat transfer coefficienthfrom the
outside of the pipe of the surroundings is given by:

hD 1. 22 

T/d^0.^25 W/m^2 K

wheredis the outside diameter of the pipe (m) andTis the temperature difference
(deg K) between the surface and surroundings.