HEAT TRANSFER 209
surroundings at 293 K from the outside of the vessel. How long does it take to heat the
liquid from 293 to 353 K and what is the maximum temperature to which the liquid can
be heated? When the liquid temperature has reached 353 K, the steam supply is turned off
for two hours and the vessel cools. How long will it take to reheat the material to 353 K?
Coil: Area 0.5m^2. Overall heat transfer coefficient to liquid, 600 W/m^2 K. Outside of
vessel: Area 6 m^2. Heat transfer coefficient to surroundings, 10 W/m^2 K.
Solution
IfTK is the temperature of the liquid at timets, then the net rate of heat input to the
vessel,UcAc
TsTUsAs
TTa DmCpdT/dtW
where the coefficient at the coil,UcD600 W/m^2 K, the coefficient at the outside of
the vessel, UsD10 W/m^2 K, the areas are: coil, AcD 0 .5m^2 , vessel, AsD 6 .0m^2 ,
the temperatures are: steam, TsD393 K, ambient,TaD293 K, the mass of liquid,
mD1000 kg and the specific heat capacity,CpD 4 .0 kJ/kg K or 4000 J/kg K.
Thus:
1000 ð 4000 dT/dtD
600 ð 0. 5
393 T
10 ð 6
T 293
and: 11 ,111 dT/dtD 376. 3 T (i)
∴ tD 11 , 111
∫T 2
T 1
dT/
376. 3 TD 11 ,111 ln[
376. 3 T 1 /
376. 3 T 2 ] (ii)
WhenT 1 D293 K andT 2 D353 K then:tD 11 ,111 ln
83. 3 / 23. 3 D 14 ,155 s(3.93 h)
The maximum temperature to which the liquid can be heated is obtained by putting
dT/dtD0in(i)togive:TD 376 .3K.
During the time the steam is turned off (for a period of 7200 s) a heat balance gives:
mCpdT/dtDUsAs
TTa
or:
1000 ð 4000 dT/dtD
10 ð 6
T 293
∴ 66 ,700 dT/dtD
293 T
Integrating:
∫T
353
dT/
293 TD 0. 000015
∫ 7200
0
dt
∴ ln
293 353 /
293 TD
0. 000015 ð 7200 D 0 .108 andTD 346 .9K.
The time taken to reheat the liquid to 353 K is then given by (ii):
tD 11 , 111
∫ 353
346. 9
dT/
376. 3 T
D 11 ,111 ln[
376. 3 346. 9 /
376. 3 353 ]D2584 s
0 .72 h
PROBLEM 9.78
A bare thermocouple is used to measure the temperature of a gas flowing through a hot
pipe. The heat transfer coefficient between the gas and the thermocouple is proportional