CHEMICAL ENGINEERING

172 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

or: 0. 00125 t^2  1. 05 t 70 D0andtD902 s

Therefore in (i): TD
290 C
0. 083 ð 902 D 364 .9K

With half the length, that is 0.050 m, and half the heat capacity, that is 7.5 g water, then
in equation (ii):

7. 5 / 1000 ð 4. 18

 dTt/dtD 0. 300 

$ð 0. 010 ð 0. 050 

TTt

or: dTt/dtD 0. 0150
TTt

The same result as before and hence the new thermometer would also read 360 K.

PROBLEM 9.40

How long will it take to heat 0.18 m^3 of liquid of density 900 kg/m^3 and specific heat
2.1 kJ/kg K from 293 to 377 K in a tank fitted with a coil of area 1 m^2? The coil is fed
with steam at 383 K and the overall heat transfer coefficient can be taken as constant at
0 .5kW/m^2 K. The vessel has an external surface of 2.5m^2 , and the coefficient for heat
transfer to the surroundings at 293 K is 5 W/m^2 K.
The batch system of heating is to be replaced by a continuous countercurrent heat
exchanger in which the heating medium is a liquid entering at 388 K and leaving at 333 K.
If the heat transfer coefficient is 250 W/m^2 K, what heat exchange area is required? Heat
losses may be neglected.

Solution

Mass of liquid in the tankD
0. 18 ð 900 D162 kg

∴ mCpD
162 ð 2100 D 340 ,200 J/deg K

Using the argument given in Problem 9.77:

340 ,200 dT/dtD 

500 ð 1

383 T 

5 ð 2. 5 

T 293

or: D 191 , 500  500 T 12. 5 TC 3663 D 195 , 163  512. 5 T

or: 664 dT/dtD 380. 8 T

The time taken to heat the liquid from 293 to 377 K is:

tD 664

∫ 377

293

dT/

 380. 8 T

D664 ln[ 

380. 8  293 /

 380. 8  377 ]D2085 s 

0 .58 h

For the heat exchanger: