CHEMICAL ENGINEERING

HEAT TRANSFER 199

The heat flow to the surroundings is:QDhoAo
T 2 To

∴ 3500 Dho
$ð 2. 0 ð 5. 0
301  293 andhoD 14 .05 W/m^2 K

If this value is halved, that isho 2 D 7 .02 W/m^2 K, then:

Q 2 D 7. 02

2 $ð 2. 0 ð 5. 0 

T 2  293 D 220. 5 

T 2  293

and: T 2 D
Q 2 / 220. 5 C 293 K.

But:
Q 2 /Q 1 D
T 1 T 22 /
T 1 T 21

∴
Q 2 / 3500 D[1073
Q 2 / 220. 5 293]/
1073  301

0. 000286 Q 2 D 0. 00130

780  0. 00454 Q 2

and: Q 2 D3473 W— a very slight reduction in the heat loss.

In this case,T 2 D
3473 / 220. 5 C 293 D 308 .7K.

PROBLEM 9.68

An open cylindrical tank 500 mm diameter and 1 m deep is three-quarters filled with a
liquid of density 980 kg/m^3 and of specific heat capacity 3 kJ/kg K. If the heat transfer
coefficient from the cylindrical walls and the base of the tank is 10 W/m^2 K and from
the surface is 20 W/m^2 K, what area of heating coil, fed with steam at 383 K, is required
to heat the contents from 288 K to 368 K in a half hour? The overall heat transfer
coefficient for the coil may be taken as 100 W/m^2 K. The surroundings are at 288 K. The
heat capacity of the tank itself may be neglected.

Solution

The rate of heat transfer from the steam to the liquid is:

UcAc 

383 TD 100 Ac 

383 T W

whereAcis the surface area of the coil.
The rate of heat transfer from the tank to the surroundingsDUTAT
T 288
whereUTis the effective overall coefficient andATthe surface area of the tank and liquid
surface. In this case:

UTATD 10 



$ð 0. 5 ð 1 C

$/ 40. 52 C 20 

$/ 40. 52 D 21 .6W/K.

∴rate of heat lossD 21. 6
T 288 W

∴net rate of heat input to the tankD 100 Ac
383 T 21. 6
T 288 W.

This is equal tomCpdT/dt, where the mean specific heat,CpD3000 J/kgK.

Volume of liquidD
75 / 100 
$/ 40. 52 ð 1 D 0 .147 m^3