CHEMICAL ENGINEERING

HEAT TRANSFER 195

∴ U 1 D 

3. 4 k^0 ð 1750.^8 /

 3. 4 Ck^0 ð 1750.^8 D 211. 8 k^0 /

 3. 4 C 62. 3 k^0

and:U 2 D 

3. 4 k^0 ð 3250.^8 /

 3. 4 Ck^0 ð 3250.^8 D 347. 5 k^0 /

 3. 4 C 102. 2 k^0

∴ [347. 5 k^0 /

 3. 4 C 102. 2 k^0 ]/[211. 18 k^0 /

 3. 4 C 62. 3 k^0 ]D 1. 60

∴ k^0 D 0. 00228

∴ U 1 D 

3. 4 ð 0. 00228 ð 1750.^8 /

 3. 4 C 0. 00228 ð 1750.^8 D 0 .136 kW/m^2 K

and the heat transfer area,AD 

1. 239 / 0. 136 D 9 .09 m^2.

PROBLEM 9.64

0 .1m^3 of liquid of specific heat capacity 3 kJ/kg K and density 950 kg/m^3 is heated
in an agitated tank fitted with a coil, of heat transfer area 1 m^2 , supplied with steam
at 383 K. How long will it take to heat the liquid from 293 to 368 K, if the tank, of
external area 20 m^2 is losing heat to the surroundings at 293 K? To what temperature
will the system fall in 1800 s if the steam is turned off? Overall heat transfer coefficient
in coilD2000 W/m^2 K. Heat transfer coefficient to surroundingsD10 W/m^2 K.

Solution

IfTK is the temperature of the liquid at timets, then:

heat input from the steamDUA
TsTD
2000 ð 1
383 Tor 2000
383 TW

Similarly, heat losses to the surroundingsD
10 ð 20 
T 293 D 200
T 293 W

and, net heat input to the liquidD 2000
383 T 200
T 293 D
824 , 600 
2200 TW

This is equal to: QDmCp dT/dt

wheremD
0. 1 ð 950 D95 kg andCpD3000 J/kg K.

∴
95 ð 3000 dT/dtD
824 , 600  2200 T

or: 129 .6dT/dtD
374. 8 T

Thus the time taken to heat from 293 to 368 K is:

tD 129. 6

∫ 368

293

dT/

 374. 8 T

D 129 .6ln 



374. 8  293 /

 374  368 D1559 s 

0 .43 h

The steam is turned off for 1800s and, during this time, a heat balance gives:

95 ð 3000 dT/dtD 

10 ð 20 

T 293

∴ 285 ,000 dT/dtD
58600  200 Tor 1425 dT/dtD
293 T