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