214 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Substituting forT 2 from (i):
104 D[413
49. 14 C 1. 809 x/
0. 157 C 0. 00628 x]/ln
1 C 0. 04 x
Solving by trial and error,xD 52 .5mm.
PROBLEM 9.82
It takes 1800 s (0.5 h) to heat a tank of liquid from 293 to 323 K using steam supplied to
an immersed coil when the steam temperature is 383 K. How long will it take when the
steam temperature is raised to 393 K? The overall heat transfer coefficient from the steam
coil to the tank is 10 times the coefficient from the tank to surroundings at a temperature
of 293 K, and the area of the steam coil is equal to the outside area of the tank.
Solution
Using the argument in Problem 9.77,
mCpdT/dtDUcAc
TsTUsAs
TTa
In this case,TsD383 K,TaD293 K,UsDUc/10 andAcDAsDA(say).
∴ mCpdT/dtDUcAc
383 T
UcAc/ 10
T 293
∴
UcAc/mCp dtDdT/
412. 3 1. 1 T
On integration:
UcAc/mCptD
∫ 323
293
dT/
412. 3 1. 1 TD
1 / 1. 1
[
ln
1 /
412. 3 1. 1 T
] 323
293
Since it takes 1800 s to heat the liquid from 293 to 333 K, then:
1800
UcAc/mCp D 0 .909 ln[
412. 3
1. 1 ð 293 ]/[412. 3
1. 1 ð 323 ]
D 0 .909 ln
90 / 57
and:
UcAc/mCp D 0 .00023071 s^1
On increasing the steam temperature to 393 K,
Heat transferred from the steamDUcAc
393 TW
Heat lost to the surroundingsD
UcAc/ 10
T 293 W
and: mCpdT/dtDUcAc
393 T
0. 1 UcAc
T 293 W
∴ dT/
422. 3 1. 1 TD
UcAc/mCp dtD 0 .0002307 dt