CHEMICAL ENGINEERING

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