CHEMICAL ENGINEERING

212 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

andinequation9.9:TmD 

70  1. 60 ToC 510. 3 /ln[70/

 1. 60 To 510. 3 ]

D 

580. 3  1. 60 To/ln[70/ 1. 60 To 510. 3 ]degK

and: 

508. 2  1. 4 To D 2. 43

580. 3  1. 60 To/ln[70/

 1. 60 To 510. 3 ]



508. 2  1. 4 To D 2. 126

508. 2  1. 40 To/ln[70/

 1. 60 To 510. 3 ]

∴ 70 /

 1. 60 To 510. 3 De^2.^126 D 8 .38 andToD 324 .2K.

and in equation (ii):TwD 510. 3  

0. 60 ð 324. 2 D 315 .8K.

PROBLEM 9.80

A reaction mixture is heated in a vessel fitted with an agitator and a steam coil of area
10 m^2 fed with steam at 393 K. The heat capacity of the system is equal to that of
500 kg of water. The overall coefficient of heat transfer from the vessel of area 5 m^2
is 10 W/m^2 K. It takes 1800 s to heat the contents from ambient temperature of 293 to
333 K. How long will it take to heat the system to 363 K and what is the maximum
temperature which can be reached? Specific heat capacity of waterD4200 J/kgK.

Solution

Following the argument of Problem 9.77 and taking ambient temperature as the initial
temperature of the mixture, 293 K, then:

net rate of heatingD
500 ð 4200 dT/dtDUcð 10
393 T
10 ð 5 
T 293 W

∴ 2 , 100 ,000 dT/dtD 3930 Uc 10 UcT 50 TC 14 , 650

D 

3930 UcC 14 , 650  

10 UcC 50 T.

∴
2 , 100 , 000 /
10 UcC 50 dT/dtD

3930 UcC 14 , 650 /
10 UcC 50 T.

∴ tD[2, 100 , 000 /
10 UcC 50 ]

∫ 333

293 dT/[^3930 UcC^14 ,^650 /

^10 UcC^50 T]

In heating from 293 to 333 K, the time taken is 1800 s and:

1800 D[2, 100 , 000 /

 10 UcC 50 ]lnf[ 

3930 UcC 14 , 650 /

 10 UcC 50 293]/

[ 

3930 UcC 14 , 650 /

 10 UcC 50 333]g

Solving by trial and error:UcD 61 .0W/m^2 K.
Thus, net rate of heating is:

2 , 100 ,000 dT/dtD
61. 0 ð 10
393 T
10 ð 5 
T 293 D 254 , 380  660 TW

or: 3182 dT/dtD 385. 4 TW(i)

∴ time for heating,tD 3182

∫ 363

293

dT/

 385. 4 T

D3182 ln[ 

385. 4  293 /

 385. 4  363 ]D3182 ln 

92. 4 / 22. 4 D4509 s 

1 .25 h