CHEMICAL ENGINEERING

HEAT TRANSFER 211

This result agrees fairly well and a mean value ofTwD545 Kis indicated.

In equation (iv):
k^0 /kD 3. 624
323  298 /
545  323 D 0. 408

∴in equation (iii) u^0.^8 D 0. 408
545 T/
T 298

WhenuD20 m/s: 10. 99 D 0. 408
545 T/
T 298 andTD 306 .8K.

PROBLEM 9.79

A hydrocarbon oil of density 950 kg/m^3 and specific heat capacity 2.5 kJ/kg K is cooled in
a heat exchanger from 363 to 313 K by water flowing countercurrently. The temperature
of the water rises from 293 to 323 K. If the flowrate of the hydrocarbon is 0.56 kg/s,
what is the required flowrate of water?
After plant modifications, the heat exchanger is incorrectly connected so that the two
streams are in co-current flow. What are the new outlet temperatures of hydrocarbon and
water, if the overall heat transfer coefficient is unchanged?

Solution

Heat lost by the oilD 0. 56 ð 2. 5
363  313 D 70 .0kW

For a flow of water ofGkg/s, heat gained by the water is:

70. 0 D

Gð 4. 18

323  293 andGD 0 .56 kg/s

i) For countercurrent flow:

T 1 D 

363  323 D40 degK, T 2 D 

313  293 D20 deg K

and from equation 9.9:

TmD 

40  20 /ln 

40 / 20 D 28 .85 deg K

In equation 9.1:

70. 0 DUAð 28 .85 andUAD 2 .43 kW/K.

ii)For co-current flow:

IfToandTware the outlet temperature of the oil and water respectively, the heat

load is:

QD 

0. 56 ð 2. 5

363 To D 

0. 56 ð 4. 18 

Tw 293 (i)

508. 2  1. 4 ToD 2. 34 Tw 685. 9

and: TwD 510. 3  0. 60 To (ii)

T 1 D 

363  293 D70 K

T 1 D

ToTw D

To 510. 3 C 0. 60 To D 

1. 60 To 510. 3 K