CHEMICAL ENGINEERING

204 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

The conduction through the firebrick is given by:

QD 400 D 

1. 5 ð 1. 0 / 0. 2

1573 T 2 (equation 9.12)

andT 2 , the temperature at the firebrick/insulating brick interface, is:

T 2 D 1579 .7K.

For the natural convection to the surroundings:

QDhoA

T 3 Ta

or: 400 D 3. 0 T^0.^25 ð 1. 0
T 3  293

but
T 3  293 DTand:

400 D 3. 0 T^1.^25

∴ TD 133. 30.^8 D 50 .1degK

and the temperature at the outer surface of the insulating brick,T 3 D
293 C 50. 1 D
343 .1 K. Thus, applying equation 9.12 to the insulating brick:

400 D 

0. 4 ð 1. 0 /x

 1519. 7  343. 1

and the thickness of the brick,xD 1 .18 m

PROBLEM 9.73

2.8 kg/s of organic liquid of specific heat capacity 2.5 kJ/kg K is cooled in a heat
exchanger from 363 to 313 K using water whose temperature rises from 293 to 318 K
flowing countercurrently. After maintenance, the pipework is wrongly connected so that
the two streams, flowing at the same rates as previously, are now in co-current flow.
On the assumption that overall heat transfer coefficient is unaffected, show that the new
outlet temperatures of the organic liquid and the water will be 320.6 K and 314.5 K,
respectively.

Solution

i) Countercurrent flow

Heat load, QD 

2. 8 ð 2. 5

363  313 D350 kW

∴water flowD 350 / 4. 18

318  293 D 3 .35 kg/s

T 1 D 

363  318 D35 deg K,T 2 D 

313  293 D20 deg K

and, from equation 9.9,TmD 

45  20 /ln 

45 / 20 D 30 .83 deg K.