CHEMICAL ENGINEERING

HEAT TRANSFER 167

For theinner lagging:r 2 D 

0. 028 C 0. 025 D 0 .053 m

r 1 D 0 .028 m

and: rmD
0. 053  0. 028 /
ln 0. 053 / 0. 028 D 0 .0392 m.

Therefore the thermal resistanceD
0. 053  0. 028 /
0. 08 ð 2 $ð 0. 0392 ð 1. 0
D 1 .2688 K/W

For theouter lagging:r 2 D 

0. 053 C 0. 040 D 0 .093 m

r 1 D 0 .053 m

and: rmD
0. 093  0. 053 /
ln 0. 093 / 0. 053 D 0 .0711 m

Therefore the thermal resistanceD
0. 093  0. 053 /
0. 04 ð 2 $ð 0. 0711 ð 1. 0
D 2 .2385 K/W

From equation 9.19:QD 

550  330 /

 0. 0004 C 1. 2688 C 2. 2385 D 62 .7W/m

PROBLEM 9.36

The temperature of oil leaving a co-current flow cooler is to be reduced from 370 to
350 K by lengthening the cooler. The oil and water flowrates, the inlet temperatures and
the other dimensions of the cooler will remain constant. The water enters at 285 K and
oil at 420 K. The water leaves the original cooler at 310 K. If the original length is 1 m,
what must be the new length?

Solution

For theoriginal cooler, for the oil: QDGoCpo
420  370

and for the water: QDGwCpw
310  285

∴
GoCp/GwCp D
25 / 50 D 0. 5

whereGoandGware the mass flows andCpoandCpwthe specific heat capacities of the
oil and water respectively.
 1 D
420  285 D135 deg K, 2 D
370  310 D60 deg K for co-current flow, and
from equation 9.9:mD
135  60 /ln
135 / 60 D 92 .5deg K

Ifais the area per unit length of tube multiplied by the number of tubes, then:

AD 1. 0 ðam^2 andinequation9.1:

GoCp 

420  370 DUa 92 .5or

GoCp/UaD 1. 85

For thenew cooler, for the oil:QDGoCpo 

420  350

and for the water, QDGwCpw
T 285