CHEMICAL ENGINEERING

HEAT TRANSFER 143

or, substituting forTw:  2 D
480 Ta D 1

Thus in equation 9.9: mD
480 Ta

Overall coefficient

The solution is now one of trial and error in that mean temperatures of both streams must
be assumed in order to evaluate the physical properties.
Inside the tubes:
a mean temperature of 320 K, will be assumed at which,

kD 0 .028 W/m K,D 0. 0193 ð 10 ^3 Ns/m^2 ,andCpD 1. 0 ð 103 J/kg K

Therefore:

hidi/kD 0. 023 

diG/^0.^8 

Cp/k^0.^4 (equation 9.61)

hið 0. 012 / 0. 028 D 0. 023

0. 012 ð 8. 0 / 0. 0193 ð 10 ^30.^8

ð 

1 ð 103 ð 0. 0193 ð 10 ^3 / 0. 0280.^4

∴ hiD 0. 0537
4. 974 ð 1030.^8
0. 6890.^4 D 41 .94 W/m^2 K

Outside the tubes:

The cross-sectional area of the tube bundleD
0. 7 ð 20

1. 5 ð 0. 012 D 0 .252 m^2 and
   hence the free flow mass velocity,G^0 D


2. 217 / 0. 252 D 0 .861 kg/m^2 s.

From Fig. 9.27:YD

1. 5 ð 0. 012 D 0 .018 m
   and therefore: G^0 maxD


2. 861 ð 0. 018 /


3. 081  0. 012 D 2 .583 kg/m^2 s
   At an assumed mean temperature of 450 K,D 0. 0250 ð 10 ^3 Ns/m^2 andkD
   0 .035 W/m K.

∴ RemaxD
0. 012 ð 2. 583 /
0. 0250 ð 10 ^3 D 1. 24 ð 103

From Table 9.3: forXD 1. 5 doandYD 1. 5 do,ChD 0 .95.

In equation 9.90:

ho 

0. 012 / 0. 035 D 

0. 33 ð 0. 95

1. 24 ð 1030.^6

1. 0 ð 103 ð 0. 0250 ð 10 ^3 / 0. 0350.^3

∴ hoD
0. 914 ð 71. 8 ð 0. 7140.^3 D 59 .3W/m^2 K

Hence, ignoring wall and scale resistances:

1 /UD 

1 / 41. 91 C 1 / 59. 3 D 4. 07 ð 10 ^2

and: UD 24 .57 W/m^2 K

Thus, in equation 9.1:

217 

Ta 290 D 24. 57 ð 6. 34

480 Ta

from which: TaD 369 .4K