HEAT TRANSFER 153
For a water outlet temperature ofTK and a mass flow ofGkg/s:
8. 25 nDGð 4. 18
T 290 kW
or GD 1. 974 n/
T 290 kg/s (i)
1 D
350 290 , 2 D
350 T
and hence in equation 9.9:
mD[
350 290
350 T]/ln[
350 290 /
350 T]
D
T 290 /ln[60/
350 T]deg K.
Considering the film coefficients:hiD 3 .2kW/m^2 K,hoD 1 .7kW/m^2 K and hence:
hioD
3. 2 ð 0. 015 / 0. 019 D 2 .526 kW/m^2 K.
The scale resistance is:
x/kD
0. 25 ð 10 ^3 / 2. 0 D 0 .000125 m^2 K/W or 0.125 m^2 K/kW
Therefore the overall coefficient, neglecting the wall resistance is given by:
1 /UD 1 /hioCx/kC 1 /ho
D
0. 5882 C 0. 125 C 0. 396 D 1 .109 m^2 K/kW orUD 0 .902 kW/m^2 K
Therefore in equation 9.1:
ADQ/UmD 8. 25 n/f 0. 902
T 290 /ln[60/
350 T]gm^2
D
4. 18 G
T 290 ln[60/
350 T]
0. 902
T 290
D 4. 634 Gln[60/
350 T]m^2 (ii)
The cross-sectional area for flowD
$/ 4
0. 0152 D 0 .000177 m^2 /tube.
Gkg/s
G/ 1000 D 0. 001 Gm^3 /s
and area/pass to give a velocity of 0.6 m/sD
0. 001 G/ 0. 6 D 0. 00167 Gm^2.
∴ number of tubes/passD
0. 00167 G/ 0. 000177 D 9. 42 G (iii)
Area per unit length of tubeD
$ð 0. 019 ð 1. 0 D 0 .0597 m^2 /m.
∴ total length of tubesD 4. 634 Gln[60/ 350 T]/ 0. 0597 D 77. 6 Gln[60/ 350 T]m
length of each tubeD 77. 6 Gln[60/ 350 T]/nm
and, substituting from (i),
tube lengthD 77. 6 ð 1. 974 nln[60/
350 T]/
n
T 290 ]
D 153 .3 ln[60/
350 T]/
T 290 m(iv)
The procedure is now to select a number of tube passesNand hencemin terms ofn
from (iii).Tis then obtained from (i) and hence the tube length from (iv). The following
results are obtained: