CHEMICAL ENGINEERING

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: