188 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Solution
The cross-sectional area of 0.025 m tubingD
$/ 4 0. 0252 D 0 .000491 m^2.
Vo l u m e fl o w o f w a t e rD
2 ð 0. 000491 D 0 .000982 m^3 /s
Mass flow of waterD
1000 ð 0. 000982 D 0 .982 kg/s
∴ Heat load,QD 0. 982 ð 4. 18
295 293 D 8 .21 kW
Surface area of 0.025 m tubingD
$ð 0. 025 ð 1. 0 D 0 .0785 m^2 /m
and: AD
0. 0785 ð 2. 5 D 0 .196 m^2
1 D
320 293 D27 deg K, 2 D
320 295 D25 deg Kand: mD
27 25 /ln
27 / 25 D 25 .98 say 26 deg K
In equation 9.1:UD 8. 21 /
0. 196 ð 26 D 1 .612 kW/m^2 K
An estimate may be made of the inside film coefficient from equation 9.221, whereT,
the mean water temperature, is 294 K.
Thus: hiD 4280
0. 00488 ð 294 1 2. 00.^8 / 0. 0250.^2
D
4280 ð 0. 435 ð 1. 741 / 0. 478 D6777 W/m^2 Kor6.78 kW/m^2 KThe scale resistance is therefore given by:
1 / 1. 612 D
1 / 6. 78 CRor:RD 0 .473 m^2 K/kWWith a water velocity of
2. 0 ð 150 / 100 D 3 .0 m/s, assuming a mean water temper-
ature of 300 K, then:
hiD 4280
0. 00488 ð 300 1 3. 00.^8 / 0. 0250.^2D
4280 ð 0. 464 ð 2. 408 / 0. 478 D10004 or 10.0kW/m^2 K∴ 1 /UD
0. 473 C 1 / 10. 0 andUD 1 .75 kW/m^2 K
For an outlet water temperature ofTK: 1 D27 deg K, 2 D
320 T deg K
and, taking an arithmetic mean: mD 0. 5
27 C 320 T D
173. 5 0. 5 T deg K.
The mass flow of waterD
0. 982 ð 150 / 100 D 1 .473 kg/s,
and the heat load, QD 1. 473 ð 4. 18
T 293 D
6. 157 T 1804 kW
∴
6. 157 T 1804 D[1. 75 ð 0. 196
173. 5 0. 5 T ]
from which: TD 294 .5K
The use of 300 K as a mean water temperature has a minimal effect on the result and
recalculation is not necessary.