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112 Applied Process Design for Chemical and Petrochemical Plants
10. Calculate the area required using Equation 10-9. (Note: disregard sign)
11. Calculate the net available area in the assumed unit, t c 15°
using only the effective tube length.
t h 81°
12. Compare values calculated in steps 10 and 11. If the cal-
culated unit is too small, re-assume a new larger unit for t c 15 0.815
step 3 or try closer baffle spacing in step 7 but do not t h 81
1
get baffles closer than / 5 the shell I.D.
Reading Figure 10-38,
13. Calculate the percent of excess area. A reasonable fig-
ure is 10—20%. F 0.32
14. Calculate the shell-side pressure drop. (Refer to the Then: t h 105 0.32(176 105) 127.7°F
later section on “Pressure Drop Relations” and Figure
Note that the arithmetic average [ / 2 (176 105) 105
1
10-140. If P is too high, reassume unit (step 3).
140°F] would be quite satisfactory for this design, because the
15. Calculate the tube-side pressure drop. (Use Figure 10-
properties do not vary significantly with temperature.
139 for the end return losses. For water in tubes, use
Figure 10-138 for tube losses. For other liquids and
1. Heat duty (6350) (176 105) (0.33)
gases in tubes, use Figure 10-137.
150,000 Btu/hr
Total pressure drop (end return tube) losses, psi.
2. Estimated unit
If the tube-side pressure drop exceeds a critical allowable
Assume: U 100
value for the process system, then recheck by either lower- LMTD 39.2
ing the flow rate and changing the temperature levels or
reassume a unit with fewer passes on tube side or more tubes
150,000
per pass. The unit must then be rechecked for the effect of A 38.2 ft 2
11002139.22
changes on heat transfer performance.
Tubes: 1-in. O.D. 14 BWG 8 ft long
Example 10-9. Convection Heat Transfer
Exchanger Design No. required 38.2
2
10.2618 ft >ft2 18 6 in.>122
See Figure 10-65. 20 tubes
The liquid bottoms from a distillation column must be
cooled from 176°F to 105°F. The cooling water is untreated Trial:
at 90°F.
1
10-inch I.D. shell with 24 1-in. tubes on 1 / 4 -in. triangu-
Operating data: Bottom flow, 6,350 lb/hr lar pitch, 4 tube passes.
Average Cp, 0.333 Btu/lb (°F)
2
Average k a , 0.055 Btu/hr (ft ) (°F/ft) 3. Log mean temperature difference (Figure 10-33),
Average , 0.404 centipoise
Average sp.gr, 0.78 176° cooling 105°
Physical properties are based on values at 140°F average 95° warming 90°
temperature. 81° 15°
Caloric fluid temperature for property evaluation can
be calculated from Equation 10-21. 4. Water rate,
The caloric value of hot liquid on the shell side is
150,000
w 30,000 lb>hr
112195 902
t h t h2 F1t h1 t h2 2
30,000
150,000 gpm 60
Rough estimate U c at cold end 262 18.3321602
138.22115°2
24
At / 4 6 tubes/pass,
150,000
U h at hot end 48.5
38.2181°2 0.546 2
Cross-sectional area/tube 0.00379 ft >tube
144
U h U c 48.5 262
C 0.815 2
U c 262 Flow area/pass = (0.00379)(6 tubes) 0.0227 ft flow area