Page 115 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition

Fluid Flow 99

Example 2-6: Water Hammer Pressure Development 1. From Table 2-4, selected velocity = 6 fps.

An 8-inch process pipe for transferring 2000 GPM of Estimated pipe diameter, d = (0.408 Q!v) 112
2
methanol of Sp Gr = 0.75 from the manufacturing plant = [ (0.4.-08) 25/6] 11 = 1.3 inch
site to a user plant location is 2,000 feet long, and the liq-
uid is flowing at 10.8 ft/ sec. Try l�inch (i.d. = 1.61), since lX-inch (i.d. =
Maximum pressure developed (preliminary solution) 1.38) is not stocked in every plant. If it is an accept-
when an emergency control valve suddenly closes: able plant pipe size, then it should be considered
and checked, as it would probably be as good pres-
sure drop-wise as the lY,-inch. The support of lX-
(2-69)
inch pipe may require shorter support spans than
the lY,-inch. Most plants prefer a minimum of l�
Since methanol has many properties similar to water: inch valves on pressure vessels, tanks, etc. The valves
at the vessels should be 11.! inch even though the
1 2
aw = 4660/ (I + Ki,, Br) 1 pipe might be IX inch The control valve system of
= 4660/(1-:-- 0.01 (24.7"')] 112 = 4175 ft/sec gate and globe valves could very well be iX inch. For
this example, use lY,-inch pipe, Schedule 40:
"For 8-inch std pipe, Br= 7.981/0.322 = 24.78
2. Linear length of straight pipe, L = 254 ft.
Time interval for pressure wave travel: 3. Equivalent lengths of fittings, valves, etc.

ts= 2L/aw = 2 (2000)/4175 = 0.95 sec (2-71)
Estimated Eq. Feet
Fittings Type (from Figure 2-20)
If the shutoff time for the valve (or a pump) is less than
0.95 seconds, the water hammer pressure will be: 10 lW'-90° Elbows 4' (10) = 40
8 L;,5"-Tees 3' (8) = 24
4 l W'-Gale Valves l ' (4) = 4
h" 11 = 4175 ( 10.8) /32.2 = 1400 ft of methanol 68 ft. Use 75 ft.
�============��==�==��=====�
= (1400)/((2.�\l)/0.75)] = 454 psi hydraulic shock

Then total pressure on the pipe system 4. No expansion or contraction losses ( except control
valve).
5. Pressure drop allowance assumed for orifice plate =
= 454 + (existing pressure from process/or pump)
5 psig.
This pressure level would most likely rupture an 8-inch Control valve loss will be by difference, trying to
Sch. 40 pipe. For a more exact solution, refer to specialty maintain minimum 60% of pipe friction loss as min-
articles on the subject. imum drop through valve, but usually not less than
10 psi.
Example 2-7: Pipe flow System With Liquid of Specific
Gravity Other Than Water 6. Reynolds number, R., = 50.6 Qp/dµ (2-49)
== so.e (25) [0.93 (62.3)]/
This is illustrated by line size sheet, Figure 2-28. (1.61) (0.91)
= 50,025 (turbulent)
Figure 2-29 represents a liquid reactor system discharg-
ing crude product similar to glycol through a flow control 7. From Figure 2-11, £/d = 0.0012 for lY,-inch steel pipe.
valve and orifice into a storage tank. The reactor is at 350
psig and 280°F with the liquid of 0.93 specific gravity and From Figure 2-3, at R,, = 50,025, read f = 0.021
0.91 centipoise viscosity. There is essentially no flashing of
liquid across the control valve.
8. Pressure drop per 100 feet of pipe:
Flow rate: 11,000 lbs/hr �P/100' = 0.0216 fp Q /d 5 (2-72)
2
2
GPM actual= 11,000/(60) (8.33) (0.93) = 23.7 = 0.0216 (0.021) (62.3) (0.93) (25) /(1.61) 5
= 1.52 psi/ 100 ft equivalent
Design rate = 23.7 (1.05) = 25 gpm