(4)
Given the following data, determine Hf using the design formula:
Pipe =.701 ft
Fuel = DF2 @ 50F
Vel = 4.0 ft/sec
Length of pipeline = 4000 ft
First, we must find "f"
Determine Reynold's Number R = V x D
Y
V = 4.0 ft/sec (Given)
D =.701
(Given)
Y = 4.9 cs = .0000527 ft/sec
92,900
Re = 4.0 ft/sec x .701 ft = 53206.8 or 5.3 x 104
.0000527
From FM 5-482, Figure C-8 f = .0216
Hf= .0216 x 4000 x 16 = 1382.4 ft
64.4 x .701
45.1444
Hf= 30.62 ft
(5)
Given the following data, determine Hf using the field modified formula:
Pipe = 8.415 inches
Fuel = DF2 @ 50F
Q = 700 GPM
Pipe Length = 4000 ft
First, you must find "f"
Determine Reynold's Number Re = 3160 x Q
dxk
3160 = constant
Q = 700 GPM (Given)
d = 8.415 (Given)
k = 4.9 cs (FM 5-482, Figure. C-7)
Re = 3160 x 700 = 2212000
8.415 x 4.9 41.234
QM5203
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