(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 10**4

.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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