(c) Now, let's solve a problem. Given the following data, determine Reynold's number
using the design data formula.
Velocity = 540 ft/min
ID = 6.415 inches
Fuel = DF2 @ 50 F
Find V in ft/sec
540 ft x 1 min = 9 ft/sec
min 60 sec
Find D in feet
6.415 in x 1 ft = .534 ft
12 in
Find Y in ft/sec squared (FM 5-482, Figure C-7)
4.9 cs = .0000527ft2/sec
92,900
Solve for R
9 ft/sec x .534 ft = 91195.446
.0000527 ft2/sec
Finally, convert to a scientific notation
91195.446 = 9.12 x 104
(d)
Now, let's solve a problem using the field data formula.
Given the following data determine Re using the field data formula.
Q = 350 GPM
d = 6.415 in
K= JP4 @ 50F
Solve for R
3160 x 350
=
1106000 = 144881
6.415 x 1.19
7.63385
Convert into scientific notation
144881 = 1.45 x 105
(e) Now that you know how to find Reynold's number, it is time to interpret it and put it to good
use in designing a military pipeline. Quite simply put, Reynold's number is used to determine the type
of flow in a pipeline, and is broken down into three categories:
QM5203
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