(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 10**4

(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 10**5

(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

1-15

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