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Old Aug 18th 2017, 03:51 PM   #1
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Equivalent length of helical coil

Hello,

I am trying to calculate the equivalent length of a copper winding with several mitered joint extensions on both ends in order to divide it into equally flowing sections.

Through some searching online I found a site that lists Appendix A-29 from the Crane Flow of Fluids paper. I've seen some other calculations online that use just the number in the K coefficient to calculate the equivalent length of fittings and bends. So my question is should I use the formula KB = (n - 1)(0.25pi fT r/d + .5K) + K to calculate the equivalent length of the coil section since it is a series of 90 degree bends?

My thought is, if you just use the number in the K coefficient, then giving that a value of x in the above formula you can factor out the fT and get:

(L/D)eq = (n - 1) (.25pi r/d + .5x) + x

My math so far for this is as follows, any help on everything I am wrong on would be appreciated:

Copper size = .5" X .75" X .100" wall tube = .3 X .55 inside dimensions
Equivalent Diameter = 4A / P = 4*.3*.55/(.3+.3+.55+.55) = .3882

(L/D)eq for mitered 90 = 60

Leq = 60 * .3882 = 23.2941

winding r/d = 3.625/.3882 = 9.337
(L/D)eq for 10 = 30

(L/D)eq = (60 - 1) (.25pi * 3.625/.3882 + .5 * 30) + 30
(L/D)eq = 1347.6681
Leq = 2645.6725 * .3882 = 523.2123

Segments:
2", mitered 90, 4", mitered 90, 2", mitered 90, 3", mitered 90, 26", mitered 90, 2", mitered 90, 2", 60 turn coil, 2", mitered 90, 2", mitered 90, 26", mitered 90, 3", mitered 90, 2" mitered 90, 4" mitered 90, 2"

12 mitered Joints = 23.292 * 12 = 279.5294
82" of straight length
523.2123 Winding Equivalent Length

Total Equivalent Length = 885"
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