Pipeflow Example Type
1 Application (continued)
Example Next
evaluate the head loss, HL, both the major and the minor head
loss. HL = (
f L/D) (V2/2g) + KL (V2/2g) = [
f L/D + KL ] (V2/2g)
where f =
0.03, L = 60 ft, D = 0.1 ft, KL = 12 Note:
Since the friction factor, f, is given, there is no need for iteration. Now V
= Q/A = Q
/ ( π D2/4 ) = 4 Q / (π D2) =
400 Q / π and HL = [
(0.3)(60)(1/10) + 12 ] [400 Q / π ]2 [ 1/64.4 ] Recall
from conservation of energy: HP =
HL So 1.7628/Q
= [ (0.3)(60)(1/10) + 12 ] [400
Q / π ]2 or
Q3 = (
1.7628) / ( 7551.889 ) Q
= 0.0615 ft3 /
sec (result) Click
here to return to section on Pipe Flow Types. |
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