Diameter D = 0.3 m
Head loss hf/L = 0.008
Relative roughness ks/D = 0.0006
A* = D1.5 (g hf/L )0.5 = (0.3)1.5 (9.81 × 0.008)0.5 = 0.046
ν = 1.14 × 10-6
Q = -2.22 A* D log [ (ks/D)/(3.7) + (1.78 ν/A*)]
Q = -2.22 × 0.046 × 0.3 × log [ (0.0006/3.7) + (1.78 × 1.14/(106 × 0.046 ) ) ]
Q = - 2.22 × 0.046 × 0.3 × (-3.68)
Q = 0.113 m3/s = 113 L/s ANSWER.
Three energy balance equations and continuity equation:
zA = zD + pD/γ + fAD (LAD /DAD ) VAD2 / (2g)
zB = zD + pD/γ + fBD (LBD /DBD ) VBD2 / (2g)
zD + pD/γ = fDC (LDC /DDC ) VDC2 / (2g)
VAD AAD + VBD ABD = VDC ADC
AAD = 0.7845 ft2
ABD = 0.1963 ft2
ADC = 0.7854 ft2
State continuity equation as follows: VAD AAD + VBD ABD - VDC ADC = X
The solution is by trial and error.
Assume pD/γ and calculate resulting velocities using energy equations; then plug velocities and areas
in continuity equation until X = 0.
First assume VAD = 0.
Then pD/γ = 60.0 ft.
Start by assuming a somewhat lessr value:
pD/γ = 50 ft.
Eventually, choose pD/γ = 39 ft.
pD/γ = 39 ft: X = -0.27
pD/γ = 38.9 ft: X = 0.006
The final velocities are:
VAD = 3.008 fps.
VBD = 4.555 fps.
VDC = 4.139 fps.
QAD = 0.7854 × 3.008 = 2.362 cfs.
QBD = 0.1963 × 4.555 = 0.894 cfs.
QDC = 0.7854 × 4.139 = 3.250 cfs.
pD = 38.9 ft × 62.4 lb/ft 3 = 2,371 lb/ft2