- With Fig. 5-6: Q= 3.4 cfs.
With ONLINE HAZEN-WILLIAMS, Q = 3.213 cfs.
- The line losses are: hL = (hL/L) L = 0.005 × 1200 = 6 ft.
From Table 5-3: Kb = 0.09
From Table 5-3: Ke = 0.03
From Table 5-3: KE = 0.72
Kt = 0.30
hL = 6 + [V2/(2g)] (Kb + Ke + KE + Kt)
hL = 6 + [32/(2 × 32.17)] (0.09 + 0.03 + 0.72 + 0.30) =
hL = 6 + [32/(2 × 32.17)] (1.14)
hL = 6.16 ft
The head delivered to the turbines is: ht = 3250 - 1575 - 6.16 = 1668.84 ft.
The discharge: Q = V A = V [(π/4) D2] = 3 × (0.7854 × 1 2) = 2.36 cfs.
The power: P (HP) = 62.4 (lb/ft3) 2.36 (ft3/s) 1668.84 (ft) / 550
P = 446.8 HP
P = 446.8 HP × 0.746 = 333 KW
- 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 = 0.3491 ft2
ABD = 0.5454 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.
Assume pD/γ = 90.0 ft: X = - 4.3
Assume pD/γ = 70.0 ft: X = - 2.12
Assume pD/γ = 43.0 ft: X = - 0.053
Assume pD/γ = 42.3 ft: X = - 0.00
Final velocities:
VAD = 5.057 fps.
VBD = 5.118 fps.
VDC = 5.801 fps.
QAD = 0.3491 × 5.057 = 1.765 cfs.
QBD = 0.5454 × 5.118 = 2.791 cfs.
QDC = 0.7854 × 5.801 = 4.556 cfs.
pD = 42.3 ft × 62.4 lb/ft 3 = 2,639.5 lb/ft2
- The head loss through either pipes 1 or 2 is the same.
V1/V2 = [(f2 L2 D1) / (f1 L1 D2)] 1/2
V1/V2 = [(0.03 × 3200 × 1) / (0.02 × 3000 × 1.3333)] 1/2 = 1.0955
(D1/D2)2 = (1/1.3333)2 = 0.5625
Q1 = Q / { { 1 / [(V1/V2) (D1/D2)2]} + 1 }
Q1 = 20 / { [ 1 / (1.0955 × 0.5625)] + 1 } = 7.625 cfs.
Q2 = Q / { [(V1/V2) (D1/D2)2] + 1}
Q2 = 20 / [(1.0955 × 0.5625) + 1 ] = 12.375 cfs.
Q1 + Q2 = 20 cfs = Q OK!
V1 = Q1 / A1 = 7.625 / [(π/4) 12] = 9.708 fps.
V2 = Q2 / A2 = 12.375 / [(π/4) 1.33332] = 8.863 fps.
hL, 1 = f1 (L1/D1) [V12/(2g)]
hL, 1 = 0.02 (3000/1) [9.7082/(2 × 32.17)] = 87.888 ft.
hL, 2 = f2 (L2/D2) [V22/(2g)]
hL, 2 = 0.03 (3200/1.3333) [8.8632/(2 × 32.17)] = 87.907 ft.
hL, 1 = hL, 2 OK!