Title HPGE 24 Solutions.pdf ✅

QC = 0.118 m3 s Q = AV 0.118 m3 s = π 4 (0.5 m) 2Vc Vc = 0.601 m s ▣ 6. Determine the headloss in pipe E. [SOLUTION] Compute for the flow in pipe E QC = QD + QE QD = 0.118 m3 s − QE Since pipes D and E are parallel, then they should have the same headloss hLD = hLE ( 8fLQ 2 π 2gD5 ) D = ( 8fLQ 2 π 2gD5 ) E (600 m) (0.118 m3 s − QE) 2 (0.3 m) 5 = (800 m)QE 2 (0.3 m) 5 QE = 0.055 m3 s Compute for the head loss: hLE = 8(0.02)(800 m) (0.055 m3 s ) 2 π 2g(0.3 m) 5 hLE = 1.63 m ▣ 7. Determine the difference in elevation of the two reservoirs. [SOLUTION] Energy equation from left to right: z1 = z2 +∑hL ∑hL = (0.02)(300 m) (1 m s ) 2 (0.2 m) (2 × 9.81 m s 2 ) + (0.02)(300 m) (0.601 m s ) 2 (0.5 m) (2 × 9.81 m s 2 ) + 1.63 m = 3.38 m H = 3.38 m
SITUATION 3. Consider a pipe that branches out into three parallel pipes then rejoins at a junction downstream. The pressure drop between the junctions is 300 kPa. The properties of the parallel pipes are listed below. Pipe Friction factor Length (m) Diameter (mm) A 0.02 720 144 B 0.015 1080 240 C 0.018 1440 480 ▣ 8. Determine the flow in pipe A. [SOLUTION] Since the pipes are parallel, the headloss in each pipe is the same as the pressure head drop. hL = ΔP γ 300 kPa 9.81 kN m3 = 8(0.02)(720 m)Q 2 π 2g(0.144 m) 5 Q = 0.0399 m3 s ▣ 9. Determine the flow in pipe B. [SOLUTION] 300 kPa 9.81 kN m3 = 8(0.015)(1080 m)Q 2 π 2g(0.24 m) 5 Q = 0.135 m3 s ▣ 10. Determine the flow in pipe C. [SOLUTION] 300 kPa 9.81 kN m3 = 8(0.018)(1440 m)Q 2 π 2g(0.48 m) 5 Q = 0.603 m3 s