Title HPGE 23 Solutions.pdf ✅

23 Hydraulics: Three Reservoir Problem Solutions SITUATION 1. For the three-reservoir problem shown, QA = 0.42 m3/s ▣ 1. Determine the head at the junction. [SOLUTION] Energy equation from A to J: zA = hJ + hLA 80 m = hJ + 8(0.02)(1800 m) (0.42 m3 s ) 2 (0.4 m) 5 hJ = 28.78 m ▣ 2. Determine the flow in pipe B. [SOLUTION] Since the elevation of B is greater than hJ , the flow of water is from B to J. Energy equation from B to J: zB = hJ + hLB 50 m = 28.78 m + 8(0.025)(2000 m)Q 2 (0.5 m) 5 Q = 0.4 m3 s ▣ 3. Determine the flow in pipe C. [SOLUTION]
Flow in pipe C is the sum of the flows in pipes A and B: QC = QA + QB QC = 0.42 m3 s + 0.4 m3 s QC = 0.82 m3 s ▣ 4. Determine the elevation of reservoir C. [SOLUTION] Energy equation from J to C: hJ = zC + hLC 28.78 m = zC + 8(0.03)(4000 m) (0.82 m3 s ) (0.8 m) 5 zC = 8.44 m SITUATION 2. For the three-reservoir system shown, reservoir A supplies water to reservoir B and C at 10 L/s and 15 L/s. If the pressure at the junction is 110 kPa, determine: ▣ 5. The diameter of pipe A. [SOLUTION] hJ = 40 m + 110 kPa 9.81 kN/m3 = 51.213 m hA = hJ + hLA 80 m = 51.213 m + 10.67(1800 m) (0.025 m3 s ) 1.85 (1101.85)D4.87 D = 157 mm
▣ 6. The diameter of pipe B. [SOLUTION] hJ = hB + hLB 51.213 m = 50 m + 10.67(2000 m) (0.01 m3 s ) 1.85 (1001.85)D4.87 D = 225 mm ▣ 7. The diameter of pipe C. [SOLUTION] hJ = hC + hLC 51.213 m = 30 m + 10.67(4000 m) (0.015 m3 s ) 1.85 (901.85)D4.87 D = 175 mm SITUATION 3. Determine the rate of flow, in m3/s, into or out of each reservoir in the pipe system shown in the table. For the pipes, the roughness coefficient n = 0.011. Pipeline Length (m) Diameter Elevation (m) 1 1200 300 30 2 900 200 24 3 1500 150 15 ▣ 8. Discharge in m3 s of pipe 1. [SOLUTION] Formula for head loss hL = 10.29n 2LQ 2 D 16 3 K = 10.29n 2L D 16 3 K1 = 10.29(0.011) 2 (1200 m) (0.3 m) 16 3 = 918.48
K2 = 10.29(0.011) 2 (900 m) (0.2 m) 16 3 = 5988.02 K3 = 10.29(0.011) 2 (1500 m) (0.2 m) 16 3 = 46288.38 Assuming that Q1 = Q2 + Q3 Energy equation from 1 to 2: 30 m = 24 m + hL1 + hL2 hL1 + hL2 = 6 m Energy equation from 1 to 3: 30 m = 15 m + hL1 + hL3 hL1 + hL3 = 15 6 − hL1 = hL2 = K2Q2 2 Q2 = √ 6 − hL1 K2 Similarly, Q3 = √ 15 − hL1 K3 Q1 = √ hL1 K1 Q1 = Q2 + Q3 √ hL1 K1 = √ 6 − hL1 K2 + √ 15 − hL1 K3 hL1 = 1.744 m Q1 = √ 1.744 K1 Q1 = 0.0436 m3 s ▣ 9. Discharge in m3/s of pipe 2. [SOLUTION]