Chapter 0 – Steady-State Hydraulics
Surge Analysis and the Wave Plan Method
Supplementary Material: Example Problems and Solutions
Chapter 0 – Problem 0.17
0.17 Consider the pipeline described in the previous questions 0.15 and 0.16. Suppose there is a pump with 100 kW of useful power located close to reservoir A, lifting water to reservoir B; compute the resulting steady-state flowrate.
Solution:
The energy equation describing steady-state flow for this pipeline is:
Energy (per unit weight) at reservoir A + Energy (per unit weight) added by pump which is same as the pump energy head – the frictional headloss in the pipeline = Energy (per unit weight) at reservoir B. This can be written as: EA + EP – ΔH = EB, where EP is the energy per unit weight added by the pump.
The useful pump power PU = γ Q EP, where γ is the specific weight of water and Q is the flowrate in m3/s. For a PU of 100 kW or 100000 W, and γ = 9810 N/m3, the pump energy head EP = 10.194/Q.
50 + 10.194/Q – (10.67 L Q1.852)/(C1.852 D4.87) = 100
Solve for Q iteratively: Q = 0.167 m3/s
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