Supplementary Material, Appendix 0 – A0.47

APPENDIX 0 – Pre-requisite: Steady State Hydraulics

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Surge Analysis and the Wave Plan Method

Supplementary Material: Example Problems and Solutions

Appendix 0 – Problem 47

A0.47 Consider the pipeline described in the previous question. Suppose there is a 100 kW useful power pump located close to reservoir A lifting water to reservoir B, compute the resulting steady state flowrate in the pipeline.

Solution:

Energy equation describing steady state flow for this pipeline is:

Energy (per unit weight) at source A + Energy (per unit weight) added by pump which is same as pump energy head – Frictional headloss in the pipeline = Energy (per unit weight) at source B

E_A + E_P – ΔH = E_B, where E_P is energy per unit weight added by the pump.

Useful pump power P_U = γ Q E_P, where γ is specific weight of water and Q is flowrate in m³/s. For P_U of 100 kW or 100000 W, and γ = 9810 N/m³, the pump energy head E_P = 10.194/Q.

50 + 10.194/Q – (10.67 L Q^1.852)/(C^1.852 D^4.87) = 100

Solve for Q iteratively: Q = 0.167 m³/s

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