Supplementary Material, Appendix 0 – A0.57

APPENDIX 0 – Pre-requisite: Steady State Hydraulics

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Supplementary Material: Example Problems and Solutions

Appendix 0 – Problem 57

A0.57 Consider the pipeline shown in the following. Water needs to be lifted from reservoir A to reservoir B using a pump placed near reservoir A.

The water surface elevation (same as hydraulic grade) in reservoir A is 10m and the water surface elevation in reservoir B is 50m. The length of pipeline between reservoirs A and B is 2400m. The pipeline is made of HDPE (high-density polyethylene) material with a nominal diameter of 280 mm and an internal diameter of 243 mm. Manufacturer suggested Hazen-William roughness coefficient is 140.

If it is desired to lift water at a flowrate of 80 lps, what is the pump energy head needed to accomplish this, neglecting all minor losses? What is the associated pump useful power? What is the pump brake power at an efficiency of 75%?

Solution:

The energy equation for this operation is:

HA + HP – ΔH = HB, where HA is the hydraulic grade (same as energy grade) at reservoir A, HP is the energy head to be added by the pump, ΔH is the frictional headloss in the pipeline from reservoir A and reservoir B, and HB is the hydraulic grade at reservoir B.

Rearranging the terms in the above equation:

HP = (HB – HA) + ΔH

The required pump head should overcome the static lift between the two reservoirs (i.e., HB – HA) and the frictional headloss (ΔH) in the pipeline associated with the desired flowrate and other pipeline characteristics.

Using Hazen-William equation, ΔH = 24.81m

Static lift HB – HA = 40m

The required pump head HP = 64.81m

Pump useful power PU = γ Q HP, where γ is specific weight of water and Q is flowrate, both in standard units.

γ = 9810 N/m3, Q = 0.08 m3/s, HP = 64.81m

Useful power PU = γ Q HP  = 50.9 kW

Brake power PB = PU/η, where pump efficiency η is expressed in decimal value. The specified pump efficiency is 75%, therefore η = 0.75.

Pump brake power PB = 50.9/0.75 = 67.9 kW


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