Chapter 0 – Steady-State Hydraulics
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Supplementary Material: Example Problems and Solutions
Chapter 0 – Problem 0.27
0.27 Consider the pipeline shown in the following figure. Water needs to be lifted from reservoir A to reservoir B using a pump placed near reservoir A.
The water surface elevation (which is the same as the hydraulic grade) in reservoir A is 10 m and the water surface elevation in reservoir B is 50 m. The length of pipeline between reservoirs A and B is 2400 m. The pipeline is made of HDPE (high-density polyethylene) material with a nominal diameter of 280 mm and an internal diameter of 243 mm. The manufacturer-suggested Hazen-William roughness coefficient is 140.
What pump energy head is required to achieve a flowrate of 80 lps, neglecting all minor losses? What is the useful power of this pump? What is the pump brake power at an efficiency of 75%?
The energy equation for this operation is:
H A + H P – ΔH = H B , where H A is the hydraulic grade (which is the same as the energy grade) at reservoir A, H P is the energy head added by the pump, ΔH is the frictional headloss in the pipeline between reservoir A and reservoir B, and H B is the hydraulic grade at reservoir B.
Rearranging the terms in the above equation:
H P = (H B – H A ) + ΔH
The required pump head should overcome the static lift between the two reservoirs (i.e., H B – H A ) and the frictional headloss (ΔH) in the pipeline associated with the desired flowrate and other pipeline characteristics.
Using the Hazen-William equation, ΔH = 24.81 m
The static lift H B – H A = 40 m
The required pump head H P = 64.81 m
The pump useful power P U = γ Q H P , where γ is the specific weight of water and Q is the flowrate, both in standard units:
γ = 9810 N/m 3 , Q = 0.08 m 3 /s, and H P = 64.81 m
The useful power P U = γ Q H P = 50.9 kW
The brake power P B = P U /η, where η is the pump efficiency expressed as a decimal value. The specified pump efficiency is 75%, therefore η = 0.75.
Pump brake power P B = 50.9/0.75 = 67.9 kW
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