Chapter 0 – Steady-State Hydraulics
Surge Analysis and the Wave Plan Method
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%?
Solution:
The energy equation for this operation is:
HA + HP – ΔH = HB, where HA is the hydraulic grade (which is the same as the energy grade) at reservoir A, HP is the energy head added by the pump, ΔH is the frictional headloss in the pipeline between 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 the Hazen-William equation, ΔH = 24.81 m
The static lift HB – HA = 40 m
The required pump head HP = 64.81 m
The pump useful power PU = γ Q HP, where γ is the specific weight of water and Q is the flowrate, both in standard units:
γ = 9810 N/m3, Q = 0.08 m3/s, and HP = 64.81 m
The useful power PU = γ Q HP = 50.9 kW
The brake power PB = PU/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 PB = 50.9/0.75 = 67.9 kW
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