University of New Orleans

ENCE 4322

Chapter 8
Problem 8.1
Determine the water horsepower, break horsepower and motor horsepower for a pump operating under the following
conditions: water flow of 490 gpm is to be pumped against a total head of 110 ft; the pump efficiency is 75%; and
the motor driving the pump has an efficiency is 85%
Solution:
Water horsepower = Q H/3957
= (490) (110)/3957
= 13.6 hp
Break horsepower = Water horsepower/Epump
= 13.6/0.75
= 18.1 hp
Motor horsepower = break horsepower/Emotor)
= 18.1/0.85
= 21.3 hp
Problem 8.2
A pump is to be located 6 ft above a wet well and must lift 600 gpm of water another 52 ft to a storage reservoir
through a piping system consisting of 1,250 ft of 6-in DIP pipe (C = 110), 2 globe valves (open), and 2 medium
sweep elbows. Determine the total dynamic head (TDH) for this water pumping system:
Solution:
Hs = Hsuction + Hd
TDH = Hs + Hf = Hsuction + Hd + Hf
Where
Hs = static head, ft
Hsuction = suction lift head, ft
Hd = discharge head, ft
TDH = total dynamic head, ft
Hf = friction loss head, ft
Hs = Hsuction + Hd
= 6 ft + 52 ft
= 58 ft
From Appendix 17 – Equivalent length of pipe to headloss in fittings:
Equivalent 6-in pipe length for 2 globe valves (open) = 2 x 150 ft = 300 ft
Equivalent 6-in pipe length for 2 medium sweep elbows = 2 x 14 ft = 28 ft.
Total equivalent length of 6-in pipe = 1250 + 300 + 28 = 1,578 ft
From Nomogram (C= 110) in Appendix 14 at Q = 600 gpm
Hf = 40 ft of water per 1000 ft of 6-in DIP pipe
= (40/1000) (1,578)
= 63.12 ft
TDH = Hs + Hf
= 58 + 63.12 = 121.12 ft