Consider a desired pump operating condition which adds 35 psi at a flow rate of 500 gpa to water. Ignore any changes in Kinetic or potential energy and assume isothermal flow (ie. Internal energy is constant, this is normal assumption we make when analyzing fluid flows in piping systems). Apply the energy balance only across the pump starting with full energy balance and simplifying to solve for the pump head symbolically. Then begin careful of unit, solve for pump head in feet. Given erosion 500 gpm P=62 lbs/ft^3 A- 10.5 ft B-100.8 ft C- 50.1 ft D-80.7 ft

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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Consider a desired pump operating condition which adds 35 psi at a flow rate of 500 gpa to water.
Ignore any changes in Kinetic or potential energy and assume isothermal flow (ie. Internal energy is
constant, this is normal assumption we make when analyzing fluid flows in piping systems). Apply the
energy balance only across the pump starting with full energy balance and simplifying to solve for the
pump head symbolically. Then begin careful of unit, solve for pump head in feet.
Given erosion
500 gpm
P=62 lbs/ft^3
A- 10.5 ft
B- 100.8 ft
C- 50.1 ft
D-80.7 ft
Transcribed Image Text:Consider a desired pump operating condition which adds 35 psi at a flow rate of 500 gpa to water. Ignore any changes in Kinetic or potential energy and assume isothermal flow (ie. Internal energy is constant, this is normal assumption we make when analyzing fluid flows in piping systems). Apply the energy balance only across the pump starting with full energy balance and simplifying to solve for the pump head symbolically. Then begin careful of unit, solve for pump head in feet. Given erosion 500 gpm P=62 lbs/ft^3 A- 10.5 ft B- 100.8 ft C- 50.1 ft D-80.7 ft
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