I\'m told a pump produces an increment pressure of 500 inches of water, gauge. T

Question

I'm told a pump produces an increment pressure of 500 inches of water, gauge. The pump is a horizontal pipe and using the Mechanical Energy Equation for steady state flow, I know the kinetic and potential energy terms are zero, and friction is negligible. Leaving me with (change in Pressure)/(density)=(work of pump) I'm having trouble using unit conversions to convert (500 inches of water) / ( 1 g/cm^3) into a unit of work. Please help and explain unit conversions. I'm told a pump produces an increment pressure of 500 inches of water, gauge. The pump is a horizontal pipe and using the Mechanical Energy Equation for steady state flow, I know the kinetic and potential energy terms are zero, and friction is negligible. Leaving me with (change in Pressure)/(density)=(work of pump) I'm having trouble using unit conversions to convert (500 inches of water) / ( 1 g/cm^3) into a unit of work. Please help and explain unit conversions. (change in Pressure)/(density)=(work of pump) I'm having trouble using unit conversions to convert (500 inches of water) / ( 1 g/cm^3) into a unit of work. Please help and explain unit conversions.

Explanation / Answer

Specific work = (Head)*g

Unit of specific work = Joule/Kg

Head = 500 inches of water = 12.7 m of water

Specific work = 12.7*9.81 =124.6 Joule/Kg

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