Civil Engineering Question:

Answer #1The input power, that is, the power required to operate the pump should be stated in Hp (horsepower) on the pump's nameplate. It can also be calculated by the 3-phase power equation:

P(in Hp) = VI(1.7c) = Rated Voltage x Rated Current x 1.73/ %Efficiency

If this is a consumer grade pump that operates on 120Vac, then the equation becomes P = VI, simply multiply the operating voltage, 120 x current (which is the number followed by the letter "A".

The output power, which really is not technically power, but rated in Gpm (gallons per minute), or caPAcity should also be on the nameplate. If you have the make, model, and (not necessarily needed) the serial number (also on the nameplate) you could call the manufacturer's customer service dept. As an application engineer, I have contacted countless manufacturers’, and service dept's for assistance. It is now big deal to them, they will be happy to answer your questions.

Answer #2If it is the absorbed power of the pump to allow you to select a suitable motor then this is a function of the generated head, flow rate, specific gravity of the fluid and the pump efficiency from the OEM / pump performance curve.

For those of us not in the US or using English units then the power can also be specified in kW.

In simple terms Power (kW) = Flow x Head x SG / (367 x Effy)

Where Flow is in cubic meters / hour, head is in meter, Specific Gravity (SG) being a dimensionless unit - cold clean water SG = 1 and efficiency as a % from the curve or other means.

If you are trying to evaluate wire to water efficiency then you can set efficiency in the equation above to 1 and then ratio the input (measured power) to output power (calculated water horse power) to obtain the overall efficiency.