Could you please clarify Power? The 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(1.73) = 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 nameplate might look something like this: ___________________ || | Make: xxxxxx | | Model: xxxxxx | | Volt: xxx | | Amp: xxx | | %Eff: xx | || |__________________| The output power, which really isn't 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 manufacturer's service dept's for assistance. It is now big deal to them, they will be happy to answer your questions. If 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, flowrate, specific gravity of the fluid and the pump efficency 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 metres / hour, head is in metre, Specific Gravity (SG) being a dimensionless unit - cold clean water SG = 1 and efficency as a % from the curve or other means. If you are trying to evalaute wire to water efficency then you can set efficency in the equation above to 1 and then ratio the input (measured power) to output power (calculated water horse power) to obtain the overall efficency. Read more: http://wiki.answers.com/Q/How_do_you_calculate_the_power_of_a_centrifugal_pump#ixzz1Y1VaQvjY
The general formula for the power input to a liquid is expressed by the formula: P = ρ.Q.g.H/η When Q is in m3/h and H in m, g = 9.81 m/s2, the liquid power in kW, is P = ρ.Q.H/(367.η) where ρ is the density in kg/L (=1 for water). 367 = 3600 (s/h) ÷ 9.81 (m/s2). Thus your formula is OK. Electric Motor Absorbed Power Estimation Very often one needs to know the power absorbed by say, a pump or a carrier or fan or some other equipment driven by an electric motor. If one has a three-phase electric power meter and one measures the three phase currents and voltages then the power can be calculated directly. Most often however all one has is the current reading from the MCC panel ammeter, the method outlined below can be used to estimate the power absorbed by the driven machine. P = η·√3·V·I·cosφ where P is absorbed power in watts η is motor efficiency V is applied voltage
I is absorbed current in amps cosφ is the power factor The applied voltage is usually known or can easily be measured, the current we have, our problems are the efficiency and the power factor. Electric motor catalogs usually state the efficiency and power factor at full load and at various part loads(for example at no load, 25%, 50%, and 75%).
Curves can be fitted to these points (as shown), but our problem is that we don't know the absorbed power and in fact that is precisely what we want to calculate: an iterative solution is required; ie.
guess the absorbed power
calculate efficiency and power factor from fitted curves
calculate power from
P = η·√3·V·I·cosφ
use this as new estimate of power to recalculate efficiency and power factor
repeat until solutions converge
