Water Pump Energy Consumption

This was discussed at length in a thread started 15 Sept.

For your 2nd question, have a think about it!

The power rating (as noted on its name plate) of an electric motor is its output or shaft power. The actual amount of energy that it consumes will depend on the applied load. Less load will result in lower power consumption.

Contrary to what may be intuitive, the energy consumed by a motor rated at 1kW that is used for one hour at full load will not be 1kWh but will be somewhat more.

For example, a single phase 240v motor with a plate rating of 1kW and pf of 0.8 will draw roughly 8 amps or 1.5kW at full load, and will therefore consume 1.5kWh if run at full load for 1 hour.

You can generally determine this from the name plate by doing a bit of math.

For a single phase motor - Power (consumed) = Volts x amps x pf

For a 3 phase motor - Power (consumed) = √3 x Volts x amps x pf.

pf is often shown as cosφ on the plate.

But for OP's question - what is mentioned as "power(Kw or Hp)" in electric water pump? Is this input power to the motor regardless of flow rate or head? - we have no way of knowing. It could be the electric power, the shaft power into the pump or even conceivably the hydraulic power. The person who does (or should) know is whoever gave the figure, and OP should ask him. This was covered in the earlier thread.

To calculate electrical power (at rated output), pf doesn't come into it. Divide the rated power by the efficiency, not the pf. As it happens, efficiency 67% is about right for a 1kW motor, so electrical power 1.5kW is a fair estimate. BTW 1kW is not a standard rating, at least not for 3-phase motors. If it is for 1-phase I'll stand corrected.

pf comes in when calculating amps, but to get shaft power from volts and amps, need to multiply by pf (and efficiency) so it cancels.

In case it helps the OP, the actual shaft power into the pump is likely to be a bit less than motor rating (as it will fall between 2 standard ratings and the higher should be selected) and it varies with where the pump operates on its Q/H curve.

"To calculate electrical power (at rated output), pf doesn't come into it. Divide the rated power by the efficiency, not the pf."

Hmmm.. To quote some info straight from page 3 of this motor manufacturer's website " Single-phase electrical power is calculated as follows: Electrical power = Voltage x Current x Power Factor - 3-phase electrical power calculation in a circuit is slightly different: Electrical power = 1,732 • Voltage • Current • Power Factor".

This Allen-Bradley site concurs with the above.

The motor's efficiency is simply a ratio of it's mechanical output power divided by its electrical input power. The efficiency of a motor generally peaks at around 75% full load and then drops off, so using efficiency to calculate power consumption would wrongly suggest that said power consumption would be less at full load than at 3/4 load. It is, however, correct to use efficiency when calculating output power, multiplying the motor input power by the efficiency as a decimal will give a close approximation of output power.

You're correct that 1kW is not a standard motor, the number was used purely as an example to make the numbers less complex.

To quote some info straight from page 3 of this motor manufacturer's website " Single-phase electrical power is calculated as follows: Electrical power = Voltage x Current x Power Factor - 3-phase electrical power calculation in a circuit is slightly different: Electrical power = 1,732 • Voltage • Current • Power Factor".

That's correct, but it doesn't relate to output (shaft) power. It just tells us electrical input kW = KVA*pf.

Allen-Bradley could easily cause confusion IMHO. Sticking to 1-phase, as that covers everything we're discussing, and can we use V for volts as I mix up E for volts with Efficiency?

He says (1st formula) To Find Amperes when HP is known I = 746*HP/V/Eff/pf. He doesn't make it immediately clear, but HP here is shaft power.

He then says To Find Amperes when kW is known I = 1000*kW/V/pf. Eff is not in there, because kW here is electrical input. Confusing or what?

Further down he puts HP formula the other way round, giving output HP = V*I*Eff/746. This confirms HP is output in 1st formula. To be fair, he never uses kW for shaft power, so if kW is always read as electrical kW, it's OK. Fine if you're US based, but for metric types, need to keep a clear head!

The motor's efficiency is simply a ratio of its mechanical output power divided by its electrical input power. I know, that's the point of my posts!

The efficiency of a motor generally peaks at around 75% full load and then drops off, so using efficiency to calculate power consumption would wrongly suggest that said power consumption would be less at full load than at 3/4 load. When calculating electrical power at 3/4 load, you divide the 3/4 load, not the rated output, by efficiency, so input power is still lower, although efficiency is worse.

I thought that's what you'd done with the 1kW, I wasn't suggesting you don't know standard motor ratings .

The actual work done is pressure x volume/sec. If the pump is just building up head, any power consumed is wasted.

To be pedantic, the actual hydraulic power is pressure x volume/sec. The work done is pressure x volume.

Correct, that's what I meant to say.