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 whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8 | reply to djr777
Re: Pool questions....pump, power usage, how to find a leak? 9.3 amps x 240v x power factor= ???? watts
The power factor of the pump is likely 0.8 or less. This means that it is likely using less energy, and not costing nearly that much, although its still not a bargain.
55.80 x 0.8 pf = $44.64 a month | |
4 edits | Edit: Removed post | |
whizkid3Premium,MVM
join:2002-02-21
Queens, NY
kudos:8
4 edits | said by x_:Lower power factors results in more power required not less. Power factor has nothing to do with efficiency. Power factor (excluding non-linear components) is lower where more reactive power is required. A motor has coils and and power must be stored in its magnetic field. This power flows on the wires, shows up in a reading of amps, but it does no work and is not used. It simply flows from the grid into the magnetic field and back every cycle.
quote:
If the pump's electrical usage is 9.3A x 240v that's 2,232W
The formula 9.3A x 240v is 2,232 Volt-Amps (VA) is missing its power factor term. A volt-amp is a measure of Apparent Power, and inlcudes reactive (stored) power and real power, which is the power used by the pump motor. Real power can only be calculated by knowing the power factor, (pf). So,
9.3 amps x 240v x power factor= X watts
Lets say the power factor is 0.8; a fair estimate. The pump motor then uses only: 9.3 amps x 240v x 0.8 = 1785.6 watts
There is innefficiency, but its not in the reactive power necessary to run the motor, but rather in losses such as friction, heat losses, magnetization losses, and more. The efficency, 'N', is nessasary to calculate the actual power delivered to the mechanical load - the pump shaft.
If the efficiency were 75%, for example, then: 1785.6 watts x 75% = 1339.2 Watts,
or 1339.2 / 746 hp = 1.795 horsepower delivered to the pump shaft.
Note that the reactive power does indeed flow in the wires, the distribution system, the breaker, the motor, etc. All have to be sized to handle it. It shows up in any reading or specification of amps; like the one djr777 got from the phone call to Hayward pump tech support. But the reactive power does not do any work; does not supply power to either the pump shaft; and in a residence you do not pay money to the electic utility for this power. | | |
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1 edit | You're correct.
I got this backwards when I was reading it:
quote:
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load in the conventional fashion, and is subject to the usual distributed losses in the production and transmission processes.
»en.wikipedia.org/wiki/Power_factor
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