I have an AC motor. It is running for 1 hours and I read the current every one minute and it reads 2A.

So how to know how much the energy my motor already consumed? In this case, voltage is steady at 220V and not knowing the power factor(or make an assumptions that it is constant).

  • \$\begingroup\$ Knowing that PF is constant doesn't actually deliver a value for what PF is. \$\endgroup\$ – Andy aka Jul 7 '17 at 13:34
  • \$\begingroup\$ Ah, too bad. So I do need to use a wattmeter to do the job? \$\endgroup\$ – Naufal B Jul 7 '17 at 13:42

For single-phase AC, power is voltage X current X power factor. Power factor is a number between zero and one that indicates how closely the voltage and current waveforms coincide with each other. The power factor for full load may be marked on the motor. A typical motor will have a high power factor, perhaps 0.8 to 0.9 at full load, but a low power factor, perhaps less than 0.2 when it is lightly loaded. If the full-load current is marked on the motor, you can compare the measured current with the full-load current to estimate hoo heavily the motor is loaded. With no load, the motor current will likely be in the area of 1/3 of full-load current.

  • \$\begingroup\$ My motor's full load current has 2.5Amp marked on the motor. So it is running 80% load I think, and how to get the power factor from it? If I assume it has PF of 1 will it has 0.8 power factor? \$\endgroup\$ – Naufal B Jul 7 '17 at 13:48
  • \$\begingroup\$ Sorry I misread your question and explained what you already know about power factor. Is the motor power or full-load power factor marked on the motor? \$\endgroup\$ – Charles Cowie Jul 7 '17 at 13:55
  • \$\begingroup\$ Yes it has motor power marked on it, but I forgot the value I think it is around 400 to 500 W. \$\endgroup\$ – Naufal B Jul 7 '17 at 14:01
  • \$\begingroup\$ The power marked on the motor is the mechanical output power at rated load. Mech. Power = V X A X efficiency X pf. Eff X Pf = Mech. Power / (V X A). You could estimate power factor as the square root of (Eff X Pf). Estimate percent operating load assuming no-load current is 2.5/3. Estimate pf drops from full-load pf to 0.2 as the load drops from 100% to zero. \$\endgroup\$ – Charles Cowie Jul 7 '17 at 14:12
  • \$\begingroup\$ Thanks, I will calculate the pf value from the formula you provide. \$\endgroup\$ – Naufal B Jul 7 '17 at 14:23

I am assuming that there is a steady load on the motor since the current is at 2A every time you check it. You mentioned the voltage is at 220V, so this gives you an operating power of 440W. (Power = Voltage x Current)

After the motor has been run for 1 hour, it consumes 440Wh of energy. (Energy = Power x Time)

  • 2
    \$\begingroup\$ Are you using the newly discovered metric hour that has 1000 seconds instead of the more conventional 3600? You do need to account for PF as well. \$\endgroup\$ – Andy aka Jul 7 '17 at 13:34
  • \$\begingroup\$ You've calculated VA and VAh. You still need to multiply by the power factor to calculate W and Wh. \$\endgroup\$ – Transistor Jul 7 '17 at 13:47
  • \$\begingroup\$ You guys are right, no PF was provided so I was assuming ideal case where PF = 1. \$\endgroup\$ – circuitpatrol Jul 7 '17 at 13:49

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