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For an 3 phase AC motor, the full load current of a 3 phase motor is: $$I= \frac{P(KW)}{V*\sqrt3 *Pf}$$

when I was studying power factor correction, I came across some concepts like motor motor loaded at 25%, 50%, 75%, 100%.

What means a motor loaded at 25%, 50%,... and how the calculations for example of the current change in this case.

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  • \$\begingroup\$ Your equation should also include a term for efficiency. \$\endgroup\$
    – mkeith
    Feb 21, 2021 at 8:45
  • \$\begingroup\$ It's percetage of the motors rated (nominal) power. The percentages are valid for current also (or very close). \$\endgroup\$
    – user208862
    Feb 21, 2021 at 9:08
  • \$\begingroup\$ Ask yourself what load the motor is driving and how much torque that load takes, compared to the motor's full load torque rating. Since it's an induction motor, you can focus on torque and current rather than power, with reasonable accuracy. \$\endgroup\$ Feb 21, 2021 at 13:29

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The equation should be $$I= \frac{P}{V*\sqrt3 *Pf *Eff}$$

Where P is the output power in Watts, V is the line-to-line voltage, Pf is the power factor and Eff is the efficiency.

What is meant by 50% load is that the motor output power is 50 percent of full load output power.

Both power factor and efficiency will change a bit when you go from full load to 75 percent to 50 percent to 25 percent. So the equation is still valid, but you need to use the Eff and Pf specified for the reduced load condition. I don't know if there is a simple way to calculate the new Eff and Pf.

To a first approximation, current and power will scale together linearly. So at half power, the motor current will be half of the full load current. But if you want a more exact calculation you will need to use Eff(50) and Pf(50).

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  • \$\begingroup\$ Thanks @mkeith for your answer. \$\endgroup\$
    – user137684
    Feb 21, 2021 at 9:23

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