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Hi currently have a 3phase 50HP motor with an 80% Power factor

I need to bring it up 95% power factor.

What capacitor size would i need to achieve this.

If someone can show the formulas they used that would be great!

Thanks

This is a link to the motor https://inventory.powerzone.com/item/55179/used-50-hp-vertical-electric-motor-reliance

575 volts, 48amps

If we assumed a power factor of 80%. how would i go about correcting it to 95% power factor?

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  • \$\begingroup\$ You need to know the current and voltage. There are a lot of online calculators, downloadable instructions explanations etc. \$\endgroup\$
    – user80875
    Jul 3, 2020 at 0:13
  • \$\begingroup\$ @CharlesCowie Can you link to one? \$\endgroup\$
    – Drew
    Jul 3, 2020 at 0:15
  • \$\begingroup\$ This Eaton guide looks useful, and here are a zillion possible links to calculators \$\endgroup\$
    – Russell McMahon
    Jul 3, 2020 at 3:08
  • \$\begingroup\$ The very first result (of the zillion) that @RussellMcMahon provides gives the same answer i get below (129uF). Here is that specific link. \$\endgroup\$ Jul 3, 2020 at 3:33

1 Answer 1

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First, use the motor nameplate data to find kva (apparent power). Subscript 3 below indicates 3 phase.

$$ S_3 = \frac{HP * 746}{power factor * efficiency}$$

But, in your case the nameplate gives rated current as 48A, so

$$ S_3 = 48A * 575 * \sqrt3 = 47.8kva$$

$$ P_3 = S_3 * power factor = 47.8 * 0.80 = 38.24kW $$

So,

$$ Q_3 = \sqrt{S_3^2-P_3^2} = 28.7kvar $$

Now, to get to your desired 0.95 power factor you need to provide some of that kvar from your capacitor.

$$ {desired power factor angle} = cos^-{^1}(0.95) = 18.2⁰$$

$$18.2⁰ = tan^-{^1}(Q_3/P_3)$$

$$18.2⁰= tan^-{^1}(Q_{target}/38.24kW) $$ So, $$Q_{target} = 12.6kvar $$

That means you need a capacitor that will supply an additional (28.7-12.6) = 16.1kvar 3-phase to the motor so that only 12.6kvar comes from the source.

Since, $$X_C = \frac{V^2}{Q}$$ $$X_C = \frac{575^2}{16.1kvar} = 20.5Ω $$

From this, we can find C (assuming 60Hz) as $$ C = \frac{1}{2*π*f*X_C} = 129uF $$

EDIT: I noticed that bxjockey had given us nameplate amps so we can directly calculate apparent power. That changed the subsequent results which i have now corrected.

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  • \$\begingroup\$ By the way, that value of C would be for wye connection. If you connect the caps in delta you need to use 1/3 that value (43uF) to produce same kvar. \$\endgroup\$ Jul 3, 2020 at 13:05

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