How to calculate reactive power for 3 phase induction motor

[ws0619]

Hi all,
I would like to install Capacitor Bank into my control panel board which contains of few 10HP induction motors. Currently I'm just able to measure the Ampere for every induction motor by using current meter.
How to calculate the reactive power for 10HP 3phase 4pole induction motor by just using Amp value?

One of my 10HP motor:
Voltage: 415V
Amp: 7.8A (measure)
I am using 10HP Teco motor with 4pole and 1450rpm.
Please enlighten me.
Thanks!

 

[russ_watters]

It can't be calculated from that information. Can you read the motor nameplate? What is the model number? Full load amps?

 

[Windadct]

This will really depend on how the motors are loaded as well - a 10HP motor that is mechanically delivering 10HP has much better PF than the same motor with only 7HP pf load. Really the only way to do this IMO is with measurements - and if the load on the motor varies, you really only want to compensate for the "best case" - not the worst.
Starting motors on a system with excessive online capacitance provided too much inrush and can over time damage the motor.
You can add the cap to each motors line - so they are switched on an off with the motor - but this also causes large inrush currents - not something I would just throw together. In larger systems they actually us a controller that measures the current and PF and then switches in the appropriate capacitance...
Lastly - if you are getting a KVAR charge on your electric bill the utility may offer some engineering - they would prefer to sell you energy - not VARs because they can charge more for them.

 

[russ_watters]

This will really depend on how the motors are loaded as well - a 10HP motor that is mechanically delivering 10HP has much better PF than the same motor with only 7HP pf load.

I'm a bit unclear on this point. My thought was that while you are certainly correct that the PF depends on the load, that's just because the active power depends on the load. That would leave the reactive power constant and make the the PF (the ratio of the active to total) vary. The OP actually just asked about the reactive power. I am uncertain of this, but I do know that motors are generally provided constant size/output capacitors for PF correction. Still:

Really the only way to do this IMO is with measurements - and if the load on the motor varies, you really only want to compensate for the "best case" - not the worst.

Agreed. Another reason besides what you said is the utilities generally don't want leading power factor, so the capacitors should be sized to not quite compensate for all of the reactive power.

Lastly - if you are getting a KVAR charge on your electric bill the utility may offer some engineering - they would prefer to sell you energy - not VARs because they can charge more for them.

Indeed the bill itself probably provides some insight. But depending on how big of a facility it is, it may not be enough information to size the capacitors.

Either way, yes: a true power meter attached to the motor is preferred here. Second choice would be to find the manufacturer's rating.

 

[alphysicist]

To calculate the reactive power for a 3-phase induction motor, you will need to use the formula P = √3 x V x I x sin(θ), where P is the reactive power, V is the voltage, I is the current, and θ is the angle between the voltage and current. In this case, since you have measured the current (7.8A) and know the voltage (415V), you can plug those values into the formula to calculate the reactive power for your 10HP motor. However, it is important to note that this formula assumes a unity power factor, meaning that the angle between the voltage and current is 0 degrees. If your motor has a different power factor, you will need to factor that in to your calculation. Additionally, the specific characteristics of your motor (such as the type of winding and the type of load it is driving) can also affect the reactive power. It may be helpful to consult with a professional or refer to the motor's specifications to get a more accurate calculation.