Increase power consumption with a capacitor

[eastoak]

Homework Statement

A motor attached to a 120 V/60 Hz power line draws an 7.60 A current. Its average energy dissipation is 840 W.

How much series capacitance needs to be added to increase the power factor to 1.0? (in micro F)

Homework Equations

Don't know which to use at all.

The Attempt at a Solution

An equation would be just a helpful thanks.

 

[berkeman]

Define power factor. (that goes into #2).

Are you sure they are asking for a series capacitor?

 

[eastoak]

Yeah. That is exactly how the question is worded. There isn't a diagram so I assume it is in series.

I think the question asks what would a capicitance on a capacitor be if it were to increase the energy dissipation to 1680W.

 

[eastoak]

The answer happens to be 431 micoF. Anyone know how that was solved? z=(Rc^2+Xc^2)^1/2 ? omega=2(3.14)60 ?

 

[antonantal]

I modeled the motor as an inductance with a series resistance. Here is how you can solve it:

Since the reactive component of the motor model (the inductance) doesn't dissipate power it means that those 840W are dissipated by the series resistance of the motor, and you can find this resistance with the formula P = R*I^2.

The total impedance of the motor can be found by Z = U/I

As you said Z=(R^2+X^2)^1/2. You have Z and R so you can find X.

The power factor is defined as the ratio between active power and apparent power, apparent power being the module of the complex power (which has the active power as real part and the reactive power as the imaginary part). So in order to have the power factor 1 you need to have zero reactive power. This means that the reactance of the motor (the X you calculated) must be equal and opposed to the reactance of the series capacitor you have to add.

 

[antonantal]

Here is a good overview of the power factor: http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html

 

[eastoak]

I'll take a look at the link.

thanks for the reply.

 

[rockstar]

not series parallel

you are using the cap to store the active power from inductor effect of the motor windings and supply it back as reactive power. so the capacitor has to be in parallel

 

[berkeman]

you are using the cap to store the active power from inductor effect of the motor windings and supply it back as reactive power. so the capacitor has to be in parallel

Thank you! It just made now sense at all in series.