# motor and power factor



## Rei (Mar 20, 2010)

sample question:

The name-plate of a 1-phase, 4-pole induction motor has the following data:

HP=0.5

Voltage=230 V

Frequency=60 Hz

FL current= 2.9A

power factor=0.71

rpm=1610

Calculate the current taken by the motor and it's power factor if a capacitor of 15.9 uF is connected across the motor while delivering the rated output as above.

The calculation used lagging power factor of the given current 2.9A above which I'm not so sure why. A motor doesn't necessary has lagging power factor, right?


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## yellowjacket03 (Mar 20, 2010)

That is a safe assumption for an induction motor.



Rei said:


> sample question:
> The name-plate of a 1-phase, 4-pole induction motor has the following data:
> 
> HP=0.5
> ...


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## yellowjacket03 (Mar 20, 2010)

My response wasn't very clear. The only possible excitation in an induction motor is the stator input. The induction motor must operate at a lagging power factor.


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## Art (Mar 21, 2010)

what is a motor?

COILS of wire

what is the operating mechanism of an INDUCTion motor?

it's an inductor(s) or wound coils, albiet moving in relation to each other

it can ONLY be LAGGING...

Z = jwL (ignoring resistance)

V = Zi = jwL i

the phase angle ( tan^-1 ) for any positive number (wL) is a positive angle, meaning that V leads i by that angle...


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## CGlade (Mar 24, 2010)

Rei said:


> sample question:
> The name-plate of a 1-phase, 4-pole induction motor has the following data:
> 
> HP=0.5
> ...


Do you have the solution to this problem? Just curious.....


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## DK PE (Mar 24, 2010)

Not sure if you are wondering if there is a solution supplied from the source of the problem or not but I get new PF is 0.95 lagging and line current reduced to 2.15 A.

I worked pretty quickly so wouldn't mind a verification.


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## Flyer_PE (Mar 24, 2010)

^I got 2.16 Amps and a PF of 0.95. I figure we're within rounding error.


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## Art (Mar 24, 2010)

I come up with

pf ~ 0.9518

i ~ 2.1633

I calc'ed the power triangle:

P ~ 0.71 x 230 x 2.9 ~ 473.6 W

S ~ 2.9 x 230 ~ 667 VA

W ~ j sin(arc cos 0.71) x S ~ j469.7 VAR

calc'ed new C reactive power added

~ V^2/Z ~ 230^2 / (1/j(2 Pi 60 15.9 x10^-6)) ~ -j317.1 VAR

new W total ~ j469.7 - j317.1 ~ j152.6 VAR

new pf ~ cos(arctan W/P) ~ cos(arctan 152.6/473.6) ~ 0.95 (lagging, ie, VAR is positive), P stays the same...same real work ~ same real power

i = P/(V pf) = 473.6 / (230 0.95) ~ 2.16 A


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## pelaw (Apr 6, 2010)

Question: For this problem, if an inductor with same kVARs was placed in series, instead of a cap in ||, the apparent power would be reduced just the same?


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## Flyer_PE (Apr 6, 2010)

Assuming an ideal inductor (0 resistance), the pf would decrease which would result in an _increase_ in apparent power.


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## pelaw (Apr 6, 2010)

So what results if the same capacitor, 15.9, was plugged in series with the motor, between the 230V supply and the motor? The current into motor gors down to 0.2A?


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## Flyer_PE (Apr 6, 2010)

^Never gave much thought to a capacitor in series. It's a simple circuit though. I'll try and fiddle with it tonight after my kid is in bed.


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## CLTEE49 (Apr 10, 2010)

Ive seen questions like this relating to the motor efficiency. Just so Im correct - I want to verify that...

Improving the power factor *to the motor* (with a parrallel capacitor) does nothing to the power output of the motor. Thus the power factor of the system improves - power to the motor increases - motor power output remains the same and this reduces the motors efficiency.

Improving the power factor *of the motor* (say its a 50mva motor with pf = .75 vs. a 50mva motor with pf = .90) increases the value of P=Sxpf. This means that you are improving the motors *output power* - and if you have the same input power - would increase efficiency.

Right??


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## pelaw (Apr 10, 2010)

I don't think that is quite right. In pf correction problems real power P is kept contant. Motor efficiency is dictated by the motor itself, the load, and the voltage. Not by the circuit. So, regardless of whether there is a capacitor or not in this problem, the motor needs certain real power P. The capacitor redices the current needed to produce that constant real power P. It does that by reducing the magnetic power deflection, the Q, the kVARs. Therefore it reduces S, which in turn reduces current.


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## DK PE (Apr 10, 2010)

CLTEE49 said:


> Ive seen questions like this relating to the motor efficiency. Just so Im correct - I want to verify that...Improving the power factor *to the motor* (with a parrallel capacitor) does nothing to the power output of the motor. Thus the power factor of the system improves - power to the motor increases - motor power output remains the same and this reduces the motors efficiency.
> 
> Improving the power factor *of the motor* (say its a 50mva motor with pf = .75 vs. a 50mva motor with pf = .90) increases the value of P=Sxpf. This means that you are improving the motors *output power* - and if you have the same input power - would increase efficiency.
> 
> Right??


I think you need to think separately re: power factor and efficiency. In your first statement, I think you are on the right path, but then take a wrong turn. Adding a parallel capacitor to improve power factor does not increase the motor's efficiency... at least to first order. There is a very, very slight second order effect that improving the power factor reduces the current supplied in the branch circuit to motor which results in less voltage drop, which may result in a slight improvement in efficiency but it is a stretch at best. Your last statement of first paragrah



CLTEE49 said:


> ... factor of the system improves - power to the motor increases -


really should be VARS to the motor decrease, not power to the motor increases.

In your second paragraph I'm not sure I follow but try this... if I replace a 50HP motor with another 50 HP motor, the REAL power input could go up or down, depending only on efficiency. The VARs required could also go up or down, depending on power factor.

Also, you seem to be always assuming full load, remember the input power is very dependent on what the output power the load requires. Hope I helped


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## Rei (Apr 11, 2010)

CGlade said:


> Do you have the solution to this problem? Just curious.....


I=2.17&lt;-17.76


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## CLTEE49 (Apr 12, 2010)

> I think you need to think separately re: power factor and efficiency. In your first statement, I think you are on the right path, but then take a wrong turn. Adding a parallel capacitor to improve power factor does not increase the motor's efficiency... at least to first order. There is a very, very slight second order effect that improving the power factor reduces the current supplied in the branch circuit to motor which results in less voltage drop, which may result in a slight improvement in efficiency but it is a stretch at best. Your last statement of first paragrah
> 
> 
> CLTEE49 said:
> ...



So power factor has nothing to do with motor efficiency... at least not in the first order.


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## Flyer_PE (Apr 12, 2010)

CLTEE49 said:


> So power factor has nothing to do with motor efficiency... at least not in the first order.



Efficiency is a factor of real power out vs. real power in. Power factor doesn't play in the equation.

Using a capacitor to correct power factor will reduce line current and thus voltage drop. The power factor of the motor itself will not change, only the power factor of the system comprised of the motor with the capacitor bank in parallel.


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## CLTEE49 (Apr 14, 2010)

Flyer_PE said:


> CLTEE49 said:
> 
> 
> > So power factor has nothing to do with motor efficiency... at least not in the first order.
> ...


I thought the constant/core/no load losses in a motor (the eddy and hysterisis losses) were imaginary. These are added to the I^2*R losses to get the total power loss. If you reduce these constant losses that improves the power factor and increases efficiency.


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## daniel2_718 (May 6, 2010)

CLTEE49 said:


> Flyer_PE said:
> 
> 
> > CLTEE49 said:
> ...


Well your on the right track by considering the motor model, but look at it this way:

A simple motor model is comprised of inductive reactances and pure resistances. For a given load, and therefore a particular amount of slip, you can combine all of the inductances and resistances to a lumped (Thevenin) impedance (R+JX). The ratio of the reactive part of that impedance to real part of the impedance will determine the power factor of the motor. As you can see, by adding a pf correction capacitor at the motor terminals you are not changing the pf or the efficiency of the motor. However, you are changing the power factor of the system as seen by the source. For a given load, the motor will still require the same real input power, and convert that power, minus losses, to output power. By using a pf correction capacitor, you are reducing the amount of vars the source must supply to the load, and therefore the overall line current reduces. An example can be seen by the example that kicked this discussion off:

0.5 hp is equal to 372.8 W - which is the output power of the motor

However, the input power is the full load amps * Voltage * pf = 2.9 * 230 * 0.71 = 473.57 W

eff = 372.8/473.57 = .787 =&gt; 78.7% eff

after the pf correction the input power is still 473.57 W, and the output power is still 0.5 hp = 372.8 (same efficiency), however the line current is reduced because the "S", or the apparent power, is reduced from 2.9*230 = 667 kVA to 2.16*230 = 497.55 kVA (hypotenuse of the power triangle after reduction in Q from pf capacitor)

I'm trying to put together a comprehensive concept there, but who knows how understandable it is.. hopefully you get something out of it.

Dan


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## HornTootinEE (May 7, 2010)

If we are talking efficiency, wouldn't it be correct to say that the idea behind pf correction is supply system efficiency improvement, not motor efficiency improvement? motor efficiency is governed by it's own internal design, the supply system efficiency is related to it's own design also. By adding a capacitor, we are designing our supply system to work more efficiently (less power loss) with the given load demanded by the customer...


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