Power Factor Question

[kwired]

I forot to put in the 1.732 for three phase, which I believe was probably meant in the original question. That changes to 2106, not 1216.

Ok after a little reasearch, been trying to use wrong formulas I come up with same answer as Besoeker in posts 5 and 14.

here is formula:

P = V x A x PF (x 1.73 for three phase)

everything is known except for PF so we need to solve for PF

1576w = 240v x 5.892a x 1.73 x PF

can be re written as

1576w = 2446 x PF

divide both sides by 2446

.64 = PF

The motor has to be three phase otherwise the power factor will come out above 1.0.

Unfortunately .64 was apparently not one of the possible answers where this question came from

Efficiency is not needed to find the answer to this problem.

We should be able to find (output) horsepower by using efficiency in a similar formula to what we were trying to use before.

HP = V X A X eff x pf (x 1.73 for three phase) / 746 watts per horsepower

HP = 240 x 5.892 x .86 x .64 / 746

HP = about 1 hp. (this would be output power at shaft)

 

[rattus]

Efficiency does not enter into the equation which is:

P = V*I*PF*1.73

then,

PF = 1576W/(240V*5.9A*1.73) = 0.643

which is reasonabe for a lightly loaded motor.

 

[perryallen]

For What It's Worth

For What It's Worth

I received a reply from the creator of the question in the original post.

I'm not saying I agree with it but I am posting for clarifying the question originator's intent.


The link takes you to his audio explanation (exam 2 #37): http://www.electricianmath.com/me_audio_support08.htm

Basically, his formula is as follows:

PF= W x eff / I x E

PF= 1576w x .86 / 5.892A x 240v (single phase)

PF= .958

 

[rattus]

No!

No!

I received a reply from the creator of the question in the original post.

I'm not saying I agree with it but I am posting for clarifying the question originator's intent.


The link takes you to his audio explanation (exam 2 #37): http://www.electricianmath.com/me_audio_support08.htm

Basically, his formula is as follows:

PF= W x eff / I x E

PF= 1576w x .86 / 5.892A x 240v (single phase)

PF= .958

Efficiency is the ratio of mechanical power at the shaft to real electrical power drawn by the motor.

PF is the ratio of real electrical power (watts) to the apparent power (volt-amperes) drawn by the motor.

The original question is vague since it does not specify that the motor is 3-phase which it must be in order for the numbers to make sense. It is also confusing since it specifies the efficiency which is not needed.

 

[perryallen]

Efficiency is the ratio of mechanical power at the shaft to real electrical power drawn by the motor.

PF is the ratio of real electrical power (watts) to the apparent power (volt-amperes) drawn by the motor.

The original question is vague since it does not specify that the motor is 3-phase which it must be in order for the numbers to make sense. It is also confusing since it specifies the efficiency which is not needed.

Agreed. It's simple. The question and math behind it is flawed.
Should simply be: PF= Watts / va or watts= E x I x P.F.

Time to put a fork in this one.

 

[steve66]

I actually agree with the author's answer as given in post #24. (Although it would have been nice if he had told us if it were single phase.)

Rattus would have been right if the question had asked "What is the power factor of the circuit?"

But the question was "What is the power factor of the motor?" I believe the definition of a "motor's power factor" normally includes the efficiency of the motor. So in this case, the answer looks right.

 

[rattus]

Wrong:

Wrong:

I actually agree with the author's answer as given in post #24. (Although it would have been nice if he had told us if it were single phase.)

Rattus would have been right if the question had asked "What is the power factor of the circuit?"

But the question was "What is the power factor of the motor?" I believe the definition of a "motor's power factor" normally includes the efficiency of the motor. So in this case, the answer looks right.

PF is commonly equated to cos(phi). Efficiency does not appear in this equation.

 

[steve66]

PF is commonly equated to cos(phi). Efficiency does not appear in this equation.

I won't say where I got this, but they reference IEEE, so it must be right :

The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load over the apparent power in the circuit.

So it really depends on what we define as the "load". If the "motor" is the load, you are right. But if we define the "mechanical load" as the load, then the equation given in post 24 is correct.

 

[david luchini]

I won't say where I got this, but they reference IEEE, so it must be right :

The power factor of an AC electric power system is defined as the ratio of the real power flowing to the load over the apparent power in the circuit.

So it really depends on what we define as the "load". If the "motor" is the load, you are right. But if we define the "mechanical load" as the load, then the equation given in post 24 is correct.

The load given in the OP was "measured" by a watt meter. This is electrical power, not mechanical power. As such the Example is incorrect.

 

[steve66]

The load given in the OP was "measured" by a watt meter. This is electrical power, not mechanical power.

Right. That's why we have to use the efficiency of the motor to calculate the mechanical power.

Again, I would agree with you if they just asked for the "power factor", or if they asked for the power factor of the circuit.

But I believe the exact question was "What is the power factor of the motor?" To me, it makes sense that the power factor of the motor would be the ratio of useful output power, to the input volt-amps.

I can't find a reference, but I believe I've seen motor power factor defined that way elsewhere.

Steve

 

[david luchini]

Right. That's why we have to use the efficiency of the motor to calculate the mechanical power.

Again, I would agree with you if they just asked for the "power factor", or if they asked for the power factor of the circuit.

But I believe the exact question was "What is the power factor of the motor?" To me, it makes sense that the power factor of the motor would be the ratio of useful output power, to the input volt-amps.

I can't find a reference, but I believe I've seen motor power factor defined that way elsewhere.

In the example, the only load information given for the circuit is for the single motor. The "circuit power factor" and the "motor power factor" are therefore the same thing.

But the power factor of the motor still has nothing to do with the ratio of electrical power to mechanical power. As you have noted above, the ratio of electrical power to mechanical power, depends on efficiency.

Contrary to the red highlighted section above, the power factor of the motor is the ratio of the input power (watts), to input volt-amps.

If you want to compare input volt-amps to output mechanical power, you would use the power factor to convert the input volt-amps of the motor to the input watts of the motor W = VA * PF. With the input watts, you could then determine the output mechanical power using the efficiency by HP = W * Eff/746. But as you can see, the motor power factor is NOT a direct factor in determining input VA to output mechanical power.

 

[rattus]

Simply put:

Simply put:

It is understood by most that "load" means the electrical impedance of the motor under a given set of conditions. Let us not create a lot of confusion by expanding the definition.

How would you do power factor correction with such a definition??

 

[kingpb]

For the record, motor efficiency is defined as the ratio of useful power output divided by its total input power; Pout/Pin

For this question, as it is stated, I believe the author was wanting you to make the assumption the measured power in watts or current in amps is the output power. Simply because it is generally assumed that measured power and current is at the input, and efficiency only comes into play when output power is measured. Therefore, because it was given, it is assumed to be the output power. But, I can see both sides of the argument.

However, the voltage given as 240V, I would have to assume 2-pole single phase unless given some other data, as that, in the US, is a standard single phase voltage. I don't see how the voltage can be assumed to be three phase.

I think this falls into test taking skills area. Sometimes the context of the question will lead you into making the correct assumptions, i.e. is this a motor training course, or a power training course? If motor related, I'd say the author wants you to use the efficiency, especially if you are not showing your work, and the correct answer is one of the choices. If you show your work, or the answers are "pick the closest to the right answer", then some information may be extraneous, or you can at least state your assumptions.

 

[Besoeker]

The load given in the OP was "measured" by a watt meter. This is electrical power, not mechanical power.

Agreed. You can measure mechanical power at the motor output shaft but, from my experience, it isn't a particularly simple thing to do.
We have had to do it on occasion but it has been done on a test bed with a torque tube and accurate speed measurement.

However the question does, as you say, mention a Wattmeter which infers electrical input power. Motor efficiency is quite irrelevant. So we have input power and input VA.
And an impossible answer.
Bad question and bad multiple choice answers in my opinion.

 

[david luchini]

Bad question and bad multiple choice answers in my opinion.

Amen.

 

[charlie b]

Bad question and bad multiple choice answers in my opinion.

Agreed. I looked through the web site, trying to come across something that does appear on Mike Holt's web site, and that is an invitation to send feedback, when you come across something that you believe is incorrect. I did not find it.

 

[charlie b]

Just FYI: I sent an email to the owner of the web site from which the test question was obtained. I invited him to add his own comments, if he wishes.

 

[rattus]

Agreed. You can measure mechanical power at the motor output shaft but, from my experience, it isn't a particularly simple thing to do.
We have had to do it on occasion but it has been done on a test bed with a torque tube and accurate speed measurement.

However the question does, as you say, mention a Wattmeter which infers electrical input power. Motor efficiency is quite irrelevant. So we have input power and input VA.
And an impossible answer.
Bad question and bad multiple choice answers in my opinion.

Hear! Hear!