motor efficiency and power factor

T

teeg123

Member
Location
Maine
Occupation
Journeyman Electrician
here is the question
What is the efficiency and power factor of a one horsepower, single phase, 208-volt motor?
36.80%
38.90%
40.70%
80.00%

formula for efficiency is n= (output power/ input power) *100
formula for power factor (PF)= active power(P) / apparent power (S)
because I got this wrong, I know the correct answer, but I just can't figure how they got it?
apparent power formula for single phase is S=V*I, S=208*8.8=1830.4
active power formula P=V×I×cosϕ P=208*8.8*cosϕ. this is where I get lost. I think I need to know power factor to get efficiency ?

thanks folks
 
Induction motor efficiency is dealing with what is happening inside that motor when running at nameplate load. Power factor is more about how the facility's electrical infrastructure handles the lagging reactive current that the motor causes.

You can have crappy power factor and still have high effieency. Like with overbuild of the conductors and transformer secondaries. Or with something as easy as a capacitor across the line.

So motor efficiency as stated on nameplates is concerned with all the losses ending at the peckerhead.

Power factor is your problem to handle and it all depends on the specifics of how you design your circuit

teeg123 said:
think I need to know power factor to get efficiency
Click to expand...
 
it would appear that the question has some missing information that is needed to correctly answer it. the answer they have is 40.70%
this was on a practice test, I can't see them putting a question like this on the actual state exam.
 
teeg123 said:
the answer they have is 40.70%
Click to expand...
You can get this answer by taking the NEC Table 430.248 full load current for a 1 HP 208V motor of 8.8A, and multiply with voltage to get 208 * 8.8 = 1830W. That would be input apparent power. Then 1 HP = 745.7W. The ratio is 745.7 / 1830 = 40.7%.

However, I would not call this effiency. I guess you could call it NEC worst-case apparent efficiency, as it is using apparent input power instead of real input power, and the NEC design full load current should be a worst case motor effiency that you wire for; actual motors you install should have a full load current that is no more than 8.8A. If the power factor were 100%, apparent power would equal real power, and it would be worst-case efficiency. But I understand that motor power factor is never 100%.

As to how to get a motor power factor out of the information provided, no clue.

Cheers, Wayne
 
You need the voltage, the current, and the power factor or power factor angle to answer the question.

1 horsepower is defined as 746W. That is your 'power out'.

Apparent power is simply voltage * current, which you've calculated as 208 * 8.8A. And it seems like 746 / (208 * 8.8) = 40.7% which is supposedly the correct answer...but that is not the correct answer because motors have a power factor.
 
teeg123 said:
What is the efficiency and power factor of a one horsepower, single phase, 208-volt motor?
Click to expand...

teeg123 said:
What is the efficiency and power factor of a one horsepower, single phase, 208-volt motor?
Click to expand...
Is the answer supposed to be simply the Eff and PF combined, or do you need each individual component.

I would say you needed to find Eff X PF as found by Wayne in post #6 paragraph 1.
Very straight forward, don't read more into questions than there is. Don't assume information not provided.
 
Thanks much, the 1 horsepower=746 part that helps make sense.
I really appreciate all the help
 
if you have a 1 1/2 hp single phase motor at 115v, continuous duty with a service factor of 1.25 do you apply the 125% twice I would think not.
table 430.248 gets you 20amps *125%(continuous)=25amps
do you just ignore the service factor or is it 25amp*125% (for service) =31.25

thanks to eveyone
 
teeg123 said:
here is the question
What is the efficiency and power factor of a one horsepower, single phase, 208-volt motor?
Click to expand...
Impossible to answer this as written without some other assumptions being involved.

in most real world situations you will have at very least the motor rated voltage and rated current. But that alone only tells you the VA rating. You need at least one of kW rating, power factor, or efficiency and you can then calculate whatever else is missing.

As far as the answer selections, 80% is poor efficiency, the other choices are beyond poor and probably beyond horrible.

If those selections are power factor - any of them might be possible depending on motor loading conditions. The 80% is probably still kind of a poor power factor if motor is operating at full load.
 
kwired said:
Impossible to answer this as written without some other assumptions being involved.
Click to expand...
The assumption you are making is that the question wants two separate answers; one for Efficiency and the other for Power Factor. The possible choices clearly show they want a single answer of Eff X PF.
 
jim dungar said:
The assumption you are making is that the question wants two separate answers; one for Efficiency and the other for Power Factor. The possible choices clearly show they want a single answer of Eff X PF.
Click to expand...
Maybe. What would you use that result for in the real world?

I have me a cheat sheet that I open whenever I am dealing with power factor/PF correction with different formulas and none of them have a direct Eff x PF, at least not for anything I ever need to find.

Say you had 90% efficiency and also a 90% power factor what do you do with the product of that? .9 x .9 =.81?
winnie said:
Apparent power is simply voltage * current, which you've calculated as 208 * 8.8A. And it seems like 746 / (208 * 8.8) = 40.7% which is supposedly the correct answer...but that is not the correct answer because motors have a power factor.
Click to expand...
(y)

kW = hp x .746 / efficiency = kVA x PF PF = kW / kVA

Nothing in these formulas puts Eff x PF together.

Efficiency is nothing more than ratio of input kW to output kW regardless of what the power factor may be or what you may have corrected power factor of the supply line to.
 
In my experience on J test they are looking for the base 746.
On one test it was a simple question. One HP= ________ watts.
On another note so easy however based on 746 it's easy to solve.
What is the amp load of a 3hp single phase motor at 240v with an efficiency of 83%.

Now using 746, and ohms law and basic math you can solve for a basic answer.
Try it and post your results.

Find the key word in the question
3 HP
Single phase
240v
83%
Then do the math
Remember what they want you to know.
746w= 1 HP
 
Joethemechanic said:
Without knowing PF, you can't answer that.

PF and efficiency are two totally different things
Click to expand...
The above question was on the Oklahoma state exam. I agree there different however that is not what they want. It's apparent( no pun intended) the OP is studying for a test.

Since PF is needed and not there you can not use PF to find the answer they're looking for. You must only use what's given to find the answer choice provided.

The question I posted kinda falls in line with what the OP was asking and what the book answer had. On test I learned not to over analyze them other try to find the correct choice based info given.

Sorry for any confusion.
 
Tulsa Electrician said:
The above question was on the Oklahoma state exam. I agree there different however that is not what they want. It's apparent( no pun intended) the OP is studying for a test.

Since PF is needed and not there you can not use PF to find the answer they're looking for. You must only use what's given to find the answer choice provided.

The question I posted kinda falls in line with what the OP was asking and what the book answer had. On test I learned not to over analyze them other try to find the correct choice based info given.

Sorry for any confusion.
Click to expand...
But if you wanted an answer based on unity, why use an induction motor as a load? Why not use an incandescent lamp? Or even reword it to ask what the active power contribution is to the circuit amperage
 
You must log in or register to reply here.
Top