3 Phase Induction motor results

[andy257]

Hi Guys

As part of my uni report i had to do an investigation on the operation of a star and delta connected induction motor. I have collected my results and have compiled a spreadsheet with some graphs.

If anyone on here could take a look at the graphs to see if these look normal to them that would be great. All i ask is for you to take a look. Particularly the efficiency graph i seem to be generating electricity.

1000%+ efficiency what the F***.
Also what unit is slip measured in? % or something else?

My power factor graph looks odd also.


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Yes this is homework, i do not ask for answers i can do that myself. All i ask is if you could look at the results and see if they coincide with what should be happening. Perhaps someone in industry who tests motors for a living would be great.

Thanks

Andy

p.s - i have tried google

 

[akg]

how did the efficiency became > 100 (200,400..1400!!) ??

 

[andy257]

how did the efficiency became > 100 (200,400..1400!!) ??

Thats what i would like to know myself.

The equation i have used to calculate efficiency is

(power out / power inn) *100

since one measurment is in KW and the other is in W i have not multiplied by 100 like the above equation, it give the same result.

 

[Nigel Goodwin]

how did the efficiency became > 100 (200,400..1400!!) ??

Thats what i would like to know myself.

The equation i have used to calculate efficiency is

(power out / power inn) *100

since one measurment is in KW and the other is in W i have not multiplied by 100 like the above equation, it give the same result.

You should use both in either W or KW, it's not exactly hard to convert!. Power out is ALWAYS going to be smaller than power in, so the result will NEVER be greater than 100% - or even get very close to it?.

 

[akg]

how did the efficiency became > 100 (200,400..1400!!) ??

Thats what i would like to know myself.

The equation i have used to calculate efficiency is

(power out / power inn) *100

since one measurment is in KW and the other is in W i have not multiplied by 100 like the above equation, it give the same result.

both shld use same unit , W or KW

 

[akg]

ohh...nigel ..already done...

 

[andy257]

how did the efficiency became > 100 (200,400..1400!!) ??

Thats what i would like to know myself.

The equation i have used to calculate efficiency is

(power out / power inn) *100

since one measurment is in KW and the other is in W i have not multiplied by 100 like the above equation, it give the same result.

You should use both in either W or KW, it's not exactly hard to convert!. Power out is ALWAYS going to be smaller than power in, so the result will NEVER be greater than 100% - or even get very close to it?.

DOW - my mistake, misted a zero off.

Ok they are now in same format and the efficiency is alot better however i still see that in some cases i am getting more out than i put in. Perhaps i have discovered the solution to the worlds power problems, lol.

130% efficiency cant be right though??????????

 

[akg]

Ok they are now in same format and the efficiency is alot better however i still see that in some cases i am getting more out than i put in. Perhaps i have discovered the solution to the worlds power problems, lol.
130% efficiency cant be right though??????????

:lol: :lol: u'll soon become famous ....

 

[JimB]

I think the problem lies with your initial measurements.

The measured power factor is way too low.
Calculating the apparent power (1.73 * V * I) and dividing that into the wattmeter reading should give the actual power factor. Doing a quick calculation I got some PF figure which were higher than your measured values, but I am still not sure they are correct.

Your method seems OK, I think the problem is with the measurements.

The efficiency problem has been discussed already!

Slip can be expressed in percent, or per unit. As you have calculated it it is per unit.

JimB

 

[andy257]

I think the problem lies with your initial measurements.

The measured power factor is way too low.
Calculating the apparent power (1.73 * V * I) and dividing that into the wattmeter reading should give the actual power factor. Doing a quick calculation I got some PF figure which were higher than your measured values, but I am still not sure they are correct.

Your method seems OK, I think the problem is with the measurements.

The efficiency problem has been discussed already!

Slip can be expressed in percent, or per unit. As you have calculated it it is per unit.
JimB

Thanks Jim some good info.

am i right in thinking the power factor will be lagging up until a certain value of torque (load) and then become leading. My graph somewhat resembles the "V shape" which is what i think it should look like anyway.

cheers

Andy

 

[JimB]

Andy

I am not an induction motor expert, but conventional wisdom and text books tells me that they always have lagging power factor, unless they are driven at greater than synchronous speed (say by a diesel engine) then they will have a leading power factor.

As a generaliation, a lightly loaded small motor will have a power factor of about 0.5, and a fully loaded large motor will have a PF of 0.85 to 0.9.
Low load give lower PF and large load gives a higher PF, and for some reason bigger motors seem to have a better PF.

JimB