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How to Calculate motor KW given actual RPM , Voltage and FrequencyHelpful Member!(3) 

yinthenovice (Mechanical) (OP)
30 Jul 10 12:09
Hi Guys,

I have set up a test rig in which I use a power supply to vary the Voltage and Frequency to a 3.73KW AC motor. I am also able to measure the shaft RPM using a tachometer. I just need to know of a way to calculate the power based on this info.

Background:

 I just needed to get the response of my load over a wide range of speeds so I kept the input frequency constant at 30Hz (for no good reason really) and varied the voltage. Now I have a bunch of voltage, frequency and RPM readings and am having trouble finding motor power.

I tried to find motor torque and find power that way, based on this old eng-tips post i found but not very realistic answers :
"
Derive the torque from rated speed and rated HP.
Actually I believe the slip formula is a little oversimplified.
Rated slip is synchronous speed minus rated speed.
Example: 1800 RPM - 1750 RPM = 50 RPM slip.
(1800 RPM - Measured RPM)/Slip RPM = % HP loading.
Caveat. This depends on the nameplate value for motor speed being accurate. I have not used this method in the field and have no idea of the precision of the nameplate speed as reported by the manufacturer. A small error in reported speed will result in a large error in calculated HP."  
Helpful Member!  jraef (Electrical)
30 Jul 10 12:43
Power as in Power Consumtion or Power Delivery?

Power consumption really has nothing to do with RPM or frequency directly, Power is A * V * PF (* 1.732 for 3 phase). So if you just get a good meter that can read the PF, you're golden. A kW meter does that.

Now if you want to know the power delivered by motor, that's a slightly different situation. You will also need to know the efficiency of the motor, in fact the efficiency of the entire circuit (depending on how you are varying the frequency and where you measure the values). Motor delivered power then would be A * V * PF * eff (* 1.732 for 3 phase). Shaft power would also have to account for mechanical losses as well.


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Helpful Member!  dpc (Electrical)
30 Jul 10 12:53
You might want to have a look at this paper by Stan Zocholl:

http://www.selinc.com/WorkArea/DownloadAsset.aspx?id=2430

If that link doesn't work, go to www.selinc.com and search for a paper titled Induction Motors Part 1 - Analysis.  

You can get a rough approximation of motor kW input using the Steinmetz model if you compute the slip.  

But if you need something closer than 10% or so, you'd be better off with a kW meter as Jeff suggested already.   

David Castor
www.cvoes.com

Helpful Member!  electricpete (Electrical)
30 Jul 10 12:57
In addition if you're going from the input side, you'd like to measure power factor to take a stab at input power (which doesn't get you to output without efficiency).

Sounds like you were varying voltage in attempt to vary speed? I imagine you didn't get a very wide speed range.

Regarding the slip method, on a rough basis slip is proportional to output power with full-load slip (determined by nameplate) corresdponding to full load output power (also from nameplate) and output power proportional to slip in between.

There are some caveats as you have mentioned and the calc will inevitably be rough in most real-world situations.  As a rough approximation  in the linear region of torque speed curve for small slips:
P ~ s*V^2 / Rrotor
We see there is voltage squared dependence which must be taken into account if V deviates from nameplate.

We can also see Rrotor plays a role, which suggests temperature plays a role. Exactly what temperature was assumed in nameplate conditions I don't know... was a source of previous discussion here:
http://www.eng-tips.com/viewthread.cfm?qid=8576&page=306

Also nameplate RPM is not exact to begin with... NEMA has some spec about 5rpm... either accurate to within 5 rpm or rounded to nearest 5 rpm (I don't remember which... there is a factor of 2 difference in those two tolerances of course)

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dpc (Electrical)
30 Jul 10 12:58
I should have added that you also need to know the slip at rated load.  This should be on the motor nameplate.  
yinthenovice (Mechanical) (OP)
2 Aug 10 12:40
jraef: I was hoping to get power delivered , and I just needed a ballpark number. Since this is a one off thing , I had hoped not to pick up a multimeter with power factor. Also efficiency is just a guess at best. I did plug my power supply into a scope and according to my labview software its only 3 degrees out of phase so i go off that. efficiency is anyone's guess at this point.

@electricpete, i was having trouble with the slip method , maybe because I just plugged in the formula. But now that I look at it- you say slip is proportional to power , but isn't it inversely proportional ? because maximum torque/power is developed at rated RPM and this is where we have minimum slip so aren't they inversely proportional ? I was going off their being proportional and was getting lower power for higher speeds. Shouldn't we have high slip at low speed and lower slip at higher speed .check the image i attached. And you're right about varying frequency giving finer resolution of speeds :) guess that's why they have VFDs not VVDs.





 
electricpete (Electrical)
2 Aug 10 13:41

Quote:

@electricpete, i was having trouble with the slip method , maybe because I just plugged in the formula. But now that I look at it- you say slip is proportional to power , but isn't it inversely proportional ? because maximum torque/power is developed at rated RPM and this is where we have minimum slip so aren't they inversely proportional ? I was going off their being proportional and was getting lower power for higher speeds. Shouldn't we have high slip at low speed and lower slip at higher speed

we have three variables:  Slip, Power (output), and speed.   Slip and power are proportional to each other.  Speed varies in opposite direction of the first two (although I wouldn't say "inversely proportional").

We can have:
low output power , low slip, high speed
or
hi output power, high slip, low speed

"High" slip is high  relative to other slips in the range of normal operation which would all be lower. These points would all be far to the right of the curve with s<0.1, speed > 90%.  

I have attached graph which may help.
"because maximum torque/power is developed at rated RPM and this is where we have minimum slip"
This is the part that seemed tripping you up...I'm not sure why you say minimum slip corresponds to max torque/power.
 

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yinthenovice (Mechanical) (OP)
2 Aug 10 14:31
aaahhh , the fog of ignorance begins to lift... you see I'm using a 3450 RPM rated motor, and I'm varying it from 300RPM up to 2000RPM. so I'm on the left side of that graph's peak while you were on the right. I was out of the range of 'normal operation'. Which would explain why things are all wonky, especially since the torque doesn't dip all the way down to zero on the left side. I'd need to know at least one speed and torque combination to predict all the others.

Thank you electricpete.  

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