Determining motor efficiency

[spacebeer]

Hi,

I need to obtain efficiency of a electric motor of 250kW, working with a VFD. I have the basic specifications of the nameplate:
- Power: 250 kW
- Current: 475 A
- Tension: 400 V
- FReq: 50Hz
- Cos fi: 0,80
- Rot. vel.: 1000 rpm

I also have the actual consumed power, measured: 183,5 kW

I also have the instant information given by the VFD:
- Rot. vel.: 998 rpm
- Current: 347,6 A
- Torque: 62,1%

Does anyone knows if the current gived by the VFD is at the outlet?
Can i assume that if the rotation speed is nominal, thus 50Hz, the tension at the outlet of the VFD is 400V?

Thanks in advance!

 

[jraef]

[ol ]
[li]Motors are often provided with an efficiency rating in their data sheets, contact the motor mfr.[/li]
[li]The current reading on a VFD display is the motor current. It is of little value in determining the energy without knowing the motor power factor at the moment you read the current, but many VFDs will provide you with a kW output value, that is more valuable.[/li]
[li]If by "consumed power" you are reading this value, that is only the motor's consumed power, it does not include the losses in the VFD, which are more difficult to attain accurately.[/li]
[li]Yes, at 50Hz the output voltage will be 400V.[/li]
[/ol]
Perhaps if you state your reasoning for the question and be more clear about the source of the kW value you stated, we can better help you find what you seek. From the looks of it, you might a[pre][/pre]lready have what you need.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[spacebeer]

Thanks jraef,

The mfr no longer exists...

The motor power factor measured at the inlet of the VFD (witch i supose is the same as the outlet) was 0,93.

I supose the losses in the VFD are low...2 to 3%?

Well, i am trying to calculate e actual efficiency of the motor. Considering the nominal characteristics (nameplate), i can obtain a nominal torque: P = 2.pi/60 x rpm x Torque
So, Torque = 2387,3 N.m

I need this value to calculate the actual torque given by the VFD as %. So using the value given by the VFD i can obtain the actual torque: 2387,3 x 0,621 =1482,5 N.m

With this i can calculate the actual net or mechanical power, with the same equation: P = 2.pi/60 x 998 x 1482,5 = 154,9 kW
Now, using the values given by the VFD and assuming the cos fi measured (0,93) i did the following:
P(elect) = sqr(3) x 400 x 347,6 x 0,93 = 224 kW
Eff = P(mec) / P(elect) = 154,9 / 224 = 69%

This seems very very low....is there anything wrong with my calculations?

 

[itsmoked]

The motor power factor measured at the inlet of the VFD (witch i suppose is the same as the outlet) was 0,93.

Is NOT a true statement. They have pretty much nothing in common if you're trying to use them as equivalent.

I believe you use the instantaneous kW value given by the VFD display directly for your torque calculation. Do not include power factor since that has already been involved in the VFD's power reading.

If you want the actual motor efficiency then you'd need some way to confirm the actual mechanical output of the motor. All you really have here is the power the motor is receiving, not the actual mechanical power the motor is delivering. Think dynamometer.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

"I also have the actual consumed power, measured: 183,5 kW"

Where did this come from???


Motor efficiency at 50Hz and from 50% to 100% load typically runs in the low 90's.

The whole VFD setup likely runs in the 90% to 95% range.

 

[spacebeer]

Thank you both for your answers.

Dear itsmoked,
I will try to otain the power from the VFD. But do you agree that i may use de %torque given by the VFD as a % of the nominal torque, to know the actual torque?
About the dynamometer, i do not know to measure it without dismanteling the motor and put it in a test bench...no easy when you're talking abaou a 250kW motor running for production...

Dear LionelHutz,
The power was measured by a power quality analyser. But, forget this because it was an isolated measurement and it wasn't coincident with the data collected from the VFD, so it isn't relevant for this calculatios (i think...)
The motor efficiency depends mostly on the efficiency rating of the motor. Not only by the load. For instance you can have IE3 motors with 95% at 50% load.

 

[jraef]

So it appears what you are REALLY after is the motor shaft torque then? You should have said that in the beginning, could have saved a lot of bytes of data here.

Torque follows current very closely. So if you look at the output current if the VFD as a percentage of the motor nameplate Full Load Current, that will give you a fairly accurate representation of the output torque at that moment. Remember, the nameplate kW rating of the motor is the MECHANICAL power rating, not electrical. So you already know the full load torque rating if the motor by applying the simple torque formula, and a VFD works to make the motor capable of putting out constant torque at any speed. So whatever torque it is providing at any moment as demanded by the load will be a percentage of FLT and represented by the percentage of FLC.

Also, many of the better VFDs will actually calculate and display the output torque, did you check?


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[GroovyGuy]

Hi SpaceBeer,
Based on the NP data that you provided above, the motor is 95% efficient. This is very typical for a motor of this size.
I would expect the ASD to be approx 96% efficient, perhaps a little higher if it is a cheap & nasty 6-pulse type PWM drive w/o an isolation transformer, input-filter or output-filter.

Q1) Does anyone knows if the current gived by the VFD is at the outlet?
A1) The current displayed on the ASD is almost certainly the motor current, and not the ASD input current.

Q2)Can i assume that if the rotation speed is nominal, thus 50Hz, the tension at the outlet of the VFD is 400V?
A2)Under normal circumstances the ASD will keep v/f constant. I would expect the voltage to be 400V at 50hz. Can you not check this with a DVM?

What is the nature of the load, constant-torque or variable torque?
GG


The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty.
Winston Churchill

 

[spacebeer]

Dear jraef
What i want to know for real is the actual efficiency. The torque is only a way to calculate it. Let me correct you: above i mentioned that what the VFD gives is a % of torque and not % of currente. The current is given in Amps (367,6 A). The previous calculations that i made considered what you are saying now: the nameplate indicates the mechanical power.
So from what you said about FLT and FLC, i can conclude that in my calculations, when i use the %torque given by the VFD and multiply by the nominal torque i obtain the actual torque. Correct?

Dear GroovyGuy,
I agree when you say that the reference efficiency of the motor is 95% and it it a baseline, but i'm very interested in knowing the actual efficiency...this motors have almost 30 years.
The load is variable torque and constant speed.

Thank you both

 

[GroovyGuy]

Hi SpaceBeer,
I lied, based on the motor's NP data that you provided, the efficiency is only 94.94%, and not 95% as I had suggested earlier.
Motor Efficiency Calculation:
[li] Sin= 475A x 1.732 x 400V = 329kVA[/li]
[li]Pin = 0.8 x 329 = 263kW[/li]
[li]Losses = 263 - 250 = 13.3kW[/li]
[li]eff = 94.94%[/li]
Unless the motor has been re-wound during it's 30 year life, I would go with the 95% efficiency calculation. I would not put too much faith in the readings obtained from the ASD's operator interface [ie I don't believe they are going to give you the level of accuracy that you are looking for].

As far as the ASD is concerned, what is the make & model #? Most ASD manufacturer's publish their efficiency ratings. Are there any input reactors or output filters attached to the ASD? This can add to the losses.

What about the motor feeder losses? A long feeder can add another 1 - 2% in losses.

If you are really serious wrt obtaining the overall efficiency of this system, me thinks you might need an accurate power monitor.

Regards,
GG


The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty.
Winston Churchill

 

[LionelHutz]

Look up a paper called Motors with Adjustable Speed Drives: Testing Protocol and Efficiency Standard

I'm not sure you'll find it, but the search should bring up other links with similar results. In the paper, their testing showed,

A 50hp motor lost about 1-2% efficiency running on a VFD compared to line voltage.
The efficiency of a 50hp VFD and motor system was a little over 90% for 50% to 100% loading when at 100% speed.
The efficiency of a large VFD is typically around 97% to 98% for the range of 50% to 100% load and 50% to 100% speed.

Now, keep in mind that this is VFD only efficiency. Add in line and/or load filters and the system efficiency will go down according to the losses of these filters.

Now, on to another point. In your first post you wrote the motor is running at 1000rpm and now you just wrote "The load is variable torque and constant speed."

I think something has been lost in translation here. Otherwise, you have been wasting energy using the VFD since you are indicating the motor always operates at 1000rpm.

 

[spacebeer]

Dear GroovyGuy,
Agree again with everything that you said. Understood perfectly.
If the motor was re-wounded, do you think a 2% loss as a usual value?

Dear LionelHutz,
I will search for the document. Undestand and agree with all that you say.
About de VFD working at nominal speed, my opinion is the same: why did they installed it? (it's a client of mine, not my motor )

Thank you both and i will try a last time to take a look at all information that the VFD gives, but i'm more driven to consider the nominal efficiency for my calculations, as GroovyGuy said...the issue is that for this size, IE3 motors are quite the same efficiency...

 

[GroovyGuy]

Hi SpaceBeer,
I don't have the experience in motor re-winds to guess what impact a re-wind might have on the motor's efficiency. But if I had to guess, I would say that an decrease in efficiency of 2% would be extreme and not likely.
If you are hot to trot on determine the efficiency of the various bits and pieces, I still would recommend that you get your mits on an (accurate) 3-phase power monitor.
Regards,
GG

The pessimist sees difficulty in every opportunity. The optimist sees the opportunity in every difficulty.
Winston Churchill

 

[spacebeer]

I have a 3-phase power monitor, but with it i i don't see how i can get de motor efficiency...