New IE4 Motor - Pulling more Amps but less kW

[Richo106]

Hi All

Apologies for a random but maybe a very simple question, last week we replaced an IE2 200kW motor for an IE4 200kW motor

The FLC on the existing IE2 was 318A and the FLC on the new IE4 motor is 347A at 415V - both 4 pole - controlled via VSD

The new IE4 motor pulls quite a bit more Amps than the existing IE2 motor

However when we checked the energy monitor over the weekend, the new motor used 80kWh's less over the same period and runs on average 10kw lower.

Could someone please explain how we are pulling more Amps but using less energy (kw)

Any advice/information would be greatly appreciated!

 

[wayne440]

Do you have the power factor figures for before and after the change?

 

[LittleInch]

You're quoting FLC - Full load Current.

What's the actual running amps?

IE4 are designated Super high Efficiency compared to "High efficiency" IE2 so not too surprising you're using less kW

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[edison123]

Unless IE4 motor has a poor pf, I don't see how its FLC can be higher than IE2.

Post both motors nameplates.

Muthu

http://www.edison.co.in

 

[waross]

Edison:
Consider a 1740 RPM motor replaced with a 1760 RPM motor driving a centrifugal fan or pump.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[RRaghunath]

The PF gets lower and the starting current inrush gets higher as the motor efficiency increases from IE2 to IE4.
Hence, no surprise that the motor draws higher current and it is because of the increased VAR consumption of the higher efficiency motor.

R Raghunath

 

[Richo106]

Thank you all for the comments, some very helpful information

I am going to upload the 2 motor name plates and VSD HIM information

 

[Richo106]

New Motor:

https://files.engineering.com/getfi...8a-8eb2-9a51707e98c8&file=New_motor_(VSD).jpg

https://files.engineering.com/getfi...2-019c1b3adb00&file=New_Motor_(nameplate).jpg

Old Motor:

https://files.engineering.com/getfi...a-a54d588da039&file=Old_Motor_(nameplate).jpg

https://files.engineering.com/getfi...7d-a633-da8016cc1c96&file=Old_Motor_(VSD).jpg

Unfortunately the old nameplate has no PF on it

Hope this helps

 

[che12345]

" #1. The FLC on the existing IE2 was 318A and the FLC on the new IE4 motor is 347A at 415V - both 4 pole - controlled via VSD. #2. The new IE4 motor pulls quite a bit more Amps than the existing IE2 motor. #3. However when we checked the energy monitor over the weekend, the new motor used 80kWh's less over the same period and runs on average 10kw lower".
I have the following opinion for your consideration.
1. check the voltage, current and power-factor at the same kW output.
2. Info: with the same kW, lower pf will have a higher current.
3. Based on IE2 318A at 415V, the motor is around 200kW. A IE4 motor of higher efficiency is (unlikely?) to achieve average 10kW lower. Based on: 10kW/200kW=0.05 or 5%. There could (be?) some slight differences in the kW loading over the one week period. However: if 5% kW deduction is ascertained, replacement from IE2 to IE4 would be beneficial, even with a higher initial cost. BTW: electrical bill is based on kW, NOT on the current value.
Che Kuan Yau (Singapore)

 

[edison123]

Old motor pf X eff = 0.875

New motor pf X eff = 0.803

Hence, the higher current in the new motor for the same KW load.

Since the new motor is much more efficient than the old one, you have lower KWHR consumption.


Muthu

http://www.edison.co.in

 

[Richo106]

Thanks again

Is it as simple just to do the 3 phase power formula to show this?

I would just like to see it as numbers to make it clear in my mind - apologies

 

[edison123]

KW = 1.732*KV*Amps*PF*Eff

Muthu

http://www.edison.co.in

 

[Richo106]

Thank you Edison

Last question

How did you calculate this:

Old motor pf X eff = 0.875

New motor pf X eff = 0.803

 

[edison123]

From old motor nameplate, plug in KW, KV and A to get (PF*Eff) (can't separate them)

From new motor nameplate, plug in PF and Eff to get the product.


Muthu

http://www.edison.co.in

 

[Richo106]

KW = 1.732*KV*Amps*PF*Eff

New Motor:
1.732 x 0.415 x 301 x 0.803 = 174kw

Old Motor:
1.732 x 0.415 x 243 x 0.875 = 152kw

This still shows the new motor using higher kw

 

[edison123]

My replies are based on 200 KW nameplates you provided.
The PF and eff change for loads other than 200 KW. See the new nameplate.

Muthu

http://www.edison.co.in

 

[Richo106]

Yes these 'like for like' motors that we have swapped

I am just looking for an explanation using numbers to show we are making a saving even though pulling more Amps

I appreciate all your help with this!

 

[edison123]

It's obvious they are not identical motors. They are of different classifications from different eras.

Muthu

http://www.edison.co.in

 

[Richo106]

Yes they are different classifications in motors, we was just after the calculation/formula to show/prove we are making savings in kwh's to justify the new energy efficient motors.

We want to do a survey around site to see which will be the most cost effective motors to do next

I might be being daft but I still can't get 0.875 from the figures on the old motor name plate, 200kw, 318A, 415V

 

[jraef]

When you installed the new motor, did you redo the Autotune procedure on the VFD?


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden