intended Motor efficiency 2

[Feg]

Hi Guys,

Is there any way to telling the intended efficiency figure of a motor if the figure is not shown on the plate. The only details I have are 45kw 1470 rpm .87 p.f 80.0 amps 400v 50hz Torque class 16. I have tried a few companies and they do not seam to have the old catalogues which would give this information, this motor is about 13 years old.

Thanks in Advance

 

[dpc]

At full load (80 Amps), your motor is drawing 400 x 80 x 1.732 = 55.42 kVA.

At the stated power factor of 0.87, this = 55.42 x 0.87 = 48.22 kW electrical input.

This equates to an efficiency of 45 kW/48.22kW = 93.3%

Seems a little high - but this is the theoretical calculation based on the nameplate data you have provided.

 

[ozmosis]

Feg
If you post the motor make then there might be some around who might have the actual details.

 

[waross]

I get the same figure (93.3%) as dpc based on the nameplate data. 45kw/KVA x PF

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Feg]

Thanks Waross,

How accurate do you think this is, bearing in mind the motor could be EFF1 EFF2 or maybe even EFF3.

Thanks

 

[waross]

This is as accurate as the information on the nameplate. To get better accuracy you will have to lab test the motor and verify the nameplate data. I would just accept the nameplate data. It is probably close enough for most purposes.
Remember that the efficiency will change with both loading and applied voltage. Best efficiency is often at about 50% to 75% loaded. A slight change in voltage may change the efficiency by a greater percentage than the margin of error on the nameplate.


Bill
--------------------
"Why not the best?"
Jimmy Carter