Jump to content


Photo

Replacing Oversized Motor


  • Please log in to reply
3 replies to this topic

#1 broc028

broc028

    Junior Member

  • Members
  • PipPip
  • 2 posts

Posted 01 May 2009 - 08:06 AM

I am starting out on a project to look at possible savings from replacing some oversized motors.


Motor Input kW has been calculated from:
kW=(Volts x Amps x 1.73 x P.F.)/1000

Motor Load has been calculated from:
Load=kW/(Rated HP x 0.746)

All values are measured.If I have a 10 H.P. motor running at 50% load, how do I justify replacing it with a 5 H.P. motor.

#2 marke

marke

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 2,627 posts
  • Gender:Male
  • Location:Christchurch, New Zealand

Posted 01 May 2009 - 08:38 AM

Hello broc028

I am not sure where your "load" formula has come from? it does not make sense to me, but I may be looking at it the wrong way!!

I believe that you are trying to express the load as a percentage, but the rated power of the motor is the shaft power (mechanical output power).
The shaft power at any load is the electrical input power x the efficiency at that load, so this needs to be taken into account.

The efficiency can best be determined from the motor data sheet.
Small motors have a lower efficiency than large motors.
Typically, the efficiency is maximum around 75% load, so sometimes you are better to have a larger motor operating at 75% load, than a smaller motor running at 100% load.
If the efficiencies are the same, there is not advantage in either motor except perhaps for the starting current.

Operating a motor at less than full load will reduce the winding temperature and thereby extend the operating life of the motor.

Best regards,
Mark.

#3 broc028

broc028

    Junior Member

  • Members
  • PipPip
  • 2 posts

Posted 01 May 2009 - 08:49 AM

Just to clarify

Load = electrical energy input/rated mechanical energy output
Load = input kW/(Rated H.P. x 0.746/Full Load Efficiency) x 100 (Express as a percentage)

I have been using this formula for a while now, got it from some U.S. DOE papers.

Is there any kW savings to be found by replacing an oversized motor?

Thanks!

#4 marke

marke

    Posting Freak

  • Moderator
  • PipPipPipPipPipPip
  • 2,627 posts
  • Gender:Male
  • Location:Christchurch, New Zealand

Posted 02 May 2009 - 10:01 AM

I often see reference to "replacing oversized motors" as a means of saving energy, but the reality is that often, the oversized motor will operate at a higher efficiency than the correctly sized fully loaded motor. - It depends on the efficiency curves of the actual motors in question.
For any given motor size, there is a wide range of efficiencies, so it is possible to save energy by using a motor of the same size and higher efficiency.
Many of those advocating the replacement of oversized motors believe that the motor draws rated power irrespective of load, or that the efficiency drops away as he load falls. The reality is that as the load reduces, the efficiency typically rises to a maximum at 75% load and only really begins to drop away below 50% load.

Compare the curves of the motors you are loking at and see if you can find motors that have a higher efficiency at equal shaft load. If you can, then you have a replacment candidate. If you con not, then leave well alone.

Best regards,




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users