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Replacing Oversized Motor |
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 May 1 2009, 08:06 AM
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#1
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Junior Member  Group: Members Posts: 2 Joined: 1-May 09 Member No.: 5,342  |
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. |
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May 1 2009, 08:38 AM
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#2
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Posting Freak Group: Moderator Posts: 2,336 Joined: 24-April 02 From: Christchurch, New Zealand Member No.: 1 |
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. -------------------- Mark Empson administrator
Skype Contact = markempson | phone +64 274 363 067 LMPForum | LMP Tech Warehouse | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Mark Empson Website |
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May 1 2009, 08:49 AM
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#3
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Junior Member Group: Members Posts: 2 Joined: 1-May 09 Member No.: 5,342 |
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! |
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May 2 2009, 10:01 AM
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#4
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Posting Freak Group: Moderator Posts: 2,336 Joined: 24-April 02 From: Christchurch, New Zealand Member No.: 1 |
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, -------------------- Mark Empson administrator
Skype Contact = markempson | phone +64 274 363 067 LMPForum | LMP Tech Warehouse | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Mark Empson Website |
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