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kW Savings From Replacing Oversized Motor 1

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broc028

Industrial
Oct 2, 2008
11
Could anyone explain to me the kW savings from replacing an oversized motor.

I have calculated the kW Input from
(V x I x 1.73 x P.F.)/1000
and calculated load from:
Load = kw Input/(Rated H.P. x 0.746)/F.L. Efficiency.

If I have a 10hp motor running at 50% load, how do I justify replacing it with, for example a 6hp motor.

Thanks!!
 
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edison - I'm guessing maybe you didn't read starkopete's link. It gives efficiency over a range of loads for each motor. I agree I could not apply the full-load efficiency toward calculating losses at half load. But I ceratinly can apply the half-load efficiency toward claculating the losses at half load. That is what the 92.3 is... efficiency at 50% load of the 10 hp motor.

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Hi Pete;
I believe that I made a blunder. I will defer to you on this.
edison123;
The no load losses are less on the 5 HP than on the 10 HP, but the load losses follow a square law and the I2R is greater on the smaller motor for a given current.
Although the no load losses are less for the 5HP motor, the load losses for a 10 HP motor running 50% loaded will be about 1/4 of the full load losses. That more than compensates for the greater no load losses.
(Did I get it right this time Pete?)

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
I know this is great fun, and good practice, but why not just give the manufacturer your requirements, and ask "what do you have which will effiecently do this at the lowest operating cost?"
 
No reason at all why not. The best comparison uses the data from existing motor (which may or may not be available) against the curve of replacement motor. And lower operating costs alone do not justify replacement... you've gotta pay back. And you've gott have mechanical interchangeability as edison said.

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lol. I've done that myself lately.

Bill I think you are right that in general going from 10hp motor driving 5hp load down to 5hp motor driving 5hp load, the no-load losses increase but the load losses decrease.

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I believe that in most normal installations, a 10 HP motor will pull a 7.5 HP load cheaper than a 7.5 HP motor.
Going from 10 HP to 5 HP, I would recommend running the numbers.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Well if that don't beat all! I have been laboring under the understanding that a partially loaded motor was a power wasting inefficient situation. Now we're saying run a motor a few hp bigger than needed for better efficiency..

Where's my Nola.

Keith Cress
kcress -
 
Hi Keith, for a very lightly loaded motor you are right. but for a motor loaded over 75%, there is a sweet spot. Depending on the efficiency curves of the individual motors and the available sizes, you can't always take advantage of the sweet spot.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Keith - I came into the thread with the same assumption as you (a motor oversized by a factor of 2 would have to be wasting energy). Starkopete showed something different for two motors that appear to be from the same product line, differeing only in horsepower. So the lesson imo is it's tough to make generalizations. As retred said, look at the invidivudal motor data whenver possible.

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As retred said, look at the invidivudal motor data whenver possible.
Sorry - waross said that too... and a little more. Didn't mean to leave that out

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Last comment - Keith - I know I didn't say anything you didn't already know.

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itsmoked said:
... Where's my Nola.
Actually Keith, this is one of the reasons why the Nola circuit didn't pan out as the panacea of energy savings. People were ASSuming that a motor running unloaded was wasting a lot of energy and it really isn't true a lot of the time, especially since motor efficiencies have been regulated to increase.

But the scammers still try to capitalize on the basic instinct people have to believe that.


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Well. That's why I like associating with other engineers. It's fun to have solid truths being smashed and replaced by actual reality.

This is right up there with wye-delta starters suck, VFDs can make three phases from a single phase, and centrifugal pumps work less with outlets restricted - not more.

Now I can add: over sized motors are not necessarily less efficient.

Thanks folks!


Keith Cress
kcress -
 
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