motor speed/power

[Lc85]

We have a 200kW 315L motor 6 pole 974 rpm not driven from an inverter. If we were to use an inverter to lower the speed to 780rpm how would we calculate the new power roughly before a test was performed?

Thanks

 

[Skogsgurra]

Possible output power is proportional to speed. At least as long as you are in a speed range where cooling isn't impaired. But, if you increase speed above rated speed (seems to be 974 RPM in your case) the maximum available output power stays constant.

Gunnar Englund

http://www.gke.org

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Half full - Half empty? I don't mind. It's what in it that counts.

 

[Lc85]

So if 200kw is proportional to 974rpm then We just proportion the power to the new speed so in our case roughly 160kw?

Thanks

 

[Skogsgurra]

Yes, that is correct. It is only when you get down to 40 - 50% of rated speed that you need to think about the reduced cooling.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jraef]

Are you asking about motor mechanical power capability of the machine, or electrical consumed (absorbed) power? Because if it is consumed power, that is completely load dependent. For example if the load is centrifugal (quadratic) like a pump or fan, the power required by the LOAD will drop at the CUBE of the speed change. The total absorbed power will be that, plus any losses in the VFD.

"Will work for (the memory of) salami"

 

[Skogsgurra]

Yes, as Jeff says. What your load does, we cannot know without knowing the load characteristics. I was talking about available power from the motor.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[Lc85]

The load is a fan. It is mechanical output power.

Thanks

 

[Skogsgurra]

OK, then the load will consume a lot less at 80% of rated speed. A little more than half the original power, actually.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[Lc85]

so the power of 160kw at a speed of 780rpm is still valid?

Thanks

 

[Skogsgurra]

The 160 kW is the possible power. The power that you can load the motor with at 80% of rated speed.

The actual power (the power consumed by the fan at 80% of rated speed) is a lot less. The affinity laws are used to calculate actual consumed power and that results in Reduced Speed Power = Original Power *(Speed Ratio)³ so that P₈₀=P₁₀₀*0.8³=200*0.51=103 kW

Have a look at

http://www.engineeringtoolbox.com/affinity-laws-d_408.html

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jraef]

Slight modification;

"The 160kW is the possible MECHANICAL power of the motor."

Now see why we use still use Horsepower? For us, I would say that the possible mechanical power of the motor at that speed would be 214HP, but the kW that is consumed by the motor to move the air at 80% speed will be a factor of the affinity law, based on the actual kW consumed at full speed.

So aside from my half joking reference to units of measurement, you cannot really calculate the consumed kW based on speed and motor rated kW alone, because you don't know what the motor power consumption really was even at full speed. The only vald formula is P_fs (Power at full speed) x .512 (the cube of 80%). If you have not measured the actual kW at full speed, you can only guess. An educated guess that will result in a "no worse than..." answer, but still a guess.

"Will work for (the memory of) salami"

 

[Skogsgurra]

Jeff, the pond that separates us also separates the thinking somewhat. To us, there is no difference between electrical and mechanical power. They are both expressed as J/s or W. Horse-powers do not exist in the (European) drives world.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jraef]

I ate some horse meat steaks in Italy once, is that why? You guys ate them all?

"Will work for (the memory of) salami"

 

[Skogsgurra]

Yes, "No Worse Than" it should be. But the 200 kW was all there was to use as input data. And no other number available later, either.


BTW, there is a slight difference in the definition of the US and European HP.

US: 550 foot-lbf/second = 550*0.3048*0.45359 Nm/s = 760.39827600 = 760 W
Europe: 75 newton-meters/second = 75*9.80665 = 735.49875 = 736 W

Now, for some reason, that isn't entirely correct since all seem to agree that 1 US HP equals 745.7 W.

Lost in translation? Or lost due to friction in the NBS/NIST? Or doesn't the (Swedish) definition of an inch (25.4 mm) apply? With one inch = 25.4, one foot sould correspond to 12*25.4 mm = 0.3048 m.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.