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Power calculations

Power calculations

Power calculations

(OP)
I was wondering if I have the ratings for a motor @ full load, say 106 Amps @ 230 V (60 Hz), and I measure that we are using 26% (around 27 Amps) of the full load amps, can I take that percentage and apply it to my original full load KWh calculations? This is for a cost savings project.
Thank you in advance.

RE: Power calculations

That might be a reasonable approximation above 75% current.

At lower power levels and with slow motors, you need to consider the presence of no-load magnetizing current which is in quadrature with load-component of current (add by SRSS).  The load-component varies with linearly with load and the no-load component is constant regardless of load.

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RE: Power calculations

(OP)
Pete would you explain a little further? I am still very green in my professional career.

RE: Power calculations

Let's say
I_FL = 100A.
I_NL = Imagnetizing = 25A
You measure Imeasured = 60A
Question:  What is the actual load

Model: I = I_magnetizing^2 + P * ILoad^2
  where P is fraction of output horsepower 0 to 1.

Call ILoad at full load = ILoadFL

ILoadFL = sqrt(100^2 - 30^2) = 96.8 A

Call ILoad at the time of measurement ILoadMeas

ILoadMeas = sqrt(60^2 - 25^2) = 54.5A.

P = ILoadMeas/ILoadFL = 54.5 / 96.8 = 56 % of full load.

The above answer is better than using straight ratio of measured current (60/100) and good enough for most purposes but still includes a few approximations.  

The tough part will be determining no-load current.  If you have run the motor uncoupled and measured current, that is the no-load current.  Otherwise you have to estimate (guess).  For 2-pole motor maybe I_NL = 20% of FLA.  For 6-pole motor maybe 30%.   20-pole motor maybe 50%

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Eng-tips forums: The best place on the web for engineering discussions.

RE: Power calculations

Correction
Model: I = sqrt(I_magnetizing^2 + P * ILoad^2)

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Power calculations

Reposted with corrections.
Let's say
I_FL = 100A.
I_NL = Imagnetizing = 25A
You measure Imeasured = 60A
Question:  What is the actual load

Model: I = sqrt(I_magnetizing^2 + P * ILoad^2)
  where P is fraction of output horsepower 0 to 1.

Call ILoad at full load = ILoadFL

ILoadFL = sqrt(100^2 - 25^2) = 96.8 A

Call ILoad at the time of measurement ILoadMeas

ILoadMeas = sqrt(60^2 - 25^2) = 54.5A.

P = ILoadMeas/ILoadFL = 54.5 / 96.8 = 56 % of full load.

The above answer is better than using straight ratio of measured current (60/100) and good enough for most purposes but still includes a few approximations.  

The tough part will be determining no-load current.  If you have run the motor uncoupled and measured current, that is the no-load current.  Otherwise you have to estimate (guess).  For 2-pole motor maybe I_NL = 20% of FLA.  For 6-pole motor maybe 30%.   20-pole motor maybe 50%

=====================================
Eng-tips forums: The best place on the web for engineering discussions.

RE: Power calculations

(OP)
Thank you very much!!!

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