Power calculations 1

[jnixon21]

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.

 

[electricpete]

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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[jnixon21]

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

 

[electricpete]

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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[electricpete]

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

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[electricpete]

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%


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[jnixon21]

Thank you very much!!!