AC Motor Current draw

[mariodileep]

Please help me understand this formulation.
say the motor is a three phase ac induction motor (asynchronous).
with the following specifications
575 V, 2 HP, 1740 RPM, Efficiency nominal/full load = 0.86
PF (full load Power Factor) = 0.78 PF (50% Load Power Factor) = 0.57

The formula to calculate the current is
I = 746*HP / (1.73*V*Eff*PF)

At full load the current drawn is ~ 2.2 amps
At half load the current drawn is ~ 3 amps.

it does not make sense? I think I am missing something here, does this mean at no load it will draw high current?
please point out where I am going wrong. Thanks in advance

 

[jmontero]

If the load is 50%, you should use 1HP for the calculations instead of 2.

So, at half load, the current drawn should be ~1.5A

Try to be perfect. You'll never make it, but you will be very, very good.

 

[electricpete]

No, it doesn't make sense. The P in the equation represents the actual load, not nameplate. For 50% load, estimate your current from above equation using p= 1hp, not 2hp

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(2B)+(2B)' ?

 

[mariodileep]

Thanks for your response. Its clear now.

 

[waross]

Another issue is the phase angle of the current. Normally as the load decreases the phase angle increases and the power factor degrades. If you use the power factor and Pythagoras' theorem to resolve the in phase current and the quadrature current, take half the in phase current and then use Pythagoras again to calculate the new current at half load, you still may not be exact but you will be much closer to the actual current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Did you notice that the 50% power factor was given in the original post? No need to estimate it.


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(2B)+(2B)' ?

 

[waross]

Hi Pete. I understood the question as why the relationship was not linear, not what is the power factor at 50% load. I was trying to present a fairly simple equation that will come close to reconciling the full load current with the 1/2 load current.
I presented this more as an approximation and explanation than an in-depth rigorous solution that may leave some readers lost in the numbers.
Yours
Bill

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

There is a non-linear relationship, quite true and it is captured in the 0.78pf and the 0.57pf. And the op has in fact applied these pf numbers to come up with:
I_100 =746*HP /(1.73*V*Eff*PF)= (746*2/(1.73*575*0.86*0.78) )=2.2
I_50 =746*HP /(1.73*V*Eff*PF)=(746*2/(1.73*575*0.86*0.57) ) = 3.0

What's wrong with this calculation has nothing to do with the value of pf plugged in and everything to do with the value of HP plugged in. jmontero nailed it straightaway. The only reason I chimed in is that our posts were so close in time that I never saw his.


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[electricpete]

For OP, by the way, if there is a different value of 50% efficiency available, that should of course also be used in the P_50 calculation.

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[electricpete]

By the way, I'm not trying to prevent you or anyone from adding info. I just thought you had misunderstood the op and would appreciate a heads up. If not, my apologies, feel free to add whatever info you think will help.

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