Electrical Machines 6th ed by Wildi

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eksor_PE

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I'm studying for the upcoming October PE-Power exam. I am using Wildi's Electrical Machines as one of my references. I am confused on one of the sections, which is Chapter 13.12 on Estimating the currents in an induction motor. In Example 13-4, the full-load current of the 3-phase induction motor was calculated with the following: I = ((600x500))/2300. Why the sqrt(3) was not included if this is a 3-phase calculation? Is this an error?

 
That is confusing. I'm not sure where the 600 comes from either. I've always calculated current for a 3PH motor to be: (746 x HP) / (1.73 x V x eff x pf)

 
I think the 600 just comes from combining up the efficiency, PF, √3, and KW to HP conversions into an empircal constant given in equation 13.5 on page 277. If you want to make some assumptions simialr to Kovz above, say efficiency of 0.95 and Pfactor of 0.9 the KVA in then is (500*0.746)/(0.95 *0.9) = 436kVA input and therefore line current = 436kVA/ (√3 *2300) = 110 Amps (a little off the 130 solution but of course depends on motor size etc.

Note this is for a large motor... if you would choose parameters for say a 10 HP motor of 90% efficiency and 80% PF, the above would give right on the 130 Amps.

I guess for completeness, the 600 is approximately ~ 746/(√3 *0.8 *0.9)

If you are comfortable with the above I don't think it is important to use the Wildi equation.

 
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I think the 600 just comes from combining up the efficiency, PF, √3, and KW to HP conversions into an empircal constant given in equation 13.5 on page 277. If you want to make some assumptions simialr to Kovz above, say efficiency of 0.95 and Pfactor of 0.9 the KVA in then is (500*0.746)/(0.95 *0.9) = 436kVA input and therefore line current = 436kVA/ (√3 *2300) = 110 Amps (a little off the 130 solution but of course depends on motor size etc.

Note this is for a large motor... if you would choose parameters for say a 10 HP motor of 90% efficiency and 80% PF, the above would give right on the 130 Amps.

I guess for completeness, the 600 is approximately ~ 746/(√3 *0.8 *0.9)

If you are comfortable with the above I don't think it is important to use the Wildi equation.


Agreed. And good explanation.

 
I think the 600 just comes from combining up the efficiency, PF, √3, and KW to HP conversions into an empircal constant given in equation 13.5 on page 277. If you want to make some assumptions simialr to Kovz above, say efficiency of 0.95 and Pfactor of 0.9 the KVA in then is (500*0.746)/(0.95 *0.9) = 436kVA input and therefore line current = 436kVA/ (√3 *2300) = 110 Amps (a little off the 130 solution but of course depends on motor size etc.

Note this is for a large motor... if you would choose parameters for say a 10 HP motor of 90% efficiency and 80% PF, the above would give right on the 130 Amps.

I guess for completeness, the 600 is approximately ~ 746/(√3 *0.8 *0.9)

If you are comfortable with the above I don't think it is important to use the Wildi equation.


Thank you for the explanation. I guess my concern was the usage of said formula by not knowing the equivalent of the "600" value and plugging the sqrt(3) for the 3-phase system.

 
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