Wildi: Ind. Motor

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Sthabik PE

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If i change the P (hp) to P (watt) and use the formula P=sq.rt(3) x VI, the value of current will be 93.6 A.

This is different from the answer in the example as, I=600Ph / E.

Am i missing something here.

Cuz in exam for these kind of question, the first thing i will remember is P=sq.rt(3) x VI.

Please help!

image.png

 
I hear ya...  I remember being annoyed with this as well, I think I saw the exact same thing/equation in Graffeo?

What I would say, is I would expect any type of question regarding motor current to be an NEC lookup; which in your case, using the NEC I see the FLC = 118A.   I think most of the NCEES sample exam questions involving motors were direct lookups from the NEC.

 
Sdhabik said:
If i change the P (hp) to P (watt) and use the formula P=sq.rt(3) x VI, the value of current will be 93.6 A.

This is different from the answer in the example as, I=600Ph / E.

Am i missing something here.

Cuz in exam for these kind of question, the first thing i will remember is P=sq.rt(3) x VI.

Please help!

View attachment 12506
Click to expand...
In your calculation you did not consider power factor and efficiency. You just used the real power (W), that's why you got 93.6 Amps . You have to convert that to apparent power(VA). 600 constant from Wildi's book includes the power factor & efficiency. 

 

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