Wildi: Ind. Motor

[Sthabik PE]

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!

 

[a4u2fear]

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.

 

[EE2PE]

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

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.