Find the field Current of synchrous Motor

[kduff70]

I have been working problems for the PE test in October and came across this problem in Kalpan  practice problem. Do anybody have a good ideal on how to solve this problem. Kaplan has a formula but it doesn't seem to work .  I have attached the problem . any help on this one will be greatly appreciated

 

[kduff70]

I attach the Kaplan solution to help make sense of the question

View attachment SKM_364e17091116110.pdf

 

[rg1]

I will do it myself in the evening, but meanwhile the process can be excitation voltage at rated values can be found. This value of Eo is proportional to 6A. (assuming no saturation). Now find Eo with 90 leading rated current. The excitation current for this Eo can be calculated by unitary method , that must be the answer. By seeing the options answer should be d- 9.6A because motor is overexcited to work as capacitor.

 

[rg1]

I will do it myself in the evening, but meanwhile the process can be excitation voltage at rated values can be found. This value of Eo is proportional to 6A. (assuming no saturation). Now find Eo with 90 leading rated current. The excitation current for this Eo can be calculated by unitary method , that must be the answer. By seeing the options answer should be d- 9.6A because motor is overexcited to work as capacitor.

A small correction. It is not at rated current while working as capacitor; it is working at same reactive power that means at reactive component of 125A. The answer may not be 9.5A.

 

[kduff70]

RG1 any better understanding of the problem?

 

[Omer]

the answer provided by the book is not clear at all.

as suggested by @rg1 , you need to calculate the excitation voltage based on the rating values first:

Eg= Vt-I*X

=460/sqrt(3) - [email protected] * 2i= 461@-25

at the rated condition the VAR generated=

sqrt(3)*460*125*sin (36.9)= 60 KVAR

since the problem state capacitor like with negligible mechanical load and same reactive power, then the total apparent power will be the same as reactive power of 60 KVA.

the current in this case is

60000/(sqrt(3)*460)=75@90

then again you calculate the excitation voltage based on the new current

=460/sqrt(3) - 75@90 * 2i= 415 V

now since excitation voltage is directly proportional to the field current, the new field current will be:

6*415/461=5.4 

 

[rg1]

RG1 any better understanding of the problem?

I am sorry did not get time last evening.Thanks Omer.  @Omer has got it right.(Not verified the calculations).

Now coming to better explanation, let me try-

Any synchronous machine ( motor or Genr) can be represented by its equivalent ckt. which has an internal generated Voltage-Eo ( which is generated by the action of field current- in this case 6A). There is one more Voltage which is terminal voltage of the Machine-In this case it is 460/sqrt3 ( Single phase is better understandable). The synchronous reactance is the internal impedance ( like internal impedance of any other machine ). The whole game of this chapter revolves around these three things. Rest is application of network theory. When the question gives us field current- it points out something towards Eo. In real sense Eo is not always proportional to field current due to saturation of the magnetic ckt of the machine. But because no other info is given, we tend to assume that Eo is proportional to Field current. So first we find Eo by applying KVL Vt-IX=Eo which is 461(Omars calculations) for a field current of 6A.

Next the questions asks field current for another operating point of the same machine at same terminal Voltage- ie. working as capacitor at same reactive power of 60KVAR. So we again find new Eo by using new current ( 75 leading Voltage by 90 deg); that comes out to be 415V so a lesser Eo will need lesser Field current  of 5.4A.

I do not know if it has made sense to you. If you tell me how much you know or exactly where is the doubt, I can explain in a different way taking ahead from there.

 

[kduff70]

thank you this was the way I thought this problem would have work out by finding the Eg  but the way  the problem was work on the solution had me confused

 

[rg1]

thank you this was the way I thought this problem would have work out by finding the Eg  but the way  the problem was work on the solution had me confused

yup. I do not know , but yes, there is lot of confusion around here. 

 

[FPar]

the answer provided by the book is not clear at all.

as suggested by @rg1 , you need to calculate the excitation voltage based on the rating values first:

Eg= Vt-I*X

=460/sqrt(3) - [email protected] * 2i= 461@-25

at the rated condition the VAR generated=

sqrt(3)*460*125*sin (36.9)= 60 KVAR

since the problem state capacitor like with negligible mechanical load and same reactive power, then the total apparent power will be the same as reactive power of 60 KVA.

the current in this case is

60000/(sqrt(3)*460)=75@90

then again you calculate the excitation voltage based on the new current

=460/sqrt(3) - 75@90 * 2i= 415 V

now since excitation voltage is directly proportional to the field current, the new field current will be:

6*415/461=5.4 

Great job Omer! Thanks!