Electric Machinery and Power System Fundamentals (Chapman) Problem 7-7

[the_r0b]

I don't quite understand why they solved Part (c) of the following problem that asks for the Air-Gap power P_ag the way they did.



They compute the air gap power in part (c) P_ag using the Thevinin equivalent circuit of the rotor and the line current solved for in part (a). See below:



The stator copper loss is calculated to be 1180 W and the stator core loss is given as 200 W.

The line current as solved is 42.3∠-25.7

My question is:
If I saw this problem on the exam, I would just use the formula P_ag = Pin - P_SCL - P_Core =√3*208*42.3*cos(25.7) - 1180 W - 200 W =12.35 kW, which is off the solution's P_ag of 12.54 kW by nearly 200 W.

What is the difference between these solutions? I was using the figure below from the PE Power Exam Handbook when solving this myself and am wondering why the air gap power isn't just input power minus the stator losses.


Thank you in advance!

 

[Zach Stone P.E.]

Hi @the_r0b, you raise an interesting question.

I find two things interesting about this.

1. The difference in the air gap power (PAG) calculated using the book's method, vs calculated using P_AG = P_in - P_SCL - Pcore is the same value as given core losses of P_core = 200W.

2. Although the problem gave the value of the stator core loss (P_core = 200W), it did NOT give the value of the core loss resistance (Rc). Notice that the magnetizing branch in the solution only includes the magnetizing reactance (Xm).

If Rc was included in parallel when calculating Zf in the circuit, it would decrease the overall value of Zf (impedances in parallel decrease the overall equivalent impedance, unlike impedances in series which increase the overall impedance).

If Zf is actually smaller if Rc was included in the circuit (and given in the problem), then the value of the line current (IA) would increase according to ohm's law (I=V/Z).

if IA is actually larger due to Rc being included in the circuit, then the input power would also be larger from Pin = √3•IA•VL•PF.

I'm willing to bet if we had the actual value of Rc and included it in the circuit, then the math would come out exact.

I'm guessing this is more of a "close approximation" of the air gap power and other circuit values without knowing the value of the core resistance.

 

[the_r0b]

Hi @the_r0b, you raise an interesting question.

I find two things interesting about this.

1. The difference in the air gap power (PAG) calculated using the book's method, vs calculated using P_AG = P_in - P_SCL - Pcore is the same value as given core losses of P_core = 200W.

2. Although the problem gave the value of the stator core loss (P_core = 200W), it did NOT give the value of the core loss resistance (Rc). Notice that the magnetizing branch in the solution only includes the magnetizing reactance (Xm).

If Rc was included in parallel when calculating Zf in the circuit, it would decrease the overall value of Zf (impedances in parallel decrease the overall equivalent impedance, unlike impedances in series which increase the overall impedance).

If Zf is actually smaller if Rc was included in the circuit (and given in the problem), then the value of the line current (IA) would increase according to ohm's law (I=V/Z).

if IA is actually larger due to Rc being included in the circuit, then the input power would also be larger from Pin = √3•IA•VL•PF.

I'm willing to bet if we had the actual value of Rc and included it in the circuit, then the math would come out exact.

I'm guessing this is more of a "close approximation" of the air gap power and other circuit values without knowing the value of the core resistance.

Thanks, Zach! The only way this makes sense to me is if there is some losses that aren't accounted for so I think you are probably correct. Another interesting thing that didn't add up to me is further in step (f) they subtract the Core losses from P_Conv in addition to the friction losses P_Mech.




This formula double-counts the core losses, does it not? I don't think any of my other AC machines references solve output power this way. I can only conclude either the author is mistaken or they are accounting for the missing Rc Resistive core loss here that you mentioned.

 

[Zach Stone P.E.]

Thanks, Zach! The only way this makes sense to me is if there is some losses that aren't accounted for so I think you are probably correct. Another interesting thing that didn't add up to me is further in step (f) they subtract the Core losses from P_Conv in addition to the friction losses P_Mech.

View attachment 27133
View attachment 27134

This formula double-counts the core losses, does it not? I don't think any of my other AC machines references solve output power this way. I can only conclude either the author is mistaken or they are accounting for the missing Rc Resistive core loss here that you mentioned.

That is interesting.

I'm assuming that by Pmech, Chapman is referring to the mechanical losses from friction, windage, and stray losses, or: Pmech = Pf+w + Pmisc

If Chapman is calculating the output power (Pout) from the converted power (Pconverted), then yes I agree the core losses (Pcore) would already have been accounted for when converting from input power (Pin) to air gap power (PAG), along with also account for the stator copper losses (PSCL) according to the induction motor power flow diagram from his book.

Looking at the power flow diagram, it should just be:

Pout = Pconverted - (Pf+w + Pmisc)
Pout = Pconverted -Pmech