#### the_r0b

##### New member

- Joined
- Feb 27, 2022

- Messages
- 2

- Reaction score
- 0

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!

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!