Synchronous Generator Homework: Finding Required Field Current and Input Torque

[Brianrofl]

Homework Statement

The Attempt at a Solution

First up, part a.

The first one is really simple. The generator is working at its rated voltage of 13.8kV, right? So, would the required field current not be just 225 Amps?

Next, part b.

To sort of explain what's going on, I first found the saturated reactance of the generator by using the data from the table (for 175 field amps). Then, I draw out a per-phase circuit using 13.8kV for the voltage source, the reactance Xs, and the load in Y-connection. I used 13.8kV because the instructions say the generator operates at rated voltage, correct? [Divide by sqrt 3 to convert to line-to-neutral voltage].

Using that, the current is found easily.

Which then translates into finding the input torque rather easily.

I'd appreciate any input. I think I know how to solve this problem but it seemed too easy, and there are just so many things you can make mistakes on doing 3 phase systems that I just want to make sure.

Thanks.

 

[cnh1995]

The first one is really simple. The generator is working at its rated voltage of 13.8kV, right? So, would the required field current not be just 225 Amps?

I believe the excitation will be higher. For 225A field current, "open circuit" voltage is 13.8kV. This voltage will sag once you connect the load, due to the synchronous impedance. From the OC and SC data given, you can calculate synchronous impedance Zs and use E=V_terminal+I*Zs.
You can calculate current I from the load. For the value of E(generated emf), you can find the field current required by interpolation-extrapolation technique using Voc vs field current table.

 

[Brianrofl]

Ok, I'll work it out again and see what I get. Are there any issues with my part b?

 

[cnh1995]

Ok, I'll work it out again and see what I get. Are there any issues with my part b?

Your approach looks right to me. I haven't verified the calculations, but I'm sure they are correct.

 

[Brianrofl]

Your approach looks right to me. I haven't verified the calculations, but I'm sure they are correct.

I'm really getting hung up on one thing. In the problem, he states: "while operating with the rated voltage." Wouldn't that just mean that the generator voltage is the 13.8kV line-to-line?

I'm referencing a lecture problem where when solving for the field current, you're given the voltage at the load and the power at the load. Using those two values, you can solve for the current, and then find the voltage dropped at the synchronous impedance. You then add the load voltage to this impedance voltage and find the voltage Ea supplied by the generator. In my homework problem's case, wouldn't my voltage Ea just be 13.8kV line-to-line?

I did try putting the voltage given (13.8kV) across the load, then calculated current, then found the generator voltage from there, which gave me a field current of ~300 amps.

So many nit picky things in this material, and I really hate turning in problems for credit. I'd so much rather just be given a solution and figure out how to do it.

 

[cnh1995]

Wouldn't that just mean that the generator voltage is the 13.8kV line-to-line?

It is terminal voltage of the generator. When you connect a load, this voltage reduces due to armature reaction. To bring it back to the rated value, field excitation is raised.

 

[Brianrofl]

It is terminal voltage of the generator. When you connect a load, this voltage reduces due to armature reaction. To bring it back to the rated value, field excitation is raised.

Like this?

Where I find my new value of Ea at which the voltage across the terminals is still 13.8kV?

 

[cnh1995]

Like this?

View attachment 98562

Where I find my new value of Ea at which the voltage across the terminals is still 13.8kV?

Yes.