# Three phase induction generator on a single phase supply

Discussion in 'The Projects Forum' started by dec335, Oct 11, 2010.

1. ### dec335 Thread Starter New Member

Oct 11, 2010
3
0
Hi,
Im really stuck on a problem for my final year project in my taught Masters degree and any suggestions what might be going wrong would be greatly appreciated.
Im trying to run a three phase delta connected induction generator on a single phase supply using a Steinmetz connection. The Steinmetz connection is a single capacitor connected across phase B or C depending on which direction the machine is running with phase A connected to the single phase supply. Ive been using the book Distributed Generation by Chan and Lai to get the formulas shown below. It says in the book for a balanced machine the following applies.

Yp = 1/Zp =|Yp|∠-ϕp
Where Yp is the positive sequence admittance of the generator and ϕp is the positive sequence impedance angle.

B1 = 2Yp sin⁡(2π/3  ϕp)
B2 = 2Yp sin⁡(ϕp - π/3)
B = 1/Xc
C1 = B1/(2.π.f)
C2 = B2/(2.π.f)

I ran the machine as an induction motor first and a certain calculated speed it had an impedance of Z = 124.26∠60 (62.2 + j107.6) and this gave a value of Yp as 0.0085∠-60, B1 = 0 and B2 = -0.0139388. A modular value of B2 was used to calculate a positive capacitance value and this was calculated as 0.00004439 which is equivalent to 44μF. This capacitance was placed on the motor across phase B and it worked really well. It was able to run with a load and all the phases were in balance with the voltages and currents very close together. (The figures have been rounded up for these examples as I have been using an excel sheet that has much longer values in each cell)

I tried to run the machine as an induction generator and a certain calculated speed it had an impedance of Z = 104.41∠120 (-52.26 + j90.4)(the real value is negative as it has a negative slip) and this gave a value of Yp as 0.00958∠-120, B1 = -0.016588 and B2 =0. A modular value of B1 was used to calculate a positive capacitance value and this was calculated as 0.00005282 which is equivalent to 52μF. This capacitance was placed on the machine across phase B and it wouldnt work properly. There was a lot of imbalance in the currents and voltages.
Would anyone have any idea of why it is not working for the induction generator the same as for the induction motor as all the literature I have read has said that an induction generator behaves exactly the same as an induction motor but it is just run faster? I have been doing tests by trial and error but I have to give a reason why it is not working. If any more information is required, just post a question and I will answer it.
Thanks
Declan

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2. ### Dyslexicbloke Active Member

Sep 4, 2010
420
19
'O' ... now I'm worried. I just asked a similar question, I think, with only a tiny insight into the issue if this is anything to go on.

I was hoping to use a 3 phase motor to generate single phase output but in island mode ....
I'v been told its possible but havnt a clue how.
Should I follow this thread or are the two problems not the same?

Sorry to be asking rather than helping but you clearly have a far greater understanding than I do.
(Any answer to my comment should be in 'My Thread' so this one isnt hijacked)

Al

3. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,852
968
You need another capacitor for the generator scheme. It's a C-2C format. The second cap is twice the value of the first. The output comes off the C cap and the one in your drawing becomes the 2C cap.

give'er another try

4. ### dec335 Thread Starter New Member

Oct 11, 2010
3
0
Kermit2,
Thanks for the reply. It is the best one that I have gotten so far on any of the forums. I have read the paper before that you posted on Dyslexicbloke query about the small hydro scheme. That is for island operation. Do you think it would work for my machine seeing it is connected to the grid? Have you ever seeing it working yourself by any chance? I'm going to give it a try when I head back into the lab on Friday.
Declan

Feb 5, 2010
3,852
968