Home › Electrical Engineering Forum › General Discussion › Three phase induction generator on a single phase supply
- This topic has 0 replies, 1 voice, and was last updated 14 years, 3 months ago by dec335.
-
AuthorPosts
-
2010/10/11 at 8:28 pm #10240dec335Participant
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
-
AuthorPosts
- You must be logged in to reply to this topic.