Open Delta

[sdilucca]

Why does an open delta provide 58% power instead of 2/3 the power of a regular delta?

 

[mpparent]

Think about what the difference is between the open delta xfmr, and it's line currents compared to what the "delta currents" would be. Then compare that to a regular delta. A standard delta's line currents differ by root 3 of the "delta currents". Make sense?

Mike

 

[sdilucca]

What I know about open delta is:

I-phase = I-line

V-phase = V-line

For regular delta:

1.73*I-phase = I-line

V-phase = V-line

Is this correct?

 

[waross]

In a Closed Delta the line current is shared by two windings.
When used in Open Delta, the transformers must be derated 86.6% to avoid overloading. After you have derated the transformers 86.6% you multiply by 2/3.

Look at it another way;
One 33 Kva transformer = 33 KVA
Two 33 1/3 Kva transformers, derated for open delta service.
66 2/3 Kva x 86.6% = 57.7 Kva
Three 33 1/3 Kva transformers = 100 Kva.

 

[cuky2000]

In an open-delta (V) connection, the line current is limited to the phase current of I_line/[√]3 to avoid transformer overload. The maximum power available is:

a) Two transformer (Open Delta): P_V=[√]3E_linex I_line/[√]3 = E_linex I_line.

b) Three transformer (Closed Delta): P_[Δ]=[√]3E_linex I_line .
P_V/ P_[Δ]= 1/[√]3= 57.7%

 

[sdilucca]

thanks everyone... I also found a great reference in the book "Principles of Electric Machines and Power Electronics" pg.75-78. It's all about those sneaky phase shifts.

 

[mpparent]

You should give Cuky a star for the answer. I suggest you draw out the circuit and work it out long-hand. If you assume a voltage of 1<0, 1<-120 and 1<120 and impedances of 1<0, you can easily determine what the power is on an open delta system. Then use the simple 3 phase power equation for a closed delta system. You'll then find the ratio Cuky gave you.

Mike