|> | As to the open delta connection, assuming a balanced secondary load of | 100% |> | power factor, one of the transformers will see an 86.6% lagging PF and | one |> | will see 86.6% leading. If we assume the case of 2 - 1kVA transformers, | we |> | see that the maximum power available will be 2 kVA X .866 = 1.732 kVA. | If we |> | add the third transformer, however, the power factor in all units is now |> | 100%, for a capacity of 3 kVA, as against 1.732 kVA for two | transformers. |> | Adding 50 % more transformer capacity gives 73.2% more bank capacity. |> | |> | That is the story on the "reactive power" in the open delta connection. |> | Obviously, for anything other than unity PF, the voltage drops on the | two |> | transformers will be different, causing an unbalance which would not | exist |> | with the full delta configuration. |>
|> In cases of getting single phase from three phase, I've seen these wiring |> configurations: |>
|> * * * |> / \ -or- / \ / \ |> * *---* * *---*---* * |> | | | | | | |>
|> * * |> / \ / \ |> *---*---* |> | | | |>
| | I haven't! These look like questions out of an engineering text, but I can't | imagine why anybody would actually hook anything up that way!! No matter how | you slice it, a single phase load is a single phase load.
Some of these are taken from actual generator specifications. See the single phase columns on the right side of the tables on page 6 of the folloring:
| Think about it; if I take any of those contraptions and put single phase | into the secondary, will I get three phase out of the primary? (Hint: | NO!)Why not? Now, understanding that transformers are reversible, do these | hookups REALLY convert 3 phase to single phase? NO. They just use 3 phases | worth of hardware to get a single phase output. Might as well have used a | single phase transformer to begin with.
There really is single phase output from three phase input. The catch is that the current phase remains single. Half the load is on one phase and the other half the load is split on the other 2 phases at what appears to me to be a 0.5 power factor, one leading, one lagging.
|> Wouldn't the first one be putting a 50% power factor on two phases? But | it |> seems the 2nd one wouldn't really be any better, other than sharing the | load. |> But all load on 2 of the phases is always going to be at 60 degrees (one |> leading, one lagging). | | | All three configurations are ridiculous.
But are used in real practice.
|> Also, in WYE configuration, if a single phase load is put on phases A-B | in: |>
|> -A |> \ |> N--C- |> / \ |> -B |>
|> wouldn't there be an 86.6% power factor on A and B as passed back through |> a delta primary to the supply? Would a delta-delta with a zig-zag neutral |> be a way around this or would the imbalance distort the neutral? | | The secondaries will see the 86.6 PF, but in the delta primary you will see | one winding at 100% PF and two windings with no load.
If the single phase connection is between A and B (getting 208, 400, or 480 volts depending on whether you are using 208Y/120, 400Y/231, or 480Y/277), then there would be current on the A-N winding, and on the B-N winding.
Now look at the delta primary for that:
\ / \ /
The horizontal winding across the top is on the same core as C-N, so I will label it as "Cp". The left side (backslashes) is on the same core as A-N, so I will label it as "Ap". The right side (slashes) is on the same core as B-N, so I will label it as "Bp".
Since there is current on A-N, and B-N on the secondary, that current is thus present on the "Ap" and "Bp" windings. The phase angle of the current is vertical in this diagram (because the line from A to B in the secondary vector diagram is vertical). So the currents on "Ap" and "Bp" are vertical. That would mean one is leading and one is lagging, by 30 degrees. So I say the power factor on the delta primary is 0.866, and current is divided between "Ap" and "Bp" (none on "Cp").
Now, where it gets interesting is if that delta primary is attached to a wye secondary of another delta-wye transformer upstream. Since two phases of the delta have current (load), that means on the upstrea wye, two legs have one each of those loads, and one leg has both. Because a delta is connected to as wye, there is a 30 degree difference between the winding angles. Now you will end up with most of the current on one phase with a power factor of 1, but some (I'll have to go draw the vectors and figure out the amounts) on the other two phases with 60 degree (0.5 pf) angles, one leading and one lagging.
That's what I say will be happening.