3-Phase Delta Currents:
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jtester
Senior Member
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- Las Cruces N.M.
Re: 3-Phase Delta Currents:
Yes you will. You have two parallel paths between the two phases of your system. One path is a single transformer, the other path is two transformers in series. The current splits depending on power factor of the load, which phase the load is hooked to, etc., but the general rule of thumb is 2/3 of the current goes thru the single transformer, and 1/3 of the current goes thru the other path, namely the two transformers in series.
Jim T
Yes you will. You have two parallel paths between the two phases of your system. One path is a single transformer, the other path is two transformers in series. The current splits depending on power factor of the load, which phase the load is hooked to, etc., but the general rule of thumb is 2/3 of the current goes thru the single transformer, and 1/3 of the current goes thru the other path, namely the two transformers in series.
Jim T
Ed MacLaren
Senior Member
Re: 3-Phase Delta Currents:
With only one load connected across any two lines, the RMS current in each phase of the secondary winding is 58% of the line current (taken by the single load).
Ed
[ December 05, 2004, 01:18 PM: Message edited by: Ed MacLaren ]
If you mean that the secondary is delta connected, it is not really possible to "apply a load to only one secondary winding".
With only one load connected across any two lines, the RMS current in each phase of the secondary winding is 58% of the line current (taken by the single load).
Ed
[ December 05, 2004, 01:18 PM: Message edited by: Ed MacLaren ]
Re: 3-Phase Delta Currents:
Touche' Ed, but you knew what I meant. I thought there would be current in all windings, but it seems that the current division would be controlled by the leakage reactance and resistance of the xfrmr. It seems that a 2:1 division would be more likely. Your example seems to be for an ideal transformer in which case 58% is plausible. I don't do vectors on Sunday!
[ December 05, 2004, 01:40 PM: Message edited by: rattus ]
Touche' Ed, but you knew what I meant. I thought there would be current in all windings, but it seems that the current division would be controlled by the leakage reactance and resistance of the xfrmr. It seems that a 2:1 division would be more likely. Your example seems to be for an ideal transformer in which case 58% is plausible. I don't do vectors on Sunday!
[ December 05, 2004, 01:40 PM: Message edited by: rattus ]
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: 3-Phase Delta Currents:
I disagree with the example that Ed used. The current will split by the square root of three, as he showed, if we are talking about 3 phase currents. As a general rule, for the single phase, the current will split by the ratio of 2:1 if the transformer impedances are equal. Twice the impedance in one path as in the other.
I disagree with the example that Ed used. The current will split by the square root of three, as he showed, if we are talking about 3 phase currents. As a general rule, for the single phase, the current will split by the ratio of 2:1 if the transformer impedances are equal. Twice the impedance in one path as in the other.
Re: 3-Phase Delta Currents:
jtester, I think Ed's explanation is correct for ideal xfrmrs which are quite scarce. The two "unloaded" phase voltages in series add up to the "loaded" voltage. This is equivalent to two identical windings in parallel in which case the current would split equally. But, there would be reactive components which must cancel. It seems that the two currents would have equal real parts and equal but opposite imaginary parts. Too much work to compute that on a Sunday, unless I can get my demo PSPICE to work.
jtester, I think Ed's explanation is correct for ideal xfrmrs which are quite scarce. The two "unloaded" phase voltages in series add up to the "loaded" voltage. This is equivalent to two identical windings in parallel in which case the current would split equally. But, there would be reactive components which must cancel. It seems that the two currents would have equal real parts and equal but opposite imaginary parts. Too much work to compute that on a Sunday, unless I can get my demo PSPICE to work.
Ed MacLaren
Senior Member
Re: 3-Phase Delta Currents:
Hey Jim, with just a single phase load connected, I'm thinking you may be right.
It would be interesting to hook up three small transformers with the secondaries connected delta and check that out.
Does anyone else have any ideas on this?
Ed
Hey Jim, with just a single phase load connected, I'm thinking you may be right.
It would be interesting to hook up three small transformers with the secondaries connected delta and check that out.
Does anyone else have any ideas on this?
Ed
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: 3-Phase Delta Currents:
Much as I'd like to take credit, this idea of the 2:1 split is published in many transformer manuals, particularly one handed out to power engineers in the '70's. It's called DISTRIBUTION TRANSFORMER MANUAL and it's by GE. I don't know if it is still available.
It talks about different general approximations for loadings of the various single and three phase configurations.
Jim T
Much as I'd like to take credit, this idea of the 2:1 split is published in many transformer manuals, particularly one handed out to power engineers in the '70's. It's called DISTRIBUTION TRANSFORMER MANUAL and it's by GE. I don't know if it is still available.
It talks about different general approximations for loadings of the various single and three phase configurations.
Jim T
Re: 3-Phase Delta Currents:
A Little Thought Experiment:
Consider an ideal, 3-phase xfrmr in the delta configuration, and define the angle of phase voltage A to be zero.
Now disconnect the phase A secondary and connect a resistive load, 2R, across phase A. The load current is:
Va/2R, at angle zero.
Now connect another resistor of value 2R across the phase B and C secondaries.. This voltage is equal to Va in magnitude and phase. Then the load current is:
Va/2R at angle zero.
Now, reconnect the phase A secondary. Now the total load current is
Va/R at angle zero.
So, it would seem that in this special, hypothetical case which cannot be demonstrated experimentally, that the currents in all three windings would be equal to half the total load current.
What think you?
A Little Thought Experiment:
Consider an ideal, 3-phase xfrmr in the delta configuration, and define the angle of phase voltage A to be zero.
Now disconnect the phase A secondary and connect a resistive load, 2R, across phase A. The load current is:
Va/2R, at angle zero.
Now connect another resistor of value 2R across the phase B and C secondaries.. This voltage is equal to Va in magnitude and phase. Then the load current is:
Va/2R at angle zero.
Now, reconnect the phase A secondary. Now the total load current is
Va/R at angle zero.
So, it would seem that in this special, hypothetical case which cannot be demonstrated experimentally, that the currents in all three windings would be equal to half the total load current.
What think you?
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: 3-Phase Delta Currents:
Rattus
If I understand your description right, you have connected 3 single phase loads across all 3 phases, one across each phase, AB, BC, and AC. That would give you the equivalent of a balanced 3 phase load. When you say you are connecting a resistance to phase A, what do you connect the other end to, B, C, or what?
I suggest that you disconnect phase A and connect the single phase load across BC. By looking at the two paths between Phases B and C you can see one path is across transformer BC and another is across the two transformers BA and AC in series. Remember that Phase A is disconnected, actually Phase A doesn't enter ito it at all, you can connect it or not.
The impedance of the three transformers will determine the current split, if all 3 transformers are the same size, have the same impedance, the current would split 2/3 thru the transformer BC and 1/3 thru the two in series, BA and AC.
Rattus
If I understand your description right, you have connected 3 single phase loads across all 3 phases, one across each phase, AB, BC, and AC. That would give you the equivalent of a balanced 3 phase load. When you say you are connecting a resistance to phase A, what do you connect the other end to, B, C, or what?
I suggest that you disconnect phase A and connect the single phase load across BC. By looking at the two paths between Phases B and C you can see one path is across transformer BC and another is across the two transformers BA and AC in series. Remember that Phase A is disconnected, actually Phase A doesn't enter ito it at all, you can connect it or not.
The impedance of the three transformers will determine the current split, if all 3 transformers are the same size, have the same impedance, the current would split 2/3 thru the transformer BC and 1/3 thru the two in series, BA and AC.
Re: 3-Phase Delta Currents:
tjester,
Only 2 loads. If we label the three phases A, B, & C, then I disconnect one of the secondaries, say AB, and let us call this isolated secondary A'B'. Then I connect a load, 2R, across A'B', and I connect a second load, 2R, across AB. Line C is not connected. Since Va'-Vb' = Va - Vb in magnitude and phase, the currents in the two loads will be equal in magnitude and phase. Then if I connect A to A' and B to B', the load is now R, and the windings AC and CB act as one to provide half the current, and the winding AB provides the other half.
Since this is an IDEAL xfmr, no resistance, no leakage reactance, no iron loss, there is nothing to impede the current. As I say, you cannot prove this in the lab because such xfrmrs are not available. The 1/3, 2/3 rule of thumb applies to real xfrmrs, not ideal xfrmrs.
tjester,
Only 2 loads. If we label the three phases A, B, & C, then I disconnect one of the secondaries, say AB, and let us call this isolated secondary A'B'. Then I connect a load, 2R, across A'B', and I connect a second load, 2R, across AB. Line C is not connected. Since Va'-Vb' = Va - Vb in magnitude and phase, the currents in the two loads will be equal in magnitude and phase. Then if I connect A to A' and B to B', the load is now R, and the windings AC and CB act as one to provide half the current, and the winding AB provides the other half.
Since this is an IDEAL xfmr, no resistance, no leakage reactance, no iron loss, there is nothing to impede the current. As I say, you cannot prove this in the lab because such xfrmrs are not available. The 1/3, 2/3 rule of thumb applies to real xfrmrs, not ideal xfrmrs.
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