The Two Legged Wye:
- Thread starter rattus
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Re: The Two Legged Wye:
OK, OK, I'll bit!!
With reference to a "wye" secondary transformer, the L-L kva is 1/3 of the kva rating of the transformer. I.e., a 45kva 480v-208y/120 transformer can have a single phase loading of no more than 15kva. So a 3ph transformer can by applied to a single phase load by using two of the phases, such as X1-X2 with a maximum load of 15kva without any problems.
Dave T
OK, OK, I'll bit!!With reference to a "wye" secondary transformer, the L-L kva is 1/3 of the kva rating of the transformer. I.e., a 45kva 480v-208y/120 transformer can have a single phase loading of no more than 15kva. So a 3ph transformer can by applied to a single phase load by using two of the phases, such as X1-X2 with a maximum load of 15kva without any problems.
Dave T
Re: The Two Legged Wye:
Still waiting for someone to prove the 57.7%.
Dave, I would argue that each transformer can deliver 15KVA under any load conditions. In the case of a single L-L load on the wye, the two transformers can deliver 30KVA, but this apparent power must be computed at the transformer, not at the load. This goes back to the discussion on the Oregon Fudge Factor, OFF. Remember, our primary concern is to correctly calculate the transformer loading.Originally posted by templdl:
With reference to a "wye" secondary transformer, the L-L kva is 1/3 of the kva rating of the transformer. I.e., a 45kva 480v-208y/120 transformer can have a single phase loading of no more than 15kva. So a 3ph transformer can by applied to a single phase load by using two of the phases, such as X1-X2 with a maximum load of 15kva without any problems.
Dave T
Still waiting for someone to prove the 57.7%.
- Location
- Mission Viejo, CA
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- Professional Electrical Engineer
Re: The Two Legged Wye:
Edit Add : I should have shown "Base" as Sqrt 3 x (Sqrt 3xI) x E = 3IE for a "balanced" secondary load.
[ March 23, 2005, 10:51 AM: Message edited by: rbalex ]
Edit Add : I should have shown "Base" as Sqrt 3 x (Sqrt 3xI) x E = 3IE for a "balanced" secondary load.
[ March 23, 2005, 10:51 AM: Message edited by: rbalex ]
Re: The Two Legged Wye:
Bob,
Thanks for your response, but I must nitpick:
You have implied a 1:1 turns ratio, then the secondary and primary phase currents would be Iline/sqrt(3).
(OK, I see you fixed it)
However, the factor is still 0.577, the inverse of sqrt(3). It appears that this factor does apply to the two-legged wye and provides the allowable connected load as claimed by Jim D. previously. His numbers had demonstrated that already.
Now if we connect two single phase loads directly across the two secondaries, the ratio becomes 0.667 because we now have 240V instead of 208V driving the loads.
In either case though, the apparent power computed at the transformers must be the same, and it is. Whether you add these apparent powers or not is immaterial.
And what is the ratio of the two ratios?
0.667/0.577 = 1.15, the O.F.F.
It is apparent to me that the O.F.F. is just a different approach to the problem of transformer loading. It is just as valid as 57.7% or any other factor in use. To say it has no basis is to bury one's head in the sand. It may not be in wide use and may never be, but it is still valid. At least give the Oregon guys credit for coming up with a killer trick question.
[ March 23, 2005, 11:16 AM: Message edited by: rattus ]
Bob,
Thanks for your response, but I must nitpick:
You have implied a 1:1 turns ratio, then the secondary and primary phase currents would be Iline/sqrt(3).
(OK, I see you fixed it)
However, the factor is still 0.577, the inverse of sqrt(3). It appears that this factor does apply to the two-legged wye and provides the allowable connected load as claimed by Jim D. previously. His numbers had demonstrated that already.
Now if we connect two single phase loads directly across the two secondaries, the ratio becomes 0.667 because we now have 240V instead of 208V driving the loads.
In either case though, the apparent power computed at the transformers must be the same, and it is. Whether you add these apparent powers or not is immaterial.
And what is the ratio of the two ratios?
0.667/0.577 = 1.15, the O.F.F.
It is apparent to me that the O.F.F. is just a different approach to the problem of transformer loading. It is just as valid as 57.7% or any other factor in use. To say it has no basis is to bury one's head in the sand. It may not be in wide use and may never be, but it is still valid. At least give the Oregon guys credit for coming up with a killer trick question.
[ March 23, 2005, 11:16 AM: Message edited by: rattus ]
- Location
- Mission Viejo, CA
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- Professional Electrical Engineer
Re: The Two Legged Wye:
I worked on it too late last night. When I scanned it this morning my jaw dropped. I already "knew the answer" and I just couldn't have made that fundamental an error .
I reread specs' original statement . To say there is no basis at all for Oregon factor may be inaccurate, but if it is to be generally applied to all line-to-line, single-phase loads on a three-phase system, as the statement seems to imply, it is definitely gross overkill.
I worked on it too late last night. When I scanned it this morning my jaw dropped. I already "knew the answer" and I just couldn't have made that fundamental an error
.I reread specs' original statement . To say there is no basis at all for Oregon factor may be inaccurate, but if it is to be generally applied to all line-to-line, single-phase loads on a three-phase system, as the statement seems to imply, it is definitely gross overkill.
Re: The Two Legged Wye:
Bob,
I think the O.F.F. applies to any single line to line load on a wye. With a single line to line load on a delta, all three windings share the load but not equally. We have covered this in an earlier topic.
To set the record straight, I do not advocate the adoption of this method. I merely claim it to be valid.
OK, I am keeping count. We now have four who believe the O.F.F. is valid. We still have a number of ostriches out there though.
Bob,
I think the O.F.F. applies to any single line to line load on a wye. With a single line to line load on a delta, all three windings share the load but not equally. We have covered this in an earlier topic.
To set the record straight, I do not advocate the adoption of this method. I merely claim it to be valid.
OK, I am keeping count. We now have four who believe the O.F.F. is valid. We still have a number of ostriches out there though.
- Location
- Mission Viejo, CA
- Occupation
- Professional Electrical Engineer
Re: The Two Legged Wye:
That is, if the effect of each and every, single-phase, line-to-line load on a three-phase wye system were calculated that way it would indicate 15.4 % more total load than the system would actually need to supply - even if it were balanced.
Only in the bizarre case where all single-phase, line-to-line loads were connected to the same two wye connected windings would this phenomenon occur. This is one very big reason we use balanced three-phase systems in the first place - wye or delta. For one additional winding (50% increase) we get a 73.2% increase in total system capacity. (1.732/1.5 = 1.154, another way to derive the infamous factor ).
I fully realize you understand the math, so we are now going back to the day-to-day practical - no one competent would design power systems with that level of imbalance; therefore the Oregon factor is merely arbitrarily imposed minutiae.
I understand that - it's why I qualified my statement to ALL single-phase, line-to-line loads on a three-phase system. The Oregon factor statement implies, as a generalized statement, that any and all such loads create a serious imbalance on a wye system. That is neither true nor likely.
That is, if the effect of each and every, single-phase, line-to-line load on a three-phase wye system were calculated that way it would indicate 15.4 % more total load than the system would actually need to supply - even if it were balanced.
Only in the bizarre case where all single-phase, line-to-line loads were connected to the same two wye connected windings would this phenomenon occur. This is one very big reason we use balanced three-phase systems in the first place - wye or delta. For one additional winding (50% increase) we get a 73.2% increase in total system capacity. (1.732/1.5 = 1.154, another way to derive the infamous factor
).I fully realize you understand the math, so we are now going back to the day-to-day practical - no one competent would design power systems with that level of imbalance; therefore the Oregon factor is merely arbitrarily imposed minutiae.
Re: The Two Legged Wye:
The important number to look at is 0.866 (cos 30 degrees) (phase shift in unbalanced transformers)
For the two legged wye?
2080W/VA of load, and 2400kVA of transformers (2*1200)
2080VA load divided by 2400VA total capacity = 0.866.
For the open delta lets say we have a 30kVA(30kW) load.
Transformers are 0.577*30kVA=17.31kVA (?times two)
Total transformer capacity is 2*17.31=34.62
30kVA load divided by 34.62kVA capacity= 0.866
1 / 0.866 = 1.154 Hey look at that!
The important number to look at is 0.866 (cos 30 degrees) (phase shift in unbalanced transformers)
For the two legged wye?
2080W/VA of load, and 2400kVA of transformers (2*1200)
2080VA load divided by 2400VA total capacity = 0.866.
For the open delta lets say we have a 30kVA(30kW) load.
Transformers are 0.577*30kVA=17.31kVA (?times two)
Total transformer capacity is 2*17.31=34.62
30kVA load divided by 34.62kVA capacity= 0.866
1 / 0.866 = 1.154 Hey look at that!
Re: The Two Legged Wye:
Engy,
Although the voltage and current are out of phase in the secondaries, that is not the reason for the discrepancy. For a given current of any phase angle, the apparent power computed at the load is,
208V x Iload.
However, the apparent power computed at the two transformers is,
2 x 120V x Iload = 240V x Iload.
This is inherent in the 120/208 system, not a matter of current lead or lag.
The phase shift between current and voltage in each secondary will determine the amount of real power in watts to the load, but has no effect on the apparent power in VA. In the case of a resistive load, the PF is 87%.
Engy,
Although the voltage and current are out of phase in the secondaries, that is not the reason for the discrepancy. For a given current of any phase angle, the apparent power computed at the load is,
208V x Iload.
However, the apparent power computed at the two transformers is,
2 x 120V x Iload = 240V x Iload.
This is inherent in the 120/208 system, not a matter of current lead or lag.
The phase shift between current and voltage in each secondary will determine the amount of real power in watts to the load, but has no effect on the apparent power in VA. In the case of a resistive load, the PF is 87%.
- Location
- Wisconsin
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- PE (Retired) - Power Systems
Re: The Two Legged Wye:
rattus,
It is precisely because of the phase angle between the voltage and the current in a wye configuration. That is why I have called the O.F.F. (I think that's what we're calling it) is a trick with numbers.
rattus,
It is precisely because of the phase angle between the voltage and the current in a wye configuration. That is why I have called the O.F.F. (I think that's what we're calling it) is a trick with numbers.
Re: The Two Legged Wye:
I was trying to point out that the derating factor is actually .866 vs .577 in the original open delta question.
.577 will get you the size of EACH of the two transformers for the open delta.
2x.577 = 1.154 (Reciprocal of 0.866 if your not from oregon)
.577, .866 and 1.154(OFF) are all fine to use if you know what specific case to apply them to. (The main problem with the OFF, is we still seem to not have an exact wording of the test question)
I have other fudge factors I use such as 1.203 and 2.776 They are for a specific purpose, I know when to use them, but wouldn't expect all to know where. It's great to use only one calculation in place of two or three. Other people have other fudge factors. The problem with oregon seems to be that they expect everyone to know and use their fudge factor?
I was trying to point out that the derating factor is actually .866 vs .577 in the original open delta question.
.577 will get you the size of EACH of the two transformers for the open delta.
2x.577 = 1.154 (Reciprocal of 0.866 if your not from oregon)
.577, .866 and 1.154(OFF) are all fine to use if you know what specific case to apply them to. (The main problem with the OFF, is we still seem to not have an exact wording of the test question)
I have other fudge factors I use such as 1.203 and 2.776 They are for a specific purpose, I know when to use them, but wouldn't expect all to know where. It's great to use only one calculation in place of two or three. Other people have other fudge factors. The problem with oregon seems to be that they expect everyone to know and use their fudge factor?
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Re: The Two Legged Wye:
According to:
Standard Handbook for Electrical Engineers
Tenth edition - copyright 1969
McGraw-Hill, Donald Fink editor
Sec. 11-26 The open-delta-connection, or V-connection
"Since full-line currents flow in the windings out of phase with the transformer voltages, the normal capacity of the open-delta bank is reduced to 57.7% of it's delta rating."
According to:
Standard Handbook for Electrical Engineers
Tenth edition - copyright 1969
McGraw-Hill, Donald Fink editor
Sec. 11-26 The open-delta-connection, or V-connection
"Since full-line currents flow in the windings out of phase with the transformer voltages, the normal capacity of the open-delta bank is reduced to 57.7% of it's delta rating."
Re: The Two Legged Wye:
?of it?s delta rating?
11-26 discusses a full delta where one winding is lost and the open delta is used as an emergency expedient or temporary measure with the intention of completing the delta?(which would also apply to having an open delta on purpose
Two 25kVA transformers in an open delta would have a 75kVA ?delta rating?
Full Delta: 75 kVA * .577 = 43.3kVA
Open Delta: 50kVA * 1.5 * 0.577 = 43.3kVA
(1.5 used to get "delta rating")
(1.5*0.577= 0.866)
?of it?s delta rating?
11-26 discusses a full delta where one winding is lost and the open delta is used as an emergency expedient or temporary measure with the intention of completing the delta?(which would also apply to having an open delta on purpose
Two 25kVA transformers in an open delta would have a 75kVA ?delta rating?
Full Delta: 75 kVA * .577 = 43.3kVA
Open Delta: 50kVA * 1.5 * 0.577 = 43.3kVA
(1.5 used to get "delta rating")
(1.5*0.577= 0.866)
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