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- Mission Viejo, CA
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- Professional Electrical Engineer
Re: Transformer Loading:
I guess I should add that I'm ignoring "cooling" methods.
I guess I should add that I'm ignoring "cooling" methods.
Transformer Loading:
- Thread starter rattus
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- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: Transformer Loading:
You are mixing the concepts of instantaneous power and average power. Specifically, when you try to add IaVa to IbVb, and get (10)(120) + (10)(120) = 2400, you are using instantaneous values and calling it an average power. That does not work ? you can?t mix the two, as you have said before, and as you have quoted a professor has having confirmed.
Power is constant in a balanced three-phase system. That is why we can use the following general process for calculating a value for service or feeder current.
</font>
Inherent in this process is the presumption of an at least ?roughly balanced? system. If it is not balanced, then this process fails us. This process only works because the VA produced in each transformer winding ? none one of the three being constant, but rather each varies instant by instant ? when added to the VA produced in the other two windings, gives a formula for the total VA in which the variable ?time? is eliminated. Power is constant only in the sum of the three balanced phases. Power is not constant in any single phase 120 volt load, nor in any single phase 208 volt load.
You are taking the 10 amps of each of two transformer secondary windings, multiplying it by the 120 volts across each winding, and getting a total of 2400 VA. That is invalid, specifically because the total apparent power in this configuration is not constant. At the instant when one of the two windings is producing a maximum VA, the other winding is producing less than the maximum VA. The total of the two is not a simple arithmetic sum of 1200 plus 1200. We usually add the VA of all loads arithmetically, but again that only works because we are dealing with balanced three-phase systems. For the three-phase transformer that feeds only a single load, and that load being a single phase 208 volt load, the overall system is not balanced, so the total VA will vary moment to moment.
If you wish to calculate the total VA produced by a transformer with a single-phase 208 volt load, you must either use phasor (vector) addition or use fundamental trigonometry. When you do either, you will get an answer of 2080 VA, not 2400 VA.
Rattus, your logic is completely flawed. The value 2400 has no place in this discussion. There is a fundamental truth that you are missing: conservation of energy. Here is the truth:
{ASIDE: This statement disregards losses, but the losses are small. Loses would not account for the difference between 2080 and 2400 anyway, so it is OK to disregard them. }
You are mixing the concepts of instantaneous power and average power. Specifically, when you try to add IaVa to IbVb, and get (10)(120) + (10)(120) = 2400, you are using instantaneous values and calling it an average power. That does not work ? you can?t mix the two, as you have said before, and as you have quoted a professor has having confirmed.
Power is constant in a balanced three-phase system. That is why we can use the following general process for calculating a value for service or feeder current.
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Take the current drawn by each single phase 120 volt load, and multiply by 120.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Take the current drawn by each single phase 208 volt load, and multiply by 208.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Take the current drawn by each three phase 208 volt load, and multiply by 208 x 1.732.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Add up all these values.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Divide by 208 x 1.732 to get the total current.</font>
Inherent in this process is the presumption of an at least ?roughly balanced? system. If it is not balanced, then this process fails us. This process only works because the VA produced in each transformer winding ? none one of the three being constant, but rather each varies instant by instant ? when added to the VA produced in the other two windings, gives a formula for the total VA in which the variable ?time? is eliminated. Power is constant only in the sum of the three balanced phases. Power is not constant in any single phase 120 volt load, nor in any single phase 208 volt load.
You are taking the 10 amps of each of two transformer secondary windings, multiplying it by the 120 volts across each winding, and getting a total of 2400 VA. That is invalid, specifically because the total apparent power in this configuration is not constant. At the instant when one of the two windings is producing a maximum VA, the other winding is producing less than the maximum VA. The total of the two is not a simple arithmetic sum of 1200 plus 1200. We usually add the VA of all loads arithmetically, but again that only works because we are dealing with balanced three-phase systems. For the three-phase transformer that feeds only a single load, and that load being a single phase 208 volt load, the overall system is not balanced, so the total VA will vary moment to moment.
If you wish to calculate the total VA produced by a transformer with a single-phase 208 volt load, you must either use phasor (vector) addition or use fundamental trigonometry. When you do either, you will get an answer of 2080 VA, not 2400 VA.
Re: Transformer Loading:
Rattus I understand all that but all the 1.154 does is make the 208 appear to be 240 right?
What does the 208 verses 240 have to do with apparent or real power it is what it is coming off the lugs of the transformer 208 or 240 in this case it would be Wye 208 or 240 delta.Then if the branch circuit had a inductor or capacitance in it would have a power factor which you would use your arithmetic to figure real power.
I didn't understand what you said above because line to neutral would be 120 and line to line would be 208? 1.73 x the power with 208.
Ronald
[ March 09, 2005, 12:52 PM: Message edited by: ronaldrc ]
Rattus I understand all that but all the 1.154 does is make the 208 appear to be 240 right?
What does the 208 verses 240 have to do with apparent or real power it is what it is coming off the lugs of the transformer 208 or 240 in this case it would be Wye 208 or 240 delta.Then if the branch circuit had a inductor or capacitance in it would have a power factor which you would use your arithmetic to figure real power.
I didn't understand what you said above because line to neutral would be 120 and line to line would be 208? 1.73 x the power with 208.
Ronald
[ March 09, 2005, 12:52 PM: Message edited by: ronaldrc ]
Re: Transformer Loading:
rattus
I think there have been 23 responses to you question. If you take out your response the remaining 16 say that your are wrong. What does it take for you to get that. Is everyone else
wrong and you are the only one right?
Your post
be correct.
To associate an angle with the R is incorrect.
That is the per phase power. If you check out the
sine wave drawing that ron posted you will see that Phase A voltage is at peak when Phase B is at a negative value of about -.5 x 120. The load on the transformer is occuring at different time intervals. This is what the guys have been saying.
You can not add 1200 VA and 1200 VA and get 2400 VA.
There is a difference between the Line Voltage and the phase voltage.
Charlie B said
[ March 09, 2005, 04:36 PM: Message edited by: bob ]
rattus
I think there have been 23 responses to you question. If you take out your response the remaining 16 say that your are wrong. What does it take for you to get that. Is everyone else
wrong and you are the only one right?
Your post
Resistance load does not have an angle associated with it. If you had XL or XC, then an angle would
be correct.
To associate an angle with the R is incorrect.
You may have this partly correct. If the load amps is 10 amps the VA per phase is 120 x 10 = 1200 VA.
That is the per phase power. If you check out the
sine wave drawing that ron posted you will see that Phase A voltage is at peak when Phase B is at a negative value of about -.5 x 120. The load on the transformer is occuring at different time intervals. This is what the guys have been saying.
You can not add 1200 VA and 1200 VA and get 2400 VA.
Well obviously this is incorrect also. Since the load is resistive the Apparent Power (VA) and the Real Power (Watts) is the same value since the power factor is 1.
I would not say that. It should be 208 volts x 10.
There is a difference between the Line Voltage and the phase voltage.
I can not believe you said that.
Charlie B said
Thats right on the mark
[ March 09, 2005, 04:36 PM: Message edited by: bob ]
Re: Transformer Loading:
Bob,
This is not a popularity contest. It is a search for truth.
You are missing some basic points regarding phase angles and putting words in my mouth that I did not utter. I have presented a detailed scenario which you obviously did not follow. All phase angles should be referenced to the positive x-axis, not to each other.
Now I believe that at least one moderator has agreed with me. His vote should count for something.
The difference of opinion occurs I believe arises because the power industry relies on standard formulas which are quite valid, but in this case one must approach the problem from the basics.
I have invited you to have a private debate on this matter, but you have so far declined. Instead you make blanket statements questioning my credibility. Are you ready for the challenge?
[ March 09, 2005, 02:58 PM: Message edited by: rattus ]
Bob,
This is not a popularity contest. It is a search for truth.
You are missing some basic points regarding phase angles and putting words in my mouth that I did not utter. I have presented a detailed scenario which you obviously did not follow. All phase angles should be referenced to the positive x-axis, not to each other.
Now I believe that at least one moderator has agreed with me. His vote should count for something.
The difference of opinion occurs I believe arises because the power industry relies on standard formulas which are quite valid, but in this case one must approach the problem from the basics.
I have invited you to have a private debate on this matter, but you have so far declined. Instead you make blanket statements questioning my credibility. Are you ready for the challenge?
[ March 09, 2005, 02:58 PM: Message edited by: rattus ]
Re: Transformer Loading:
Thanks Bob
Good to know someone checked my drawing out it took me a good while to draw that one.
I have been in the trade for over 30 years and I never really understood 3 phase but now I believe I have a handle on it.I knew enough to trouble shoot and hook up machines but not the mechanics of how it was produced.
Whats so funny is most of this knowledge came from what Rattus had told to me in private post.
If you will examine the graph not only is B phase neg. But B phase and C phase are both neg. at 50 volts and this is added together and at that point the positive peak of A phase is canceled out to Zero.
Edited because I'm not sure thats right weres Ed?
Ronald
[ March 09, 2005, 03:43 PM: Message edited by: ronaldrc ]
Thanks Bob
Good to know someone checked my drawing out it took me a good while to draw that one.
I have been in the trade for over 30 years and I never really understood 3 phase but now I believe I have a handle on it.I knew enough to trouble shoot and hook up machines but not the mechanics of how it was produced.
Whats so funny is most of this knowledge came from what Rattus had told to me in private post.
If you will examine the graph not only is B phase neg. But B phase and C phase are both neg. at 50 volts and this is added together and at that point the positive peak of A phase is canceled out to Zero.
Edited because I'm not sure thats right weres Ed?
Ronald
[ March 09, 2005, 03:43 PM: Message edited by: ronaldrc ]
Re: Transformer Loading:
Dag gone Rattus we can't tell whats what when you holler like that. I not saying that you are wrong but I haven't seen anyone on here perfect yet.I will be the first to admit its the sharpest crowd I personally have ever discussed this field with of course being from Tennessee that might not be saying much.
When I screw up I admit it theres a few on here that don't.I hope you don't join that crowd.There I've made another enemy, but just being honest.
Dag gone Rattus we can't tell whats what when you holler like that. I not saying that you are wrong but I haven't seen anyone on here perfect yet.I will be the first to admit its the sharpest crowd I personally have ever discussed this field with of course being from Tennessee that might not be saying much.
When I screw up I admit it theres a few on here that don't.I hope you don't join that crowd.There I've made another enemy, but just being honest.
- Location
- Mission Viejo, CA
- Occupation
- Professional Electrical Engineer
Re: Transformer Loading:
Since I appear to be the ?culprit? moderator, I guess I?ll add another comment.
A great deal of the difference is a matter of approach, the engineer?s vs. the physicist?s. On a day-to-day basis, I?m usually interested in the ?practical.? If I had to do all my work deriving the calcs from Maxwell?s equations I still wouldn?t be finished with my first project, begun in the late 60?s. Nevertheless, occasionally it is not only valuable, but also necessary, to review the true ?physics? of an issue.
I said in another post, for an ideal transformer, the throughput in the circuit under discussion would be 2080Ws and 1200VARs, assuming a purely resistive load. That is what properly installed metering would indicate ? whether on the primary or secondary of the transformer and whether the primary was wye or delta connected.
Most of us would say that is 2400VA. I deliberately avoided it. Watts and VARs have true, physical meaning: the rate energy is used and rate that energy is stored and returned to the system from the magnetic fields. In this case, the 1200VARs would be cycled between the magnetic fields of the two involved windings. They would never ?see the light of day? outside the transformer.
VA does not have real physical meaning ? but it IS practical. With only a couple of provisos, relatively high load power-factors and loads ?relatively balanced? on all three phases, everything works fine because we don?t use ?ideal? transformers and the ones we do use have the tolerance to accommodate minor differences.
Occasionally, the loads are seriously imbalanced or they have terrible power factors or both. In those cases, we have to review the effect on the transformer, and I believe that is some of what rattus is getting at; the need of that review is something that I believe most of us are already in agreement.
I believe the open question is whether the transformer loading is effectively Vphase x Iphase in VA regardless of the load configuration. Assuming that Vphase is the measured voltage across the winding and Iphase the measured current through it, my opinion is a very qualified ?yes? for a given single winding. Does this mean the transformer under discussion ?sees? 2400VA? I don?t know, I can only say that since power transformers are current limited regardless of the output voltage it is the equivalent effect of 1200VA on each of two windings.
Since I appear to be the ?culprit? moderator, I guess I?ll add another comment.
A great deal of the difference is a matter of approach, the engineer?s vs. the physicist?s. On a day-to-day basis, I?m usually interested in the ?practical.? If I had to do all my work deriving the calcs from Maxwell?s equations I still wouldn?t be finished with my first project, begun in the late 60?s. Nevertheless, occasionally it is not only valuable, but also necessary, to review the true ?physics? of an issue.
I said in another post, for an ideal transformer, the throughput in the circuit under discussion would be 2080Ws and 1200VARs, assuming a purely resistive load. That is what properly installed metering would indicate ? whether on the primary or secondary of the transformer and whether the primary was wye or delta connected.
Most of us would say that is 2400VA. I deliberately avoided it. Watts and VARs have true, physical meaning: the rate energy is used and rate that energy is stored and returned to the system from the magnetic fields. In this case, the 1200VARs would be cycled between the magnetic fields of the two involved windings. They would never ?see the light of day? outside the transformer.
VA does not have real physical meaning ? but it IS practical. With only a couple of provisos, relatively high load power-factors and loads ?relatively balanced? on all three phases, everything works fine because we don?t use ?ideal? transformers and the ones we do use have the tolerance to accommodate minor differences.
Occasionally, the loads are seriously imbalanced or they have terrible power factors or both. In those cases, we have to review the effect on the transformer, and I believe that is some of what rattus is getting at; the need of that review is something that I believe most of us are already in agreement.
I believe the open question is whether the transformer loading is effectively Vphase x Iphase in VA regardless of the load configuration. Assuming that Vphase is the measured voltage across the winding and Iphase the measured current through it, my opinion is a very qualified ?yes? for a given single winding. Does this mean the transformer under discussion ?sees? 2400VA? I don?t know, I can only say that since power transformers are current limited regardless of the output voltage it is the equivalent effect of 1200VA on each of two windings.
hardworkingstiff
Senior Member
- Location
- Wilmington, NC
Re: Transformer Loading:
Rattus,
What you have said makes perfect sense to me.
Didn't the NEC get away from "watts" and go to "VA" in order to make sure the conductors (and I would guess transformers) can handle the apparent power?
Rattus,
What you have said makes perfect sense to me.
Didn't the NEC get away from "watts" and go to "VA" in order to make sure the conductors (and I would guess transformers) can handle the apparent power?
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Re: Transformer Loading:
rattus,
Are you taking the long way to say that an open wye transformer bank cannot support the same three phase loading as a "closed" wye?
Going back to your March 4 17:18 post.
The wye transformer bank sees a 20.8 ohm phase to phase load identical to 2 separate 12 ohm phase to neutral loads. The effective load is 2080VA.
p = phase
n = neutral
w = winding
pp = phase to phase
pn = phase to neutral
a = phase a
c = phase c
t = total
Case 1, 20.8 ohm load phase to phase
Vpp = 208V
Vw = Vpn = Vpp/1.73 = 208V/1.73 = 120V
Ip = Vpp/Zpp = 208/20.8 = 10A
Iw = Ip = 10A
VAw = Vw*Iw = 120*10 = 1200VA
VAt = Vpp*Ip = (Vw*1.73)*Iw = 2080VA
Case 2, 12 ohm load phase to neutral
Vpn = 120V
Vw = Vpn = 120V
Ip = Vpn/Zpn = 120/12 = 10A
Iw = Ip = 10A
VAw = Vw*Iw = 120*10 = 1200VA
VAt = Vwa*Iw + Vwc*Iw = (Vwa+Vwc)*Iw = Vpp*Iw = 208*10 = 2080VA
Assuming these are full load values, a closed wye transformer would be: VAt = 3*VAw = 3*1200 = 3600VA. But because this is an open wye arrangement the effective three phase rating is only 57.77% of 3600VA, or 2080VA. But in neither case is any winding overloaded.
Now please explain how load power factor comes into any of the transformer VA calculations. Or, how total VA is a fictitious number.
rattus,
Are you taking the long way to say that an open wye transformer bank cannot support the same three phase loading as a "closed" wye?
Going back to your March 4 17:18 post.
The wye transformer bank sees a 20.8 ohm phase to phase load identical to 2 separate 12 ohm phase to neutral loads. The effective load is 2080VA.
p = phase
n = neutral
w = winding
pp = phase to phase
pn = phase to neutral
a = phase a
c = phase c
t = total
Case 1, 20.8 ohm load phase to phase
Vpp = 208V
Vw = Vpn = Vpp/1.73 = 208V/1.73 = 120V
Ip = Vpp/Zpp = 208/20.8 = 10A
Iw = Ip = 10A
VAw = Vw*Iw = 120*10 = 1200VA
VAt = Vpp*Ip = (Vw*1.73)*Iw = 2080VA
Case 2, 12 ohm load phase to neutral
Vpn = 120V
Vw = Vpn = 120V
Ip = Vpn/Zpn = 120/12 = 10A
Iw = Ip = 10A
VAw = Vw*Iw = 120*10 = 1200VA
VAt = Vwa*Iw + Vwc*Iw = (Vwa+Vwc)*Iw = Vpp*Iw = 208*10 = 2080VA
Assuming these are full load values, a closed wye transformer would be: VAt = 3*VAw = 3*1200 = 3600VA. But because this is an open wye arrangement the effective three phase rating is only 57.77% of 3600VA, or 2080VA. But in neither case is any winding overloaded.
Now please explain how load power factor comes into any of the transformer VA calculations. Or, how total VA is a fictitious number.
Re: Transformer Loading:
Jim, let me quote from a post by grlsound:
"After consulting with not less than THREE electrical engineers, I think I may have a glimmer of an idea as to where, and why, one might use that elusive factor. This example is of a purely resistive single phase load, DELTA connected to a WYE source. (ie: transformer) I am told that when this connection happens, the current and voltage become out of phase by 30 degrees. (phasor diagrams in many engineering books). This appears to be related to something similar to power factor. The resistive load will not add to the phase shift, will not subtract from that out of phase voltage and current, but WILL NOT be able to generate its rated wattage due to the current phase shift away from the voltage. Please do not ask me WHY that phase shift occurs. I do not know. I do know that both of my electrical engineering books say that it does occur. That elusive factor is the reciprocal of the cosine of 30 degrees."
grlsound reports that three, count them, EEs agree that the wye phase currents and wye phase voltages are out of phase by 30 degrees. This is should be obvious from the vector diagram.
If we have a 30 degree shift between Vphase and Iphase, then the PF = cos(30) = 0.87. Then,
Preal = 2 x 10 x 120 x 0.87 = 2080W delivered by the transformer. PF in the load is 1.0.
In your critique of my example, you left the angles off my 12 Ohm loads. That is what makes the currents equivalent. No, they are not resistors, they are impedances.
I ask again, why should the load be different on two transformers driving the same current through 2 impedances to neutral as opposed to driving this current through a single impedance line to line? Give me a solid reason for that, and I will concede that I am dead wrong and don't even know the difference between resistance and impedance.
Now, for the real world, you will never see such an imbalanced load, so the standard methods work just fine.
Rattus,
BSEE, MSEE
Jim, let me quote from a post by grlsound:
"After consulting with not less than THREE electrical engineers, I think I may have a glimmer of an idea as to where, and why, one might use that elusive factor. This example is of a purely resistive single phase load, DELTA connected to a WYE source. (ie: transformer) I am told that when this connection happens, the current and voltage become out of phase by 30 degrees. (phasor diagrams in many engineering books). This appears to be related to something similar to power factor. The resistive load will not add to the phase shift, will not subtract from that out of phase voltage and current, but WILL NOT be able to generate its rated wattage due to the current phase shift away from the voltage. Please do not ask me WHY that phase shift occurs. I do not know. I do know that both of my electrical engineering books say that it does occur. That elusive factor is the reciprocal of the cosine of 30 degrees."
grlsound reports that three, count them, EEs agree that the wye phase currents and wye phase voltages are out of phase by 30 degrees. This is should be obvious from the vector diagram.
If we have a 30 degree shift between Vphase and Iphase, then the PF = cos(30) = 0.87. Then,
Preal = 2 x 10 x 120 x 0.87 = 2080W delivered by the transformer. PF in the load is 1.0.
In your critique of my example, you left the angles off my 12 Ohm loads. That is what makes the currents equivalent. No, they are not resistors, they are impedances.
I ask again, why should the load be different on two transformers driving the same current through 2 impedances to neutral as opposed to driving this current through a single impedance line to line? Give me a solid reason for that, and I will concede that I am dead wrong and don't even know the difference between resistance and impedance.
Now, for the real world, you will never see such an imbalanced load, so the standard methods work just fine.
Rattus,
BSEE, MSEE
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
Re: Transformer Loading:
My formulas work just as well with Z as with R. And yes I can imagine several possible real world load profiles that could unbalance a wye transformer as in your example.
What does the phase angle between the current and the voltage have to do with calculating the VA loading of a transformer? The only affect I can visualize is the resultant neutral current. But the specific value of the neutral current does not come into play when analyzing the loading of each individual winding.
And I can believe that there are at least three EEs that do not understand power relationships. Even as an EE myself, I would be out of my area of expertise in a discussion on electronics.
My formulas work just as well with Z as with R. And yes I can imagine several possible real world load profiles that could unbalance a wye transformer as in your example.
What does the phase angle between the current and the voltage have to do with calculating the VA loading of a transformer? The only affect I can visualize is the resultant neutral current. But the specific value of the neutral current does not come into play when analyzing the loading of each individual winding.
And I can believe that there are at least three EEs that do not understand power relationships. Even as an EE myself, I would be out of my area of expertise in a discussion on electronics.
Re: Transformer Loading:
rasttus posted
. He reports that the engineers said with a delta connected load the current and voltage is out of phase by 30 degrees. You then say that the wye phase current and wye phase voltage are out of phase by 30 degrees.
Neither of you are correct. In a balanced delta connected load the line current is sqrt of 3 x phase current and it is out of phase with the component phase current by
30 degrees. You can't have it both ways.
Sence the phase shift is out I assume the following is out also.
[ March 10, 2005, 08:58 AM: Message edited by: bob ]
rasttus posted
You have misquoted grlsound.
. He reports that the engineers said with a delta connected load the current and voltage is out of phase by 30 degrees. You then say that the wye phase current and wye phase voltage are out of phase by 30 degrees.
Neither of you are correct. In a balanced delta connected load the line current is sqrt of 3 x phase current and it is out of phase with the component phase current by
30 degrees. You can't have it both ways.
Sence the phase shift is out I assume the following is out also.
Edit to remove the term balanced from the post.
[ March 10, 2005, 08:58 AM: Message edited by: bob ]
Re: Transformer Loading:
You have suggested that I consult with someone more knowledgeable than I, and I know a number of them. I have consulted with two of them, both geniuses, and they say I am right.
Rattus
No, I did not misquote, this is a cut and paste. And, he did not say it was a balanced delta either. It was a "single phase load, DELTA connected". The word "balanced" does not appear in the quote. If you continue to misread the text, you will never understand this. You appear to be sniping, trying to find some fault where none exists.
You have suggested that I consult with someone more knowledgeable than I, and I know a number of them. I have consulted with two of them, both geniuses, and they say I am right.
Rattus
Re: Transformer Loading:
Stiff, my thought is that apparent power is a convenient way to tally up loading without having to bother with power factor and separating the single phase and three phase loads. Don't know what the NEC says about it, but it was taught in the EE schools 50 years ago.
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