ptonsparky
Tom
- Occupation
- EC - retired
Post 11 assumes the breaker will trip at exactly 80 amps in his calculations.
Safely Maximize Power from a 100amp 3-Phase Delta Panel
- Thread starter EricJ
- Start date
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- Massachusetts
ptonsparky
Tom
- Occupation
- EC - retired
Which make no sense and why I asked about it in post 12.
Why? Because he doesn't care about reality, only the theoretical at this point.
The problem is the assumption that a breaker will trip at some arbitrary level is not theory. The theory problem I see here is that the OP is not seeing that L- L loads are all served by 240V windings and will only see 240V, not 120 or 208.
Roger
- Location
- Massachusetts
Do you know that? I do not.
Further I still do not know what the OP means by 'carrying the load safely'.
Safely to who, to what standard and in what circumstances? I could safely carry 100 amps on 14 AWG under the right conditions.
ptonsparky
Tom
- Occupation
- EC - retired
Ok, imaginary, hypothetical?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170318-0946 EDT
EricJ:
Finally in post #11 you presented your question in a form that can be answered.
That the panel is from a wild leg delta has nothing to do with the question except that the source is a delta. Actually the source could be a wye and it does not change the answer.
The answer is obviously 80*138*3 = 11040*3 = 33120 W. Note 138 is a rounded value. The exact value is closer to 138.5641 . Because of this rounding some of the values below are not as close as you might want.
Two different resistive loads will accomplish this or some combination.
The possible load configurations are balanced delta or wye.
In delta each resistor is R = 240^2/11040 = 5.22 ohms. In wye each is R = 138^2/11040 = 1.73 ohms. The two resistance values should be related by 3. So 1.73*3 = 5.18 ohms. Close but not real close.
If I do the calculation with the full resolution of my HP 32S, then the results are:
138.5641...
80*this = 11085.1252...
Rd = 240/Ipp = 5.1962... ohms.
Rw = 138...../80 = 1.7321... ohms.
1.7321...*3 = 5.1962... . Now the check is closer.
.
EricJ:
Finally in post #11 you presented your question in a form that can be answered.
That the panel is from a wild leg delta has nothing to do with the question except that the source is a delta. Actually the source could be a wye and it does not change the answer.
The answer is obviously 80*138*3 = 11040*3 = 33120 W. Note 138 is a rounded value. The exact value is closer to 138.5641 . Because of this rounding some of the values below are not as close as you might want.
Two different resistive loads will accomplish this or some combination.
The possible load configurations are balanced delta or wye.
In delta each resistor is R = 240^2/11040 = 5.22 ohms. In wye each is R = 138^2/11040 = 1.73 ohms. The two resistance values should be related by 3. So 1.73*3 = 5.18 ohms. Close but not real close.
If I do the calculation with the full resolution of my HP 32S, then the results are:
138.5641...
80*this = 11085.1252...
Rd = 240/Ipp = 5.1962... ohms.
Rw = 138...../80 = 1.7321... ohms.
1.7321...*3 = 5.1962... . Now the check is closer.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170318-1957 EDT
EricJ:
What are the classes required for a BSEE today? Back when I went to school we had to cover basics. And then there were two specialty areas, power and electronics-communications.
The broad outline of common courses was:
English, math, chemistry, physics, drawing, chem-met, economics, statics, dynamics, strength of materials, fluid mechanics, heat engines, machine design, dc machines and circuits, ac machines and circuits, electricity and magnetism, electromechanics, electrical measurements, ac apparatus, electrical design, electronics and electron tubes. This totaled 120 credit hours. Then there were another 20 credit hours in your speciality.
Although I have lived within a mile or two of my school ever since graduation I have little contact with with the University as a whole or the EE department. I have no idea what a present class program looks like. Some classes that were basic requirements for us have probably been eliminated and others watered down.
Today I think 4 years is too short for a good basic education in EE, 6 years might be a better number for a BSEE. Thinking, reasoning, and an education in basic concepts is what I expect of an electrical engineer. Just being able to plug numbers into an equation does not mean the resulting answer is valuable or useful. One has to understand how a particular equation relates to a real world problem.
.
EricJ:
What are the classes required for a BSEE today? Back when I went to school we had to cover basics. And then there were two specialty areas, power and electronics-communications.
The broad outline of common courses was:
English, math, chemistry, physics, drawing, chem-met, economics, statics, dynamics, strength of materials, fluid mechanics, heat engines, machine design, dc machines and circuits, ac machines and circuits, electricity and magnetism, electromechanics, electrical measurements, ac apparatus, electrical design, electronics and electron tubes. This totaled 120 credit hours. Then there were another 20 credit hours in your speciality.
Although I have lived within a mile or two of my school ever since graduation I have little contact with with the University as a whole or the EE department. I have no idea what a present class program looks like. Some classes that were basic requirements for us have probably been eliminated and others watered down.
Today I think 4 years is too short for a good basic education in EE, 6 years might be a better number for a BSEE. Thinking, reasoning, and an education in basic concepts is what I expect of an electrical engineer. Just being able to plug numbers into an equation does not mean the resulting answer is valuable or useful. One has to understand how a particular equation relates to a real world problem.
.
kwired
Electron manager
- Location
- NE Nebraska
If one had 100 amps of 120 volt load on A, 100 amps of 120 volt load on C and 100 amps of 208 volt load on B, I guess the first approach taken sort of works, the breaker won't care what voltages are or how well the source can handle that type of loading, but it will still see 100 amps per line and will be fine with that, at least before we get into how long it can hold with a continuous 100 amp load.
Reality is you will have very little if any 208 load connected to the high leg, you may have 120 volt loads connected to A and C, you may have unbalanced 240 volt loads connected between any two phase conductors, and your balanced three phase loads will each draw a factor of 1.732 of the total amps from each phase conductor. The ghost phase of an open delta will not provide any VA to the system, the other two phases will each have to contribute more VA to a balanced three phase load then they would if it were a full delta system.
Reality is you will have very little if any 208 load connected to the high leg, you may have 120 volt loads connected to A and C, you may have unbalanced 240 volt loads connected between any two phase conductors, and your balanced three phase loads will each draw a factor of 1.732 of the total amps from each phase conductor. The ghost phase of an open delta will not provide any VA to the system, the other two phases will each have to contribute more VA to a balanced three phase load then they would if it were a full delta system.
Woot we have an answer!! Thank you guys for staying with me.
L-G was given just to make it clear what kind of power we have
Just theory
This would tell us what type of transformer is used correct? But if we assume the transformer has no limitations would this matter?
Exactly! I suppose I choose 80 amps because it leaves room on the breaker for other things, changes in temperature, etc. If a client asked you to load as many 240V resistive loads onto a panel of this sort (100amp 3ph high leg delta w/ oversized transformer) as he could and there were also small L-N loads and power outlets and the such how much would you add? More? less? I feel like 20 amps of head room on each leg is appropriate for whatever else might get plugged in or used.
ahhhh I see!!
Good to know but I really wanted to assume that the transformer doesn't present any limitations. Lets assume for example it is way over sized.
I understand that L-L voltages are 240, that is the reason why I made all the loads 240 and not 120 or 208.
Haha I remember reading about the guys near Switzerland that shoot little particles around a 17 mile underground tunnel at speeds close to that of light. They need insane amount of current and obviously don't have unlimited amounts of copper. I'm pretty sure they do something like that. I think liquid nitrogen is involved
Thanks gar. I see now. You use the mid point of the delta.
But now for the second half of the question:
How would the loads be wired such that they take full advantage of the available power?
Theres 3 different L-L connections. AB, BC, and AC. C is clearly the high leg.
Now if we loaded just one of the LL connections with all 33120w I'm pretty certain the theoretical 80amp breakers would flip. We need to balance the loads across all three.
Would each connection be loaded to 33120/3 = 11040w? Or would the BC and AC be loaded slightly more and the AB slightly less since the C leg has more potential?
L-G was given just to make it clear what kind of power we have
Just theory
This would tell us what type of transformer is used correct? But if we assume the transformer has no limitations would this matter?
Exactly! I suppose I choose 80 amps because it leaves room on the breaker for other things, changes in temperature, etc. If a client asked you to load as many 240V resistive loads onto a panel of this sort (100amp 3ph high leg delta w/ oversized transformer) as he could and there were also small L-N loads and power outlets and the such how much would you add? More? less? I feel like 20 amps of head room on each leg is appropriate for whatever else might get plugged in or used.
ahhhh I see!!
Good to know but I really wanted to assume that the transformer doesn't present any limitations. Lets assume for example it is way over sized.
I understand that L-L voltages are 240, that is the reason why I made all the loads 240 and not 120 or 208.
Haha I remember reading about the guys near Switzerland that shoot little particles around a 17 mile underground tunnel at speeds close to that of light. They need insane amount of current and obviously don't have unlimited amounts of copper. I'm pretty sure they do something like that. I think liquid nitrogen is involved
Thanks gar. I see now. You use the mid point of the delta.
But now for the second half of the question:
How would the loads be wired such that they take full advantage of the available power?
Theres 3 different L-L connections. AB, BC, and AC. C is clearly the high leg.
Now if we loaded just one of the LL connections with all 33120w I'm pretty certain the theoretical 80amp breakers would flip. We need to balance the loads across all three.
Would each connection be loaded to 33120/3 = 11040w? Or would the BC and AC be loaded slightly more and the AB slightly less since the C leg has more potential?
kwired
Electron manager
- Location
- NE Nebraska
You have no choice but to connect 120 volt loads in an unbalanced fashion. If you have more then 100 amps of 120/240 single phase the only way you can steal more power from the third phase is with transformation. Otherwise figure all your balanced three phase load per phase then add unbalanced loads to whichever phase they are connected to. Unless a majority of your load is true three phase load, these systems are typically pretty unbalanced with the high leg being the one that doesn't carry much load.
ptonsparky
Tom
- Occupation
- EC - retired
The C leg has more potential, voltage to ground. It does not have the ability to carry more current because of that volt to ground, if anything it has less because the transformer serving it is smaller. Open Delta. C leg is not normally used to feed L-N loads. Some one else can explain why, because it is above my pay grade. Throw Volt to ground away. Forget it. Don't reference it.
If the load is 240v single phase why cant you put a breaker between A-B or C-B? (In this diagram B is the highleg) Wouldnt that use the third leg?
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
You can. That is the only "normal" use of the B leg, namely to feed three phase loads. 240V single phase loads can connect that way too
B to neutral, at 208V, is not recommended because of transformer limitations.
The limit in using the B leg at all is set by the size of the transformers if the delta source is not symmetric.
mobile
ptonsparky
Tom
- Occupation
- EC - retired
"B. to myself B. to myself B." Even after more than 20 years of moving the wild leg to B, I still think of it as at the C position. :slaphead:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
170318-1449 EDT
EricJ:
Suppose as your question now stands based on post #11 that 80 A is the limit per leg and that you fully load the single phase center tapped source with a resistive load of 240 V at 80 A, 19200 W, 3 ohms, then there is no way for you to extract more power from this source.
To do so would require additional current from one or both of the already fully loaded legs and that fails your 80 A limit.
When you specify a specific current as a limit on each line of a balanced 3 phase source, then the only way to get maximum power from that source is with a balanced resistive load.
You can use any crazy wiring combination of resistors in this load. They just have to combine in some way that produces a balanced resistive load on the source.
Once assembled this mess of resistors can be replaced by an equivalent circuit of just three resistors wired as either a delta load or a wye load.
Now suppose that instead of the source being 3 wires it is changed to 6 wires, then with the same 80 A current limit per wire I can load each of the three phases individually to 19200 W per phase, or a total of 57600 W. What I have really done is use 2 times the copper to get 57600 W whereas with the 1 times copper I was able to get 33255 W.
Note: my power transfer per pound of copper was greater by 1.15 times for the 3 wire approach vs 6 wires. This is one reason why 3 phase power transfer is better than 1 phase.
.
EricJ:
Suppose as your question now stands based on post #11 that 80 A is the limit per leg and that you fully load the single phase center tapped source with a resistive load of 240 V at 80 A, 19200 W, 3 ohms, then there is no way for you to extract more power from this source.
To do so would require additional current from one or both of the already fully loaded legs and that fails your 80 A limit.
When you specify a specific current as a limit on each line of a balanced 3 phase source, then the only way to get maximum power from that source is with a balanced resistive load.
You can use any crazy wiring combination of resistors in this load. They just have to combine in some way that produces a balanced resistive load on the source.
Once assembled this mess of resistors can be replaced by an equivalent circuit of just three resistors wired as either a delta load or a wye load.
Now suppose that instead of the source being 3 wires it is changed to 6 wires, then with the same 80 A current limit per wire I can load each of the three phases individually to 19200 W per phase, or a total of 57600 W. What I have really done is use 2 times the copper to get 57600 W whereas with the 1 times copper I was able to get 33255 W.
Note: my power transfer per pound of copper was greater by 1.15 times for the 3 wire approach vs 6 wires. This is one reason why 3 phase power transfer is better than 1 phase.
.
Limiting neut to 100 A
36 kva
3 feeders x 86.6 A
2 x 50 x 240 = 24 kva
2 x 50 x 120 = 12 kva
upsize neut by ~17%
46.6 kva
3 feeders x 100 A
2 x 57.7 x 240 = 27.7 kva
2 x 57.7 x 120 = 18.9 kva
if limiting mains to 80 A
33.3 kva
3 feeders x 80 A
2 x 46.2 x 240 = 22.2 kva
2 x 46.2 x 120 = 11.1 kva
fyi sqrt3 x 80 x 240 = 33.3 kva
36 kva
3 feeders x 86.6 A
2 x 50 x 240 = 24 kva
2 x 50 x 120 = 12 kva
upsize neut by ~17%
46.6 kva
3 feeders x 100 A
2 x 57.7 x 240 = 27.7 kva
2 x 57.7 x 120 = 18.9 kva
if limiting mains to 80 A
33.3 kva
3 feeders x 80 A
2 x 46.2 x 240 = 22.2 kva
2 x 46.2 x 120 = 11.1 kva
fyi sqrt3 x 80 x 240 = 33.3 kva
For strictly 240V resistive loads it makes no difference which is the wild leg, but FWIW Code requires B to be the wild leg on premises wiring. Some POCO require C to be the wild leg for metering purposes and Code permits this, but C must be flopped to B immediately thereafter.
For calculating total load you should use the conventional formula:
80A × 240V × √3 = 33,255VA
And being purely resistive loads, that converts directly into Watts (VA × PF = W).Loads are the same as voltage in the respect that they require two points having a voltage differential. In balancing the loads, it will be 33,255VA ÷ 3 = 11,085VA per LL pair. What you will want to note for later discussion is that 11,085VA ÷ 240V = 46.2A and 46.2A × 2 = 92.4A which is significantly more than 80A. Can you tell us why?
Sweet I get it now. Thanks for sticking with me.
I guess the high leg was throwing me off, but it's really just a delta with one of the coils center tapped to allow greater utility.
Thanks Smart$
I'll take a stab at it. I think this is because each leg is 120 degrees out of phase from the other. If you have a load LL between leg A and leg B, leg A will supply some current and leg b will also supply some current. So even though the total current across the load is 92.4 amps, the individual legs will only see 80 max.
Am I in the ballpark Smart$?
I guess the high leg was throwing me off, but it's really just a delta with one of the coils center tapped to allow greater utility.
Thanks Smart$
I'll take a stab at it. I think this is because each leg is 120 degrees out of phase from the other. If you have a load LL between leg A and leg B, leg A will supply some current and leg b will also supply some current. So even though the total current across the load is 92.4 amps, the individual legs will only see 80 max.
Am I in the ballpark Smart$?
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