Single phase load on 208/120Y calculation req'd by Oreg
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Re: Single phase load on 208/120Y calculation req'd by Oreg
Several days ago I opened my big mouth when I should not have done so. Now I have to eat a little crow, so to speak!
While the 1.154 is necessary to change amperage on a single phase load to 208 volt single phase when used on a three phase wye system, I do not believe it is commonly used in the field. No where can I find that an author uses this in teaching load calculations. I do believe that it is necessary.
I will attempt to post a future comment when I have time to expalin this method further.
Several days ago I opened my big mouth when I should not have done so. Now I have to eat a little crow, so to speak!
While the 1.154 is necessary to change amperage on a single phase load to 208 volt single phase when used on a three phase wye system, I do not believe it is commonly used in the field. No where can I find that an author uses this in teaching load calculations. I do believe that it is necessary.
I will attempt to post a future comment when I have time to expalin this method further.
Re: Single phase load on 208/120Y calculation req'd by Oreg
fsimmons
I stick my foot in my mouth daily I think its part of being human.I'm still trying to dig it out from my last post.
But are you saying that if I know what the amperage is at 240 volts I can use 1.154 as a constant multiplier to find my amperage at 208 if so that makes sense.
A 6000 watt resistor no pf draws 25 amp. at 240 Vac single phase.
If I where going to use this same resistor on 208 single phase all I need to do is multiply 25 x 1.154 which equals 28.85 amps? That makes everything fit.
I edited this after I posted the reply below if you reverse my formula and divide 25 amp. instead of multiplying. 25/1.154=21.66 which would be the current at 208 volts.
Ronald
[ February 28, 2005, 11:41 PM: Message edited by: ronaldrc ]
fsimmons
I stick my foot in my mouth daily I think its part of being human.I'm still trying to dig it out from my last post.
But are you saying that if I know what the amperage is at 240 volts I can use 1.154 as a constant multiplier to find my amperage at 208 if so that makes sense.
A 6000 watt resistor no pf draws 25 amp. at 240 Vac single phase.
If I where going to use this same resistor on 208 single phase all I need to do is multiply 25 x 1.154 which equals 28.85 amps? That makes everything fit.
I edited this after I posted the reply below if you reverse my formula and divide 25 amp. instead of multiplying. 25/1.154=21.66 which would be the current at 208 volts.
Ronald
[ February 28, 2005, 11:41 PM: Message edited by: ronaldrc ]
Re: Single phase load on 208/120Y calculation req'd by Oreg
I stick my foot in my mouth daily I think its part of being human.I'm still trying to dig it out from my last post.
Well I did it again, Someone was courteous enough to E-mail me a private message and point out a mistake I made in my figures.Hes is right and I make this mistake a lot when figuring wattage/voltage to amps.
when I figured the math 6000/240=25 amps. and 6000/208=28.85 amps. but you can't do it that way because when the voltage goes down the amp. draw goes down.And a 208 that produces 6000 watts would have to have a be lower ohm resistance.
So scrap that idea a again Thanks for pointing that out.
Ronald
I stick my foot in my mouth daily I think its part of being human.I'm still trying to dig it out from my last post.
Well I did it again, Someone was courteous enough to E-mail me a private message and point out a mistake I made in my figures.Hes is right and I make this mistake a lot when figuring wattage/voltage to amps.
when I figured the math 6000/240=25 amps. and 6000/208=28.85 amps. but you can't do it that way because when the voltage goes down the amp. draw goes down.And a 208 that produces 6000 watts would have to have a be lower ohm resistance.
So scrap that idea a again Thanks for pointing that out.
Ronald
Re: Single phase load on 208/120Y calculation req'd by Oreg
I will take a chance and see if this will format the same using cut and paste from a Microsoft Word Document. If not, maybe someone can help me with being able to take my document and put it on this forum so that it makes sense.
I am of the opinion that there are four ways to do this problem at the very minimum! The method that some writers use is to take the total vA load of all three phases totaled together and divide by 1.732 X 208. If you have a totally balanced load, no problem but this is highly unlikely.
The second method I see a lot is to take phase A as a worse case scenario and apply the line to neutral on phase A and N, the line to line single phase load on line A & B, and the three phase load and apply to A, B & C. Then they total up the phase A load and this will be the worse possible case without balancing the system and divide that vA load by 120 volts. While this is much easier and quicker, it is not the most accurate and in some cases it would require a larger service than necessary to meet Minimum NEC requirements.
The third method is to monkey around with incorrect math trying to equate the difference between 208 and 230 volts when calculating the single phase 208 volt loads but the fact remains that this is comparing apples with oranges. There is no relationship between 120 volt or 240 volts in a line to line 208 volt load. While I am not a vector math person, I have friends that are and we are convinced that the phase shift between the delta primary and the wye secondary does not change the fact that a single phase line to line load is a direct result of turns ratio. A 6,000 vA load is a 6,000 vA load if it is rated at 208 volts. When you apply to a different voltage then you have to determine the resistance of the load and if my memory is correct take the voltage squared over the resistance to fine the new volt-ampere rating.
In 1990 the NEC first showed a 208 volt column in Table 430.148 and prior to this time we had to make an adjustment for the currents listed in the 230 volt column when we were using a 208 volt motor.
The more I look at what the State of Oregon is doing, they are trying to make an adjustment to make the answers on a sheet of paper comparing the amp method to the volt-ampere method, work out correctly.
The more I play with these numbers the more I am convinced that the following method is to be used.
Calculate the following loads:
2 ? 15 kW heating units ? 3 phase 208v
1- 8 kW water heaters ? 3 phase 208v
4 ? 1,600 vA soft drink machines ? 1 phase 120v
1 ? 1,200 vA copy machine ? 1 phase 120v
1 ? 10 HP Motor ? 3 phase 208v
2 - 5 HP Motors ? 3 phase 208v
1 - Special Appliance 22.6 Amp ? 3 phase 208v
60 ? receptacles @ 180vA each ? 1 phase 120v
5 ? 12 kW feeders ? 1 phase 208v
**********************************************
Amperage Method
Work all of the three phase equipment first.
A B C N
41.63 41.63 41.63 0
41.63 41.63 41.63 0
22.20 22.20 22.20 0
30.8 30.8 30.8 0
16.7 16.7 16.7 0
16.7 16.7 16.7 0
22.6 22.6 22.6 0
192.26 192.26 192.26 0
Add in the single phase line to line loads:
57.69 57.69 0 0
0 57.69 57.69 0
57.69 0 57.69 0
57.69 57.69 0 0
0 57.69 57.69 0
173.07 230.76 173.07 0
Add in the single phase line to neutral 120 volt loads:
13.33 0 0 13.33
0 0 13.33 13.33
13.33 0 0 13.33
0 0 13.33 13.33
10 0 0 10
30 30 30 30
66.66 30 56.66 93.32
TOTALS
431.99A 453.02A 421.99A 93.32A
************************************************
Volt-ampere Method
Work all of the three phase equipment first:
A B C N
5,000 5,000 5,000 0
5,000 5,000 5,000 0
2,666.6 2,666.6 2,666.6 0
3,699 3,699 3.699 0
2,005 2,005 2,005 0
2,005 2,005 2,005 0
2,712 2,712 2,712 0
23,087.6 23,087.6 23,087.6 0
Add in the single phase line to line loads:
6,000 6,000 0 0
0 6,000 6,000 0
6,000 0 6,000 0
6,000 6,000 0 0
0 6,000 6,000 0
18,000 24,000 18,000 0
Add in the single phase line to neutral 120 volt loads:
1,600 0 0 1,600
0 0 1,600 1,600
1,600 0 0 1,600
0 0 1,600 1,600
1,200 0 0 1,200
3,600 3,600 3,600 3,600
8,000 3,600 6,800 4,800
TOTALS
49,087.6 50,687.6 47,887.6 11,200
If you take the single phase line to line load on phase A (23,087.6) and divide by 120 you get 192.3 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
If you take the line to line single phase load on phase A (18,000) and divide by ? the line to line voltage which is 104v, you will get 173.07 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
If you take the single phase line to neutral load on phase A (8,000) and divide by 120 you get 66.66 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
That being said, I believe that you have to break out your loads in the following order and then add together your totals.
First list all three phase loads in category #1, Secondly list all single phase line to line loads in category #2 and Thirdly list all line to neutral 120 volt loads into Category #3.
I welcome your comments. If you care to contact me directly my e-mail address is:
{Moderator's Note: Edited to remove personal contact information. If you wish to contact this person directly, then start by sending a Private Message.}
[ March 03, 2005, 06:51 PM: Message edited by: charlie b ]
I will take a chance and see if this will format the same using cut and paste from a Microsoft Word Document. If not, maybe someone can help me with being able to take my document and put it on this forum so that it makes sense.
I am of the opinion that there are four ways to do this problem at the very minimum! The method that some writers use is to take the total vA load of all three phases totaled together and divide by 1.732 X 208. If you have a totally balanced load, no problem but this is highly unlikely.
The second method I see a lot is to take phase A as a worse case scenario and apply the line to neutral on phase A and N, the line to line single phase load on line A & B, and the three phase load and apply to A, B & C. Then they total up the phase A load and this will be the worse possible case without balancing the system and divide that vA load by 120 volts. While this is much easier and quicker, it is not the most accurate and in some cases it would require a larger service than necessary to meet Minimum NEC requirements.
The third method is to monkey around with incorrect math trying to equate the difference between 208 and 230 volts when calculating the single phase 208 volt loads but the fact remains that this is comparing apples with oranges. There is no relationship between 120 volt or 240 volts in a line to line 208 volt load. While I am not a vector math person, I have friends that are and we are convinced that the phase shift between the delta primary and the wye secondary does not change the fact that a single phase line to line load is a direct result of turns ratio. A 6,000 vA load is a 6,000 vA load if it is rated at 208 volts. When you apply to a different voltage then you have to determine the resistance of the load and if my memory is correct take the voltage squared over the resistance to fine the new volt-ampere rating.
In 1990 the NEC first showed a 208 volt column in Table 430.148 and prior to this time we had to make an adjustment for the currents listed in the 230 volt column when we were using a 208 volt motor.
The more I look at what the State of Oregon is doing, they are trying to make an adjustment to make the answers on a sheet of paper comparing the amp method to the volt-ampere method, work out correctly.
The more I play with these numbers the more I am convinced that the following method is to be used.
Calculate the following loads:
2 ? 15 kW heating units ? 3 phase 208v
1- 8 kW water heaters ? 3 phase 208v
4 ? 1,600 vA soft drink machines ? 1 phase 120v
1 ? 1,200 vA copy machine ? 1 phase 120v
1 ? 10 HP Motor ? 3 phase 208v
2 - 5 HP Motors ? 3 phase 208v
1 - Special Appliance 22.6 Amp ? 3 phase 208v
60 ? receptacles @ 180vA each ? 1 phase 120v
5 ? 12 kW feeders ? 1 phase 208v
**********************************************
Amperage Method
Work all of the three phase equipment first.
A B C N
41.63 41.63 41.63 0
41.63 41.63 41.63 0
22.20 22.20 22.20 0
30.8 30.8 30.8 0
16.7 16.7 16.7 0
16.7 16.7 16.7 0
22.6 22.6 22.6 0
192.26 192.26 192.26 0
Add in the single phase line to line loads:
57.69 57.69 0 0
0 57.69 57.69 0
57.69 0 57.69 0
57.69 57.69 0 0
0 57.69 57.69 0
173.07 230.76 173.07 0
Add in the single phase line to neutral 120 volt loads:
13.33 0 0 13.33
0 0 13.33 13.33
13.33 0 0 13.33
0 0 13.33 13.33
10 0 0 10
30 30 30 30
66.66 30 56.66 93.32
TOTALS
431.99A 453.02A 421.99A 93.32A
************************************************
Volt-ampere Method
Work all of the three phase equipment first:
A B C N
5,000 5,000 5,000 0
5,000 5,000 5,000 0
2,666.6 2,666.6 2,666.6 0
3,699 3,699 3.699 0
2,005 2,005 2,005 0
2,005 2,005 2,005 0
2,712 2,712 2,712 0
23,087.6 23,087.6 23,087.6 0
Add in the single phase line to line loads:
6,000 6,000 0 0
0 6,000 6,000 0
6,000 0 6,000 0
6,000 6,000 0 0
0 6,000 6,000 0
18,000 24,000 18,000 0
Add in the single phase line to neutral 120 volt loads:
1,600 0 0 1,600
0 0 1,600 1,600
1,600 0 0 1,600
0 0 1,600 1,600
1,200 0 0 1,200
3,600 3,600 3,600 3,600
8,000 3,600 6,800 4,800
TOTALS
49,087.6 50,687.6 47,887.6 11,200
If you take the single phase line to line load on phase A (23,087.6) and divide by 120 you get 192.3 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
If you take the line to line single phase load on phase A (18,000) and divide by ? the line to line voltage which is 104v, you will get 173.07 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
If you take the single phase line to neutral load on phase A (8,000) and divide by 120 you get 66.66 Amperes. Same Math!! Same for Phase B & C. ?WORK THE MATH?
That being said, I believe that you have to break out your loads in the following order and then add together your totals.
First list all three phase loads in category #1, Secondly list all single phase line to line loads in category #2 and Thirdly list all line to neutral 120 volt loads into Category #3.
I welcome your comments. If you care to contact me directly my e-mail address is:
{Moderator's Note: Edited to remove personal contact information. If you wish to contact this person directly, then start by sending a Private Message.}
[ March 03, 2005, 06:51 PM: Message edited by: charlie b ]
Re: Single phase load on 208/120Y calculation req'd by Oreg
The formula given is amps=kva/208x1.154.
In the 2 phase system for PF = 1 the KVA =
2 x Ip x Vp. Therefore Ip = KVA/(2 x Vp)
It happens that 2 x 120 and 208 x 1.154 both = 240. IF you take an example of kva = 12 kva on
Phase A and 12 kva on Phase B, Using the formula
Amps = KVA/(2 x Vp) = 24/(2 x 120) = 100 amps.
Using the Oregon formula Amps = 24/(208 x 1.154)
= 100 amps.
It appears that the inspector has found a factor that provide the correct answer but its use is not readly clear.
If the load is not balanced the formula is incorrect.
The formula given is amps=kva/208x1.154.
In the 2 phase system for PF = 1 the KVA =
2 x Ip x Vp. Therefore Ip = KVA/(2 x Vp)
It happens that 2 x 120 and 208 x 1.154 both = 240. IF you take an example of kva = 12 kva on
Phase A and 12 kva on Phase B, Using the formula
Amps = KVA/(2 x Vp) = 24/(2 x 120) = 100 amps.
Using the Oregon formula Amps = 24/(208 x 1.154)
= 100 amps.
It appears that the inspector has found a factor that provide the correct answer but its use is not readly clear.
If the load is not balanced the formula is incorrect.
Re: Single phase load on 208/120Y calculation req'd by Oreg
The question that everyone is trying to understand is how to determine the volt-ampere rating of a 208 volt single phase load on a three phase wye system when the load is given in amps and you are performing your calculation using the volt-ampere method.
I have talked with Stan Penrose, acting Chief of Oregon regarding this matter and he is adament that you have to take the amp draw, example in point a 8.8 amp load and multiply it by 208 times 1.154 in order to get the correct volt-amperes. His method is incorrect as the vA load is what it is. 8.8 X 208v = 1,830.4 end of story! At that point you simply split it into half and apply 915.2 vA to phase A and 915.2vA to phase B (or C) and your done.
Mr Penrose thinks that the correct vA should be 8.8 X (208 X 1.154) for an answer of 2,112.28 vA and he is dead wrong.
There is no relationship to 120 volts in this question. 1.154 X 208 = 240.032.
This originally started when Mr. Penrose marked test questions wrong based on his belief that the only way to determine the correct vA load is by using "the Penrose formula"!! Since he writes the test monthly and grades the test, there is no checks and balances in this process.
I have had students in my class that have failed State Exams and had to go to his review and listen to him go into a long explanation on this subject and everyone go away with only one thing in common, do it the "Penrose method". For sure, no one understood why. I have yet to have someone explain to me how this is possible.
Please, someone help me put this myth to bed!
The question that everyone is trying to understand is how to determine the volt-ampere rating of a 208 volt single phase load on a three phase wye system when the load is given in amps and you are performing your calculation using the volt-ampere method.
I have talked with Stan Penrose, acting Chief of Oregon regarding this matter and he is adament that you have to take the amp draw, example in point a 8.8 amp load and multiply it by 208 times 1.154 in order to get the correct volt-amperes. His method is incorrect as the vA load is what it is. 8.8 X 208v = 1,830.4 end of story! At that point you simply split it into half and apply 915.2 vA to phase A and 915.2vA to phase B (or C) and your done.
Mr Penrose thinks that the correct vA should be 8.8 X (208 X 1.154) for an answer of 2,112.28 vA and he is dead wrong.
There is no relationship to 120 volts in this question. 1.154 X 208 = 240.032.
This originally started when Mr. Penrose marked test questions wrong based on his belief that the only way to determine the correct vA load is by using "the Penrose formula"!! Since he writes the test monthly and grades the test, there is no checks and balances in this process.
I have had students in my class that have failed State Exams and had to go to his review and listen to him go into a long explanation on this subject and everyone go away with only one thing in common, do it the "Penrose method". For sure, no one understood why. I have yet to have someone explain to me how this is possible.
Please, someone help me put this myth to bed!
Re: Single phase load on 208/120Y calculation req'd by Oreg
Mr. P. must be computing this first as two separate loads on 120V each which would yield his result. He then equates these two separate loads to a single, 208V load, and puts in the fudge factor to achieve the desired result. Wrong, wrong, wrong, as I have pointed out earlier.
What would he do with a delta config?
I presume the Chief Inspector is of no help. Maybe a note from some eminent authority, say someone like Mike Holt?
Mr. P. must be computing this first as two separate loads on 120V each which would yield his result. He then equates these two separate loads to a single, 208V load, and puts in the fudge factor to achieve the desired result. Wrong, wrong, wrong, as I have pointed out earlier.
What would he do with a delta config?
I presume the Chief Inspector is of no help. Maybe a note from some eminent authority, say someone like Mike Holt?
Re: Single phase load on 208/120Y calculation req'd by Oreg
I CAN'T STAY QUIET ANY LONGER. TEACHING AND INSPECTING IN OREGON FOR LONGER THAT I WANT TO ADMIT,I BELIVE MR SIMMONS IS (RIGHT ON) WITH CALCS. . I WOULD LIKE TO HEAR WHAT MIKE HOLT HAS TO SAY ABOUT THIS. THIS IS NOT THE ONLY PROBLEM WITH SALEM AND THERE (NEW MATH ?). ANY COMENTS FROM SOME OF YOU THAT REALLY KNOW!!!
I CAN'T STAY QUIET ANY LONGER. TEACHING AND INSPECTING IN OREGON FOR LONGER THAT I WANT TO ADMIT,I BELIVE MR SIMMONS IS (RIGHT ON) WITH CALCS.
. I WOULD LIKE TO HEAR WHAT MIKE HOLT HAS TO SAY ABOUT THIS. THIS IS NOT THE ONLY PROBLEM WITH SALEM AND THERE (NEW MATH ?). ANY COMENTS FROM SOME OF YOU THAT REALLY KNOW!!!Ed MacLaren
Senior Member
Re: Single phase load on 208/120Y calculation req'd by Oreg
Re: Single phase load on 208/120Y calculation req'd by Oreg
CCJersey posted
will give the correct answer in amps. This is the
same as the formula Ip = KVA/(2 x Vp) that I posted.
CCJersey posted
What we are talking about "I think" is taking phase A & B + a neutral (208 volts) to a single phase panel with loads such as lighting. The load is connected Phase to Neutral. If the loads are balanced then the formula amps=kva/208x1.154
will give the correct answer in amps. This is the
same as the formula Ip = KVA/(2 x Vp) that I posted.
Re: Single phase load on 208/120Y calculation req'd by Oreg
What about this? Perhaps Mr. P. is claiming that the apparent power, in KVA, delivered by the xfrmr is the summation of the phase voltages x the phase currents. After all, the xfrmr does not know whether loads are connected in wye or delta config. This makes a bit of sense to me, but I think it is clumsy to use a special formula for that.
What about this? Perhaps Mr. P. is claiming that the apparent power, in KVA, delivered by the xfrmr is the summation of the phase voltages x the phase currents. After all, the xfrmr does not know whether loads are connected in wye or delta config. This makes a bit of sense to me, but I think it is clumsy to use a special formula for that.
Re: Single phase load on 208/120Y calculation req'd by Oreg
Well I think I am going to have to eat a little more crow because I have made a mistake again.
It's almost like I have set a trend every since I started opening my big mouth on this web site.
One think for sure, I'm never to big to say I'm wrong. I think there is more to this story that has not been told and my friend Garrett will be posting another bit of information that we all need to take a close look at!
It's going to be an eye opener to all of us!
Keep Tuned In, the horse is not dead yet!
Well I think I am going to have to eat a little more crow because I have made a mistake again.
It's almost like I have set a trend every since I started opening my big mouth on this web site.
One think for sure, I'm never to big to say I'm wrong. I think there is more to this story that has not been told and my friend Garrett will be posting another bit of information that we all need to take a close look at!
It's going to be an eye opener to all of us!
Keep Tuned In, the horse is not dead yet!
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: Single phase load on 208/120Y calculation req'd by Oreg
A comment from Bob and a question asked twice by Rattus (with no response yet from Specs) have led me to the following suspicion: We (meaning all of us) don?t know what we are talking about! By that I am saying that we do not really know what question it is that requires that 1.154 factor.
Consider two panels. Both are rated 120/240 volt, 225 amp, single phase. Both have a connected load of (let us say) 42 KVA. What is the current on each panel? (Did everyone get the answer of 175 amps?) More to the point, do you do the calculation differently for one panel than for the other panel?
Now I will tell you what you did not know before: Panel 1 is fed from the secondary of a single phase, 480 volt to 120/240 volt transformer. Panel 2 is fed from the two of the three legs of a three phase, 480 volt to 120/208 volt transformer. I repeat my question: What is the current on each panel? Once again, do you do the calculation differently for one panel than for the other panel?
Is this what we are dealing with: a poorly worded test question that incompletely describes the configuration?
A comment from Bob and a question asked twice by Rattus (with no response yet from Specs) have led me to the following suspicion: We (meaning all of us) don?t know what we are talking about!
By that I am saying that we do not really know what question it is that requires that 1.154 factor.Consider two panels. Both are rated 120/240 volt, 225 amp, single phase. Both have a connected load of (let us say) 42 KVA. What is the current on each panel? (Did everyone get the answer of 175 amps?) More to the point, do you do the calculation differently for one panel than for the other panel?
Now I will tell you what you did not know before: Panel 1 is fed from the secondary of a single phase, 480 volt to 120/240 volt transformer. Panel 2 is fed from the two of the three legs of a three phase, 480 volt to 120/208 volt transformer. I repeat my question: What is the current on each panel? Once again, do you do the calculation differently for one panel than for the other panel?
Is this what we are dealing with: a poorly worded test question that incompletely describes the configuration?
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