Page 9 of 13 FirstFirst ... 7891011 ... LastLast
Results 81 to 90 of 122

Thread: Single phase load on 208/120Y calculation req'd by Oreg

  1. #81
    Join Date
    Nov 2003
    Posts
    9

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Charlie B:

    Please allow me to try and put forth a question that everyone can understand.

    If I am performing a load calculation on a three phase 208/120 volt system and I desire to use the volt-ampere method of calculation instead of using the amp method, how do I show my loads for a 208 single phase load.

    Do I simply multiply my amp load times 208 to get my vA numbers or do I have to plug in some factor, possibly 1.154 times the line to line 208 volt to get the correct answer.

    Another way to put it would be, if I have a 8.8 amp load rated at 208 volt single phase and I am attempting to convert it to vA would the answer be 8.8 X 208 = 1,830.4, half on phase A and the other half on phase B or would the answer be 8.8 X (208X1.154)= 2112.28 with half on phase A and the other half on phase B.

    As you can see the vA load is increased and the mystery to me is we go to NEC 220.54 & 220.55 of the 2005 NEC, NEC 220.18 & 220.19 of the 2002 NEC and we are allowed to actually lower the ratings of the Dryers and Ranges using the basis of twice the maximum number connected between any two phases for single-phase dryers and ranges fed from a 3-phase, 4-wire feeder or service.

    This just seems strange to me! Because we are placing a single phase motor load on a three phase system we have to increase the vA rating and yet on Dryers and Ranges we can lower the total vA rating?????????

    I might go on to say that several months ago I drove to Salem, Oregon and talked with the individual that is writing and grading part II of our Supervisor Test and he indicated to me that it was because of the phase shift that the 1.154 X line to line voltage had to be used when converting amps to volt-amperes on a wye three phase system, where a single phase 208 volt load was applied.

    I trust this will explain the question. There is nothing more to the question than How do I convert amps to volt-amperes for a single phase load, 208 line to line on a three phase wye, 208/120 volt system!

    I also would like to know why none of the well known authors speak to this issue or maybe I need someone to point me to such material!

    I do not wish to go on and on being negative on this subject but I really need to understand the facts to effectively teach my students.

  2. #82
    Join Date
    Feb 2003
    Location
    Tennessee
    Posts
    1,566

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Most all have been taught and understand that when we want the load in amps. if we have the Kw or Kva of a single phase load we divide.
    Amps.=Kw/E

    I think the constant of 1.154 could be put to good use to find the amount that a load draws at 208 Vac. if the load was designed for a working voltage 240 Vac. applied to it.

    The wattage output will not be as high used on the 208 Vac. but you can use this constant and not have to go through arithmetic of finding the resistance in ohms of the load.

    A 6,000 watt heater draws 25 amps. at 240 volts its wattage output is 6,000 watts.
    25=6,000/240



    A 6,000 watt heater draws 21.8 amps at 208 volts its wattage output is 4,534 watts.
    21.8=6000/(240/1.154)



    Ronald

  3. #83
    Join Date
    Feb 2003
    Posts
    898

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    fsimmons,
    I agree with what you said earlier - "this myth needs to be put to bed".

    if I have a 8.8 amp load rated at 208 volt single phase and I am attempting to convert it to vA would the answer be 8.8 X 208 = 1,830.4, half on phase A and the other half on phase B
    It seems to me that some of the confusion that is developing in this thread is caused by the idea that a va or kva load can be divided between two conductors, in the same way that a current can.
    One cannot attribute half of a va load to one conductor of a circuit, and the other half to another conductor, any more than one could with a voltage.
    A va or kva load can only be associated with a circuit, not a conductor.

    The total kva load on a transformer is the sum of the kva loads on all of the circuits supplied by the transformer, however, as someone noted in another thread, a transformer is current limited, that is, an overload is caused only by excess current in the windings.


    he indicated to me that it was because of the phase shift that the 1.154 X line to line voltage had to be used when converting amps to volt-amperes on a wye three phase system, where a single phase 208 volt load was applied.
    That is plain wrong. Phase shift does not affect a va or kva calculation.

    The "usefulness" of the volt-amp unit in rating an alternator or transformer is to "emphasize" the loading effect of any current in the windings, whether it is in-phase or completely (90 deg) out-of-phase.

    Ed
    (edited typo)

    [ March 04, 2005, 09:36 AM: Message edited by: Ed MacLaren ]

  4. #84
    Join Date
    Nov 2004
    Posts
    3,172

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Fsimmons,

    I would coach my students to understand the basics in this matter rather than relying on some obscure formula. For example,

    Apparent power, in VA, delivered by the transformer is 120 x 2 x I.

    Apparent power, in KVA, consumed by the load is 208 x I.

    If the students know this, they should be able to make the right choice on the state exam especially if they have been warned about the magic fudge factor.

    This discrepancy occurs because 120 + 120 = 208 when combined vectorially as opposed to albegraic addition where the sum is 240 as in a 120/240 single phase service.

    A tempest in a teapot now that we understand the question, I think.
    Don't mess with B+!
    (Signal Corps. Motto)

  5. #85
    Join Date
    Nov 2004
    Posts
    3,172

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Ed,

    Again, I beg to differ. The apparent power in this case is provided by two legs of a wye. From the transformers' point of view, each is providing 120 x I VA, and that is of prime importance. This computation should be used when sizing the transformer.

    However, the load's point of view, it is only receiving 208 x I VA. It feels slighted. How come? It is because the phase angles of the phase voltages cause the vector sum to be 208 instead of 240. PF, as you indicate, is not a factor.

    I think we have stumbled onto a little known pitfall one might enounter when computing transformer loading with unbalanced loads.
    Don't mess with B+!
    (Signal Corps. Motto)

  6. #86
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    17,140

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Originally posted by rattus: The apparent power in this case is provided by two legs of a wye. From the transformers' point of view, each is providing 120 x I VA, and that is of prime importance.
    I would say that it is of no importance. It fits into the category of “true, but irrelevant.” The load taken by a single phase 208 volt load is indeed shared equally between two legs of a WYE secondary. But so long as we make a reasonable effort to balance the loads among the three phases, we do not ever need to know the exact amount of load that any given secondary winding is supplying to any given load.

    I use a Panel Schedule with built-in demand factors (and the occasional 25% add on factor) to calculate the total load on a transformer. The panel schedule computes the VA that each load takes from each phase. This is a “background” part of the process of sizing a transformer; it does not get any of my attention. I only need to check to see that the load is reasonably balanced, and then to look at the total load.

    However, the load's point of view, it is only receiving 208 x I VA. It feels slighted. How come? It is because the phase angles of the phase voltages cause the vector sum to be 208 instead of 240.
    It does not feel slighted. If the load is rated for 208 volts, and if it gets 208 volts, then it is getting what it needs.

    I again suggest that the confusion here comes from the following three concepts:
    </font>
    • <font size="2" face="Verdana, Helvetica, sans-serif">A load can be rated for 208 volts. A load can be rated for 240 volts. A load can be rated to handle any voltage in a given range (e.g., from 208 to 240 volts).</font>
    <font size="2" face="Verdana, Helvetica, sans-serif"></font>
    • <font size="2" face="Verdana, Helvetica, sans-serif">A panelboard can be rated for 120/208 volts (these are 3 phase boards). A panel can be rated for 120/240 volts (these are single phase boards).</font>
    <font size="2" face="Verdana, Helvetica, sans-serif"></font>
    • <font size="2" face="Verdana, Helvetica, sans-serif">A power source (e.g., transformer secondary) can supply 208 volts. A power source (e.g., transformer secondary) can supply 240 volts.</font>
    <font size="2" face="Verdana, Helvetica, sans-serif">We will never understand the purpose (if any exists) of the 1.154 factor until we know for certain what the question uses as its given configuration, in terms of the rating of the load, the rating of the panelboard, and the rating of the power source.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  7. #87
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    17,140

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Originally posted by fsimmons: . . . if I have a 8.8 amp load rated at 208 volt single phase and I am attempting to convert it to vA would the answer be 8.8 X 208 = 1,830.4, half on phase A and the other half on phase B . . . ?
    My responses are “Yes,” and “Don’t bother.”

    Yes, the load is 1830.4 VA.

    Don’t bother assigning half to Phase A and the other half to Phase B. You can add up all the loads in this fashion, and can assign loads to each phase as you go, but that is a tedious process. It is better to let Excel do the math for you. Keep in mind that the only purpose for doing such a tedious calculation would be to verify that the loads are balanced among the three phases. There are easier ways to go about that task. As far as the total load is concerned (e.g., to size the transformer), you do not need to add up the load phase by phase. Just add up the total load, and apply the applicable demand factors.
    Charles E. Beck, P.E., Seattle
    Comments based on 2014 NEC unless otherwise noted.

  8. #88
    Join Date
    Mar 2005
    Posts
    8

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Hello to the Forum,
    I am new to the site, and to the State of Oregon. I was referred to this forum by a colleague who is experiencing the same difficulties as I, mainly obtaining the Oregon State Electrical Supervisors License. The reluctance of Salem to admit out of state journeymen to apply is very frustrating. It cannot be fair and balanced to see that the very same person who I had to argue with in the verification process is now going to be the person who will evaluate my test score. And to make things even more skewed, this is the person who created the test.
    That said, the question of how to calculate a single phase load on a wye connected circuit has always been, until now a very straight forward question. Charlie B has it exactly right and there is no apparent reason to use any factor to adjust apparent power.
    Is this factor being used in any of the apprentice programs? Is anyone here from JATC? Do they use this in their training material?
    To carry this a bit further, I can only wonder what the folks in the manufacturing industry such as Allen Bradley, Megger, Square D, Dobble, and United Labs, are going to think when the realize that they have to rewrite their engineering standards to fit this previously unheard of factor.
    If the State of Oregon electrical licensing program is trying to find "Qualified" electricians to fill the need for Journeymen and Supervisors within the state, what use is it to test for ones ability to use obscure methods of calculation? It is not covered in the NEC See Examples D5(a) and D5(b). And I cannot find an educational or apprentice program that uses it. What standard is the state testing to, shouldn’t it be the NEC? And if it isn’t then the state is obliged to make its standards known.
    If my qualification to be an electrical supervisor is my ability to use obscure reasoning to solve problems then how does that help the state maintain a strict standard for the application of the NEC to the electrical services being installed and maintained throughout the state? The issue that is most important to being a qualified electrician at any level is whether she or he is using the same codes and standards as the rest of the industry. The use of 1.154 is not any standard that I or any of my learned colleagues are aware of.
    I would ask Governor Ted Kulongoski if the current licensing process is a fair and balanced system and if it is the best way to attract a healthy and diversified group of experts to lead the electrical industry within the state? Creating obscure conditions for licensing that require some kind of insider knowledge such as that being taught by, “those in the know”, is anti competitive.

    [ March 04, 2005, 08:29 PM: Message edited by: ray94553 ]

  9. #89
    Join Date
    Feb 2005
    Posts
    12

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    Hi All, Sorry to take so long to get back on this thread, but I wanted to make sure that my ducks AND electrons were in a row. So with that in mind I will once again enter into this morass and try to make sense of the muddle that has been presented here.

    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.

    As a further confirmation of the validity of that factor, try a little math. Consider three 12000W resistive loads connected to a three phase Wye source in a balanced configuration. ( 12Kva on a-b, 12Kva on b-c, and 12Kva on c-a) The accepted way to calculate a BALANCED three phase load on a three phase system is to apply the formula: total va/(voltage [line to line] * sqrt(3)). SO if the total load is 12Kva*3 then= 36000va is the total load. 36000/(202*sqrt(3))=99.926 Amps on each line conductor to the transformer. OK so far?

    Now we look at each load separately and use the common formula: va / voltage [line to line]. 12000/208 = 57.692 Amps on each line conductor back to the transformer. If this is completed around the wye we would have 57.692 Amps * 2 = 115.385 amps on each line conductor back to the transformer.

    We have a slight difference, or not so slight depending on your point of view. If the total line amperage is multiplied by the cosine of 30 degrees { 115.385*cos(30)} we get 99.926 amps on each leg. If we DIVIDE 115.385 by the reciprocal of cos(30), we get 115.385/cos(30) = 99.926 amps.

    It would appear that the old square root of three formula that we were all taught to use compensates for that phase shift. Even better is to use the formula: square root of [A squared + B squared + C squared] ie:{ sqrt( sqr(57.692A)+ sqr(57.692A) + sqr(57.692A) } we get 99.926 amps. Too scary.


    This elusive factor equals so many things is what makes it hard to pin down.

    1.1547 = reciprocal[cos(30)]
    1.1547 = 240 / 208 (not exact, but pretty close)
    1.1547 = (2/3)*sqrt(3)
    1.154 = sunspot activity on Rigal4

    This formula factor DOES seem to have its place if one is trying to closely engineer a load.
    In my OPINION, the code has always seemed to work off apparent Watts, or va, NOT true Watts. This would allow a fudge factor for less efficient motors, transformers and such. Are we now going to base our calculations off TRUE Watts? Wow, what a can-O-worms.


    I would have to agree with FSimmons that the way this information has been presented here in Oregon is backwards. If a fundamental change is noted in science, it is first disseminated to the professionals who teach those fundamentals, and then LATER is tested upon.

    Is there anyone on this forum who sits on a committee, or code panel that deals with calculations for the code??? What is the intent of the code in this issue??

    Ed, The horse is still alive and well.

    disclaimer: this is an attempt to find a reason. I STILL don't know for sure.
    Garrett
    Let\'s Boogie in A, Really Loud!

  10. #90
    Join Date
    Feb 2003
    Location
    Alabama
    Posts
    2,621

    Re: Single phase load on 208/120Y calculation req'd by Oreg

    grlsound
    I think you have a contradiction in your post.
    The accepted way to calculate a BALANCED three phase load on a three phase system is to apply the formula: total va/(voltage [line to line] * sqrt(3)). SO if the total load is 12Kva*3 then= 36000va is the total load. 36000/(202*sqrt(3))=99.926 Amps on each line conductor to the transformer.
    This is correct for a delta 3 phase load.

    Now we look at each load separately and use the common formula: va / voltage [line to line]. 12000/208 = 57.692 Amps on each line conductor back to the transformer. If this is completed around the wye we would have 57.692 Amps * 2 = 115.385 amps on each line conductor back to the transformer.
    You stated above correctly that the line current
    was 99.9 amps. now you have changed it to 115 amps. You can not add currents directly. It is a vector addition. 2 * 115 is incorrect. You can add the vectors or use the ole sqrt of 3.
    57.7 x 1.73 = 99.8 amps as you said first.
    If the total line amperage is multiplied by the cosine of 30 degrees { 115.385*cos(30)} we get 99.926 amps on each leg. If we DIVIDE 115.385 by the reciprocal of cos(30), we get 115.385/cos(30) = 99.926 amps
    I'm not sure where you are going with this. You seem to be doing the same math problem over and over.
    You caculations are for a balanced delta connected load.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •