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three-phase power transmission, circuits, etc.

Discussion in 'Mathematics and Physics' started by PG1995, Dec 10, 2013.

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Well, dont worry too much about the source type, delta or wye, when considering the circuit with no neural wire.

    The reason for the higher currents could be because the differential voltage is higher: 208vac line to line as opposed to 120vac line to neutral.
    But to make a better assessment i think we need to consider different kinds of loads, including a fully 'balanced' load. To find out for sure, do another problem with a balanced load and compare. Note that the wye source voltage neutral is zero volts and so is the balanced load 'fake' neutral.
     
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  2. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    I don't really follow your suggestion. You are suggesting me to solve another problem for a balanced load but I don't see why. In case of a balanced load, neutral won't even be carrying any current; in other it's not even necessary. In my humble opinion, we only needed to compare 'with neutral' and 'without neutral' cases which we did and found the results without reaching any conclusion so far. Please guide me where I'm going wrong. Thanks.
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    Please note i was making a small point.

    The point is, you saw the currents higher with no neutral. I was simply saying that will not be the case if the load is balanced, and also that there may be loads that produce other unusual results.
    You had looked at one specific case where there was one L, one C, and one R, and of specific values.
    To find out more, you may have to try other load values and other load types.
    For some simple quick examples: Two C's and one R, two L's and one R, two C's and one L, etc.
    It may turn out interesting :)
     
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  4. dave

    Dave New Member

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  5. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    Every new problem can teach us something new if we are solving for the sake of learning and not for the sake of good grade. We still need to balance our learning habits. I know that your suggestion is good and I will learn more by solving cases of different loads but my time doesn't really allow this right now. So, let's take another bite at the problem we were doing.

    The circuit in this example has been solved without any neutral.

    We solved the same circuit using a neutral here.

    This data shows the line currents with and without a neutral.

    I also tried to summarize a new set of data. I believe that it might be helpful. From this set of data it looks like that more real power is delivered to the load without a neutral. What do you think? Thank you.

    Regards
    PG
     

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  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    Well, if you have calculated say 100 watts with a neutral and 200 watts without a neutral, then what can we say? If you did the calculation correctly (i dont know if you checked the phase error yet) then since 200 watts is greater than 100 watts we would say that without a neutral there is more power dissipated. Simple right? But you must ensure that all the calculations were done correctly first, including the phase angles.
     
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  7. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    I know that you also mentioned about the phase error in post #35 and #37 but I have checked the calculations again and everything looks okay to me now. This is the final correct version. You were saying that sign got swapped in final equation for I3. Please guide me with this error because I'm not able to see it. We can talk about other things later. Thanks

    Regards
    PG
     

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    Last edited: May 30, 2015
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Well i am sure it is just an algebra error.
    I really think it is better though to convert everything to rectangular first, then go from there, saving the polar form for the very end. It is usually not a good idea to mix forms because errors happen too easily.

    You know you could also verify you solution by simulation right? You should always have a back up method to check your first method because we as humans always make errors and the only way to check is to do the problem using a different method, which may just be a simulation. You should be doing this with everything you calculate. I cant stress this enough.
     
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  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Here is the complete worked out solution and verified using simulation...

    Va=Ampl
    Vb=-(sqrt(3)*j*Ampl)/2-Ampl/2
    Vc=(sqrt(3)*j*Ampl)/2-Ampl/2

    and after replacing Ampl with 120 we have:
    Va=120
    Vb=-(sqrt(3)*j*120)/2-120/2
    Vc=(sqrt(3)*j*120)/2-120/2

    The needed equations are again:
    eq1=Va-j*5*I1=0
    eq2=-Vb-10*(I2-I3)=0
    eq3=-Vc+Vb-10*(I3-I2)+j*10*I3=0

    and with the transformed source voltages replacing Va,Vb,Vc we have:
    120-5*j*I1=0
    -10*(I2-I3)+20*3^(3/2)*j+60=0
    -10*
    (I3-I2)+10*j*I3-40*3^(3/2)*j=0

    Solving the last two for I2 and I3 we get:
    I2=6*(sqrt(3)+1)*(j+1)
    I3=6*(j+sqrt(3))

    and solving the first one for I1 we get:
    I1=-24*j

    Then we have the currents:
    ia=i1=-24*j
    ib=i3-i2=-6*(sqrt(3)*j+1)
    ic=-i3=-6*(j+sqrt(3))

    so we have the current in the neutral (direction source neutral to load neutral):
    In=-ia-ib-ic=6*(sqrt(3)+1)*(j+1)+24*j

    and expand that a little we get:
    In=2*3^(3/2)*j+30*j+2*3^(3/2)+6

    and grouping real and imaginary parts we have:
    In=(2*3^(3/2)+30)*j+2*3^(3/2)+6

    which in floating point would be approximately:
    In=40.3923*j+16.3923

    so the real and imaginary parts are:
    RP=16.3923
    IP=40.3923

    and the amplitude is the square root of the sum of the squares of real and imag parts:
    Iampl=sqrt(RP^2+IP^2)=43.5918

    and the phase shift is the two argument inverse tangent function:
    PhaseRads=atan2(IP,RP)=atan2(40.3923,16.3023)=1.187191
    PhaseDegrees=PhaseRads*180/pi=68.02 degrees

    and a quick measurement yields 68.4 degrees phase shift subject to some small graphical error.

    When i use a little more accuracy for the phase shift i get 67.9113 degrees.

    You know what else is interesting, if you swap two phases you probably get a different result entirely. If you swap B and C for example, you get a different result for the neutral current.
     
    Last edited: May 31, 2015
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  10. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    Everything else was correct and nothing had been swapped. The mistake occurred because I didn't take the negative of Ia+Ib+Ic. :(

    [​IMG]

    It also means that my other calculations were correct because the line currents had been calculated correctly. We will get back to what we were trying to discuss.

    Regards
    PG
     

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    Last edited: Jun 1, 2015
  11. PG1995

    PG1995 Active Member

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    Hi again

    The circuit in this example has been solved without any neutral.

    We solved the same circuit using a neutral here.

    Q1:
    This data shows the line currents with and without a neutral.

    I also tried to summarize a new set of data. I believe that it might be helpful. From this set of data it looks like that more real power is delivered to the load without a neutral. What do you think?

    Q2:
    Personally I don't have any experience with 3 phase loads. If my 'set of data' in Q1 above is correct then it leads us to conclude that 3 phase load which has neutral terminal differ from the one which doesn't have one assuming everything else is the same, right? Thank you.

    Regards
    PG
     

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  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yes that's right, for unbalanced loads.
    If the load is balanced though then the center should be zero also. If there is a neutral, then no neutral current flow. If there is no neutral, then obviously no neutral current flow but still zero volts.
    You might also note how the system with no neutral acts like a delta with a wye load.
     
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  13. PG1995

    PG1995 Active Member

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    Thank you, MrAl.

    Q1: So, you are saying that for an unbalanced load more real power is delivered without a neutral (in case of Y-Y system) assuming everything else is kept the same like the phase impedances. Right?

    Q2: From the discussion above, it can be said that a wye 3 phase load which has neutral terminal differ from the one which doesn't have one, right?

    Yes, I agree with you. If the load is balanced then neutral isn't required and also having it would make it any difference.

    Q3: Please confirm this. Perhaps, you are saying that I should also consider the case of delta source and wye load without a neutral (or, we cay say that where neutral cannot be incorporated because delta source doesn't come with a neutral terminal). Thank you.

    Regards
    PG
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Q1:
    I never said that. I simply said that the loads with and without a neutral differ from one another, unless the load is balanced.
    Consider the case of one 10k resistor phase B to neutral, one 10k resistor phase C to neutral, and one 10 ohm resistor from phase A to neutral. The current in the 10 ohm resistor is 120/10=12 amps with a neutral, and will be much less with two 10k resistors in parallel in 'series' with it. That means a great reduction in 'real' power.
    Consider the case with three unequal capacitors, one on each phase to neutral. There's no real power anyway so 'real' power can not increase.

    Q2:
    You seem to be asking this question over and over again. I keep saying YES, except when the load is balanced. What dont you understand about this?


    Q3:
    Maybe we should save this one for later until you start to get the idea of the others.
    When you have three phases with no neutral, it's almost like a delta even if it is a wye because you have three differnt phases with no neutral for both cases. The neutral of the wye has no effect because you are not allowed to use it, so it looks like a delta although the voltage levels may be different. This is just a simple idea, no complications...you have three phases separated by 120 degrees in each system, with no neutral in either system, so they are similar, so the basic analysis would be the same. But there could be real life differences where impedances and regulation come into play. For now, just think of they as similar when there is no neutral.

    The bottom line of any of this is analysis. Analysis of a system with a particular load compared to the analysis of another system with a different load, or with a neutral or without a neutral, etc. You can do a few examples to confirm any of your suspicions.
    I'll try to post a few tomorrow if i can.
    You should understand that people dont start to understand three phase systems or any other system for that matter by black magic -- they think, they do analyses, they reflect upon those analyses, they draw conclusions.
    When a new drug is designed it is not sure if it works as intended, so there are tests done, and these tests are done not one at a time but perhaps 25 or 50 or even 100 at a time to see the reaction in various situations. If they had to do this one at a time it would take 100 years to test a single drug.
    When we want to know something, the best way is to do analysis and think about what it takes to answer the question by way of that analysis. For us, we dont have to have 100 pipettes just to experiment, we can use the computer for all of the calculations, which means we can come up with 1000 trials in less than a second. Comparing results, we can draw significant conclusions.
    For this problem we could probably generalize the load into one R, one C, and one L for each phase, then vary the three for each phase little by little, then maybe do some graphs or just print some numerical data. What i think we would see for unbalanced loads with no neutral is either one dominant phase or two that combine to dominate, and either of these would tell us the significant aspects of that particular load. The less dominance we see the more closely the system will resemble the system that actually does have a neutral i think, unless the loads are very much different in character.
     
    Last edited: Jun 3, 2015
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  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again PG,

    I am about set up to do multiple experiments now, as it took a little while to prepare this mostly because i had other things coming up that had to be taken care of, like car, house, etc.

    Here are some somewhat simple results, where each set is a set of 5 experiments, where the start is always with the original problem set load values, and the loads are changed as noted. What isnt shown yet is the sets where there is no neural, but hopefully i'll get to that next.

    All results are shown as Amplitude of the neutral current, and phase angle of the neutral current as we did before.

    Set 1:
    Adding a parallel 0.1 Farad cap to all three phases, four times in a row:
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406

    Adding a 100 ohm resistor in parallel to phase A, B, and C, four times in a row:
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406
    43.585339 67.900406

    Adding a 100 ohm parallel resistor to phase A only, four times in a row:
    43.585339 67.900406
    43.148205 69.376960
    42.740302 70.882714
    42.362475 72.416403
    42.015534 73.976555

    Adding a 100 ohm resistor in parallel to phase A and B only, four times in a row:
    43.585339 67.900406
    44.332602 69.124274
    45.099416 70.307048
    45.884803 71.449806
    46.687824 72.553680

    Note that adding anything in parallel to all three phases (one additional parallel element each on each phase) does nothing to the neutral current, but adding to only one or two phases changes the neutral current.

    If not anything else, at least this gives you more examples to try yourself.
    Note that when a phase load is changed the additional element is added in parallel, then another is added in parallel (so now there are two of the same in parallel) and then another is added in parallel (now three additional elements in parallel( and finally a fourth element in parallel. It is always the same type of element, either a cap or resistor, but we can do inductors too later.
    We always start with the single element for each phase as in the original example with only one inductor, one resistor, and one capacitor, and so you will recognize the amplitude and phase angle of the current for each set starting computation.
     
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  16. PG1995

    PG1995 Active Member

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    Thanks a lot, MrAl, and I'm sorry for getting back to you a little late.

    I would need your help to clarify few points from your previous post.

    Do you mean to say that an element is added in parallel position four times?

    Would would put the element across green dots or red dots like shown here for phase A?

    \\

    Did you miss something in the text in red? Thank you.

    Regards
    PG
     

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  17. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Let me explain using your drawing.

    The load for phase A for example is connected across the two RED dots. This is the same as the two GREEN dots for this circuit but later if we remove the neutral it wont be the green dots just the red dots.

    So first we do a calculation with no extra components.
    Next, if we are to connect 100 ohm resistors, we first connect 100 ohm across the two red dots, but do the same on all three phases if doing all three phases, then do a calculation and that calculation leads possibly a different amplitude and phase angle.
    Next, we connect another 100 ohm resistor in parallel with the first one, which gives us a total of 50 ohms (two 100 in parallel) across the red dots, and the same on all three phases if we are doing all three phases the same.
    Next, we connect still yet another 100 ohm resistor in parallel to that, so that puts a 50 ohm in parallel with a 100 ohm.
    Next, and finally, we connect a fourth 100 ohm resistor across that, so we end up with phase A (and other phases) with only 25 ohms across it as well as whatever was originally there (like the inductor from the original problem).

    The only change in the above is if we only do one phase, like phase A, then we only connect resistors across phase A and no other phases. If doing two phases, then we only connect resistors across those two phases, and if doing all three phases we connect them across all three phases. So doing all three phases we would do:
    1. 100 ohm across phase A, another 100 ohm across phase B, another across C. That's three different resistors.
    2. Next, another 100 ohm across each phase, which would be three more resistors unless we combine them in parallel and only use one 50 ohm resistor for each phase.
    3. Next, another resistor on each phase.
    4. Finally, one more resistor across each phase so each phase now has a total of 25 ohms across it (four 100 ohm resistors in parallel is 25 ohms).

    If we use capacitors we do it with caps instead of resistors.
    After each change, we do another calculation and record the resulting amplitude and phase angle.

    The whole idea is to generate more examples, so by stepping the loads to each phase we can get a variety of examples (and tests) to work with. Doing them by hand should lead to the same results as the posted values.

    I have it set up now where i can add any element across any phase and get a result in as long as it takes to enter the values. If the values are stepped, then i can show several results after entering the base values and the step value.
    I did this so we could look at more examples without having to calculate every single one by hand which takes forever :)
    I could probably rig it up to do the circuit with no neutral too, given another 15 minutes or so.
     
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