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Insights AC Power Analysis – Part 2, Network Analysis - Comments

  1. Mar 30, 2016 #1

    anorlunda

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  2. jcsd
  3. Mar 30, 2016 #2

    ProfuselyQuarky

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    Part 1 was amazing! Can't wait to dig into this one :smile:
     
  4. Mar 30, 2016 #3
    Very interesting topic I never gave much thought to. Looking forward to the third part.
     
  5. Mar 30, 2016 #4

    anorlunda

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    We all take infrastructure for granted as long as it's working.
     
  6. Mar 30, 2016 #5

    anorlunda

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    Jim Hardy suggested a mechanical analogy that did not make it into the article.

    Imaging a cylindrical rotating shaft. How do we know if it is transmitting power from one end to the other?

    Stop the rotation, and paint a dotted line along the top surface. Then rotate again with you riding on the shaft like a rider on a horse. The shaft appears stationary to you. If the shaft is transmitting power from my end to the far end, it will twist and I can see the twist because of the dotted line. If the far end ships power to my end, the shaft twists the other way. Twist is another way of saying that the angle at my end is different than the angle at the far end.

    To increase the twist, i make the RPM at my end slightly faster than at the far end for a short duration. To do that, I must temporarily put in more power at my end, then return the power to the previous value.

    In the steady state, when twist is not changing. I can state with certainty that the RPM at my end is identical to RPM at the far end, and I don't need a computer to figure that out.

    It would be possible to send the same power down the shaft at different RPMs, so making the twist stop changing and making the RPMs return to normal, are two independent goals.

    What happens if load at the far end increases? Both ends of the shaft will start slowing down. I'll have to put more power in at my end, until the angle stops changing and the RPM is back to normal. When I get there, I should see that the twist has increased, and that the extra amount of power I need to achieve that, exactly matches the increase in load power.

    Which angle is zero? It doesn't matter. Twist is the same, no matter what angle you choose to be zero.

    A shaft has only two ends, but the power grid has many nodes with an angle at each node, so the analogy is not perfect. Analogies never are. The analogy is also primarily DC, and it does nothing to help us understand VARs, unless we start adding springs, flywheels, and oscillators making a simple analogy complicated.
     
  7. Mar 30, 2016 #6

    donpacino

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    I like those two articles. They provides a good explanation of a concept that can be very confusing.

    Have you thought about writing on piece on the effects of an increase in distributed loads such as windmills, solar panels, home battery banks (for off peak energy use) and plug in hybrid/electric vehicles on the grid? Particularly their effect on the operation and operation philosophy of grid control companies?

    The reason I ask is I feel with your background you might be able to tackle that problem.
     
  8. Mar 30, 2016 #7

    anorlunda

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    Thanks for the kind words.

    I'll give that some thought. If I did write such an article, it will cover both engineering and economics. As you can already see perhaps, discussion of the grid is inherently multidisciplinary.
     
  9. Mar 30, 2016 #8

    jim hardy

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    Sure glad we have a genuine power system guy here . I've only looked over the shoulders of some.

    Electrical, mechanical, control theory(downright right burly math - search on "Power System Stabilizer"), and a rapidly increasing share of computer science.
     
    Last edited: Mar 31, 2016
  10. Apr 1, 2016 #9
    Could you please explain a bit more the proposition that flow of reactive power is proportional to the magnitude of voltage difference between nodes?
     
  11. Apr 1, 2016 #10

    cnh1995

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    Let me see if I understood it correctly.
    Suppose a purely resistive load R is fed from an ac supply V through a transmission line with reactance X(negligible resistance). Transmission line reactance X is fixed. Now if load R is increased, more current is drawn from the supply. This causes a voltage drop across X and load voltage is reduced, increasing the difference Vsource-Vload. If the load is decreased, voltage across the load increases and Vsource-Vload decreases, reducing the reactive power flow.
     
  12. Apr 1, 2016 #11
    Thank you for your comment. But if resistance of load is responsible for reactive power flow, what is responsible for flow of active power?
     
  13. Apr 1, 2016 #12

    donpacino

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    imagine an AC voltage source with an inductor and resistor in series. The inductor will resist change in current.
    You can analyze this in the time or frequency domain. Looking at it in the frequency domain, the inductor will impart a phase shift on that signal. The larger the inductor is in relation to the resistor, the closer that phase shift will move to 90 degrees. Keep in mind at 90 degrees, the power would be purely reactive. At 0, the power would be purely real (at zero it would just be a resistor).

    A phase shift translates to a time delay in the time domain. with no time delay the sin waves will be the same (pure real power. With some time delay the power will lag slightly, because one signal lags, the difference between the 2 voltages will increase, proportionally to the reactive power (the closer to 90 degrees you get, the greater the instantaneous difference, and the greater the reactive power).

    note: i personally think that power is driven by the voltage and current, and the difference between voltages. But thats how I think of it. Voltage is MUCH easier to measure than current.
     
  14. Apr 1, 2016 #13

    cnh1995

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    The difference between angles of Vsource and Vload i.e. δ determines the active power. At δ=90°, you get maximum active power. But that reduces the terminal voltage much below the permissible limit and satbility is compromised. So, δ is kept 15-20°.
     
  15. Apr 1, 2016 #14
    What you wrote is helpful. If i understand your point, the voltage difference is due to a time difference between generator voltage and load voltage. But this phase/time difference is the cause of the real power as well according to the elegant analysis of the article which i don't understand
     
  16. Apr 1, 2016 #15

    donpacino

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    ahhhh. so I answered the wrong question... my apologies.. I'll try to get you an answer

    edit. to be honest. I'm having trouble coming up with a simple explanation. Einstein said if you can't explain it simply, you don't understand it well enough. I guess I need to hit the books.
     
    Last edited: Apr 1, 2016
  17. Apr 1, 2016 #16

    cnh1995

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    Screenshot_2016-04-01-21-31-34.png Screenshot_2016-04-01-21-32-19.png Screenshot_2016-04-01-21-32-19.png Screenshot_2016-04-01-21-32-57.png
    Source voltage is 10V. You can see increase in load angle (by increasing the load) increases the active power (from bottom to top) but reduces the load voltage.
     
  18. Apr 1, 2016 #17

    anorlunda

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    You can see a tutorial here, http://www.pjm.com/~/media/training...son5-power-flow-on-ac-transmission-lines.ashx

    and a derivation of the equations here, http://home.eng.iastate.edu/~jdm/ee553/DCPowerFlowEquations.pdf

    The article gives ##P=\frac{V_A\cdot V_B}{X}\sin(\theta)##.

    The equation for Q is ##Q=\frac{V_A\cdot (V_A-V_B)}{X}\cos(\theta)##
    but where ##V_A## and ##\cos(\theta)## are both close to one,
    the approximate equation is ##Q=\frac{(V_A-V_B)}{X}##

    Note that both of these equations use only ##V_A##, ##V_B##, and ##X##. We don't need to know the load resistance. That is why I try to discourage thinking about a load R at all. Focus on ##V_A##, ##V_B##, and ##X##

    I could assume that the load is some computer controlled device that manipulates stuff to draw the P and Q that it wants. If the load P and Q do not vary with ##V_B## then the load is not an impedance or a resistance. The load is just a load and how the load varies with voltage or frequency we don't know, but we don't need to know to calculate P and Q as above. So once again, I urge forget the load R.
     
  19. Apr 8, 2016 #18
    What you wrote is very helpful. I want to ask you something. Suppose that we have a generator that supplies electric power to a factory. When the workers of the factory start a motor, in order for a magnetic field to be created at the stator of the motor the stator of the generator is demagnetized. So the field current of the generator is increased in order to compensate. Is this correct?
     
  20. Apr 8, 2016 #19

    anorlunda

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    • It is rare for one generator to supply one factory. Think of the whole grid providing the supply.
    • The generator field is part of the voltage regulator loop. It will probably increase because extra VARs will be needed.
    • The generators main response to the motor will be to increase its angle.
    • Power generated does not come from the field. It is theoretically possible (but impractical) to use permanent magnets for the field. So you should stop thinking more power more field.
     
  21. Apr 8, 2016 #20
    Thank you for your reply. Small grids with only one or two generators running in parallel are very rare but are the rule in merchant and military navy. At these small grids the proposition that the flow of reactive power is proportional to voltage difference is still valid?
     
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