Is this problem/solution accurate?

Discussion in 'Homework Help' started by ElectricMagician, Jul 26, 2012.

  1. ElectricMagician

    Thread Starter Member

    Jul 26, 2012
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    0
    Our professor posted the solution to a problem and I really have a problem understanding it.

    1. The problem asks us to adjust a capacitor to modify the sending end. I don't understand how can a circuit element modify the voltage source?

    2. In the solution at some point (second line) it says Ic = ..... = jIc, and it continues with that assumption. How is that possible?

    Is this problem and its solution accurate?

    Thank you very much for your help!

    problem and solution are attached
     
  2. mlog

    Member

    Feb 11, 2012
    276
    36
    1. The capacitor does not "modify" the voltage source in the sense that it changes the output of some generator. The capacitor changes the VARs in the system, which I call reactive power. The capacitor allows the generator output to be changed.

    The capacitor affects the current flowing through the line, so it affects the voltage on the line. If the generator is to maintain its output voltage at a certain value, say 6400 volts, then it will have to adjust its voltage and phase angle. (Generators are like motors in that both can have current out of phase with voltage.)

    You weren't asked to calculate the original (pre-capacitor) generator voltage as part of the problem, but if you had done so, you would have found the source voltage was 7.1 kV. Compare that to new 6.4 kV with the capacitor installed at the load end. The capacitor reduced the magnitude of the line current from 31 A to 19 A, because the generator no longer has to supply a large reactive current.

    2. Someone got a little sloppy when writing the equation. The Ic on the far left side is a complex variable. For all practical purposes, the capacitor has no real component of current, so the current would be of the form "0 + jIc" or simply jIc as written on the second line.

    Notice the load current is lagging, i.e. inductive and has a negative value of imaginary current (-j25). The capacitor has a positive value of imaginary current. That's why the capacitor is able to "cancel out" the lagging inductive component.

    As far as I can tell, the solution is accurate, although when I solved the quadratic equation, I got Ic=28.83 rather than 28.33. It looks like the solution either has a subtraction error or a typo.
     
  3. ElectricMagician

    Thread Starter Member

    Jul 26, 2012
    57
    0
    Hi mlog,

    Thank you very much for your generous help. I understand it better now, but I still don't get how the notation in the solution is accurate


    As I understand (assuming re Ic = 0) then
    jIc = j × (re Ic + j im Ic) = -1× im Ic

    This, of course, is a different value than if we use "jIc" to mean j(im Ic)

    So both notations are not equal.

    Now on the 2nd line it says
    j V÷Xc = jIc << I think by "jIc" he really meant j(im Ic)

    but on line 4 this jIc is actually used to mean the first jIc described above (since the j is separated and Ic is solved for)

    I'm surprised that this confusion in notation does not change the solution, not even by a "j" coefficient.
     
    Last edited: Jul 27, 2012
  4. mlog

    Member

    Feb 11, 2012
    276
    36
    As I said, in the second line, someone (the professor?) got sloppy with notation. In the first part, the "Ic" was intended as a complex number, which could possibly have both a a real and an imaginary part (even though we all agree in this problem there is no real component of Ic). Later on in the same line, he used Ic to represent a value. Ic could have meant a complex number or a magnitude. It was used in both ways, and I can understand why it confused you.

    When he says "jIc," the Ic is a value or a number. The "j" in front represents that it is imaginary. In this problem, the sloppiness didn't matter, because Ic the magnitude (28.83) was the same value as the imaginary part (28.83). Good for you that you noticed the professor's error. When I saw it, I knew what he meant and ignored it.
     
  5. ElectricMagician

    Thread Starter Member

    Jul 26, 2012
    57
    0
    Thank you very much.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Attached is a solution based purely on the phasor diagram method of solving for the required capacitance to give equal source and load voltage magnitudes. No complex algebra used. Just trigonometric relationships. Thought it might be of interest.
     
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