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Thread: Line Reactor Ratings

  1. #11
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    Quote Originally Posted by topgone View Post
    Why go through the trouble of testing for x/r when you can just test for x and then r?
    For the test on x, you impress a suitable voltage and measure the current and apply the necessary computations and get x. For r, you need a voltage source with near zero frequency and use a wheatstone bridge to get the resistance.
    I'm not looking to test for the x/r of the reactor. I'm looking to determine if there is a standard for how series reactors are given their rating. Typically a standard of this nature would define a range for which the x/r ratio of the source used for testing needs to fall in. You can't stamp a rated let-thru on the equipment without knowing the 65kA source impedance and x/r ratio.

    If you know the source x/r ratio that it was tested at, you can then calculate the source impedance (in terms of r and x) and then from there the only unknown variable in the equation would be the r of the reactor and it could be calculated. I was hoping for a workaround for calculating r of the reactor without having access to the test report.

  2. #12
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    Quote Originally Posted by ish1284 View Post
    I'm not looking to test for the x/r of the reactor. I'm looking to determine if there is a standard for how series reactors are given their rating. Typically a standard of this nature would define a range for which the x/r ratio of the source used for testing needs to fall in. You can't stamp a rated let-thru on the equipment without knowing the 65kA source impedance and x/r ratio.

    If you know the source x/r ratio that it was tested at, you can then calculate the source impedance (in terms of r and x) and then from there the only unknown variable in the equation would be the r of the reactor and it could be calculated. I was hoping for a workaround for calculating r of the reactor without having access to the test report.
    you don't need the source x/r, only its Z

    why do you need it anyways?
    for SC calc of this nature Z is sufficient

    from the data it can't be calculated
    you know the Xl = 2 Pi 60 67 uH = 25.2584 1ph, 43.7488 3ph
    you know total Z = 480/9415 = 50.9825 3ph
    you know source Z = 480/65000 = 7.3846 mOhm 3ph
    you know reactor Z = total - source = 43.5979
    problem is reactor Xl > reactor Z

    we can assume the 9415 includes X and R, since that is the reactor output

    try another approach
    system Vdrop at 9415 = 69.5262 v, so at reactor v = 410.4738
    reactor i = 410.4738/43.5979 x 1000 = 9414.9912

    current without reactor R = 480/(system Z + Xl) = 9387.2004 (lower since no R contribution)
    new source vdrop = 9387.2004 x 7.3846/1000 = 69.3209 or 410.6791 at reactor

    S = sqrt3 (9415 x 43.5979/1000) (9415) = 6693.7023 kva
    Q = sqrt3 (9387.2004 x 43.7488/1000) (9387.2004) = 6677.2701 kvar
    P = sqrt(6693.7023^2 - 6677.2701^2) = 468.7371

    pf = 468.7371/6693.7023 = 0.0700, 85.9845 deg, x/r = 14.2453

    R = 3.0711 mOhm

  3. #13
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    vdrop at 590.6 amps ( 590.6 = 491000/(480 x sqrt3) )

    590.6 x 43.5979/1000 = 25.75 or 5.36%, likely a rated 5% reactor

    power loss ~ 3 reactors x (3.0711/1000 x 590.6) [R x I = vdrop across reactor] x 590.6 [I thru reactor]= 3.2 kw

  4. #14
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    IEEE paper on reactors
    my calcs match
    x/r is basically irrelavent
    http://www.ewh.ieee.org/soc/pes/swit...S10HVCBa11.pdf

    pg 55 shows required and optional nameplate data
    x/r is not even optional

    in school reactors were always considered 0 + Xj
    the delta X to Z is very small

    pg 32 https://www.cedengineering.com/userf...lculations.pdf

    When the ratio of the reactance to the resistance (X/R ratio) of the system impedance is greater than 4, negligible errors (less than 3%) will result from neglecting resistance.

    On systems above 600 volts, circuit X /R ratios are usually greater than 4 and resistance can generally be neglected in short circuit current calculations. However, on systems below 600 volts, the circuit X/R ratio at locations remote from the supply transformer can be low and the resistance of circuit conductors should be included in the short circuit current calculations. Because of their high X/R ratio, rotating machines, transformers and reactors are generally represented by reactance only, regardless of the system voltage, except transformers with impedances less than 4%.
    Last edited by Ingenieur; 07-10-18 at 08:48 PM.

  5. #15
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    reactor terminology
    x/r = Q or Q factor or Quality factor

  6. #16
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    Quote Originally Posted by ish1284 View Post
    I'm not looking to test for the x/r of the reactor. I'm looking to determine if there is a standard for how series reactors are given their rating. Typically a standard of this nature would define a range for which the x/r ratio of the source used for testing needs to fall in. You can't stamp a rated let-thru on the equipment without knowing the 65kA source impedance and x/r ratio.

    If you know the source x/r ratio that it was tested at, you can then calculate the source impedance (in terms of r and x) and then from there the only unknown variable in the equation would be the r of the reactor and it could be calculated. I was hoping for a workaround for calculating r of the reactor without having access to the test report.
    This is how understand the nameplate re your reactor:
    1) Your short-circuit available current from your system is 65kA.
    2) You add a reactor to allow a lower let-through current of 9,415 A.
    3) Computing for the XL of your reactor will yield = 2 x pi x 60 x 67 x 10^-6 = 0.025258 ohms.
    4) We need to compute for the equivalent X of the system (no reactor) = (.48)^2/(1.732x65,000x480/1e6) = 0.00426 ohms.
    5) The equivalent X with the reactor in and the let-through is 9,415A = (.48)^2/(1.732x9415x480/1e6) = 0.02943 ohms.
    6) the diference in X's is the value of XL of the reactor = .02943 - 0.00426= 0.02517 ohms~ which is very near your 0.025258 computed from 67uH.

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