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## Line Reactor Ratings

Line reactor nameplate ratings are typically given in terms of rated inductance and let thru current at a given available fault current at the line side of the reactor at a particular voltage (in my example it's 67uH with a let thru of 9416A with 65kA @ 480VAC applied to the line side of the air-core reactor).

I'm running a calculation where it's important to know both the x and the r of the reactor. I can calculate the x based on the 67uH inductance. However, in order to calculate r, I believe I would need to know the x/r ratio of the 65kA at the line side of the reactor so I can "back calculate" from the 9416A let thru rating. I don't have a test report available.

Does anyone know the typical test x/r ratios applied to a line reactor in order to give its nameplate rating (i.e. the x/r ratio of the 65kA on the nameplate)?

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do they give v drop at rated i ?

data sheet?
make mfg?
wire gauge? length?

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No they do not, attached is a photo of the nameplate:

Originally Posted by ish1284
Line reactor nameplate ratings are typically given in terms of rated inductance and let thru current at a given available fault current at the line side of the reactor at a particular voltage (in my example it's 67uH with a let thru of 9416A with 65kA @ 480VAC applied to the line side of the air-core reactor).

I'm running a calculation where it's important to know both the x and the r of the reactor. I can calculate the x based on the 67uH inductance. However, in order to calculate r, I believe I would need to know the x/r ratio of the 65kA at the line side of the reactor so I can "back calculate" from the 9416A let thru rating. I don't have a test report available.

Does anyone know the typical test x/r ratios applied to a line reactor in order to give its nameplate rating (i.e. the x/r ratio of the 65kA on the nameplate)?
Last edited by ish1284; 07-09-18 at 12:21 PM. Reason: Not legible photo

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can't see the kvar?

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X = 2 Pi 60 67 uH = 25.2584 mOhm

i fault reactive = 480/2X = 9501.7876 A, rated 9416
the difference is r
Z = 480/9416/2 = 25.4885 mOhm
R = sqrt(Z^2 - X^2) = 3.4174 mOhm

X/R = 7.39
pf = 0.134

I think, let me look at it some more

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Originally Posted by ish1284
No they do not, attached is a photo of the nameplate:

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Your calculated x/r seems pretty low and not sure what the factor of 2 you have in there is for. Looks like you're missing a sqrt(3) factor.

I have a test report for a smaller model made by the same mfg and the x/r came out to be ~65. I'd expect something closer to that.

The x/r ratio of the 65kA it is rated for also plays a role in the let thru current on the other side. The 480/z that you calculated below also includes the source impedance. The impedance of the source needs to be split out from the impedance of the reactor. X/R of a line reactor should be pretty high because it's primarily reactance.

Originally Posted by Ingenieur
X = 2 Pi 60 67 uH = 25.2584 mOhm

i fault reactive = 480/2X = 9501.7876 A, rated 9416
the difference is r
Z = 480/9416/2 = 25.4885 mOhm
R = sqrt(Z^2 - X^2) = 3.4174 mOhm

X/R = 7.39
pf = 0.134

I think, let me look at it some more

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Originally Posted by ish1284
Your calculated x/r seems pretty low and not sure what the factor of 2 you have in there is for. Looks like you're missing a sqrt(3) factor.

I have a test report for a smaller model made by the same mfg and the x/r came out to be ~65. I'd expect something closer to that.

The x/r ratio of the 65kA it is rated for also plays a role in the let thru current on the other side. The 480/z that you calculated below also includes the source impedance. The impedance of the source needs to be split out from the impedance of the reactor. X/R of a line reactor should be pretty high because it's primarily reactance.
I looked at it as a 480 ph-ph
this would involve 2 reactors (the 480 with be 1ph, but I guess you could use sqrt3)
this equates to a ph ang of 83 deg
what is the current rating?

S fault ~ 54.0400 Mva
1ph source Z ~ 480/(1.732 x 65000) = 4.2635 mOhm, 7.3846 3ph
1ph reactor Xl = 25.2584, 3ph 43.7488

total Z = 480/9415 = 50.9825 3ph, 1ph 29.4847
reactor Z = total - source = 43.5979 3ph
but reactor X > Z so R can't be calculated

it appears that rounding the ratings prevents calc R
but it looks like x/r >100 is a safe assumption

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Originally Posted by Ingenieur
I looked at it as a 480 ph-ph
this would involve 2 reactors (the 480 with be 1ph, but I guess you could use sqrt3)
this equates to a ph ang of 83 deg
what is the current rating?

S fault ~ 54.0400 Mva
1ph source Z ~ 480/(1.732 x 65000) = 4.2635 mOhm, 7.3846 3ph
1ph reactor Xl = 25.2584, 3ph 43.7488

total Z = 480/9415 = 50.9825 3ph, 1ph 29.4847
reactor Z = total - source = 43.5979 3ph
but reactor X > Z so R can't be calculated

it appears that rounding the ratings prevents calc R
but it looks like x/r >100 is a safe assumption
I think we've gotten a little off topic from the original question. I'm not looking to make an assumption. I'm looking to see if there's a standard testing procedure that defines the test x/r ratio required rate a series reactor. Similar to the test x/r ratios required to rate circuit breakers (i.e. UL489).

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Originally Posted by ish1284
I think we've gotten a little off topic from the original question. I'm not looking to make an assumption. I'm looking to see if there's a standard testing procedure that defines the test x/r ratio required rate a series reactor. Similar to the test x/r ratios required to rate circuit breakers (i.e. UL489).
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.