Line Reactor Ratings

[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)?

 

[Ingenieur]

do they give v drop at rated i ?

data sheet?
make mfg?
wire gauge? length?

 

[ish1284]

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

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)?

 

[Ingenieur]

can't see the kvar?

 

[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

 

[ish1284]

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

My bad, kVAR is 491

 

[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.

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

 

[Ingenieur]

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

 

[ish1284]

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).

 

[topgone]

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.

 

[ish1284]

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.

 

[Ingenieur]

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

 

[Ingenieur]

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

 

[Ingenieur]

IEEE paper on reactors
my calcs match
x/r is basically irrelavent
http://www.ewh.ieee.org/soc/pes/switchgear/minutes/2010-1/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/userfiles/Introduction to Short Circuit Current Calculations.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%.

 

[Ingenieur]

reactor terminology
x/r = Q or Q factor or Quality factor

 

[topgone]

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.