Line Reactors in Short Circuit/Arc Flash Studies

[ish1284]

Hello I am trying to educate myself a bit on line reactors. If the only information I have on a line reactor is that it is rated 250A and 5%, is this enough information to calculate the uH of the line reactor? Or do I need the vendor data to obtain the uH value? What does the 5% rating represent? The purpose of a line reactor is to reduce short circuit current without reducing the load voltage, correct?

The background behind this question is that I am modeling remote power panels (RPPs) which are fed by line reactors. I am trying to determine if it is worth modeling the reactors in calculating short circuit and arc flash incident energy values.

 

[petersonra]

Hello I am trying to educate myself a bit on line reactors. If the only information I have on a line reactor is that it is rated 250A and 5%, is this enough information to calculate the uH of the line reactor? Or do I need the vendor data to obtain the uH value? What does the 5% rating represent? The purpose of a line reactor is to reduce short circuit current without reducing the load voltage, correct?

The background behind this question is that I am modeling remote power panels (RPPs) which are fed by line reactors. I am trying to determine if it is worth modeling the reactors in calculating short circuit and arc flash incident energy values.

the 5% number relates to how much impedance is added to the circuit by the inductor. Off hand I don't recall exactly what the 5% represents. Most of the inductors I have seen the uH value is readily available from the data sheets.

I would think it could change your calculations similar to the way adding a longer run of wire would.

 

[Ingenieur]

Does the reactor have a rated kva or voltage?
the 5% is a pu Z
if v is 480
kva S = 1.732 x 480 x 250 = 208 kva
S = 208 kva = (250)^2 x Z base
Z base = 3.3255 Ohm
Z pu = Zact/Zbase

Z act = 0.1663 Ohm = 2 x Pi x 60 Hz x L
L = 0.44 mH or 440 uH

Without the data it may have to be done on a 1 ph base
v = 277
no 1.732 in S equation

 

[Jraef]

the 5% number relates to how much impedance is added to the circuit by the inductor. Off hand I don't recall exactly what the 5% represents. Most of the inductors I have seen the uH value is readily available from the data sheets.

I would think it could change your calculations similar to the way adding a longer run of wire would.

The 5% value of impedance is what is ADDED to the already existing impedance of the circuit. So if I have an existing circuit that will draw 250A at 480V, Z = V/I so the Z of that circuit is 1.92ohms. So if you add a 5% reactor, that is 5% of 1.92 ohms, so the inductive reactance (X_L) of that reactor = .096 ohms. If I want to know the _uH from that then, the formula is
L (in Henrys) =X_L/2¶F (Frequency in Hz) = 255 _uH, for what that's worth.

 

[Ingenieur]

https://www.ab.com/support/abdrives/documentation/techpapers/Line Reactors and AC Drives1.pdf

The first rule is make sure you have a high enough amp rating. In terms of the impedance value, you willusually find that 3% to 5% is the norm with most falling closer to 3%. A 3% reactor is enough to provideline buffering and a 5% reactor would be a better choice for harmonic mitigation if no link choke ispresent. Output reactors, when used, are generally around 3%. This % rating is relative to the load ordrive where the reactor impedance is a % of the drive impedance at full load. Thus a 3% reactor will drop3% of the applied voltage at full rated current. To calculate the actual inductance value we would use thefollowing formula. L =XL/(2¶FL) Where L is inductance in Henrys, XL is inductive reactance or impedancein Ohms and F is the frequency. In general Frequency will be the line frequency for both input and outputreactors

it's vdrop at rated i
5%
if 480 vac
drop = 24 vac
Z = 24/250 = 0.096 Ohm
L = 0.096/(2 Pi 60) = 255 uH

when given as a source pu Z for a power system it as I described in my previous post

 

[Ingenieur]

After looking at it
same thing on a single ph basis for both methods
440 uH / 1.732 = 255 uH

 

[ron]

The purpose of a line reactor is to reduce short circuit current without reducing the load voltage, correct?
The background behind this question is that I am modeling remote power panels (RPPs) which are fed by line reactors. I am trying to determine if it is worth modeling the reactors in calculating short circuit and arc flash incident energy values.

Generally they are used to reduce short circuit current. They however can produce some voltage drop, but not when sized properly.

You need to model them for a short circuit study, because it is needed to get accurate results. Short circuit current will be lower and arc flash energy will likely be higher.

 

[Sajid khan]

As i know ,The line reactor is used for harmonic reduction when used at the input side of the VFD and if used at the output side is to protect the motor from fast varing voltage (dv/dt) due to fast switching of IGBTS. A 5% reactor is better than 3% due to more capability of mitigating the line harmonics.

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