Reactor Rating
Reactor Rating
(OP)
I am trying to find the rating of a reactor connected to a medium voltage system for fault current calculations. the reactor has specs as follows:
Reactor connected to a 4.16KV bus
300A
Z = 6%
Reactor connected to a 4.16KV bus
300A
Z = 6%






RE: Reactor Rating
Generally the 6% Z means it will have 6% voltage drop (assuming rated voltage) at the rated current of 300 A.
You will need to estimate the X/R ratio based on typical reactor data that you should be able find on-line.
These reactor ratings can be confusing, at least to me.
David Castor
www.cvoes.com
RE: Reactor Rating
Thanks
RE: Reactor Rating
What you can have is rated Mvar. For an SLD, it would be 300*0.06*4.16/sqrt(3) which computes to 43.2 kvar.
Why was my first answer RFd?
Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.
RE: Reactor Rating
Thank you for your help. I reallu appreciate it.
RE: Reactor Rating
That way you don't screw up your dimensional analysis :)
RE: Reactor Rating
My first answer was "Yes?", which was removed for some reason. The 'Yes?' indicated that it wasn't very clear what you asked for and that we needed to know more about the question.
It still isn't very clear what you actually want to know. Is it reactive power or something else? Your Kvar = VLL*Irated*Z leakage/sqrt(3) is wrong. There is no VLL applied to the reactor, only a voltage drop, which is six percent of the VLN at rated current.
There should not be any need to refer to an IEEE book, violet or whatever colour, for a basic calculation like this.
Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.
RE: Reactor Rating
Since this is rated at 300A at 4.16kV then can we say it is 3*300*4.16kV = 3744kVA or 3.744MVA base
Then, ca we say Z=6% at 3.744MVA base?
6% = [(3.744 x 10^6)/ (4160)^2] * Zohms
Zohms = 0.06 * (4160)^2 / (3.744 x 10^6)
Zohms = 0.277 ohms (3 phase)
RE: Reactor Rating
@ThePunisher: I think you did a mistake by multiplying the value by 3 because the 4.16KV is line-to-line. I think you have to multiply by Sqrt(3) instead.. or you can still multiply by three but use the line-to-ground (phase) voltage.. right ?
RE: Reactor Rating
Since this is rated at 300A at 4.16kV then can we say it is 3*300*4.16kV = 3744kVA or 3.744MVA base
NYET!
The equation should be 3*300*4.16kV/Sqrt3 = 2.16 MVA.
6% of that is 129.6 kVAR.
Gunnar gave a single phase figure of 43.2 kVAR which, amazingly, if you multiply by 3 gives a figure of 129.6 kVAR.
RE: Reactor Rating
RE: Reactor Rating
When you say that "Since the reactor impedance is represented in per unit (percentage) then there is an MVA rating envolved. I need to know what is that value so that I can convert the percentage impedance to my Base", you have forgotten that percents were invented long before electricity and p.u. math and that there is no absolute need for p.u. as soon as percents are used.
The percent notation was originally used to represent the voltage drop of a transformer when fully loaded. It still is, and is recognized as Uk in percent on most transformer nameplates.
The same notation is used for reactors. A 2 or 4 % commutation reactor, for example. Or, in your case, a 6 % reactor, which simply means that you will have a 6 % voltage drop across the reactor when current is 300 A.
Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
RE: Reactor Rating
What one need in short-circuit diagram is Xr[p.u.].
The resistance of a limiting reactor is about 3% so may be neglected and Xr=Zr.
The voltage drop across the reactor is:
Du=Ir*Xr[Ir A Xr ohm Du Volts].
Voltage drop Du%=Du/VL-N*100[ per phase!] then Du=Du%/100*VL-L/sqrt(3)
If Vrated reactor=VL-L system
VL-N= VL-L/sqrt(3) rated system voltage.
Ir*Xr= Du%/100*VL-L/sqrt(3) then:
Xr=Du%/100*VL-L/sqrt(3)/Ir and multiplying and dividing by VL-L we shall get:
Xr=Du%/100*VL-L^2/Sd where Sd= throughput power= SQRT(3)*Ir*VL-L
Zb=Vbase^2/Sbase usually Vbase=VL-L
Xr[p.u.]= Xr/Zbase Xr[p.u.]= Du%/100*VL-L^2/Sd/VL-L^2*Sbase
Xr[p.u.]= Du%/100*/Sd/Sbase. You don't need inherent power Se=3*Du*Ir in Xr calculation.
Sd=sqrt(3)*4.16*0.300=2.16 MVA
Sbase=100 MVA
Xr[p.u.]=6/100*100/2.16=2.778
RE: Reactor Rating
Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
RE: Reactor Rating
The designers could have shown a 1.5mH inductor in the circuit. It's quite factual but doesn't give a sense of purpose for the device.
I'm in Australia, where the % is shown on all power transformer nameplates. Is the same done in North & South America?
RE: Reactor Rating
RE: Reactor Rating
involved in Machine Design and Engineering and for a while even I was cable Manufacturer. But, for more than 20 years, I am Power Station designer so I was forced to understand the Electric Power Engineers' problems.
I like to use X and R in ohms and not in p.u. But, for rapid approximation, using p.u. is more useful- mainly in multi-voltage systems. So, if I well understood, this was backer86 intention.
RE: Reactor Rating
RE: Reactor Rating
%Z is used to represent the condition of a line to line short circuit at the transformer terminals. In this case, the Z of the transformer is the total circuit. This is used to determine the symmetrical short circuit current, also called the available short circuit current.
The voltage drop across the reactor may be at close to a right angle to the voltage across the load, so that the the 6% voltage drop of the reactor may cause much less voltage drop across the load.
I would expect a 6% reactor to limit the symmetrical current to 1/0.06% of rated current. Under normal conditions, the voltage reduction at the load may be less than 1% at full load current.
Bill
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"Why not the best?"
Jimmy Carter
RE: Reactor Rating
RE: Reactor Rating
The regulation describes the voltage drop of a transformer at full load. The impedance describes the operation with the terminals shorted.
Yes, if you neglect the resistance, which it is probably safe to do. But the 6% voltage drop at full load current will be almost at right angles to the voltage of a resistive load and will have little effect on the load voltage. If I did the math right, a 6% reactive voltage drop will drop the voltage across a resistive load less than 0.2%
I was responding to the use of impedance percentages to calculate voltage drops under normal operating conditions.
Transformer and reactor impedances describe short circuit conditions when the impedance is the total circuit.
Bill
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"Why not the best?"
Jimmy Carter
RE: Reactor Rating
100/6 = 16.67 times normal = 300*16.67 = 5000 amps
Also it means the reactor is rated .48 ohms
2400 volts/.48 = 5000 amps