Reactor Rating 3

[backer86]

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%

 

[dpc]

This is a series-connected current-limiting reactor, three phase?

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

http://www.cvoes.com

 

[backer86]

Yes it is a series current-limiting reactor in a three phase system. However, those values are given in a single line diagram of a power distribution system. I need to find the rated MVA to change the impedance value to my Base.

Thanks

 

[Skogsgurra]

I'm afraid there is no rated MVA.

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

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[backer86]

So you are saying that Rated Kvar = VLL*Irated*Z leakage/sqrt(3) ?? why did you multiply by Z ??? and I thought rated values are based on line-to-line values so I think you have to multiply by sqrt(3). according to IEEE violet book, kVA = sqrt(3)IRatedVRated, LL !!! Is that correct ?

Thank you for your help. I reallu appreciate it.

 

[sibeen]

If you're computing a var, which is volt amp reactive, it is probably a good idea to have a reactive term in the equation.

That way you don't screw up your dimensional analysis

 

[Skogsgurra]

I showed the one-phase kvar. If you need all three phases, just take it times three.

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

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[ThePunisher]

Z =6% should have an MVA base 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)

 

[backer86]

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.

@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 ?

 

[sibeen]

As Gunnar has taken pains to explain, you don't use the L-L voltage in the calculation as the 6% is the voltage drop of the L-N voltage.


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.

 

[backer86]

That's what I am talking about. I think 2.16 MVA is the correct answer. Thank you

 

[Skogsgurra]

Backer

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

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[7anoter4]

First of all I agree with Skogsgurra.
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

 

[Skogsgurra]

Thank you for the elaboration, 7/4. But I wonder, does the p.u. thinking really make things clearer in a case like this?


Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[sibeen]

I'm with Gunnar, here. The % rating of the choke is given for clarity and ease of use.

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?

 

[magoo2]

It is standard practice in North America to show the % IZ on the nameplate of power and distribution transformers.

 

[7anoter4]

I agree with you, Skogsgurra. I am, myself, graduated in Electrical Machines and Apparatus and for 10 years I was
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.

 

[backer86]

7anoter4 Thank you very much for the help. You got exactly what I needed.

 

[waross]

I got caught on that one Gunnar. In North America, Regulation represents the voltage drop across a transformer when it is heavily loaded. The R of the transformer and the load may predominate.
%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
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

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.

True if the current is resistive, but fault current is usually mostly reactive.