Current Limiting Reactor Sizing 2

[nightfox1925]

This is in relation with a feasibility study I am conducting with regards to options of mitigating high transformer in-rush currents.

I have a downstream 3 phase transformer, delta-wye, with 36A primary at 208VAC with an in-rush currents of Ia=220Arms, Ib=310Arms and 310Arms. This transformer is supplied from a 480-208/120V Delta-wye transformer

Based from the above values, the highest in-rush per phase is at approximately 12x transformer primary FLA. I wish to limit the in-rush to 175.29A per phase. I checked an ABB swtichgear handbook for sizing a current limiting reactor and considered the rated reactor current Ir = 26A,
through kVA of the reactor as SD= 1.732*26*120=5.4038kVA
kVA to be limited, SK"1 = (310A*208)/1.732 = 37.2275kVA
Desired kVA, SK"2 = (175.29A*208)/1.732 = 21.0503kVA

Therefore, the required percentage voltage drop of the reactor (Vr) = (1.1*100%*SD)* [(SK"1-SK"2)/(SK"1*SK"2)]

With this I am getting Vr = 12.271% or 13%
With this Vr=13%, the current is limited to 170.8A

The Voltage Drop at the reactor is also given by:
Ur = (%Vr * Vl-n)/ (1.732*100%)
= (0.13*120)/(1.732) = 9.0VAC

The voltage variation of the system with the insertion of this reactor will also be checked by calculation.

I wish to consult the experts on reactors or anybody who used this ABB switchgear handbook formula if I am in the proper way of applying the formula.

Secondly, are current limiting reactors also effective in mitigating asymmetrical transformer in-rush currents?

Thank you.

 

[burnt2x]

nightfox1925,
Can you send a diagram?

 

[nightfox1925]

Hello burnt2x,

Please find below simple sketch for your reference. Our objective is to mitigate the amount of in-rush within the UPS overload capability (150% for 10 seconds).

This reactor may be designed to either stay on-line or be fitted with shorting contactor. It is assume that the other smaller continuous loads are running and this rectifier input transformer is energized.

 

[burnt2x]

nightfox1925,
Here's my take on this problem:
1. With your output trafo rated at 50kVA, the UPS needed to supply this output trafo should have been bigger. A rule of thumb I have been using is to size the UPS 120% of the ouput trafo,1.2 X 50 = 60kVA. Since it's there now, the needed mitigation thought of is installing a series reactor to limit the inrush.
2. From your diagram's reactor insertion point, The receiving voltage Vr (at T2 input) can be computed as equal to the sending end voltage Vs minus the voltage drop across the proposed reactor, vectorially of course. To make computations easy, we will consider a perfect inductive reactance X(inductor coil):
Vs = Vr - j I (X)
We have to specify the desired allowable voltage drop, else we will have problems with the downstream UPS's. Let's choose a percent VD = 5%.
So:
Vr = 0.95Vs
Using I = 175.3 amps:
Vs = 0.95Vs + j 175.3X
120 = 0.95(120) + j 175.3X
6 = j175.3X
and X = 0.034 ohms and VD = 6 volts
Hence:

    reactor VA = 175.3 X 6
    reactor VA = 1,052 kVA per phase
               = 3.156 kVA

At a normal load of 26 amperes:
VD = I x X
VD = 26 x 0.034
VD = 1.23 volts; I guess you don't need to short the reactor if it introduces 99% VD.
Hope this helps.

 

[bacon4life]

Wouldn't the magnetizing inrush be much shorter than the 10 second overload limit? Seems like the short circuit characteristics of the UPS would be more appropriate.

Something else to consider is the effect of the DC and the higher on the reactor and the UPS.

In the voltage drop equation, is it really Vln/1.732 instead of Vll/1.732? Either way, this seems like a rather large amount of voltage regulation.

 

[nightfox1925]

Thanks burnt2x,

In the calculation you have presented (ohm's law), I may also consider the effect of the existing cable in Y meters and the load power factor at maximum loading such that:

If the the cable impedance is Rc + JXc, Xr = reactor reactance and power factor is cos theta,

Total voltage drop formula using the approximate approach is

Vd = I*Length*[(Rc*Cos theta)+ j(Xc + Xr)sin theta] volts

and Vreceiving = Vsending - Vd

The question is, what is the P.F. to be considered, is it maximum load P.F. or a low power factor (similar to motor starting maybe)?

Where I is the limiting current. Is the reactor going to be rated for the maximum continuous kVA or the equivalent transient kVA or both?

------------------------------------------------------------
bacon4life, its Vln/1.732 this is derived from the formula:

%Drop Vr = [(Volts Drop Vr)*1.732/Vln]* 100%

You have mentioned about DC component of the inrush. If current value used in the calculation already account for the total inrush, the DC component is already inclusive right?

 

[burnt2x]

nightfox1925,
I doubt if the cable inductance for a very short length (I am assuming)could impact your results! Just give it a try (add whatever impedances there maybe). Excel can surely shorten you calculation time.
PF for transformer inrush should be very low. And I also forgot to consider the duration of the inrush! I think you can reduce the reactor kVA since the the inrush lasts for a few seconds only.
Glad to be of help.

 

[nightfox1925]

Actually based from the waveform output of the measurements done at site, the inrush dissipates and settles to steady state for a maximum time of 0.0018 seconds.

Is it safe for me to assume an in-rush power factor of 10-15%?

I was having this idea that the cable size was designed with a predetermined voltage drop which will add up to the voltage drop resulting with placing the reactor. I agree with you that it will definitely be very less and the size may have been significantly dictated by ampacity rather than voltage drop.

By the way, is the reactor supposed to be specified with a short circuit rating as well?

 

[bacon4life]

In case anyone needs it, the ABB manual is can be downloaded at

http://rs178.rapidshare.com/files/90331371/abb_manual_10e.pdf

In the example, Un stands for voltage Nominal, which is 10 kV Line to line rather than Line to neutral. Thus all the equations above need the factor of sqrt (3) changed.

In reference to the DC component, if it can be simplified to a DC voltage in series with a switch, a resistor and an inductor, then the DC current is given by:

I = V/R * (1- exp(-R*t/L)

Adding larger pure inductance does not change the magnitude of the DC, but does make the decay significantly longer.

For higher order harmonics, the impedance of the reactor would go up with frequency. I am not sure what happens to the impedance of the rectifier or the inverter at higher frequencies.

If anyone else missed it like me, more information is in thread238-226471.

 

[nightfox1925]

Thanks burnt2x and bacon4life.

I have another query. In sizing the reactor, do I have to use the RMS value or the PEAK value of inrush currents?

PEAK value = sqrt(2) * RMS value

 

[nightfox1925]

Hi to all,

I have made some calculations using both the approach using the ABB switchgear manual as posted by bacon4life and the simple approach done by burnt2x with the perception that both should be able to provide me the same reactor ohmic values...its seems it's not and the simple approach done by burnt2x is yielding me smaller reactor impedance.

With a maximum inrush current of 310A per phase, the UPS is 50kVA plus 150% maximum overload capability, the "other continuous loads" are consuming 17.097kVA.

The downstream rectifier is having an input rated for 26A, 208VAC, 3 phase with a maximum input voltage tolerance of +10% and -15%.

hence using the method illustrated by ABB swgr. manual:

Desired in-rush limit = (1.5*50)-(17.097*1.25)
S"k2 = 53.62875kVA
= 148.858A at 208VAC, 3 phase

Throughput kVA of the reactor, SD

SD = 1.732*VL*Ir where Ir is 26A (rectifier cont. amps)
= 9.367kVA

Maximum Inrush kVArms (S"k1)= 1.732*310*208
= 111.6826kVA

Therefore minimum % voltage drop at the reactor (%VDr)
%VDr = 1.1*SD*[(S"k1-S"k2)/(S"k1*S"k2)]*100%
= 9.987% or 10%
is less than maximum 15% rectifier input voltage tolerance

Therefore: VDr = [(%VDr)*(VL)]/(1.732*100%)
= (10%*208)/(1.732*100%)
= 12volts

Therefore VDr = Ir*X
X = Vdr/Ir = 12/26 = 0.461ohms

Using the simple method shown by burnt2X,

Assuming I set the voltage drop at 10% to match what I have calculated above, then

Vs = 0.90Vs + jIrX with Ir = 148.858A (limiting current)
120 = 0.90(120) + j(148.858*X)
12 = 148.858X

X = 0.0806 ohms

The big difference is due to the Ir value.

The Ir value on the ABB method equates to the rated continuous current which is 26A.

The Ir value on the simple approach is 148.858A, the simple approach appears to make sense but I am trying to substantiate this with another calculation method for further confidence.

I hope the experts here would shed light on my doubts. Thank you.

 

[bacon4life]

Taking a step back and ignoring that the inrush has harmonics, the impedance of the circuit is:
208V/310A/1.732=0.39 ohms

Adding .461 ohms gives 141 A whereas the .08 reactor only reduces the current slightly.

As for peak or RMS, what do you think the UPS is responding to? The labels on the oscillography in the other thread weren't readable.

 

[burnt2x]

nightfox1925,

Formulas! Formulas!

I beg to disagree on your computations. Actually, ABB formula yields the same values as my formula (they are using the same principle). Let's go back to your posted diagram to be clear on this: From your diagram, you specied:

S_D= 63.154 kVA = "maximum acceptable inrush"(your terminology); Network voltage = 208 V; and I chose V_D = 5% (you could have choosen 3%, 6%, 8%, or 10%), the rated percent voltage drop.
Using ABB formula you get:

X_D =(5)(208)²/(100 . 63,154) =0.034 ohms.
That is the same value you get using the formula I posted.

I understand that when we design, we want this design to withstand all possible conditions this design will be subjected to. RMS current represents the net heating effect of current on electrical equipment, thereby RMS values determine the thermal rating of the equipment. I believe the oscillographhy you posted is not germaine to finding the thermal rating of your design. Those graphs only helps us know relationships of values wrt timestamps of other monitored parameters.

If you feel you are a loss when understanding effects of harmonics on heating of equipment, try this link:

http://ecmweb.com/mag/electric_effects_harmonics_power/

 

[nightfox1925]

Burnt2X, the difference on how you made the calculations and mine with the same formula is on the aspect of the currents being used.

I used Ir as 26A, while you used Ir as 148.858Amps. I giess I made an error of using the transformer rated input current of 26A instead of the limiting current of 148.858A.

Assuming I used 26A and yielded 0.461ohms, then the voltage drop during current limitation will be:

Vdr = 148.858 x 0.461 = 68.62 volts which will be 33% may not be able to magnetize the rectifier input transformer at all.

Thanks for the comment Burnt2X

In this regard, can I say that the rated current of the reactor should be the required current limitation and not the actual steady state maximum continuous input current of the rectifier which is in this case is 26A?

On the ABB formula, I used Ir = 26A in the calculation of the reactor throughput kVA (SD). The problem is that if I am going to use the required limiting current on the formula, I will end up with an unrealistic % voltage drop rating of the reactor. I think I am confusing myself with the ABB method, I would appreciate anybody will tkae me out of this confusion I am into.

 

[burnt2x]

I used S_D = 63.154 kVA. The current at 208V will be 175.3A, not 148.858amps.
I_r is the current rating of the reactor. By this it means the level of current that it limits at a specified voltage drop. In your problem, you said you want to limit the inrush current to 175.3 A. That's your I_r. The actual inrush will be smaller after installing the reactor because the voltage impressed on the downstream equipment will be smaller. Don't use the values of current where there is no reactor yet, that's key.
Remember that you are exposing the reactor to a value of current inrush for a very short slice of time. VD might be great for some milliseconds but should improve immediately. By that time, your UPS OL trip counter will have reset because there is no more overcurrent sensed.

 

[nightfox1925]

Thank you burnt2x for your clarification.

I am a little concerned about the ABB formula and determining the appropriate reactor %voltage drop which is:

%VDr = (1.1)*(SD)*{[(S"k1-S"k2)/(S"k1*S"k2)]*100%---Fmla 1

wherein SD = Throughput kVA of the reactor which is the
the product of the line-to-neutral voltage
(VL/sqrt(3))and the rated current Ir

S"k1 = short circuit kVA power of the grid system
(in our case, I may equate this to the in-
rush kVA)

S"k2 = Reduced kVA (or in our case, is the limiting
kVA)

The Ir in this formula as I perceive is the rated maximum continuous current that will flow into the reactor which is in my original post and diagram is 26A. Therefore, my SD will be:

SD = 1.732*26*208 = 9.367kVA ------three phase
S"k1 = 1.732*310*208 = 111.67936kVA----three phase
S"k2 = [(1.5*50)-11.846]= 63.154 kVA

Plugging these values in the above ABB formula will result to %VDr = 7.089% or say 8%. Again from ABB:

VDr = [(%VDr)*(VL)]/[sqrt(3)*100%]--------Fmla 2
= [(8%)(*208)]/[1.732*100%]
= 9.6V

Xr = VDr/Ir-------Fmla 3

But now, I will use the current limiting current to attain the ohmic value for that given voltage drop.

Ir = (63,154)/(1.732*208) = 175.3A

therefore, Xr = 9.6/175.3A = 0.054 ohms

Burnt2x, is my approach and understanding using the ABB formula correct or I missed something? If yes, which part of the calculation?

I am laso trying to figure out the derivation of the ABB formula as shown in fmla 1 above...if anybody has an idea how it was dervived, I will appreciate it much.

 

[burnt2x]

You are correct. In the ABB formula, S*k1 is the inrush kVA before the reactor is inserted and S*k2 is the desired kVA after the reactor is inserted.
In my formula, I failed to rightly compute for the percent VD! This is a case of putting my "ASS" in front of "U" and "ME" ("ASSUME"D)

 

[nightfox1925]

Thanks Burnt2x and bacn4life for patiently guiding me in my understanding.

Hmmm I like the sense of humor.