Harmonic Amplification Calculation

[RiverBeaver]

Hello all,

I'm familiar with calculating resonance for adding PF correction capacitors in industrial plants. My question is, knowing what the resonant frequency will be, and knowing the existing harmonic currents, how do I manually predict the actual amplification factor on these currents that are close to the resonant frequency? I know the roll off of the resonance bell curve will influence the neighboring harmonics, but I can't find any references on how to calculate how much. I can't afford sim software BTW.

For example:
1000 kVA transformer, 5.3 %Z
100 kVAR capacitor
h.res = 13.7th harmonic
13th harmonic = 3 amps without caps in system
What is the expected new 13th harmonic current with the 100 kVAR cap inserted in the system?

Thanks a lot,
RB.

 

[7anoter4]

There are some missing data as:
Transformer rated[ low] voltage[ULV].
Capacitor rated voltage[Uc]=480 V
Note: The transformer, the cables and the load reactance included also parasite capacities which are not evident for first harmonic but for 13 could be significant.
I neglected these capacities and also all the resistances.
1) How was load reactance when the 3 A was measured:
Let say ULV=480 V [0.48 kV]
Fundamental harmonic frequency 60 c/sec
Let say Sload=0.1*Strf=100 kVA and pf=0.8.
Then Iload=0.1/sqrt (3)/ULV=0.1203 kA
If resonance frequency is 13.7 then for this harmonic XLh=XCh
Q [MVAr] =0.1; w=2*pi ()*frq=377
C=Uc^2/Q/w =0.48^2/0.1/377=1151.3 microF
XCh=1/C/w/h h=13.7 XCh=0.16817 ohm
XLh=XCh=0.16817 ohm
XL=XLh/h=0.012275 ohm
Note: this is approx. equal to transformer reactance in parallel with the load reactance:
Xload=ULV/sqrt (3)*SQRT (1-PF^2)/Iload =1.38 ohm
Xtrf=ULV^2/Strf*5.3/100=0.01221 ohm
XL=Xtrf*Xload/ (Xtrf+Xload) =0.01221*1.38/ (0.01221+1.38) =0.0122
When no capacity is connected the transformer reactance and the load reactance are in series.
XLnocap=Xload+Xtrf=1.38+0.01221=1.39 ohm
At 13th harmonic XLhnocap. =1.39*13=18.3 ohm.
and the 13th harmonic EMF will be:
Vh=sqrt (3)*Ih*XLhnocap. =94.2 V
2) If the shunt capacity is connected then will be two circuits as shown.
XL13=XL*13=0.1587 X13Cap=1/C/w/13=0.1772 ohm
The new current will be:
I13new=Vh*(XL13-XL13Cap)/13/ (Xtrf+Xload)*(13*Xload-X13Cap)-Xload^2*13^2) =5.44 A
Usually one would test this value instead calculate it. It is my opinion.
Regards

 

[7anoter4]

Sorry! Something is rotten in my calculation kingdom! I have to revise entire calculation.

 

[RiverBeaver]

Thank you 7anoter4 for your detailed response, this was helpful. I haven't had a chance to work through the calculation yet but will once you update it.

RB.

 

[jghrist]

The voltage harmonic is calculated by multiplying the harmonic current times the source impedance in parallel with the capacitor reactance.

The source impedance is Z_s = R + j·2·pi·f·L
The capacitor impedance is -j·XC = 1j/(2·pi·f·C)

f is the harmonic frequency = h·fundamental = 13·60 = 780 Hz

 

[7anoter4]

1) First of all I agree with you, jghrist, naming the harmonic order as harmonic frequency it was not correct.
2) Second, we don't know the source of the 13th harmonic and I suppose it is the transformer inrush current.
3) The 3 A current was measured when no capacity was connected [see the O.P.] so the harmonic circuit was, in someway, to be closed by a load impedance.
4) It seems to me the current calculated when taken into consideration only the source impedance, neglecting the load impedance even when the capacity was connected is the not the actual current will flow through the load impedance. Am I wrong?
Best Regards

 

[7anoter4]

Yes, of course I was wrong! The harmonic current was measured when no load was connected but on the MV side, so you are right jghrist, only transformer impedance is to be considered and for the case when the capacitor was connected –still no load-the transformer reactance will be in parallel connected as you said. In this case the solution is less complicate and the current after capacitor connection will be 0.3 A.
Let say MV=11 KV
For fundamental [h=1] Xcap=1/C/w=1/ 0.001151294/377=2.3 ohm on LV side.
Transferred at MV side will be Xcap*(MV/LV)^2=2.3*(11/0.48)^2=1210 ohm
The transformer impedance will be Xtrf=MV^2/Strf*xk/100 where MV=11KV,Strf=1 MVA, xk%=5.3 so Xtrf=6.413 ohm
Now for h=13 Xcap13=Xcap/13=1210/13=93.1 ohm [actually -93.1 ohm]
And Xtrf13=Xtrf*13=6.413*13=83.37 ohm
The equivalent [parallel] reactance will be Xtrf13*Xcap13/ (Xtrf13-Xcap13) =799.1 ohm
The EFM will be sqrt (3)*Xtrf13*3=433.2 V and the 13th harmonic current [after capacitor was connected] will be 433.2V/sqrt (3)/799.1ohm=0.31 A
Thank you, jghrist!
Best Regards

 

[RiverBeaver]

I should have clarified in the original post:

3A 13th harmonic current is a load current measured on the secondary side with no PF caps in system. Vs = 600V.
Lets make up a load current harmonic spectrum for example:
I1 = 100A,
I5 = 18A,
I7 = 13A
I11 = 5A
I13 = 3A

I made up this example for simplicity, but often it is the case that there are some caps in the system which cannot easily be turned off for testing i.e. several switched with process motors. Predicting the amplification from adding capacitors to a system which already has some harmonic amplification would add significant complexity to the calculation I believe.

Thanks,
RB.