How can i calculate the x/r ratio of YNd11 transformer ? 1

[Roleci]

Hello;

We have a 125 MVA YNd11 380/33 kV transformer on the field.Low voltage side is connected to ground through a zig zag grounding transformer without a resistor.I want to calculate the x/r ratio of a L-G fault in order to find the decaying time of DC component.As far as I know there is no zero sequence impedance on delta side.So in this case should I only use the positive and negative sequence impedances of the power transformer together with zero sequence impedance of the grounding transformer to calculate the x/r ratio of fault ?
Thanks in advance.

Grounding transformer zero sequence impedance is 10+j90 ohm.

 

[RRaghunath]

I think you are right!

 

[krisys]

Go to the basic tests, namely measure the copper loss and Iron losses, by energising/shorting the desired pairs. Separate out the components. Thus you can derive the X/R ratios of the respective pairs.

The above descriptions are conceptual. You have to put some efforts for the arithmetic.

 

[7anoter4]

On the name plate has to be noted the short-circuit impedance [uk% -percent] and full load losses[Pload] and no-load losses[Po].The copper losses[Pcu] is the difference between these losses.
Pcu=Pload-Po
So the resistance-at rated temperature-will be [3*Irated^2*Rcu=Pcu] Rcu=(Pcu/3/Irated^2).
The impedance will be Z=Vrated^2/S*uk%/100
An approximate calculation may be as following:
According to IEC 60076-5 Tab.1 for 125 MVA the minimum uk%=12.5%
Z=380^2/125*12.5%=144.4 ohm
Let's say η[eff.]=99% Pload=0.01*125=1.25 MW ≈Pcu
Irated=125/√3/380=0.19 kA
R=11.5 ohm
X=√(Z^2-R^2)=144 ohm
Usually Xo=(0.7-1)*X1
Xo=100-144 ohm

 

[prc]

7anoter4, I have to correct:
From the rating plate or test report, get the % impedance- Z pu
From test report note the full load loss in kW
R in pu = Full Load losses in KW/ rated transformer KVA
X in pu = Square root of ( square Z- square r)
In large rated transformers r is very low and hence even Z/R can be used as X/R .
I think zero sequence X need not be considered for above ratio. Not sure. Zero sequence impedance will be approx. 0.9 of positive sequence impedance.

 

[7anoter4]

Thank you, prc.

 

[jghrist]

I think zero sequence X need not be considered for above ratio. Not sure. Zero sequence impedance will be approx. 0.9 of positive sequence impedance.

Correct. Z0 of the 380/33 kV transformer is not in the ground fault circuit.

 

[7anoter4]

If the line-to-ground fault it is in high voltage supply line.

 

[Roleci]

and full load losses[Pload] and no-load losses[Po].The copper losses[Pcu] is the difference between these losses.
Pcu=Pload-Po
So the resistance-at rated temperature-will be [3*Irated^2*Rcu=Pcu] Rcu=(Pcu/3/Irated^2).
The impedance will be Z=Vrated^2/S*uk%/100
An approximate calculation may be as following:
According to IEC 60076-5 Tab.1 for 125 MVA the minimum uk%=12.5%
Z=380^2/125*12.5%=144.4 ohm
Let's say η[eff.]=99% Pload=0.01*125=1.25 MW ≈Pcu
Irated=125/√3/380=0.19 kA
R=11.5 ohm
X=√(Z^2-R^2)=144 ohm
Usually Xo=(0.7-1)*X1
Xo=100-144 ohm]

Thanks for the answer.I noticed you divide the R value by 3 since what you found is the total resistance of 3 phases.But you don't divide the X value ? Why is that ? The other thing is that you calculate the high voltage side x and r values.But I want to calculate the secondary side fault x/r ratio.Can I use those values for a low voltage side fault ?

 

[7anoter4]

Sorry for the delay.[It seems to me I am the busiest grandfather].
Z calculated with formula Z=VL-L^2/S it is per phase impedance.
If phase-to-phase VL-L=√3 x Iphase x Zphase Zphase=VL-L/(√3xI)=VL-L^2/(√3xIxVL-L)=
VL-L^2/S
If X1HV it is the reactance viewed from 380 kV side then X1LV=(WLV/WHV)^2*X1HV
The same R1LV=(WLV/WHV)^2*R1HV
WHV=high voltage winding number of turns per phase[380/√3 kV]
WLV=low voltage windings number of turns per phase[13.2 kV]
WLV/WHV≈13.2/(380/√3)
However it is only viewed from inner side [let's say" as in star connection"].Viewed from outside [delta connection] ZLV=13.2^2/125
RLV=RHV*(13.2/380)^2 XLV=XHV*(13.2/380)^2
The Zo impedance viewed from 13.2 kV side is infinite[∞]

 

[Roleci]

.
Z calculated with formula Z=VL-L^2/S it is per phase impedance.
If phase-to-phase VL-L=√3 x Iphase x Zphase Zphase=VL-L/(√3xI)=VL-L^2/(√3xIxVL-L)=
VL-L^2/S
If X1HV it is the reactance viewed from 380 kV side then X1LV=(WLV/WHV)^2*X1HV
The same R1LV=(WLV/WHV)^2*R1HV
WHV=high voltage winding number of turns per phase[380/√3 kV]
WLV=low voltage windings number of turns per phase[13.2 kV]
WLV/WHV≈13.2/(380/√3)
However it is only viewed from inner side [let's say" as in star connection"].Viewed from outside [delta connection] ZLV=13.2^2/125
RLV=RHV*(13.2/380)^2 XLV=XHV*(13.2/380)^2
The Zo impedance viewed from 13.2 kV side is infinite[∞]]

Thank you very much for answering my questions in your busy time So in this case looks like I can ignore high voltage side impedance since it's effect is too low for a low voltage side fault due to division by square of transformer winding ratio.Can I only use grounding transformer x/r ratio for dc component decaying time calculation ?

 

[prc]

1) Full load loss means not Po+Pcu, but only PCu.
2) Pcu is not I2R alone but add 5-30 % for taking care the stray and eddy losses. So the correct way to find Rpu is load losses in kW/ rating in KVA.
3) Impedance in pu is the Pu voltage to be applied on any side to get rated current on both windings with the other side in shorted condition.
4) X in pu = square root of (squareZ -sqaure R) all in pu.

 

[7anoter4]

Since 380kV System and 125 MVA transformer impedances viewed from 13.2 kV are less than 1 ohm I also think you may use only Zig-Zag transformer Zo impedance.