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Zero Sequence Equiv. Impedance of 2-Autotransformer in parallel 1

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GRAEE

Electrical
Jan 15, 2010
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We have a two-50 MVA autotransformer, 138kV-Wye Grounded/69kV-Wye Grounded/13.8kV-Delta, operating in parallel with %IZ, Zps=10.3%, Zpt=0.2%
and Zst=5.5%. Only the 138 & 69kV sides are connected in parallel.

I'm in the process of determining the appropriate Zero Sequence Impedance Equivalent Circuit.

I'll appreciate if somebody could give me your opinion or idea on this query.


thanks///

galen1119

 
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Im trying to verify with you guys the correct equivalent circuit of zero sequence for these 2-paralleled autotransformer as i can't come up with the correct result using the SynerGEE software.

This will be used to calculate the SLG fault and for setting the relays.

Need your help on this problem.
 
galen1119:

See attached for zero seq diag for the auto transformer type you're looking at (credit to ABB T&D Reference Book, formerly known as Westinghouse T&D Book).
The impedances shown on your diag are typically quoted for pos seq, not zero. In my experience the zero seq is typically 0.85 to 0.95 times the positive, but this may be inherent in the design specifics from our company. the transformers I am comparing to are similar in size and voltages but our impedance is much lower, around 5 to 6 %.
If you need to calculate L-G faults on the delta, it will be trickier. But I do know our delta tertiaries are grounded on one corner, and I'm not sure how to calculate that.
One more tip: There's no mention on your diag of higher ratings when cooled by fans. If it has fans, ensure you're quoting impedance on the correct base. Our factory impedances are usually quoted based on the rating without fans.
Hope this helps,
 
 http://files.engineering.com/getfile.aspx?folder=cbaaef79-6968-4f78-86fd-696589a5ec5c&file=Auto_Tfmr_Zero_Seq.pdf
ichrist,

Thanks for the reference...But the impedances i'm referring for are the Zps=primary to secondary impedance, Zpt = primary to tertiary impedance, and Zst = secondary to tertiary impedance. I believe that we will convert first the impedances using the formulas: %Zp = 1/2(%Zps + %Zpt -%Zst)-for primary side; %Zs = 1/2(%Zps + %Zst - %Zpt) - for secondary side; and %Zt = 1/2(%Zpt + %Zst - %Zps) - for tertiary side.

Assuming that we will use 50 MVA as our FA rating, what would be the equivalent zero sequence impedance circuit when there is a L-G fault at the 69 kV side of two-paralleled autotransformers? Note that the tertiaries are left ungrounded, meaning to say that no loads are connected.

When drawing the zero sequence impedance circuit, the tertiary impedance will have a mutual impedance and thats why it will be connected to the source impedance. Thats the case when only 1 autotransformer is used. How about when there will be 2-autotransformers connected in parallel at 69 kV side? What will be the appropriate zero sequence circuit?

thanks
 
galen1119,

Formulas that you wrote are used with ohm impedanse (not percentages).
Also you wrote 2 different nominal voltage for tetriary side (in first post it is 13.8 kV and in schemes that you give it is 13.2 kV)

I will calculate with this datas:

uk_ps = 10.5 [%]
uk_st = 10.31 [%]
uk_pt = 5.6 [%]

Un_p = 138 [kV]
Un_s = 69 [kV]
Un_t = 13.2 [kV]

Sn_p = 50 [MVA]
Sn_s = 50 [MVA]
Sn_t = 20 [MVA]


All of the following impedance is when you look from primary side

So first off all you should calculate ohm impedanses of transofrmers:

Zd_ps = uk_ps/100 * Un_p^2 / Sn_p % [ohm]
Zd_st = uk_st/100 * Un_p^2 / Sn_t % [ohm]
Zd_pt = uk_pt/100 * Un_p^2 / Sn_t % [ohm]

Then you calculate separate impedances of windings:

Zd_p = 0.5 * (Zd_ps + Zd_pt - Zd_st) % [ohm]
Zd_s = 0.5 * (Zd_ps + Zd_st - Zd_pt) % [ohm]
Zd_t = 0.5 * (Zd_pt + Zd_st - Zd_ps) % [ohm]

Inverse impedances are equal to direct:

Zi_p = Zd_p
Zi_s = Zd_s
Zi_t = Zd_t

Zero impedanse are almost egual to direct impedanses only if primary (secondary) side are grounded directly without any impedance :

Z0_p ~ Zd_p
Z0_s ~ Zd_s
Z0_t ~ Zd_t

Zd_p = Zi_p = Z0_p = -2.43 ohm
Zd_s = Zi_s = Z0_s = 42.42 ohm
Zd_t = Zi_t = Z0_t = 55.75 ohm

Tertiary sides are connected to grounds on zero equivalent schemes because zero current cant go outside triangle.

Equivalent schemes are in attach.

Check everything one more time by yourself or google it.

Milovan Milosevic
 
MilovanSRB,

Sorry for the inconsistencies of the data.

Thanks for the support & enlightenment.

The zero sequence equivalent impedance of the two paralleled autotransformers is being supplied by infinite bus. One thing that i wanted to clarify: How about if the 138 kV source is not infinite, how does the Z0_t will be connected, does it have a return path to the source?(assuming the tapping point having an available fault current for:SLG=2,465 Amps; L-L=2,843 Amps; & 3P=3,283 Amps



 
galen1119,

If there is zero sequence impedanse on 138 kV side of network you should connect this impedance before everything on scheme. Same for direct and inverse equivalent scheme.

Milovan Milosevic
 
galen1119,

It is correct if you assume that there is no zero voltage source in network (and this is more or less ussualy true).

If this data that you wrote for short circuit currents (L-G , L-L and L-L-L) of network are OK then you can calculate direct, inverse and zero impedance of network using 3 equations.

Milovan Milosevic
 
MilovanSRB,

Thank you so much for sharing and for the support.

Looking forward to discuss with you in the future if i have another query.

galen1119

 
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