Convert Transformer Impedances to different voltage level

[Tinh123]

Hi, I'm very new at this and need your advises. I was recently asked to determine the fault current contribution of a group of generators locates at a bus within our system. This group of generators was represented as a single equivalent source in Aspen software and connected to a 345 kV system through the 13.8/345 kV Y-Delta (Delta lags) transformer. The issue is that how can i represent this same equivalent generator and transformer if it was connected to a completely different voltage levels (i.e. The original generator would be connected to 145 kV system through a 18/145 kV transformer). I was able to represent the generator equivalent impedance at 18 kV through the use of this equation Z2 = Z1*(V1/V2)^2 (assumed same MVA base), but have much difficult in represent the 345/13.8 kV transformer impedances in the 145/18 kV transformer scheme. Below are some of the data:

Original Data: (13.8 - 345 kV network)
Generators Equivalent Impedances (connected at 13.8 kV bus)
Subtransient = Transient = Ps = .001 + j.200
ns = .003 + j.2005
zs = .004 + j.2007

Transformer Impedances (13.8/345 kV Xfmr)
Zero Seq = Pos Seq = .0005 + j.0300

How can i correctly translate the transformer impedances to 145/18 kV system, assume that MVA bases are the same.

Thanks for any advice. I have been struggling with this for couple of days. Even without the translation of impedance process, is there any other way that I can calculate fault current contribution of the 345/13.8 kV system at 145/18 kV one. Thanks for all the help.

 

[jghrist]

What do you mean by equivalent generator and transformer? Are you going to overexcite the 13.8 kV generator to generate at 18 kV? Generate at 13.8 kV and step it up with a 18-145 kV transformer? Use a different generator rated 18 kV but with the same pu impedance?

A 345-13.8 kV transformer cannot be represented as a 145-18 kV transformer. The ratio is wrong.

 

[davidbeach]

There is a good discussion of this in Protective Relaying: Principles and Applications by J. Lewis Blackburn. This book is a very good reference for many topics related to protection and is one of the core reference volumes I would recommend for anyone working in the protection field.

 

[Tinh123]

jghrist/davidbeach: Truly appreciate for looking into this.

I'm looking into a hypothetical situation for which I have to determine the fault current contribution of a generator that is connected to a 345 kV network through a 13.8/345 kV
Y-Delta transformer. Assume the same generator was being used (same MVA rating, etc)at different location, how do I represent this generator impedances if it was connected to an 18 kV bus instead of the 13.8 kV bus, and the equivalent transformer impedances of the 18-145 kV instead of the original 13.8 - 345 kV system. If I can convert the generator and transformer impedances as seeking above, I can use these values and entered into Aspen to determine the fault current contribution of this generator at the new location. This is the only method I know of, and with limited engineering knowledge, any suggestion would help me tremendously. Please let me know if you need more info. The data of the original system has been included in the previous message. One more question, If I know the fault current contribution at the original setting (let assume 40 kA), is there any way I can approximate the fault current contribution at the new location based on voltages, etc? Thanks.

 

[jghrist]

If you are talking about a hypothetical 18 kV generator with the same characteristics as the 13.8 kV generator and a hypothetical 145-18 kV transformer with the same characteristics as the 345-13.8 kV transformer, then use the same per unit impedances.

 

[DanDel]

Tinh123, first of all, I seriously doubt that the Zero-sequence impedance (Z0) equals the Positive-sequence impedance (Z1) for your transformer, you probably mean Negative-sequence (Z2) equals Positive-sequence. For a delta-wye transformer, Z0 is infinite.

If you were to use your generator on an 18kV system it would have to be an 18kV generator, with different impedances than it has now.

Your impedances are not listed as pu, percent, or ohms, and which voltage if the transformer impedances are in ohms. You didn't even list the transformer kVA.

If they are in pu or percent, you can calulate the fault current directly, just use the different voltage, and assume that the pu impedance values are the same for both generators.

You can always calculate the maximum possible fault current through your transformer by dividing the FLA by the pu value of the percent impedance (%Z), which can be calulated from your Z1, once you determine if it is in pu, percent, or ohms (and, if so, which voltage system).

 

[davidbeach]

For a delta-wye transformer, Z0 is infinite.

No, that is not true. Z₀ will be of the same order of magnitude as Z₁, but not necessarily exactly the same. In a delta wye transformer, there is an open circuit between the system on the delta side and the transformer's Z₀, but if the wye side is grounded the transformer Z₀ is connected to ground in the zero sequence network, allowing zero sequence current to flow on the wye side. If the transformer's Z₀ were actually infinite there would be no zero sequence current on the wye side.

 

[DanDel]

davidbeach, obviously, zero-sequence current can flow in the wye winding of a delta-wye transformer. The infinite inpedance is seen when modeling the system as symmetrical (Positive, Negative, and Zero-sequence) components, where no zero-sequence current flow can travel through the transformer.

 

[davidbeach]

Not an infinite impedance. It is there, it is measurable, and it is finite. It may also be open circuited, but being open circuited doesn't change its impedance. That would be like saying that a piece of wire laying on the floor with no connections has in infinite impedance because no current is flowing through it.

 

[DanDel]

davidbeach, without going into a tangent, and in the context of Tinh123's question, about calculating generator contributions through a delta-wye transformer, would you agee that the Z0 is infinite?

 

[Tinh123]

Apologize for the confusion.

Here are the somewhat factual data of the generator and the transformer I mentioned.

There are actually 5 generators at the original location but it is represented in Aspen as a single generator with 5 built-in units. Each generator is rated at 100 MVA, and that is the base MVA of our system.

Each of the unit has the following characteristic:
Subtransient = Transient = Z1 = .004+ j.150 (all in per unit)
Z2 = .012 + j.145
Z0 = .006 + j.100

As I have mentioned earlier, these units were represented as a single generator connect to 13.8 kV bus, and it has an equivalence (fictitious) 2-winding Y-Delta, Delta lags 13.8/345 kV transformer. The transformer has the following data in Aspen:

Z1 = Z0 = .005 + j.020 per unit with MVA base of 100 MVA

It also has MVA1 = MVA2 = MVA3 = 0 (Aspen indicates these as MVA ratings of the transformer). But since this is a fictitious transformer, I don't think I should be concerned about this. Aside from that, I've seen several 2 winding xfmrs with these MVA ratings equal to zero as well.

The question is: If these generators were to connect to the 18 kV bus and step up to 145 kV system through the 18/145 kV 2-winding Y-Delta, delta lags transformer then what is the fault current contribution of these generators if the fault was to be placed on the 145 kV bus ???

My strategy is that if i can somehow represent these generators and transformer impedances through some conversion method, then i can enter these impedance data directly into Aspen and should get a fault current contribuion by these generators immediately. After some search, I found some reading which suggests the i can convert the generator equivalence impedance to different voltage level, but I have no way of representing the transformer impedance at different voltage level (i.e. 18/145 instead of 13.8/345 kV).

I need your help in understanding how can I translate the transformer impedance.

If the conversion of the transformer impedance to a significant different voltage level is not applicable, then why is this, and is there any other method out there which can help to approximate the fault current contribution from these generators at the new location (18/145 kV system).

My objective is to find the approximate fault current contribution and somehow reasoning it out of the logicality of the methodology that I use I have been troubled by this for the past several days and appreciate for all your help. Thanks.

p.s. All data are in per unit, 100 MVA system. Also, the generators I mentioned are belong to other utility and I have no other data than the ones I mentioned above which were taken from Aspen case.

 

[Tinh123]

I forgot to mention that I can convert the generation impedances using the following formula:

Z_new = Z_old * (V_old ^2)/(V_new ^2)

(Assume both old and new location has the same MVA base)

 

[DanDel]

I'm not sure how it is done in Aspen, but in ETAP, SKM, or EDSA, you would have inputs for nameplate and/or pu values, base MVA, and voltages. If all the data is in pu and on the same base (and are all assumptions anyway), then just change the voltages as you want, this will change the fault current for that voltage system.

Using your equation: "Z2 = Z1*(V1/V2)^2 (assumed same MVA base)" is really just used for changing a base quantity (voltage and/or current), as in: Zpunew = Zpuold(baseVold/baseVnew)^2 x (baseMVAnew/baseMVAold)

The calulation for fault current at any point will involve the system voltage at that point, which you can change for your conversion.

 

[Tinh123]

I have done further analysis and came up with some answer. The results are a bit strange and I wonder if it makes sense or not. Given the same set of generation step up to 345 kV instead of 138 kV, the fault current contribution from this same capacity generator at 138 kV lelvel is almost twice as high as that of the fault contribution at 345 kV. I anticipated a higher fault contribution at lower voltage, but didn't think it would be that significant. Any idea?

 

[DanDel]

The fault current may be almost twice as high at 138kV, but is the fault MVA twice as high?

You have a certain amount of fault MVA that the generator system can output, whatever the voltage. You have a certain pu transformer impedance, which does not change in your model, just the voltages. And you have a certain amount of fault MVA on the secondary of the transformer, which should be close to the original fault MVA.

The fault current, when calculated out from the fault MVA using the new voltage, will certainly be different.

 

[tinfoil]

I agree with DanDel. I would assume the current, in amps, to be a function of P=I*V, so the current at 138 should be 345/138 times that of the 345kV scenerio, assuming identical p.u. impedances between the genration and the fault.

You must be modelling the transformers differently to get "the fault current contribution from this same capacity generator at 138 kV lelvel is almost twice as high as that of the fault contribution at 345 kV"

You may want to review your per-unit theory if this result is not intuitive to you.