Effect of utility X/R ratios on system

[mull982]

I have recently been running fault current calculations/studies and had been wondering what effect the utility X/R ratio had on the results.

I'll first start by asking how a modern software package converts the inputed utility information to use for calcualation with the rest of the system. When the system model (load) it will have a specifiec combined impedance (thevinin) and X/R ratio. This is then combined with the utility fault contribution for the SS analysis on the system. Most of the time the utility contribution is entered in avaliable fault current value with an associated X/R ratio. Does the software then take this current and X/R value and use the utility voltage to convert the current value to a p.u. impedance value based on the MVA base entered for the system. Does it then combine the utility and system impedance values for use in any fault current calculation with the system?

Seeing that what I said above is true I am looking to see the effect that an X/R ratio has on the utility contribution. For instance lets say we are analysing a system that has a specified impedance and X/R value. We then add a utility fault contribution to this sytem that has a specifed fault contribution in amps with an associated X/R ratio. Now lets say that we keep the utility fault vaule the same but change the X/R ratio to a higher value. Would this represent a worse case?

From what I can tell for a given utility fault current raising the utilty X/R ratio will make the combined Uitlity&system impedance lower and therefore result in higher fault magnitudes for faults anywhere in the system for a fixed voltage.

Is this correct that raising the utility X/R ratio will result in worse case faults?

 

[kingpb]

The X/R ratio is going to be used to calculate the R and X short circuit impedance values in % or p.u. at some base. By knowing the MVAsc value and X/R value. This information would need to be know for 3-ph and 1-ph. Sometimes the 1-ph is not readily known and it will be assumed that they are the same. Most times the 3-ph KAsc is greater than the 1-ph KAsc so this is a reasonable assumption, however, it is possible for the 1-ph to be greater, but not very often.

The larger the X/R, the greater the KAsc is going to be, since it is nothing more than the components of Z=R+jX. Determining the magnitude of Z, it becomes clear that R is less of a consequence as X increases, because Z magnitude is determined by the sq root of the sum of the squares.

Z = V^2/MVA so knowing MVAsc, Vsys, and X/R you can determine the other values needed for calculations.

Doing a little quick math you can easily see how the increase in X/R, i.e. X greater than R, affects the impedance Z. Say if R = 1 and X = 2 then X/R is 2, and Z = 2.24. Now have R = 1 and X = 10, X/R = 10, and the magnitude of Z is essentially 10, or the same as X, and X is what affects short circuit values. So as X goes up, so does Isc.

 

[mull982]

Doing a little quick math you can easily see how the increase in X/R, i.e. X greater than R, affects the impedance Z. Say if R = 1 and X = 2 then X/R is 2, and Z = 2.24. Now have R = 1 and X = 10, X/R = 10, and the magnitude of Z is essentially 10, or the same as X, and X is what affects short circuit values. So as X goes up, so does Isc.

If X increases and therefore the overall impedance increases, wont an increased impedance result in a lesser fault current or Isc for a given voltage? So therefore as X increases X/R increases and thus overall impdeance increases resulting in a lesser Isc?

 

[jghrist]

You need to consider two effects of X/R. First is the effect on the fault current magnitude. Second is the effect on circuit breaker interrupting capacity.

If you are given a utility current magnitude and X/R, then the software will calculate the magnitude of Z from the current magnitude and the voltage (Z = V/I) taking the units into account with proper multiplication or division by 1000 if necessary, and taking 1? pr 3? quantities into accounty wiith proper application of sqrt(3). The utility R will be added to the distribution system R to the point of the fault. Same with X. Then the total R & X will be converted to Z with sqrt(R? + X?). The fault magnitude is I = V/Z. So, if you increase the utility X/R, the effect on total Z depends on the distribution system R and X and the relative values of the utility R and X. The total Z will be higher (for a given utility fault current magnitude) if the utility X/R is the same as the distribution X/R. Think of putting two arrows (representing the utility and distribution system Zs) end to end. If the arrows are line up at the same angle, the diatance from the end of one to the tip of the other is greatest.

The second effect of X/R is that circuit breakers are rated at a maximum X/R ratio and if the fault X/R is higher, then the fault duty has to be adjusted to compare with the breaker rating. This adjustment is needed because fault current is higher initially, with a dc component that decreases with time (asymmetric vs symmetric fault). The increase of the initial current is higher if X/R is higher. Now things really get tricky. For low voltage circuit breakers, you don't just take the overall X/R ratio. You have to get an equivalent X by analyzing the system without any resistance, then do the same thing to get an equivalent R for a system without any X, then use this X/R ratio.

 

[mull982]

You need to consider two effects of X/R. First is the effect on the fault current magnitude. Second is the effect on circuit breaker interrupting capacity.

If you are given a utility current magnitude and X/R, then the software will calculate the magnitude of Z from the current magnitude and the voltage (Z = V/I) taking the units into account with proper multiplication or division by 1000 if necessary, and taking 1? pr 3? quantities into accounty wiith proper application of sqrt(3). The utility R will be added to the distribution system R to the point of the fault. Same with X. Then the total R & X will be converted to Z with sqrt(R? + X?). The fault magnitude is I = V/Z. So, if you increase the utility X/R, the effect on total Z depends on the distribution system R and X and the relative values of the utility R and X. The total Z will be higher (for a given utility fault current magnitude) if the utility X/R is the same as the distribution X/R. Think of putting two arrows (representing the utility and distribution system Zs) end to end. If the arrows are line up at the same angle, the diatance from the end of one to the tip of the other is greatest.

The second effect of X/R is that circuit breakers are rated at a maximum X/R ratio and if the fault X/R is higher, then the fault duty has to be adjusted to compare with the breaker rating. This adjustment is needed because fault current is higher initially, with a dc component that decreases with time (asymmetric vs symmetric fault). The increase of the initial current is higher if X/R is higher. Now things really get tricky. For low voltage circuit breakers, you don't just take the overall X/R ratio. You have to get an equivalent X by analyzing the system without any resistance, then do the same thing to get an equivalent R for a system without any X, then use this X/R ratio.

Thanks for the good info. Yes I am familiar with the X/R ratio having an effect on the aysmmetrical fault current value and the breaker rating.

Here is a situation that is confusing me that may be related to X/R rations:

I am modeling a system that has two utility sources through a main-tie-main arrangement. The tie is always closed so only on of the utility sources is supplying power to the system at any given time.

The first source has 7800A @ X/R=30 available, and the second source has 2400A @ X/R=5.4 avaliable. Both these currents are at 12.47kV.

With the first source having an avaliable fault current of almost 3x the second fault source I would expect all the bus fault vaules downstream in the system to have a fault vaule/magnitude which is aprox 3x greater when fed from source 1 then when fed from source 2. What I notice however is that when the system is supplied by source 2 the fault vaules at the 480V buses downsteam are only slightly less than the vaules when fed from the higher source 1? I would think they would be 3x less when fed from source 2.

For example at one of the 480V buses downstream the fault current when fed from source one is aprox 10kA and when fed from source 2 the fault current is aprox 9kA. Can this be right? Why for such a difference in fault current between utility sources can they end up so close downstream at a lower voltage? Does this have something to do with the utility source X/R ratio's?

Thanks

 

[rcwilson]

The utility feed is at 12.47 kV, you are looking at the 480V short circuit levels so there is a transformer in between. The transformer impedance has the most effect on the 480V short circuit.

12.47 kV x 2400 x 1.732 /1000 = 51.84 MVA Short Circuit from one source.

12.47 kV x 7800 x 1.732 /1000 = 168.5MVA Short Circuit from the other source.

Your fault level at 480 V is = 480 x 10 kA x 1.732/1000 = 8.3 MVA That means the transformer and cable impedance is about equivalent to a 8.75 MVA Short Circuit.

Using the other results:
Fault level at 480 V is = 480 x 9 kA x 1.732/1000 = 7.48 MVA That means the transformer and cable impedance is about equivalent to a 8.74 MVA Short Circuit.

Both results show the transformer and cables are equivalent to an 8.74 MVA let through from an infinite bus.

If you had an infinite bus say 500,000 kA short circuit available at 12.47 kV, the largest fault level at 480V will be that 8.74 MVA of your transformer impedance. In current that's 10, 520 amps.

Remember the short circuit amps or MVA is just another way of describing the impedance of the power system. When we calculate short circuit current we add all the series impedances and divide into the voltage. Just because one of the impedances has changed by a factor of three doesn't mean all of them have.

Note that the above calculations ignored the X/R ratios for simplicity to make a point.

 

[rcwilson]

Short Explanation of MVA method of SC calculation used above.

Short Explanation of MVA method of SC calculation used above.

My previous post calculated short circuit current using the MVA method, which is a quick, easy to remember way of coming up with symmetrical short circuit currents that are accurate enough for most applications.

Each component of the power system has a Short Circuit MVA defined as the short circuit that could flow through the element if it was connected to an infinite bus, i.e. it was the only impedance in the system.

Transformer's MVAsc-tx = kVA/(%Z x 10) ex. 1000 kVA, 5% Z. MVAsc-tx = 1000 /( 10 x5 )= 20 MVA.

Utility MVAsc-ut = MVA given or if only amps are given = kV x Amps x 1.732 x 1000.

Cable MVA = (kV)^2 /Z.
example: a 0.005 ohm cable circuit at 480V, MVAsc = (0.48 x 0.48)/0.005 ohm = 46MVA

MVAsc add as admittances (like resistors in parallel): MVA total = 1/([1/MVAtx] + [1/MVAutil] + [1/MVAcable])

Since the final MVA at 480 was 8.3 MVA and the Utilty 1 was 168.5 MVA, the transformer and cable together have to be the difference:

MVAtx = 1 / (1/8.3 - 1/168.5) = 1/(0.1205 -0.00594) = 1/0.1145 = 8.73 MVA sc for the transformer.

 

[mull982]

rcwilson

I follow what you are doing in concept but am still trying to figure out the MVA concept.

I have always learned these calcs be converting everything into a p.u. impedance and then dividing the base MVA by the p.u. impedance to come up with a fault vaule. Same concept I guess just a different way of arriving at the result. I've also done these calcs with 1pu voltage dividing by the combined system p.u. impdeance to arrive at fault current.

It looks like you are just using a method that avoids having to convert all devices to a p.u. impedance based off of the system base.

To make things simpiler Let's just look at the fault values on the primary and secondary of the transformer this way we can ignore cable impdeances. The transformer is a 1000kVA 12.47kV primary - 480V secondary with a 5.79%Z.

From this maximum let through current of transformer is 1000kVA/(.0579/.480V/1.73)= 20,798A. This is the maximum current that the transformer will let through for an infinite bus.

I ran the fault calcs for the two cases and here is what I found.

Case 1: Fault current at primary of transformer=8.4kA. Fault current at secondary of transformer=19kA

Case 2: Fault current at primary of transformer=3.8kA. Fault current at secondary of transformer=17.2kA

I'm curious to see how your MVA method works for just considering the primary fault and transformer impedance.

I'm thinking that we can take the primary fault current and convert it to an MVA value. Using the transformer value as the base we can take the transormer kVA/Fault MVA to come up with an upstream system impedance. We can then add this upstream system impedance to the transformer system impedance and divide this combined impedance by the transformer kVA to come up with the secondary fault current in KVA.

So for example using case 1:

Fault MVA = 12.47kV x 8.4kA x 1.73 = 181.2MVA
Source Imp = xfmr MVA/fault MVA = 1/181.2 = .00551
Transformer impedance at 5.79% = .0579 (using transformer kVA as base)
Combined source impedance + transformer impedance = .0051 + .0579 = .0634
Fault KVA at secondary of xfmr = Xfmr kVA/combined impedance = 1000kVA / .0634 =15772.87kVA
Secondary fault current = Fault kVA/.480V/1.73 = 15772.87/.480/1.73= 18,994A.

The vaule of 18,994A that I came up with is very close to the 19kA calculated by the software. I suspect I would also get an answer that was close to the software by doing the second case.

I guess the point I see here is that the transformer impedance is very large compared to the upstream fault impedance so it is the transformer impedance that limits the fault current on the secondry. So even though primary fault current is almost 3x as high the transformer impedance is dominant and thus results in secondary fault currents that are close?

 

[rcwilson]

MVA Method

MVA Method

Note that you only need to convert the utility or upstream short circuit data to its equivalent MVA. You don't need to change base or adjust it to the transformer rating.

Using your example: 1000 kVA, 5.79%Z: MVA tx = 1MVA/0.0579= 17.27 MVAsc.

Case 1 MVA utility = 181.2 MVA

Transformer secondary MVA sc = (MVA tx x MVA utility)/(MVA tx + MVA utility) = (17.27x 181.2)/(17.27+181.2)= 15.77 MVA. Isc @ 480 V = 15.77 MVA x 1000 / (0.480kV x 1.732) = 18,968 A

Same answer, fewer steps.

(Note this is the same formula as before, only written differently. Same as two resistors in parallel, R = (R1xR2)/(R1 + R2) ).

Your conclusion about the transformer being the major factor for the secondary short circuit is correct.

I use the MVA method because it is easy to remember and there are no complicated formulas.

When I have time, I'll post links to previous threads giving Moon Yuen's original IEEE paper describing the method.

Here is a webpage based on that article. http://www.arcadvisor.com/faq/about_mva_method.html