Sample from the Short Circuit Studies DVD series – Per Phase Analysis



NEW FOR 2017!  4 DVD training program taught by Jim Phillips, P.E. $199.   Learn how to perform short circuit analysis / calculations and equipment adequacy evaluations. Understand the importance of X/R ratios, symmetrical vs. asymmetrical faults and how series ratings work. Many calculation examples are used to illustrate how to perform an analysis. The methods are based on the IEEE Violet book and is loaded with many in class examples and problems for a hands on learning experience.


Video Transcript

In a per phase analysis, what that actually involves is you have a three-phase power system. And the assumption is the event on all three phases, the result of whatever it is you’re modeling on all three phases, is the same. It’s balanced. So a three-phase short circuit, you’re considering that, well, whatever the current is, tens of thousands of amps, it will be the same on phase A, on phase B, and on phase C.

So what do you do? You just analyzed one phase. It’s not aligned to ground calculation, it’s a per phase calculation. So you analyze what’s on one phase, and then what you determine on that one phase is also the answer for the other phases. And so when we’re performing the calculations, like I had up here on the screen, the voltage that we’re looking at across one phase is the line to neutral voltage.

So for the per phase analysis, the example that I have for the per phase analysis, I have aligned the neutral voltage on the left side of this diagram, 277 volts, that’s lined to neutral from a 480 system. And then I have what amounts to the equivalent impedance of the source, of the utility system. Some you may remember the term Thevenin equivalent impedance.

Sorry about that. You thought you got away from that one. There’s actually a use for it. For those of you that are not familiar with that term, what the Thevenin equivalent impedance is, it’s basically taking the impedance of everything out there in the utility world, and series and parallel combinations, and reducing it down to just a point equivalent impedance that represents everything out there.

And so that’s what the 0.0041 ohms is in the upper left. That’s basically the equivalent impedance of what the utility system appears like, electrically, at the beginning of the study, usually at your service entrance or your interconnection point. Then the next one, the transformer, even though the transformers are normally in percent, I have this in ohms just for the sake of this particular problem, 0.0085 ohms.

And then we have a conductor on the secondary side, a set of conductors, 0.0034 ohms. And then we have a load on the far right side. The impedance of the load, 2 ohms.

So, a per phase analysis just using basic Ohm’s law, voltage, current, and impedance, if you were to take 277 volts and divide by these four impedances, the source being 0.0041, the transformer being 0.0085, the conductor point 0.0034, and the load 2 ohms, and you add all of this together, 277 volts divided by 2.016 ohms, the current that would flow would be 137 amps. That’s a load curve. That would flow in all three phases using this per phase analysis.

That’s the load current. But what happens, looking at this diagram, near the load just downstream from the overcurrent device on the far right side, if we have a short circuit? The short circuit, the way we model it, we look at the short circuit as a zero impedance path. We consider this to be the bolted connection.

We don’t want any additional impedance because, for this type of study, we want the maximum worst case short circuit current. So we don’t want to factor in any additional impedance. We say it’s a bolted fault, a bolted short circuit.

So now what happens, looking at the right side of this, the load impedance is still there. 2 ohms are still there. But the current that’s going to flow from the source is not going to go through the 2 ohms. Current takes the path of least resistance, literally. That’s how it works.

And so now what’s going to happen, as far as the analysis, we still have 277 volts. But now we just divide by the source, the transformer, and the conductor impedance. The load impedance doesn’t factor into this.

So 277 volts divided by just the small impedances of the source, the transformer, and the conductor, the resulting current is quite large. 17,313 amps. So what you need to do is consider, will the protective device operate?

And not only will it operate, will it operate safely? Does the protective device have a sufficient or an adequate interrupting rating to be able to handle the predicted 17,313 amps? That’s what a short circuit study is all about.

The short circuit study is to mathematically model the power system with all the different impedances. The impedance of the utility system, the impedance of transformers, of conductors, to mathematically model all that, and to be able to calculate or predict what would be the maximum short circuit current at different locations for equipment. And you compare the calculated short circuit currents with the device’s, or the equipment’s, interrupting or withstand rating. That’s what a short circuit study is all about.

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