Short Circuit

[Mike01]

X/R what exactly is the X/R ratio consist of and what is the best way to calculate the X/R for your system. I know this is an integral part of short circuit studies but I never really understood that part of it how it is calculated, also how do you know what the phase angle of your service is? On the secondary of xfmrs does it always reset back to 30 degrees just curious. Thanks in advance.

 

[ron]

In an AC system, everything (cables, equipment loads, etc) have a real and imaginary part of its impedance R=resistance X=reactance

The utility starts with a number at their generator, and by the time it gets to your transformer, it could be just about anything.

 

[Mike01]

Customer Side

Customer Side

Is there ever a reason to know the X/R for the customer side of the service? also this would have to be calculated for the secondary of each stepdown xfmr as that would act like a new point in the distirbution? is this correct??

 

[kingpb]

You are asking quite a complex question my friend, no pun intended (e.g. Z = R + jX is a complex number).

The X/R ratio is the reactance over resistance. Knowing the R and X components provides you with the ability to know the impedance, Z. Z is a vector which means it has magnitude and direction (e.g. angle). You determine the impedance of the system at your fault location by combining the network components to the fault point, usually by doing the R and X networks independently. The impedance combines by the same rules as reistors in parallel and series. This is quite tedious to do by hand since it takes the form of using matrices. Computer programs do all this for you, and is much, much faster. X/R values are needed to determine bus bracing and momentary withstand ratings of devices.

I would recommend looking at the Cutler Hammer Consulting Application Guide, under system Analysis section. It would be a helpful tool to start with. You should be able to download it from the CH website.

 

[mayanees]

... X/r

... X/r

Mike,

Consider the X/R ratio as a measure of the strength of an electrical system, whether it's a reported Utility value, or the measured value on a new transformer. A high X/R value indicates more inductance than resistance - or put in a more simplistic way - lots of copper.

And yes the value changes throughout the distribution system, based on the parameters of the system.

Circuit breakers are tested at a given X/R ratio, and their ratings are reported based on that value - which is usually a fairly high value. If they are applied in an environment where the X/R ratio is higher than what they've been tested and listed at, their performance is downgraded.

JM

 

[Mike01]

Agreed

Agreed

Thanks, for all the responses I understand the notion of using complex numbers, and somewhat understand vectors (learning more and more everyday) and was just curious. I know there are computer programs that due this with a lot more speed and accuracy (provided the information put in is good garbage in garbage out) I have used these before but I just like to understand it a little more I was going to try on a small one-line to accomplish this by hand just so I know where it comes from is there any good books or other reading material you would suggest to help digest this as I venture foreword, I guess my thing is if you do not understand where it came from or how to do it (longhand) you should not be allowed to use a program that does it for you because you will never be able to glance over a generated report and notice a flaw, I once saw a S.C. study someone did where they put in information printed out a report and submitted it and for some reason on the secondary of a step down xfmr. Feeding a receptacle panel the schedule / one-line / report indicated more available fault current at the panel than the xfmr could let thru if in an infinite bus situation this was later corrected but the person who looked over the report either just missed it completely or did not review it completely, he should have seen the large number and then questioned were it came from unfortenly it was assumed the software did it so it must be right. I for one prefer to do most things on a pad and paper but hey I must agree time is money, and pad and paper for some things are definitely time consuming.

 

[mikehughes8]

I was reading the IEEE buff book on short circuit calculations. It got incredible involved. It was talking about asymetrical vs. symetrical waves not to mention countless equations. Needless to say, my need to understand short circuit calculations in any detail has died a miserable death.

 

[charlie b]

. . . you should not be allowed to use a program that does it for you because you will never be able to glance over a generated report and notice a flaw. . . .

I would call that an extraordinarily astute observation. I congratulate you on this level of logical reasoning.

The principal reason that there is any significant ?fault current? in the first place comes from the principle that current cannot change instantaneously in an inductor. Put another way, when you try to change the current in an inductor, a voltage is created across the inductor, and the orientation of that voltage is such as to oppose the change in current. If your studies have included calculus, the expression for voltage is given as the amount of inductance times the rate of change of current (V = L dI/dt).

Consider a circuit with only an AC voltage source and a load that is purely inductive. Consider what happens when you suddenly place a wire across the terminals of the inductor. The current that the source had been driving through the inductor now wants to take the lower impedance path through the wire. That would deprive the inductor of its source of current, and it wouldn?t like that. So it creates a voltage across its terminals, in an effort to oppose the change in its current.

Look at the circumstances happening in the ?short circuit wire.? Now what do you have? You have the original voltage source driving current through that wire. You also have a new voltage source, the inductor, and over the next few moments, perhaps as long as a second or two, that new voltage source will also be driving current through the short circuit wire. So you see that the total fault current is higher than the original source could have supplied on its own. This is the sense in which we say that motors (themselves being inductive loads) will contribute to the total available fault current.

Please note also that the original source will itself be primarily and inductive device. Whether it be a generator or a transformer, it is constructed out of coils of wire. So the process I describe above also happens within the source. That is why it supplies more current during a fault than it would during normal loading conditions.

Finally, let?s get back to the question of X/R. First of all, it refers both to the X/R value associated with the utility transformer and the X/R value of the power system. When you look at the power system as a whole, you can model it as a ?Thevenin Equivalent Impedance.? Have you studied that concept yet? It means you can replace the entire facility with a single impedance value, one that consists of a X value and an R value. The ration of X/R tells you, essentially, how much of the load is inductive, how much, for example, is motors. That in turn will give you an idea of the amount of current these inductive loads, these motors, would contribute to a fault. That is why X/R is one of the ways we model a system, and one of the terms we use to analyze a fault situation.

 

[mull982]

Look at the circumstances happening in the ?short circuit wire.? Now what do you have? You have the original voltage source driving current through that wire. You also have a new voltage source, the inductor, and over the next few moments, perhaps as long as a second or two, that new voltage source will also be driving current through the short circuit wire. So you see that the total fault current is higher than the original source could have supplied on its own. This is the sense in which we say that motors (themselves being inductive loads) will contribute to the total available fault current.[/SIZE][/FONT]

Is what you described above what is called Subtransient Reactance X"dv? I have heard of this term before but never really understood where it came from, however your excellent explanation above helps explain where this X"dv comes from. I have also noticed that a typical value for this X"dv seems to be about 16.7% with induction motors. Any particular explanation as to why this is the typical value?

 

[Mike01]

Thanks

Thanks

charlie b, thanks for your response it sheds more light on the topic I somewhat understand the "Thevenin Equivalent Impedance" I have read quite a bit about it and can say I am beginning to understand it more and more ther is also something else I have been researching also call "The norton therom". as for my "studies" as of the end of this year I will have completed all my you could say "pre-requsite" (from a community college) courses for the study of electrical engineering (power engineering) and approx. 1 year from now beginning my studies a local college that is accrediated and supports this program. It's funny looking around tho and seeing people 10 years younger than me starting school I just got a late start.

 

[charlie b]

Is what you described above what is called Subtransient Reactance X"dv?

It is related.

A power system is comprised of power sources, resistive loads, inductive loads, and capacitive loads. You add the last three of these things to get a total ?impedance.? Once you have the system built, the ?total impedance? is set, and it does not change unless you add or remove loads. So when the fault occurs, the impedance seen at the fault point does not change during the event. That is the same as saying that the ?reactance? seen at the fault point does not change during the event, since ?reactance? and ?impedance? are two ways to describe the same thing. In fact, the two are reciprocals of each other.

We all know that ?Ohm?s Law? relates voltage, current, and impedance. I just said that the impedance does not change during a fault. But the voltage and current do change, and they change in a way that does not seem to follow Ohm?s Law anymore. How can that be, if the impedance is remaining constant? The reason is that the system suddenly has a bunch of new voltage sources (i.e., the motors that are contributing to the fault current). The system would also have a bunch of new current sources, if by chance it includes a bunch of capacitive loads.

Now our analytical tools (e.g., for load flow analyses or short circuit calculations) rely heavily on Ohm?s Law. So how can we make it work? We model the system as though it did have different impedance values at different times. We alter the model, to avoid having to add voltage and current sources, and we do it by assigning the varying characteristics to the loads.

There are three separate time intervals that are analyzed. The long term, or ?steady state? conditions, would be those that would continue forever, or at least until the overcurrent protection elements take action to terminate the event. During that period, the system is modeled as having the very same impedance values as it had before the fault took place. I would call this the third time interval of interest.

The first time interval lasts less than one cycle. During that time, the motors contribute most of their energy to the fault point. Also during that time, the ?apparent impedance? is very low, the fault current high. The system is modeled during this interval as having a ?subtransient reactance,? meaning a value of reactance that is in effect in an interval that is faster than the transient period.

The second time interval, obviously, is between the other two. I do not recall the exact time duration, but I think it is on the order of a few cycles, but it is certainly not as long as a half second. The current contributions from the motors has gone down a great deal, because the simple act of transferring energy to the fault point will have caused them to slow down. The system is modeled during this interval as having a ?transient reactance.?

 

[charlie b]

I somewhat understand the "Thevenin Equivalent Impedance." I have read quite a bit about it and can say I am beginning to understand it more and more. There is also something else I have been researching also call "The Norton Theorem."

The two are very closely related, and if you know the values for one you can quickly calculate the values for the other.

Both are methods of modeling a complete power system, as seen from any specific point along the system you care to pick. The system will have some combination of voltage sources, current sources, and impedances. The Thevenin Equivalent is a single voltage source in series with a single impedance. It is ?equivalent? in the following sense: Start with the original power distribution system, and connect something else to it. Measure the current that the system supplies to the new something, and measure the voltage as seen by the new something. Now build a system that comprises nothing more than the voltage source and the impedance value from the Thevenin Equivalent. To it, you attach the same ?new something,? and you take the same measurements. You will get the same values both times. So the bottom line is that you can put a ?big black box? around either system (the original and the Thevenin Equivalent), and not look into the box to discover what is really inside, and you will not know which you have. That makes it easy to analyze the system, because you are dealing with only two numbers (the voltage value and the impedance value), instead of perhaps thousands of values associated with a large number of components that make up the original system.

The Norton Model is an ?equivalent model? in the same context. It is comprised of a single current source and a single impedance value, with the two connected in parallel.

 

[mull982]

Put another way, when you try to change the current in an inductor, a voltage is created across the inductor, and the orientation of that voltage is such as to oppose the change in current.

You also have a new voltage source, the inductor, and over the next few moments, perhaps as long as a second or two, that new voltage source will also be driving current through the short circuit wire.

Let me ask another question pertaining to the same subject material. If I am using an in line Reactor (basically an Inductor) on a circuit to soft start (reactance start) a motor or some other application such as a CLR what happens with the voltage created across the inductor in this case? Does this newley created voltage drive an additional current on the inrush startup of a motor? If a CLR is beig used on a circuit to lower fault duty ratings of equipment, wouldn't this additional induced voltage just create additional current during a fault? Using a CLR is there was a fault after the CLR wouldn't the fault be a combination of the fault current plus additional current driven by the induced voltage in the reactor.

 

[charlie b]

Someone else is going to have to answer that one, mull982. I do not know what a CLR is, and I am not well versed in motor starting schemes. Sorry.

 

[kingpb]

If I may, just to clairfy something charlie said, is that the system impedance doesn't change unless you add or subtract loads or equipment, which is true, however I think what needs to be clarified is that at the point in which the impedance is calcualted it will not change. The impedance will be different at every point in the power system and that is why you must re-calculate the impedance at each bus you are interested in analyizing. Hence, computer models. I remember doing hand calcs in college, thank goodness I never have had to resort to them since.

As far as printouts go, it takes an engineer with experience in power system analysis to be able to pick out errors or problems. Otherwise junk input = junk output.

A reactor used for reduced voltage motor starting is related to an autotransformer, where the reactance (inductor) reduces the voltage and thereby reduces inrush current. (The principle of voltage down/current up does not apply during a motor start.) However, torque capability is also decreased, and can be a problem if a lot of starting torque is required.

 

[charlie b]

. . . however I think what needs to be clarified is that at the point in which the impedance is calculated it will not change.

Quite true. I was careful to say that as well, but I had not given it any extra emphasis. Thanks for the clarification.

So when the fault occurs, the impedance seen at the fault point does not change during the event.