Short-circuit currents represent a tremendous amount of destructive energy, which can be released through electrical systems under fault conditions. Baseline short-circuit studies should be performed when the facility electrical system is first designed, and then updated when a major modification or renovation takes place — but no less frequently than every five years. Major changes would be considered a change in feed by the electric utility, a change in the primary or secondary system configuration within the facility, a change in transformer size or impedance, a change in conductor lengths or sizes, or a change in the motors that are energized by the system.

Every electrical system confines electrical current flow to selected paths by surrounding the conductors with insulators of various types. Short-circuit current is the flow of electrical energy that results when the insulation barrier fails and allows current to flow in a shorter path than the intended circuit.

Fig. 1. Short-circuit current through an impedance.

In normal operations, as shown in Fig. 1, the impedance of the electrical load limits the current flow to relatively small values. However, a short-circuit path bypasses the normal current-limiting load impedance, resulting in excessively high current values that are restricted only by limitations of the power system itself, and by the impedances of the conductive elements that still remain in the path between the power source and the short-circuit point (Fig. 2).

Using basic Ohm's Law (E = I × Z or I = E ÷ Z) as a guide, it's obvious that if the voltage remains constant and the impedance suddenly decreases, approaching zero, then the current must simultaneously increase, approaching infinity, to satisfy Ohm's Law.

Fig. 2. Short-circuit current through two impedances.

There are three basic sources of short-circuit current: the electric utility, motors, and on-site generators. Obviously, the largest source is the electric utility, although the high- and medium-voltage lines leading to the facility do have finite impedances, as does the utility service transformer. The second largest source is from motors within a facility.

With today's high fault currents, it's more important than ever to protect electrical equipment from extremely high current levels. Otherwise, the equipment will explode as it attempts to interrupt the fault. But for many, fault current calculations have always been difficult to get a handle on, until now.

Here's a new method to calculate short-circuit currents, one we like to call the “Easy Way kVA Method.” You can use in it in place of the abstract “per-unit” method of short-circuit calculations from the past. With the kVA method, you can easily visualize what currents will flow and where they will flow, and you can calculate them using an inexpensive handheld calculator in moments, regardless of the complexity of the electrical power system.

This method is simple because there are no awkward “base” changes to make, because kVAs are the same on both the primary and secondary sides of every transformer. Best of all, you only need one calculation to determine the short-circuit values at every point within the entire electrical power system. With the old per-unit method, you needed a separate calculation for each point in the system.

You can obtain short-circuit kVA values from your electrical utility company, but short-circuit power is also protected by generators and motors. The kVA produced by a motor is equal to its starting inrush current. Likewise, the kVA produced by a generator is equal to its kVA nameplate rating divided by its nameplate subtransient reactance rating (Xd).

For example, suppose we have a 1,000kVA generator with a subtransient rating of 0.15. It would instantaneously produce 6,667kVA (1,000 ÷ 0.15). Or, suppose we have a 100-hp motor with subtransient rating of 0.17. It would instantaneously produce 588kVA (100 ÷ 0.17).

Now suppose this motor and generator connects to the same bus. Then, the short-circuit power available at that bus is the sum 6,667kVA plus 588kVA, or 7,255kVA. If the electrical utility is rated to deliver 100,000kVA to this same bus, then the total short-circuit power available at that bus is 107,255kVA.

Using the kVA method also greatly simplifies the calculation of short-circuit power attenuation (or holdback) provided by reactors, transformers, and conductors. For example, a 2,000kVA, 7% impedance transformer will pass through its windings a maximum of 28,571kVA of power (2,000 ÷ 0.07), if infinite power flows to one side of its windings. If instead of an infinite current source, the above bus connects to this transformer, then the amount of power that will be “let through” the transformer is the reciprocal of the sum of the reciprocals of the two, or 1 ÷ ( [1 ÷107,254] + [1 ÷ 28,571] ), or 22,561kVA. You can determine transformer impedance, reactor impedance, or cable size with the kVA method quickly enough to make “what-if” calculations.

Comparisons over several years have found results of the kVA method to be accurate within 3% of computer calculations using expensive software, so you can even use the kVA method as a “check” on the input and output of a computer calculation. This is an excellent benefit because standard engineering procedure requires you to check calculations using a different method from the one originally used.

tnx mr.Faisal REHMAN Thebo

This link will be very helpful regarding short-circuit calculations;

http://nptel.ac.in/courses/108107028/22

well come dear Amir abbas