Fault Level You need to specify what type of "Fault Level".Do you mean:1. Fault level internal to an Alternator/Generator/Battery.2. Maximum Short Circuit Current available at the input to a Main switchboard. 3. Maximum Fault level internal to a Main switchboard. 4. Fault level at input to a Distribution sub-board. 5. Fault level internally at a Distribution sub-board. 6. Fault level at the input to a load device = motor/welder/light fitting etc. 7. Fault level internal to a load device = motor/welder/light fitting etc. You need to specify the following:1. Supply type = DC or AC. 2. If DC the Internal resistance of the supply battery or Generator.3. If AC the Voltage, frequency, number of phases, at all stages of the transmission network. 4. Resistances of cables if DC supply. 5. Reactance/impedance of lines/cables at supply frequency if AC supply. 6. If AC, then reactance of transformers etc. in the transmission network from the Supply. There are other factors to consider, but the above are the main considerations.Once these are known, the "Fault levels" at all parts of the network may be calculated...

The percentage impedance of a transformer is the volt drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage.

It is also the percentage of the normal terminal voltage required to circulate full-load current under short circuit conditions 

Importance of fault level is that the protective devices installed should be capable to withstand the fault current and break the specified fault current level. Method to calculate fault level is -The rated Volt ampere of transformer is taken as base VA in MVA.

Fault MVA = Base MVA Base Impedance

Suppose one transformer is a 11kv/433V, 1600KVA, impedance=5%Then Base MVA = 1.6 MVA Base Impedance = 5/100Fault MVA = 1.6x100 = 32MVA 5Fault current on LV side is 32x1000x1000 = 42669.16 A 1.732 x 433

Fault current on HV side is 32x1000x1000 = 1679.61 A 1.732 x 11000 So LV breaker should be capable of breaking 43KA for a specified time (1 sec or 3 sec)And bus bar should be able to withstand 43KA for the specified time.The LV cables should be capable to take the required fault current till the breaker opens(breaker opening time).

Significance of transformer impedance

VOLTAGE, CURRENT OR power, and temperature-we're accustomed to seeing those characteristics (especially the first two) as rating information for any electrical equipment. Transformers used in electric power distribution systems are no exception. But in addition to those familiar properties, transformer ratings include one other item: percent impedance. We don't see that term used for anything other than transformers or reactors (which are similarly constructed but don't involve a ratio between input and output voltages).

Why do transformers have an impedance rating and why in "percent"? The simplest answer to the first question is that a transformer is always interposed between a primary (upstream) source of energy and a secondary or downstream load that makes use of that energy. Hence, the transformer constitutes impedance-to current flow from source to load. The higher that impedance, the more voltage drop will occur for a given load current. If a downstream fault occurs, its effect on the circuit will be lessened to the extent that transformer impedance increases. The circuit designer therefore needs to know the value of transformer impedance.

Does the impedance rating in percent equate to percent voltage drop associated with the flow of rated current? Not necessarily; that depends upon the power factor in the secondary circuit. Remember that in a series a-c circuit the voltages are phasors, at phase angles with one another depending upon the inductance and capacitance involved.

In any inductive apparatus, impedance is a complex variable made up of both resistance and reactance. Since current flow through resistance always results in I^sup 2^R loss energy that is wasted as heat, transformer design aims at minimizing the resistive component. Therefore, in most voltage drop calculations, transformer impedances can be treated as entirely reactive. The terms percent impedance and percent reactance are often used interchangeably.

When the secondary load is a large motor, its impedance will also be largely reactive. Therefore, calculating voltage drop by simple arithmetic is often accurate enough. With a small transformer, or with loads and connecting cable that are more resistive, voltages must be viewed vectorially, and the result can be quite different.

A transformer's percent impedance is evaluated by a simple test. The secondary terminals are short-circuited. A low voltage is then applied to the primary terminals, and increased until the current measured in the short-circuited secondary reaches the rated ampere value. The impedance is then the ratio of that primary voltage to the rated voltage; multiplying that by 100 gives the impedance in percent.

In addition to its role in voltage drop calculation, that figure determines the maximum fault current the transformer is capable of delivering to a downstream short-circuit. For example, consider a 1,000 kVA transformer with a 5% impedance rating and a 480 volt secondary. Rated full-load secondary current would be 1,200 amperes (current = kVA/[ 1.73 times 0.48 kV]). The maximum short-circuit current would be 1,200 divided by per unit impedance of 0.05. or 24,000 amperes.

Looking at that another way: If 5% of rated primary voltage results in 100% rated current in the short-circuited secondary, 100% of rated primary voltage would cause 20 times as much secondary current. In this example, that's 20 times 1,200, or 24,000. Of course, that's only a theoretical maximum.

Under fault conditions in a real system, impedance of cables and other circuit components upstream from the transformer primary would drop the primary voltage below 100%.

Transformers voltage ratings are typically at full load.

For instance, A 24 VAC, 10A transformer will have a terminal voltage of 24 when it is feeding 10 amps to a load. Since the transformer windings have some resistance, the transformer designer has to wind the transformer to put out more than 24 volts, since some of the voltage will be lost, dropped across the resistance of the secondary windings.

But, according to Ohm's law, the voltage dropped across a resistance is proportional to the current (E=IR). If we take away the 10A load, there is no current, and therefore no winding voltage drop! The excess voltage the designer built in now appears at the terminals. This is the no-load voltage. In my example above, when we remove the 10A load, the output voltage of the transformer might rise to 26.4V. We would say the no-load voltage of that transformer is 26.4V

The ratio of full-load voltage to no-load voltage is called the transformer's "regulation factor". It is calculated as:

(no-load voltage - full-load voltage) / full-load voltage * 100.

Ours is: ((26.4 - 24) / 24) * 100 = 10%.