How to Calculate Fault MVA of a Transformer?

Whenever a dead short-circuit fault occurs in the transformer circuit, the transformer delivers a very large current. Under this condition, the apparent power delivered by the transformer becomes very high. The maximum MVA that the transformer can deliver during this fault condition is the fault MVA of the transformer.

Calculation of Fault Current of a Transformer

Suppose, the secondary winding of the transformer becomes solidly short-circuited. Under this condition, we apply the rated voltage to the primary winding. The current taken by the transformer from the source depends on the internal impedance of the transformer. It is equal to the applied primary voltage divided by the impedance of the transformer. Obviously, the impedance of the transformer referred to the primary side in this case.

Therefore, we can write the short-circuit current as

$I_{sc} = \frac{V_p}{Z_p}\cdot\cdot\cdot(i)$​​

Here, Z_p is the impedance of the transformer referred to the primary side in this case.

Under this condition, the total MVA taken from the source is V_p×I_sc​. This represents the apparent power drawn by the transformer during the fault. Obviously, this is the fault MVA that the transformer takes from the source during a dead short-circuit fault.

Calculation of Percentage Impedance

Let us now apply a gradually increasing voltage to the primary winding. The secondary winding remains short-circuited. At a certain applied voltage, the rated secondary current flows through the short-circuit path. Obviously at that condition primary takes the rated primary current from the source. It is needless to say, this applied voltage is much smaller than the rated voltage of the transformer.

If we divide this applied voltage by the rated primary current, we obtain the impedance of the transformer Z_p (referred to primary). Therefore, the impedance of the transformer referred to the primary side can be written as

$Z_p=\frac{V_{psc}}{I_p}\cdot\cdot\cdot(ii)$

Here, I_p​ is the rated primary current. The term V_psc​ represents the applied primary voltage that produces the rated primary current when the secondary winding is short-circuited.

The voltage required to establish the rated primary current in the transformer, while keeping the secondary winding short-circuited, is called the short-circuit voltage. If we express this voltage with respect to the rated primary voltage, it is known as the per-unit impedance of the transformer. When we express the same quantity as a percentage of the rated primary voltage, it is called the percentage impedance of the transformer.

Calculation of Fault MVA of a Transformer

From the relation (i) and (ii), we can write,

$I_{sc}=\frac{V_p}{\frac{V_{psc}}{I_p}}\Rightarrow I_{sc}=\frac{V_p\times I_p}{V{psc}}$

This equation shows that the short-circuit current is equal to the product of the primary voltage and primary current divided by the short-circuit voltage. Therefore,

$\text{ \;\;\,\,}V_p\times I_{sc}=V_p\times \frac{V_p\times I_p}{V{psc}}\\\\\Rightarrow V_p\times I_{sc}\times 10^{-6}\text{ MVA}\\\\=V_p\times \frac{V_p\times I_p}{V{psc}}\times 10^{-6}\text{ MVA}$

This shows that the product of primary voltage and fault current. Therefore, this term represents the fault MVA of the transformer. Therefore, we can write

$\text{Fault MVA}= \frac{V_p\times I_p\times 10^{-6}}{\left(\frac{V_{psc}}{V_p}\right)} \text{ MVA}$

Here, $\frac{V_{}psc}{V_p}$​​ represents the ratio of short-circuit voltage to rated voltage. This ratio is nothing but the per-unit impedance of the transformer.

Therefore, we can write$$\text{Fault MVA} = \frac{\text{Rated MVA}}{\text{Per-unit impedance of the transformer}}.$$

Thus, the fault MVA of a transformer is equal to the rated MVA divided by the per-unit impedance of the transformer. This relation shows that a transformer with lower impedance will produce a higher fault MVA during a short-circuit condition.

The per-unit impedance of a transformer can be expressed as percentage impedance if we multiply it by 100. Therefore,

$$\text{Percentage Impedance} = \text{Per-unit Impedance} \times 100$$

Therefore, we can ultimately write the fault MVA in terms of percentage impedance. We know that percentage impedance is equal to per-unit impedance multiplied by 100. Using this relation, we can write

$$\text{Fault MVA} = \frac{\text{Rated MVA}}{\text{Per-unit Impedance}}$$Since$$\text{Per-unit Impedance} = \frac{\%Z}{100},$$we obtain$$\text{Fault MVA} = \frac{\text{Rated MVA}}{\left(\dfrac{\%Z}{100}\right)}.$$Therefore,$$\text{Fault MVA} = \frac{\text{Rated MVA} \times 100}{\%Z}.$$Thus, the fault MVA of a transformer is equal to the rated MVA multiplied by 100 and divided by the percentage impedance of the transformer.

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