# Earth fault protection scheme.

Discussion in 'General Electronics Chat' started by dileepchacko, Aug 4, 2008.

1. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
Hi All

This problem related to Earth fault protection scheme for Transformers. The Fig.1 is Line to Line short circuit fault, Fig.2 is Line to ground short circuit fault and Fig.3 is Line to line to Ground short circuit faults. Please find out the current flowing through the Resistor R7( fault current ) for 3-cases.

Note: A. The transfromer ratio is 1:10
B. The transformer is Star-Star winding type.

C. See the attachement for refering the figer ( fig.1, fig.2, and fig.3 )

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2. ### dileepchacko Thread Starter Active Member

May 13, 2008
102
1
As we knows in normal condition ( There is no fault in transformer )

The sum of the all current in earth point ( primary side and secondary side ) will be zero.

Ir + Iy + Ib = 0 = In ( In = neutral current or balance current ----------------------------(1)

Case:1 ( refer fig: 1 in attachment )

Short circuit of R phase and Y phase on secondary side of the transformer.

In this case Ir will split in to two component, Ir1 and Ir2, and Iy will split in to two component, Iy1 and Iy2.

Ir1 and Iy1 will circulate in the RY phase loop, and remaining current Ir2 and Iy2 will flow through the Neutral point. Now the equation (1) becomes.

Ir∆ + Iy∆ + Ib = In + ( Ir2 + Iy2 ) ---------------------------------------------(2)

Ir∆ = Now actual Ir has been changed ( in faulty condition )
Iy∆ = Change in Iy ( in faulty condition )

Case:2 ( refer fig: 2 in attachment )

Short circuit of R phase and ground on secondary side of the transformer.

In this case Ir will split into two components, Ir1 and Ir2. Here Ir2 will flow throw the ground.
And no change in remaining currents ( Iy and Ib )

The equation (1) becomes Ir∆ + Iy + Ib = In + Ir2 -------------------------------------------(3)

Ir∆ = Now actual Ir has been changed ( in faulty condition )

Case:3 ( refer fig: 3 in attachment )
Short circuit of R phase, Y phase, B phase and ground on secondary side of the transformer.

In this case Ir will split into 3 components Ir1, Ir2 and Ir3
Iy will split into 3 components Iy1, Iy2 and Iy3
Ib will split into 3 components Ib1, Ib2 and Ib3

Now equation (1) becomes

Ir∆ + Iy∆ + Ib∆ = In + ( Ir2 + Iy2 + Ib2)

Ir∆ = Now actual Ir has been changed ( in faulty condition )
Iy∆ = Change in Iy ( in faulty condition )

Ib∆ = Change in Ib ( in faulty condition )

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