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Through Fault Estimation Using Gaussian Elimination Method


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Through Fault Simulation for 3 Banks of Transformer
Using Gaussian Elimination Method



V(in KV) = ;

change to decimal % Z1 = ;

change to decimal % Z2 = ;

change to decimal % Z3 =

I1 = ;

I1 - I2(T1) = ; I2 = ;

I2 - I3 (T2) = ;

I3 (T3) =

In parallel transformer , make sure the impedance of all transformers are equal or close enough. The reason is in order for each of the transformer to share the load equally. As you can see from above load sharing Transformer T1 with impedance Z1 (9.2% Z) is drawing more load current compare to Transformer T2 with impedance Z2 (9.7% Z) and Transformer T3 with impedance Z3 (9.5%).

I1 + I2 + I3 = Equation 1


I1 + I2 + I3 = Equation 2


I1 + I2 + I3 = Equation 3




GIVEN DATA

Given Z1 = 0.092

Given Z2 = 0.097

Given Z3 = 0.095

VL-G = 303

Now use the factory test report short circuit impedance value to estimate the short circuit for each transformer.

Transformer 1, factory test report short circuit impedance value is Z1 = 0.092 or 9.2 %

Transformer 2, factory test report short circuit impedance value is Z2 = 0.097 or 9.7 %

Transformer 3, factory test report short circuit impedance value is Z3 = 0.095 or 9.5 %

To validate your answer substitute the values of I1, I2, and I3 to your original equation 1 , equation 2, and equation 3. The answer must satisfy all the equations or with very small error.

Answer, Loop Current I1 = 9,606.66 Bus fault along segment HG to ground
Transformer Bank 1 XF 1 GB through fault current; I 1 - I 2 = 3,293.48


Answer, Loop Current I2 = 6,313.18 Bus fault along segment GF to ground
Transformer Bank 2 XF 2 FC through fault current; I 2 - I 3 = 3,123.71


Answer, Loop Current I3 = 3,189.47 Bus fault along segment FE to ground
Transformer Bank 3 XF 3 ED through fault current; = I 3

Result of Pivot 1, which means making column 1 = [ 1, 0, 0 ]

Result of Pivot 1, eliminating and in column 1.

I1 + I2 + I3 = Equation 1

Divide Equation 1 by 10


I1 + I2 + I3 = Equation 2. Eliminating


I1 + I2 + I3 = Equation 3 . Eliminating


Result of Pivot 2, which means making column 2 = [ CI2, 1, 0 ] where C is the constant of I2.

Result of Pivot 2 eliminating in equation 3 above.
Then converting in equation 2 to one (1) in column 2.


ANSWER: I3 = = .
Using back substitution , substitute I3 = in equation 2 below, you will get I2 = .
Finally substitute I3 = , I2 = in equation 1 below, you will get I1 = .

I1 + I2 + I3 = Equation 1


I1 + I2 + I3 = Equation 2


I1 + I2 + I3 = Equation 3



Important definition to remember about row echelon. The process of reducing augmented matrix in row echelon form is called Gaussian elimination

1. The first nonzero entry in each row is always 1
2. A row with more leading zero entries compare to the previous row must be located below.
3. A row with all zero entries must be below the rows having nonzero entries. Gaussian elimination will not work properly if one of the definition is violated.

Operation 1 - The order in which any two equations are written may be interchanged. Operation 2 - Both sides of the equation may be multiplied by the same nonzero real number. Operation 3 - A multiple of one equation may be added to another equation.

Circuit Analysis Pattern 3 Circuit Analysis Pattern 2 Wheatstone Bridge

Circuit Analysis Pattern 1 4 Transformer Bank Fault Simulation

KVL & KCL Circuit Analysis Transformer Max. Available Fault Current

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