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Bus Fault Simulation of a four (4) Parallel Transformers
Interactive Calculator







NOTE: If the transformer is out of service use Z = 1e9 (1,000,000,000) to simulate open circuit , transformer current should be almost zero.

Singe Resistor Z1 = 10 Z2 = 1e9 Z3 = 1e9
Z4 = 1e9 I = V / Z = 30.3 A Total current
Two Parallel Resistor Z1 = 10 Z2 = 10 Z3 = 1e9
Z4 = 1e9 I = V / Z = 60.6 A Total current
Three Parallel Resistor Z1 = 10 Z2 = 10 Z3 = 10
Z4 = 1e9 I = V / Z = 90.9 A Total current
Four Parallel Resistor Z1 = 10 Z2 = 10 Z3 = 10
Z4 = 10 I = V / Z = 121.2 A Total current


I 1 + I 2 + I 3 + I 4 = Equation 1

I 1 + I 2 + I 3 + I 4 = Equation 2

I 1 + I 2 + I 3 + I 4 = Equation 3

I 1 + I 2 + I 3 + I 4 = Equation 4

For SCENARIO ANALYSIS, assume all transformers have the same % Z value. You only need one through fault current from one transformer to get the constant voltage = IZ across the four banks of transformer. Then you can do quick adjustment of voltage to approximate the known through fault current from one faulted transformer. You can get this through fault current from PMU data, DFR, or SEL 87T relay.

Z 1 = 10 Ω Z 2 = 30 Ω

Z 3 = 10 Ω Z 4 = 10 Ω

V L-G = 303 V Voltage adjustment to approximate the known through fault current. Next you apply the appropriate % Z of each transformer and you will get the approximate through fault current for the remaining three banks of transformer.

Practice : Z 1 = 0.092 Ω Z 2 = 0.092 Ω
Z 3 = 0.092 Ω Z 4 = 0.092 Ω
V L-G = 303 V

Given PMU through fault for Transformer T2 = 1054 A. What is the adjusted voltage to simulate a 1054 A through fault assuming all transformers have the same % Z of 0.092? ANSWER = 97 V. Using the 97 V as your voltage source to let through fault current of 1054 A flow through Transformer T2. Now you are ready to estimate the through fault current passing through Transformer T1, T3, and T4 using their actual nameplate impedance value.

What is the estimate thru fault current for T1 at Z1 = 0.096 ; T3 at Z3 =0.065 ; T4 at Z4 = 0.0623

Using back substitution to find value of I 1, I 2 and I 3.



INITIAL CONDITION :
Answer, I 1 = 101.00 XF 1 Load current; I 1 - I 2 = 30.30
Answer, I 2 = 70.70 XF 2 Load current; I 2 - I 3 = 10.10
Answer, I 3 = 60.60 XF 3 Load current; I 3 - I 4 = 30.30
Answer, I 4 = 30.30 XF 4 Load current; I 4

Result of Gaussian elimination

I 1 + I 2 + I 3 + I 4 = Equation 1
I 1 + I 2 + I 3 + I 4 = Equation 2
I 1 + I 2 + I 3 + I 4 = Equation 3
I 1 + I 2 + I 3 + I 4 = Equation 4

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.



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