Question

Question: In Problem 9.1 and Figure 9.17, let be replaced by , keeping the rest of the data to be the same. Repeat (a) Problems 9.1, (b) 9.2, and (c) 9.3.

Short answer

Answer

(a) Draw the per unit zero sequence diagram as shown below.

Draw the per unit negative sequence diagram as shown below.

Draw the per unit positive sequence diagram as shown below.

(b) The zero-sequence Thevenin’s circuit is shown below.

The negative sequence Thevenin’s equivalent circuit is shown below.

The positive sequence Thevenin’s equivalent circuit is shown below.

(c) The sub transient fault current in per unit and kA are$\left[\begin{array}{c}8.177\angle -90°\\ 8.177\angle -150°\\ 8.177\angle -30°\end{array}\right]PU$andrespectively.$\left[\begin{array}{c}9.4417\angle -90°\\ 9.441\angle -150°\\ 9.441\angle -30°\end{array}\right]KA$

Step-by-step solution

Write the given data from the question:

Write the rating of the generators.

The power rating of generator is , the voltage rating is , the reactance and are equal to , and the value of is .

The power rating of generator is , the voltage rating is , the reactance and are equal to , and the value of is .

The power rating of generator is , the voltage rating is , the reactance and are equal to , and the value of is .

The power rating of generator is , the voltage rating is , the reactance is , is , and the value of is .

Write the rating of the transformer.

The power rating of transformer is , the voltage rating is , and the value of is .

The power rating of transformer is , the voltage rating is, and the value of is .

The power rating of transformer is , the voltage rating is, and the value of is .

The power rating of transformer is , the voltage rating is, and the value of is .

Write the rating of transmission.

The voltage rating of transmission line is , reactance is , and is .

The voltage rating of transmission line is , reactance is , and is .

The voltage rating of transmission line is , reactance is , and is .

The neutral reactance of the generator 3,.

The base MVA, .

The base voltage for generator, .

The base voltage for generator, .

The pre fault voltage, .

Determine the formulas to draw the zero, positive, and negative sequence diagram of the system, Thevenin’s equivalent for each sequence network, and calculate the sub transient fault current in kA and per unit.

The equation to calculate the reactance on the new base value can be mathematically presented as shown below.

Calculate the new value of the reactance and draw the sequence diagram.

(a)

Consider the single line diagram of the system.

As the base old MVA and KV values of generator 1 are same as the new value, the reactance of generator 1 remains the same.

As the base old MVA and KV values of generator 2 are same as the new value, the reactance of generator 1 remains the same.

Calculate the new value of the reactance of generator 3.

For positive and negative sequence reactance, substitute for , for , for , for , and for into equation (1).

For zero sequence reactance, Substitute for , for , for , for , and for into equation (1).

Calculate the reactance of neutral of generator 3, . Substitute for into the equation.

Calculate the per unit reactance for generator 4.

For zero sequence, substitute for , for , for , for , and for into equation (1).

For positive sequence, substitute for , for , for , for , and for into equation (1).

For negative sequence, substitute for , for , for , for , and for into equation (1).

Calculate the per unit reactance for transformers.

As the base old MVA and KV values of transformer 1 are the same as the new value, the reactance of generator[l1] 1 remains the same.

As the base old MVA and KV values of transformer 2 are the same as the new value, the reactance of generator 2 remains the same.

For transformer 3, substitute for , for , for , for , and for into equation (1).

For transformer 4, substitute for , for , for , for , and for into equation (1).

Calculate the base impedance.

Substitute for and for into equation (2).

Calculate the positive sequence reactance for line 1-2.

Substitute for and for into equation (3).

Calculate the positive sequence reactance for line 1-3 and 2-3.

Substitute for and for into equation (3).

Calculate the zero-sequence reactance for line 1-2.

Substitute for and for into equation (3).

Calculate the zero-sequence reactance for lines 1-3 and 2-3.

Substitute for and for into equation (3).

Draw the per unit zero sequence diagram as shown below.

Draw the per unit negative sequence diagram as shown below.

Draw the per unit positive sequence diagram as shown below.

[l1]Which generator?

Determine the Thevenin’s equivalent for each sequence network.

b)

Assume bus 1 is a faulty bus and draw the zero-sequence network as as shown below.

Reduce the above figure for the zero-sequence impedance from bus 1.

Calculate the equivalent zero-sequence impedance.

The zero-sequence Thevenin’s circuit is shown below.

Draw the negative sequence network as shown below.

Reduce the above figure for the negative-sequence impedance from bus 1.

Calculate the equivalent negative sequence impedance.

The negative sequence Thevenin’s equivalent circuit is shown below.

Similarly, the positive sequence impedance is , and the positive sequence Thevenin’s equivalent circuit is shown below.

Calculate the sub transient fault current in per unit and kA.

c)

Calculate current in the positive sequence circuit.

Substitute for and for into equation (4).

Calculate the base current.

Substitute for , and for into equation (5).

Since the fault is symmetrical, zero and negative sequence currents are zero.

Calculate the sub transient fault current in per unit.

Substitute for and for into equation (6).

Calculate the sub transient current in kA.

Substitutefor and for into equation (7).

Hence, the sub transient fault current in per unit and kA are and respectively.