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Faults at bus n in Problem 9.8 are of interest (the instructor selects , or ). Determine the Thevenin equivalent of each sequence network as viewed from the fault bus. Prefault voltage is 1.0 per unit. Prefault load currents and D–Y phase shifts are neglected.

Short Answer

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Answer

zero sequence Thevenin equivalent circuit is shown below.

The positive sequence Thevenin equivalent circuit is shown below.

The negative sequence Thevenin equivalent circuit is shown below.

Step by step solution

01

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 is ,is and the value of is .

Write the rating of transformers.

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 reactance of each line, is and is .

The base MVA,

The base voltage for generator side,

The prefault fault voltage,

02

Determine the formulas to calculate the reactance of the zero, positive and negative sequence network.

The equation to calculate the reactance on the new base value is given as follows.

…… (1)

Here is the per unit value on new base values, is the per unit value on old base values, it the old voltage rating, is the new voltage rating, is the new power rating and is the old power rating.

The equation to calculate the base impedance is given as follows.

…… (2)

Here, is the base value, is the power rating, is the voltage rating.

The equation to calculate the new per unit impedance of the transmission line is given as follows.

…… (3)

03

Calculate the reactance of the zero, positive and negative sequence network.

Calculate the new value of the reactance of generator 1.

For positive and negative sequence reactance.

Substitute for , for , for , for , for into equation (1).

For zero-sequence reactance.

Substitute for , for , for , for , for into equation (1).

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

Calculate the new value of the reactance of transformer 1.

Substitute for , for , for , for , for into equation (1).

Calculate the new value of the reactance of transformer 2.

Substitute for , for , for , for , for into equation (1).

Calculate the base impedance.

Substitute for and for into equation (2).

Calculate the positive and negative sequence reactance for line.

Substitute for and for into equation (3)

Calculate the zero-sequence reactance for line.

Substitute for and for into equation (3)

Draw the zero-sequence network.

Draw the positive sequence network.

Draw the negative network.

Calculate the zero-sequence equivalent impedance.

The zero sequence Thevenin equivalent circuit is shown below.

Calculate the positive-sequence equivalent impedance.

The positive sequence Thevenin equivalent circuit is shown below.

Calculate the negative-sequence equivalent impedance.

The negative sequence Thevenin equivalent circuit is shown below.

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Most popular questions from this chapter

Question: Consider the system shown in Figure 9.18. (a) As viewed from the fault at F, determine the Thevenin equivalent of each sequence network. Neglect phase shifts. (b) Compute the fault currents for a balanced three phase fault at fault point F through three fault impedances ZFA=ZFB=ZFC=j0.5per unit. Equipment data in per-unit on the same base are given as follows:

Synchronous generators:

G1X1=0.2X2=0.12X0=0.06G2X1=0.33X2=0.22X0=0.066

Transformers:

T1X1=X2=X0=0.2T2X1=X2=X0=0.225T3X1=X2=X0=0.27T4X1=X2=X0=0.16

Transmission lines:

L1X1=X2=0.14X0=0.3L1X1=X2=0.35X0=0.6

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.


In order of frequency of occurrence of short-circuit faults in three-phase power systems, list those: ________, ________, ________, ________.

Repeat Problem 9.24 for a bolted line-to-line fault.

The results in Table 9.5 show that during a phase a single line-to-ground fault the phase angle on phase a voltages is always zero. Explain why we would expect this result.

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