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Chapter 9: Q9-5P (page 578)

Question: Equipment ratings for the four-bus power system shown in Figure 7.14 are given as follows:

Generator

Generator

Generator

Transformer

Transformer

Transformer

Each line:

The inductor connected to generator neutral has a reactance of . Draw the zero, positive, and negative-sequence reactance diagrams using a base in the zone of generator. Neglect transformer phase shifts.

Short Answer

Expert verified

Answer

The per unit zero-sequence network is as shown below.

The per unit negative-sequence network is as shown below.

The per unit positive-sequence network is as 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 and are equal to , and the value of is .

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

The neutral has reactance of .

Write the rating of the transformer.

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

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

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

Write the reactance of the line.

The positive sequence impedance of line,.

The zero-sequence impedance of line,.

The base MVA, .

The base voltage, .

02

Determine the formulas to draw the zero, positive and negative sequence diagram of the system.

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

…… (1)

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

The equation to calculate the base impedance is .

…… (2)

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

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

…… (3)

Here, is the reactance on old base values and is the base impedance.

03

Draw the zero, positive and negative sequence diagram of the system.

Consider the line diagram for the system.

Calculate the new value of the reactance for generator 1.

Calculate the zero-sequence reactance.

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

Calculate the positive and negative sequence reactance.

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

Calculate the new value of the reactance for generator 2.

Calculate the zero-sequence reactance.

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

Calculate the positive and negative sequence reactance.

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

The new value of the reactance for generator 3 is given below.

Calculate the new value of the reactance for transformer 1.

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

Calculate the new value of the reactance for transformer 2.

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

Calculate the new value of the reactance for transformer 3.

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 per unit zero-sequence network as shown below.

Draw the per unit negative-sequence network as shown below.

Draw the per unit positive-sequence network as shown below.

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Most popular questions from this chapter

Consider the one line diagram of a simple power system shown in Figure 9.20. System data in per-unit on a 100 MVAbase are given as follows:

Synchronous generators:

G1              100 MVA              20 kV               X1=X2=0.15                  X0=0.05G2              100 MVA             20 kV              X1=X2=0.15                  X0=0.05

Transformers:

T1                     100 MVA            20/220 kV            X1=X2=X0=0.1aT2                     100 MVA            20/220 kV            X1=X2=X0=0.1

Transmission lines:

L12               100 MVA         220 kV      X1=X2=0.125                X0=0.3L13               100 MVA         220 kV      X1=X2=0.15                X0=0.35L23               100 MVA         220 kV      X1=X2=0.25                X0=0.7125

The neutral of each generator is grounded through a current-limiting reactor of æ 0.08333 per unit on a100 MVA base. All transformer neutrals are solidly grounded. The generators are operating no-load at their rated voltages and rated frequency with their EMFs in phase. Determine the fault current for a balanced three-phase fault at bus 3 through a fault impedance ZF=0.1per unit on a100 MVA base. Neglect Δ-Yphase shifts.

Consider a simple circuit configuration shown in Figure 9.24 to calculate the fault currents I1,I2 androle="math" localid="1655895881788" I with the switch closed.

(a) Compute E1 andE2 prior to the fault based on the pre fault voltage V=1∠0°and then, with the switch closed, determineI1,I2 and I.

(b) Start by ignoring pre fault currents, withE1=E2=1∠0° . Then superimpose the load currents, which are the pre fault currents, I1=-I2=1∠0°. Compare the results with those of part (a).

Repeat Problem 9.43 for a bolted single line-to-ground fault at bus 1.

Using the bus impedance matrices determined in Problem 9.47, verify the fault currents for the faults given in Problems 9.4(b), 9.4(c), and 9.19 (a through d).

The system shown in Figure 9.28 is the same as in Problem 9.48 except that the transformers are now Y–Yconnected and solidly grounded on both sides. (a) Determine the bus impedance matrix for each of the three sequence networks. (b) Assume the system to be operating at nominal system voltage without prefault currents when a bolted single-line-toground fault occurs on phase A at bus3. Compute the fault current, the current out of phase C of machine2 during the fault, and the line-toground voltages at the terminals of machine 2during the fault.
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