Lab 3 Het 560 Final

SWINBURNE UNIVERSITY OF TECHNOLOGY

FACULTY OF ENGINEERING AND INDUSTRIAL SCIENCE

HET 560 POWER SYSTEM OPERATIONS AND CONTROL

LAB 3 – CIRCUIT BREAKER SELECTION

(SEMESTER 1, 2012)

PREPARED BY:

TRISTA YUDHITIA BINTORO (4204751)

POWER SYSTEM OPERATION AND CONTROL (SEM1,2012)

LAB 3 – CIRCUIT BREAKER SELECTION 1 | P a g e

I. OBJECTIVES

1) To compute short circuit currents when a fault occurs in power systems

2) To compute the normal load current to calculate the rated continuous current.

3) To select suitable circuit breakers

II. INTRODUCTION

The American National Standards Institute (ANSI) defines a circuit breaker as: “A

mechanical switching device, capable of making, carrying and breaking currents

under normal circuit conditions. Also capable of making and carrying for a specified

time and breaking currents under specified abnormal circuit conditions, such as those

of a short circuit.”

The standard circuit breaker are based on symmetrical interrupting current and

usually is calculated by considering the symmetrical fault current at the system

location and the symmetrical interrupting current is chosen equal or above the

calculated current.

If X/R ratio at the system located is less than 15, the breakers are capable of

interrupting the dc offset in addition to the sub-transient fault current. If X/R is

greater than 15, the dc offset may not decay to a sufficiently low value. And it X/R is

unknown, the calculated fault current should not be greater than 80% of the breaker

interrupting capability.

To calculate the circuit breaker rating in generator, we need to consider the sub-

transient fault current and hence the sub-transient machine reactance is used

depending on circuit breaker.

In this experiment, the circuit breaker will be selected based on following ratings:

Rated maximum voltage, rated continuous current and rated short circuit

current.

III. EQUIPMENTS

a) PowerWorld Simulator



IV. PRELIMINARIES

Figure 1. single line diagram

Length 2 = 50km

The positive sequence reactance data per unit value:

Zbase =(V baseL-L)2/S base 3p

= (230)2 / 1x105

= 529 ohm

Ybase = 1/ Zbase

= 0.00189 S

Z1= 0.008+j0.5Ω/km;

Z0 = 0.2+j1.5/km;

Y1 = j3.3x10-6 S/km



Impedance Per unit

Z1 = Z1*L1 = 1.2+j7.5 Ω Z1,Zbase = 0.0023+j0.0142 p.u

Z2 = Z1*L2= 4+j25 Ω Z2/Zbase = 0.00756+j0.047 p.u

Z3 = Z1*L3= 3.2+j20 Ω Z3/Zbase = 0.006+j0.0378 p.u

Z4= Z1*L4= 1.2+j7.4 Ω Z4/Zbase = 0.0023+j0.0142 p.u

Z5 = Z1*L5= 4+j25 Ω Z5/Zbase = 0.00756+j0.047 p.u

Admittance Per unit

Y1 = Y1*L1 = j0.0000495 Y1/Ybase = j0.0262 p.u

Y2 = Y1*L2 = j0.000165 Y2/Ybase = j0.087 p.u

Y3 = Y1*L3 = j0.000132 Y3/Ybase = j0.0698 p.u

Y4 = Y1*L4 = j0.0000495 Y4/Ybase = j0.0262 p.u

Y5 = Y1*L5 = j0.000165 Y5/Ybase = j0.087 p.u

XT1 = 0.1

XT2 = 0.05

XG1-1 = 0.12 pu, XG1-2 = 0.14 pu, XG1-0 = 0.05 pu, P = 80 MW, Q=60 MVAR

XG2-1 = 0.06 pu, XG2-1 = 0.07 pu, XG2-0 = 0.025 pu, P = 180 MW, Q=80

MVAR



XTH,3 = j(0.12+0.1+0.0142)//j(0.047+0.02469+0.05+0.06)

= j0.1023

= 0.1023<90

assuming Vf = 1

𝐼𝐹 =𝑉𝑓

𝑋𝑡ℎ= 9.77 < −90° pu,.

Ibase = 𝑆𝑏𝑎𝑠𝑒3 𝑝ℎ𝑎𝑠𝑒

√3 𝑉𝑏𝑎𝑠𝑒 𝐿𝐿 Ibase =

100 ×1000

√3 ×230 = 251.022A

Ifault at 3_1 = 2453.78𝐴

Circuit breaker selection for bus 3

Given that we have now calculated out fault current, we need to examine what rating to

determine the value of circuit breaker. The maximum operating voltage, the X/R ratio, the

rated continuous current and soon.

-The first consideration is we know that the operating voltage maximum is 230kV

-The standard circuit breaker are based on symmetrical interrupting current and usually

is calculated by considering the symmetrical fault current at the system location and the

symmetrical interrupting current is chosen equal or above the calculated current. Then

the symmetrical interrupting current equal to 2.45kA or above

-As Normal load current = 00 ×1000

√3 ×230 = 1004.9 A,

Hence, 30 % of Normal load current + Normal load current = (1004.9 × 30÷ 100) +

1004.9 = 1.305kA

Examining the lines at bus 3, the rating is 400 MVA, therefore the CB needs to be able to

carry a continuous current of 1.305kA

The breaker was compared and chosen from Table 7.10 in the book by Glover & Sarma

and therefore, a 230 kV CB is selected with a continuous rating of 1600A and a breaking

rating of 31.5kA



V. PROCEDURES

1. PowerWorld Simulator was run and figure 1 was drawn.

2. Positive sequence reactance data per unit was stored and simulated.

3. Fault Analysis was run to find the fault current at bus 3 as stated in preliminary.

The result was compared.

4. The maximum operating voltage was determined based on the data in each

location.

5. A normal load current was computed following the step: Edit Mode->solve Right

click the the desired transmission line or transformer -> “Add New Field Around

Line/Xfmr” -> choose any Pos (Pos 1, Pos 2, ..Pos 4 )-> on the Type of field, choose

Amp. Flow -> OK.

The normal load current was then calculated into the rated continuous current for

at least 30% larger.

6. The fault current on each location was computed by following these steps:

6.1 Run Mode->Simulate->solve and animate->stop simulation/animation. (To

see the power flow)

6.2 Tool-> Fault Analysis-> choose which line that is needed to computed -

>Fault Type-> Single Phase to ground->Data Type Shown->Amp. Repeat

again by using fault type->3 phase balanced.

6.3 The highest values of fault currents bus to bus (i.e bus 3 and bus 4) were

computed and taken as consideration for maximum short circuit current at

any type of fault.

7. The data from PowerWorld Simulator was then compared with table 7.10 from

textbook “Power System and Analysis” by Glover & Sarma.

8. Circuit breaker was chosen based on these 3 following citeria:

Rated voltage was larger than maximum operating voltage

Rated continuous current was at least 30% larger than the normal load

current.

Rated short circuit was larger than the maximum fault current for any type

of fault types.

X/R value was assumed to be less than 15

The cost was taken into account when choosing the circuit breaker



VI. RESULT AND ANALYSIS

PowerWorld Simulator was run and the transmission line, transformer, and machine was

created as shown in figure 2. The data were computed based on the preliminary result.

Figure 2. PowerWorld Simulator single line diagram

The system was tested for a bolted three-phase to ground short circuit at bus 3 to confirm

the preliminary, and results were obtained as shown below in figure 3. The fault current is

1615.03A or 6.34pu.

Figure 3. PowerWorld Simulator single line diagram- 3 phase to ground fault at bus 3



CIRCUIT BREAKER SELECTION

The circuit breaker must have the rated continuous current is 30% larger than the normal

load current. To find the normal load current, PowerWorld Simulator was run by

following the IV.PROCEDURE : step 5 and given the result in figure 4. The normal load

current in each bus to bus is shown in table 1.

Figure 4. normal load current in single line diagram

Bus to Bus Normal Load Current

(Amp)

Rated continuous current

(30% of normal load

current)(Amp)

1-2 5371.06 6982.378

2-3 179.53 233.389

3-4 122.71 159.523

4-5 90.99 118.287

4-6 239.34 311.142

5-6 140.92 183.196

6-7 531.9 691.47

Table 1. Normal load current and the rated continuous current for each bus

The fault current was observed by using fault analysis menu from PowerWorld Simulator

(refer to IV.Procedure step 6). The circuit breaker must have a rated short circuit current

larger than maximum fault current for any type of fault at the bust where the breaker is



located. Short circuit current for all fault types (single-line –to ground, line to line, 3 phase

balanced, double line to ground) in figure 5-figure 8.

Figure 5. Single line to ground fault current at bus 1

Figure 6. Line to line fault current at bus 1

Figure 7. 3 phase balanced fault current at bus 1

Figure 8. double line to ground fault current at bus 1



The simulation was performed and the result obtained in each bus for each case is shown

in table 2.

Table 2. Short circuit current fault

FAULT BUS

Fault types and fault current Highest

fault

current

(kA)

Single line-

to-ground

Fault

(Amps)

Line-to-

line Fault

(Amps)

Balanced

three-phase

Fault

(Amps)

Double line-

to-ground

Fault

(Amps)

BUS 1 FAULT CURRENT 1550.850 16786.80 45198.8 1296.760 45.1988

Fault Current-Lines (phase A)

Bus 2-1 4361.19 12775.56 6692.23 12962.1



Bus 1-2 5537.36 10934.82 21520.71 10966.64

Bus 3-2 120.6 456.70 511.85 464.37



Bus 2-3 200.4 445.99 1206.41 452.99

Bus 4-3 91.31 296.07 529.75 316.80



Bus 3-4 134.29 275.55 970.92 294.92

Bus 5-4 96.17 125.50 277.37 144.71

Bus 6-4 152.89 156.37 307.11 173.38



Bus 4-5 91.13 123.38 591.72 141.75

Bus 6-5 90.56 251.65 753.18 356.94



Bus 5-6 249.26 251.47 414.17 256.46

Bus 7-6 608.92 562.5 608.92 573.85



Bus 6-7 588.97 557.27 591.15 539.21



1. Circuit breaker selection for bus 1- bus 2

Based on single diagram on figure 2, the power flow is run from bus 1 to bus 2. There is

one circuit breaker is placed on higher voltage side, which is 230kV. Hence, it can be

concluded that the circuit breaker have the maximum operating system of 230kV. By

refering to table 1 and table 2, we obtained the rated continuous current and the

maximum fault current occurred on any type of faults value.

Comparing the table 7.10 of textbook by Glover & Sarma, the suitable circuit

breakers located at bus 1 and bus 2 are the one with the specifications table 3

below:

Nominal

Voltage

Class

(Kv,rms)

Rated

Max

Voltage

(kV,

rms)

Rated

Voltage

Range

Factor

(K)

Rated Values

Rated

Continuous

Current at

60Hz

(Amp,rms)

Rated

short

circuit

current

(kA,rms)

Rated

Interrupting

time(cycle)

Rated

permissible

tripping

delay (sec)

230 242 1.0 3000 63 3 1

Related Required Capabilities

Max Symmetrical Interrupting

Capability

3 sec short-time

current carrying

capability

Closing and latching

capability 1.6K times

rated short circuit current

(kA)

63 63 101

Table 3 – circuit breaker rating



2. Circuit breaker selection for bus 2- bus 3 (line 2)

Based on single diagram on figure 2, the power flow is run from bus 2 to bus 3

the maximum operating system in line 1 is 230kV

From table 1, the rated continuous current is 233.389 A

From table 3, the maximum fault current occurred at any type of fault is: 21.52kA

Hence, the suitable circuit breaker have the following ratings (table 4) and placed on each

side of bus 2 and bus 3 lines.

Nominal

Voltage

Class

(Kv,rms)

Rated

Max

Voltage

(kV,

rms)

Rated

Voltage

Range

Factor

(K)

Rated Values

Rated

Continuous

Current at

60Hz

(Amp,rms)

Rated

short

circuit

current

(kA,rms)

Rated

Interrupting

time(cycle)

Rated

permissible

tripping

delay (sec)

230 242 1.0 1600 31.5 3 1

Related Required Capabilities

Max Symmetrical Interrupting

Capability

3 sec short-time

current carrying

capability

Closing and latching

capability 1.6K times

rated short circuit current

(kA)

31.5 31.5 50

Table 4– circuit breaker rating






Hence, the suitable circuit breakers have the same ratings with circuit breaker on bus2-

bus3 which is 230kV class circuit breaker. (more specification rating, refer to table 4).

The circuit breaker is placed on each side of bus 3 and bus 4 lines.



4. Circuit breaker selection for bus 4- bus 5 (Line 3)

Based on single diagram on figure 2, the power flow is run from bus 5 to bus 4.




Hence, the suitable circuit breakers have the same ratings with circuit breaker on

bus2-bus3 which is 230kV class circuit breaker with 1600 rated continuous current.

(for more specification rating, refer to table 4). The circuit breaker is placed on each

side of bus 4 and bus 5 lines.







bus3 which is 230kV class circuit breaker. (for more specification rating, refer to table 4).








bus3 which is 230kV class circuit breaker. (for more specification rating, refer to table 4).




7. Circuit breaker selection for bus 6- bus 7 (Transformer 2)

Based on single diagram on figure 2, the power flow is run from bus 7 to bus 6 and the

circuit breaker is placed on higher voltage side which is 230kV. Hence,





bus3 which is 230kV class circuit breaker. (For more specification rating, refer to table 4).

The circuit breaker is installed on the higher voltage side.

VII. DISCUSSION/CONCLUSION

From simulation results and preliminary calculations, analysis was made. The circuit

breakers located on the transmission lines and transformer 2 have the same rating

because of the same base and the maximum fault current is less than 31.5kA while the

rated continuous current is less than 1600. The cost has been considered efficiently.

However, for transformer 1, the circuit breaker has the bigger value due to the maximum

fault current occurred. We have to use the circuit breaker that interrupts 63kA. However,

in the real application, a lower rated (i.e 63kV is changed to 31.5Kv) one could be used if

available to save more costs.

All circuit breakers that were selected has the rated maximum voltage is greater than

operating voltage, the rated continues current is 30% larger than normal load current

and a rated short-circuit current larger than the maximum fault current for any type of

fault at the bus where the breaker is located.

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Lab 3 Het 560 Final