Consequences of Switch Shunt on
an Unsymmetrical Fault in Power System
Chanuan Uakarn1
and Chart Rithirun2
Electrical Engineering Dept, Faculty of Engineering,
Kasem Bandit University, Thailand
1
cuakarn@gmail.com
2
chart20052548@yahoo.com
Abstract - This paper presents to calculate
the magnitude of current, voltage at the bus
location of any on the unsymmetrical faults
in three-phase power systems. For power
system analysis in per-unit, they are on
without switched shunt and with switched
shunt into the bus and make solved powerflow fault calculation program Power
World Simulator PWS-16. The results show
the active, reactive and appearance power
factor, phase angle in a real time
comparison with the solved power flow in
accordance with standard IEC-909 without
switched shunt in the consequence of switch
shunt on an unsymmetrical fault in power
system.
location. As such the computation of fault
current is greatly simplified by the use of
sequence networks besides the above controls,
the power flow program can be used to
investigate the consequence of switching in or
out lines, transformers, loads, and generators.
Proposed system changes to meet future load
growth, including new transmission, new
transformers, and new generation can also be
investigated. Power-flow design studies are
normally conducted by trial and error. Using
engineering
judgment,
adjustments
in
generation levels and controls are made until
the desired equipment loadings and voltage
profile are obtained.
II. THEORY
Keywords - Unsymmetrical Faults, Solved
Power Flow, IEC-909, Switched Shunt
I. INTRODUCTION
Short circuits occur in three-phase power
systems are unsymmetrical faults, which may
consist of unsymmetrical short circuits, the
unsymmetrical faults occur as single line-toground faults, line-to-line faults, double lineto-ground faults, and balanced three-phase
faults [1]. The path of the fault current may
have either zero impedance, which is called a
bolted short circuit, or nonzero impedance.
Other types of faults include one-conductor
open and two-conductors open, which can
occur when conductors break or when one or
two phases of a circuit breaker inadvertently
open. When an unsymmetrical fault occurs in
an otherwise balanced system, the sequence
networks are interconnected only the fault
Control of power flow: The following
means are used to control system power flows:
1. Prime mover and excitation control of
generators
2. Control of tap-changing and regulating
transformers
3. Switching of shunt capacitor bank, shunt
reactors, and static var systems
A simple model of generator operating
under balance steady-state conditions is the
Thevenin equivalent shown in fig. 1. Vt is the
generator terminal voltage, Eg is the excitation
voltage, δ is the power angle, and Xg is the
positive sequence synchronous reactance
[2, 3].
International Journal of the Computer, the Internet and Management Vol.23 No.2 (May-August, 2015) pp. 54-59
54
Chanuan Uakarn and Chart Rithirun
+
Eg = E g
δ
However, when δ is less than 15o, the increase
in P is much larger than the decrease in Q.
From the power-flow standpoint, an increase in
prime-mover power corresponds to an increase
in P at the constant-voltage bus to which the
generator is connected. The power-flow
program computes the increase in δ along with
the small change in Q.
Q
P
+
jXg
-
Vt = Vt
0
0
-
Fig 1. Generator the Venin Equivalent
Equation (5) shows that reactive power
output, Q, increases when the excitation
voltage, Eg, increases. From the operational
standpoint, when the generator exciter output
increases while holding the prime-mover
(1)
power constant, the rotor current increases. As
the rotor current increases, the excitation
And the complex power delivered by the
voltage, Eg, also increases, causing an increase
generator is
in generator reactive power output Q. There is
also a small decrease in δ required to hold P
constant in (4). From the power-flow
standpoint, an increase in generator excitation
corresponds to an increase in voltage
(2) magnitude at the constant-voltage bus to which
the generator is connected. The power-flow
program computes the increase power Q
suppied by the generator along with the small
change in δ.
(3)
The consequence of adding a shunt
The real and reactive power delivered are capacitor in the fig. 2 bank to a power system
then
bus. The system is modeled by its Thevenin
equivalent. Before the capacitor bank is
connected, the switch, SW, is opened and the
(4) bus voltage equals to Eth. After the bank is
connected, the switch, SW, is closed, and the
capacitor current, Ic, leads the bus voltage, Vt,
by 90o.The phasor diagram shows that Vt is
(5) larger than Eth when SW is closed. From the
power-flow standpoint, the addition of a shunt
Equation (4) shows that the real power P capacitor bank to a load bus corresponds to the
increases when the power angle δ increases. addition of a negative reactive load, since a
From an operational standpoint, when the capacitor absorbs negative reactor power. The
prime mover increases the power input to the power-flow program computes the increase in
generator while the excitation voltage is held bus voltage magnitude along with the small
constant, the rotor speed increases. As the change in δ. Similarly, the addition of a shunt
rotor speed increases, the power angle δ also reactor corresponds to the addition of a
increases, causing an increase in generator real positive load, wherein the power-flow program
power output P. There is also a decrease in computes the decrease in voltage magnitude
reactive power output Q, Given by (5). [4, 5, 6].
The generator current is
International Journal of the Computer, the Internet and Management Vol.23 No.2 (May-August, 2015) pp. 54-59
55
Consequences of Switch Shunt on an Unsymmetrical Fault in Power System
Transformer impedance and admittance:
SW
+
jXth
Eth
_
Rth
+
Vt C
_
IC
IC
Eth
jXthIC
RthIC
Vt
a) Equivalent Circuit.
b) Phasor Diagram with SW Closed.
Base power = 100MV; base voltage at bus
1, 3 = 15 kV; base voltage at bus 2, 4, 5
= 345 kV Fault impedance :
Fig 2. Consequence of Adding a Shunt Capacitor Bank
to a Power System Bus.
III. EXPERIMENT
A capacitor’s Mvar rating is based on an
assumed voltage of 1.0 per unit.
V =1.05+j0.0 pu.
3
3
4
T2
80 km. 345 kV
1
Line 3
322 km.
400 kVA
15/345kV
800 MVA
15 kV
345 kV
G
Line 1
345 kV
400 MVA
15 kV
Line 2
G
5
T1
160 km.
Open Power World Simulator case study.
This case is identical to fig. 4, 5, and 6 except
that a 200-Mvar shunt capacitor bank has been
added at bus 2. Initially this capacitor is
opened. Click on the capacitor’s circuit to
close the capacitor and then select Single
Solution to solve the case. The capacitor
increases the bus 2 voltage from 0.834 per unit
to a more acceptable 0.959 per unit. Notice
that the amount of reactive power actually
supplied by the capacitor is only 184 Mvar.
This discrepancy arises because capacitor’s
reactive varies with the square of the terminal
voltage,
V =1.0+j0.0 pu.
1
1
2
800 kVA
345/15 kV
Bus 2
2
200 MVAR
80 MW+j40 MVAR
800 MW+j280 MVAR
Fig 3. This Five-Bus Power System is Modeled
in Fault Calculation.
Power:
The figure shows a single-line diagram of a
five-bus power system. Input data is given in
following:
Bus voltage:
and
Line impedance and line admittance:
As shown in Fig. 3 at bus 1, to which a
generator is connected, it is the swing bus. Bus
1, to which a generator and a load are
connected, is a voltage-controlled bus. Buses
2, 4, and 5 are load buses [1].
The elements of [Ybus] are computed from
Fig. 3. Since buses 1 and 3 are not directly
connected to bus 2,
Mutual admittance:
,
International Journal of the Computer, the Internet and Management Vol.23 No.2 (May-August, 2015) pp. 54-59
56
Chanuan Uakarn and Chart Rithirun
,
One
-37 Mvar
,
Five
Four
One
Five
A
Four
MVA
Three
A
MVA
160 Mvar
1.005 pu
-4.432 Deg
A
A
MVA
MVA
1.036 pu
-2.685 Deg
MVA
1.050 pu
-0.415 Deg
1.005 pu
-4.432 Deg
A
A
MVA
MVA
1.036 pu
-2.685 Deg
80 MW
40 Mvar
1.050 pu
-0.415 Deg
0.959 pu
-19.779 Deg
Two
0.0 Mvar
Fig 4. This Five-Bus Power System is Modeled
in Power World Simulator Case without Switch Shunt.
One
A
Five
Four
MVA
A
Three
slack
1.000 pu
0.000 Deg
A
A
MVA
MVA
1.000 pu
0.000 Deg
Fig 6. This Five-Bus Power System is Modeled
in Load Condition in Power World Simulator Case
with Switch Shunt.
TABLE I
POSITIVE AND NEGATIVE SEQUENCE
160 Mvar
MVA
1.000 pu
0.000 Deg
184.0 Mvar
520 MW
MVA
A
-36 Mvar
800 MW
280 Mvar
This five-bus power system is modeled in
Power World Simulator case fig. 4 and 5. For
larger networks, most of the elements of the
elements of the [Ybus] can be saved in a Matlab
Format file by first right clicking within the
Ybus matrix to display the local menu, and
than selecting “Save [Ybus] in Matlab Format”
from the menu. Finally, notice that no flows is
showing on the one-line, because the nonlinear
power-flow equations have not yet been
solved. We cover the solution of these
equations next in Table I, II.
800 MW
280 Mvar
385 MW
Two
Where half of the shut admittance to bus is
included in Y22 (the other half is located at the
other ends of these lines).
160 Mvar
slack
1.000 pu
0.000 Deg
80 MW
40 Mvar
520 MW
A
-36 Mvar
520 MW
slack
0.959 pu
-19.779 Deg
385 MW
Three
MVA
,
,
Determine the consequence of adding a 200
Mvar shunt capacitor bank at bus 2 on the
power system in the fig. 4, 5, and 6.
A
MVA
A
1.000 pu
0.000 Deg
Self admittance:
A
MVA
385 MW
1.000 pu
0.000 Deg
80 MW
40 Mvar
1.050 pu
0.000 Deg
Two
800 MW
280 Mvar
200.0 Mvar
Fig 5. This Five-Bus Power System is Modeled
in Initial Condition in Power World Simulator Case
with Switch Shunt.
International Journal of the Computer, the Internet and Management Vol.23 No.2 (May-August, 2015) pp. 54-59
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Consequences of Switch Shunt on an Unsymmetrical Fault in Power System
TABLE II
ZERO SEQUENCE
Fig 9. IEC-909, Fault Calculation (+Zero Sequence)
without Switched Shunt
Fig 7. Solved Power-Flow Fault Calculation
without Switched Shunt.
Fig 10. IEC-909, Fault Calculation (+Zero Sequence)
with Switched Shunt
IV. CONCLUSIONS
The results are from the test bus standard 5
bus Power World Simulator case to calculate
the current fault Asymmetry. Compared to
having switched shunt and not having
switched shunt when the fault in a
differentiation in power system of Fig. 7, 8, 9,
and 10, respectively. According to the
requirements of IEC-909 program, PWS 16 is
used to find short-circuit current value. Open
circuit and current asymmetry have the correct
value and fast. The analysis of power systems
determine the CT, PT, precisely to prevent
power system, Electrical system will be stable.
Fig 8. Solved Power-Flow Fault Calculation
with Switched Shunt.
International Journal of the Computer, the Internet and Management Vol.23 No.2 (May-August, 2015) pp. 54-59
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Chanuan Uakarn and Chart Rithirun
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(Arranged in the order of citation in the
same fashion as the case of Footnotes.)
[1]
[2]
[3]
[4]
[5]
[6]
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Eriksson, L., Saha, M.M., and
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