EE 1352 Monograph UnitV
EE 1352 Monograph UnitV
EE 1352 Monograph UnitV
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VI semester EEE Branch (2006-2010 Batch)
Unit – V – Power System Stability
Syllabus :
Basic concepts and definitions – Rotor angle stability – Voltage stability –
Mid Term and Long Term stability – Classification of stability – An elementary view
of transient stability – Equal area criterion – Factors influencing transient stability
– Numerical integration methods – Euler method – Modified Euler method –
Runge-Kutta methods.
5.1 Introduction
The transient behavior of a power system resulting from major disturbances such
as a fault followed by switching operations, sudden rejection of load or generation etc.
A major disturbance upsets the balance between mechanical input and electrical output of
the generator with the result that some generators may accelerate while others may
decelerate. The rotor angles will undergo wide variations and in this process the
“synchronism” of the system gets affected.
A necessary condition for satisfactory power system operation is that all synchronous
machine remain in synchronism or in step. This aspect of stability is influenced by the
dynamics of generator rotor angles and power angle relationships.
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It is the ability of inter connected synchronous machines of a power system to remain
in synchronism. The stability problem involves the study of the electromechanical
oscillations inherent in power systems. A fundamental factor in this problem is the
manner in which the power outputs of synchronous machines vary as their motor
oscillate.
It is usual to characterize the rotor angle stability phenomena in terms of the following
two categories.
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The mechanical torque Tm and electrical torque Te are considered positive for
synchronous generator. These means that Tm is the resultant shaft torque which tends to
accelerate the rotor in the positive θm direction of rotation as shown in fig. 9.1 (a).
Under steady state operation the generator Tm and Te are equal and the accelerating
torque Ta is zero. In this case there is no acceleration or deceleration of the rotor masses
which include the rotor of the generator and prime mover are said to be in synchronism
with the other machines operating at synchronous speed in the power system.
The electrical torque Te corresponds to the net air gap power in the machine and
thus accounts for the total output power of the generator plus I2R losses in the armature
winding. In the synchronous motor the direction of power flow is opposite to that in the
generator. Accordingly for a motor both Tm and Te are reversed in sign as shown in fig.
9.1 (b). Te is then corresponds to the air gap power supplied by the electrical system to
drive the rotor while Tm → the counter torque of the load and the rotational losses
tending to retard the rotor. In our discussion Tm is considered to be constant since the
prime mover is controlled by governors.
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dθ m
From equation (3) we can say the rotor angular velocity is constant and
dt
dδ m dδ m
equals the synchronous speed only when is zero. Therefore represents the
dt dt
deviation of the rotor speed from synchronism and the units of measure are mechanical
radians per second.
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1 1
Jω sm
2
Mω sm
H= 2 = 2 MJ / MVA
S mach S mach
where Smach three phase rating of the machine in MVA
2H
M = S mach MJs/mach.rad
ω sm
Substitute this in equation (7) we get
2 H d 2δ m P P − Pe
= a = m . . . . (8)
ω sm dt 2
S mach S mach
δm → expressed in mech. rad.
ωsm → expressed in mech. rad./sec.
We can write the above equation as
2 H d 2δ
= Pa = Pm − Pe p.u. . . . . . (9)
ω s dt 2
δ and ωs are in mech (or) electrical degrees or radians.
Thus for a system with a frequency of f herts,
2 H d 2δ
= Pa = Pm − Pe
2πf dt 2
H d 2δ
= Pa = Pm − Pe perunit.......(10)
πf dt 2
when δ is in electrical degrees.
Equation (10) is called the swing equation o the machine is the fundamental
equation which governs the rotational dynamics of the synchronous machine in stability
studies.
When swing equation is solved we obtain the expression for δ as a function of
time. A graph of the solution is called the swing curve of the machine and inspection of
the swing curves of all the machines of the system will show whether the machines
remain in synchronism after a disturbance.
Multimachine system
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S mach
where H system − H mach = Machine inertia constant in system base.
S system
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Solved problems :
Example : A 50 Hz, four pole turbogenerator rated 100 MVA, 11 kV has an inertia
constant of 8.0 KJ/MVA. a) Find the stored energy in the rotor at synchronous speed.
b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW,
find the rotor acceleration elec.deg./sec2 neglecting mechanical and electrical losses.
Solution :
Stored energy in MJ
H = ————————————
Machine rating in MVA
b) Pa = Pm – Pe = 80 – 50 = 30 MW
Md 2δ
Pa =
dt 2
H 8
M = S mach = x100 = 0.089MJ − s / elec. deg ree
180 f 180 x50
d 2δ d 2δ Pa 30
Pa = M 2 ⇒ 2 = = = 337.08elec. / deg ree
dt dt M 0.089
d 2δ
Acceleration = 337.08elec. deg ree / sec 2
dt 2
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Typical values of H
Type of machine Inertia constant H
in MJ/MVA
Water wheel generator
Slow speed < 200 rpm 2.3
High speed > 200 rpm 2.3
Synchronous capacitor
Large 1.25
Small 1.00
Synchronous motor with load varying from
2.00
1.0 to 5.0 and higher for flywheels
Turbine generator
Condensing 1800 rpm 9.6
3000 rpm 7.4
Non condensing 3000 rpm 4.1
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I a 1 1 1 I a 0
I = 1 a 2 a I
b a1
I c 1 a a I a 2
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∴ I a = I a 0 + I a1 + I a 2
I b = I a 0 + a 2 I a1 + aI a 2
I c = I a 0 + aI a1 + a 2 I a1
Given that, I a 0 = 350∠90° = 0 + j 350
I a1 = 600∠ − 90° = 0 − j 600
I a2 = 250∠90° = 0 + j 250
∴ aI a1 = 1∠120° x600∠ − 90° = 600∠30° = 519.62 + j 300
a 2 I a1 = 1∠240° x600∠ − 90° = 600∠150° = −519.62 + j 300
aI a 2 = 1∠120° x 250∠90° = 250∠210° = −216.51 − j125
a 2 I a 2 = 1∠240° x 250∠90° = 250∠330° = 216.51 − j125
I a = I a 0 + I a1 + I a 2 = j 350 − j 600 + j 250 = 0
I b = I a 0 + a 2 I a1 + aI a 2 = j 350 − 519.62 + j 300 − 216.51 − j125 = −736.13 + j 525 = 904.16∠145° A
I c = I a 0 + aI a1 + a 2 I a 2 = j 350 + 519.62 + j 300 + 216.51 − j125 = 736.13 + j 525 = 904.16∠35° A
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The total complex power flowing into a three phase circuit through three lines a,b
S = P + jQ = Va I a + Vb I b + Vc I c
* * *
and c is
Where Va, Vb and Vc are voltages to neutral at the terminals and Ia, Ib and Ic are
the currents flowing into the circuit in the three lines. A neutral connection may or may
not present.
In matrix notation,
* T *
I a Va I a
S = [Va Vb Vc ] I b = Vb I
b
I c Vc I c
Where the conjugate of a matrix is composed of elements that are the conjugates
of the corresponding elements of the original matrix.
Now use the symmetrical components of the voltages and currents in the above
equation.
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Va 1 1 1 Va 0 I a 1 1 1 I a0
V = 1 a 2 a Va1 and I b = 1 a 2
a I a1
b
Vc 1 a a 2 Va 2 I c 1 a a 2 I a 2
Va I a
V = AV , and I = AI
b b
Vc I c
∴ S = [ AV ] x[ AI ]
T *
The reverse rule of matrix algebra states the transpose of the product of two
matrices is equal to the product of the transposes of the matrices in reverse order.
According to this rule,
[ AV ] T = V T AT
∴ S = V T AT A * I *
Noting that AT = A and a and a2 are conjugates we get,
*
1 1 1 1 1 1 I a0
S = [Va 0 Va1 Va 2 ] 1 a 2 a 1 a a 2 I a1
1 a a 2 1 a 2 a I a 2
1 0 0
A A = 30 1 0
T *
0 0 1
*
I a0
S = 3[Va 0 Va1 Va 2 ] I a1
I a 2
So complex power is
Va I a* + Vb I b* + Vc I c* = 3V0 I 0* + 3V1 I 1* + 3V2 I 2*
If the symmetrical components of currents and voltages are known, the power
consumed by a three phase circuit can be computed directly from the sequence
components. Thus the above relation is very useful for computing power in unbalanced
power system.
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Sequence Impedance
In any part of circuit, the voltage drop caused by current of a certain sequence
depends on the impedance of the part of the circuit to current of that sequence. The
impedance of any section of a balanced network to current of one sequence may be
different from impedance to current of another sequence.
The sequence impedance of an equipment or component of a power system are
the positive, negative and zero sequence impedances. They are defined as follows :
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The impedance of a circuit element when positive sequence currents alone are
flowing is called positive sequence impedance. Similarly, when only negative sequence
currents are present, the impedance is called negative sequence impedance. When only
zero sequence currents are present the impedance is called zero sequence impedance.
The impedance of any element of a balanced circuit to current of one sequence may be
different from impedance to current of another sequence.
The single phase equivalent circuit of power system (impedance or reactance
diagram) formed using the impedances of any one sequence only is called the sequence
network for the particular sequence. Therefore the impedance or reactance diagram
formed using positive sequence impedance is called positive sequence network. Similarly
the impedance or reactance diagram formed using negative sequence impedance is called
negative sequence network. The impedance or reactance diagram formed using zero
sequence impedance is called zero sequence network.
The sequence impedances and networks are useful in the analysis of
unsymmetrical faults in the power system. In unsymmetrical fault analysis of a power
system, the positive, negative and zero sequence networks of the system are determined
and then they are interconnected to represent the various unbalanced fault conditions.
Each sequence network includes the generated emfs and impedances of like sequences.
Also, the sequence network carries only the currents of like sequence.
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Note : The positive and negative sequence currents are balanced currents and so
they will not pass through neutral reactance.
The reactances in positive sequence network is subtransient, transient or
synchronous depending on whether subtransient, transient or steady state conditions are
being studied. Under no load condition the emf Ea is the induced emf per phase. Under
load or fault condition Ea is replaced by Eg’ for transient state and Ea is replaced by Eg”
for subtransient state.
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The zero sequence network is a single phase network and assumed to carry only
the zero sequence current of one phase. Hence the zero sequence current of one phase,
must have an impedance 3Z0 + Zg0.
With reference to fig. 3, the equations for the phase-a component voltages are,
Va1 = E a − I a1 Z 1
Va 2 = − I a 2 Z 2
Va 0 = − I a 0 Z 0
The zero sequence network of generator when the neutral is solidly grounded (i.e.
directly grounded) and when the neutral is ungrounded are shown in Fig. 4 and Fig. 5
respectively. In these cases there is no change in positive and negative sequence network.
Fig. 4 : Zero sequence network of a generator Fig. 5 : Zero sequence network of a generator
when the neutral is solidly grounded. when the neutral is ungrounded.
Note : The sequence networks of synchronous motor is same as that of generator when
the directions of currents in the sequence network of generator are reversed.
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The positive, negative and zero sequence impedances of transmission lines are
represented as a series impedance in their respective sequence networks as shown in
fig. 6.
When the applied voltage is balanced, the positive and negative sequences of
linear, symmetrical, static devices are identical. Therefore in a transformer the positive
and negative sequence impedances are identical. Even though the zero sequence
impedance may slightly differ from positive and negative sequence impedance, it is
normal practice to assume the zero sequence impedance as equal to positive or negative
sequence impedance. [For all types of transformers the series impedances of all
sequences are assumed equal].
Note : When the neutral of star connection is grounded through reactance Zn
then 3Zn should be added to zero sequence impedance of transformer to get the
total zero sequence impedance.
Let, Z1 = Positive sequence impedance of transformer.
Z2 = Negative sequence impedance of transformer.
Z0 = Zero sequence impedance of transformer.
The positive, negative and zero sequence impedances of transformer are
represented as a series impedance in their respective sequence networks as shown in
fig. 7
The zero sequence network of the transformer depends on the type of connections
(Y or Δ) of the primary and secondary windings and also on the grounding of neutral in Y
connection.
The following general observations can be made for zero sequence currents in
transformers.
1. When magnetizing current is neglected, the primary winding will carry
current only if there is a current flow in the secondary winding.
Therefore the zero sequence current can flow in the primary winding of
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a transformer only if there is a path for zero sequence current in
secondary winding or vice-versa.
2. If the neutral point in the Y connected winding is not grounded then
there is no path for zero sequence current in star connected winding.
3. The zero sequence current flows in the star connected winding and in
the lines connected to the winding only when the neutral point is
grounded.
4. The zero sequence current can circulate in the delta connected winding
but the zero sequence current cannot flow through the lines connected
to the delta connected winding.
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Sequence impedances and networks of loads
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In balanced Y or Δ connected loads the positive, negative and zero sequence
impedances are equal. When the neutral point of star connected load is grounded through
a reactance Zn then 3Zn is added to the zero sequence impedance of load to get the total
zero sequence impedance of load.
Let, ZL1 = Positive sequence impedance of load.
ZL2 = Negative sequence impedance of load.
ZL0 = Zero sequence impedance of load.
The positive, negative and zero sequence impedances of transmission lines are
represented as a shunt impedance in their respective sequence networks as shown in
fig. 8.
The zero sequence network of the 3 phase load depends on the type of connection,
i.e. Y or Δ connection. The zero sequence current will flow in network only if a return
path exists for it. The zero sequence network for various types of loads are shown in
table below :
Table 2 : Zero sequence networks of loads
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Review Questions (2 marks)
1. How do short circuits occur on a power system ?
2. Distinguish between symmetrical and unsymmetrical short circuits.
3. What are the applications of short-circuit analysis?
4. Define short circuit capacity of a power system.
5. What are the different types of fault which occur in a power system?
6. Define symmetrical fault.
7. Name any two methods of reducing short circuit current.
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9. In the system configuration of figure, the system impedance data is given below :
Transient reactance of each generator = 0.15 p.u.
Leakage reactance of each transformer = 0.05 p.u.,
Z12 = j0.1, Z13 = j0.12, Z23 = j0.08 p.u.
For a solid 3 phase fault on bus 3, find all bus voltages and S C currents in each
component.
Assume prefault voltages to be 1 p.u. and prefault currents to be zero.
10. For a fault (solid) location shown in figure, find the short circuit currents in
line 1-2 and 1-3. Prefault system is on no-load with 1 p.u. voltage and prefault
currents are zero. Use ZBUS method and compute its elements by the current injection
technique.
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