Download Power System Analysis(EEE-601) (PUT Paper

Power System Analysis(EEE-601)
(PUT Paper solution)
1. Attempt any Four parts of following
(5 marks x 4 = 20)
(a).(Per Unit value of a given quantity is the ratio of the actual value in any given unit to the
base value in the same unit. the main advantage in using the pu system of computations is that
the result that comes out of the sum, product, quotient, etc. of two or more pu values is
expressed in per unit itself.
In an electrical power system, the parameters of interest include the current, voltage, complex
power (VA), impedance and the phase angle. Of these, the phase angle is dimensionless and the
other four quantities can be described by knowing any two of them. Thus clearly, an arbitrary
choice of any two base values will evidently fix the other base values.
Normally the nominal voltage of lines and equipment is known along with the complex power
rating in MVA. Hence, in practice, the base values are chosen for complex power (MVA) and line
voltage (KV). The chosen base MVA is the same for all the parts of the system. However, the base
voltage is chosen with reference to a particular section of the system and the other base voltages
(with reference to the other sections of the systems, these sections caused by the presence of the
transformers) are then related to the chosen one by the turns-ratio of the connecting transformer.
If Ib is the base current in kilo amperes and Vb, the base voltage in kilovolts, then the base MVA is,
Sb = (VbIb). Then the base values of current & impedance are given by
Base current (kA), Ib= MVAb/KVb = Sb/Vb
Base impedance, Zb = (Vb/Ib)=(kVb
Hence the per unit impedance is given by
Zpu = Zohms/Zb = Zohms (MVAb/KVb)
In 3-phase systems, KVb is the line-to-line value & MVAb is the 3-phase MVA.
[1-phase MVA = (1/3) 3-phase MVA].
Changing the base of a given pu value:
It is observed from equation that the pu value of impedance is proportional directly to the base
MVA and inversely to the square of the base KV. If Zpu,new is the pu impedance required to be
calculated on a new set of base values: MVAb,new & kVb, new from the already given per unit
impedance Zpu,old, specified on the old set of base values, MVAb,old & kVb,old , then we have
Z pu ,new  Z pu ,old *
MVAb ,new kVb ,old 2
*(
)
MVAb ,old kVb,new
On the other hand, the change of base can also be done by first converting the given pu impedance
to its ohmic value and then calculating its pu value on the new set of base values.
Merits and Demerits of pu System
Following are the advantages and disadvantages of adopting the pu system of computations in
electric power systems:
Merits:
 The pu value is the same for both 1-phase and & 3-phase systems
 The pu value once expressed on a proper base, will be the same when refereed to either side
of the transformer. Thus the presence of transformer is totally eliminated
 The variation of values is in a smaller range (0.9 nearby unity). Hence the errors involved
in pu computations are very less.
 Usually the nameplate ratings will be marked in pu on the base of the name plate ratings,
etc.
Demerits:
 If proper bases are not chosen, then the resulting pu values may be highly absurd (such as
5.8 pu, -18.9 pu, etc.). This may cause confusion to the user. However, this problem can be
avoided by selecting the base MVA near the high-rated equipment and a convenient base
KV in any section of the system.
(b)
Current limiting reactors are large coils wound for high self-inductance and very low
resistance.
Application:
 In order to limit the short circuit current to a value which the circuit breaker can
handle
 For the purpose of protecting power plant or power system network
Methods of locating reactors: A reactor may be located in such a way that while there is no
large voltage drop across it under normal conditions of operation. The reactors may be located in
the following positions.
 Generator Reactors: In this method reactors may be inserted in series with each generator

In this method reactors may be inserted in series with each feeder.

Bus bar reactors

Tie-bar reactors
(c) Symmetrical components can be expressed as
 I ao 
1 1
 I   1 1 a
 a1  3 
 I a 2 
1 a 2
1  I a 
a 2   I b 
a   I c 
Therefore,
0
 I ao 
1 1 1  10030 
 I   1 1 a a 2  50300 0 

 a1  3 

2
0

 I a 2 
1 a
a   30180 
a  1120 0
0
 I ao  57.9843.29 
 I   18.9724.96 0 

 a1  
 I a 2   81.874.69 0 
(d) The impedance diagram on single-phase basis for use under balanced conditions can be easily
drawn from the SLD. The following assumptions are made in obtaining the impedance diagrams.
Assumptions:
1. The single phase transformer equivalents are shown as ideals with impedances on appropriate
side (LV/HV),
2. The magnetizing reactances of transformers are negligible,
3. The generators are represented as constant voltage sources with series resistance or reactance,
4. The transmission lines are approximated by their equivalent π-Models,
5. The loads are assumed to be passive and are represented by a series branch of resistance or
reactance
6. Since the balanced conditions are assumed, the neutral grounding impedances do not appear in
the impedance diagram.
Example system
As per the list of assumptions as above and with reference to the system of figure 2, the
impedance diagram can be obtained as shown in figure 3.
Reactance Diagram
With some more additional and simplifying assumptions, the impedance diagram can be
simplified further to obtain the corresponding reactance diagram. The following are the
assumptions made.
Additional assumptions:
 The resistance is often omitted during the fault analysis. This causes a very
negligible error since, resistances are negligible,
 Loads are omitted
 Transmission line capacitances are ineffective.
 Magnetizing currents of transformers are neglected.
Example system
As per the assumptions given above and with reference to the system of figure 2 and figure 3, the
reactance diagram can be obtained as shown in figure 4.
(e) Figure 1 shows a single line diagram of a 3-phase system. The percentage reactance of each
alternator is based on its capacity. Find the short-circuit current that will flow in to a complete
3-phase short-circuit at fault point F.
(f) A three phase transmission line operating at 33kV and having a resistance and reactance of
10Ω and 30Ω respectively is connected to a generating station busbar through a 15MVA stepup transformer which has a reactance of 0.06pu. Busbar is connected to two generators, one
10MVA having 0.15pu reactance another 5 MVA having 0.075pu reactance. Calculate the short
circuit MVA and fault current when three phase short circuit occurs (a) at the high voltage
terminals of the transformer (b) at the load end of the transmission line.
2. Attempt any Two parts of following
(10 marks x 2 = 20)
(a). (i) Single line to ground (LG) fault analysis:
An unloaded balanced three-phase synchronous generator with neutral grounded through an
impedance Zn is shown in Figure 1. Suppose a single line to ground fault (LG) occurs on phase
‘a’ though impedance Zf.
LG fault on phase ‘a’ of an unloaded generator
Since the generator is unloaded, the following terminal
conditions exist at the fault point:
Va  Z f I a
Ib  0
Ic  0
Substituting, I b  I c  0 , the symmetrical components
of currents can be calculated as
Figure 1: L-L-G fault
Figure 2: L-G fault on phase 'a'
I a0 
1 1 1   I a 
 I   1 1 a a 2   0 
 a1  3 
 
 I a 2 
1 a 2 a   0 
The values if the symmetrical components of fault current I a are
1
I a 0  I a1  I a 2  I a
3
The voltage of phase 2 can be expressed in terms of symmetrical components
Va  Va 0  Va1  Va 2
Therefore,
Va  Ea  ( Z 0  Z1  Z 2 ) * I a 0
Also,
3Z f I a 0  E a  ( Z 0  Z1  Z 2 ) * I a 0
Ea
Z 0  Z1  Z 2  3Z f
Therefore fault current
3E a
I f  I a  3I ao 
Z 0  Z1  Z 2  3Z f
I ao 
Note that, for solidly grounded generator, Zn
Figure 3: Symmetrical network equivalent
= 0 and for bolted fault Zf = 0.
Extending the above concept to the analysis of LG fault in a power system, the Thevenin’s
equivalent circuit (as seen from the fault point) is obtained, individually for the three sequence
networks.
For the positive sequence network Vth, the open circuit pre-fault voltage at the fault point, and
Z1th, the positive sequence Thevenin’s equivalent impedance as seen from the fault point are
determined. For negative and zero sequence networks, only the Thevenin’s equivalent
impedances Z2th and Z0th, respectively are calculated. The three Thevenin’s equivalent networks
are then connected in series.
(ii). Figure shows a double line to ground (LLG) fault on phases ‘b’ and ‘c’ through an
impedance Zf on an unloaded three phase generator. The terminal conditions at the fault point
are
Vb  Vc  Z f I f  Z f ( I b  I c )
I a  I a1  I a 2  I a 0  0
Vb and Vc can be written as,
Vb  Va 0  a 2Va1  aVa 2
Vc  Va 0  aVa1  a 2Va 2
Since Vb= Vc, therefore
Va1=Va2
Substituting Ib and Ic in terms of their sequence
components, voltage of phase b can
be expressed as,
Vb  Z f ( I a 0  a 2 I a1  aI a 2  I a 0  aI a1  a 2 I a 2 )
Vb  Z f (2 I a 0  (a 2  a )( I a1  I a 2 ))
Vb  Z f (2 I a 0  ( I a1  I a 2 ))
Since, 1  a  a 2  0 and I a  I a1  I a 2  I a 0  0 hence,
Vb  3Z f I a 0
Further substituting Vb,
Vb  3Z f I a 0  Va0  (a 2  a)Va1  Va0  Va1
Substituting Vao and Va1, the zero sequence component of,
( E  Z1 I a1 )
I a0  a
( Z 0  3Z f )
I a2 
( Ea  Z1 I a1 )
Z2
Ea
Z 2 ( Z 0  3Z f )
I a1 
( Z1 
( Z 0  Z 2  3Z f )
Fault current,
I f  3I ao
Positive, negative and zero phase sequence components are called symmetrical
(b) (i)
components of the original unbalanced system. The subscript 1, 2 & 0 are used to show positive,
negative and zero sequence components respectively.



Let us express the symmetrical components of R-Phase in terms of phase currents I R , I Y & I B
From fig:












I R  I R1  I R 2  I R 0        (i )





I Y  I Y 1  I Y 2  I Y 0  a 2 I R1  a I R 2  I R 0      (ii )

I B  I B1  I B 2  I B 0  a I R1  a 2 I R 2  I R 0        (iii )
Zero sequence current: (i) + (ii) + (iii)






I R  I Y  I B  I R1 (1  a 2  a)  I R 2 (1  a  a 2 )  3 I R 0        (iv )
Since,
(1  a  a 2 )  0
Therefore,

1   
I R0  ( I R  I Y  I B )
3
Positive Sequence Current: multiply equation (ii) by a & equation (iii) by a2 & then (i) + (ii) +
(iii)



1 
I R1  ( I R  a I Y  a 2 I B )
3
Negative sequence of current: multiply equation (ii) by a2 & equation (iii) by a then adding to (i)



1 
I R2  ( I R  a 2 I Y  a I B )
3
Thus, Symmetrical components can be expressed as
 I ao 
1 1 1   I a 
 I   1 1 a a 2   I 
 a1  3 
 b 
 I a 2 
1 a 2 a   I c 
Similarly, symmetrical components of voltage,
 Eao 
1 1
 E   1 1 a
 a1  3 
 Ea 2 
1 a 2
1   Ea 
a 2   Eb 
a   Ec 
(ii) The per unit values of positive, negative and zero sequence reactance of a network at fault
are 0.16. 0.14 and 0.2. Determine the fault current if fault is double line to ground.
I a1 
( Z1 
Ea

Z 2 ( Z 0  3Z f )
( Z 0  Z 2  3Z f )
I a0 
1
 1.01729
0.14(0.2  3 * 0)
(0.16 
(0.2  0.14  3 * 0)
( Ea  Z1 I a1 ) (1  0.16 *1.01729)

 4.186
(Z 0  3Z f )
(0.2  3 * 0)
(c) An 11 kV, 25MVA synchronous generator has positive, negative and zero sequence
reactances of 0.12, 0.12 and 0.08 per unit respectively. The generator neutral is grounded
through a reactance of 0.03pu. A single line-to-ground fault occurs at the generator terminals.
Determine the fault current and line-to-line voltages. Assume that the generator was unloaded
before the fault.
3.
Attempt any Two parts of following
(10 marks x 2 = 20)
(a). The line admittances of a 4-bus power system for a load flow solution is given in table 1.
The schedules of active and reactive power are in table 2. Determine the voltages at the end of
first iteration using Gauss-Siedel method.
Table-1
Bus code
Admittance
1-2
2-j8
1-3
1-j4
2-3
0.666-j2.664
2-4
1-j4
3-4
2-j8
Table-2
Bus Code
P
Q
V
Remarks
1
1.06
Slack
2
0.5
0.2
1+j0
PQ
3
0.4
0.3
1+j0
PQ
4
0.3
0.1
1+j0
PQ
(b) With the help of load flow studies, we determine the node voltage magnitude (V), phase angle
(δ), nodal active (P) and reactive (Q) power injection. Load flow study help us in determination
of line power flow and losses as well. Load flow studies determines all these terms in
compliance with Kirchhoff’s law. we can also determine the best size and as well as the most
favorable locations for the power capacitors both for improvement of the p.f and also raising the
bus voltage of electrical network. This also helps us to determine the best locality as well as
optimal capacity of the proposed generating stations, substations or new lines.
Classifications of the buses: Buses are classified on the behalf of information or quantity
known and unknown on then. Active power (P), Reactive power (Q), voltage magnitude (V) and
phase angle (δ) are the parameters to be evaluated.
(i)
(ii)
(iii)
Generating bus or voltage controlled bus: This is also called p-v bus. P and V are
known in this bus and unknown parameters are Q & δ.
Load bus: This is also called p-q bus. These are the load buses on which loads are
present. quantity know on these buses are P & Q while V and δ are unknown.
Slack bus or Swing bus: This is also known as reference bus. Reference bus is chosen
before starting the load flow calculations. V & δ are considered as 1pu and 0 degree
respectively.
Sr Bus name
Known
Unknown
no
1 Generator bus
P, V
Q, δ
2 Load bus
P, Q
V, δ
3 Swing bus
V, δ
P, Q
4 Voltage controlled bus P, V
Q, δ
(c)
Fast Decoupled is a very fast method of obtaining load flow solution. This is actually an
extension of Newton’s method formulated in polar coordinates. The equation of load flow
studies using Newton-Raphson’s method can be expressed in polar coordinates as
 P   H N    
Q    M L   E       (i )
  
  E 
Where, H, N, M & L are the elements. Since changes in real ΔP is less sensitive to changes in
reactive power ΔQ is less sensitive to the changes in angle Δδ, then
 P   H 0    
Q    0 L   E       (ii )
  
  E 
Equation (2) is decoupled equation, which can be expanded as,
P  H        (iii )

E
      (iv )
 E
Pp
Q p

& L pq 
Eq
 q
E q
Q  L
H pq
n
Pi   Ei E k Yik cos[ ik   i   k ]
k 1
4. Attempt any Four parts of following
(5 marks x 4 = 20)
(a) The laws of rotation developed in section.3 are applicable to the synchronous machine. From
I *  T
I*
d 2 m
 T      (i)
dt 2
Here T is the net torque of all torques acting on the machine, which includes the shaft torque (due to
prime mover of a generator or load on a motor), torque due to rotational losses (friction, windage and
core loss) and electromagnetic
torque.
Let
Tm = shaft torque or mechanical
torque corrected for rotational
losses
Te = Electromagnetic or electrical
torque
For a generator Tm tends to
accelerate the rotor in positive
direction of rotation as shown in
Figure 1. It also shows the
corresponding torque for a motor
with respect to the direction of
rotation.
The accelerating torque for a
generator is given by:
Ta  Tm  Te      (i)
Under steady-state operation of the generator, Tm is equal to Te and the accelerating torque is
zero. There is no acceleration or deceleration of the rotor masses and the machines run at a
constant synchronous speed. In the stability analysis in the following sections, Tm is assumed to
be a constant since the prime movers (steam turbines or hydro turbines) do no act during the
short time period in which rotor dynamics are of interest in the stability studies.
Now equation (i) has to be solved to determine  m as a function of time. Since  m is measured
with respect to a stationary reference axis on the stator, it is the measure of the absolute rotor
angle and increases continuously with time even at constant synchronous speed. Since machine
acceleration /deceleration is always measured relative to synchronous speed, the rotor angle is
measured with respect to a synchronously rotating reference axis.
Let
 m   m  sm * t      (iii )
Where,  sm is the synchronous speed in mechanical rad/s and  m is the angular displacement in
mechanical radians. Taking the derivative of (iii), we get
d m d m

  sm
dt
dt
d 2 m d 2 m

     (iv )
dt 2
dt 2
Substituting equation (iv) in (i)
d 2 m
I
 Ta  Tm  Te
dt 2
Multiplying by  m on both sides we get
m I
d 2 m
  mTa   m (Tm  Te )
dt 2
d 2 m
 Pm  Pe  Pa      (v)
dt 2
Where, M is the angular momentum, also called inertia constant, Pm is shaft power input less
rotational losses, Pe is electrical power output corrected for losses and Pa is the acceleration
power. M depends on the angular velocity  m , and hence is strictly not a constant, because  m
deviates from the synchronous speed during and after a disturbance. However, under stable
conditions  m does not vary considerably and M can be treated as a constant. (v) is called the
“Swing equation”. The constant M depends on the rating of the machine and varies widely with
the size and type of the machine. Another constant called H constant (also referred to as inertia
constant) is defined as
Energy _ stored _ at _ Synchrnous _ speed
H
MJ/MVA
Machine _ rating
M
(b)
The first electric power system was a dc system built by Edison in 1882. The subsequent
power systems that were constructed in the late 19th century were all dc systems. However
despite the initial popularity of dc systems by the turn of the 20th century ac systems started to
outnumber them. The ac systems were though to be superior as ac machines were cheaper than
their dc counterparts and more importantly ac voltages are easily transformable from one level to
other using transformers. The early stability problems of ac systems were experienced in 1920
when insufficient damping caused spontaneous oscillations or hunting. These problems were
solved using generator damper winding and the use of turbine-type prime movers.
The stability of a system refers to the ability of a system to return back to its steady state when
subjected to a disturbance. As mentioned before, power is generated by synchronous generators
that operate in synchronism with the rest of the system. A generator is synchronized with a bus
when both of them have same frequency, voltage and phase sequence. We can thus define the
power system stability as the ability of the power system to return to steady state without losing
synchronism.
Usually
power
system
stability
is
categorized
into Steady
State, Transient and Dynamic Stability.
Steady State Stability studies are
restricted to small and gradual changes in
the system operating conditions. In this
we basically concentrate on restricting
the bus voltages close to their nominal
values. We also ensure that phase angles
between two buses are not too large and
check for the overloading of the power equipment and transmission lines. These checks are
usually done using power flow studies.
Transient Stability involves the study of the power system following a major disturbance.
Following a large disturbance the synchronous alternator the machine power (load) angle
changes due to sudden acceleration of the rotor shaft. The objective of the transient stability
study is to ascertain whether the load angle returns to a steady value following the clearance of
the disturbance.
The ability of a power system to maintain stability under continuous small disturbances is
investigated under the name of Dynamic Stability (also known as small-signal stability). These
small disturbances occur due random fluctuations in loads and generation levels. In an
interconnected power system, these random variations can lead catastrophic failure as this may
force the rotor angle to increase steadily.
(c) Given 1HP=746W
25000HP=(25000*746)/1000=18650kW
P=4
Rating of machine, G =20MVA
H=9.0kWsec/kVA or 9.0 MJ/MVA
Mechanical power input= 18650kW
(i)
K. E energy stored in the rotor
Electric power developed=15000kW
We know that, H= stored energy(MJ)/Machine rating (MVA)
K E stored in the rotor=H*Machine rating= 9*20=180MJ
(ii)
The acceleration :
The acceleration power, PAG=Mechanical power input-electrical power output
PAG= 18650-15000=3650kW=3.65 MW
d 2 P
3.65
New accelerating power   2  AG 
M 180 * 50
dt
GH
20 * 9
Where, M 

180 f 180 * 50
(d) Transient stability assessment of an single machine infinite bus (SMIB) system is possible
without resorting to actual solution of the swing equation, by a method known as equal–area
criterion. In a SMIB system, if the system is unstable after a fault is cleared, δ(t) increases
indefinitely with time, till the machine loses synchronism. In contrast, in a stable system, δ(t)
reaches a maximum and then starts reducing as shown in Fig.1.
Mathematically,
d  (t )
0
dt
d  (t )
 0 , for a long time after the fault
Sometime after the fault is cleared in a stable system and
dt
is cleared in an unstable system. Consider the swing equation
d 2 m
M
 Pm  Pe  Pa
dt 2
d 2 m Pa

M
dt 2
Multiplying both side by 2
d  (t )
dt
d (t ) d 2 m
d (t ) Pa
2
*
2
*
2
dt
dt
M
dt
P
d d 2
d (t ) a
[( ) ]  2
*
dt dt
dt
M
Now Integrating both side,
{
d 2
2
} 
dt
M


Pa d
0
Or
For stability,
d
 0 , this means
dt
d
2

dt
M


Pa d
0
2 
Pa d  0
M  0
The integral gives the area under the Pa – δ curve. The condition for stability can be, thus stated
as follows: A SMIB system is stable if the area under the Pa – δ curve, becomes zero at some
value of δ. This means that the accelerating (positive) area under Pa – δ curve, must equal the
decelerating (negative) area under Pa – δ curve. This condition is called equal area criterion.
SUDDEN CHANGE IN MECHANICAL INPUT:
Consider the SMIB system shown in Fig. 2. The electrical power transferred is given by
Pe  Pm Sin
Where,
Pm 
E g' V
xd'  xe
Under steady state Pm = Pe. Let the
machine be initially operating at a
steady
state
angle
δo,
at
synchronous speed ωs, with a
mechanical input Pmo, as shown in
Fig. 3 ( point a).
If there is a sudden step increase in
input power to Pm1 the accelerating
power is positive (since Pm1 > Pmo)
and power angle δ increases. With
increase in δ, the electrical power Pe
increases, the accelerating power
decreases, till at δ = δ1, the electrical
power matches the new input Pm1.
Fig. 2: SMIB System
Fig. 3: Equal area criterion–sudden change in mechanical input
The area A1, during acceleration is given by
1
A1   ( Pm1  Pe )d = accelerating area
0
A1  Pm1 (1   0 )  Pmax (cos 1  cos  0 )
At b, even though the accelerating power is zero, the rotor is running above synchronous speed.
Hence, δ and Pe increase beyond b, wherein Pe < Pm1 and the rotor is subjected to deceleration.
The rotor decelerates and speed starts dropping, till at point d, the machine reaches synchronous
speed and δ = δmax. The area A2, during deceleration is given by
 max
A2  
1
 max
A2  
1
( Pe  Pm1 )d = deceleration area
( Pe  Pm1 )d  Pmax (cos  max  cos  1 )  Pm1 ( max   1 )
By equal area criterion A1 = A2. The rotor would then oscillate between δ0 and δmax at its
natural frequency. However, damping forces will reduce subsequent swings and the machine
finally settles down to the new steady state value δ1 (at point b). Stability can be maintained only
if area A2 at least equal to A1, can be located above Pm1. The limiting case is shown in Fig.4,
where A2 is just equal to A1.
Here δmax is at the intersection of Pe and Pm1. If the
machine does not reach synchronous speed at d, then
beyond d, Pe decreases with increase in δ, causing δ to
increase indefinitely. Applying equal area criterion to
Fig.4 we get
A1  A2
Pm1 ( max   0 )  Pmax (cos  0  cos  max )
Substituting Pm1  Pmax sin  max , we get
Fig. 4: Maximum increase in mechanical power
( max   0 ) sin  max  cos  max  cos  0        (i)
Equation (i) is a non-linear equation in δmax and can be solved by trial and error or by using any
numerical method for solution of non-linear algebraic equation (like Newton- Raphson, bisection
etc). From solution of δmax, Pm1 can be calculated. Pm1 – Pmo will give the maximum possible
increase in mechanical input before the machine looses stability.
(e) Since we are interested in finding out the maximum time that the circuit breakers may take
for opening, we should be more concerned about the critical clearing time rather than clearing
angle. Furthermore, notice that the clearing angle is independent of the generalized inertia
constant H. Hence we can comment that the critical clearing angle in this case is true for any
generator that has a d-axis transient reactance of 0.20 per unit. The critical clearing time,
however, is dependent on H and will vary as this parameter varies.
To obtain a description for the critical clearing time, let us consider the period during which the
fault occurs. We then have Pe = 0. We can therefore write from,
d 2  s

Pm
dt 2 2 H
Integrating the above equation with the initial acceleration being zero, we get
t 

d
  s Pm dt  s Pmt
0
dt
2H
2H
Further equation will lead to,
t
s
0
2H
 
Replacing  by  cr and t by t cr
t cr 
s
Pm tdt 
4H
Pm t 2   0
4H
( cr   0 )
 s Pm
(f) The equal area criterion can be used to determine the critical fault clearing angle. The point
by point solution (or step-by-step) method is used to determine the critical fault clearing time
associated with this angle. A Swing curve (δ Vs t) is plotted for each machine to know its
tendency to remain stable. We shall illustrate a general method for one machine connected to an
infinite bus. However, it can be applied t0 every machine of a multi machine system.
The following assumptions are made in the equation of swing equation by point by point method.
(i)
The accelerating power Pa and angular acceleration α are constant from the middle of
the precedding at the middle of present interval considered. Both of these values are
calculated at the beginning of this interval.
(ii)
The angular velocity ω, computed at the middle of an interval, remains constant from
the beginning to the end of the interval considered.
These assumptions are not strictly correct, since δ is changing continouslyand both Pa and ω
depend upon δ. However, if time increment Δt is made smaller, the computed curve
approximates the actual curve.
The accelerating at t  (n  1) is  ( n1) . Over the region the constant acceleration  ( n1) there is an
increment of angular velocity from 
3
( n )
2
to 
w
 n 1 
 n 1 
w
1
n
2
Again, the angular velocity w
n
1
2
1 .
( n )
2
n
1
2
w
n
3
2
t
Pa ( n 1)
M
w
3
n
2

Pa ( n 1)
M
* t
remains constant from t = (n-1) to t=n. The displacement angle
increase from  ( n 1) to  n during the time interval t .
From the velocity/time curve (Fig.1)
w
1
n
2

 n   ( n1)
t
  n   ( n 1)  w
n
Similarly,
w
3
n
2

* t
1
2
 n 1   ( n  2 )
t
  n 1   ( n 2)  w
n
3
2
* t
The change in δ over any interval = value of ω for that interval * time of interval
But the change in δ over the nth interval, that is from t  n to t  n  1 is given by
( ) n   n   n 1  w 1 (t )
n
2
The change in δ over (n-1)th interval, that is from t  n  1 to t  n  2 is given by
( ) n 1   n 1   n  2  w 3 (t )
n
( ) n  ( n 1 )  w
1
n
2
(t )  w
3
n
2
( ) n  ( n1 ) 
2
(t ) 
Pa ( n1)
M
(t ) 2
Pa ( n1)
(t ) 2
M
Equation shows that the change in torque angle during a given interval is equal to the change in
torque angle during the preceding interval plus the accelerating power at the beginning of the
interval times (t ) 2 / M . This equation forms the basis of the numerical solution by point-bypoint method.
5. Attempt any Two parts of following
(10 marks x 2 = 20)
(a) When lightning injects surge currents into power lines, they move down the conductors as
traveling waves. Understanding the nature of traveling waves helps in predicting the voltage and
current levels that occur on power systems during lightning strikes. Surges travel at 99% the
speed of light on overhead circuits — about 1000 ft/msec (300 m/msec). The inductance L and
capacitance C uniformly distributed along the line determine the velocity and the relationship
between voltage and current. The velocity in a unit length per second is based on L and C per
length unit as
1
v
LC
The voltage and current are related by Z, the surge impedance,
V  ZI
Where, Z is a real value, typically 300
to 400W. The distributed inductance
and capacitance determines the surge
impedance
Fig.1: Forward and backward traveling waves passing each other
Z  60
L
2h
 60 ln
C
r
Where,
h, average conductor height
r, conductor radius
When surge voltages of like
polarity pass over each other, the
voltages add as shown in Fig.1.
The currents subtract; current is
charge moving in a direction.
Charge of the same polarity
flowing in opposite directions
cancels.
V  VF  VB  Z ( I F  I B )
Fig.2: Traveling wave reflection at an open point.
I  Z (I F  I B )
A surge arriving at an open point in the circuit reflects, sending back a voltage wave equal to the
incoming wave (see Fig.2). The voltage doubles. Think of the incoming wave as a stream of
electrons. When these electrons hit the open point, they stop, but the pileup of electrons at the
end forces the electrons back in the direction they came from (like charges repel each other).
While the voltage doubles, the current cancels to zero as the return wave counters the incoming
wave.
Voltage doubling at open points is an important consideration for the protection of distribution
insulation. It is a very important consideration when protecting underground cables. It also
comes into play when placing arresters to protect a substation. When a surge hits a short circuit,
the voltage drops to zero as we expect. The ground releases opposing charge that cancels the
voltage, creating a reverse wave traveling back toward the source. Now the current doubles. Two
cases are very close to a short circuit: a conducting surge arrester and a fault arc. Both of these
act like a short. They generate a voltage canceling wave.
Line taps, capacitors, open points, underground taps — any impedance discontinuities — cause
reflections. In the general case, the voltage wave reflected from a discontinuity is
VR  aVI
Where, a is the reflection coefficient.
Z  ZA
a B
ZB  ZA
For an open circuit (ZB = infinite), the
reflection coefficient, a = 1, and for a
short circuit (ZB = 0), a = –1.
The voltage at the discontinuity is the
Fig.3: Traveling wave reflection at a surge-impedance discontinuity where ZB < ZA.
sum of the incoming wave
and the reflected wave: VT  VI  VR  VI  aVI  bVI
Where, b is the refraction coefficient:
ZB  ZA
2Z B

ZB  ZA ZB  ZA
V
V
IF  F & IB   B
Z
Z
When a traveling wave hits a split in the line, a
portion of the wave goes down each path and
another wave reflects back toward the source as
shown in Figures 4. Use VT  bVI to calculate the
voltage at the split, just use a new ZB’ as the
parallel combination of ZB and ZC,
Z B'  ( Z B || Z C ) . For equal surge impedances (ZA
= ZB = ZC), the total voltage is 2/3 of the
incoming wave.
To a traveling wave, a capacitor initially appears
b  1 a  1
Fig.4: Traveling wave reflections at a split
as very small impedance, almost a short circuit.
Then, as the capacitor charges, the effective
impedance rises until it becomes an open circuit with the capacitor fully charged. For a traveling
wave section terminated in a capacitor, the voltage ramps up to double the voltage as
t
VT  VI (1  e ZC )
An inductance initially appears almost as an open circuit. As the inductor allows more current to
flow, its impedance drops until it is equivalent to a short circuit. For a traveling wave section
terminated in an inductor, the voltage initially doubles then decays down to zero as
Z
 t
L
VT  VI (1  e )
A cable section attached to an overhead circuit behaves similarly to a capacitor. It is like an
open-circuited traveling wave section with a low surge impedance. In fact, a capacitor can be
modeled in a traveling wave program such as EMTP as a traveling wave section with a surge
impedance of Dt/2
C and a travel time of Dt/2, where Dt is small (usually the time step of the simulation) (Dommel,
1986). Likewise, we may model an inductor as a traveling wave section shorted at the end with a
surge impedance equal to 2L/Dt and a travel time of Dt/2.
Corona impacts traveling waves by modifying the surge impedance. When lightning strikes an
overhead distribution circuit, tremendous voltages are created. The voltage stress on the
conductor surface breaks down the air surrounding the conductor, releasing charge into the air in
many small streamers. A high-voltage transmission line may have a corona envelope of several
feet (1 m), which increases the capacitance. This decreases the surge impedance and increases
coupling to other conductors and slopes off the front of the wave. Normally, distribution line
insulation flashes over before significant corona develops, so we can neglect corona in most
cases. Pre-discharge currents, an effect similar to corona, may also form between conductors.
When the voltage between two parallel conductors reaches the flashover strength of air (about
180 kV/ft or 600 kV/m), a sheet of many small streamers containing pre-discharge currents flows
between the conductors. This is a precursor to a complete breakdown. The pre-discharge current
delays the breakdown and relieves voltage stress between the conductors.
While the pre-discharge currents are flowing, the resistance across the gap is 400 W/ft or 1310
W/m of line section (Wagner, 1964; Wagner and Hileman, 1963; Wagner and Hileman, 1964).
Normally, on overhead distribution circuits, the weakest insulation are at poles, so in most cases,
the circuit flashes over before the circuit enters the pre-discharge state. Pre-discharge currents do
make midspan flashovers less likely and may help with using arresters to protect lines.
Normally, for modeling lightning in circuits or cables, we can ignore corona and pre-discharge
and use lossless line sections (wire resistance and damping is ignored). This is normally
sufficient since only short sections need to be modeled for most situations.
A multi-conductor circuit has several modes of propagation; each has different velocity and
different surge impedance. In most cases, these multi-conductor effects are ignored, and we
assume each wave travels on each conductor at 95 to 100% of the speed of light.
For multi-conductor circuits, the coupling between circuits is important. The mutual surge
impedance determines the voltage on one conductor for a current flowing in another as V2 =
Z12I1.
(b) The velocity in a unit length per second is based on L and C per length unit as
1
v
LC
The voltage and current are related by Z, the surge impedance,
V  ZI
Where, Z is a real value, typically 300
to 400W. The distributed inductance
and capacitance determines the surge
impedance
L
2h
Z  60
 60 ln
C
r
Where,
Fig.1: Forward and backward traveling waves passing each other
h, average conductor height
r, conductor radius
When surge voltages of like
polarity pass over each other, the
voltages add as shown in Fig.1.
The currents subtract; current is
charge moving in a direction.
Charge of the same polarity
flowing in opposite directions
cancels.
V  VF  VB  Z ( I F  I B )
I  Z (I F  I B )
Fig.2: Traveling wave reflection at an open point.
A surge arriving at an open point in the circuit reflects, sending back a voltage wave equal to the
incoming wave (see Fig.2). The voltage doubles. Think of the incoming wave as a stream of
electrons. When these electrons hit the open point, they stop, but the pileup of electrons at the
end forces the electrons back in the direction they came from (like charges repel each other).
While the voltage doubles, the current cancels to zero as the return wave counters the incoming
wave.
(c) Bewlwy’s Lattice diagram: A horozontal ine is drawn to represent the line & two vertical
lines at the ends on which equal intervals of the time T i. e marked as shown. The diagram
begins at the top left corner at the source and proceeds along the line OT. The attenuation a is
also shown for one travel. At the open one, The wave reflects completely as a. To the right is
marked 2a which is the voltage at the open end after one travel time-arrows shown the progress
of the wave. At T, the wave reflects and reaches the source at time 2T with an amplitude a2 and
since source voltage remains constant at +1. The wave arriving at the source is reflected
negatively and shown as –a2 when this reaches the open end at time 3T. It has the value of (-a3).
Again , at the end this wave doubles and a reflection (-a3) travels back to the source. The total
volatge at the open end is now (a+a-a2-a3)=2(a-a3) which is sum of all voltages marked in the
inclined lines up to time 3T. Proceeding in this manner, we obseve that in order to keep the
source volatge at +1, continued reflections have to tape place with –ve sign at source & +ve
signs at open end.