1
00:00:00,110 --> 00:00:07,110
Friends, today we continue our discussion
on introduction of the power system stability
2
00:01:09,560 --> 00:01:10,799
problem.
3
00:01:10,799 --> 00:01:17,799
Today, I will discuss something more about
inertia constant H, we introduced last time.
4
00:01:19,220 --> 00:01:26,220
I will also discuss certain assumptions that
commonly made in stability studies. We will
5
00:01:30,189 --> 00:01:37,189
discuss small signal stability and voltage
stability concepts; we shall be dealing with
6
00:01:42,409 --> 00:01:49,409
these topics in depth at later stage. Today
our intention is just to give you basic concepts
7
00:01:54,530 --> 00:02:01,530
related to a small signal stability and voltage
stability.
8
00:02:06,530 --> 00:02:13,530
Last time, we have derived this swing equation,
the swing equation is very important for analysis
9
00:02:17,090 --> 00:02:24,090
of the stability of a power system, this equation
is used for transitivity analysis and this
10
00:02:24,840 --> 00:02:31,840
is the basis for small signal stability analysis
also. This equation has the form the coefficient
11
00:02:34,610 --> 00:02:41,610
is 2H upon omega s ah d2 delta by dt2 equal
to Pm minus Pe. Initially when the equation
12
00:02:44,000 --> 00:02:51,000
was written, the coefficient was M, the inertia
constant. However, we define a another constant
13
00:02:56,220 --> 00:03:03,220
known as the inertia constant H, let us see
why we prefer inertia constant H for analyzing
14
00:03:08,810 --> 00:03:15,810
the stability of the system.
15
00:03:22,830 --> 00:03:29,830
The inertia constant H has the desirable property,
that its value unlike that of M does not vary
16
00:03:31,959 --> 00:03:38,959
greatly with the rated MVA and speed of the
machine but instead has a characteristic value
17
00:03:41,280 --> 00:03:48,280
or set of values for each class of machines.
I had very clearly, very in particularly told
18
00:03:51,060 --> 00:03:56,519
you last time that the coefficient M varies
over a wide range depending upon the size
19
00:03:56,519 --> 00:04:00,549
of the machine, speed of the machine, type
of the machine.
20
00:04:00,549 --> 00:04:07,549
However, the inertia constant H has some characteristic
values and in that respect H is similar to
21
00:04:15,970 --> 00:04:22,970
per unit reactance of machines as you all
know that the per unit reactance of the synchronous
22
00:04:24,979 --> 00:04:31,580
machines have some characteristic values.
Similarly, the inertia constant H also has
23
00:04:31,580 --> 00:04:38,580
certain characteristic values and sometimes
in the absence of the specific knowledge about
24
00:04:38,770 --> 00:04:45,770
inertia constant H, we can assume some characteristic
value and do our calculations, manufacturers
25
00:04:48,539 --> 00:04:55,539
specify or provide the value of H on the MVA
rating of the machine.
26
00:05:12,180 --> 00:05:19,180
This diagram shows the typical value of the
inertia constant H for turbo generators. On
27
00:05:26,830 --> 00:05:33,279
this axis, we have marked the rating of the
machine in MVA and on this axis we have marked
28
00:05:33,279 --> 00:05:40,279
the inertia constant H in mega joules per
MVA. Now machines, where turbo generators
29
00:05:40,779 --> 00:05:47,779
are of different type and different speeds,
if you take a turbo generator whose lateral
30
00:05:48,080 --> 00:05:55,080
speed is 800 rpm a powerful machine then the
variation of the inertia constant is shown
31
00:05:55,619 --> 00:06:02,619
by this graph, that is when the rating varies
as something like about 10 mega, 10 MVA to
32
00:06:04,409 --> 00:06:11,409
say 100 MVA, the variation of H is
in the range starting from about 9.5, it goes
to 6.5.
33
00:06:18,210 --> 00:06:25,210
Similarly, for a turbo generator whose speed
is 3600 rpm , the variation is seen in this
34
00:06:32,800 --> 00:06:39,800
graph, it varies from 4 to as much as 7 or
7.5. Another graph which is shown here is
35
00:06:41,159 --> 00:06:48,159
the known condensing type of steam turbines,
the inertia constant again varies but over
36
00:06:48,399 --> 00:06:55,399
a certain range.
37
00:07:00,619 --> 00:07:07,619
Now this graphs shows the characteristic value
of inertia constant H for a hydro generator.
38
00:07:11,869 --> 00:07:18,869
Again on the x axis, we have marked the rating
of the machine in MVA and y axis, the inertia
39
00:07:20,740 --> 00:07:27,740
constant H in mega joules per MVA. Here, you
see actually that again you find that the
40
00:07:30,550 --> 00:07:37,389
machines having different speeds do have different
value of inertia constant H, it increases
41
00:07:37,389 --> 00:07:44,389
with increase in the MVA rating of the machine
but still you can see actually it lies in
42
00:07:46,649 --> 00:07:49,529
a certain range.
43
00:07:49,529 --> 00:07:56,529
Now I would like to know from you what this
inertia constant H is whether it is the inertia
44
00:07:57,759 --> 00:08:03,229
constant of the prime over or whether it is
inertia constant of the rotor of the synchronous
45
00:08:03,229 --> 00:08:09,809
generator or it is inertia constant of the
prime over and rotor grouped together, who
46
00:08:09,809 --> 00:08:16,809
will reply this? The question I imposed is,
that I am talking of inertia constant H for
47
00:08:23,110 --> 00:08:30,110
the synchronous machine. This inertia constant
whether it corresponds to the total inertia
48
00:08:30,619 --> 00:08:37,619
constant of the turbine plus generator rotor
or only generator rotor or only turbine.
49
00:08:40,680 --> 00:08:47,680
Yes you are correct, the inertia constant
is the total inertia constant of the system
50
00:08:51,829 --> 00:08:58,829
that is for the rotating body okay. Now, here
I will just give you a small example suppose
51
00:09:04,709 --> 00:09:11,709
in a power plant we have two machines okay
of different ratings and the information which
52
00:09:13,940 --> 00:09:20,940
is provided to us is about the inertia constant
on the MVA rating, if this two machines are
53
00:09:23,870 --> 00:09:30,870
tightly coupled. So that they form a coherent
group, coherent group means the machines will
54
00:09:32,779 --> 00:09:39,779
swing together then we can the place this
machine by equivalent machine.
55
00:09:40,170 --> 00:09:47,170
Now just take the example, let us say that
unit 1, there are two units are there in the
56
00:09:48,500 --> 00:09:55,500
plant, unit 1 is 500 MVA and its is a constant
H1 is specified as 5 mega joules per MVA,
57
00:10:12,240 --> 00:10:19,240
unit 2 7000 MVA its inertia constant say H2
is 3mega joules per MVA because the manufacturer
58
00:10:41,560 --> 00:10:47,329
provide the information about this H constant
on its on MVA rating okay.
59
00:10:47,329 --> 00:10:54,329
These two machines are in the same plant,
we want to find out the equivalent inertia
60
00:10:54,720 --> 00:11:01,720
constant of the2 machines, how do we compute
the equivalent inertia constant for that,
61
00:11:04,870 --> 00:11:10,420
we need one more information that on both
MVA basis. We want to find out the equivalent
62
00:11:10,420 --> 00:11:15,949
inertia constant. Let us assume that we want
to find out the equivalent inertia constant
63
00:11:15,949 --> 00:11:22,949
H on base MVA equal to 100 okay. Then the
calculations proceed in a straight forward
64
00:11:34,889 --> 00:11:41,889
manner, first we find out the total kinetic
energy stored in the two machines, that is
65
00:11:43,470 --> 00:11:50,470
total kinetic energy will be equal to, take
the basic definition for the first machine
66
00:12:04,009 --> 00:12:11,009
it is going to be H that is 5, MVA rating
is 500 plus second machine H is 3 and MVA
67
00:12:17,519 --> 00:12:24,519
rating is 1000. This comes out to be 5500
mega joules.
68
00:12:28,459 --> 00:12:35,459
Therefore, “H” I can say equivalent
on base MVA equal to 100 is 5500 divided by
100 because base MVA is 100, it comes out
69
00:13:09,149 --> 00:13:16,149
to be 55 mega joules per, this is the way
one can find out the equivalent inertia constant
70
00:13:21,319 --> 00:13:28,319
of a number of machines which are running
in the same plant and are coherent. Now before
71
00:13:37,910 --> 00:13:44,910
we talk further about the different types
of stability, I would like to tell the the
72
00:13:49,569 --> 00:13:56,569
commonly made assumptions for stability analysis.
73
00:13:59,430 --> 00:14:04,309
The first assumption which is very commonly
made is the mechanical power input remains
74
00:14:04,309 --> 00:14:11,309
constant during the period of the transients,
if you look into the swing equation, our swing
75
00:14:18,680 --> 00:14:25,680
equation is 2H divided by omega s d2 delta
by dt2 equal to Pm minus Pe okay. But this
76
00:14:37,589 --> 00:14:44,519
assumption which I am telling you here is
that this Pm will be assumed to remain constant
77
00:14:44,519 --> 00:14:50,629
during the transient okay.
78
00:14:50,629 --> 00:14:57,629
The logic for making this assumption is that
in any any prime over whether is a stream
79
00:15:07,720 --> 00:15:14,720
turbine or is a hydro turbine right, the mechanical
power input is governed by governor and by
80
00:15:19,189 --> 00:15:26,189
supplementary control action which you normally
call a set point control. Now the governors
81
00:15:30,540 --> 00:15:37,540
sense the speed deviation and actuate the
turbine walls. So that more steam enters when
82
00:15:40,279 --> 00:15:47,279
the speed drops and when speed increases the
walls will be closed to decrease the speed
83
00:15:49,300 --> 00:15:56,300
decrease the speed by decreasing the steam
input to the machine.
84
00:15:56,499 --> 00:16:03,499
In practice, the speed deviation is very small
during the transient stability analysis until
85
00:16:09,009 --> 00:16:15,309
unless the machine looses synchronism, the
machine deviation is very the speed deviation
86
00:16:15,309 --> 00:16:22,309
is very very small and therefore governors
do not act okay and therefore, this assumption
87
00:16:23,990 --> 00:16:30,559
is a very very valid assumption in all stability
studies. We will assume this Pm to remain
88
00:16:30,559 --> 00:16:32,480
constant during the transient period.
89
00:16:32,480 --> 00:16:39,480
The second very important assumption is, we
neglect damping or asynchronous power in the
90
00:16:55,290 --> 00:17:02,290
stability studies and you can say the damping
or asynchronous power is negligible. In fact
91
00:17:04,140 --> 00:17:11,140
when the machine is in dynamic condition,
it develops a synchronous power as well as
92
00:17:12,399 --> 00:17:19,399
asynchronous power. This asynchronous power
results into damping, now the swing equation
93
00:17:20,179 --> 00:17:27,179
which I have derived, in this swing equation
we have not accounted for damping.
94
00:17:27,199 --> 00:17:34,199
Suppose, I want to modify this swing equation
including damping aspect then I have to add
95
00:17:34,750 --> 00:17:41,750
one more term here and that becomes minus
D times d delta by dt. The damping torque
96
00:17:47,610 --> 00:17:53,370
will always act in a direction opposite to
the dragging torque, okay. So that is the
97
00:17:53,370 --> 00:17:58,440
accelerating torque at any instant of time
will now becomes Pm minus Pe minus D times
98
00:17:58,440 --> 00:18:05,440
d delta by dt but to simplify the analysis
we ignore this damping term and therefore,
99
00:18:09,440 --> 00:18:16,440
the second assumption is the damping or asynchronous
power is negligible.
100
00:18:16,720 --> 00:18:23,720
The third assumption which is made the synchronous
machine can be represented by constant voltage
101
00:18:32,710 --> 00:18:39,710
source behind a transient reactance. This
assumption is, is not very truly applicable
102
00:18:48,960 --> 00:18:55,960
however, for classical stability studies when
you make this assumption the results obtained
103
00:18:57,140 --> 00:19:04,140
are not very far away from the actual results.
As we will see when we talk about, the detail
104
00:19:07,549 --> 00:19:14,549
model of the synchronous generator right,
we can represent this synchronous generator
105
00:19:14,919 --> 00:19:21,919
model in more detail while for performing
classical stability analysis, this assumption
106
00:19:24,149 --> 00:19:31,149
can be made and this results into a very simplified
model or simplified representation of synchronous
107
00:19:31,799 --> 00:19:32,600
generator.
108
00:19:32,600 --> 00:19:38,990
However actually in our studies, we will develop
detail mathematical model of the synchronous
109
00:19:38,990 --> 00:19:45,990
generator and this assumption will need not
be made at that time. Another very important
110
00:19:49,470 --> 00:19:56,470
assumption which is made is the mechanical
angle of the synchronous machine rotor coincides
111
00:19:57,960 --> 00:20:04,960
with the electrical phase angle of the voltage
behind transient reactance, okay as here we
112
00:20:06,169 --> 00:20:13,169
have assumed that a synchronous machine can
be represented by a constant voltage behind
113
00:20:13,590 --> 00:20:20,590
transient reactance okay and further assumption
which is made is the phase angle of this voltage
114
00:20:21,940 --> 00:20:28,940
coincides with the rotor angle
115
00:20:37,169 --> 00:20:43,970
The last and the most important assumption
here is the synchronous power may be calculated
116
00:20:43,970 --> 00:20:50,970
from a steady state solution of the network
to which the machines are connected because
117
00:21:06,200 --> 00:21:13,200
in our swing equation, we have to compute
the Pe , that is the electrical power output
118
00:21:15,340 --> 00:21:22,340
from the machine this is the synchronous power
okay and for computing the synchronous power,
119
00:21:23,640 --> 00:21:30,640
we assume that the network is in steady state
okay, although when the rotors of the synchronous
120
00:21:32,880 --> 00:21:39,880
machines are oscillating right, the network
is not in steady state in the real sense that
121
00:21:41,950 --> 00:21:48,309
is the voltage and the frequency also is deviating
slightly depending upon the but since this
122
00:21:48,309 --> 00:21:54,010
quantity is very small as compared to the
power frequency and the computations can be
123
00:21:54,010 --> 00:22:01,010
conveniently done by assuming that the network
is in steady state condition.
124
00:22:09,130 --> 00:22:16,130
Now while classifying the power system stability
we have classified into 2 categories angle
125
00:22:18,610 --> 00:22:25,610
stability and voltage stability, angle stability
was further classified into different categories.
126
00:22:26,899 --> 00:22:33,899
Now, we will just look quickly, how we define
the small signal stability the small signal
127
00:22:42,120 --> 00:22:49,120
stability is the ability of the power system
to maintain synchronism under small perturbations,
128
00:22:49,169 --> 00:22:56,169
this is a important point, that is the machine
or synchronous machine has the ability to
129
00:23:01,279 --> 00:23:08,279
maintain synchronism, this is primary requirement
for stability under small perturbations or
130
00:23:09,220 --> 00:23:13,830
small disturbances.
131
00:23:13,830 --> 00:23:20,830
The small disturbances are small variation
in loads and generations, any power system
132
00:23:25,880 --> 00:23:32,880
loads keep on changing continuously therefore
this is one example of small disturbance,
133
00:23:38,090 --> 00:23:45,090
the second important aspect here is the disturbances
are considered to be sufficiently small for
134
00:23:45,590 --> 00:23:52,590
linearization of the system equations to be
permissible for purpose of analysis. As I
135
00:23:52,960 --> 00:23:59,960
have told you earlier the swing equation is
a non-linear differential equation, okay but
136
00:24:04,419 --> 00:24:11,419
when we consider small perturbations this
equation can be linearized around a nominal
137
00:24:12,440 --> 00:24:14,159
operating condition.
138
00:24:14,159 --> 00:24:21,159
Once the equation is a linear equation, we
can make use of linear control theory as well
139
00:24:31,000 --> 00:24:38,000
as we can get the closed form solution of
the differential equation once it is a non-linear
140
00:24:38,130 --> 00:24:45,130
differential equation I have to solve by applying
numerical technique however, if the differential
141
00:24:45,260 --> 00:24:49,730
equation is a linear differential equation,
we know the closed form solution for the equation
142
00:24:49,730 --> 00:24:56,730
and once the differential equations are linear
we can use a linear control theory to design
143
00:24:57,450 --> 00:25:04,450
controllers for the system. Now here, I would
like to take one example how do we linearize
144
00:25:09,809 --> 00:25:13,409
a non-linear differential equation.
145
00:25:13,409 --> 00:25:20,409
Our non-linear differential equations which
we will consider is, same as the swing equation
146
00:25:26,190 --> 00:25:33,190
derived okay. Now, when I say that we linearize
this equation around a nominal operating condition
147
00:25:46,330 --> 00:25:53,330
or point. The nominal operating condition
here can be characterized by the initial value
148
00:25:54,470 --> 00:26:01,470
of the angle delta. Let us say, machine is
operating where delta is equal to delta naught
149
00:26:01,529 --> 00:26:08,529
and we will assume here that Pe is the electrical
power is given by this equation Pmax sin delta
150
00:26:13,679 --> 00:26:20,419
okay.
151
00:26:20,419 --> 00:26:27,419
Now under steady state operating condition
Pm is equal to Pe that is Pmax sin delta O,
152
00:26:31,889 --> 00:26:37,840
that is I substitute the value of delta equal
to delta O, that will give me the electrical
153
00:26:37,840 --> 00:26:44,840
power output and under steady state condition
these 2 powers are equal. Now what we do is
154
00:26:46,620 --> 00:26:53,620
that we give an incremental change or we make
an incremental change in delta, let us say
155
00:26:55,080 --> 00:27:02,080
that delta is equal to delta naught plus delta
okay.
156
00:27:04,700 --> 00:27:11,700
Let us substitute, this in our equation, swing
equation it becomes 2H upon omega s d2 delta
157
00:27:22,669 --> 00:27:29,669
naught plus delta divided by dt2, Pm we assume
to be constant it remains Pm, Pmax sin I put
158
00:27:50,500 --> 00:27:57,500
here delta naught plus delta, okay. Now you
expand this equation and write down Pm minus
159
00:28:07,000 --> 00:28:14,000
Pmax sin delta O cos delta, okay plus plus
cos delta O, sin delta okay. Now let us look
160
00:28:52,539 --> 00:28:59,539
the at this equation and since we know that,
delta delta is a incremental change, small
161
00:29:00,679 --> 00:29:01,480
one.
162
00:29:01,480 --> 00:29:08,480
So that cos delta delta is very nearly equal
to 1, sin delta delta is very nearly equal
163
00:29:17,519 --> 00:29:24,519
to delta delta good okay. We will make this
substitution in this equation now let us write
164
00:29:42,500 --> 00:29:49,500
down what is the value of this derivative.
The derivative of delta O is 0 and therefore
165
00:29:50,600 --> 00:29:57,600
this derivative term becomes d2 delta delta
divided by dt two equal to Pm minus Pmax into
166
00:30:10,179 --> 00:30:17,179
sin delta O cos delta delta we have made it
one, okay and I put plus here plus cos delta
167
00:30:30,490 --> 00:30:37,490
O delta delta
168
00:30:40,210 --> 00:30:47,210
Now Pm minus sine P max sin delta O, Pmax
sin delta O is equal to Pm therefore this
169
00:30:48,470 --> 00:30:55,470
becomes now equal to minus Pm cos delta O
delta delta okay or now I can write down my
170
00:31:08,710 --> 00:31:15,710
differential equation as 2H upon omega s,
d2 delta delta by dt2 plus Pm. Pmax right,
171
00:31:25,909 --> 00:31:32,909
we will write in Pmax here cos delta O, this
is also Pmax delta delta. This differential
172
00:31:53,720 --> 00:32:00,720
equation is a second order linear differential
equation, do you agree, why it is second order
173
00:32:04,309 --> 00:32:06,830
linear differential equation?
174
00:32:06,830 --> 00:32:13,830
Because, here we do not have delta delta or
any term, it is trigonometrical term or non-linear
175
00:32:20,399 --> 00:32:27,399
term. So far this term is concerned, this
coefficient is concerned this is a constant
176
00:32:29,539 --> 00:32:36,539
because Pmax is known, delta O is known Pmax
cos delta O is known. Pmax cos delta naught
177
00:32:54,950 --> 00:33:01,950
is known as the synchronizing power coefficient,
this term is known as synchronizing
power coefficient
178
00:33:27,090 --> 00:33:34,090
and may be denoted by the symbol Sp, that
is Sp is equal to P max cos delta naught.
179
00:33:42,889 --> 00:33:49,889
So that we can write down the swing equation
as 2H divided by omega s d2 delta delta by
180
00:33:56,279 --> 00:34:03,279
dt2 equal plus Sp delta delta.
181
00:34:09,370 --> 00:34:16,370
Further this is generally written in the form,
sorry is written in the form d2 delta delta
182
00:34:28,480 --> 00:34:35,480
by dt2 plus omega s, Sp divided by 2H delta
delta equal to 0, in case Sp is positive the
183
00:34:47,710 --> 00:34:54,710
solution of this equation will be similar
to, similar to the solution of a simple harmonic
184
00:35:00,340 --> 00:35:07,340
motion. It is going to be a sinusoidal, in
case Sp is negative then it gives you a exponentially
185
00:35:15,080 --> 00:35:21,280
increasing value of delta as a function of
time. Therefore primary requirement for the
186
00:35:21,280 --> 00:35:27,000
system to be stable here stable under small
perturbations is that this synchronizing power
187
00:35:27,000 --> 00:35:30,920
coefficient should be positive.
188
00:35:30,920 --> 00:35:37,920
Now, when we write the equation for simple
harmonic motion the equation is written as
189
00:35:39,470 --> 00:35:46,470
d2 x by dt2 plus omega n square x equal to
0, where omega n is the natural frequency
190
00:35:53,640 --> 00:36:00,640
of oscillation. Now here by comparing the
coefficients, we can write down that omega
191
00:36:02,210 --> 00:36:09,210
n is equal to square root of omega s, Sp divided
by 2H. This will be in radians electrical
192
00:36:20,570 --> 00:36:27,570
radians, we call it electrical radians per
second.
193
00:36:31,570 --> 00:36:38,570
Now further, you can see that the synchronizing
power coefficient is the slope of the power
194
00:36:41,840 --> 00:36:48,840
angle characteristic at the operating condition,
that is if you find out dPe divided by d delta
195
00:37:01,660 --> 00:37:08,660
it comes out to be Pmax cos delta O and therefore
dP by d delta is the slope of the power angle
196
00:37:19,610 --> 00:37:26,610
characteristic and our primary requirement
is that the synchronizing power coefficient
197
00:37:27,660 --> 00:37:34,660
should be positive for the system to be stable
under small disturbances, and that is why
198
00:37:42,160 --> 00:37:49,160
when I talk to you about the stable and non-stable
operating points, where this was our power
199
00:37:50,260 --> 00:37:56,150
angle characteristic P versus delta.
200
00:37:56,150 --> 00:38:03,150
This was your mechanical input line, these
are the operating condition and this is the
201
00:38:07,600 --> 00:38:12,630
stable operating condition because at this
point the slope of the power angle curve is
202
00:38:12,630 --> 00:38:19,630
positive. I shall take one small example to
find out what is the frequency of oscillations,
203
00:38:35,700 --> 00:38:42,700
which we normally come across.
204
00:38:44,810 --> 00:38:51,810
Let us take a case where a synchronous machine
as H equal to 5 mega joules per MVA. Further
205
00:38:58,270 --> 00:39:05,270
we assume that Pe equal to 2 sin delta expressed
in per unit 2 is your Pmax now. Let delta
206
00:39:23,880 --> 00:39:30,880
equal to delta naught equal to 60 degree,
okay. Then Sp comes out to be equal to 2 cos
207
00:39:39,670 --> 00:39:46,670
delta naught which is equal to 1 okay. Let
us take system frequency as 50 hertz, f equal
208
00:39:55,120 --> 00:40:02,120
to 50 hertz, omega n natural frequency of
oscillation of the rotor can be computed by
209
00:40:11,100 --> 00:40:18,100
substituting the values of small s, omega
s is 314 for 50 hertz system, Sp has come
210
00:40:20,510 --> 00:40:27,510
out to be equal to one synchronizing torque
coefficient 2 into 5H is 5.
211
00:40:32,490 --> 00:40:39,490
This quantity is square root of 31.4, if you
calculate it comes out to be equal to 5.6
212
00:40:45,530 --> 00:40:52,530
electrical radians per second. Frequency of
oscillation in hertz can be obtained as omega
213
00:41:01,450 --> 00:41:08,450
n by 2 phi which comes out to be in this case
0.89 hertz. This example is given here to
214
00:41:16,680 --> 00:41:23,680
give you the frequency of oscillations we
come across, system frequency is 50 hertz
215
00:41:25,530 --> 00:41:32,530
but when the rotors oscillate the frequency
of oscillation depends upon initial operating
216
00:41:34,440 --> 00:41:41,440
condition which determines the synchronizing
power coefficient, it depends upon the initial
217
00:41:42,040 --> 00:41:49,040
constant H of the machine and it also depends
upon the system frequency okay and it is of
218
00:41:49,300 --> 00:41:56,300
the order of 1 hertz in this case but it does
vary depending upon the value of this parameters
219
00:41:56,590 --> 00:42:03,590
but the variation is in the range of point
5 to 2 hertz normally.
220
00:42:12,060 --> 00:42:19,060
Now we have defined the small signal stability
and small signal stability is the ability
221
00:42:24,830 --> 00:42:31,830
of the synchronous machines to remain in synchronism
under small perturbations. Now when the synchronous
222
00:42:34,260 --> 00:42:41,260
machines rotors are oscillating it develops
an electrical power delta T, when the rotors
223
00:42:46,260 --> 00:42:53,260
are oscillating it produces a electrical torque
we call delta T. This torque can be decomposed
224
00:42:53,460 --> 00:43:00,460
into two components, one is called synchronizing
torque, another is called damping torque.
225
00:43:06,130 --> 00:43:13,130
The synchronizing torque is in phase with
speed deviation, I am sorry there is a correction
226
00:43:15,810 --> 00:43:22,810
it is in phase with phase with angle deviation
delta delta and the damping torque is produced
227
00:43:24,480 --> 00:43:31,480
in phase with speed deviation. when the rotor
is oscillating and if the oscillations are
228
00:43:32,750 --> 00:43:39,750
sinusoidal in nature then we can represent
this torque as a phaser in a plane where,
229
00:43:41,410 --> 00:43:48,410
delta delta is the x axis and delta omega
is the y axis
230
00:44:06,190 --> 00:44:13,190
Look here, actually in this diagram the plane
is x axis is marked as delta delta y axis
231
00:44:15,040 --> 00:44:22,040
is delta omega and in case, the delta T the
change in electrical torque lies in the second
232
00:44:28,040 --> 00:44:35,040
quadrant then you have 2 components, 1 component
is damping torque component, another is the
233
00:44:36,640 --> 00:44:41,920
synchronized torque component. Damping torque
component is positive while synchronizing
234
00:44:41,920 --> 00:44:48,920
torque component comes out to be negative
235
00:44:59,880 --> 00:45:06,880
In case delta T lies in the first quadrant
of the delta delta delta omega plane, then
236
00:45:13,780 --> 00:45:20,780
both synchronizing torque and damping torques
are positive for stability of the system both
237
00:45:28,530 --> 00:45:35,530
synchronizing and damping torque should be
positive. Okay I will discuss these aspects
238
00:45:35,770 --> 00:45:42,770
slight in more detail when we talk about the
small signal stability problem.
239
00:45:54,360 --> 00:46:01,360
The small signal stability can be classified
into 2 categories, one is called non oscillatory
240
00:46:02,150 --> 00:46:09,150
instability not a stability instability, it
can be non oscillatory instability this occurs
241
00:46:10,800 --> 00:46:17,800
because of insufficient synchronizing torque,
another is oscillatory instability this is
242
00:46:19,860 --> 00:46:26,860
due to insufficient damping torque or by due
to unstable control action.
243
00:46:38,780 --> 00:46:45,780
Now we can see the response of the system
under small perturbations, under 2 conditions,
244
00:46:46,390 --> 00:46:52,990
one is that synchronous machine has constant
field voltage there is no voltage regulator
245
00:46:52,990 --> 00:46:58,220
acting, another case is where the synchronous
machine is provided with a voltage regulator
246
00:46:58,220 --> 00:47:04,520
where we call this as one as a constant field
voltage another is with excitation control
247
00:47:04,520 --> 00:47:11,520
. Now suppose a situation is like this where
you have a negative synchronizing torque while
248
00:47:17,670 --> 00:47:24,670
positive damping torque the response of the
rotor to small perturbation will be of this
249
00:47:26,390 --> 00:47:33,390
form, starts from 0 and these oscillations
are oscillations are like damp because damping
250
00:47:33,640 --> 00:47:38,790
is positive but since synchronizing torque
coefficient is negative it increases monotonically.
251
00:47:38,790 --> 00:47:45,790
Therefore, this represents a case of insufficient
synchronizing torque.
252
00:47:47,380 --> 00:47:54,380
Here is another case where synchronizing torque
coefficient is positive but damping torque
253
00:47:56,850 --> 00:48:03,850
coefficient is negative, the response of the
rotor or del to perturbation starts from 0,
254
00:48:06,370 --> 00:48:13,370
you can see actually that the oscillations
are growing. Okay, therefore this is the case
255
00:48:14,890 --> 00:48:21,890
of oscillatory instability and this is primarily
due to lack of damping torque or here I can
256
00:48:27,260 --> 00:48:34,260
say that the system has negative damping.
In any power system, we will come across different
257
00:48:49,520 --> 00:48:56,520
modes of oscillations, a power system is a
large interconnected system okay.
258
00:48:57,700 --> 00:49:04,700
We come across a local modes, inter area modes,
control modes and torsional modes. The local
259
00:49:06,690 --> 00:49:13,690
modes are the one where a synchronous machines
located in a power plant oscillate with respect
260
00:49:14,580 --> 00:49:21,580
to rest of the system and the frequency is
in the range of .7 to 2 hertz. We have been
261
00:49:24,150 --> 00:49:31,150
taken the example we found that frequency
came out to .89 okay. Similarly, another case
262
00:49:31,540 --> 00:49:38,540
comes where a group of machines of a power
system oscillate with another group of machines
263
00:49:40,220 --> 00:49:47,220
connected by a weak tie line and these group
of machines will form one area another group
264
00:49:49,940 --> 00:49:56,860
forms, another area and therefore we say this
is inter area oscillation and when the group
265
00:49:56,860 --> 00:50:01,760
of machines are considered the inertia constant
H is large and the frequency of oscillation
266
00:50:01,760 --> 00:50:04,980
is in the range of point 2 to 1 hertz.
267
00:50:04,980 --> 00:50:11,450
So for control modes are concerned the frequency
of the different modes depend upon the actually
268
00:50:11,450 --> 00:50:17,960
control phenomena therefore, no specific frequency
is specified. The torsional modes is another
269
00:50:17,960 --> 00:50:24,960
phenomena which occurs in the steam turbine
and the different frequencies which we come
270
00:50:28,000 --> 00:50:35,000
across are for typicals mass system it comes
out be 16.3, 24.1, 30.3 and 44 hertz.
271
00:50:39,430 --> 00:50:46,430
We will devote about two lectures to deal
with about the torsional modes in detail at
272
00:50:46,930 --> 00:50:53,930
that time it will be clear, how the different
masses which are mounted on the same shaft
273
00:50:53,990 --> 00:51:00,990
will result into different modes of sub synchronous
oscillations and I will conclude here, my
274
00:51:05,560 --> 00:51:12,560
today is presentation by summarizing what
we have done.
275
00:51:12,720 --> 00:51:19,720
We have addressed to 3pecific aspects, one
is inertia constant H its importance, second
276
00:51:25,050 --> 00:51:32,050
aspect we have considered here is the basic
assumptions which are made instability analysis
277
00:51:33,290 --> 00:51:38,970
and the third aspect we have studied is the
small signal, stability, its definitions ,we
278
00:51:38,970 --> 00:51:45,970
have also talked about how to linearize a
non-linear differential equations and how
279
00:51:47,300 --> 00:51:54,300
to compute the natural frequency of oscillations
through an example. We will cover rest of
280
00:51:57,460 --> 00:52:04,460
the ah aspects of the introduction to power
system stability in the next lecture.