Mechanical Dynamics, the Swing

Equation, Units
1.0 Preliminaries
The basic requirement for generator operation
is that they must remain in synchronism.
This means that all generators must have
mechanical speeds so as to produce the same
electrical speed.
Electrical speed and mechanical speed are
related as a function of the number of
machine poles, p, or pole-pairs, p/2.
If p=2, as in Fig. 1, then there is one magnetic
rotation for every one mechanical rotation,
i.e., the stator windings see one flux cycle as
the rotor turns once.

Fig. 1
1

If p=4, as in Fig. 2, there are two magnetic

rotations for every one mechanical rotations,
i.e., the stator windings see two flux cycles as
the rotor turns once.

Fig. 2
Therefore, the electrical speed, e, will be
greater than (if p4) or equal to (if p=2) the
mechanical speed m according to the
number of pole-pairs p/2, i.e.,
e

p
m
2

(1)

The adjustment for the number of pole-pairs

is needed because the electrical quantities
(voltage and current) go through one rotation
for every one magnetic rotation.
2

So to maintain synchronized electrical

speed (frequency) from one generator to
another, all generators must maintain
constant mechanical speed. This does not
mean all generators have the same
mechanical speed, but that their mechanical
speed must be constant.
All two-pole machines must maintain
m=(2/2)e=(2/2)377 =377rad/sec
We can also identify the mechanical speed of
rotation in rpm according to
N m m

rad 60 sec/ min

sec 2 rad/rev

(2)

Substituting for m from (1), we get:

N m e

2 rad 60 sec/ min

p sec 2 rad/rev

(3)

Using this expression, we see that a 2 pole

machine will have a mechanical synchronous
speed of 3600 rpm, and a 4 pole machine will
have a mechanical synchronous speed of
1800 rpm.

2.0 Causes of rotational velocity change

Because of the synchronism requirement, we
are concerned with any conditions that will
cause a change in rotational velocity.
But what is a change in rotational velocity?
It is acceleration (or deceleration).
What are the conditions
acceleration (+ or -)?

that

cause

To answer this question, we must look at the

mechanical system to see what kind of
forces that are exerted on it.
Recall that with linear motion, acceleration
occurs as a result of a body experiencing a
net force that is non-zero. That is,
a

(4)

F
m

where a is acceleration (m/sec2), F is force

(newtons), and m is mass (kg). Here, it is
important to realize that F represents the sum
4

of all forces on the body. This is Newtons

second law of motion.
The situation is the same with rotational
motion, except that here, we speak of torque
T (newton-meters), inertia J (kg-m2), and
angular acceleration A (rad/sec2) instead of
force, mass, and acceleration. Specifically,

(5)

T
J

Here, as with F in the case of linear motion, T

represents the net torque, or the sum of all
torques acting on the rotational body.
It is conceptually useful to remember that the
torque on a rotating body experiencing a force
a distance r from the center of rotation is
given by

T r F

(6)

where r is a vector of length r and direction

from center of rotation to the point on the

body where the force is applied, F is the

applied force vector, and the operation is
5

the vector cross product. The magnitudes are

related through
(7)

T rF sin

where is the angle between r and F . If the

force is applied tangential to the body, then
=90 and T=rF.

Lets consider that the rotational body is a

shaft connecting a turbine to a generator,
illustrated in Fig. 3.
TURBINE

GEN
SHAFT

Fig. 3
For purposes of our discussion here, lets
assume that the shaft is rigid (inelastic, i.e., it
does not flex), and lets ignore frictional
torques.
What are the torques on the shaft?
From turbine: The turbine exerts a torque in
one direction (assume the direction shown
in Fig. 3) which causes the shaft to rotate.
6

This torque is mechanical. Call this torque

Tm.
From generator: The generator exerts a
torque in the direction opposite to the
mechanical torque which retards the motion
caused by the mechanical torque. This
torque is electromagnetic. Call this torque
Te.
These two torques are in opposite directions.
If they are exactly equal, there can be no
angular acceleration, and this is the case
when the machine is in synchronism, i.e.,
(8)

Tm Te

When (8) does not hold, i.e., when there is a

difference
between
mechanical
and
electromagnetic
torques,
the
machine
accelerates (+ or -), i.e.,. it will change its
velocity. The amount of acceleration is
proportional to the difference between Tm and
Te. We will call this difference the accelerating
torque Ta, i.e.,
Ta Tm Te

(9)

The accelerating torque is defined positive

when it produces acceleration in the direction
of the applied mechanical torque, i.e., when it
increases angular velocity (speeds up).
Now we can ask our original question (page 4)
in a somewhat more rigorous fashion:

Given that the machine is initially

operating in synchronism (Tm=Te), what
conditions can cause Ta0?

There are two broad types of changes: change

in Tm and change in Te. We examine both of
these carefully.
1. Change in Tm:
a. Intentionally: through change in steam
valve opening, with Tm either increasing
or decreasing.
b. Disruption in steam flow: typically a
decrease in Tm causing the generator to
experience negative acceleration (it
would decelerate).
2. Change in Te:
8

a. Increase in load: this causes an

increase in Te, and the generator
experiences negative acceleration.
b. Decrease in load: this causes a
decrease in Te,
and the generator
experiences positive acceleration.
All of the above changes, 1-a, 1-b, 2-a, and 2b are typically rather slow, and the
generators turbine-governor will sense the
change in speed and compensate by changing
the steam flow appropriately.
There is a third way that Te can change, that
is not slow.
c.Faults: We discuss this in the next
section.
3.0 Generator under faulted conditions:
qualitative
Consider the circuit of Fig. 4.

V0

E
Fig. 4

Here, the voltage E represents the internal

voltage of a synchronous machine and the
voltage V0 represents the terminal voltage of
the machine. We are assuming balanced
conditions and therefore we utilize the perphase equivalent circuit for analysis of the
three phase machine.
Assuming a round-rotor machine, we may
apply S V I , express I in terms of the two
voltages using Ohms law, and then take the
real part to show that the steady-state real
power supplied at the machine terminals is
given by
*

Pe

EV
sin
X

(10)

Lets assume that a three-phase fault occurs

at the machine terminals, so that V=0.
10

Then clearly, by (10), Pe=0.

Recall that torque and power are related by
Te

Pe
m

(11)

And so if Pe=0, it must be the case also that

Te=0.
By (9), then Ta=Tm, which means that all
mechanical torque is being used to accelerate
the machine. This is a very severe situation in
that the machine will accelerate at a very
high rate.
Of course, faults at the machine terminals are
very
rare
(although
they
do
occur
occasionally). Most faults are not so severe in
that they occur somewhere in the network
rather than at the machine terminals. But
even for network faults, the voltage V at the
machine terminals is reduced in magnitude,
causing Pe and therefore Te to reduce, causing
an imbalance between Tm and Te and
therefore a non-zero accelerating torque Ta.

11

There are two main influences on the amount

of overspeed seen by a synchronous
generator under faulted conditions.

The amount of reduction in Te: The greater

is the electrical distance between the fault
point in the network and the machine
terminals, the less will be the reduction on
V, and consequently, the less will be the
reduction on Pe (see (10)) and also Te (see
(11)). The fault location is something we
cannot control of course. But there is
another way to prevent reduction in V, and
that is through excitation control. Todays
excitation systems are very fast responding
so that terminal voltage reduction is
sensed and field current is boosted within
just a few cycles following a faulted
condition.

Minimize the amount of time that Te is

reduced: This is achieved by removing the
faulted condition very quickly. EHV
protection systems are typically able to
12

sense and clear a fault within 4 cycles

(4/60=.0667 seconds).

This discussion shows that the mechanical

dynamics associated with the acceleration of
the generator is intimately related to the
effect on Te of the fault. Such effects can only
be properly ascertained by analysis of the
network before, during, and after the faulted
condition.
In the next section, we will therefore derive
the relationship between the mechanical
dynamics and the electric network.
4.0 Derivation of swing equation
We begin with
convenience.
A

(5),

repeated

here

for
(5)

T
J

where we recall that T is the net torque on

the rotating body.
We will write the angular acceleration in
terms of the angle , which is here defined as
13

the absolute angle, in radians. It gives the

position of a tic-mark on the shaft relative to a
fixed point on the generator frame as
illustrated in Fig. 5.

aphase

Fig. 5
We

can express angular acceleration as

A d / dt , i.e., angular acceleration is the 2 nd
time derivative of . Noting that Ta is the net
torque on the turbine-generator shaft, we
have
2

(12)

T
a
J

Here, J is the moment of inertia of the

combined turbine-generator set, in kg-m2.
14

We also define R as the rated mechanical

angular velocity of the shaft, in rad/sec and
note that
R

e
p/2

(13)

where p is the number of poles, as before.

This allows us to define a synchronously
rotating reference frame as:
ref R t

(14)

where is the initial angle at t=0 and allows

us to position our reference frame wherever it
might be convenient for us. We will position it
so that it is numerically equal to the angle of
the
magnetomotive
force
(mmf)
corresponding to the a-phase terminal voltage
v (t )
( v (t ) is the time-domain representation
corresponding to the phasor we have called V
)1. This mmf is composed of the mmf
produced by the rotor and the mmf produced
a

We assume here that the a-phase terminal voltage differs from the internal voltage only by
the effect of armature reaction, i.e., we neglect leakage flux and armature winding
resistance. More discussion on this topic is provided in the book by Machowski, Bialek, and
Bumby, Power System Dynamics and Stability, (1997) pp 61-63.

15

by all three phase currents (typically called

the mmf of armature reaction).
Note from (14), that
(15)

ref R

The implication of (15) is that the reference

speed is constant, no matter what happens to
the rotor.
Lets define the rotor mechanical torque
angle, m. This is the angle by which the rotor
leads the synchronously rotating reference.
Since the rotor is in phase with the mmf it
produces, this angle is also the angle of the
rotor mmf.
Conveniently, recall that each (time varying)
mmf is assumed to induce a voltage in the aphase windings. These voltages are denoted
by e(t) and va(t) (or by their phasor
representations E and V ). Since
The voltages each lag their respective mmfs
by 90,

16

 The rotor mmf leads the a-phase terminal

voltage by m,
then

leads

by m.

Since the rotor position is designated by ,

and the reference position is designated by
ref, we have that
(16)

m ref

The relation between the three defined angles

is illustrated in Fig. 6.

ref

aphase

Fig. 6
From (16) we can write that
ref m

17

(17)

Substituting (14) into (17) yields

Rt m

(18)

Observe that under steady-state conditions,

increases linearly with time in exactly the
same way that ref increases with time, and
therefore m is constant, and so we might
rewrite (18) as
(t ) Rt m

(18a)

However, under transient conditions, because

of rotor acceleration, m = m(t), and we can
rewrite (18) as
(t ) Rt m (t )

(18b)

Considering the transient condition, by taking

the first derivative of (18b), we have:
(t ) R m (t )

(19)

Differentiating again results in

(t ) m (t )

(20)

Substituting (20) into (12), repeated here for

convenience,
18

(12)

T
a
J

results in
T
m (t ) a
J

Jm (t ) Ta

(21)

We observe at this point that all of what we

have done is in mechanical radians, and
because we have focused on a 2-pole
machine, the angles in electrical radians are
the same. However, we want to accommodate
the general case of a p-pole machine. To do
so, recall (1), repeated here for convenience:
e

p
m
2

(1)

Differentiating, we have
e (t )

p
m (t )
2

(22)

which is just
p
e (t ) m (t )
2

Substitution of (23) into (21) results in

19

(23)

2
e (t ) Ta
p

(24)

From now on, we will drop the subscript e

on and with the understanding that both
are given in electrical radians. Therefore (24)
becomes:
2 J
(t ) Ta
p

or since

(25)

,
2J
(t ) Ta Tm Te
p

(26)

Equation (26) is one form of the swing

equation. We shall derive some additional
forms in what follows.
5.0 A second form of the swing equation
Because power system analysis is more
convenient in per-unit, lets normalize (26) by
dividing by a base torque chosen to be
TB

SB3
R

(27)

where SB3 is a chosen 3-phase MVA rating.

Dividing both sides of (26) by TB results in
2 JR
T
(t ) a Tau
pS B 3
TB
20

(28)

We can express the kinetic energy WK of the

turbine-generator set, when rotating at R, as
Wk

1
JR2
2

(29)

where the units are watt-seconds or joules.

Solving (29) for J results in
J

2Wk
R2

(30)

Substituting (30) into (28) yields

2

2Wk

R2 R
(t ) Tau
pS B 3

(31)

Simplifying:
4Wk
(t ) Tau
pS B 3 R

(32)

Lets write one of the 2s in the numerator as

a in the denominator, and group it with p
and R, yielding
2Wk
(t ) Tau
p

SB3
R
2

(33)

Recalling that R is the mechanical reference

speed, the reason for the last step is
apparent, because we can now identify what
21

is inside the brackets in the denominator as

the electrical reference speed, which we can
denote as Re. This would be, in North
America, 377 rad/sec. Thus, (33) becomes
2Wk
(t ) Tau
S B 3Re

(34)

Now define the inertia constant:

H

Wk
SB3

(35)

Here, when SB3 has units of MVA, and Wk has

units of MW-sec (or Mjoules), then H has units
of MW-sec/MVA or seconds.
When SB3 is chosen as the generator MVA
rating, H falls within a fairly narrow range. I
have pulled some numbers from the Appendix
D of your text to illustrate.
Unit

Smach

WK

Hmach

Hsys

=Wk/Sma =Wk/Ssys
ch

H1

23.5
22

2.61

Ssys=10
0
0.235

H9

86

233

2.71

2.33

615

3166

5.15

31.7

25

125.4

5.02

1.25

F11

270

1115

4.13

11.15

F21

911

2265

2.49

22.65

CF1-HP

128

305

2.38

3.05

CF1-LP

128

787

6.15

7.87

N1

76.8

281.7

3.67

2.82

N8

1340

4698

3.51

47.0

SC1

25

30

1.2

0.3

SC5

75

89.98

1.2

0.9

H18
F1

Notes:
1. On machine base, H ranges 1-7, but on
system base, it ranges w/ machine size.
2. Cross-compound machines (side-by-side
turbines, same steam, different gens) have
a high LP H because of large blades required
by low pressure steam.
23

3. Synchronous condensers have no turbine

and therefore small H.
6.0 A third form of the swing equation
Recall (34):
2Wk
(t ) Tau
S B 3Re

(34)

Substitution of Wk=HSB3 (from (35)) results in

2 HS B 3
(t ) Tau
S B 3Re

(36)

2H

(37)

or
Re

(t ) Tau

Equation (37) is equation 2.17 in our text.

Some clarifications:
A.

Comments on Re:

1. It
is
the
rated
radian/frequency (377).

electrical

2. Your text is confusing on this.

Use of R in eq. 2.13, 2.14, 2.15,
2.17: it should be Re.
24

B.
H must be given on the same base as
SB3 used to normalize the right-hand side
torque.
C.
You can convert Hs from one base to
another as follows:
Wk H mach S mach H sys S sys

H sys H mach

S mach
S sys

(38)
(39)

7.0 Comments on Inertia

We make three additional comments about
representing inertia in the swing equation.
7.1 Use of M for inertia
Another quantity often used in the literature
for inertia is M, the angular momentum at
rated speed, where
(40)

M J R

We can see how M is related to the kinetic

energy according to the following. The kinetic
energy of the rotor at speed R is
Wk

1
J R2
2
25

(41)

Solving for J in (40),

J

(42)

M
R

Substituting (42) into (41) yields

Wk

1 M 2 1
R M R
2 R
2

(43)

Wk
SB3

(35)

Also, from

and substituting (43) into (35) results in

1
M R
H 2
S B3

(44)

which, when solved for M, results in

M

2 HS B 3
R

(45)

Two additional issues to note here:

1. A different M is sometimes used in the
literature to denote the mechanical starting
time. This is the total time required to
accelerate the unit from standstill to rated
speed R if rated torque (Tau=1.0) is applied
as a step function at t=0. I will denote this
26

as T4, nomenclature that is consistent with

the Anderson & Fouad text (see page 450).
Kundur, in his book on page 132, shows that
this time, in seconds, is given by
T4 2 H

(46)

where H is given on the machine base.

2. The three constants M, H, and Wk, are

defined at the particular angular velocity of
R. However, the machine speed m does
deviate from R during the transient
conditions for which we are interested to
study. Therefore, to be rigorous, we should
define M, H, and Wk relative to the machine
speed m(t) so that M, H, and Wk vary with
time.
However,
this
considerably
complicates the swing equation, and does
so with negligible improvement in accuracy,
since m, although time varying during
disturbance conditions, does not deviate
much from R. On the other hand, the
moment of inertia J is an actual constant,
27

i.e., it is a function of only the machine

geometry and mass and does not depend on
speed.

7.2 W-R Squared

Another constant that is often used by
manufacturers (and it will be, usually, what
you get from a manufacturer) is the W-Rsquared, denoted WR2, which is the moment
28

of inertia expressed in English units of lb(m)ft2:

WR 2 [mass of rotating parts][radius of gyration]2

lb(m)

ft

(47)

The radius of gyration is a root-mean-square

average distance of all parts of the rotating
object from its axis of rotation.
The conversion of units may be obtained so
that the moment of inertia, J, in kg-m2, is
0.4536 kg 0.3048 m
J WR lb(m)ft

lb(m)
ft

0.0421 WR 2

(48)

We may also relate the kinetic energy at rated

speed, Wk, to WR2, by substituting (48) into
the expression for Wk:
Wk

1
1
J R2 (0.0421WR 2 ) R2 0.2105(WR 2 )( R2 )
2
2

(49)

where Wk is given in joules. If we wanted to

write (49) as a function of RPM instead of
rad/sec, where r=2nR/60,
Wk 0.2105(WR 2 )(2nR2 / 60) 2.31 10 4 (WR 2 )(nR2 )

(50)

where again Wk is in joules (this is the same

as the equation at the top of pg 16 in your
29

text). Expressing Wk in Mjoules=MW-sec, eqts.

(49) and (50) become
Wk 2.105 10 8 (WR 2 )( R2 ) 2.31 10 10 (WR 2 )( nR2 )

(51)

7.3 Summing up
Remember, inertia should account for all
masses on the shaft. This will always be the
turbine and generator, but it may or may not
include an exciter (depends on whether the
machine utilizes a rotating exciter or not and
whether that rotating exciter is mounted on
the same shaft or not).
Also remember that we have five forms in
which inertia can be expressed:
J, Wk, H, M, and WR2
You should be able to convert from any one
form to any other form.

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