Chapter 3. Generator and Transformer Models;
The Per-Unit System
3.1 Introduction
1. Use of the "per phase" basis to represent a three-phase balanced system.
2. Use the π model to describe transmission lines (see Chapter 5).
3. Simple models for generators and transformers to study steady state balanced
operation.
4. One-line diagrams to represent a three phase system.
5. The per-unit system and the impedance diagram on a common MVA base.
3.2 Synchronous Generators
•
Note the coils aa', bb', and cc' are 120o apart. Axis of a-coil is the x-axis as shown.
• These are
Fr
"concentrated" full
pitch windings. In
δr
real machines the
a
Fsr
windings are
n
distributed among
many slots, and often
c'
b'
are not full pitch.
ωt
• We assume the
ψ
r
r'
windings produce a
sinusoidal mmf
ψ
around the rotor
periphery (angle
F
around the airgap
s
c
b
with respect to
winding axis) .
• Assume the rotor is
m
a'
excited with DC
current I f producing
a flux φ which
rotates with the rotor
at speed ω . At time t the rotor would have moved an angle ωt .
The flux linkage for an N-turn a-winding will be maximum ( Nφ ) at ω t = 0 . The
flux linkage with the a-winding will go to 0 at ω t = π / 2 . The flux linkage will go to
( − Nφ ) at ω t = π , and back to 0 at ω t = 3π / 2 . When ω t = 2π the flux linkage will
Rotor position shown at
•
w t = p /2
be back to ( Nφ ) . Thus a whole cycle… This repeats every revolution of the rotor.
•
•
The flux linkage with winding "a" is thus given by: λa = Nφ cos ωt
Using Faraday's law, the voltage induced in phase "a" is given by:
dλ
= ω N φ sin ω t
dt
= Emax sin ωt
ea = −
π

= Emax cos ω t − 
2

Where Emax = ω Nφ = 2π fNφ , since 2π / 2 ≈ 4.44 , the rms value of the generated
voltage is given by:
E = 4.44 fNφ
Fr
a
δr
Fsr
n
c'
b'
r
ωt
ψ
r'
The frequency is a
function of speed and the
number of poles, thus:
P n
f =
where n is
2 60
the speed in rpm (the
synchronous speed) and
P is the number of poles
(always an even
number).
ψ
If the phase "a" is
connected to a load, then
Fs
b
c
a current ia will flow.
Depending on the load,
this current will have a
m
a'
phase angle, say ψ (see
figure) lagging the
generated voltage ea
which is along the xRotor position shown at w t = p /2
axis. Again, this is
shown in the figure as the line "mn". N.B. The line "mn" is attached to and rotates with
the rotor. Be careful also that the "Flux" vectors are spatial while the voltage and current
vectors are phasors (in time).
The same is true for phases "b" and "c" but they will "lag" the voltage in phase "a" by
120 and 240 degrees respectively.
2
Since ea ∝ sin (ωt ) , then we have:
ia = Imax sin ( ωt − ψ )
ib = I max sin ( ω t −ψ − 120o )
ic = I max sin ( ωt − ψ − 240o )
Since mmf is proportional to the current, we then have:
Fa = Fm sin ( ωt − ψ )
Fb = Fm sin ( ω t −ψ − 120o )
Fc = Fm sin ( ωt − ψ − 240o )
We now take components of these phasors along the line "mn" and in quadrature with it.
Along "mn" we have:
F1 = Fm sin ( ω t −ψ ) cos ( ωt − ψ ) + Fm sin (ω t −ψ − 120o ) cos ( ω t −ψ − 120o ) +
Fm sin (ω t − ψ − 240o ) cos (ω t −ψ − 240o )
1
Using the identity sin α cos α = sin ( 2α ) the above equation becomes:
2
F1 =
Fm 
sin 2 (ωt − ψ ) + sin 2 ( ω t −ψ − 120o ) + sin2 ( ωt − ψ − 240o ) 

2
It is noted that this expression is the sum of three balanced phasors hence is equal to zero.
Next we consider the components of the mmf perpendicular to "mn":
F2 = Fm sin ( ω t −ψ ) sin ( ω t −ψ ) + Fm sin ( ωt − ψ − 120 o ) sin ( ωt − ψ − 120o ) +
Fm sin (ω t − ψ − 240 o ) sin ( ωt − ψ − 240o )
Using the identity sin 2 α =
F2 =
1
(1 − cos2α ) we have:
2
{
}
Fm 
3 − cos2 (ωt −ψ ) + cos2 (ω t − ψ − 120o ) + cos2 (ω t − ψ − 240o ) 


2
The three cosine terms add to zero (balanced phasors) thus we have:
3
FS =
3
Fm
2
Thus the result of having three pulsating single phase fluxes produce (when applied
symmetrically) a constant flux perpendicular to line "mn" and rotates at the same speed
n as the rotor. This flux is called the "armature reaction" to the field of the rotor Fr . the
various fields are shown for one phase (say phase "a") in the diagram below: Note that Fs
is perpendicular to line "mn" and rotates with it at the same speed. The fact that Fs
(which is proportional to Ia) is perpendicular to line "mn" indicates that that we can
theorize that the armature current Ia is producing a reactive voltage drop parallel to line
"mn" due to inductance in the machine.
Run the command "rotfield" in a Matlab window to see a demo of the rotating field.
Fr
Fsr
n
Ear
Esr
Fs
jX ar Ia
V
R a Ia
m
E
jX l Ia
Ia
Figure 3.2 Page 53
θ is the angle between V and I a it is the power factor angle
δ is the angle between E and V it is the power angle
First the rotor field Fr produces the no-load generated voltage E (at zero armature
current). Note that E lags Fr by 90 degrees. E is called the "excitation" voltage. It is
directly proportional to the field current.
The voltage/current phasors for phase "a" are shown above lagging the flux diagram by
90 deg. Note the above diagram is a "hybrid" combining spatial and temporal vectors.
4
Second, assume the armature carries a current to a load, now the armature reaction flux Fs
is produced. This is perpendicular to line "mn". The two fluxes (due to rotor and
armature) combine together to form the "resultant" flux Fsr. The resultant flux induces
the generated on-load emf Esr. The armature mmf Fs induces the voltage Ear known as
the armature reaction voltage.
In all cases, each mmf produces a voltage lagging the mmf by 90 degrees.
Note that the voltage Ea r leads Fs (hence Ia) by 90 degrees. Thus we can theorize an
inductor model for this relationship with reactance Xar i.e. Ear = jX ar I a . Xar is known as
the reactance of armature reaction. Thus we now have the circuit equation:
E = Esr + jX ar I a
The terminal voltage V is found by considering the armature resistance and leakage
reactance thus:
E = V + Ra + j ( X l + X ar ) I a
Which can be simplified to:
E = V + [ Ra + jX s ] I a
where X s = X l + X ar is known as the synchronous reactance.
The angle between V and Ia is θ the power factor angle. The power angle is the angle
between E and V called δ. The circuit model of the machine is as shown below:
The resistance Ra is much smaller than
the synchronous reactance Xs and is often
neglected. This is shown in the simplified
one line diagram below where the
machine is shown connected to an infinite
E
jX s
Ia
V
Ra
Ia
jX s
+
+
E
V
-
-
Load
bus. An infinite bus has a voltage V (usually assumed at an angle zero) which does not
change no matter how the generator is controlled or excited.
A figure of merit of a generator is the percent voltage regulation VR defined as:
VR =
Vnl − Vrated
E − Vrated
×100 =
×100
Vrated
Vrated
5
The smaller the VR the better. In a real generator, the regulation is a few percent.
3.3 Steady-State Characteristics, Cylindrical Rotor
Power factor control. Most generators are connected to a large power grid. This is an
infinite bus (its voltage, angle and frequency are constant). Assuming the generator has
small leakage reactance and small armature resistance, then its model is shown below:
We have two ways of calculating the power (per
phase):
E
jX s
V
Ia
1. P1φ = V I a cosθ
2. P1φ =
E V
sin δ
Xs
Equating these two equations we have E sin δ = X s I a cos θ . Assuming the power is
constant, then from equation 1 we have: I a cos θ = const. This locus is shown as the
vertical dashed line in the figure below.
From equation 2, assuming the power is a
constant we have: E sin δ = const . This locus is
the horizontal dashed line shown.
I a3
E3
E2
E1
Note that for this load, the minimum armature
δ
I
V
current is I a 2 when the power factor of the
q
generator is unity. Note that E2 is directly above
I
the voltage V . Thus I a 2 and V are in phase,
producing unity power factor. If the excitation is
increased, the emf of the generator is increased to
some value say E1 as shown. Clearly the voltage of the generator leads the current hence
it is like an inductor consuming vars (lagging power factor). On the other hand, if the
excitation is reduced below that for unity power factor, the emf of the generator is
smaller, say E3 which lags I a 3 . Now the power factor of the generator is "leading", i.e.
it is like a capacitor (current leading voltage). A plot of the variation of I a as a function
of field current is known as a V-curve. A family of such curves, one at each load is
known as the V-curves of the generator. The minimum point on each curve corresponds
to unity power factor. To the right (higher excitation) is the lagging power factor area
and to the left is the leading power factor zone.
1
a2
1
a1
6
Power angle characteristics. Consider the machine connected to an infinite bus and
neglecting armature resistance, the machine model is shown below:
E
S 3φ = 3VI a*
But we have Ia =
jX s
Ia
V
E ∠δ − V ∠0
thus for complex power:
Z s ∠γ
2
E V
V
S 3φ = 3
∠γ − δ − 3
∠γ thus the real and imaginary powers are given by:
Zs
Zs
2
EV
V
P3φ = 3
cos (γ − δ ) − 3
cos γ
Zs
Zs
2
E V
V
Q3φ = 3
sin (γ − δ ) − 3
sin γ
Zs
Zs
and if we neglect the armature resistance we have: (Note that γ = 90o and Z s → X s )
EV
V
sin δ and Q3φ = 3
( E cos δ − V ) . Thus the real power is a function of
Xs
Xs
the power angle δ and varies as sin δ . The maximum power is given by:
E V
Pmax(3φ ) = 3
.
Xs
The angle δ starts at zero when the machine does not deliver any power. As the
machine is loaded, the power delivered increases and the angle δ increases. As the angle
δ reaches 90 o , maximum power is reached. If prime mover power is increased beyond
this point, the machine will speed up and loose synchronism. This means the tie between
the machine and the power grid can handle only the maximum power shown above. In
real operation, the power angle is very small of the order of 10 degrees and less. The
power angle δ can "swing" by larger amounts under transient conditions.
P3φ = 3
The equation for Q shows that Q3φ ; 3
V
Xs
( E − V ).
Thus if E > V the generator
delivers reactive power and the generator is said to be overexcited. If E < V , the
generator is underexcited and the reactive power delivered to the bus is negative, thus the
generator is receiving reactive power from the bus. Generators are usually operated in
the overexcited mode since they need to supply the system with reactive power: most our
domestic loads are reactive.
Example 3.1
7
A 50-MVA, 30-kV, three phase, 60 Hz synchronous generator has Xs = 9 Ω per phase
and a negligible resistance. The generator is delivering rated power at 0.8 power factor
lagging at the rated terminal voltage to an infinite bus.
(a) Determine the excitation voltage per phase E and the power angle δ.
(b) With the excitation held constant at the value found in (a), the driving torque is
reduced until the generator is delivering 25 MW. Determine the current and the
power factor.
(c) If the generator is operating at the excitation voltage of part (a), what is the steadystate maximum power the machine can deliver before losing synchronism? Also, find
the armature current corresponding to this maximum power.
The Matlab program follows:
Ra=0; Xs = 9; Zs = Ra + j*Xs;
S = 40 + j*30; V = 30/sqrt(3); % MW and kV
Ia1 = conj(S)*1000/(3*conj(V)); % Amp
Ia1M = abs(Ia1), Ia1ang=angle(Ia1)*180/pi
E1 = V + Zs*Ia1*0.001; % kV
E1M = abs(E1), delta1 = angle(E1)*180/pi % kV, deg
disp('(b)')
P = 25; % MW
delta2 = asin(P*Xs/(3*abs(E1)*V)); delta2d = delta2*180/pi
E2 = E1M*(cos(delta2) +j*sin(delta2)); % kV
Ia2 = (E2 - V)*1000/Zs; Ia2M = abs(Ia2), Ia2ang=angle(Ia2)*180/pi
PF = cos(angle(Ia2))
disp('(c)')
Pmax = 3 *E1M*V/Xs % kW
E3 = E1M*(cos(pi/2) +j*sin(pi/2)); % kV
Ia3 = (E3 -V)*1000/Zs; % A
Ia3M = abs(Ia3), Ia3ang=angle(Ia3)*180/pi
PF = cos(angle(Ia3))
Ia1M =
962.2504
Ia1ang =
-36.8699
E1M =
23.5584
delta1 =
17.1027
(b)
delta2d =
10.5914
Ia2M =
807.4918
Ia2ang =
-53.4284
PF =
0.5958
(c)
Pmax =
136.0147
8
Ia3M =
3.2489e+003
Ia3ang =
36.3239
PF =
0.8057
Example 3.2
The generator of example 3.1 is delivering 40 MW at a terminal voltage of 30 kV.
Compute the power angle, armature current, and power factor when the field current is
adjusted for the following excitations:
(a) The excitation voltage is decreased to 79.2 percent of the value found in example 3.1.
(b) The excitation voltage is decreased to 59.27 percent of the value found in example
3.1.
(c) Find the minimum excitation below which the generator will lose synchronism.
The Matlab program follows:
Ra=0; Xs = 9; Zs = Ra + j*Xs;
V = 30/sqrt(3); P = 40;
E1M = 0.792*23.558;
delta1 = asin(P*Xs/(3*abs(E1M)*V)); delta1d = delta1*180/pi
E1 = E1M*(cos(delta1) +j*sin(delta1));
Ia1 = (E1 - V)*1000/Zs; Ia1M = abs(Ia1), Ia1ang=angle(Ia1)*180/pi
PF = cos(angle(V/Ia1))
disp('(b)')
E2M = 0.5925*23.558;
delta2 = asin(P*Xs/(3*E2M*V)); delta2d = delta2*180/pi
E2 = E2M*(cos(delta2) +j*sin(delta2));
Ia2 = (E2 - V)*1000/Zs; Ia2M = abs(Ia2), Ia2ang=angle(Ia2)*180/pi
PF = cos(angle(Ia2))
disp('(c)')
E3M = P *Xs/(3*V*1)
E3=E3M*(cos(pi/2)+j*sin(pi/2));
Ia3 = (E3 -V)*1000/Zs;
Ia3M = abs(Ia3), Ia3ang=angle(Ia3)*180/pi
PF = cos(angle(Ia3))
delta1d =
21.7975
Ia1M =
769.8005
Ia1ang =
-0.0283
PF =
1.0000
(b)
delta2d =
29.7592
Ia2M =
962.7210
9
Ia2ang =
36.9072
PF =
0.7996
(c)
E3M =
6.9282
Ia3M =
2.0728e+003
Ia3ang =
68.1986
PF =
0.3714
Notes: in (a) the generator is operating at unity power factor. In both (b) and (c) the
generator is under-erexcited and is receiving reactive power (power factor of load is
leading, i.e. capacitive load and the generator is inductive.) In the real world the load is
usually inductive (lagging power factor) and the generator is over-excited, delivering
reactive power.
Example 3.3
For the generator of example 3.1, construct the v-curve for the rated power of 40 MW
with varying field excitation from 0.4 power factor leading to 0.4 powr factor lagging.
Assume the open-circuit characteristic in the operating reagion is given by E = 2000If.
The Matlab program follows:
P = 40;
% real power, MW
V = 30/sqrt(3)+ j*0;
% phase voltage, kV
Zs = j*9;
% synchronous impedance
ang = acos(0.4);
theta=ang:-0.01:-ang;
% Angle from 0.4 leading to 0.4 lagging pf
P= P*ones(1,length(theta));
% generates P array of same size
Iam = P./(3*abs(V)*cos(theta));
% current magnitude kA
Ia = Iam.*(cos(theta) + j*sin(theta));
% current phasor
E = V + Zs.*Ia;
% excitation voltage phasor
Em = abs(E);
% excitation voltage magnitude, kV
If = Em*1000/2000;
% field current, A
plot(If, Iam), grid
xlabel('If, A'), ylabel('Ia, kA')
text(3.4, 1, 'Leading pf'), text(13, 1, 'Lagging pf')
text(9, .71, 'Upf')
10
2
1.8
1.6
Ia, kA
1.4
1.2
1
Leading pf
Lagging pf
0.8
Upf
0.6
2
4
6
8
10
If, A
12
14
16
18
Note that the over-excited generator (load is inductive, generator is capacitive) is the
normal way to operate the generator since most loads are inductive. If the load becomes
capacitive, then the generator is operated in the under-excited mode.
11