Lesson 33:
Synchronous Machines
Per phase equivalent circuit

The per phase equivalent can thus be drawn
Ias
+
+
vas
-
terminal voltage stator winding
(line-to-neutral)
resistance

vas
-
excitation or
generated voltage
synchronous
reactance
This circuit would constitute one phase of a
three-phase source (either Y or D connected).
Per phase equivalent circuit

This circuit would constitute one phase of a
three-phase source (either Y or D connected).
Y-connected source
D-connected source
Excitation voltage

The excitation (or generated) voltage is derived
from the term
P

Lsf I f e cos  mt 
2

so it has an rms value of
Lsf I f e
2

The angle of Ea is not necessarily 0°, because
it is related to the mechanical position. We will
call it d, the power (or torque) angle such that
Excitation voltage

Notice that the excitation voltage is a function of
 the
dc rotor field current If
 inductance Lsf which models flux linkage between the
rotor and stator windings (a fixed quantity)
 the electrical frequency e (a fixed quantity)

Therefore, if we wish to control the excitation
voltage (and the generator output voltage) we
can do so by controlling the dc field current If.
Real power

Grouping the real and imaginary components
VE
S1P  P1P  jQ1P  - s a sin(d ) 
Xs

 Vs2 Vs Ea

j
cos(d ) 
Xs
 Xs

The total real power PT dissipated by our
machine is given
P3P  3P1P  -3
Vs Ea
sin(d )
Xs
which is a function of the power angle d.
Power vs. angle d

Plotting power as a function of d we obtain the
following
Vs Ea
P3P  3P1P  -3
sin(d )
Xs

What does the power angle d physically
represent?
Power angle d = 0°

With no load, the rotor’s net magnetic field
aligns exactly with the stator’s rotating magnetic
field.
 With
no load, no power is being dissipated and torque
exerted on the rotor is zero.
stator field (F ) and
m
s
rotor field (Fr) align,
hence d = 0°
Fs
Fr
d = 0°
Power angle d < 0º (motor)

With a load, the rotor’s net magnetic field lags
behind the stator’s rotating magnetic field.
 The
angular difference results in a positive torque,
hence positive power dissipation.
m
rotor field (Fr) lags
stator field (Fs)
hence d < 0°
Fs
d = -20°
d = -20°
Fr
Power angle d > 0º (generator)

With a negative load (prime mover driving the
rotor), the rotor’s net magnetic field leads
stator’s rotating magnetic field.
 The
angular difference results in a negative torque,
hence negative power dissipation.
m
Fr
d = +20°
d = +20°
Fs
rotor field (Fr) leads
stator field (Fs)
hence d > 0°
Torque

Since we have assumed 100% efficiency the
electrical power must equal the mechanical
power at the shaft
Vs Ea
P3P  -3
sin(d )  Pmech  Tem
Xs
thus

Tdev 
Since
Pmech
m
2
P
m  e
we conclude that
Tdev  -
P 3Vs Ea
sin(d )
2 e X s
vas
Ns turns
-
Example Problem 1
I as
+
+
A 208-Vrms (line), 60E =Hz,
generator has
V = V Ð 0°
E Ð d Y-connected
_
Xs= 100 W, rs = 0 W. At rated terminal
voltage, the machine
delivers 1.5-Arms (line) to a 0.707 pf lagging load.
r
jX I
a. Determine the machine
current.
+
b. Determine the excitation
(or generator) voltage Ea.
+
V = V Ð 0°
E = L I w Ðd _
c. Determine the power (torque) angle d.
a
as
a
as
a
sf f
as
s
as
e
s
s
-
j100 Ω
I as = 1.5 A
+
Ea
+
_
V as = VsÐ 0°
n
Example Problem 1 (continued)
d. If the generator has 4 poles, what is the speed of the
prime mover.
e. Determine the developed torque.
f. Suppose the load current increases to 3-Arms at a 0.707
lagging pf. If we have the same terminal voltages, how
must the field current change?
Example Problem 2
An Arleigh-Burke Class destroyer has a 3-phase, Yconnected, 4-pole, 60 Hz synchronous generator rated to
deliver 3.75 MVA at a 0.8 lagging pf with a line voltage of
450 V. The machine stator resistance is negligible and the
synchronous reactance is equal to 0.04 Ω. The constant
Lsf = 0.06 H. The actual system load on the machine
draws 2 MW at 0.8 lagging pf. Assume that the voltage
regulator has automatically adjusted the field current to
match the terminal voltage to its rated value.
a. What is the rated speed?
b. Determine the reactive and apparent power delivered by
the machine.
Example Problem 2 (continued)
c. Find the current drawn from the machine using the
terminal voltage as the reference phasor.
d. Determine the excitation voltage and identify the power
angle.
e. Find the field current.
Example Problem 3
Consider the synchronous machine from our previous
example where rs ≈ 0 Ω, Xs = 0.04 Ω, Lsf = 0.06 H, and the
machine is operated in a system where the rated line
voltage is 450 V and the frequency of the system is 60 Hz.
Suppose we are given that the prime mover supplies 14
kN-m of torque to the rotor shaft and the exciter is
delivering 25 A DC to the field winding. From the previous
problem, we determined that the phase voltage amplitude
450 𝑉
was
= 259.8 V and the rated speed was 188.5 rad/s.
3
a. Determine the excitation voltage.
b. Find the current Ĩas
c. Compute how much real and reactive power the
machine is delivering to the system.