Chapter 4: AC Machinery Fundamentals
4-1. The simple loop is rotating in a uniform magnetic field shown in Figure 4-1 has the following
characteristics:
B = 0.5 T to the right r = 01
. m
l = 0.5 m ω = 103 rad/s
(a) Calculate the voltage etot ( t ) induced in this rotating loop.
(b) Suppose that a 5 Ω resistor is connected as a load across the terminals of the loop. Calculate the
current that would flow through the resistor.
(c) Calculate the magnitude and direction of the induced torque on the loop for the conditions in (b).
(d) Calculate the electric power being generated by the loop for the conditions in (b).
(e) Calculate the mechanical power being consumed by the loop for the conditions in (b). How does
this number compare to the amount of electric power being generated by the loop?
ωm
c r
d
N vab S
vcd
b
a
B
B is a uniform magnetic
field, aligned as shown.
SOLUTION
(a) The induced voltage on a simple rotating loop is given by
eind ( t ) = 2 rωBl sin ω t (4-8)
eind ( t ) = 2 ( 0.1 m )(103 rad/s)(0.5 T )( 0.5 m ) sin103t
eind ( t ) = 5.15 sin103t V
(b) If a 5 Ω resistor is connected as a load across the terminals of the loop, the current flow would be:
eind 5.15 sin 103t V
i (t ) = = = 1.03 sin 103t A
R 5Ω
(c) The induced torque would be:
τ ind ( t ) = 2 rilΒ sin θ (4-17)
τ ind ( t ) = 2 ( 0.1 m )(1.03 sin ω t A )( 0.5 m )( 0.5 T ) sin ω t
τ ind ( t ) = 0.0515 sin 2 ωt N ⋅ m, counterclockwise
(d) The instantaneous power generated by the loop is:
103
, P ( t ) = eindi = (5.15 sin ω t V )(1.03 sin ω t A ) = 5.30 sin 2ωt W
The average power generated by the loop is
1
Pave = 5.30 sin 2ω t dt = 2.65 W
T T
(e) The mechanical power being consumed by the loop is:
( )
P = τ indω = 0.0515 sin 2 ω t V (103 rad/s ) = 5.30 sin 2ω t W
Note that the amount of mechanical power consumed by the loop is equal to the amount of electrical power
created by the loop. This machine is acting as a generator, converting mechanical power into electrical
power.
4-2. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14
poles operating at frequencies of 50, 60, and 400 Hz.
SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical
frequency is
120 f e
nm =
P
The resulting table is
Number of Poles f e = 50 Hz f e = 60 Hz f e = 400 Hz
2 3000 r/min 3600 r/min 24000 r/min
4 1500 r/min 1800 r/min 12000 r/min
6 1000 r/min 1200 r/min 8000 r/min
8 750 r/min 900 r/min 6000 r/min
10 600 r/min 720 r/min 4800 r/min
12 500 r/min 600 r/min 4000 r/min
14 428.6 r/min 514.3 r/min 3429 r/min
4-3. A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot
of the windings. All coils in each phase are connected in series, and the three phases are connected in ∆.
The flux per pole in the machine is 0.060 Wb, and the speed of rotation of the magnetic field is 1800 r/min.
(a) What is the frequency of the voltage produced in this winding?
(b) What are the resulting phase and terminal voltages of this stator?
SOLUTION
(a) The frequency of the voltage produced in this winding is
nm P (1800 r/min )( 4 poles )
fe = = = 60 Hz
120 120
(b) There are 12 slots on this stator, with 40 turns of wire per slot. Since this is a four-pole machine,
there are two sets of coils (4 slots) associated with each phase. The voltage in the coils in one pair of slots
is
E A = 2π N C φ f = 2π ( 40 t )( 0.060 Wb )( 60 Hz ) = 640 V
There are two sets of coils per phase, since this is a four-pole machine, and they are connected in series, so
the total phase voltage is
104
, Vφ = 2 ( 640 V ) = 1280 V
Since the machine is ∆-connected, VL = Vφ = 1280 V .
4-4. A three-phase Y-connected 50-Hz two-pole synchronous machine has a stator with 2000 turns of wire per
phase. What rotor flux would be required to produce a terminal (line-to-line) voltage of 6 kV?
SOLUTION The phase voltage of this machine should be Vφ = VL / 3 = 3464 V . The induced voltage per
phase in this machine (which is equal to Vφ at no-load conditions) is given by the equation
E A = 2π N C φ f
so
EA 3464 V
φ= = = 0.0078 Wb
2π N C f 2π ( 2000 t )(50 Hz )
4-5. Modify the MATLAB program in Example 4-1 by swapping the currents flowing in any two phases. What
happens to the resulting net magnetic field?
SOLUTION This modification is very simple—just swap the currents supplied to two of the three phases.
% M-file: mag_field2.m
% M-file to calculate the net magetic field produced
% by a three-phase stator.
% Set up the basic conditions
bmax = 1; % Normalize bmax to 1
freq = 60; % 60 Hz
w = 2*pi*freq; % angluar velocity (rad/s)
% First, generate the three component magnetic fields
t = 0:1/6000:1/60;
Baa = sin(w*t) .* (cos(0) + j*sin(0));
Bbb = sin(w*t+2*pi/3) .* (cos(2*pi/3) + j*sin(2*pi/3));
Bcc = sin(w*t-2*pi/3) .* (cos(-2*pi/3) + j*sin(-2*pi/3));
% Calculate Bnet
Bnet = Baa + Bbb + Bcc;
% Calculate a circle representing the expected maximum
% value of Bnet
circle = 1.5 * (cos(w*t) + j*sin(w*t));
% Plot the magnitude and direction of the resulting magnetic
% fields. Note that Baa is black, Bbb is blue, Bcc is
% magneta, and Bnet is red.
for ii = 1:length(t)
% Plot the reference circle
plot(circle,'k');
hold on;
% Plot the four magnetic fields
plot([0 real(Baa(ii))],[0 imag(Baa(ii))],'k','LineWidth',2);
plot([0 real(Bbb(ii))],[0 imag(Bbb(ii))],'b','LineWidth',2);
105
, plot([0 real(Bcc(ii))],[0 imag(Bcc(ii))],'m','LineWidth',2);
plot([0 real(Bnet(ii))],[0 imag(Bnet(ii))],'r','LineWidth',3);
axis square;
axis([-2 2 -2 2]);
drawnow;
hold off;
end
When this program executes, the net magnetic field rotates clockwise, instead of counterclockwise.
4-6. If an ac machine has the rotor and stator magnetic fields shown in Figure P4-1, what is the direction of the
induced torque in the machine? Is the machine acting as a motor or generator?
SOLUTION Since τind = kB R × B net , the induced torque is clockwise, opposite the direction of motion. The
machine is acting as a generator.
4-7. The flux density distribution over the surface of a two-pole stator of radius r and length l is given by
B = BM cos ( ω m t − α ) (4-37b)
Prove that the total flux under each pole face is
φ = 2 rlBM
106
4-1. The simple loop is rotating in a uniform magnetic field shown in Figure 4-1 has the following
characteristics:
B = 0.5 T to the right r = 01
. m
l = 0.5 m ω = 103 rad/s
(a) Calculate the voltage etot ( t ) induced in this rotating loop.
(b) Suppose that a 5 Ω resistor is connected as a load across the terminals of the loop. Calculate the
current that would flow through the resistor.
(c) Calculate the magnitude and direction of the induced torque on the loop for the conditions in (b).
(d) Calculate the electric power being generated by the loop for the conditions in (b).
(e) Calculate the mechanical power being consumed by the loop for the conditions in (b). How does
this number compare to the amount of electric power being generated by the loop?
ωm
c r
d
N vab S
vcd
b
a
B
B is a uniform magnetic
field, aligned as shown.
SOLUTION
(a) The induced voltage on a simple rotating loop is given by
eind ( t ) = 2 rωBl sin ω t (4-8)
eind ( t ) = 2 ( 0.1 m )(103 rad/s)(0.5 T )( 0.5 m ) sin103t
eind ( t ) = 5.15 sin103t V
(b) If a 5 Ω resistor is connected as a load across the terminals of the loop, the current flow would be:
eind 5.15 sin 103t V
i (t ) = = = 1.03 sin 103t A
R 5Ω
(c) The induced torque would be:
τ ind ( t ) = 2 rilΒ sin θ (4-17)
τ ind ( t ) = 2 ( 0.1 m )(1.03 sin ω t A )( 0.5 m )( 0.5 T ) sin ω t
τ ind ( t ) = 0.0515 sin 2 ωt N ⋅ m, counterclockwise
(d) The instantaneous power generated by the loop is:
103
, P ( t ) = eindi = (5.15 sin ω t V )(1.03 sin ω t A ) = 5.30 sin 2ωt W
The average power generated by the loop is
1
Pave = 5.30 sin 2ω t dt = 2.65 W
T T
(e) The mechanical power being consumed by the loop is:
( )
P = τ indω = 0.0515 sin 2 ω t V (103 rad/s ) = 5.30 sin 2ω t W
Note that the amount of mechanical power consumed by the loop is equal to the amount of electrical power
created by the loop. This machine is acting as a generator, converting mechanical power into electrical
power.
4-2. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14
poles operating at frequencies of 50, 60, and 400 Hz.
SOLUTION The equation relating the speed of magnetic field rotation to the number of poles and electrical
frequency is
120 f e
nm =
P
The resulting table is
Number of Poles f e = 50 Hz f e = 60 Hz f e = 400 Hz
2 3000 r/min 3600 r/min 24000 r/min
4 1500 r/min 1800 r/min 12000 r/min
6 1000 r/min 1200 r/min 8000 r/min
8 750 r/min 900 r/min 6000 r/min
10 600 r/min 720 r/min 4800 r/min
12 500 r/min 600 r/min 4000 r/min
14 428.6 r/min 514.3 r/min 3429 r/min
4-3. A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot
of the windings. All coils in each phase are connected in series, and the three phases are connected in ∆.
The flux per pole in the machine is 0.060 Wb, and the speed of rotation of the magnetic field is 1800 r/min.
(a) What is the frequency of the voltage produced in this winding?
(b) What are the resulting phase and terminal voltages of this stator?
SOLUTION
(a) The frequency of the voltage produced in this winding is
nm P (1800 r/min )( 4 poles )
fe = = = 60 Hz
120 120
(b) There are 12 slots on this stator, with 40 turns of wire per slot. Since this is a four-pole machine,
there are two sets of coils (4 slots) associated with each phase. The voltage in the coils in one pair of slots
is
E A = 2π N C φ f = 2π ( 40 t )( 0.060 Wb )( 60 Hz ) = 640 V
There are two sets of coils per phase, since this is a four-pole machine, and they are connected in series, so
the total phase voltage is
104
, Vφ = 2 ( 640 V ) = 1280 V
Since the machine is ∆-connected, VL = Vφ = 1280 V .
4-4. A three-phase Y-connected 50-Hz two-pole synchronous machine has a stator with 2000 turns of wire per
phase. What rotor flux would be required to produce a terminal (line-to-line) voltage of 6 kV?
SOLUTION The phase voltage of this machine should be Vφ = VL / 3 = 3464 V . The induced voltage per
phase in this machine (which is equal to Vφ at no-load conditions) is given by the equation
E A = 2π N C φ f
so
EA 3464 V
φ= = = 0.0078 Wb
2π N C f 2π ( 2000 t )(50 Hz )
4-5. Modify the MATLAB program in Example 4-1 by swapping the currents flowing in any two phases. What
happens to the resulting net magnetic field?
SOLUTION This modification is very simple—just swap the currents supplied to two of the three phases.
% M-file: mag_field2.m
% M-file to calculate the net magetic field produced
% by a three-phase stator.
% Set up the basic conditions
bmax = 1; % Normalize bmax to 1
freq = 60; % 60 Hz
w = 2*pi*freq; % angluar velocity (rad/s)
% First, generate the three component magnetic fields
t = 0:1/6000:1/60;
Baa = sin(w*t) .* (cos(0) + j*sin(0));
Bbb = sin(w*t+2*pi/3) .* (cos(2*pi/3) + j*sin(2*pi/3));
Bcc = sin(w*t-2*pi/3) .* (cos(-2*pi/3) + j*sin(-2*pi/3));
% Calculate Bnet
Bnet = Baa + Bbb + Bcc;
% Calculate a circle representing the expected maximum
% value of Bnet
circle = 1.5 * (cos(w*t) + j*sin(w*t));
% Plot the magnitude and direction of the resulting magnetic
% fields. Note that Baa is black, Bbb is blue, Bcc is
% magneta, and Bnet is red.
for ii = 1:length(t)
% Plot the reference circle
plot(circle,'k');
hold on;
% Plot the four magnetic fields
plot([0 real(Baa(ii))],[0 imag(Baa(ii))],'k','LineWidth',2);
plot([0 real(Bbb(ii))],[0 imag(Bbb(ii))],'b','LineWidth',2);
105
, plot([0 real(Bcc(ii))],[0 imag(Bcc(ii))],'m','LineWidth',2);
plot([0 real(Bnet(ii))],[0 imag(Bnet(ii))],'r','LineWidth',3);
axis square;
axis([-2 2 -2 2]);
drawnow;
hold off;
end
When this program executes, the net magnetic field rotates clockwise, instead of counterclockwise.
4-6. If an ac machine has the rotor and stator magnetic fields shown in Figure P4-1, what is the direction of the
induced torque in the machine? Is the machine acting as a motor or generator?
SOLUTION Since τind = kB R × B net , the induced torque is clockwise, opposite the direction of motion. The
machine is acting as a generator.
4-7. The flux density distribution over the surface of a two-pole stator of radius r and length l is given by
B = BM cos ( ω m t − α ) (4-37b)
Prove that the total flux under each pole face is
φ = 2 rlBM
106
