(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

% Plot a closeup of the near spectrum % positive side only figure5; plotfftshiftfstep,fftshiftabsspec; title'\bfSpectrum of Output Voltage for fr = 500 Hz'; xlabel'\bfFrequency Hz'; ylab

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% 10 V

for ii = 1:length(t)

vin(ii) = 10 * sin(2*pi*50*t(ii));

end

% Now calculate vout

[vout(ii) vu(ii) vv(ii)] = vout(vin(ii), vx(ii), vy(ii));

end

% Plot the reference voltages vs time

figure(1)

plot(t,vx,'b','Linewidth',1.0);

hold on;

plot(t,vy,'k ','Linewidth',1.0);

title('\bfReference Voltages for fr = 500 Hz');

xlabel('\bfTime (s)');

ylabel('\bfVoltage (V)');

legend('vx','vy');

axis( [0 1/30 -10 10]);

hold off;

% Plot the input voltage vs time

figure(2)

plot(t,vin,'b','Linewidth',1.0);

title('\bfInput Voltage');

axis( [0 1/30 -10 10]);

% Plot the output voltages versus time

figure(3)

plot(t,vout,'b','Linewidth',1.0);

title('\bfOutput Voltage for fr = 500 Hz');

axis( [0 1/30 -120 120]);

% Now calculate the spectrum of the output voltage

spec = fft(vout);

% Calculate sampling frequency labels

len = length(t);

df = fs / len;

fstep = zeros(size(t));

for ii = 2:len/2

fstep(ii) = df * (ii-1);

fstep(len-ii+2) = -fstep(ii);

end

% Plot the spectrum

figure(4);

plot(fftshift(fstep),fftshift(abs(spec)));

title('\bfSpectrum of Output Voltage for fr = 500 Hz');

xlabel('\bfFrequency (Hz)');

ylabel('\bfAmplitude');

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% Plot a closeup of the near spectrum

% (positive side only)

figure(5);

set(gca,'Xlim',[0 1000]);

When this program is executed, the input, reference, and output voltages are:

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97

(b) The output spectrum of this PWM modulator is shown below There are two plots here, one showing

the entire spectrum, and the other one showing the close-in frequencies (those under 1000 Hz), which will have the most effect on machinery Note that there is a sharp peak at 50 Hz, which is there desired frequency, but there are also strong contaminating signals at about 850 Hz and 950 Hz If necessary, these components could be filtered out using a low-pass filter

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(c) A version of the program with 1000 Hz reference functions is shown below:

% M-file: prob3_15b.m

% M-file to calculate the output voltage from a PWM

% modulator with a 1000 Hz reference frequency Note

% that the only change between this program and that

% of part a is the frequency of the reference "fr"

% Sample the data at 20000 Hz to get enough information

% for spectral analysis Declare arrays

fs = 20000; % Sampling frequency (Hz)

t = (0:1/fs:4/15); % Time in seconds

vx = zeros(size(t)); % vx

vy = zeros(size(t)); % vy

vin = zeros(size(t)); % Driving signal

vu = zeros(size(t)); % vx

vv = zeros(size(t)); % vy

vout = zeros(size(t)); % Output signal

fr = 1000; % Frequency of reference signal

T = 1/fr; % Period of refernce signal

% Calculate vx at 1000 Hz

vx(ii) = vref(t(ii),T);

vy(ii) = - vx(ii);

end

% Calculate vin as a 50 Hz sine wave with a peak voltage of

% 10 V

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% Now calculate vout

[vout(ii) vu(ii) vv(ii)] = vout(vin(ii), vx(ii), vy(ii));

end

% Plot the reference voltages vs time

figure(1)

plot(t,vx,'b','Linewidth',1.0);

hold on;

plot(t,vy,'k ','Linewidth',1.0);

title('\bfReference Voltages for fr = 1000 Hz');

legend('vx','vy');

axis( [0 1/30 -10 10]);

hold off;

% Plot the input voltage vs time

figure(2)

plot(t,vin,'b','Linewidth',1.0);

title('\bfInput Voltage');

axis( [0 1/30 -10 10]);

% Plot the output voltages versus time

figure(3)

plot(t,vout,'b','Linewidth',1.0);

title('\bfOutput Voltage for fr = 1000 Hz');

axis( [0 1/30 -120 120]);

% Now calculate the spectrum of the output voltage

spec = fft(vout);

% Calculate sampling frequency labels

len = length(t);

df = fs / len;

fstep = zeros(size(t));

for ii = 2:len/2

fstep(ii) = df * (ii-1);

fstep(len-ii+2) = -fstep(ii);

end

% Plot the spectrum

figure(4);

% Plot a closeup of the near spectrum

% (positive side only)

figure(5);

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set(gca,'Xlim',[0 1000]);

When this program is executed, the input, reference, and output voltages are:

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101

(d) The output spectrum of this PWM modulator is shown below

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(e) Comparing the spectra in (b) and (d), we can see that the frequencies of the first large sidelobes

doubled from about 900 Hz to about 1800 Hz when the reference frequency was doubled This increase in sidelobe frequency has two major advantages: it makes the harmonics easier to filter, and it also makes it less necessary to filter them at all Since large machines have their own internal inductances, they form natural low-pass filters If the contaminating sidelobes are at high enough frequencies, they will never affect the operation of the machine Thus, it is a good idea to design PWM modulators with a high frequency reference signal and rapid switching

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103

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= 05 T to the right r = 01 m

l= 05 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

r

vab

vcd

B

B is a uniform magnetic

field, aligned as shown.

a b

c d

SOLUTION

(a) The induced voltage on a simple rotating loop is given by

( )

ind 2 sin

e t = ωr Bl ωt (4-8)

ind 2 0.1 m 103 rad/s 0.5 T 0.5 m sin103

( )

ind 5.15 sin103 V

(b) If a 5 Ω resistor is connected as a load across the terminals of the loop, the current flow would be:

( ) ind 5.15 sin 103 V

1.03 sin 103 A

5

R

Ω

(c) The induced torque would be:

( )

ind t 2rilΒsin

ind t 2 0.1 m 1.03 sin t A 0.5 m 0.5 T sin t

ind t 0.0515 sin t N m, counterclockwise

(d) The instantaneous power generated by the loop is:

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( ) ( )( ) 2 ind 5.15 sin V 1.03 sin A 5.30 sin W

The average power generated by the loop is

2 ave

1 5.30 sin 2.65 W

T

(e) The mechanical power being consumed by the loop is:

ind 0.0515 sin V 103 rad/s 5.30 sin 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 e

m

f n

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

(1800 r/min 4 poles)( )

60 Hz

m e

n P

(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

2 2 40 t 0.060 Wb 60 Hz 640 V

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105

2 640 V 1280 V

Since the machine is ∆-connected, 1280 V L=Vφ= 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=V L/ 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

2

E = π φN f

so

3464 V

0.0078 Wb

A C

E

N f

φ

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

% 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);

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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=kBR×Bnet, 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

cos

B=B ω − α t (4-37b)

Prove that the total flux under each pole face is

φ=2rlB M

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107

SOLUTION The total flux under a pole face is given by the equation

d

φ= B⋅ A

Under a pole face, the flux density B is always parallel to the vector dA, since the flux density is always

perpendicular to the surface of the rotor and stator in the air gap Therefore,

B dA

φ=

A differential area on the surface of a cylinder is given by the differential length along the cylinder (dl)

times the differential width around the radius of the cylinder (rdθ)

( )( )

dA= dl rdθ where r is the radius of the cylinder

Therefore, the flux under the pole face is

B dl r d

φ= θ

Since r is constant and B is constant with respect to l, this equation reduces to

rl B d

Now, B=B M cos (ω αt− )=B M cos θ (when we substitute θ ω α= t− ), so

rl B d

/ 2

π π

2rlB M

φ=

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4-8 In the early days of ac motor development, machine designers had great difficulty controlling the core losses

(hysteresis and eddy currents) in machines They had not yet developed steels with low hysteresis, and were not making laminations as thin as the ones used today To help control these losses, early ac motors

in the USA were run from a 25 Hz ac power supply, while lighting systems were run from a separate 60 Hz

ac power supply

(a) 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 25 Hz What was the fastest rotational speed available to these early motors?

(b) For a given motor operating at a constant flux density B, how would the core losses of the motor

running at 25 Hz compare to the core losses of the motor running at 60 Hz?

(c) Why did the early engineers provide a separate 60 Hz power system for lighting?

SOLUTION

(a) The equation relating the speed of magnetic field rotation to the number of poles and electrical

frequency is

120 e

m

f n

P

= The resulting table is

Number of Poles f e = 25 Hz

The highest possible rotational speed was 1500 r/min

(b) Core losses scale according to the 1.5th power of the speed of rotation, so the ratio of the core losses

at 25 Hz to the core losses at 60 Hz (for a given machine) would be:

1.5

1500

3600

(c) At 25 Hz, the light from incandescent lamps would visibly flicker in a very annoying way

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109

Chapter 5: Synchronous Generators

5-1 At a location in Europe, it is necessary to supply 300 kW of 60-Hz power The only power sources

available operate at 50 Hz It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator How many poles should each of the two machines have in order to convert 50-Hz power to 60-Hz power?

SOLUTION The speed of a synchronous machine is related to its frequency by the equation

120 e

m

f n

P

=

To make a 50 Hz and a 60 Hz machine have the same mechanical speed so that they can be coupled together, we see that

sync

120 50 Hz 120 60 Hz

n

2

1

6 12

5 10

P

P = = Therefore, a 10-pole synchronous motor must be coupled to a 12-pole synchronous generator to accomplish this frequency conversion

5-2 A 2300-V 1000-kVA 0.8-PF-lagging 60-Hz two-pole Y-connected synchronous generator has a

synchronous reactance of 1.1 Ω and an armature resistance of 0.15 Ω At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW The field circuit has a dc voltage of 200 V, and the maximum I F is 10 A The resistance of the field circuit is adjustable over the range from 20 to 200 Ω The OCC of this generator is shown in Figure P5-1

(a) How much field current is required to make V T equal to 2300 V when the generator is running at no load?

(b) What is the internal generated voltage of this machine at rated conditions?

(c) How much field current is required to make V T equal to 2300 V when the generator is running at rated conditions?

(d) How much power and torque must the generator’s prime mover be capable of supplying?

(e) Construct a capability curve for this generator

Note: An electronic version of this open circuit characteristic can be found in file

p51_occ.dat, which can be used with MATLAB programs Column 1 contains field current in amps, and column 2 contains open-circuit terminal voltage in volts

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