95 % 10 V. for ii = 1:length(t) vin(ii) = 10 * sin(2*pi*50*t(ii)); end % Now calculate vout for ii = 1:length(t) [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'); xlabel('\bfTime (s)'); ylabel('\bfVoltage (V)'); 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'); xlabel('\bfTime (s)'); ylabel('\bfVoltage (V)'); 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'); 96 % Plot a closeup of the near spectrum % (positive side only) figure(5); plot(fftshift(fstep),fftshift(abs(spec))); title('\bfSpectrum of Output Voltage for fr = 500 Hz'); xlabel('\bfFrequency (Hz)'); ylabel('\bfAmplitude'); set(gca,'Xlim',[0 1000]); When this program is executed, the input, reference, and output voltages are: 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. 98 (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. for ii = 1:length(t) 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. for ii = 1:length(t) vin(ii) = 10 * sin(2*pi*50*t(ii)); end 99 % Now calculate vout for ii = 1:length(t) [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'); 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'); xlabel('\bfTime (s)'); ylabel('\bfVoltage (V)'); 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'); xlabel('\bfTime (s)'); ylabel('\bfVoltage (V)'); 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 = 1000 Hz'); xlabel('\bfFrequency (Hz)'); ylabel('\bfAmplitude'); % Plot a closeup of the near spectrum % (positive side only) figure(5); plot(fftshift(fstep),fftshift(abs(spec))); 100 title('\bfSpectrum of Output Voltage for fr = 1000 Hz'); xlabel('\bfFrequency (Hz)'); ylabel('\bfAmplitude'); set(gca,'Xlim',[0 1000]); When this program is executed, the input, reference, and output voltages are: 101 (d) The output spectrum of this PWM modulator is shown below. 102 (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. 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 et tot ()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? ω m r v ab v cd B N S B is a uniform magnetic field, aligned as shown. a b c d S OLUTION (a) The induced voltage on a simple rotating loop is given by ( ) ind 2 sin et rBl t ω =ω (4-8) () ( )( )( )( ) ind 2 0.1 m 103 rad/s 0.5 T 0.5 m sin103et t= () ind 5.15 sin103 Vet t= (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 et it t R == = Ω (c) The induced torque would be: () ind 2 sin trilΒ τθ = (4-17) () ( )( )( )( ) ind 2 0.1 m 1.03 sin A 0.5 m 0.5 T sin tt t τωω = () 2 ind 0.0515 sin N m, counterclockwisett τω =⋅ (d) The instantaneous power generated by the loop is: 104 () ( )( ) 2 ind 5.15 sin V 1.03 sin A 5.30 sin WPt e i t t t ωω ω == = The average power generated by the loop is 2 ave 1 5.30 sin 2.65 W T Ptdt T ω == (e) The mechanical power being consumed by the loop is: () () 22 ind 0.0515 sin V 103 rad/s 5.30 sin WPt t τω ω ω == = 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. S OLUTION 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 e f = 50 Hz e f = 60 Hz e f = 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? S OLUTION (a) The frequency of the voltage produced in this winding is ()() 1800 r/min 4 poles 60 Hz 120 120 m e nP f == = (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 AC ENf πφ π == = 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 [...]... 0 .15 + j1 .1 + 6. 667 ∠30° 1. 829 Ω Therefore, the magnitude of the phase voltage is Vφ = I A Z = (18 6 A ) (6. 667 Ω ) = 12 40 V and the terminal voltage is VT = 3 Vφ = 3 (12 40 V ) = 214 8 V (b) Armature current is I A = 18 6 − 30° A , and the phase voltage is Vφ = 12 40∠0° V Therefore, the internal generated voltage is E A = Vφ + RAI A + jX S I A E A = 12 40∠0° + ( 0 .15 Ω ) (18 6 − 30° A ) + j (1. 1 Ω) (18 6 − 30°... 3 = 13 77 V The load is ∆-connected with three impedances of 20∠30° Ω From the Y-∆ transform, this load is equivalent to a Y-connected load with three impedances of 6. 667 ∠30° Ω The resulting per-phase equivalent circuit is shown below: 11 2 0 .15 Ω j1 .1 Ω IA + EA + - Vφ Z 6. 667 ∠30° - The magnitude of the phase current flowing in this generator is IA = EA 13 77 V 13 77 V = = = 18 6 A RA + jX S + Z 0 .15 ... j (1. 1 Ω) (18 6 − 30° A ) E A = 13 77 6. 8° V The resulting phasor diagram is shown below (not to scale): E = 13 77 6. 8° V A θ V = 12 40∠0° V φ IA = 18 6 -30° (c) The efficiency of the generator under these conditions 3can be found as follows: POUT = 3 Vφ I A cos θ = 3 (12 40 V ) (18 6 A )( 0.8) = 554 kW PCU = 3 I A2 RA = 3 (18 6 A ) ( 0 .15 Ω ) = 15 .6 kW 2 PF&W = 24 kW Pcore = 18 kW Pstray = (assumed 0) PIN... = P 10 00 kVA = = 2 51 A at an angle of – 36. 87° 3 VL 3 ( 2300 V ) The phase voltage of this machine is Vφ = VT / 3 = 13 28 V The internal generated voltage of the machine is E A = Vφ + RAI A + jX S I A E A = 13 28∠0° + ( 0 .15 Ω )( 2 51 − 36. 87° A ) + j (1. 1 Ω )( 2 51 − 36. 87° A ) E A = 15 37∠7.4° V (c) The equivalent open-circuit terminal voltage corresponding to an E A of 15 37 volts is VT ,oc = 3 (15 27... Vφ = 12 40∠0° V I = 18 6 -30° A I′ A (e) The new impedance per phase will be half of the old value, so Z = 3.333∠30° Ω The magnitude of the phase current flowing in this generator is IA = EA 13 77 V 13 77 V = = = 335 A R A + jX S + Z 0 .15 + j1 .1 + 3.333∠30° 1. 829 Ω Therefore, the magnitude of the phase voltage is Vφ = I A Z = ( 335 A )(3.333 Ω ) = 11 17 V and the terminal voltage is VT = 3 Vφ = 3 (11 17 V... convert 50-Hz power to 60 -Hz power? SOLUTION The speed of a synchronous machine is related to its frequency by the equation nm = 12 0 f e 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 nsync = 12 0 (50 Hz ) 12 0 (60 Hz ) = P P2 1 P2 6 12 = = P 5 10 1 Therefore, a 10 -pole synchronous motor must be coupled to a 12 -pole synchronous generator... mover must be capable of supplying 17 5 kW Since the generator is a two-pole 60 Hz machine, to must be turning at 360 0 r/min The required torque is τ APP = (e) PIN = ωm 17 5.2 kW = 465 N ⋅ m  1 min  2π rad  ( 360 0 r/min )    60 s  1 r  The rotor current limit of the capability curve would be drawn from an origin of Q=− 3Vφ 2 XS 3 (13 28 V ) = −4 810 kVAR 1. 1 Ω 2 =− The radius of the rotor current... E A XS = 3 (13 28 V ) (15 37 V ) = 5 567 kVA 1. 1 Ω The stator current limit is a circle at the origin of radius S = 3Vφ I A = 3 (13 28 V )( 2 51 A ) = 10 00 kVA A MATLAB program that plots this capability diagram is shown below: % M-file: prob5_2.m % M-file to display a capability curve for a % synchronous generator % Calculate the waveforms for times from 0 to 1/ 30 s Q = -4 810 ; DE = 5 567 ; S = 10 00; % Get... plot(real(s_curve),imag(s_curve),'b','LineWidth',2.0); hold on; plot(real(r_curve),imag(r_curve),'r ','LineWidth',2.0); % Add x and y axes 11 1 plot( [ -15 00 15 00],[0 0],'k'); plot( [0,0],[ -15 00 15 00],'k'); % Set titles and axes title ('\bfSynchronous Generator Capability Diagram'); xlabel('\bfPower (kW)'); ylabel('\bfReactive Power (kVAR)'); axis( [ -15 00 15 00 -15 00 15 00] ); axis square; hold off; The resulting capability diagram is shown below: 5-3 Assume... 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: ratio = (c) 15 00 360 0 1. 5 = 0. 269 or 26. 9% At 25 Hz, the light from incandescent lamps would visibly flicker in a very annoying way 10 8 Chapter 5: Synchronous Generators 5 -1 At a location in Europe, it is necessary to supply 300 kW of 60 -Hz power The . shown below: 11 3 + - E A 0 .15 Ω j1 .1 Ω 6. 667 ∠30°Z + - V φ I A The magnitude of the phase current flowing in this generator is 13 77 V 13 77 V 18 6 A 0 .15 1. 1 6. 667 30 1. 829 A A AS E I RjXZ. ( ) ( ) ( ) ( ) 12 40 0 0 .15 18 6 30 A 1. 1 18 6 30 A A j=∠°+ Ω∠−°+ Ω∠−°E 13 77 6. 8 V A =∠°E The resulting phasor diagram is shown below (not to scale): I = 18 6 ∠ -30° A V = 12 40 ∠ 0° V φ E = 13 77 ∠ 6. 8°. is ()( ) 18 6 A 6. 667 12 40 V A VIZ φ == Ω= and the terminal voltage is () 3 3 12 40 V 214 8 V T VV φ == = (b) Armature current is 18 6 30 A A =∠−°I , and the phase voltage is 12 40 0 V φ =∠°V