(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 2 Part 4 pptx

255 % the line "Ea_ar - Vt - i_a*r_a" goes negative. i_a = 0:1:37; for jj = 1:length(i_a) % Calculate the equivalent field current due to armature % reaction. i_ar = (i_a(jj) / 50) * 200 / n_f; % Calculate the Ea values modified by armature reaction Ea_ar = interp1(if_values,ea_values,i_f - i_ar); % Get the voltage difference diff = Ea_ar - Vt - i_a(jj)*r_a; % This code prevents us from reporting the first (unstable) % location satisfying the criterion. was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % Save terminal voltage at this point v_t(jj) = Vt(ii); i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); end; % Plot the terminal characteristic figure(1); plot(i_l,v_t,'b-','LineWidth',2.0); xlabel('\bf\itI_{L} \rm\bf(A)'); ylabel('\bf\itV_{T} \rm\bf(V)'); title ('\bfTerminal Characteristic of a Shunt DC Generator w/AR'); hold off; axis([ 0 50 0 120]); grid on; 256 The resulting terminal characteristic is shown below: 9-26. If the machine in Problem 9-25 is running at 1800 r/min with a field resistance R adj = 10 Ω and an armature current of 25 A, what will the resulting terminal voltage be? If the field resistor decreases to 5 Ω while the armature current remains 25 A, what will the new terminal voltage be? (Assume no armature reaction.) S OLUTION If A I = 25 A, then ( ) ( ) 25 A 0.18 AA IR =Ω = 4.5 V. The point where the distance between the A E and T V curves is exactly 4.5 V corresponds to a terminal voltage of 104 V, as shown below. 257 If R adj decreases to 5 Ω , the total field resistance becomes 29 Ω , and the terminal voltage line gets shallower. The new point where the distance between the A E and T V curves is exactly 4.5 V corresponds to a terminal voltage of 115 V, as shown below. Note that decreasing the field resistance of the shunt generator increases the terminal voltage. 9-27. A 120-V 50-A cumulatively compounded dc generator has the following characteristics: 0.21 AS RR+= Ω 1000 turns F N = 20 F R =Ω SE 20 turnsN = adj 0 to 30 , set to 10 R =Ω Ω 1800 r/min m n = The machine has the magnetization curve shown in Figure P9-7. Its equivalent circuit is shown in Figure P9-10. Answer the following questions about this machine, assuming no armature reaction. (a) If the generator is operating at no load, what is its terminal voltage? (b) If the generator has an armature current of 20 A, what is its terminal voltage? 258 (c) If the generator has an armature current of 40 A, what is its terminal voltage'? (d) Calculate and plot the terminal characteristic of this machine. S OLUTION (a) The total field resistance of this generator is 30 Ω , and the no-load terminal voltage can be found from the intersection of the resistance line with the magnetization curve for this generator. The magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a terminal voltage of 121 V. (b) If the armature current is 20 A, then the effective field current contribution from the armature current () SE 20 20 A 0.4 A 1000 A F N I N == and the () AA S IR R+ voltage drop is () ( ) ( ) 20 A 0.21 4.2 V AA S IR R+= Ω= . The location where the triangle formed by SE A F N I N and AA IR exactly fits between the A E and T V lines corresponds to a terminal voltage of 120 V, as shown below. 259 (c) If the armature current is 40 A, then the effective field current contribution from the armature current () A 6.0A 40 1000 15 SE == A F I N N and the () SAA RRI + voltage drop is ()()() V 8 20.0 A 80 =Ω=+ SAA RRI . The location where the triangle formed by A F I N N SE and AA RI exactly fits between the A E and T V lines corresponds to a terminal voltage of 116 V, as shown below. 260 A MATLAB program to locate the position where the triangle exactly fits between the A E and T V lines is shown below. This program created the plot shown above. % M-file: prob9_27b.m % M-file to create a plot of the magnetization curve and the % field current curve of a cumulatively-compounded dc generator % when the armature current is 20 A. % Get the magnetization curve. This file contains the % three variables if_values, ea_values, and n_0. clear all load p97_mag.dat; if_values = p97_mag(:,1); ea_values = p97_mag(:,2); n_0 = 1800; % First, initialize the values needed in this program. r_f = 20; % Field resistance (ohms) r_adj = 10; % Adjustable resistance (ohms) r_a = 0.21; % Armature + series resistance (ohms) i_f = 0:0.02:6; % Field current (A) n = 1800; % Generator speed (r/min) n_f = 1000; % Shunt field turns n_se = 20; % Series field turns % Calculate Ea versus If Ea = interp1(if_values,ea_values,i_f); % Calculate Vt versus If Vt = (r_f + r_adj) * i_f; % Calculate the Ea values modified by mmf due to the % armature current 261 i_a = 20; Ea_a = interp1(if_values,ea_values,i_f + i_a * n_se/n_f); % Find the point where the difference between the % enhanced Ea line and the Vt line is 4 V. This will % be the point where the line "Ea_a - Vt - 4" goes % negative. diff = Ea_a - Vt - 4; % This code prevents us from reporting the first (unstable) % location satisfying the criterion. was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % We have the intersection. Tell user. disp (['Ea_a = ' num2str(Ea_a(ii)) ' V']); disp (['Ea = ' num2str(Ea(ii)) ' V']); disp (['Vt = ' num2str(Vt(ii)) ' V']); disp (['If = ' num2str(i_f(ii)) ' A']); disp (['If_a = ' num2str(i_f(ii)+ i_a * n_se/n_f) ' A']); % Plot the curves figure(1); plot(i_f,Ea,'b-','LineWidth',2.0); hold on; plot(i_f,Vt,'k ','LineWidth',2.0); % Plot intersections plot([i_f(ii) i_f(ii)], [0 Vt(ii)], 'k-'); plot([0 i_f(ii)], [Vt(ii) Vt(ii)],'k-'); plot([0 i_f(ii)+i_a*n_se/n_f], [Ea_a(ii) Ea_a(ii)],'k-'); % Plot compounding triangle plot([i_f(ii) i_f(ii)+i_a*n_se/n_f],[Vt(ii) Vt(ii)],'b-'); plot([i_f(ii) i_f(ii)+i_a*n_se/n_f],[Vt(ii) Ea_a(ii)],'b-'); plot([i_f(ii)+i_a*n_se/n_f i_f(ii)+i_a*n_se/n_f],[Vt(ii) Ea_a(ii)],'b-'); xlabel('\bf\itI_{F} \rm\bf(A)'); ylabel('\bf\itE_{A} \rm\bf or \itE_{A} \rm\bf(V)'); title ('\bfPlot of \itE_{A} \rm\bf and \itV_{T} \rm\bf vs field current'); axis ([0 5 0 150]); set(gca,'YTick',[0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150]') set(gca,'XTick',[0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0]') legend ('Ea line','Vt line',4); hold off; grid on; 262 (d) A MATLAB program to calculate and plot the terminal characteristic of this generator is shown below. % M-file: prob9_27d.m % M-file to calculate the terminal characteristic of a % cumulatively compounded dc generator without armature % reaction. % Get the magnetization curve. This file contains the % three variables if_values, ea_values, and n_0. clear all load p97_mag.dat; if_values = p97_mag(:,1); ea_values = p97_mag(:,2); n_0 = 1800; % First, initialize the values needed in this program. r_f = 20; % Field resistance (ohms) r_adj = 10; % Adjustable resistance (ohms) r_a = 0.21; % Armature + series resistance (ohms) i_f = 0:0.02:6; % Field current (A) n = 1800; % Generator speed (r/min) n_f = 1000; % Shunt field turns n_se = 20; % Series field turns % Calculate Ea versus If Ea = interp1(if_values,ea_values,i_f); % Calculate Vt versus If Vt = (r_f + r_adj) * i_f; % Find the point where the difference between the two % lines is exactly equal to i_a*r_a. This will be the % point where the line line "Ea - Vt - i_a*r_a" goes % negative. i_a = 0:1:50; for jj = 1:length(i_a) % Calculate the Ea values modified by mmf due to the % armature current Ea_a = interp1(if_values,ea_values,i_f + i_a(jj)*n_se/n_f); % Get the voltage difference diff = Ea_a - Vt - i_a(jj)*r_a; % This code prevents us from reporting the first (unstable) % location satisfying the criterion. was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; 263 % Save terminal voltage at this point v_t(jj) = Vt(ii); i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); end; % Plot the terminal characteristic figure(1); plot(i_l,v_t,'b-','LineWidth',2.0); xlabel('\bf\itI_{L} \rm\bf(A)'); ylabel('\bf\itV_{T} \rm\bf(V)'); string = ['\bfTerminal Characteristic of a Cumulatively ' 'Compounded DC Generator']; title (string); hold off; axis([ 0 50 0 130]); grid on; The resulting terminal characteristic is shown below. Compare it to the terminal characteristics of the shunt dc generators in Problem 9-25 (d). 9-28. If the machine described in Problem 9-27 is reconnected as a differentially compounded dc generator, what will its terminal characteristic look like? Derive it in the same fashion as in Problem 9-27. S OLUTION A MATLAB program to calculate and plot the terminal characteristic of this generator is shown below. % M-file: prob9_28.m % M-file to calculate the terminal characteristic of a % differentially compounded dc generator without armature % reaction. % Get the magnetization curve. This file contains the 264 % three variables if_values, ea_values, and n_0. clear all load p97_mag.dat; if_values = p97_mag(:,1); ea_values = p97_mag(:,2); n_0 = 1800; % First, initialize the values needed in this program. r_f = 20; % Field resistance (ohms) r_adj = 10; % Adjustable resistance (ohms) r_a = 0.21; % Armature + series resistance (ohms) i_f = 0:0.02:6; % Field current (A) n = 1800; % Generator speed (r/min) n_f = 1000; % Shunt field turns n_se = 20; % Series field turns % Calculate Ea versus If Ea = interp1(if_values,ea_values,i_f); % Calculate Vt versus If Vt = (r_f + r_adj) * i_f; % Find the point where the difference between the two % lines is exactly equal to i_a*r_a. This will be the % point where the line line "Ea - Vt - i_a*r_a" goes % negative. i_a = 0:1:26; for jj = 1:length(i_a) % Calculate the Ea values modified by mmf due to the % armature current Ea_a = interp1(if_values,ea_values,i_f - i_a(jj)*n_se/n_f); % Get the voltage difference diff = Ea_a - Vt - i_a(jj)*r_a; % This code prevents us from reporting the first (unstable) % location satisfying the criterion. was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % Save terminal voltage at this point v_t(jj) = Vt(ii); i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); end; % Plot the terminal characteristic figure(1); [...]... jX1 1.8 Ω j0.5X2 { j2 .4 Ω j0.5XM 0.5ZF - ZB = R2 s 0.5 R2 2 s Forward jX2 { 0.5ZB ZF = 0.5 j30 Ω V ZF = j1 .20 Ω j1 .20 Ω j0.5XM j30 Ω ( R2 / s + jX 2 )( jX M ) R2 / s + jX 2 + jX M (50 + j 2. 40 )( j 60 ) = 28 .15 + j 24 .87 Ω 50 + j 2. 40 + j 60 R2 / ( 2 s ) jX 2 ( jX M ) R2 / ( 2 − s ) + jX 2 + jX M 27 0 { I1 { 10-1 Reverse ZB = (a) (1 .28 2 + j 2. 40 )( j 60) = 1.185 + j 2. 3 32 Ω 1 .28 2 + j 2. 40 + j 60 The input... 40 0 r/min, the slip is s= 1800 r/min − 40 0 r/min = 0.778 1800 r/min ZF = ZF = ZB = ZB = ( R2 / s + jX 2 )( jX M ) R2 / s + jX 2 + jX M (100 + j 2. 40 )( j 60 ) = 2. 96 + 100 + j 2. 40 + j 60 R2 / ( 2 s ) jX 2 j 2. 46 Ω ( jX M ) R2 / ( 2 − s ) + jX 2 + jX M (1 .28 2 + j 2. 40 )( j 60 ) = 1.90 + j 2. 37 Ω 1 .28 2 + j 2. 40 + j 60 The input current is I1 = V R1 + jX 1 + 0.5Z F + 0.5Z B I1 = 120 ∠ 0° V = 18.73∠ − 48 .7°... 0.5Z F + 0.5Z B I1 = 120 ∠ 0° V = 5 .23 ∠ − 44 .2 A 1.80 + j 2. 40 ) + 0.5 ( 28 .15 + j 24 .87 ) + 0.5 (1.185 + j 2. 3 32 ) ( PIN = VI cos θ = ( 120 V )(5 .23 A ) cos 44 .2 = 45 0 W (b) The air-gap power is PAG,F = I 12 ( 0.5RF ) = (5 .23 A ) ( 14. 1 Ω ) = 386 W 2 PAG,B = I 12 ( 0.5 RB ) = (5 .23 A ) (0.5 92 Ω ) = 16 .2 W 2 PAG = PAG,F − PAG,B = 386 W − 14. 8 W = 371 W (c) The power converted from electrical to mechanical... 0.713 lagging 10 -2 = 1.97 N ⋅ m The load torque is τ load = (g) = Repeat Problem 10-1 for a rotor slip of 0. 025 ZF = ZF = ( R2 / s + jX 2 )( jX M ) R2 / s + jX 2 + jX M (100 + j 2. 40 )( j 60 ) = 28 .91 + 100 + j 2. 40 + j 60 j 43 .83 Ω 27 1 1 min 60 s = 1.68 N ⋅ m R2 / ( 2 s ) ZB = ZB = (a) jX 2 ( jX M ) R2 / ( 2 − s ) + jX 2 + jX M (1 .28 2 + j 2. 40 )( j 60) = 1.170 + j 2. 331 Ω 1 .28 2 + j 2. 40 + j 60 The input... for Pconv , we get 0.0 42 3 I A 2 − 23 0 I A + 40 , 000 = 0 I A2 − 543 4.8 I A + 945 180 = 0 I A = 179.9 A and E A = 22 2 .4 V Therefore, the power into the dc machine is VT I A = 41 .38 kW , while the power converted from electrical to mechanical form (which is equal to the output power) is E A I A = ( 22 2 .4 V )(179.9 A ) = 40 kW The internal generated voltage E A of the dc machine is 22 2 .4 V The armature current... 0.5Z F + 0.5Z B I1 = 120 ∠ 0° V = 4. 03∠ − 59.0° A (1.80 + j 2. 40 ) + 0.5 (25 .91 + j 43 .83) + 0.5 (1.170 + j 2. 331) PIN = VI cos θ = ( 120 V )( 4. 03 A ) cos 59.0° = 24 9 W (b) The air-gap power is PAG,F = I 12 ( 0.5RF ) = ( 4. 03 A ) ( 12. 96 Ω ) = 21 0.5 W 2 PAG,B = I 12 (0.5RB ) = ( 4. 03 A ) ( 0.585 Ω ) = 9.5 W 2 PAG = PAG,F − PAG,B = 21 0.5 W − 9.5 W = 20 1 W (c) The power converted from electrical to mechanical... ac machine Therefore, E A on the dc machine will decrease to (0.99) (22 2 .4 V) = 22 0 .2 V The resulting armature current is I A,dc = VT − E A 23 0 V − 22 0 .2 V = = 23 1.7 A RA 0.0 42 3 Ω The power into the dc motor is now (23 0 V) (23 1.7 A) = 53.3 kW, and the power converted from electrical to mechanical form in the dc machine is (22 0 .2 V) (23 1.7 A) = 51 kW This is also the output power of the dc machine, the... 18.73∠ − 48 .7° A (1.80 + j 2. 40 ) + 0.5 ( 2. 96 + j 2. 46 ) + 0.5 (1.90 + j 2. 37 ) The air-gap power is PAG,F = I 12 ( 0.5RF ) = (18.73 A ) (1 .48 Ω ) = 519 .2 W 2 PAG,B = I 12 ( 0.5RB ) = (18.73 A ) ( 0. 945 Ω ) = 331.5 W 2 PAG = PAG,F − PAG,B = 519 .2 W − 331.5 W = 188 W The power converted from electrical to mechanical form is Pconv,F = (1 − s ) PAG,F = (1 − 0.778 )(519 .2 W ) = 115 .2 W Pconv,B = (1 − s ) PAG,B... (1 − s ) PAG,F = (1 − 0. 025 )( 21 0.5 W ) = 20 5 W Pconv,B = (1 − s ) PAG,B = (1 − 0. 025 )( 9.5 W ) = 9.3 W Pconv = Pconv,F − Pconv,B = 20 5 W − 9.3 W = 196 W (d) The output power is POUT = Pconv − Prot = 20 5 W − 51 W = 1 54 W (e) The induced torque is τ ind = (f) 21 0.5 W 2 rad (1800 r/min ) 1r 1 min 60 s = 1. 12 N ⋅ m POUT ωm = 1 54 W (0.975)(1800 r/min ) 2 rad 1r 1 min 60 s = 0. 84 N ⋅ m The overall efficiency... X S (51 kW )( 2. 0 Ω ) = 18.9° = sin −1 3Vφ E A 3 ( 27 7 V )( 380 V ) The new E A of this machine is thus 380∠18.9° V , and the resulting armature current is IA = E A − Vφ jX S = 380∠18.9° V − 27 7∠0° V = 74. 0∠ − 33.8° A j 2. 0 Ω The real and reactive powers are now P = 3 VT I L cos θ = 3 ( 48 0 V )( 74. 0 A ) cos 33.8° = 51 kW Q = 3 VT I L sin θ = 3 ( 48 0 V )( 74. 0 A ) sin 33.8° = 34 .2 kvar 26 8 The phasor . ( ) ( ) 50 2. 40 60 28 .15 24 .87 50 2. 40 60 F jj Zj jj + ==+Ω ++ () () () 22 22 /2 /2 M B M R s jX jX Z R s jX jX = −+ + 27 1 ()() 1 .28 2 2. 40 60 1.185 2. 3 32 1 .28 2 2. 40 60 B jj Zj jj + ==+Ω ++ . ()() 100 2. 40 60 28 .91 43 .83 10 02. 40 60 F jj Zj jj + ==+Ω ++ 27 2 () () () 22 22 /2 /2 M B M R s jX jX Z R s jX jX = −+ + ()() 1 .28 2 2. 40 60 1.170 2. 331 1 .28 2 2. 40 60 B jj Zj jj + ==+Ω ++ . ) 1 120 0 V I5 .23 44 .2 A 1.80 2. 40 0.5 28 .15 24 .87 0.5 1.185 2. 332jj j ∠° ==∠−° ++ + + + ()( ) IN cos 120 V 5 .23 A cos 44 .2 45 0 WPVI θ == °= (b) The air-gap power is () ()() 2 2 AG,