Electric Machinery Fundamentals (Solutions Manual) Part 7 pdf

With a 20% decrease, E A2 = 10,584 V , and sin 1 1 E 1 A sin sin 13, 230 ™ ™ V sin = = 27.9   35.8 ° = ° 2 2 E A 2 Therefore, the new armature current is 10,584 V 146  A2 ⎞ 10, 584 35.8 70 ° 44  0 780 14.0 A E V A I = = jX S 8.18 j ° =  ° With a 25% decrease, E A2 = 9, 923 V , and sin 1 1 E 1 A sin sin 13, 230 ™ ™ V sin = = 27.9   38.6 ° = ° 2 2 E A 2 Therefore, the new armature current is 9,923 V  A2 ⎞ 9, 923 38.6 70 ° 44  0 762 6.6 A E V A I = = jX S j 8.18 ° =  ° (f) A MATLAB program to plot the magnitude of the armature current below. I A as a function of E A is shown % M-file: prob5_28f.m % M-file to calculate and plot the armature current % supplied to an infinite bus as Ea is varied. % Define values for this generator Ea = (0.65:0.01:1.00)*13230; % Ea Vp = 7044; % Phase voltage d1 = 27.9*pi/180; % torque angle at full Ea Xs = 8.18; % Xs (ohms) % Calculate delta for each Ea d = asin( 13230 ./ Ea .* sin(d1)); % Calculate Ia for each flux Ea = Ea .* ( cos(d) + j.*sin(d) ); Ia = ( Ea - Vp ) ./ (j*Xs); % Plot the armature current versus Ea figure(1); plot(abs(Ea)/1000,abs(Ia),'b-','LineWidth',2.0); title ('\bfArmature current versus \itE_{A}\rm'); xlabel ('\bf\itE_{A}\rm\bf (kV)'); ylabel ('\bf\itI_{A}\rm\bf (A)'); grid on; hold off; 147 The resulting plot is shown below: 148 Chapter 6: Synchronous Motors 6-1. A 480-V, 60 Hz, four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the following questions: (a) What is the output torque of this motor? Express the answer both in newton-meters and in pound-feet. (b) What must be done to change the power factor to 0.8 leading? Explain your answer, using phasor diagrams. (c) What will the magnitude of the line current be if the power factor is adjusted to 0.8 leading? S OLUTION (a) If this motor is assumed lossless, then the input power is equal to the output power. The input power to this motor is ( )( = = )( ) = IN 3 co T L s P V I ⎝ The output torque would be 3 480 V 50 A 1.0 41.6 kW ⎮ OUT P = = 41.6 kW 221 N = m ⊕ LOAD ⎤ m In English units, ( ) 1800 r/min 1 min 2  rad 60 s 1 r ( )( ) 7.04 OU P T 7.04 41.6 = = kW 163 lb ft = ⊕ ⎮ LOAD n m ( ) 1800 r/min (b) To change the motor’s power factor to 0.8 leading, its field current must be increased. Since the power supplied to the load is independent of the field current level, an increase in field current increases A E while keeping the distance E A sin ™ constant. This increase in E A changes the angle of the current I A , eventually causing it to reach a power factor of 0.8 leading.  P I A 2 I A 1 ⎞ V jX I S A }  P Q  I sin ⎝ A (c) The magnitude of the line current will be EE A 1 A 2 I P 41.6 kW = = 62.5 A = L 3 PF 3 ( ) 480 V ( 0.8 ) T V 6-2. A 480-V, 60 Hz, 400-hp 0.8-PF-leading six-pole  -connected synchronous motor has a synchronous reactance of 1.1 & and negligible armature resistance. Ignore its friction, windage, and core losses for the purposes of this problem. 149 (a) If this motor is initially supplying 400 hp at 0.8 PF lagging, what are the magnitudes and angles of E A and I A ? (b) How much torque is this motor producing? What is the torque angle ™ ? How near is this value to the maximum possible induced torque of the motor for this field current setting? (c) If E A is increased by 15 percent, what is the new magnitude of the armature current? What is the motor’s new power factor? (d) Calculate and plot the motor’s V-curve for this load condition. S OLUTION (a) If losses are being ignored, the output power is equal to the input power, so the input power will be ( )( ) IN P 400 hp 746 W/hp 298.4 kW = = This situation is shown in the phasor diagram below: V ⎞ I A The line current flow under these circumstances is jX I S A A E I P 298.4 kW = = 449 A = L 3 PF 3 ( ) 480 V ( 0.8 ) T V Because the motor is  -connected, the corresponding phase current is 449 / 3 259 A I = = . The angle of A the current is cos 1  ( ) 0.80 36.87  =  ° , so I A =  E V I 259 36 = .87  A ° . The internal generated voltage E A is A S ⎞ jX A ( ) 480 0 V ( 1.1 )( 259 36.87 A ) 384 36.4 V E A = ° j  &  ° =  ° (b) This motor has 6 poles and an electrical frequency of 60 Hz, so its rotation speed is m n = 1200 r/min. The induced torque is ⎮ OUT P = = 298.4 kW 2375 N = m ⊕ ind ⎤ m ( ) 1200 r/min 1 min 2  rad 60 s 1 r The maximum possible induced torque for the motor at this field setting is 3 3 ( ) 480 V ( 384 V ) ⎮ ⎞ A V E = = ind,max 1 min 2 rad 4000 N = m ⊕ ⎤ m S X ( ) 1200 r/min  ( ) 1.1 & 60 s 1 r (c) If the magnitude of the internal generated voltage be found from the fact that E A sin ™  P = constant . ( ) = = = 2 1 1.15 1.15 E E 38 A A 4 V 441.6 V E A is increased by 15%, the new torque angle can 150 sin E 1 A sin sin 384 V ™ ™ sin = = ( ) 1 1 36   .4 31.1  ° =  ° 2 1 E A2 The new armature current is 441.6 V ⎞  A 2 V E I 480 0 V ° 4  41.6 31  .1 V ° 227 24.1 A = = =  ° 2A jX S j 1.1 & The magnitude of the armature current is 227 A, and the power factor is cos (-24.1 ° ) = 0.913 lagging. (d) A MATLAB program to calculate and plot the motor’s V-curve is shown below: % M-file: prob6_2d.m % M-file create a plot of armature current versus Ea % for the synchronous motor of Problem 6-2. % Initialize values Ea = (1:0.01:1.70)*384; % Magnitude of Ea volts Ear = 384; % Reference Ea deltar = -36.4 * pi/180; % Reference torque angle Xs = 1.1; % Synchronous reactance Vp = 480; % Phase voltage at 0 degrees Ear = Ear * (cos(deltar) + j * sin(deltar)); % Calculate delta2 delta2 = asin ( abs(Ear) ./ abs(Ea) .* sin(deltar) ); % Calculate the phasor Ea Ea = Ea .* (cos(delta2) + j .* sin(delta2)); % Calculate Ia Ia = ( Vp - Ea ) / ( j * Xs); % Plot the v-curve figure(1); plot(abs(Ea),abs(Ia),'b','Linewidth',2.0); xlabel('\bf\itE_{A}\rm\bf (V)'); ylabel('\bf\itI_{A}\rm\bf (A)'); title ('\bfSynchronous Motor V-Curve'); grid on; 151 . = 27. 9   35.8 ° = ° 2 2 E A 2 Therefore, the new armature current is 10,584 V 146  A2 ⎞ 10, 584 35.8 70 ° 44  0 78 0 14.0 . sin = = 27. 9   38.6 ° = ° 2 2 E A 2 Therefore, the new armature current is 9,923 V  A2 ⎞ 9, 923 38.6 70 ° 44  0 76 2 6.6 . values for this generator Ea = (0.65:0.01:1.00)*13230; % Ea Vp = 70 44; % Phase voltage d1 = 27. 9*pi/180; % torque angle at full Ea Xs = 8.18; % Xs (ohms)