Electric Machinery Fundamentals (Solutions Manual) Part 5 ppsx

( ) cos ⎝ 2 sin ⎝ 2 ( ) cos ⎝ sin ⎝ 2 A E = ⎞ V + R A I A  X S I A + X S I A + R A I S V ⎞ = A E 2 ( ) ⎝ X  S I A cos + R A I S sin ⎝ 2 ⎝  R A I A cos + X S I A sin ⎝ If we examine these three cases, we can see that the only difference among them is the sign of the term sin ⎝ . If ⎝ is taken as positive for lagging power factors and negative for leading power factors (in other words, if ⎝ is the impedance angle ), then all three cases can be represented by the single equation: V ⎞ = A E 2 ( ) ⎝ X  S I A cos  R A I S sin ⎝ 2 ⎝  R A I A cos  X S I A sin ⎝ A MATLAB program that calculates terminal voltage as function of impedance angle is shown below: % M-file: prob5_20.m % M-file to calculate and plot the terminal voltage % of a synchronous generator as a function of impedance % angle as PF changes from 0.85 lagging to 0.85 % leading. % Define values for this generator EA = 509; % Internal gen voltage I = 361; % Current (A) R = 0.04; % R (ohms) X = 0.695; % XS (ohms) % Calculate impedance angle theta in degrees theta = -31.8:0.318:31.8; th = theta * pi/180; % In radians % Calculate the phase voltage and terminal voltage VP = sqrt( EA^2 - (X.*I.*cos(th) - R.*I.*sin(th)).^2 ) - R.*I.*cos(th) - X.*I.*sin(th); VT = VP .* sqrt(3); % Plot the terminal voltage versus power factor figure(1); plot(theta,abs(VT),'b-','LineWidth',2.0); title ('\bfTerminal Voltage Versus Impedance Angle'); xlabel ('\bfImpedance Angle (deg)'); ylabel ('\bfTerminal Voltage (V)'); %axis([0 260 300 540]); grid on; hold off; The resulting plot of terminal voltage versus impedance angle (with field and armature currents held constant) is shown below: 134 5-21. Assume that the generator is connected to a 480-V infinite bus, and that its field current has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the armature resistance R A when answering the following questions. (a) What would happen to the real and reactive power supplied by this generator if the field flux (and therefore E A ) is reduced by 5%. (b) Plot the real power supplied by this generator as a function of the flux ⎞ as the flux is varied from 75% to 100% of the flux at rated conditions. (c) Plot the reactive power supplied by this generator as a function of the flux ⎞ as the flux is varied from 75% to 100% of the flux at rated conditions. (d) Plot the line current supplied by this generator as a function of the flux ⎞ as the flux is varied from 75% to 100% of the flux at rated conditions. S OLUTION (a) If the field flux in increase by 5%, nothing would happen to the real power. The reactive power supplied would increase as shown below. V T E A1 E A2 Q Q 2 Q 1 G Q sys I A1 I A2 Q  I sin ⎝ A ⎞ V jX I S A The reactive power 135 (b) If armature resistance is ignored, the power supplied to the bus will not change as flux is varied. Therefore, the plot of real power versus flux is (c) If armature resistance is ignored, the internal generated voltage E A will increase as flux increases, but the quantity expression A E sin ™ will remain constant. Therefore, the voltage for any flux can be found from the = ⎣ ⎞ ⎞ A E ⎧ ⎟ E Ar ⎨ ⎞ r ⎠ and the angle ™ for any E A can be found from the expression ™ = sin  1 ⎣ E sin ™ ⎞ ⎧ ⎧ Ar ⎟ r ⎠ ⎨ E A where ⎞ is the flux in the machine, ⎞ is the flux at rated conditions, r E Ar is the magnitude of the internal generated voltage at rated conditions, and ™ is the angle of the internal generated voltage at rated r conditions. From this information, we can calculate I A for any given load from equation = E A  ⎞ V I A jX S and the resulting reactive power from the equation Q = 3 ⎞ V I A sin ⎝ where ⎝ is the impedance angle, which is the negative of the current angle. Ignoring generated voltage at rated conditions is E A = V ⎞ + jX S I A R A , the internal 277 0 ( ) 0.899 ( 565.3 31.8 A ) E A = j ° + &  ° 136 E A 695 38.4 = V ° so E Ar = 461 V and ™ = 5 r .2 7 ° . A MATLAB program that calculates the reactive power supplied voltage as a function of flux is shown below: % M-file: prob5_21c.m % M-file to calculate and plot the reactive power % supplied to an infinite bus as flux is varied from % 75% to 100% of the flux at rated conditions. % Define values for this generator flux_ratio = 0.90:0.01:1.00; % Flux ratio Ear = 695; % Ea at full flux dr = 38.4 * pi/180; % Torque ang at full flux Vp = 277; % Phase voltage Xs = 0.899; % Xs (ohms) % Calculate Ea for each flux Ea = flux_ratio * Ear; % Calculate delta for each flux d = asin( Ear ./ Ea .* sin(dr)); % Calculate Ia for each flux Ea = Ea .* ( cos(d) + j.*sin(d) ); Ia = ( Ea - Vp ) ./ (j*Xs); % Calculate reactive power for each flux theta = -atan2(imag(Ia),real(Ia)); Q = 3 .* Vp .* abs(Ia) .* sin(theta); % Plot the power supplied versus flux figure(1); plot(flux_ratio,Q/1000,'b-','LineWidth',2.0); title ('\bfReactive power versus flux'); xlabel ('\bfFlux (% of full-load flux)'); ylabel ('\bf\itQ\rm\bf (kVAR)'); grid on; hold off; 137 When this program is executed, the plot of reactive power versus flux is (d) The program in part (c) of this program calculated I A as a function of flux. A MATLAB program that plots the magnitude of this current as a function of flux is shown below: % M-file: prob5_21d.m % M-file to calculate and plot the armature current % supplied to an infinite bus as flux is varied from % 75% to 100% of the flux at rated conditions. % Define values for this generator flux_ratio = 0.75:0.01:1.00; % Flux ratio Ear = 695; % Ea at full flux dr = 38.4 * pi/180; % Torque ang at full flux Vp = 277; % Phase voltage Xs = 0.899; % Xs (ohms) % Calculate Ea for each flux Ea = flux_ratio * Ear; % Calculate delta for each flux d = asin( Ear ./ Ea .* sin(dr)); % Calculate Ia for each flux Ea = Ea .* ( cos(d) + j.*sin(d) ); Ia = ( Ea - Vp ) ./ (j*Xs); % Plot the armature current versus flux figure(1); plot(flux_ratio,abs(Ia),'b-','LineWidth',2.0); title ('\bfArmature current versus flux'); xlabel ('\bfFlux (% of full-load flux)'); ylabel ('\bf\itI_{A}\rm\bf (A)'); grid on; 138 hold off; When this program is executed, the plot of armature current versus flux is 5-22. A 100-MVA 12.5-kV 0.85-PF-lagging 50-Hz two-pole Y-connected synchronous generator has a per-unit synchronous reactance of 1.1 and a per-unit armature resistance of 0.012. (a) What are its synchronous reactance and armature resistance in ohms? (b) What is the magnitude of the internal generated voltage E A at the rated conditions? What is its torque angle ™ at these conditions? (c) Ignoring losses in this generator, what torque must be applied to its shaft by the prime mover at full load? S OLUTION The base phase voltage of this generator is V ⎞ ,base 12, 500 / 3 7217 V = = . Therefore, the base impedance of the generator is 2 ( ) 2 Z base 3 V ⎞ ,base 3 7217 V 1.56 = = = & S base 100, 000, 000 VA (a) The generator impedance in ohms are: ( ) 0.012 ( 1.56 ) 0.0187 R = & = & A ( ) 1.1 ( 1.56 ) 1.716 X = & = & S (b) The rated armature current is S I I = = 100 = MVA 4619 = A 3 3 ( ) 12.5 kV A L T V The power factor is 0.8 lagging, so I A 4619 36.87 =  A ° . Therefore, the internal generated voltage is = + E V + I I A A ⎞ A S A R jX 139 7217 0 ( 0.0187 )( ) 4619 36.87 A ( 1.716 )( ) 4619 36.87 A E A = ° + &  ° j + &  ° E A 13, 590 27.6 V = ° Therefore, the magnitude of the internal generated voltage E A = 13,590 V, and the torque angle ™ = 23 ° . (c) Ignoring losses, the input power would equal the output power. Since ( )( ) and OUT P 0.85 100 MVA 85 MW = = 120 120 ( ) 50 Hz n f e = = = sync P 3000 r/min 2 the applied torque would be ⎮ ⎮ = = 85, 000, 000 W 270, = 000 N m ⊕ app ind ( ) 3000 r/min ( 2  rad/r )( 1 min/60 s ) 5-23. A three-phase Y-connected synchronous generator is rated 120 MVA, 13.2 kV, 0.8 PF lagging, and 60 Hz. Its synchronous reactance is 0.9 & , and its resistance may be ignored. (a) What is its voltage regulation? (b) What would the voltage and apparent power rating of this generator be if it were operated at 50 Hz with the same armature and field losses as it had at 60 Hz? (c) What would the voltage regulation of the generator be at 50 Hz? S OLUTION (a) The rated armature current is S I I = = 12 = 0 MVA 5249 = A 3 3 ( ) 13.2 kV A L T V The power factor is 0.8 lagging, so I A 5249 36.8 = 7  A ° . The phase voltage is 13.2 kV / 3 = 7621 V. Therefore, the internal generated voltage is = + E V + I I A A ⎞ A S A R jX 7621 0 ( ) 0.9 ( 5249 36.87 A ) E A j = ° + &  ° E A 11,120 19.9 V = ° The resulting voltage regulation is 11,120  7621 RV 100% 45.9% = ⋅ = 7621 (b) If the generator is to be operated at 50 Hz with the same armature and field losses as at 60 Hz (so that the windings do not overheat), then its armature and field currents must not change. Since the voltage of the generator is directly proportional to the speed of the generator, the voltage rating (and hence the apparent power rating) of the generator will be reduced by a factor of 5/6. 5 ( ) 13.2 kV 11.0 kV = = T V ,rated 6 S rated 5 ( ) 120 MVA 100 MVA = = 6 140 . ( ) 0.899 ( 56 5.3 31.8 A ) E A = j ° + &  ° 136 E A 6 95 38.4 = V ° so E Ar = 461 V and ™ = 5 r .2 7 ° . program is executed, the plot of armature current versus flux is 5- 22. A 100-MVA 12 .5- kV 0. 85- PF-lagging 50 -Hz two-pole Y-connected synchronous generator has a per-unit synchronous. OUT P 0. 85 100 MVA 85 MW = = 120 120 ( ) 50 Hz n f e = = = sync P 3000 r/min 2 the applied torque would be ⎮ ⎮ = = 85, 000, 000