Electric Machinery Fundamentals (Solutions Manual) Part 4 ppsx

E A 2825 10.9 = V ° Problems 5-11 to 5-21 refer to a two-pole Y-connected synchronous generator rated at 470 kVA, 480 V, 60 Hz, and 0.85 PF lagging. Its armature resistance R A is 0.016 & . The core losses of this generator at rated conditions are 7 kW, and the friction and windage losses are 8 kW. The open-circuit and short-circuit characteristics are shown in Figure P5-2. 127 Note: An electronic version of the saturated open circuit characteristic can be foun in file p52_occ.dat , and an electronic version of the air-gap characteristi can be found in file p52_ag_occ.dat . These files can be used wit MATLAB programs. Column 1 contains field current in amps, and column contains open-circuit terminal voltage in volts. An electronic version of th short circuit characteristic can be found in file p52_scc.dat . Column contains field current in amps, and column 2 contains short-circuit termina current in amps. 5-11. (a) What is the saturated synchronous reactance of this generator at the rated conditions? (b) What is the unsaturated synchronous reactance of this generator? (c) Plot the saturated synchronous reactance of this generator as a function of load. S OLUTION (a) The rated armature current for this generator is S I I = = 47 = 0 kVA 565 A = 3 3 ( ) 480 V A L T V The field current required to produce this much short-circuit current may be read from the SCC. It is 0.534 A 3 . The open circuit voltage at 0.532 A is 880 V 4 , so the open-circuit phase voltage (= 508 V. The approximate saturated synchronous reactance X S is E A ) is 880/ 3 = 3 If you have MATLAB available, you can use the file p52_scc.dat and the interp1 function to look up this value as shown below. Note that column 1 of p52_scc contains field current, and column 2 contains short-circuit terminal current. load p52_scc.dat if = interp1(p52_scc(:,2),p52_scc(:,1),565) if = 128 S X 508 V 0.899 = = & 565 A (b) The unsaturated synchronous reactance X Su is the ratio of the air-gap line to the SCC. This is a straight line, so we can determine its value by comparing the ratio of the air-gap voltage to the short-circuit current at any given field current. For example, at SCC is 532 A. I F = 0.50 A, the air-gap line voltage is 1040 V, and the X Su ( ) 1040 V / 3 1.13 = = & 532 A (c) This task can best be performed with MATLAB. The open-circuit characteristic is available in a file called p52_occ.dat , and the short-circuit characteristic is available in a file called p52_scc.dat . Each of these files are organized in two columns, where the first column is field current and the second column is either open-circuit terminal voltage or short-circuit current. A program to read these files and calculate and plot X S is shown below. % M-file: prob5_11c.m % M-file to calculate and plot the saturated % synchronous reactance of a synchronous % generator. % Load the open-circuit characteristic. It is in % two columns, with the first column being field % current and the second column being terminal % voltage. load p52.occ; if_occ = p52(:,1); vt_occ = p52(:,2); % Load the short-circuit characteristic. It is in % two columns, with the first column being field % current and the second column being line current % (= armature current) load p52.scc; if_scc = p52(:,1); ia_scc = p52(:,2); % Calculate Xs if = 0.001:0.02:1; % Current steps vt = interp1(if_occ,vt_occ,If); % Terminal voltage ia = interp1(if_scc,ia_scc,If); % Current Xs = (vt ./ sqrt(3)) ./ ia; 0.534 4 If you have MATLAB available, you can use the file p52_occ.dat and the interp1 function to look up this value as shown below. Note that column 1 of p52_occ contains field current, and column 2 contains open-circuit terminal voltage. load p52_occ.dat vt = interp1(p52_occ(:,1),p52_occ(:,2),0.534) vt = 880.400 129 % Plot the synchronous reactance figure(1) plot(If,Xs,'LineWidth',2.0); title ('\bfSaturated Synchronous Reactance \itX_{s} \rm'); xlabel ('\bfField Current (A)'); ylabel ('\bf\itX_{s} \rm\bf(\Omega)'); grid on; The resulting plot is: 5-12. (a) What are the rated current and internal generated voltage of this generator? (b) What field current does this generator require to operate at the rated voltage, current, and power factor? S OLUTION (a) The rated line and armature current for this generator is S I I = = 47 = 0 kVA 565 A = 3 3 ( ) 480 V A L T V The power factor is 0.85 lagging, so I A 565.3 31.8 = A  ° . The rated phase voltage is V ⎞ = 480 V / 3 = 277 V. The saturated synchronous reactance at rated conditions was found to be 0.450 & in the previous problem. Therefore, the internal generated voltage is = + + A A E V ⎞ A I S A R jX I 277 0 ( ) 0. ( 016 565.3 31.8 A ) ( ) 0.899 ( 565.3 31.8 A ) E A = ° + &  ° j + &  ° E A 509 30.5 = V ° (b) This internal generated voltage corresponds to a no-load terminal voltage of ( ) 3 509 = 881 V. From the open-circuit characteristic, the required field current would be 0.535 A. 5-13. What is the voltage regulation of this generator at the rated current and power factor? S OLUTION The voltage regulation is 130 V V   VR ,nl ,fl 10 T T 0% 881 = ⋅ 480 = 100% 83.5 ⋅ % = T V ,fl 480 5-14. If this generator is operating at the rated conditions and the load is suddenly removed, what will the terminal voltage be? S OLUTION From the above calculations, T V will be 881 V. 5-15. What are the electrical losses in this generator at rated conditions? S OLUTION The electrical losses are 2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 565 A 0.016 15.3 kW 5-16. If this machine is operating at rated conditions, what input torque must be applied to the shaft of this generator? Express your answer both in newton-meters and in pound-feet. S OLUTION To get the applied torque, we must know the input power. The input power to this generator is equal to the output power plus losses. The rated output power and the losses are ( )( ) OUT P 470 kVA 0.85 400 kW = = 2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 565 A 0.016 15.3 kW F& P = W 8 kW co P re = 7 kW st P ray = (assumed 0) = + + + + = IN OUT P P CU P F&W P Therefore, the applied torque is core P stray P 430.3 kW ⎮ IN P = = 430.3 kW 2280 N = m ⊕ APP ⎤ m ( ) 1800 r/min 2 r  ad 1 min 7.04 P 1 r 60 s 7.04 ( 430.3 kW ) or ⎮ APP 1800 r/min 1680 lb ft = = = ⊕ m n 5-17. What is the torque angle ™ of this generator at rated conditions? S OLUTION From the calculations in Problem 5-12, ™ = 30.5 ° . 5-18. Assume that the generator field current is adjusted to supply 480 V under rated conditions. What is the static stability limit of this generator? ( Note: You may ignore R A to make this calculation easier.) How close is the full-load condition of this generator to the static stability limit? S OLUTION At rated conditions, E A 509 30.5 = V ° . Therefore, the static stability limit is 3 ( ) 3 277 V ( ) 509 V P ⎞ A V E = = = MAX X S 0.899 & 471 kW The full-load rated power of this generator is reasonably close to the static stability limit. Normal generators would have more margin than this. 131 5-19. Assume that the generator field current is adjusted to supply 480 V under rated conditions. Plot the power supplied by the generator as a function of the torque angle ™ . ( Note: You may ignore calculation easier.) R A to make this S OLUTION We will again ignore R A to make this calculation easier. The power supplied by the generator is 3 ⎞ V E P sin A 3 ( ) 277 V ( 509 V ) ( ) sin 471 kW sin ™ ™ ™ = = = S X G 0.899 & The power supplied as a function of the torque angle ™ may be plotted using a simple MATLAB program: % M-file: prob5_19.m % M-file to plot the power output of a % synchronous generator as a function of % the torque angle. % Calculate Xs delta = (0:1:90); % Torque angle (deg) Pout = 561 .* sin(delta * pi/180); % Pout % Plot the output power figure(1) plot(delta,Pout,'LineWidth',2.0); title ('\bfOutput power vs torque angle \delta'); xlabel ('\bfTorque angle \delta (deg)'); ylabel ('\bf\itP_{OUT} \rm\bf(kW)'); grid on; The resulting plot is: 5-20. Assume that the generator’s field current is adjusted so that the generator supplies rated voltage at the rated load current and power factor. If the field current and the magnitude of the load current are held constant, how will the terminal voltage change as the load power factor varies from 0.85 PF lagging to 0.85 PF 132 leading? Make a plot of the terminal voltage versus the impedance angle of the load being supplied by this generator. S OLUTION If the field current is held constant, then the magnitude of E A will be constant, although its angle ™ will vary. Also, the magnitude of the armature current is constant. Since we also know R A , X S , and the current angle ⎝ , we know enough to find the phase voltage ⎞ V , and therefore the terminal voltage T V . At lagging power factors, ⎞ V can be found from the following relationships: E A ⎝ ™ ⎝ I A By the Pythagorean Theorem, ⎝ jX I V ⎞ S A I R AA ( ) 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 ⎝ At unity power factor, ⎞ V can be found from the following relationships: E A ™ I A V ⎞ jX I S A R I AA By the Pythagorean Theorem, ( 2 2 = + ) 2 A S E V ⎞ 2 A X I ( ) 2 V ⎞ = A E  X S I A At leading power factors, ⎞ V can be found from the following relationships: E A jX I S A ⎝ I A ™ ⎝ By the Pythagorean Theorem, I R A A ⎝ ⎞ V 133 . stray P 43 0.3 kW ⎮ IN P = = 43 0.3 kW 2280 N = m ⊕ APP ⎤ m ( ) 1800 r/min 2 r  ad 1 min 7. 04 P 1 r 60 s 7. 04 ( 43 0.3 . is 130 V V   VR ,nl ,fl 10 T T 0% 881 = ⋅ 48 0 = 100% 83.5 ⋅ % = T V ,fl 48 0 5- 14. If this generator is operating at the rated conditions and. 47 = 0 kVA 565 A = 3 3 ( ) 48 0 V A L T V The power factor is 0.85 lagging, so I A 565.3 31.8 = A  ° . The rated phase voltage is V ⎞ = 48 0