Electric Machinery Fundamentals (Solutions Manual) Part 2 ppt

(a) This generator is Y-connected, so L A I I = . At rated conditions, the line and phase current in this generator is I I 1000 kVA 251 A = = P = = at an angle of –36.87 ° A L ( 3 3 2300 ) V L V = = . The internal generated voltage of the machineThe phase voltage of this machine is ⎞ T / V V 3 1328 V is = + E V + I I A A ⎞ A S A R jX 1328 0 ( 0.15 )( 251 36.87 A ) ( 1.1 )( 251 36.87 A ) E A = ° + &  ° j + &  ° E A 1537 7.4 V = ° The input power to this generator is equal to the output power plus losses. The rated output power is ( )( ) OUT P 1000 kVA 0.8 800 kW = = 2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 251 A 0.15 28.4 kW F& P = W 24 kW co P re = 18 kW st P ray = (assumed 0) = + + + + = IN OUT P P CU P F&W P core P stray P 870.4 kW η OUT P 100% 800 kW 100% 91.9% = ⋅ = ⋅ = IN P 870.4 kW (b) If the generator is loaded to rated kVA with lagging loads, the phase voltage is φ V = 1328 0  ° V and the internal generated voltage is E A 1537 7.4 V = ° . Therefore, the phase voltage at no- load would be V ⎞ 1537 0 V = ° . The voltage regulation would be: 1537  1328 RV 100% 15.7% = ⋅ = 1328 (c) If the generator is loaded to rated kVA with leading loads, the phase voltage is φ V the internal generated voltage is = 1328 0  ° V and = + E V + I I A A ⎞ A S A R jX 1328 0 ( ) 0. ( 15 251 36.87 A ) ( 1.1 )( 251 36.87 A ) E A = ° + & ° j + & ° E A 1217 11.5 = V ° The voltage regulation would be: 1217  1328 RV 100% 8.4% = ⋅ =  1328 (d) If the generator is loaded to rated kVA at unity power factor, the phase voltage is V φ = 1328 0  ° V and the internal generated voltage is = + E V + I I A A ⎞ A S A R jX 115 1328 0 ( 0.15 )( ) 251 0 A ( 1.1 )( ) 251 0 A E A = ° + & ° j + & ° E A 1393 11.4 = V ° The voltage regulation would be: 1393  1328 RV 100% 4.9% = ⋅ = 1328 (e) For this problem, we will assume that the terminal voltage is adjusted to 2300 V at no load conditions, and see what happens to the voltage as load increases at 0.8 lagging, unity, and 0.8 leading power factors. Note that the maximum current will be 251 A in any case. A phasor diagram representing the situation at lagging power factor is shown below: E A ⎝ ™ ⎝ I A By the Pythagorean Theorem, ⎝ jX I V ⎞ S A I R AA 2 ( ) = + cos + sin ( ) 2 + co ⎝ ⎝ s  sin ⎝ 2 ⎝ A A E V ⎞ R A I S A X I S A X I A R S I 2 ( ) =  cos s  in 2 co ⎝ ⎝  s s  in ⎝ ⎝ ⎞ A S V E A X I A S R I A A R I S A X I A phasor diagram representing the situation at leading power factor is shown below: E A jX I S A ⎝ I A I R ™ ⎝ A A ⎝ ⎞ V By the Pythagorean Theorem, 2 ( = + cos  sin ) ( ) 2 + co ⎝ ⎝ s + sin ⎝ 2 ⎝ A A E V ⎞ R A I S A X I S A X I A R S I 2 ( ) =  cos s + in 2 co ⎝ ⎝  s s + in ⎝ ⎝ ⎞ A S V E A X I A S R I A A R I S A X I A phasor diagram representing the situation at unity power factor is shown below: E A ™ I A V ⎞ 116 jX I S A R I AA By the Pythagorean Theorem, ( 2 2 = + ) 2 A S E V ⎞ A X I 2 ( ) =  2 ⎞ A S V E A X I The MATLAB program is shown below takes advantage of this fact. % M-file: prob5_4e.m % M-file to calculate and plot the terminal voltage % of a synchronous generator as a function of load % for power factors of 0.8 lagging, 1.0, and 0.8 leading. % Define values for this generator EA = 1328; % Internal gen voltage I = 0:2.51:251; % Current values (A) R = 0.15; % R (ohms) X = 1.10; % XS (ohms) % Calculate the voltage for the lagging PF case VP_lag = sqrt( EA^2 - (X.*I.*0.8 - R.*I.*0.6).^2 ) - R.*I.*0.8 - X.*I.*0.6; VT_lag = VP_lag .* sqrt(3); % Calculate the voltage for the leading PF case VP_lead = sqrt( EA^2 - (X.*I.*0.8 + R.*I.*0.6).^2 ) - R.*I.*0.8 + X.*I.*0.6; VT_lead = VP_lead .* sqrt(3); % Calculate the voltage for the unity PF case VP_unity = sqrt( EA^2 - (X.*I).^2 ); VT_unity = VP_unity .* sqrt(3); % Plot the terminal voltage versus load plot(I,abs(VT_lag),'b-','LineWidth',2.0); hold on; plot(I,abs(VT_unity),'k ','LineWidth',2.0); plot(I,abs(VT_lead),'r ','LineWidth',2.0); title ('\bfTerminal Voltage Versus Load'); xlabel ('\bfLoad (A)'); ylabel ('\bfTerminal Voltage (V)'); legend('0.8 PF lagging','1.0 PF','0.8 PF leading'); axis([0 260 1500 2500]); grid on; hold off; The resulting plot is shown below: 117 5-5. Assume that the field current of the generator in Problem 5-2 has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor. (You may ignore the effects of answering these questions.) R A when (a) What is the torque angle ™ of the generator when supplying rated current at unity power factor? (b) When this generator is running at full load with unity power factor, how close is it to the static stability limit of the machine? S OLUTION (a) The torque ™ angle can be found by calculating E A : = + E V + I I A A ⎞ A S A R jX 1328 0 ( 0.15 )( ) 251 0 A ( 1.1 )( ) 251 0 A E A = ° + & ° j + & ° E A 1393 11.4 = V ° Thus the torque angle ™ = 11.4 ° . (b) The static stability limit occurs at ™ = ° 90 . This generator is a very long way from that limit. If we ignore the internal resistance of the generator, the output power will be given by P A V E sin = 3 ⎞ ™ X S and the output power is proportional to sin ™ . Since sin 11.4 °= 0.198 , and sin 90 1. °= stability limit is about 5 times the current output power of the generator. 00 , the static 5-6. A 480-V 400-kVA 0.85-PF-lagging 50-Hz four-pole  -connected generator is driven by a 500-hp diesel engine and is used as a standby or emergency generator. This machine can also be paralleled with the normal power supply (a very large power system) if desired. (a) What are the conditions required for paralleling the emergency generator with the existing power system? What is the generator’s rate of shaft rotation after paralleling occurs? 118 (b) If the generator is connected to the power system and is initially floating on the line, sketch the resulting magnetic fields and phasor diagram. (c) The governor setting on the diesel is now increased. Show both by means of house diagrams and by means of phasor diagrams what happens to the generator. How much reactive power does the generator supply now? (d) With the diesel generator now supplying real power to the power system, what happens to the generator as its field current is increased and decreased? Show this behavior both with phasor diagrams and with house diagrams. S OLUTION (a) To parallel this generator to the large power system, the required conditions are: 1. The generator must have the same voltage as the power system. 2. The phase sequence of the oncoming generator must be the same as the phase sequence of the power system. 3. The frequency of the oncoming generator should be slightly higher than the frequency of the running system. 4. The circuit breaker connecting the two systems together should be shut when the above conditions are met and the generator is in phase with the power system. After paralleling, the generator’s shaft will be rotating at 120 120 ( ) 50 Hz n f e = = 1500 r/ = min P m 4 (b) The magnetic field and phasor diagrams immediately after paralleling are shown below: BB E S R A I jX I S A B net A V ⎞ (c) When the governor setpoints on the generator are increased, the emergency generator begins to supply more power to the loads, as shown below: f e E A jX I S A I P P P 1 sys A V ⎞ 2 P G Note that as the load increased with E A reactive power. constant, the generator began to consume a small amount of (d) With the generator now supplying power to the system, an increase in field current increases the reactive power supplied to the loads, and a decrease in field current decreases the reactive power supplied to the loads. 119 V T I A1 E A1 E A2 E A3 QQ QQ 3 G 2 1 Q sys I A2 I ⎞ V jX I S A A3 V T I A2 A2 EE A1 jX I S A Q Q Q G 1 2 Q sys I A1 V ⎞ 5-7. A 13.8-kV 10-MVA 0.8-PF-lagging 60-Hz two-pole Y-connected steam-turbine generator has a synchronous reactance of 12 & per phase and an armature resistance of 1.5 & per phase. This generator is operating in parallel with a large power system (infinite bus). (a) What is the magnitude of E A at rated conditions? (b) What is the torque angle of the generator at rated conditions? (c) If the field current is constant, what is the maximum power possible out of this generator? How much reserve power or torque does this generator have at full load? (d) At the absolute maximum power possible, how much reactive power will this generator be supplying or consuming? Sketch the corresponding phasor diagram. (Assume I F is still unchanged.) S OLUTION (a) The phase voltage of this generator at rated conditions is V 13, 800 V 7967 V ⎞ = = 3 The armature current per phase at rated conditions is S 10, 000, = = 000 VA 418 A = A I 3 3 ( ) 13,800 V T V Therefore, the internal generated voltage at rated conditions is = + + A A E V ⎞ A I S A R jX I 7967 0 ( ) 1.5 ( 418 36.87 A ) ( 12.0 )( 418 36.87 A ) E A = ° + &  ° j + &  ° E A 12, 040 17.6 V = ° 120 The magnitude of E A is 12,040 V. (b) The torque angle of the generator at rated conditions is ™ = 17.6 ° . (c) Ignoring R A , the maximum output power of the generator is given by 3 3 ( ) 7967 V ( 12,040 V ) P ⎞ A V E = = = MAX X S 12 & 24.0 MW The power at maximum load is 8 MW, so the maximum output power is three times the full load output power. (d) The phasor diagram at these conditions is shown below: E A jX I S A I A R I A A V ⎞ Under these conditions, the armature current is A E V  I ⎞ 12, 040 90 V ° - 7967 0 V ° 1194 40.6 A = = = ° A 1.5 12.0 + + & A S R jX j The reactive power produced by the generator at this point is 3 sin 3 ( ) 7967 V ( 119 = = ⎞ ⎝ A Q V I 4 A ) sin ( 0 40.6 ) 18 ° .6  MVA ° R =  The generator is actually consuming reactive power at this time. 5-8. A 480-V, 100-kW, two-pole, three-phase, 60-Hz synchronous generator’s prime mover has a no-load speed of 3630 r/min and a full-load speed of 3570 r/min. It is operating in parallel with a 480-V, 75-kW, four- pole, 60-Hz synchronous generator whose prime mover has a no-load speed of 1800 r/min and a full-load speed of 1785 r/min. The loads supplied by the two generators consist of 100 kW at 0.85 PF lagging. (a) Calculate the speed droops of generator 1 and generator 2. (b) Find the operating frequency of the power system. (c) Find the power being supplied by each of the generators in this system. (d) If V T is 460 V, what must the generator’s operators do to correct for the low terminal voltage? S OLUTION The no-load frequency of generator 1 corresponds to a frequency of ( ) 3630 r/min ( 2 ) f m n P = = = nl1 120 120 60.5 Hz The full-load frequency of generator 1 corresponds to a frequency of . OUT P 1000 kVA 0.8 800 kW = = 2 ( ) = = 2 ( ) & = CU 3 A A P I 3 R 25 1 A 0.15 28 .4 kW F& P = W 24 kW co P re = 18 kW st P . 2 ( ) = + cos + sin ( ) 2 + co ⎝ ⎝ s  sin ⎝ 2 ⎝ A A E V ⎞ R A I S A X I S A X I A R S I 2 ( ) =  cos s  in 2 . Theorem, 2 ( = + cos  sin ) ( ) 2 + co ⎝ ⎝ s + sin ⎝ 2 ⎝ A A E V ⎞ R A I S A X I S A X I A R S I 2 ( ) =  cos s + in 2 co ⎝