115 (a) This generator is Y-connected, so LA II= . At rated conditions, the line and phase current in this generator is ( ) 1000 kVA 251 A 3 3 2300 V AL L P II V == = = at an angle of –36.87 ° The phase voltage of this machine is / 31328 V T VV φ == . The internal generated voltage of the machine is AAASA RjX φ =+ +EV I I ( ) ( ) ( ) ( ) 1328 0 0.15 251 36.87 A 1.1 251 36.87 A A j=∠°+ Ω∠− °+ Ω∠− °E 1537 7.4 V A =∠°E The input power to this generator is equal to the output power plus losses. The rated output power is ()() OUT 1000 kVA 0.8 800 kWP == ()( ) 2 2 CU 3 3 251 A 0.15 28.4 kW AA PIR== Ω= F&W 24 kWP = core 18 kWP = stray (assumed 0)P = IN OUT CU F&W core stray 870.4 kWPP PP P P=++++= OUT IN 800 kW 100% 100% 91.9% 870.4 kW P P η =× = × = (b) If the generator is loaded to rated kVA with lagging loads, the phase voltage is 1328 0 V φ = ∠ °V and the internal generated voltage is 1537 7.4 V A E =∠°. Therefore, the phase voltage at no-load would be 1537 0 VV φ =∠°. The voltage regulation would be: 1537 1328 VR 100% 15.7% 1328 − =×= (c) If the generator is loaded to rated kVA with leading loads, the phase voltage is 1328 0 V φ = ∠ °V and the internal generated voltage is AAASA RjX φ =+ +EV I I ()( )()( ) 1328 0 0.15 251 36.87 A 1.1 251 36.87 A A jE =∠°+ Ω∠°+ Ω∠ ° 1217 11.5 V A E =∠° The voltage regulation would be: 1217 1328 VR 100% 8.4% 1328 − =×=− (d) If the generator is loaded to rated kVA at unity power factor, the phase voltage is 1328 0 V φ = ∠ °V and the internal generated voltage is AAASA RjX φ =+ +EV I I 116 ()()()() 1328 0 0.15 251 0 A 1.1 251 0 A A jE =∠°+ Ω∠°+ Ω∠° 1393 11.4 V A E =∠° The voltage regulation would be: 1393 1328 VR 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: I A V φ E A θ δ θ θ I A R A jX S I A By the Pythagorean Theorem, () () 2 2 2 cos sin cos sin AAASA SAAS E V RI XI XI RI φ θθ θθ =+ + + − () 2 2 cos sin cos sin ASA AS AA SA VEXI RI RI XI φ θθ θ θ =− − − − A phasor diagram representing the situation at leading power factor is shown below: I A V φ E A θ δ θ θ I A R A jX S I A By the Pythagorean Theorem, ( ) () 2 2 2 cos sin cos sin AAASA SAAS E V RI XI XI RI φ θθ θθ =+ − + + () 2 2 cos sin cos sin ASA AS AA SA VEXI RI RI XI φ θθ θ θ =− + − + A phasor diagram representing the situation at unity power factor is shown below: I A V φ E A δ I A R A jX S I A 117 By the Pythagorean Theorem, ( ) 2 22 ASA EV XI φ =+ () 2 2 ASA VEXI φ =− 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: 118 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 A R when answering these questions.) (a) What is the torque angle δ 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 δ angle can be found by calculating A E : AAASA RjX φ =+ +EV I I ()()()() 1328 0 0.15 251 0 A 1.1 251 0 A A jE =∠°+ Ω∠°+ Ω∠° 1393 11.4 V A E =∠° Thus the torque angle δ = 11.4 ° . (b) The static stability limit occurs at °= 90 δ . 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 3 sin A S VE P X φ δ = and the output power is proportional to sin δ . Since sin 11.4 0.198°= , and sin 90 1.00°= , the static stability limit is about 5 times the current output power of the generator. 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? 119 (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 50 Hz 120 1500 r/min 4 e m f n P == = (b) The magnetic field and phasor diagrams immediately after paralleling are shown below: I A V φ E A jX S I A B R B S B net (c) When the governor setpoints on the generator are increased, the emergency generator begins to supply more power to the loads, as shown below: I A V φ E A jX S I A f e P 1 P 2 P sys P G Note that as the load increased with A E constant, the generator began to consume a small amount of reactive power. (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. 120 V φ E A 1 jX S I A Q sys Q G Q 2 Q 1 Q 3 E A 2 E A 3 I A 3 I A 2 I A 1 V φ E A 1 jX S I A Q sys Q G Q 2 Q 1 E A 2 I A 2 I A 1 V T V T 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 Ω per phase and an armature resistance of 1.5 Ω 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 13,800 V 7967 V 3 V φ == The armature current per phase at rated conditions is () 10,000,000 VA 418 A 3 3 13,800 V A T S I V == = Therefore, the internal generated voltage at rated conditions is AAASA RjXEV I I φ =+ + ()( )( )( ) 7967 0 1.5 418 36.87 A 12.0 418 36.87 A A jE =∠°+Ω∠− °+ Ω∠− ° 12,040 17.6 V A E =∠° 121 The magnitude of A E is 12,040 V. (b) The torque angle of the generator at rated conditions is δ = 17.6°. (c) Ignoring A R , the maximum output power of the generator is given by ()( ) MAX 3 3 7967 V 12,040 V 24.0 MW 12 A S VE P X φ == = Ω 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: jX S I A V φ I A E A I A R A Under these conditions, the armature current is 12,040 90 V - 7967 0 V 1194 40.6 A 1.5 12.0 A A AS RjX j EV I φ − ∠° ∠° == =∠° ++Ω The reactive power produced by the generator at this point is ()()( ) 3 sin 3 7967 V 1194 A sin 0 40.6 18.6 MVAR A QVI φ θ == °−°=− 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 ()() nl1 3630 r/min 2 60.5 Hz 120 120 m nP f == = The full-load frequency of generator 1 corresponds to a frequency of 122 ()() fl1 3570 r/min 2 59.5 Hz 120 120 m nP f == = The no-load frequency of generator 2 corresponds to a frequency of ()() nl2 1800 r/min 4 60.00 Hz 120 120 m nP f == = The full-load frequency of generator 2 corresponds to a frequency of ()() fl2 1785 r/min 4 59.50 Hz 120 120 m nP f == = (a) The speed droop of generator 1 is given by nl fl 1 fl 3630 r/min 3570 r/min SD 100% 100% 1.68% 3570 r/min nn n −− =×= ×= The speed droop of generator 2 is given by nl fl 2 fl 1800 r/min 1785 r/min SD 100% 100% 0.84% 1785 r/min nn n −− =×= ×= (b) The power supplied by generator 1 is given by () 1 1 nl1 sysP Ps f f=− and the power supplied by generator 1 is given by () 2 2 nl2 sysP Ps f f=− The power curve’s slope for generator 1 is 1 nl fl 0.1 MW 0.1 MW/Hz 60.5 Hz 59.5 Hz P P s ff == = −− The power curve’s slope for generator 1 is 2 nl fl 0.075 MW 0.150 MW/Hz 60.00 Hz 59.50 Hz P P s ff == = −− The no-load frequency of generator 1 is 60.5 Hz and the no-load frequency of generator 2 is 60 Hz. The total power that they must supply is 100 kW, so the system frequency can be found from the equations LOAD 1 2 PPP=+ ()() LOAD 1 nl1 sys 2 nl2 sysPP Psffsff=−+− () () () () sys sys 100 kW 0.1 MW/Hz 60.5 Hz 0.15 MW/Hz 60.0 Hzff=−+ − () () sys sys 100 kW 6050 kW 0.10 MW/Hz 9000 kW 0.15 MW/Hzff=− +− () sys 0.25 MW/Hz 6050 kW 9000 kW 100 kWf =+− sys 14,950 kW 59.8 Hz 0.25 MW/Hz f == (c) The power supplied by generator 1 is 123 () ()( ) 11nl1sys 0.1 MW/Hz 60.5 Hz 59.8 Hz 70 kW P Ps f f=−= − = The power supplied by generator 2 is () ()( ) 22nl2sys 0.15 MW/Hz 60.0 Hz 59.8 Hz 30 kW P Ps f f=−= − = (d) If the terminal voltage is 460 V, the operators of the generators must increase the field currents on both generators simultaneously. That action will increase the terminal voltages of the system without changing the power sharing between the generators. 5-9. Three physically identical synchronous generators are operating in parallel. They are all rated for a full load of 3 MW at 0.8 PF lagging. The no-load frequency of generator A is 61 Hz, and its speed droop is 3.4 percent. The no-load frequency of generator B is 61.5 Hz, and its speed droop is 3 percent. The no-load frequency of generator C is 60.5 Hz, and its speed droop is 2.6 percent. (a) If a total load consisting of 7 MW is being supplied by this power system, what will the system frequency be and how will the power be shared among the three generators? (b) Create a plot showing the power supplied by each generator as a function of the total power supplied to all loads (you may use MATLAB to create this plot). At what load does one of the generators exceed its ratings? Which generator exceeds its ratings first? (c) Is this power sharing in (a) acceptable? Why or why not? (d) What actions could an operator take to improve the real power sharing among these generators? S OLUTION (a) Speed droop is defined as nl fl nl fl fl fl SD 100% 100% nn f f nf −− =×=× so nl fl SD 1 100 f f = + Thus, the full-load frequencies of generators A, B, and C are nl,A fl,A A 61 Hz 59.0 Hz SD 3.4 11 100 100 f f === ++ nl,B fl,B B 61.5 Hz 59.71 Hz SD 3.0 11 100 100 f f === ++ nl,C fl,C C 60.5 Hz 58.97 Hz SD 2.6 11 100 100 f f === ++ and the slopes of the power-frequency curves are: 3 MW 1.5 MW/Hz 2 Hz PA S == 3 MW 1.676 MW/Hz 1.79 Hz PB S == 3 MW 1.961 MW/Hz 1.53 Hz PC S == 124 (a) The total load is 7 MW, so the system frequency is ()()() LOAD nlA sys nlB sys nlC sysPA PB PC P sffsffsff=−+−+− ( ) () ( ) () ( ) () sys sys sys 7 MW 1.5 61.0 1.676 61.5 1.961 60.5fff=−+ −+ − sys sys sys 7 MW 91.5 1.5 103.07 1.676 118.64 1.961fff=−+− +− sys 5.137 306.2f = sys 59.61 Hzf = The power supplied by each generator will be () ()( ) nlA sys 1.5 MW/Hz 61.0 Hz 59.61 Hz 2.09 MW APA Psf f=−= − = () ()( ) nlB sys 1.676 MW/Hz 61.5 Hz 59.61 Hz 3.17 MW BPB Psf f=−= − = () ()( ) nlC sys 1.961 MW/Hz 60.5 Hz 59.61 Hz 1.74 MW CPC Ps f f=−= − = (b) The equation in part (a) can be re-written slightly to express system frequency as a function of load. () () () () () () LOAD sys sys sys 1.5 61.0 1.676 61.5 1.961 60.5Pf f f=−+ −+ − LOAD sys sys sys 91.5 1.5 103.07 1.676 118.64 1.961Pf f f=− + − + − sys LOAD 5.137 313.2fP=− LOAD sys 313.2 5.137 P f − = A MATLAB program that uses this equation to determine the power sharing among the generators as a function of load is shown below: % M-file: prob5_9b.m % M-file to calculate and plot the power sharing among % three generators as a function of load. % Define values for this generator fnlA = 61.0; % No-load freq of Gen A fnlB = 61.5; % No-load freq of Gen B fnlC = 60.5; % No-load freq of Gen C spA = 1.5; % Slope of Gen A (MW/Hz) spB = 1.676; % Slope of Gen B (MW/Hz) spC = 1.961; % Slope of Gen C (MW/Hz) Pload = 0:0.05:10; % Load in MW % Calculate the system frequency fsys = (313.2 - Pload) ./ 5.137; % Calculate the power of each generator PA = spA .* ( fnlA - fsys); PB = spB .* ( fnlB - fsys); PC = spC .* ( fnlC - fsys); % Plot the power sharing versus load plot(Pload,PA,'b-','LineWidth',2.0); [...]... voltage of the generators is 416 0 V / their rated current is IA = IL = 3 = 2402 V, and since the generators are Y-connected, S 6250 kVA = = 8 67 A 3 VT 3 ( 416 0 V ) The power factor is 0.85 lagging, so I A = 8 67 − 31. 8° A Therefore, E A = Vφ + RAI A + jX S I A E A = 2402∠0° + ( 0.04 Ω )(8 67 ∠ − 31. 8° A ) + j ( 0 .75 Ω )(8 67 − 31. 8° A ) 12 6 E A = 2825 10 .9° V Problems 5 -11 to 5- 21 refer to a two-pole Y-connected... power and the losses are POUT = ( 470 kVA )( 0.85) = 400 kW PCU = 3 I A2 RA = 3 ( 565 A ) ( 0. 016 Ω ) = 15 .3 kW 2 PF&W = 8 kW Pcore = 7 kW Pstray = (assumed 0) PIN = POUT + PCU + PF&W + Pcore + Pstray = 430.3 kW Therefore, the applied torque is τ APP = or 5 - 17 τ APP = PIN ωm = 430.3 kW (18 00 r/min ) 2π rad 1r 1 min 60 s = 2280 N ⋅ m 7. 04 P 7. 04 (430.3 kW ) = = 16 80 lb ⋅ ft 18 00 r/min nm What is the torque... a frequency of ( 5. 31 MW = ( 3.0 MW/Hz ) 61 Hz − f sys ) f sys = 59.23 Hz At this point the power supplied by Generators 1 and 2 is ( ) P1 = P2 = sP1 f nl1 − f sys = ( 2.5 MW/Hz )( 61 Hz − 59.23 Hz ) = 4.425 MW The total power supplied by the generators at this condition is 14 .16 MW (c) To get each of the generators to supply 5. 31 MW at 60 Hz, the no-load frequencies of Generator 1 and Generator 2 would... 5 -13 What is the voltage regulation of this generator at the rated current and power factor? SOLUTION The voltage regulation is 13 0 3 (509 ) = 8 81 V VR = 5 -14 VT ,nl − VT ,fl 8 81 − 480 × 10 0% = × 10 0% = 83.5% 480 VT ,fl If this generator is operating at the rated conditions and the load is suddenly removed, what will the terminal voltage be? SOLUTION From the above calculations, VT will be 8 81 V 5 -15 ... S 470 kVA = = 565 A 3 VT 3 ( 480 V ) The power factor is 0.85 lagging, so I A = 565.3∠ − 31. 8° A The rated phase voltage is Vφ = 480 V / 3 = 277 V The saturated synchronous reactance at rated conditions was found to be 0.450 Ω in the previous problem Therefore, the internal generated voltage is E A = Vφ + RAI A + jX S I A E A = 277 ∠0° + ( 0. 016 Ω )(565.3∠ − 31. 8° A ) + j ( 0.899 Ω)( 565.3∠ − 31. 8°... ( 277 V )(509 V ) sin δ = ( 4 71 kW ) sin δ 0.899 Ω The power supplied as a function of the torque angle δ 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 = 5 61 * sin(delta * pi /18 0); % Pout % Plot the output power figure (1) ... are 61 Hz, then the power supplied by the generators will be ( (f (f ) ) = (2.5 MW/Hz )( 61 Hz − 60 Hz ) = 2.5 MW ) = (3.0 MW/Hz )( 61 Hz − 60 Hz ) = 3.0 MW P1 = sP1 f nl1 − fsys = ( 2.5 MW/Hz )( 61 Hz − 60 Hz ) = 2.5 MW P2 = s P 2 nl2 − fsys P3 = s P 3 nl3 − fsys Therefore the total power supplied by the generators is 8 MW (b) The maximum power supplied by any one generator is (6250 kVA)(0.85) = 5. 31 MW... 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 RA is 0. 016 Ω 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 12 7 Note: 5 -11 An electronic version of the saturated open circuit characteristic... SOLUTION At rated conditions, E A = 509 ∠30.5° V Therefore, the static stability limit is PMAX = 3 Vφ E A XS = 3 ( 277 V )(509 V ) = 4 71 kW 0.899 Ω The full-load rated power of this generator is reasonably close to the static stability limit generators would have more margin than this 13 1 Normal 5 -19 Assume that the generator field current is adjusted to supply 480 V under rated conditions Plot the power... load p52.scc; if_scc = p52(: ,1) ; ia_scc = p52(:,2); % Calculate Xs if = 0.0 01: 0.02 :1; vt = interp1(if_occ,vt_occ,If); ia = interp1(if_scc,ia_scc,If); Xs = (vt / sqrt(3)) / ia; % Current steps % Terminal voltage % Current 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, . ( ) () ( ) () ( ) () sys sys sys 7 MW 1. 5 61. 0 1. 676 61. 5 1. 9 61 60.5fff=−+ −+ − sys sys sys 7 MW 91. 5 1. 5 10 3. 07 1. 676 11 8.64 1. 961fff=−+− +− sys 5 .13 7 306.2f = sys 59. 61 Hzf = The power supplied. AAASA RjX φ =+ +EV I I ()( )()( ) 13 28 0 0 .15 2 51 36. 87 A 1. 1 2 51 36. 87 A A jE =∠°+ Ω∠°+ Ω∠ ° 12 17 11 .5 V A E =∠° The voltage regulation would be: 12 17 13 28 VR 10 0% 8.4% 13 28 − =×=− (d) If the. 1. 9 61 60.5Pf f f=−+ −+ − LOAD sys sys sys 91. 5 1. 5 10 3. 07 1. 676 11 8.64 1. 961Pf f f=− + − + − sys LOAD 5 .13 7 313 .2fP=− LOAD sys 313 .2 5 .13 7 P f − = A MATLAB program that uses this equation