EECE 370 Lecture, Spring 2008

5-1. §-2. At a location in Europe, it is necessary to supply 200 kW? of 60-Hz power. The only power sources available operate at 50 Hz. It is decided to gererate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous gererator. How many poles should each of the tyo machines have in order to convert 50-Hz power to 60-Hz power? SoLumien The speed of a synchronous machire is related to its frequeney by the equation _ 120%, - #- Pe To make a 50 Hz anda 60 H2 machire have the sare mechanical speed so that they can be coupled together, we see that _ 120(50 Hz) _ 120(60 Hz) Ht, om A f, F812 Pp 5 10 Therefore, a 10-pole synchronous toter rust be coupled to a 12-pole synchronous sererator to aceormplish this fraqueney conversion. A 2300-¥ 1000--VA 0.8-PRlasving 60-H2z two-pole Y-connected synchronous seneratoer has a synchronous reactance of 1.1 £2 and an armature resistance of 0.15 £2. At a0 Hz, its friction and windase losses are 24 KW, andits core losses are 18 kW. The field circuit has a de voltage of 200 VY, and the maximum fp is 10.4, The resistance of the field circuit is adjustable over the range from 20 te 200 £2, The OCC of this gener ator is shown in Figure P3-1, fa) How tiuch field current is required to make V, equal to 2300 V when the sererator is ranning at no load? fb) What is the internal generated voltage of this machine at rated conditions? fc) How much field current is required to make V, equal to 2300 Y when the generator is running at rated conditions? (a) How much power and torque must the generators prime mover be capable of supplying? fe) Construct a capability airve for this generat er, An electronic version of this open circuit characteristic can be found in file pol_oce. dat, which can be used with MATLAB programs, Column 1 contains field currant in amps, and column 2 contains open-circuit terminal voltage 2700 3 s A E Open-circuit terminal voltage, a 8 fs 8 i S 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Field current, A SOLUTION (a) If the no-load terminal voltage is 2300 V, the required field current can be read directly from the open-circuit characteristic It is 4.25 A. (6) This generator is Y-connected, so J, =J,. At rated conditions, the line and phase current in this gererator is L=l, =P - M000 KVA 951.4 at an angle of -36.87° ~3V, ¥3(2300 V) The phase voltage of this machine is V, = V, {V3 = 1328 V. The internal generated voltage of the machine 8 E,=V,+Rly+ Xl, E, = 132820°+ (0.15 2)(2512- 36.87° A}+ j(1.1.Q)(2512- 36.87° A) E, = 1537.27.4° V (c) The equivalent open-circuit terminal voltage corresponding to an Ey of 1537 volts is Vio = V3 (1527 V) = 2662 V From the OCC, the required field current is 5.9 A. (@) The input power to this generator is equal to the output power plus losses. The rated output power is Posy = (1000 kVA)(0.8) = 800 kW Poy = 31,2 R, = 3(251 AY (0.15 Q)=28.4kW Pray = 24 KW P,, =18 KW P stray Pay = Pun + Pan + Brea + Pare + Pay =870.4 EW = (assumed 0) Therefore the prime mover must be capable of supplying 175 kW. Since the generator is a two-pole 60 Hz. machine, to must be turning at 3600 r/min. The required torque is Pay 175.2 kW Tap = BL = =465N-m @m (3600 rnin | =) $s r (e) The rotor current limit of the capability curve would be drawn from an origin of Wy __ 3(1328 Vy ae = -4810 KVAR x, 11a Q=- The radius of the rotor current limit is px _ 3(1828 VI(1537 V) a oe 112 = 5567 KVA The stator qirrent limit is a circle at the origin of radius S=Wylg= 3(1328 V)(251 A) = 1000 kVA A MATLAB program that plots this capability diagram is shown below: & M-file; proeb5_2.m & M-file to display a capability curve for a % synchronous generator. & Calculate the waveforms for times from 0 to 1/30 s QO = -4810; DE = 5567; S = 1000; % Get points for stator current limit theta = -95:1:95; % Angle in degrees rad = theta * pi / 180; $ Angle in radians s_ curve = § .* ( cos(rad) + j*sin(rad) ); & Get points for rotor current limit orig = j*Q; theta = 75:1:105; & Angle in degrees rad = theta * pi / 190; % Angle in radians r_curve = orig + DE .* ( cos{rad) + j*sin(rad) ); % Plot the capability diagram figure (1); plot (real (s_curve),imag(s_curve),'b!','LineWidth!' ,2.0) ; hold on; plot(real (r_curve) ,imag(r_curve),'r--', 'LineWidth!,2.0); % Add x and y axes plot( [-1500 1500], [0 0],'k'); plot( [0,0], [-1500 1500],'k'); % Set titles and axes title ('\bfSynchronous Generator Capability Diagram') ; xlabel('\bfPower (kW)'); ylabel('\bfReactive Power (kVAR)') ; axis( [ -1500 1500 -1500 1500] }); axis square; hold off; The resulting capability diagram is shown below: Synchronous Generator Capability Diagram 1500 1000 Reactive Power (kKVAR) ° 8 é 1000} 1 1 | A 1 1500 =1000 500 f) 500 1000 1500 Power (kW) Assume that the field current of the generator in Problem 5-2 has been adjusted to a value of 4.5 A. (a) What will the terminal voltage of this generator be if it is connected to a A-connected load with an impedance of 20.230° 2? (b) Sketch the phasor diagram of this generator. (c) What is the efficiency of the generator at these conditions? (@) Now assume that another identical A-conrected load is to be paralleled with the first ome. What happens to the phasor diagram for the generator? (e) What is the new terminal voltage after the load has been added? (f) What must be done to restore the terminal voltage to its original value? SOLUTION (a) If the field current is 4.5 A, the open-circuit terminal voltage will be about 2385 Y, and the phase voltage in the generator will be 2385/V3 =1377 V. The load is A-connected with three impedances of 2030°. From the Y-A transform, this load is equivalent to a Y-conrected load with three impedances of 6.667.230°. The resulting per-phase equivalent circuit is shown below: W150 flia f E, 6.667.239 0° The magnctide af the ahase eurreat flowiryr in this gererator 2 “ aq" ; B, TAT? V TFT = lke & [Rt y+ 2] fOS+ fl iter 2307 1829 2 Therefore, the mesnitide of the paaze voltage is V, = 1,2 =(186 A}(6.367 ) = 1240 V ane the terminal voltage is Vp = V3 ¥,= v8 (1240 ¥)=2148 ¥ f) Armature current ia 1, =126.4- 30° 4, and the phase volage is v, = 1240_40° ¥. Therefore, the internal gener ged voltage -s E,= V¥,tklyt isl, EB, = 1240.c07+(9.15 O) (088.2 - 307 Al+ j11.1 2) 136.2- 30" A) E, = 137746.8° V The resulting phasor diagram is sown below (not to scale): E = 1377£6.8° ¥ ad _—_— __ _—— é V, ~ 1ogacer V I, = 1864-30° (c) The efficiency of foe generator nicer these condiens Azan he fond sa follows: Pay HSV, Fy aos #= 3124) V [186 A) (0.8) = 854 kW Pop = 382, = 3(186 AY (9.15 Q)= 15.6 kW ro ee a Pig = 18 KW _ Poy = Pour + Bac + Baw + Pore + Peay = 612 kW = fesaumed 0) y= Fem 100% — St EY 100% = 90.5% Pa 612 kW (@) When the new load is added, the total current flow increases at the same phase angle. Therefore, jacl, increases in length at the same angle, while the magnitude of EF, must remain constant, Therefore, E, “swings” out along the are of constant magnitude until the mew jX I, fits exactly between Vy and E,. a = 137746.8° V ¥; Vj = 124020° ¥ I = 1862-30° A I’ A (e) The new impedance per phase will be half of the old value, so Z=3.333230° 2. The magnitude of the phase current flowing inthis gererator is Ey 1377 V 1377 V l= = = Ret io +Z] [015+ f1.14+3,3332304 1,829 8 ‘< =3354 Therefore, the magnitude of the phase voltage is V, = 1,2 = (335 A)(3.333 &) = 1117 V and the terminal voltage is Vp =V2 V,= v3 (1117 V)=194 V (f) To restore the terminal voltage to its original value, increase the field current J... 5-4. Assume that the field airrent of the generator in Problem 5-2 is adjusted to achieve rated voltage (2300 Y) at full load conditions ineach of the questions below, (a) What is the efficiency of the generator at rated load? (b) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0,.8-PF- lagsing loads? (c) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with 0.8-PF- leading loads? (@) What is the voltage regulation of the generator if it is loaded to rated kilovoltamperes with unity-power- factor loads? (e) Use MATLAB to plot the terminal voltage of the generator as a function of load for all three power factors, SOLUTION (a) This generator is Y-conmeded, so 7, =Z,. At rated conditions, the line and phase current in this gererator is P 1000 kVA =l= = = 251A at f-36.87° aE BY, ¥3(2300V) wan ange 0 The phase voltage of this machire is V, = V, {J3 = 1328 V. The internal generated voltage of the machine 18 E,=V,+Rl,+iX sl, Ey = 1328.20°+ (0.15 Q)(2512 - 36.87° A) + j(1.1.2)(2512—- 36.87° A) E, =153727.4° V The input power to this generator is equal to the output power plus losses. The rated output power is Poyp = (1000 KVA)(0.8) = 800 kW Poy = 31,2R, = 3(251 AY (0.15 2) =28.4 kW Prow = 24 kW Pog = 18 KW Porgy = (assumed 0) Poy = Fou + Pons + Besse + Pooce + Poy = 870.4 kW 9 = 222 x 100% = SOO EM 100% = 91.9% B70.4kW In (6) Ifthe generator is loaded to rated kVA with lagging loads, the phase voltage is V, = 132820° V and the internal generated voltage is E, =153727.4°V. Therefore, the phase voltage at no-load would be V, =1537.20° V. The voltage regulation would be: p _ 1537-1328 1328 (c) If the generator is loaded to rated KVA with leading loads, the phase voltage is V, =132820° V and the internal generated voltage is E,=V,+R I, + X51, EK, = 1328.20°+(0.15 Q)(251236.87° A) +j (11 Q) (251236.87° A) E, =1217.211.5°V * 100% = 15.7% The voltage regulation would be: g = 1217-1328 1328 (a) If the generator is loaded to rated kVA at unity power factor, the phase voltage is ¥, =1328240° V andthe internal generated voltage is x 100% = -8.4% E,=Vyt Rly t Xgl, E, = 1328.20°+(0.15 2)(251.20° A)+ jf 1.12) (251.20° A) E, = 1393211.4° V The voltage regulation would be: Ra PP 8 100% = 4.9% (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 251.4 in any case. A phasor diagram representing the situation at lacsing power factor is shown below: E = By the Pythagorean Theorem, Eg =(V,+Ri,cos +X], sin ay +(X,1, cos8- Rl, sin8) v, = Ee - (X5I4 cos 8-R,I,sin8) —Ryl, cos8-X,I,sin@ A phasor diagram representing the situation at leading power factor is shown below: By the Pythagorean Theorem, 2 Ee =(V,+Rz,cos@— Xgl, sin8) +(Xpl,cos8+ RI, siné) V,= EC - (X,F, cos 8+R,J,siné) —R,L, cos@+X,1,sine A phasor diagram representing the situation at unity power factor is showm below: Ey x, I, 1, vy, RJ, 116 By the Pythagorean Theorem, Eg= Vv; + (Xel4) Vp= Ee -(X5la) The MATLAE program is shown below takes advantage of this fact. & M-file: probs 4de.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, 1.9, and 0.8 leading. % Define values for this generator ER = 1328; % Internal gen voltage IT = 0:38.51:251; & Current values {A} R= 0.15; = R (ohms) XK = 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.*2.*0.8 - ¥.*1.4*0.6; VT_lag = VP_lag .* sqrt (a); % Calculate the voltage for the leading PF case VP_lead = sqrt{ EA*2 - (X.*1.*0.8 + R.*I.*0.6).°2 } - R.*#L.*0.8 + 4.*1.*0.6; VT_lead = VP_lead .* sqrt(Sij; % Calculate the voltage for the unity PF case VPLunity = sqrt{ EA*™2 - (X.*I)."2 3; VTLunity = VPlunity .* sqrt{3i; # Plot the terminal voltage versus load plot(T,abs(VT_lag}),'b-', 'LineWidth',2.0) ; hold on; plot(T,abs(VT_unity),'k--','LineWidth',2.0); plot(T,abs(VT_lead) ,'r-.!','LineWidth! ,2.0) ; title ('\bfTerminal Voltage Versus Load'); xlabel ('\bfLoad {Aj'); ylabel ('\btTerminal Voltage (Vi!) ; 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: Terminal Voltage Versus Load 2500 T T , T 2400; cfg ET + 2300 2200 in Ss & 2000 Teriniinal Voltage (V) 1900 + + 1800+ 5 4 — G8 FF lagging +700}. i === 4.0PF 4 --- 0.8 PF leading 1600 1500 | u i a 60 100 150 200 250 Load (A) 5-10. A paper mill has installed three stearn generators (boilers) to provide process steam and also to use sorne its waste products as an enersy souree. Since there is extra capacity, the mill has installed thee 5-MW turbine generators to take advantage of the situation Each generator is a 4160-V 6250-kVA 0.85-PF lagging two-pole Y-comnected synchronous generator with a synchronous reactance of (1.75 £2 and an armature resistance of 0.04 ©, Generators 1 and 2 have a characteristic power-frequency slope 3, of 2.5 MW/Hz, and generators 2 and 3 have a slope of 3 MW/Hz. fa) If the no-load frequency of each of the three generators is adjusted to 61 Hz, how much power will the three machines be supplying when actual system frequency is 60 Hz? (>) What is the maximum power the three generators can supply in this condition without the ratings of one of them being exceeded? At what frequency does this limit oceur? How much power doas each generator supply at that point? fe) What would have to be dere to set all three seneratora to supply their rated real and reactive powers at an overall operating frequency of 60 Hz? f@) What vould the internal generated voltages of the three senerators be under this condition?