Electric Machinery Fundamentals (Solutions Manual) Part 3 docx

121 ( ) 3570 r/min ( 2 ) f m n P = = = fl1 120 120 59.5 Hz The no-load frequency of generator 2 corresponds to a frequency of ( ) 1800 r/min ( 4 ) f m n P = = = nl2 120 120 60.00 Hz The full-load frequency of generator 2 corresponds to a frequency of ( ) 1785 r/min ( 4 ) f m n P = = = fl2 120 120 59.50 Hz (a) The speed droop of generator 1 is given by n n   SD 1 nl fl 100% = ⋅ 3630 r/m = in 3570 r/min 100% 1. ⋅ 68% = n fl 3570 r/min The speed droop of generator 2 is given by n n   DS 2 nl fl 100% 18 = ⋅ 00 r/m = in 1785 r/min 100% 0. ⋅ 84% = n fl 1785 r/min (b) The power supplied by generator 1 is given by ( ) =  1 1P P s nl1 f sys f and the power supplied by generator 1 is given by ( ) =  2 P 2 P s nl2 f sys f The power curve’s slope for generator 1 is s P 0.1 M = = W 0.1 MW/ = Hz 1 P   nl fl 6 f f 0.5 Hz 59.5 Hz The power curve’s slope for generator 1 is s P 2 P 0.07 = = 5 MW 0.150 M = W/Hz   nl fl 60 f f .00 Hz 59.50 Hz 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 P P P ( ) =  ( + )  LOAD 1 nl1 P s sy f s f 2 nl2 s sy f sP P f ( ) ( ) ( ) ( ) =  +  100 kW 0.1 MW/Hz 60.5 Hz sys 0.15 MW/Hz 60.0 Hz sys f f ( ) ( ) 100 kW 6050 kW 0.10 MW =  /Hz ( ) sys 9000 kW + 0.15 MW  /Hz sys f f 0.25 MW/Hz f sys 6050 kW 9000 kW = + 100 kW  f sys 14, 950 kW 59.8 Hz = = 0.25 MW/Hz (c) The power supplied by generator 1 is 122 ( ) ( ) =  = (  ) = 1 1P P s nl1 f sys f 0.1 MW/Hz 60.5 Hz 59.8 Hz 70 kW The power supplied by generator 2 is ( ) ( ) =  = ( )  = 2 2P P s nl2 f sys f 0.15 MW/Hz 60.0 Hz 59.8 Hz 30 kW (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 n n f f   SD nl fl 100% = ⋅ nl fl = 100% ⋅ fl fl n f so f = f nl fl SD + 1 100 Thus, the full-load frequencies of generators A, B, and C are f fl,A f fl,B f nl,A 61 Hz 59.0 Hz = = = SD A 3.4 1 1 + + 100 100 f nl,B 61.5 Hz 59.71 Hz = = = SD B 3.0 1 1 + + f fl,C 100 100 f nl,C 60.5 Hz 58.97 Hz = = = SD C 2.6 1 1 + + 100 100 and the slopes of the power-frequency curves are: 3 MW 1.5 MW = = /Hz S PA S PB S PC 2 H z 3 MW 1.676 MW = = 1.79 Hz 3 MW 1.961 MW = = 1.53 Hz /Hz /Hz 123 (a) The total load is 7 MW, so the system frequency is ( ) =  ( ) +  ( ) +  LOAD P PA nl s A sys PB nl f f s B f sys f PC n s lC f sys f ( ) ( ) ( =  ) ( ) + (  ) ( ) +  7 MW 1.5 61.0 sys 1.676 61.5 sys 1.961 60.5 f f sys f  +  +  7 MW 91.5 1. = 5.137 f sys = 306.2 f sys = 59.61 Hz 5 sys 103.07 1.676 sys 118.64 1.961 sys f f f The power supplied by each generator will be ( ) ( ) =  = (  ) = P s nl A P A A f sys f 1.5 MW/Hz 61.0 Hz 59.61 Hz 2.09 MW ( ) ( ) =  = (  ) = P s nlB P BB f sys f 1.676 MW/Hz 61.5 Hz 59.61 Hz 3.17 MW ( ) ( ) =  = (  ) = P s nlC P CC f sys f 1.961 MW/Hz 60.5 Hz 59.61 Hz 1.74 MW (b) The equation in part (a) can be re-written slightly to express system frequency as a function of load. ( ) ( ) ( ) =  ( ) + ( )  ( ) +  LOAD 1.5 61.0 sys 1.676 61.5 P f sys 1.961 60.5 f sys f =  +  +  LOAD 91.5 1.5 sys 103.07 1.6 P f 76 sys 118.64 1.96 f =  5.137 sys 313.2 f P LOAD 1 sys f f sys = 313.2  L P OAD 5.137 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); 124 hold on; plot(Pload,PB,'k ','LineWidth',2.0); plot(Pload,PC,'r ','LineWidth',2.0); plot([0 10],[3 3],'k','LineWidth',1.0); plot([0 10],[0 0],'k:'); title ('\bfPower Sharing Versus Total Load'); xlabel ('\bfTotal Load (MW)'); ylabel ('\bfGenerator Power (MW)'); legend('Generator A','Generator B','Generator C','Power Limit'); grid on; hold off; The resulting plot is shown below: This plot reveals that there are power sharing problems both for high loads and for low loads. Generator B is the first to exceed its ratings as load increases. Its rated power is reached at a total load of 6.45 MW. On the other hand, Generator C gets into trouble as the total load is reduced. When the total load drops to 2.4 MW, the direction of power flow reverses in Generator C. (c) The power sharing in (a) is not acceptable, because Generator 2 has exceeded its power limits. (d) To improve the power sharing among the three generators in (a) without affecting the operating frequency of the system, the operator should decrease the governor setpoints on Generator B while simultaneously increasing them in Generators A and C. 5-10. A paper mill has installed three steam generators (boilers) to provide process steam and also to use some its waste products as an energy source. Since there is extra capacity, the mill has installed three 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-connected synchronous generator with a synchronous reactance of 0.75 & and an armature resistance of 0.04 & . Generators 1 and 2 have a characteristic power-frequency slope s P of 2.5 MW/Hz, and generators 2 and 3 have a slope of 3 MW/Hz. (a) 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? 125 (b) 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 occur? How much power does each generator supply at that point? (c) What would have to be done to get all three generators to supply their rated real and reactive powers at an overall operating frequency of 60 Hz? (d) What would the internal generated voltages of the three generators be under this condition? S OLUTION (a) If the system frequency is 60 Hz and the no-load frequencies of the generators are 61 Hz, then the power supplied by the generators will be ( ) ( ) =  = ( )  = 1 1P P s nl1 f sys f 2.5 MW/Hz 61 Hz 60 Hz 2.5 MW ( ) ( =  = )( )  = 2 P 2 P s nl2 f sys f 2.5 MW/Hz 61 Hz 60 Hz 2.5 MW ( ) ( =  = )( )  = 3 P 3 P s nl3 f sys f 3.0 MW/Hz 61 Hz 60 Hz 3.0 MW 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. Generator 3 will be the first machine to reach that limit. Generator 3 will supply this power at a frequency of ( ) ( ) =  5.31 MW 3.0 MW/Hz 61 Hz f sys f sys = .5 9 Hz23 At this point the power supplied by Generators 1 and 2 is = = ( )  ( ) =  ( ) = 1 2 P P 1P s nl1 f sys f 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 have to be adjusted to 62.12 Hz, and the no-load frequency of Generator 3 would have to be adjusted to 61.77 Hz. The field currents of the three generators must then be adjusted to get them supplying a power factor of 0.85 lagging. At that point, each generator will be supplying its rated real and reactive power. (d) Under the conditions of part (c) , which are the rated conditions of the generators, the internal generated voltage would be given by = + + A A E V ⎞ A I S A R jX I The phase voltage of the generators is 4160 V / 3 = 2402 V, and since the generators are Y-connected, their rated current is S I I = = 62 = 50 kVA 867 = A 3 3 ( ) 4160 V A L T V The power factor is 0.85 lagging, so I A 867 31. = 8  A ° . Therefore, = + + A A E V ⎞ A I S A R jX I 2402 0 ( 0.04 )( ) 867 31.8 A ( 0. )( ) 75 867 31.8 A E A = ° + &  ° j + &  ° 126 . 91.5 1.5 sys 1 03. 07 1.6 P f 76 sys 118.64 1.96 f =  5. 137 sys 31 3.2 f P LOAD 1 sys f f sys = 31 3.2  L P OAD 5. 137 A MATLAB . Hz 60 Hz 2.5 MW ( ) ( =  = )( )  = 3 P 3 P s nl3 f sys f 3. 0 MW/Hz 61 Hz 60 Hz 3. 0 MW Therefore the total power supplied by the generators. kVA)(0.85) = 5 .31 MW. Generator 3 will be the first machine to reach that limit. Generator 3 will supply this power at a frequency of ( ) ( ) =  5 .31 MW 3. 0 MW/Hz