Electric Machinery Fundamentals Power & Energy_12 pptx

237 (a) At full load, 100 A 0.75 A 99.25 A ALF III=−= − = , and () ( ) ( ) 240 V 99.25 A 0.14 0.05 221.1 V ATAA S EVIRR=− + = − Ω+ Ω= The actual field current will be adj 240 V 0.75 A 200 120 T F F V I RR == = +Ω+Ω and the effective field current will be () * SE 12 turns 0.75 A 99.25 A 1.54 A 1500 turns FF A F N II I N =+=+ = This field current would produce a voltage Ao E of 290 V at a speed of o n = 1200 r/min. The actual A E is 240 V, so the actual speed at full load will be () 221.1 V 1200 r/min 915 r/min 290 V A o Ao E nn E == = (b) A MATLAB program to calculate the torque-speed characteristic of this motor is shown below: % M-file: prob9_17.m % M-file to create a plot of the torque-speed curve of the % a cumulatively compounded dc motor. % Get the magnetization curve. load p96_mag.dat; if_values = p96_mag(:,1); ea_values = p96_mag(:,2); n_0 = 1200; % First, initialize the values needed in this program. v_t = 240; % Terminal voltage (V) r_f = 200; % Field resistance (ohms) r_adj = 120; % Adjustable resistance (ohms) r_a = 0.19; % Armature + series resistance (ohms) i_l = 0:2:100; % Line currents (A) n_f = 1500; % Number of turns on shunt field n_se = 12; % Number of turns on series field % Calculate the armature current for each load. i_a = i_l - v_t / (r_f + r_adj); % Now calculate the internal generated voltage for % each armature current. e_a = v_t - i_a * r_a; % Calculate the effective field current for each armature % current. i_f = v_t / (r_f + r_adj) + (n_se / n_f) * i_a; % Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve. e_a0 = interp1(if_values,ea_values,i_f); 238 % Calculate the resulting speed from Equation (9-13). n = ( e_a ./ e_a0 ) * n_0; % Calculate the induced torque corresponding to each % speed from Equations (8-55) and (8-56). t_ind = e_a .* i_a ./ (n * 2 * pi / 60); % Plot the torque-speed curves figure(1); plot(t_ind,n,'b-','LineWidth',2.0); xlabel('\bf\tau_{ind} (N-m)'); ylabel('\bf\itn_{m} \rm\bf(r/min)'); title ('\bfCumulatively-Compounded DC Motor Torque-Speed Characteristic'); axis([0 200 900 1600]); grid on; The resulting torque-speed characteristic is shown below: (c) The no-load speed of this machine is the same as the no-load speed of the corresponding shunt dc motor with adj R = 120 Ω , which is 1125 r/min. The speed regulation of this motor is thus nl fl fl 1125 r/min - 915 r/min SR 100% 100% 23.0% 915 r/min nn n − =×= ×= 9-18. The motor is reconnected differentially compounded with adj R = 120 Ω . Derive the shape of its torque- speed characteristic. S OLUTION A MATLAB program to calculate the torque-speed characteristic of this motor is shown below: % M-file: prob9_18.m % M-file to create a plot of the torque-speed curve of the 239 % a differentially compounded dc motor. % Get the magnetization curve. load p96_mag.dat; if_values = p96_mag(:,1); ea_values = p96_mag(:,2); n_0 = 1200; % First, initialize the values needed in this program. v_t = 240; % Terminal voltage (V) r_f = 200; % Field resistance (ohms) r_adj = 120; % Adjustable resistance (ohms) r_a = 0.19; % Armature + series resistance (ohms) i_l = 0:2:40; % Line currents (A) n_f = 1500; % Number of turns on shunt field n_se = 12; % Number of turns on series field % Calculate the armature current for each load. i_a = i_l - v_t / (r_f + r_adj); % Now calculate the internal generated voltage for % each armature current. e_a = v_t - i_a * r_a; % Calculate the effective field current for each armature % current. i_f = v_t / (r_f + r_adj) - (n_se / n_f) * i_a; % Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve. e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from Equation (9-13). n = ( e_a ./ e_a0 ) * n_0; % Calculate the induced torque corresponding to each % speed from Equations (8-55) and (8-56). t_ind = e_a .* i_a ./ (n * 2 * pi / 60); % Plot the torque-speed curves figure(1); plot(t_ind,n,'b-','LineWidth',2.0); xlabel('\bf\tau_{ind} (N-m)'); ylabel('\bf\itn_{m} \rm\bf(r/min)'); title ('\bfDifferentially-Compounded DC Motor Torque-Speed Characteristic'); axis([0 200 900 1600]); grid on; 240 The resulting torque-speed characteristic is shown below: This curve is plotted on the same scale as the torque-speed curve in Problem 6-17. Compare the two curves. 9-19. A series motor is now constructed from this machine by leaving the shunt field out entirely. Derive the torque-speed characteristic of the resulting motor. S OLUTION This motor will have extremely high speeds, since there are only a few series turns, and the flux in the motor will be very small. A MATLAB program to calculate the torque-speed characteristic of this motor is shown below: % M-file: prob9_19.m % M-file to create a plot of the torque-speed curve of the % a series dc motor. This motor was formed by removing % the shunt field from the cumulatively-compounded machine % if Problem 9-17. % Get the magnetization curve. load p96_mag.dat; if_values = p96_mag(:,1); ea_values = p96_mag(:,2); n_0 = 1200; % First, initialize the values needed in this program. v_t = 240; % Terminal voltage (V) r_a = 0.19; % Armature + series resistance (ohms) i_l = 20:1:45; % Line currents (A) n_f = 1500; % Number of turns on shunt field n_se = 12; % Number of turns on series field % Calculate the armature current for each load. i_a = i_l; 241 % Now calculate the internal generated voltage for % each armature current. e_a = v_t - i_a * r_a; % Calculate the effective field current for each armature % current. (Note that the magnetization curve is defined % in terms of shunt field current, so we will have to % translate the series field current into an equivalent % shunt field current. i_f = (n_se / n_f) * i_a; % Calculate the resulting internal generated voltage at % 1800 r/min by interpolating the motor's magnetization % curve. e_a0 = interp1(if_values,ea_values,i_f); % Calculate the resulting speed from Equation (9-13). n = ( e_a ./ e_a0 ) * n_0; % Calculate the induced torque corresponding to each % speed from Equations (8-55) and (8-56). t_ind = e_a .* i_a ./ (n * 2 * pi / 60); % Plot the torque-speed curves figure(1); plot(t_ind,n,'b-','LineWidth',2.0); xlabel('\bf\tau_{ind} (N-m)'); ylabel('\bf\itn_{m} \rm\bf(r/min)'); title ('\bfSeries DC Motor Torque-Speed Characteristic'); grid on; The resulting torque-speed characteristic is shown below: 242 The extreme speeds in this characteristic are due to the very light flux in the machine. To make a practical series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of 12. 9-20. An automatic starter circuit is to be designed for a shunt motor rated at 15 hp, 240 V, and 60 A. The armature resistance of the motor is 0.15 Ω , and the shunt field resistance is 40 Ω . The motor is to start with no more than 250 percent of its rated armature current, and as soon as the current falls to rated value, a starting resistor stage is to be cut out. How many stages of starting resistance are needed, and how big should each one be? S OLUTION The rated line current of this motor is 60 A, and the rated armature current is ALF III=− = 60 A – 6 A = 54 A. The maximum desired starting current is (2.5)(54 A) = 135 A. Therefore, the total initial starting resistance must be start,1 240 V 1.778 135 A A RR+= = Ω start,1 1.778 0.15 1.628 R =Ω−Ω=Ω The current will fall to rated value when A E rises to ()() 240 V 1.778 54 A 144 V A E =−Ω = At that time, we want to cut out enough resistance to get the current back up to 135 A. Therefore, start,2 240 V 144 V 0.711 135 A A RR − += =Ω start,2 0.711 0.15 0.561 R =Ω−Ω=Ω With this resistance in the circuit, the current will fall to rated value when A E rises to ( ) ( ) 240 V 0.711 54 A 201.6 V A E =−Ω = At that time, we want to cut out enough resistance to get the current back up to 185 A. Therefore, start,3 240 V 201.6 V 0.284 135 A A RR − += = Ω start,3 0.284 0.15 0.134 R =Ω−Ω=Ω With this resistance in the circuit, the current will fall to rated value when A E rises to ()() 240 V 0.284 54 A 224.7 V A E =−Ω = If the resistance is cut out when A E reaches 228,6 V, the resulting current is 240 V 224.7 V 102 A 135 A 0.15 A I − ==< Ω , so there are only three stages of starting resistance. The three stages of starting resistance can be found from the resistance in the circuit at each state during starting. start,1 1 2 3 1.628 RRRR=++= Ω start,2 2 3 0.561 RRR=+= Ω start,3 3 0.134 RR== Ω Therefore, the starting resistances are 1 1.067 R =Ω 2 0.427 R =Ω 3 0.134 R =Ω 243 9-21. A 15-hp 120-V 1800 r/min shunt dc motor has a full-load armature current of 60 A when operating at rated conditions. The armature resistance of the motor is A R = 0.15 Ω , and the field resistance F R is 80 Ω . The adjustable resistance in the field circuit adj R may be varied over the range from 0 to 200 Ω and is currently set to 90 Ω . Armature reaction may be ignored in this machine. The magnetization curve for this motor, taken at a speed of 1800 r/min, is given in tabular form below: E A , V 5 78 95 112 118 126 I F , A 0.00 0.80 1.00 1.28 1.44 2.88 Note: An electronic version of this magnetization curve can be found in file prob9_21_mag.dat , which can be used with MATLAB programs. Column 1 contains field current in amps, and column 2 contains the internal generated voltage E A in volts. (a) What is the speed of this motor when it is running at the rated conditions specified above? (b) The output power from the motor is 7.5 hp at rated conditions. What is the output torque of the motor? (c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)? (d) What is the efficiency of the motor at full load? (e) If the motor is now unloaded with no changes in terminal voltage or R adj , what is the no-load speed of the motor? (f) Suppose that the motor is running at the no-load conditions described in part (e). What would happen to the motor if its field circuit were to open? Ignoring armature reaction, what would the final steady- state speed of the motor be under those conditions? (g) What range of no-load speeds is possible in this motor, given the range of field resistance adjustments available with adj R ? S OLUTION (a) If adj R = 90 Ω , the total field resistance is 170 Ω , and the resulting field current is adj 230 V 1.35 A 90 80 T F F V I RR == = +Ω+Ω This field current would produce a voltage Ao E of 221 V at a speed of o n = 1800 r/min. The actual A E is ( ) ( ) 230 V 60 A 0.15 221 V ATAA EVIR=− = − Ω= so the actual speed will be () 221 V 1800 r/min 1800 r/min 221 V A o Ao E nn E == = (b) The output power is 7.5 hp and the output speed is 1800 r/min at rated conditions, therefore, the torque is ()( ) () out out 15 hp 746 W/hp 59.4 N m 2 rad 1 min 1800 r/min 1 r 60 s m P τ π ω == = ⋅ (c) The copper losses are 244 ( ) ( ) ( ) ( ) 2 2 CU 60 A 0.15 230 V 1.35 A 851 W AA FF PIRVI=+= Ω+ = The power converted from electrical to mechanical form is ()() conv 221 V 60 A 13,260 W AA PEI== = The output power is ()( ) OUT 15 hp 746 W/hp 11,190 WP == Therefore, the rotational losses are rot conv OUT 13,260 W 11,190 W 2070 WPP P=−= − = (d) The input power to this motor is () ()( ) IN 230 V 60 A 1.35 A 14,100 W TA F PVII=+= + = Therefore, the efficiency is OUT IN 11,190 W 100% 100% 79.4% 14,100 W P P η =× = × = (e) The no-load A E will be 230 V, so the no-load speed will be () 230 V 1800 r/min 1873 r/min 221 V A o Ao E nn E == = (f) If the field circuit opens, the field current would go to zero ⇒ φ drops to res φ ⇒ A E ↓ ⇒ A I ↑⇒ ind τ ↑ ⇒ n ↑ to a very high speed. If F I = 0 A, Ao E = 8.5 V at 1800 r/min, so () 230 V 1800 r/min 48,700 r/min 8.5 V A o Ao E nn E == = (In reality, the motor speed would be limited by rotational losses, or else the motor will destroy itself first.) (g) The maximum value of adj R = 200 Ω, so adj 230 V 0.821 A 200 80 T F F V I RR == = +Ω+Ω This field current would produce a voltage Ao E of 153 V at a speed of o n = 1800 r/min. The actual A E is 230 V, so the actual speed will be () 230 V 1800 r/min 2706 r/min 153 V A o Ao E nn E == = The minimum value of adj R = 0 Ω , so adj 230 V 2.875 A 0 80 T F F V I RR == = +Ω+Ω This field current would produce a voltage Ao E of about 242 V at a speed of o n = 1800 r/min. The actual A E is 230 V, so the actual speed will be 245 () 230 V 1800 r/min 1711 r/min 242 V A o Ao E nn E == = 9-22. The magnetization curve for a separately excited dc generator is shown in Figure P9-7. The generator is rated at 6 kW, 120 V, 50 A, and 1800 r/min and is shown in Figure P9-8. Its field circuit is rated at 5A. The following data are known about the machine: Note: An electronic version of this magnetization curve can be found in file p97_mag.dat , which can be used with MATLAB programs. Column 1 contains field current in amps, and column 2 contains the internal generated voltage E A in volts. 246 0.18 A R =Ω 120 V F V = adj 0 to 30 R =Ω 24 F R =Ω 1000 turns per pole F N = Answer the following questions about this generator, assuming no armature reaction. (a) If this generator is operating at no load, what is the range of voltage adjustments that can be achieved by changing R adj ? (b) If the field rheostat is allowed to vary from 0 to 30 Ω and the generator’s speed is allowed to vary from 1500 to 2000 r/min, what are the maximum and minimum no-load voltages in the generator? S OLUTION (a) If the generator is operating with no load at 1800 r/min, then the terminal voltage will equal the internal generated voltage A E . The maximum possible field current occurs when adj R = 0 Ω . The current is ,max adj 120 V 5 A 24 0 F F F V I RR == = +Ω+Ω From the magnetization curve, the voltage Ao E at 1800 r/min is 129 V. Since the actual speed is 1800 r/min, the maximum no-load voltage is 129 V. The minimum possible field current occurs when adj R = 30 Ω . The current is ,max adj 120 V 2.22 A 24 30 F F F V I RR == = +Ω+Ω From the magnetization curve, the voltage Ao E at 1800 r/min is 87.4 V. Since the actual speed is 1800 r/min, the minimum no-load voltage is 87 V. (b) The maximum voltage will occur at the highest current and speed, and the minimum voltage will occur at the lowest current and speed. The maximum possible field current occurs when adj R = 0 Ω . The current is ,max adj 120 V 5 A 24 0 F F F V I RR == = +Ω+Ω From the magnetization curve, the voltage Ao E at 1800 r/min is 129 V. Since the actual speed is 2000 r/min, the maximum no-load voltage is [...]... = Ea - Vt - 3.6; % This code prevents us from reporting the first (unstable) % location satisfying the criterion was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % We disp disp disp have the intersection Tell (['Ea = ' num2str(Ea(ii)) ' (['Vt = ' num2str(Vt(ii)) ' (['If = ' num2str(i_f(ii)) ' user V']); V']); A']); % Plot the curves... Vt - i_a(jj)*r_a; % This code prevents us from reporting the first (unstable) % location satisfying the criterion was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % Save terminal voltage at this point v_t(jj) = Vt(ii); i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); end; % Plot the terminal characteristic figure(1); plot(i_l,v_t,'b-','LineWidth',2.0);... Vt - i_a(jj)*r_a; % This code prevents us from reporting the first (unstable) % location satisfying the criterion was_pos = 0; for ii = 1:length(i_f); if diff(ii) > 0 was_pos = 1; end if ( diff(ii) < 0 & was_pos == 1 ) break; end; end; % Save terminal voltage at this point v_t(jj) = Vt(ii); i_l(jj) = i_a(jj) - v_t(jj) / ( r_f + r_adj); end; % Plot the terminal characteristic figure(1); plot(i_l,v_t,'b-','LineWidth',2.0); . (['Vt = ' num2str(Vt(ii)) ' V']); disp (['If = ' num2str(i_f(ii)) ' A']); % Plot the curves figure(1); plot(i_f,Ea,'b-','LineWidth',2.0);. figure(1); plot(t_ind,n,'b-','LineWidth',2.0); xlabel('f au_{ind} (N-m)'); ylabel('fitn_{m} mf(r/min)'); title ('fDifferentially-Compounded. figure(1); plot(t_ind,n,'b-','LineWidth',2.0); xlabel('f au_{ind} (N-m)'); ylabel('fitn_{m} mf(r/min)'); title ('fSeries DC Motor