EE 255 - Chapter 9b

# EE 255 - Chapter 9b - I F* = I F N SE 27 turns I A = 0.873...

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225 ( ) * SE 27 turns 0.873 A 40 A 0.473 A 2700 turns F F A F N I I I N = - = - = From Figure P9-1, this field current would produce an internal generated voltage Ao E of 197 V at a speed o n of 1200 r/min. Therefore, ( ) 227.4 V 1200 r/min 1385 r/min 197 V A o Ao E n n E = = = At A I = 60A, the internal generated voltage A E is ( ) ( )( ) 240 V 60 A 0.44 213.6 V A T A A S E V I R R = - + = - Ω = The equivalent field current is ( ) * SE 27 turns 0.873 A 60 A 0.273 A 2700 turns F F A F N I I I N = - = - = From Figure P9-1, this field current would produce an internal generated voltage Ao E of 121 V at a speed o n of 1200 r/min. Therefore, ( ) 213.6 V 1200 r/min 2118 r/min 121 V A o Ao E n n E = = = (c) The torque-speed characteristic can best be plotted with a MATLAB program. An appropriate program is shown below. % M-file: prob9_12.m % M-file to create a plot of the torque-speed curve of the % a differentially compounded dc motor withwithout % armature reaction. % Get the magnetization curve. Note that this curve is % defined for a speed of 1200 r/min. load p91_mag.dat if_values = p91_mag(:,1); ea_values = p91_mag(:,2); n_0 = 1200; % First, initialize the values needed in this program. v_t = 240; % Terminal voltage (V) r_f = 100; % Field resistance (ohms) r_adj = 175; % Adjustable resistance (ohms) r_a = 0.44; % Armature + series resistance (ohms) i_l = 0:50; % Line currents (A) n_f = 2700; % Number of turns on shunt field n_se = 27; % 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

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% current. i_f = v_t / (r_f + r_adj) - (n_se / n_f) * i_a; % Calculate the resulting internal generated voltage at % 1200 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([0 100 800 1600]); grid on; The resulting plot is shown below: Compare this torque-speed curve to that of the shunt motor in Problem 9-7 and the cumulatively- compounded motor in Problem 9-10. (Note that this plot has a larger vertical scale to accommodate the speed runaway of the differentially-compounded motor.) 9-13. A 7.5-hp 120-V series dc motor has an armature resistance of 0.2 Ω and a series field resistance of 0.16 Ω . At full load, the current input is 58 A, and the rated speed is 1050 r/min.
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## This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.

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EE 255 - Chapter 9b - I F* = I F N SE 27 turns I A = 0.873...

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