EE 255 - Chapter 9e

EE 255 - Chapter 9e - % the line "Ea_ar - Vt - i_a*r_a"...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
255 % the line "Ea_ar - Vt - i_a*r_a" goes negative. i_a = 0:1:37; for jj = 1:length(i_a) % Calculate the equivalent field current due to armature % reaction. i_ar = (i_a(jj) / 50) * 200 / n_f; % Calculate the Ea values modified by armature reaction Ea_ar = interp1(if_values,ea_values,i_f - i_ar); % Get the voltage difference diff = Ea_ar - 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); xlabel('\bf\itI_{L} \rm\bf(A)'); ylabel('\bf\itV_{T} \rm\bf(V)'); title ('\bfTerminal Characteristic of a Shunt DC Generator w/AR'); hold off; axis([ 0 50 0 120]); grid on;
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
256 The resulting terminal characteristic is shown below: 9-26. If the machine in Problem 9-25 is running at 1800 r/min with a field resistance R adj = 10 Ω and an armature current of 25 A, what will the resulting terminal voltage be? If the field resistor decreases to 5 Ω while the armature current remains 25 A, what will the new terminal voltage be? (Assume no armature reaction.) S OLUTION If A I = 25 A, then ( )( ) 25 A 0.18 A A I R = Ω = 4.5 V. The point where the distance between the A E and T V curves is exactly 4.5 V corresponds to a terminal voltage of 104 V, as shown below.
Background image of page 2
If R adj decreases to 5 Ω , the total field resistance becomes 29 Ω , and the terminal voltage line gets shallower. The new point where the distance between the A E and T V curves is exactly 4.5 V corresponds to a terminal voltage of 115 V, as shown below. Note that decreasing the field resistance of the shunt generator increases the terminal voltage. 9-27. A 120-V 50-A cumulatively compounded dc generator has the following characteristics: 0.21 A S R R + = Ω 1000 turns F N = 20 F R = Ω SE 20 turns N = adj 0 to 30 , set to 10 R = Ω Ω 1800 r/min m n = The machine has the magnetization curve shown in Figure P9-7. Its equivalent circuit is shown in Figure P9-10. Answer the following questions about this machine, assuming no armature reaction. (a) If the generator is operating at no load, what is its terminal voltage? (b)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 10

EE 255 - Chapter 9e - % the line "Ea_ar - Vt - i_a*r_a"...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online