zxing barcode reader java download V A = 180 V, (c) V A = 240 V in Software

Draw PDF 417 in Software V A = 180 V, (c) V A = 240 V

V A = 180 V, (c) V A = 240 V
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SOLUTION At no-load conditions, E A = V A The field current is given by
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IF =
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VF 240 V 240 V = = = 096 A Radj + RF 175 + 75 250
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From Figure P8-1, this field current would produce an internal generated voltage E Ao of 277 V at a speed
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no of 1200 r/min Therefore, the speed n with a voltage of 240 V would be EA n = E Ao no
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E n = A no E Ao
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If V A = 120 V, then E A = 120 V, and
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120 V n= (1200 r/min ) = 520 r/min 277 V
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(a) If V A = 180 V, then E A = 180 V, and
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180 V n= (1200 r/min ) = 780 r/min 277 V
(a) If V A = 240 V, then E A = 240 V, and
240 V n= (1200 r/min ) = 1040 r/min 277 V
8-11 For the separately excited motor of Problem 8-10: (a) What is the maximum no-load speed attainable by varying both V A and Radj (b) What is the minimum no-load speed attainable by varying both V A and Radj SOLUTION (a) The maximum speed will occur with the maximum V A and the maximum Radj The field current when Radj = 400 is:
IF =
VT 240 V 240 V = = = 0505 A Radj + RF 400 + 75 475
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 207 V at a speed
no of 1200 r/min At no-load conditions, the maximum internal generated voltage E A = V A = 240 V
Therefore, the speed n with a voltage of 240 V would be
EA n = E Ao no
E 240 V n = A no = (1200 r/min ) = 1391 r/min E 207 V Ao
(b) The minimum speed will occur with the minimum V A and the minimum Radj The field current when Radj = 100 is:
IF =
VT 240 V 240 V = = = 137 A Radj + RF 100 + 75 175
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 289 V at a speed
no of 1200 r/min At no-load conditions, the minimum internal generated voltage E A = V A = 120 V
Therefore, the speed n with a voltage of 120 V would be
EA n = E Ao no
E 120 V n = A no = (1200 r/min ) = 498 r/min E 289 V Ao
8-12 If the motor is connected cumulatively compounded as shown in Figure P8-4 and if Radj = 175 , what is its no-load speed What is its full-load speed What is its speed regulation Calculate and plot the torque-speed characteristic for this motor (Neglect armature effects in this problem)
SOLUTION At no-load conditions, E A = VT = 240 V The field current is given by
IF =
VT 240 V 240 V = = = 096 A Radj + RF 175 + 75 250
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 277 V at a speed
no of 1200 r/min Therefore, the speed n with a voltage of 240 V would be EA n = E Ao no
E 240 V n = A no = (1200 r/min ) = 1040 r/min 277 V E Ao
At full load conditions, the armature current is
I A = I L I F = 110 A - 096 A = 109 A
The internal generated voltage E A is
E A = VT I A R A = 240 V - (109 A )(021 ) = 2171 V
The equivalent field current is
IF = IF +
N SE 12 turns (109 A ) = 144 A I A = 096 A + NF 2700 turns
From Figure P8-1, this field current would produce an internal generated voltage E Ao of 290 V at a speed
no of 1200 r/min Therefore,
E 2171 V n = A no = (1200 r/min ) = 898 r/min E 290 V Ao
The speed regulation is 157
SR =
n nl nfl 1040 r/min - 898 r/min 100% = 100% = 158% 898 r/min nfl
The torque-speed characteristic can best be plotted with a MATLAB program An appropriate program is shown below
% M-file: prob8_12m % M-file to create a plot of the torque-speed curve of the % a cumulatively compounded dc motor without % armature reaction % Get the magnetization curve This file contains the % three variables if_values, ea_values, and n_0 load p81mat % First, initialize the values needed in this program v_t = 240; % Terminal voltage (V) r_f = 75; % Field resistance (ohms) r_adj = 175; % Adjustable resistance (ohms) r_a = 021; % Armature + series resistance (ohms) i_l = 0:2:110; % Line currents (A) n_f = 2700; % 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 % 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 (8-38) n = ( e_a / e_a0 ) * n_0; % Calculate the induced torque corresponding to each % speed from Equations (8-28) and (8-29) t_ind = e_a * i_a / (n * 2 * pi / 60); % Plot the torque-speed curves figure(1); plot(t_ind,n,'b-','LineWidth',20); xlabel('\bf\tau_{ind} (N-m)'); ylabel('\bf\itn_{m} \rm\bf(r/min)'); title ('\bfCumulatively-Compounded DC Motor Torque-Speed Characteristic'); axis([0 250 800 1250]); grid on;
The resulting plot is shown below:
Compare this torque-speed curve to that of the shunt motor in Problem 8-9 (Both curves are plotted on the same scale to facilitate comparison) 8-13 The motor is connected cumulatively compounded and is operating at full load What will the new speed of the motor be if Radj is increased to 250 How does the new speed compared to the full-load speed calculated in Problem 8-12 SOLUTION If Radj is increased to 250 , the field current is given by
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