Page 286 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 286

P      1 s P      1 0.778  978.3 W  217.2 W
                          conv,F       AG,F
                         P          1 s P      1   0.778  623.8 W  138.5 W
                          conv,B       AG,B
                                                     
                         P     P     P     217.2 W 138.5 W   78.7 W
                          conv  conv,F  conv,B
                 The output power is

                                                 
                         P OUT    P conv    P   rot  78.7 W 51 W   27.7 W
                 The induced torque is

                              P               354.5 W
                          ind    AG                             1.88 N m
                                                                         
                              
                               sync           1800 r/min      2 rad           1 min   
                                                  1 r      60 s 
                 Assuming that the rotational losses are still 51 W, this motor will still be able to speed up because  P conv   is
                 78.7 W, while the rotational losses are 51 W, so  there is more power than it required to cover the
                 rotational losses.  The motor will continue to speed up.
          9-4.   Use MATLAB to calculate and plot the torque-speed characteristic of the motor in Problem 9-1, ignoring
                 the starting winding.
                 SOLUTION    This  problem is best solved with MATLAB, since it involves calculating the torque-speed
                 values at many points.  A MATLAB program to calculate and display both torque-speed characteristics is
                 shown  below.    Note that this program shows the torque-speed curve for both positive and negative
                 directions of rotation.  Also, note that we had to avoid calculating the slip at exactly 0 or 2, since those
                 numbers would produce divide-by-zero errors in  Z  and  Z  respectively.
                                                                      B
                                                              F

                 % M-file: prob9_4.m
                 % M-file create a plot of the torque-speed curve of the
                 %   single-phase induction motor of Problem 9-4.

                 % First, initialize the values needed in this program.
                 r1 = 2.00;                  % Stator resistance
                 x1 = 2.56;                  % Stator reactance
                 r2 = 2.80;                  % Rotor resistance
                 x2 = 2.56;                  % Rotor reactance
                 xm = 60.5                   % Magnetization branch reactance
                 v = 120;                    % Single-Phase voltage
                 n_sync = 1800;              % Synchronous speed (r/min)
                 w_sync = 188.5;             % Synchronous speed (rad/s)

                 % Specify slip ranges to plot
                 s = 0:0.01:2.0;

                 % Offset slips at 0 and 2 slightly to avoid divide by zero errors
                 s(1)   = 0.0001;
                 s(201) = 1.9999;

                 % Get the corresponding speeds in rpm
                 nm = ( 1 - s) * n_sync;

                 % Caclulate Zf and Zb as a function of slip
                 zf = (r2 ./ s + j*x2) * (j*xm) ./ (r2 ./ s + j*x2 + j*xm);
                 zb = (r2 ./(2-s) + j*x2) * (j*xm) ./ (r2 ./(2-s) + j*x2 + j*xm);

                 % Calculate the current flowing at each slip
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