Page 223 - Solutions Manual to accompany Electric Machinery Fundamentals
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E     n
                           A  
                         E Ao  n o
                              E        240 V 
                         n     A     n o            1200 r/min   1061 r/min
                              E Ao     271.5 V 

                 The field current when  R  = 400  is:
                                        adj
                                V          240 V      240 V
                         I      T                          0.505 A
                          F
                             R   adj  R F  400     75   500 
                 From Figure P8-1, this field current would produce an internal generated voltage  E  of 167 V at a speed
                                                                                           Ao
                  n   of 1200 r/min.  Therefore, the speed n with a voltage of 240 V would be
                   o
                         E     n
                           A  
                         E Ao  n o
                              E        240 V 
                         n     A     n o           1200 r/min   1725 r/min
                              E Ao     167 V 

          8-6.   What is the starting current of this machine if it is started by connecting it directly to the power supply
                 V T  ?  How does this starting current compare to the full-load current of the motor?

                 SOLUTION  The starting current of this machine (ignoring the small field current) is

                                V    240 V
                         I      T          1260 A
                           ,start
                          L
                                R A  0.19 
                 The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current.  This
                 much current is extremely likely to damage the motor.
          8-7.   Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a
                 full-load armature reaction of 1200 Aturns.  (Assume that the armature reaction increases linearly with
                 increases in armature current.)
                 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.

                 % M-file: prob8_7.m
                 % M-file to create a plot of the torque-speed curve of the
                 %   the shunt dc motor with and without armature reaction.

                 % Get the magnetization curve.  Note that this curve is
                 % defined for a speed of 1200 r/min.
                 load p81_mag.dat
                 if_values = p81_mag(:,1);
                 ea_values = p81_mag(:,2);
                 n_0 = 1200;

                 % 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 = 0.19;             % Armature resistance (ohms)
                 i_l = 0:1:110;          % Line currents (A)

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