Page 231 - Solutions Manual to accompany Electric Machinery Fundamentals
P. 231

From Figure P8-1, this field current would produce an internal generated voltage  E  of 229 V at a speed
                                                                                           Ao
                  n   of 1200 r/min.  Therefore,
                   o
                              E        235.8 V 
                         n     A     n o            1200 r/min   1236 r/min
                              E Ao     229 V 

                 At I   = 40A, the internal generated voltage  E  is
                     A                                    A
                            E      I   V   R  R     240 V       40 A 0.21    231.6 V
                             A   T   A   A   S
                 The equivalent field current is
                                     N                14 turns
                            I  *    I   SE  I   0.96 A         40 A   0.753 A
                            F
                                 F
                                     N F  A          2700 turns
                 From Figure P8-1, this field current would produce an internal generated voltage  E  of 214 V at a speed
                                                                                           Ao
                  n o  of 1200 r/min.  Therefore,
                              E        231.6 V 
                         n     A     n o            1200 r/min   1299 r/min
                              E Ao     214 V 

                 At I A  = 60A, the internal generated voltage  E  is
                                                          A
                                                                
                            E   A  T  I  A   V   A  R S R     240 V       60 A 0.21    227.4 V
                 The equivalent field current is
                                     N                14 turns
                            I  *    I   SE  I   0.96 A         60 A   0.649 A
                                 F
                            F
                                     N F  A          2700 turns
                 From Figure P8-1, this field current would produce an internal generated voltage  E  of 196 V at a speed
                                                                                           Ao
                  n o  of 1200 r/min.  Therefore,
                              E        227.4 V 
                         n     A     n o            1200 r/min   1392 r/min
                              E Ao     196 V 

                 (c)  The torque-speed characteristic can best be  plotted  with a MATLAB program.  An appropriate
                 program is shown below.

                 % M-file: prob8_12.m
                 % 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.  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 = 0.21;             % Armature + series resistance (ohms)
                 i_l = 0:1:110;          % Line currents (A)
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