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+// FUNDAMENTALS OF ELECTICAL MACHINES 
+// M.A.SALAM 
+// NAROSA PUBLISHING HOUSE 
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.5
+
+clc;clear; // clears the console and command history 
+
+// Given data
+P = 6          // number of poles
+V_t = 230      // supply to shunt motor in V
+Z = 450        // number of conductors
+R_a = 0.8      // armature resistance in ohm
+I = 30          // supply current in A
+P_0 = 5560      // out put power in W
+I_F = 3         // current through field winding
+phi = 25*10^-3  // flux per pole in Wb
+
+// caclulations 
+A = 6                    // for lap wond A=P
+I_a = I-I_F              // armature current in A
+E_b = V_t-I_a*R_a        // back emf in V
+N = 60*A*E_b/(P*Z*phi)   // speed in rpm
+T_sh = 9.55*P_0/N        // shaft torque in N-m
+
+// display the result 
+disp("Example 5.5 solution");
+printf(" speed \n N = %.1f rpm \n", N );
+printf(" shaft torque \n T_sh = %.1f N-m \n", T_sh );