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+// FUNDAMENTALS OF ELECTICAL MACHINES 

+// M.A.SALAM 

+// NAROSA PUBLISHING HOUSE 

+// SECOND EDITION

+

+// Chapter 5 : DIRECT CURRENT MOTORS

+// Example : 5.8

+

+clc;clear; // clears the console and command history 

+

+// Given data

+I_L1 = 5     // dc shunt motor current in A

+V_t = 230    // supply voltage in V

+R_a = 0.25   // armature resistance in ohm

+R_sh = 115   // field resistance in ohm

+I_L2 = 40    // dc shunt motor current in A

+

+

+// caclulations 

+// at noload condition

+P_in1 = V_t*I_L1                        // input power in W

+I_sh = V_t/R_sh                         // shunt field current in A

+I_a1 = I_L1-I_sh                        // armature current in A

+P_acu1 = I_a1^2*R_a                     // armature copper loss in W

+P_shcu = I_sh^2*R_sh                    //shunt field copper loss in W

+P_iron_friction = P_in1-(P_acu1+P_shcu) // iron and friction losses in W

+// under load condition

+I_a2 = I_L2-I_sh                        // armature current in A

+P_acu2 = I_a2^2*R_a                     // armature copper loss in W

+P_loss = P_iron_friction+P_shcu+P_acu2  // total losses in W

+P_in2 = V_t*I_L2                        // input power in W

+P_0 = P_in2-P_loss                      // output power in W

+n = (P_0/P_in2)*100                     // efficiency in percent

+

+// display the result 

+disp("Example 5.8 solution");

+printf(" \n iron and friction losses \n P_iron_friction = %.2f W \n", P_iron_friction );

+printf(" \n efficiency \n n = %.0f percent \n", n)

+

+