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//Book Name:Fundamentals of Electrical Engineering
//Author:Rajendra Prasad
//Publisher: PHI Learning Private Limited
//Edition:Third ,2014
//Ex10_3.sce
clc;
clear;
Pin=40; // power in kW
Ps=1.5; // power in kW
Ns=100; //speed percentage value
N=40; //speed percentage value
power_loss=0.8; // power in kW
printf("\n (a)")
rotor_input_power=Pin-Ps;
s=0.04;
rotor_copper_loss=s*rotor_input_power;
mec_power_developed=rotor_input_power-rotor_copper_loss;
printf("\n Mechanical power developed by the rotor=%2.2f kW",mec_power_developed)
printf("\n Rotor copper loss=%2.2f kW \n",rotor_copper_loss)
printf("\n (b)")
motor_output_power=mec_power_developed-power_loss;
printf("\n Output of the motor=%2.2f kW \n",motor_output_power)
printf("\n (c)")
motor_efficiency=(motor_output_power/Pin)*100;
printf("\n The motor efficiency=%2.1f percentage \n",motor_efficiency)
printf("\n (d)")
new_slip=(Ns-N)/Ns;
total_rotor_copper_loss=new_slip*rotor_input_power;
printf("\n Total rotor copper loss when speed reduced to 40percentage of synchronous speed=%2.1f kW \n",total_rotor_copper_loss)
printf("\n (e)")
total_rotor_loss=total_rotor_copper_loss+power_loss;
motor_output_power=rotor_input_power-total_rotor_loss;
motor_efficiency=(motor_output_power/Pin)*100;
printf("\n Efficiency of motor when speed reduced to 40percentage of synchronous speed=%2.1f percentage",motor_efficiency)
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