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//CHAPTER 3- THREE-PHASE A.C. CIRCUITS
//Example 6
disp("CHAPTER 3");
disp("EXAMPLE 6");
//VARIABLE INITIALIZATION
v_l=3300; //in Volts
p_out=1500*735.5; //in Watts as 1 metric horsepower= 735.498W
eff=0.85;
pow_fact=0.81;
//SOLUTION
//solution (a)
p_in=p_out/eff;
disp(sprintf("(a) The motor input is %f kW",p_in/1000));
//solution (b)
I=p_in/(sqrt(3)*v_l*pow_fact);
disp(sprintf("(b) The line and phase current of the alternator is %f A",I));
//solution (c)
I_l=I;
I_ph=I_l/sqrt(3);
disp(sprintf("(c) The line current of the motor is %f A",I_l));
disp(sprintf("The phase current of the motor is %f A",I_ph));
//Answers may be different due to precision of floating point numbers
//END
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