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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 15
disp("CHAPTER 8");
disp("EXAMPLE 15");
//VARIABLE INITIALIZATION
v_t=230; //in Volts
I_a1=3.33; //in Amperes
N1=1000; //in rpm
r_a=0.3; //armature resistance in Ohms
r_f=160; //field resistance in Ohms
I_l=40; //in Amperes
phi1=1; //in Wb (phi=1 is an assumption)
phi2=(1-(4/100)); //in Wb (phi2=0.96 of phi1)
//SOLUTION
//At no load
E_a1=v_t-(I_a1*r_a);
I_f=v_t/r_f;
//At full load
I_a2=I_l-I_f;
E_a2=v_t-(I_a2*r_a);
N2=(E_a2/E_a1)*(phi1/phi2)*N1;
N2=round(N2); //to round off the value
disp(sprintf("The full load speed is %d rpm",N2));
//END
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