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//CHAPTER 10- THREE-PHASE INDUCTION MACHINES
//Example 8
disp("CHAPTER 10");
disp("EXAMPLE 8");
//VARIABLE INITIALIZATION
N_s=1200; //in rpm
p_in=80; //in kW
loss=5; //copper and iron losses in kW
f_loss=2; //friction and windage loss in kW
N=1152; //in rpm
//SOLUTION
//solution (a)
p_rotor=p_in-loss;
disp(sprintf("(a) The active power transmitted to rotor is %d kW",p_rotor));
//solution (b)
s=(N_s-N)/N_s;
cu_loss=s*p_rotor;
disp(sprintf("(b) The rotor copper loss is %d kW",cu_loss));
//solution (c)
p_m=(1-s)*p_rotor;
disp(sprintf("(c) The mechanical power developed is %d kW",p_m));
//solution (d)
p_shaft=p_m-f_loss; //output power
disp(sprintf("(d) The mechanical power developed to load is %d kW",p_shaft));
//solution (e)
eff=p_shaft/p_in;
disp(sprintf("(e) The efficiency of the motor is %f %%",eff*100));
//END
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