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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 8
disp("CHAPTER 8");
disp("EXAMPLE 8");
//250 V DC shunt machine
//VARIABLE INITIALIZATION
v_t=250; //in Volts
r_a=0.1; //armature resistance in Ohms
r_f=125; //field resistance in Ohms
p_o=20*1000; //output power in Watts
N_g=1000; //speed as generator in rpm
//SOLUTION
//machine as a generator
I_l=p_o/v_t; //load current
I_f=v_t/r_f; //field current, I_f is same as I_sh
I_ag=I_l+I_f; //Output current as generator
E_a=v_t+(I_ag*r_a); //induced emf = E_a = E_g
//machine as a motor
I_l=p_o/v_t; //full load current
I_f=v_t/r_f;
I_am=I_l-I_f; //output current as motor
E_b=v_t-(I_am*r_a); //back emf = E_b = E_m
//solution (a)
N_m=(N_g*E_b)/E_a; //Speed of motor in RPM
N_m=round(N_m); //to round off the value of N_m
disp(sprintf("(a) The speed of the same machine as a motor is %d rpm",N_m));
//solution (b)
//internal power developed as generator
//(i)
//total power developed in the armature
//=Eg.Iag
p_g=(E_a*I_ag)/1000; //to express the answer in kW divide by 1000
disp(sprintf("(b) (i) The internal power developed as generator is %.1f kW",p_g));
//(ii)
//internal power developed as motor
// is total power developed in armature
//=Em.Iam
p_m=(E_b*I_am)/1000;
disp(sprintf("(b) (ii) The internal power developed as motor is %.1f kW",p_m));
//END
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