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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 6
disp("CHAPTER 8");
disp("EXAMPLE 6");
//VARIABLE INITIALIZATION
p_o=20*746; //output power from H.P. to Watts (1 H.P.=745.699 or 746 W)
v_t=230; //in Volts
N=1150; //speed in rpm
P=4; //number of poles
Z=882; //number of armature conductors
r_a=0.188; //armature resistance in Ohms
I_a=73; //armature current in Amperes
I_f=1.6; //field current in Amperes
//SOLUTION
//solution (i)
E_b=v_t-(I_a*r_a);
w=(2*%pi*N)/60; //in radian/sec
T_e=(E_b*I_a)/w;
disp(sprintf("(i) The electromagnetic torque is %f N-m",T_e));
//solution (ii)
A=P; //since it is lap winding, so A=P and A=number of parallel paths
phi=(E_b*60*A)/(P*N*Z);
disp(sprintf("(ii) The flux per pole is %f Wb",phi));
//solution (iii)
p_rotor=E_b*I_a; //power developed on rotor
p_rot=p_rotor-p_o; //p_shaft=p_out
disp(sprintf("(iii) The rotational power is %f W",p_rot));
//solution (iv)
tot_loss=p_rot+((I_a^2)*r_a)+(v_t*I_f);
p_i=p_o+tot_loss;
eff=(p_o/p_i)*100;
disp(sprintf("(iv) The efficiency is %f %%",eff));
//solution (v)
T=p_o/w;
disp(sprintf("(v) The shaft torque is %f N-m",T));
//The answers are slightly different due to the precision of floating point numbers
//END
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