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//CHAPTER 8- DIRECT CURRENT MACHINES
//Example 5
disp("CHAPTER 8");
disp("EXAMPLE 5");
//VARIABLE INITIALIZATION
v_t=200; //in Volts
I_l=50; //in Amperes
r_a=0.1; //armature resistance in Ohms
r_f=100; //field resistance in Ohms
s_loss=500; //core and iron loss in Watts
//SOLUTION
//solution (a)
I_f=v_t/r_f; //I_sh is same as I_f and r_sh is same as r_f
I_a=I_f+I_l;
E_a=v_t+(I_a*r_a);
disp(sprintf("(a) The induced emf is %f V",E_a));
//solution (b)
arm_loss=(I_a^2)*r_a; //armature copper loss
sh_loss=(I_f^2)*r_f; //shunt field copper loss
tot_loss=arm_loss+sh_loss+s_loss;
p_o=v_t*I_l; //output power
p_i=p_o+tot_loss; //input power
bhp=p_i/735.5; //1 metric horsepower= 735.498W
disp(sprintf("(b) The Break Horse Power(B.H.P.) of the prime mover is %f H.P.(metric)",bhp));
//solution (c)
c_eff=(p_o/p_i)*100;
p_EE=E_a*I_a; //electrical power
m_eff=(p_EE/p_i)*100;
e_eff=(p_o/p_EE)*100;
disp(sprintf("(c) The commercial efficiency is %f %%, the mechanical efficiency is %f %% and the electrical efficiency is %f %%",c_eff,m_eff,e_eff));
//END
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