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clc,clear
printf('Example 5.22\n\n')
V=600
VA=60*10^3
I_sc=210
V_oc=480
I_f=10
R_a=0.2
I=VA/V //VA=V*I and alternator is single phase
I_a=I
Z_s=V_oc/I_sc //Synchronous Impedance
X_s=sqrt(Z_s^2-R_a^2) //SYnchronous Reactance
printf('Synchronous impedances is %f ohms and synchronous reactance is %f ohms\n',Z_s,X_s)
//PART (i)
phi1=acos(0.8) //and lagging
E1=sqrt((V*cos(phi1)+I_a*R_a)^2+(V*sin(phi1)+I_a*X_s)^2) //plus sign for lagging power factor
regulation1=100*(E1-V)/V
printf('\nRegulation at 0.8 lagging pf is %.2f percent ',regulation1 )
//PART (ii)
phi2=acos(1)
E2=sqrt((V*cos(phi2)+I_a*R_a)^2+(V*sin(phi2)+I_a*X_s)^2)
regulation2=100*(E2-V)/V
printf('\nRegulation at UNITY pf is %.2f percent ',regulation2 )
//PART (iii)
phi3=acos(0.6) //and leading
E3=sqrt((V*cos(phi3)+I_a*R_a)^2+(V*sin(phi3)-I_a*X_s)^2) //minus sign for leading power factor
regulation3=100*(E3-V)/V
printf('\nRegulation at 0.6 leading pf is %.2f percent ',regulation3 )
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