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+//Chapter 6:Induction Motor Drives
+//Example 18
+clc;
+
+//Variable Initialization
+
+//Ratings of the star connected Induction motor 
+f=50         // frequency in HZ
+Vl=440       // line voltage in V
+P=6          // number of poles
+N=970        // rated speed
+n=2          // ratio of stator to rotor
+Sm=0.25      // it is given the speed range is 25% below the synchronous speed which is proportional to the Slip
+
+//Parameters referred to the stator
+Xr_=0.4      // rotor winding reactance in ohm
+Xs=0.3       // stator winding reactance in ohm
+Rr_=0.08     // resistance of the rotor windings in ohm
+Rs=0.1       // resistance of the stator windings in ohm
+alpha=165    // maximum value of the firing angle in degress
+
+//Solution
+Ns=120*f/P   // synchronous speed
+Wms=2*%pi*Ns/60
+//(i) transformer turns ratio
+al=alpha*(%pi/180)
+a=-Sm/cos(al)  //since  Sm=a*math.cos(alpha)   
+m=n/a        //since a=n/m where m is the transformer ratio
+
+//(ii)When speed is 780 rpm and firing angle is 140 degrees
+N1=780           //given speed
+alpha1=140       //given firing angle
+s1=(Ns-N1)/Ns    //slip at the given speed N1
+Vd1=(3*sqrt(6)/%pi)*s1*(Vl/sqrt(3))/n
+al1=alpha1*(%pi/180)
+Vd2=(3*sqrt(6)/%pi)*(Vl/sqrt(3))/m*cos(al1)
+Rs_=Rs*(1/n)**2   //stator resistance referred to the rotor
+Rr=Rr_*(1/n)**2   //rotor resistance referred to the rotor
+Rd=0.01           //equivalent resistance of the DC link inductor
+Id=(Vd1+Vd2)/(2*(s1*Rs_+Rr)+Rd)
+T1=abs(Vd2)*Id/s1/Wms     //required torque
+
+//(iii)when speed is 800rpm and firing angle is half the rated motor torque
+N1=800           //given speed
+s=(Ns-N)/Ns      //rated slip
+x=(Rs+Rr_/s)**2+(Xs+Xr_)**2
+Trated=(3/Wms)*(Vl/sqrt(3))**2*(Rr_/s)/x          //rated torque
+T_half=Trated/2        //half rated torque
+s1=(Ns-N1)/Ns          //given slip at speed N1=800rpm
+Vd1=(3*sqrt(6)/%pi)*s1*(Vl/sqrt(3))/n
+Vd2=(3*sqrt(6)/%pi)*(Vl/sqrt(3))/m
+Id=(Vd1+Vd2)/(2*(s1*Rs_+Rr)+Rd)
+T=abs(Vd2)*Id/s1/Wms     //required torque
+
+//since the given torque is half of the rated value
+//To find the find the firing angle we assumed cos(alpha1)=-X
+//The given quadratic equation is X**2-0.772X+0.06425=0
+a = 1
+b = -0.772
+c = 0.06425
+//Discriminant
+d = (b**2) - (4*a*c)
+
+X1 = (-b-sqrt(d))/(2*a)
+X2 = (-b+sqrt(d))/(2*a)
+alpha1=-acos(X2)   //since cos(alpha1)=-X where alpha1 is radians
+alpha1=alpha1*(180/%pi)
+alpha1=180+alpha1            //required firing angle
+
+
+//Results
+mprintf("(i)Transformer ratio is:%.3f",m)
+mprintf("\n(ii)Required torque is :%.2f N-m",T1)
+//There is a slight difference in the answer for the torque due to accuracy
+mprintf("\n(iii)The half rated torque at the given speed of %d rpm is:%.2f N-m",N1,T_half)
+mprintf("\nWith a slip of s:%.1f",s1)
+mprintf("\nThe solutions for X are %.4f and %.4f",X1,X2)
+mprintf("\nFor X1:%.4f the motor is unstable so we use X2:%.4f",X1,X2)
+mprintf("\nHence the required firing angle is :%.1f °",alpha1)