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//Chapter 6:Induction Motor Drives
//Example 3
clc;
//Variable Initialization
//Ratings of the star connected Induction motor
f=50 // frequency in HZ
Vl=400 // line voltage in V
P=6 // number of poles
//Parameters referred to the stator
Xr_=2 // rotor winding reactance in ohm
Xs=Xr_ // stator winding reactance in ohm
Rr_=1 // resistance of the rotor windings in ohm
Rs=Rr_ // resistance of the stator windings in ohm
//Solution
Ns=120*f/P //synchronous speed in rpm
Wms=2*%pi*Ns/60
//(i)
Sm=-Rr_/sqrt(Rs**2+(Xs+Xr_)**2) //slip at maximum torque
x=sqrt((Rs+Rr_/Sm)**2+(Xs+Xr_)**2)
Ir_=(Vl/sqrt(3))/x //current at maximum torque
Tmax=(1/Wms)*3*Ir_**2*Rr_/Sm //maximum torque
N=(1-Sm)*Ns //range of speed
//(ii)an overhauling torque of 100Nm
Tl=100 //overhauling torque in Nm
// Tl=(3/Wms)*(Vl**2*Rr_/s)/y
// where y=(Rs+Rr_/s)**2+(Xs+Xr_)**2
a=(1/Wms)*(Vl**2*Rr_)/(-Tl) //a=s*(Rs+Rr_/s)**2+(Xs+Xr_)**2
a = 17
b = 17.3
c = 1
//Discriminant
d = (b**2) - (4*a*c)
// find two solutions
s1 = (-b-sqrt(d))/(2*a)
s2 = (-b+sqrt(d))/(2*a)
N2=(1-s2)*Ns //motor speed and we neglect s1
//slight difference in the answer due to accuracy
//(iii)when a capacitive reactance of 2 ohm is inserted in each phase stator
Xc=2 //reactance of the capacitor in ohms
Sm=-Rr_/sqrt(Rs**2+(Xs+Xr_-Xc)**2) //slip at maximum torque
x=sqrt((Rs+Rr_/Sm)**2+(Xs+Xr_-Xc)**2)
Ir_=(Vl/sqrt(3))/x //current at maximum torque
Tmax_=(1/Wms)*3*Ir_**2*Rr_/Sm //maximum overhauling torque with capacitor
ratio=Tmax_/Tmax
//Results
mprintf("(i)Maximum overhauling torque that the motor can hold is:%.1f N-m",abs(Tmax))
mprintf(" \nRange of speed is from %d to %d rpm\n",Ns,abs(N))
mprintf("\n(ii)Now s*(1+1/s)**2+16s=%d",a)
mprintf("\n Or 17s**s+17.3s+1=0")
mprintf("\nThe solutions for s are %.3f and %.3f\n",s1,s2)
mprintf("\nTherefore Motor speed is:%d rpm\n",N2)
//Note :There is a slight difference in the answer due to the decimal place"
mprintf("\n(iii) Ratio of maximum torque with capacitor and to maximum torque without capacitor is:%.2f",ratio)
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