1 files changed, 85 insertions, 0 deletions
diff --git a/3731/CH5/EX5.10/Ex5_10.sce b/3731/CH5/EX5.10/Ex5_10.sce
new file mode 100644
index 000000000..ebe71b3aa
--- /dev/null
+++ b/3731/CH5/EX5.10/Ex5_10.sce
@@ -0,0 +1,85 @@
+//Chapter 5:Dc Motor Drives
+//Example 10
+clc;
+
+//Variable Initialization
+
+
+V=220     // rated voltage in v
+N=1000    // rated speed in rpm
+Ia=175    // rated current in A
+Ra=0.08   // armature resistance in ohms
+N1=1050   // initial speed of the motor in rpm
+J=8       // moment of inertia of the motor load system kg-m2
+La=0.12   // armature curcuit inductance in H
+
+//Solution
+E=V-Ia*Ra
+Wm=N*2*%pi/60     //rated speed in rad/s
+//(a)When  the braking current is twice the rated current
+Ia1=2*Ia
+E1=N1/N*E
+x=(V+E1)/Ia1   //x=Rb+Ra
+Rb=x-Ra        //required braking resistance
+
+//(b)To obtain the expression for the transient value of speed and current including the effect of armature inductance
+//The values given below are taken from Ex-5.9
+ta=0.194                    //time constant in sec
+B=0
+tm1= %inf           //tm1=J/B and B=0 which is equal to infinity
+tm2=1.274
+K=1.967
+Trated=E*Ia/Wm             //rated torque
+Tl=0.5*Trated              //load torque is 50% of the rated torque
+Ra=Rb
+K1=N1*2*%pi/60         //initial speed in rad/s
+//Values of the coefficient of the quadratic equation for Wm
+x1=(1+ta/tm1)/ta
+x2=1/tm2/ta
+x3=-(K*V+Ra*Tl)/J/Ra/ta   
+//Values of the coefficient of the quadratic equation ia
+y1=(1+ta/tm1)/ta
+y2=1/tm2/ta
+y3=-B*V/J/Ra/ta+K*Tl/J/Ra/ta   
+
+//solving the quadratic equation
+a = 1 
+b = x1
+c = x2
+//Discriminant
+d = (b**2) - (4*a*c)
+
+alpha1 = (-b+sqrt(d))/(2*a)
+alpha2 = (-b-sqrt(d))/(2*a)
+
+K3=x3/x2
+K4=y3/y2
+
+Wm_0=K1                       ;ia_0=0
+d_Wm_dt_0=(K*ia_0-B*Wm-Tl)/J  ;d_ia_dt_0=(-V-Ra*ia_0-K*K1)/La    //Wm=K1 at t=0 and during braking rated voltage V is equal to -V
+
+a = [1,1;real(alpha1),real(alpha2)]
+b = [Wm_0;d_Wm_dt_0]
+x = inv(a)*b
+c = [1,1;real(alpha1),real(alpha2)]
+d = [-K4;d_ia_dt_0]
+y = inv(c)*d
+
+//(c)To calculate the time taken for the speed to fall to zero value
+a=-K3/x(1)                      //a=exp(-0.966*t1)
+t1=alpha1*log(a)           //take log base e on both sides
+
+
+//Results
+mprintf("(a)Hence the braking resistance is :%.3f ohm",Rb)
+mprintf("\n(b)The solutions for alpha are %.3f and %.3f",real(alpha1),real(alpha2))
+mprintf("\nWm=%.2f + A*exp(%.3f*t) + B*exp(%.2f*t)",K3,real(alpha1),real(alpha2))
+mprintf("\nia=%.2f+ C*exp(%.3f*t) + D*exp(%.2f*t)",K4,real(alpha1),real(alpha2))
+mprintf("\nWe have to find the value of A,B,C and D in the linear equation using the initial condition")
+mprintf("\nA=%.2f B=%.2f C=%.2f D=%.2f",x(1),x(2),y(1),y(2))
+mprintf("\nHence the expression for the transient value for the speed is")
+mprintf("\nWm=%.2f+%.2f*exp(%.3f*t)%.2f*exp(%.2f*t)",K3,x(1),real(alpha1),x(2),real(alpha2))
+mprintf("\nHence the expression for the transient value for the current is")
+mprintf("\nia=%.2f %.1f *exp(%.3f*t) +%.2f*exp(%.2f*t)",K4,y(1),real(alpha1),y(2),real(alpha2))
+mprintf("\n(c)Hence the time taken is :%.2f sec",real(t1))
+//There is slight difference in the answers due to accuracy