6 files changed, 68 insertions, 0 deletions
diff --git a/3811/CH10/EX10.1/Ex10_1.jpg b/3811/CH10/EX10.1/Ex10_1.jpg
new file mode 100644
index 000000000..4ccb41365
--- /dev/null
+++ b/3811/CH10/EX10.1/Ex10_1.jpg
diff --git a/3811/CH10/EX10.1/Ex10_1.sce b/3811/CH10/EX10.1/Ex10_1.sce
new file mode 100644
index 000000000..c0ed65907
--- /dev/null
+++ b/3811/CH10/EX10.1/Ex10_1.sce
@@ -0,0 +1,30 @@
+//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi
+//chapter 10
+//example 10.1
+//edition 1
+//publishing place:Thomson Learning
+clc;
+clear;
+V=208;//source voltage in volts
+p=6;//number of poles
+R1=0.6;//given resistance in ohm
+R2=0.4;//given R'2 in ohm
+Xeq=5;//given Xeq in ohm
+Td=30;//load torque of the motor in ohm
+f=60;//frequency for 3 phase line
+ns=(120*f)/p
+disp('a)To find the regenerative speed:')
+Tl=-Td//reversed load torque
+rpss=ns/60;
+omegas=(2*%pi*rpss);//angular speed
+s=(Tl*omegas*R2)/V^2;
+n=ns*(1-s);
+mprintf("The regenerative speed is %f rpm",n)
+disp('b)To calculate the regenerative speed :')
+rps=n/60;
+omega=(2*%pi*rps);
+Pd=Td*omega;
+I2=sqrt(-Pd/(3*(R2/s)*(1-s)));//to find I'2 which is taken as I2
+Ploss=3*(R1+R2)*I2'^(2)
+Pds=Pd-Ploss;
+mprintf("The power delivered to the electric supply is %f watt",Pds)
diff --git a/3811/CH10/EX10.2/Ex10_2.jpg b/3811/CH10/EX10.2/Ex10_2.jpg
new file mode 100644
index 000000000..7cd2c09d1
--- /dev/null
+++ b/3811/CH10/EX10.2/Ex10_2.jpg
diff --git a/3811/CH10/EX10.2/Ex10_2.sce b/3811/CH10/EX10.2/Ex10_2.sce
new file mode 100644
index 000000000..b8d8fe152
--- /dev/null
+++ b/3811/CH10/EX10.2/Ex10_2.sce
@@ -0,0 +1,14 @@
+//Book name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi
+//chapter 10
+//example 10.2
+//edition 1
+//publishing place:Thomson Learning
+clc;
+clear;
+Vdc=200;//voltage at the dc link in volt
+I=25;//motor current in A
+R1=0.5;//stator resistance in ohm
+Ib=3*I;
+Vb=Ib*1.5*R1;//braking voltage in volt
+d=1.5*(Vb/Vdc)^2;
+mprintf("\nThe duty ratio of the FWM is %f",d)
diff --git a/3811/CH10/EX10.3/Ex10_3.jpg b/3811/CH10/EX10.3/Ex10_3.jpg
new file mode 100644
index 000000000..7f787876a
--- /dev/null
+++ b/3811/CH10/EX10.3/Ex10_3.jpg
diff --git a/3811/CH10/EX10.3/Ex10_3.sce b/3811/CH10/EX10.3/Ex10_3.sce
new file mode 100644
index 000000000..554c51dfa
--- /dev/null
+++ b/3811/CH10/EX10.3/Ex10_3.sce
@@ -0,0 +1,24 @@
+//Book Name: Fundamentals of electrical drives by Mohamad A. El- Sharkawi
+//chapter 10
+//example 10.3
+//edition 1
+//publishing place:Thomson Learning
+clc;
+clear;
+n1=1150;//full load speed in rpm
+V=300;//terminal voltage in volt
+f=80;//frequency in Hz
+Rr=0.5;//rotor resistance of the motor in ohm
+Xeq=3;//equivalent inductive reactance in ohm
+ns=1200;//nearest synchronous speed in rpm
+rpss=ns/60;
+omegas=(2*%pi*rpss);
+s1=(ns-n1)/ns;
+T6=(V^(2)*s1)/(omegas*Rr);//torque at the point 6
+T6=ceil(T6);
+mprintf("\nThe torque developed is %d Nm",T6)//approximated value
+s6=(T6*Rr*(-omegas))/V^(2);
+mprintf("\nThe slip is %f",s6)
+n6=(-ns)*(1-s6);
+mprintf("\nThe current of the induction motor does not surge to high value when the concurrent braking is implemented")
+