8 files changed, 130 insertions, 0 deletions
diff --git a/3802/CH11/EX11.1/Ex11_1.jpg b/3802/CH11/EX11.1/Ex11_1.jpg
new file mode 100644
index 000000000..94cfdaa0f
--- /dev/null
+++ b/3802/CH11/EX11.1/Ex11_1.jpg
diff --git a/3802/CH11/EX11.1/Ex11_1.sce b/3802/CH11/EX11.1/Ex11_1.sce
new file mode 100644
index 000000000..5c1e618ea
--- /dev/null
+++ b/3802/CH11/EX11.1/Ex11_1.sce
@@ -0,0 +1,59 @@
+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex11_1.sce
+
+clc;
+clear;
+V_not=220;
+I_not=4;
+W_not=100;
+Vsc=110;
+Isc=10;
+Wsc=400;
+p=6;
+V=220;
+f=50;
+
+printf("\n (a)")
+r1=(Wsc/Isc^2)/2;
+x1=sqrt((Vsc/Isc)^2-(2*r1)^2)/2;
+r2_dash=r1;
+x2_dash=x1;
+phi_not=acosd(W_not/(V_not*I_not));
+V_not_dash=V_not-((I_not*(cosd(phi_not)-%i*sind(phi_not)))*((r1+r2_dash/4)+%i*(x1+x2_dash/2)));
+Wi=W_not-(I_not^2*(r1+r2_dash/4));
+R_not_by_2=(V_not_dash^2)/Wi;
+Y_not=(I_not)/(V_not_dash*2);
+B_not=sqrt((2*Y_not)^2-(1/R_not_by_2)^2)/2;
+X_not_by_2=1/(2*B_not);
+printf("\n  Parameters of the motor:")
+printf("\n \t r1=r2dash=%d ohm",r1)
+printf("\n \t x1=x2dash=%1.3f ohm",x1)
+printf("\n \t R0/2=%3.2f ohm",sqrt(real(R_not_by_2)^2+imag(R_not_by_2)^2))
+printf("\n \t X0/2=%2.2f ohm",sqrt(real(X_not_by_2)^2+imag(X_not_by_2)^2))
+
+printf("\n (b)")
+//From the applied parameters of equivalent circuit of the motor  stator current  is simplified
+I1=complex(1.096,-0.526)*complex(6.36,-1.92);
+I1_mag=sqrt(real(I1)^2+imag(I1)^2);
+I1_angle=atand(imag(I1)/real(I1));
+pf=cosd(I1_angle);
+P_input=1075;
+P_loss=102.87;
+P_not=P_input-P_loss;
+Ns=1000;
+s=0.04;
+Nfl=(1-s)*Ns;
+T_net=P_not/(2*%pi*Nfl/60);
+motor_input=V*I1_mag*pf;
+efficiency=(P_not/motor_input)*100;
+printf("\n  Stator current: \n\t magnitude=%1.2f V,\n\t angle=%2.2f degree",I1_mag,I1_angle)
+printf("\n  Power factor=%0.3f lagging",pf)
+printf("\n  Power output=%3.2f watt",P_not)
+printf("\n  Speed=%d r.p.m",Nfl)
+printf("\n  Torque=%1.2f Nm",T_net)
+printf("\n  Efficiency=%d percentage",efficiency)
+//Answer vary dueto roundoff error
diff --git a/3802/CH11/EX11.2/Ex11_2.jpg b/3802/CH11/EX11.2/Ex11_2.jpg
new file mode 100644
index 000000000..2359d72ba
--- /dev/null
+++ b/3802/CH11/EX11.2/Ex11_2.jpg
diff --git a/3802/CH11/EX11.2/Ex11_2.sce b/3802/CH11/EX11.2/Ex11_2.sce
new file mode 100644
index 000000000..12f2b680b
--- /dev/null
+++ b/3802/CH11/EX11.2/Ex11_2.sce
@@ -0,0 +1,20 @@
+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex11_2.sce
+
+clc;
+clear;
+t=0.5;        //pole pitch
+f=50;
+vmp=162;
+fd=100e3;
+vm=vmp*1e3/(60*60);
+pd=fd*vm;
+vs=2*t*f;
+s=(vs-vm)/vs;
+pcu=s*fd*vs;
+printf("\n  The developed power by the motor=%d kw \n",pd/1000)
+printf("\n  Secondary copper loss=%d kw \n",pcu/1000)
diff --git a/3802/CH11/EX11.3/Ex11_3.jpg b/3802/CH11/EX11.3/Ex11_3.jpg
new file mode 100644
index 000000000..df4d5adcb
--- /dev/null
+++ b/3802/CH11/EX11.3/Ex11_3.jpg
diff --git a/3802/CH11/EX11.3/Ex11_3.sce b/3802/CH11/EX11.3/Ex11_3.sce
new file mode 100644
index 000000000..cfd423c73
--- /dev/null
+++ b/3802/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,25 @@
+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex11_3.sce
+
+clc;
+clear;
+Ra=0.8;
+Va=40;
+Td=1.2;
+Ka=600;
+phi_p=0.004;
+
+printf("\n (a)")
+n=(Va/(Ka*phi_p))-(2*%pi*Ra*Td/(Ka*phi_p)^2);
+N=n*60;
+printf("\n  The speed of the motor=%d r.p.m \n",N)
+//The book answer for part(a) is wrong value
+
+printf("\n (b)")
+n=0;
+Td=(Va*Ka*phi_p)/(2*%pi*Ra);
+printf("\n  The blocked rotor torque=%d Nm \n",Td)
diff --git a/3802/CH11/EX11.4/Ex11_4.jpg b/3802/CH11/EX11.4/Ex11_4.jpg
new file mode 100644
index 000000000..e3d8885cb
--- /dev/null
+++ b/3802/CH11/EX11.4/Ex11_4.jpg
diff --git a/3802/CH11/EX11.4/Ex11_4.sce b/3802/CH11/EX11.4/Ex11_4.sce
new file mode 100644
index 000000000..5257c1654
--- /dev/null
+++ b/3802/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,26 @@
+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex11_4.sce
+
+clc;
+clear;
+P=200;
+V=100;
+N=1500;
+Ka=525;
+Ra=2;
+Pl=15;
+
+Pd=P+Pl;
+n=N/60;
+Td=Pd/(2*%pi*n);
+//n=(Va/(Ka*phi_p))-(2*%pi*Ra*Td/(Ka*phi_p)^2);
+//from this equation we get phi^2-o-0.0076*phi+2.5e-6=0;
+a=1;
+b=-0.0076;//a,b,c are coefficient values taken from the above second order equation
+c=2.5e-6;
+phi_p=(-b+sqrt(b^2-(4*a*c)))/(2*a);
+printf("\n The magnetic flux=%1.3f mWb \n",phi_p*1000)