11 files changed, 658 insertions, 0 deletions
diff --git a/1092/CH7/EX7.1/Example7_1.sce b/1092/CH7/EX7.1/Example7_1.sce
new file mode 100755
index 000000000..1d81bfcd9
--- /dev/null
+++ b/1092/CH7/EX7.1/Example7_1.sce
@@ -0,0 +1,74 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_sh = 120 ; // Shunt field resistance in ohm
+R_a = 0.1 ; // Armature resistance in ohm
+V_L = 120 ; // Line voltage in volt
+E_g1 = 125 ; // Generated voltage by dynamo A
+E_g2 = 120 ; // Generated voltage by dynamo B
+E_g3 = 114 ; // Generated voltage by dynamo C
+
+// Calculations
+// case a
+// 1:
+I_gA = ( E_g1 - V_L ) / R_a ; // Current in the generating source A ( in  A)
+I_f = V_L / R_sh ; // Shunt field current in A
+I_a1 = I_gA + I_f ; // Armature current in A for generator A
+I_L1 = I_gA ; // Current delivered by dynamo A to the bus in A
+
+// 2:
+I_gB =  ( E_g2 - V_L ) / R_a ; // Current in the generating source B ( in  A)
+I_a2 = I_gB + I_f ; // Armature current in A for generator B
+I_L2 = I_gB ; // Current delivered by dynamo B to the bus in A
+
+// 3:
+I_gC =  ( V_L - E_g3 ) / R_a ; // Current in the generating source C ( in  A)
+I_a3 = I_gC ; // Armature current in A for generator C
+I_L3 = I_gC + I_f ; // Current delivered by dynamo C to the bus in A
+
+// case b
+// 1:
+P_LA = V_L * I_L1 ; // Power delivered to the bus by dynamo A in W
+P_gA = E_g1 * I_a1 ; // Power generated by dynamo A
+
+// 2:
+P_LB = V_L * I_L2 ; // Power delivered to the bus by dynamo B in W
+P_gB = E_g2 * I_a2 ; // Power generated by dynamo B
+
+// 3:
+P_LC = V_L * I_L3 ; // Power delivered to the bus by dynamo C in W
+P_gC = E_g3 * I_a3 ; // Power generated by dynamo C
+
+// Display the results
+disp("Example 7-1 Solution : ");
+printf(" \n a: 1. I_gA = %d A \t I_f = %d A ", I_gA,I_f );
+printf(" \n       Thus,dynamo A delivers %d A to the bus and has an armature", I_gA);
+printf(" \n       current of %d A + %d A = %d \n", I_gA,I_f,I_a1 );
+
+printf(" \n    2. I_gB = %d A ", I_gB);
+printf(" \n       Thus, dynamo B is floating and has as armature & field current of %d A \n",I_f);
+
+printf(" \n    3. I_gC = %d A ",I_gC);
+printf(" \n       Dynamo C receives %d A from the bus & has an armature current of %d A\n",I_L3,I_a3);
+
+printf(" \n b: 1. Power delivered to the bus by dynamo A is : ");
+printf(" \n       P_LA = %d W ",P_LA);
+printf(" \n       Power generated by dynamo A is \n       P_gA = %d W \n",P_gA);
+ 
+printf(" \n    2. Since dynamo B neither delivers power to nor receives power from the bus, ");
+printf(" \n       P_B = %d W ",P_LB);
+printf(" \n       Power generated by dynamo B,to excite its field, is");
+printf(" \n       P_gB = %d W \n ", P_gB); 
+
+printf(" \n    3. Power delivered by the bus to dynamo C is ");
+printf(" \n       P_LC = %d W ", P_LC);
+printf(" \n       while the internal power delivered in the direction of rotation");
+printf(" \n       of its prime mover to aid rotation is \n       P_gC = %d W", P_gC );
diff --git a/1092/CH7/EX7.10/Example7_10.sce b/1092/CH7/EX7.10/Example7_10.sce
new file mode 100755
index 000000000..2ee988e51
--- /dev/null
+++ b/1092/CH7/EX7.10/Example7_10.sce
@@ -0,0 +1,70 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-10
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt
+E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt
+
+theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees
+theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees
+
+// writing given voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt
+E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt
+
+// writing given impedance(in ohm)in exponential form as follows
+Z_1 = 6 * expm(%i * 50*(%pi/180) ); // %pi/180 for degrees to radians conversion
+Z_2 = Z_1 ;
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// Calculations
+E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case a
+I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b
+E_gp1 = E_1_mag;
+P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W
+
+// case c
+E_gp2 = E_2_mag;
+P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W
+
+// case d
+// but consider +ve vlaue for P_2 for finding losses, so
+P2 = abs(P_2);
+losses = P_1 - P2 ; // Losses in the armature in W
+
+// E_r_a yields -80 degrees which is equivalent to 100 degrees, so
+theta = 100 - I_s_a ; // Phase difference between E_r and I_s in degrees
+
+check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7
+R_aT = 12*cosd(50) ; // total armature resistance of alternator 1 and 2 in ohm
+double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7
+
+// Display the results
+disp("Example 7-10 Solution : ");
+printf(" \n a: I_s = ");disp(I_s);
+printf(" \n    I_s = %.2f <%.2f A \n ",I_s_m, I_s_a );
+
+printf(" \n b: P_1 = %.f W (power delivered to bus)",P_1);
+printf(" \n    Note:Slight variation in P_1 is due slight variations in ")
+printf(" \n         phase angle of I_s,& angle btw (E_gp1,I_s)\n")
+printf(" \n    P_2 = %.f W (power received from bus)\n",P_2);
+
+printf(" \n c: Losses: P_1 - P_2 = %.f W",losses);
+printf(" \n    Check: E_a*I_s*cos(theta) = %.f W ",check );
+printf(" \n    Double check : (I_s)^2*(R_a1+R_a2) = %.f W ",double_check );
diff --git a/1092/CH7/EX7.11/Example7_11.sce b/1092/CH7/EX7.11/Example7_11.sce
new file mode 100755
index 000000000..e65ef2eea
--- /dev/null
+++ b/1092/CH7/EX7.11/Example7_11.sce
@@ -0,0 +1,91 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-11
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// writing supply voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+V_AB = 100 * expm(%i * 0*(%pi/180) ); // voltage supplied across A & B in volt
+V_BC = 100 * expm(%i * -120*(%pi/180) ); // voltage supplied across B & C in volt
+V_CA = 100 * expm(%i * 120*(%pi/180) ); // voltage supplied across C & A in volt
+
+disp("Example 7-11 : ");
+printf("\n Writing two mesh equations for I_1 and I_2 in fig.7-23a yields following\n array :"); 
+printf(" \n  I_1 \t\t I_2 \t\t V ");
+printf(" \n ____________________________________________");
+printf(" \n  6 + j0 \t -3 + j0 \t 100 + j0 ");
+printf(" \n -3 + j0 \t 3 - j4 \t -50 - j86.6 ");
+
+// Calculations
+A = [ (6+%i*0) (-3+%i*0) ; (-3+%i*0) (3-%i*4) ]; // Matrix containing above mesh eqns array
+delta = det(A); // Determinant of A
+
+// case a 
+I_1 = det( [ (100+%i*0) (-3+%i*0) ; (-50-%i*86.60) (3-%i*4) ] ) / delta ;
+// Mesh current I_1 in A 
+I_1_m = abs(I_1);//I_1_m=magnitude of I_1 in A
+I_1_a = atan(imag(I_1) /real(I_1))*180/%pi;//I_1_a=phase angle of I_1 in degrees
+
+I_2 = det( [ (6+%i*0) (100+%i*0) ; (-3+%i*0) (-50-%i*86.6) ] ) / delta ;
+// Mesh current I_2 in A 
+I_2_m = abs(I_2);//I_2_m=magnitude of I_2 in A
+I_2_a = atan(imag(I_2) /real(I_2))*180/%pi;//I_2_a=phase angle of I_2 in degrees
+
+// case b
+I_A = I_1 ; //  Line current I_A in A
+I_A_m = abs(I_A);//I_A_m=magnitude of I_A in A
+I_A_a = atan(imag(I_A) /real(I_A))*180/%pi;//I_A_a=phase angle of I_A in degrees
+
+I_B = I_2 - I_1 ; //  Line current I_B in A
+I_B_m = abs(I_B);//I_B_m=magnitude of I_B in A
+I_B_a = atan(imag(I_B) /real(I_B))*180/%pi - 180;//I_B_a=phase angle of I_B in degrees
+
+I_C = -I_2 ; //  Line current I_C in A
+I_C_m = abs(I_C);//I_C_m=magnitude of I_C in A
+I_C_a = 180 + atan(imag(I_C) /real(I_C))*180/%pi;//I_C_a=phase angle of I_C in degrees
+
+// case c
+Z_A = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line A in ohm
+Z_B = 3 * expm(%i * 0*(%pi/180) ); // Impedance in line B in ohm
+Z_C = 4 * expm(%i * -90*(%pi/180) ); // Impedance in line C in ohm
+
+V_AO = I_A * Z_A ; // Phase voltage V_AO in volt
+V_AO_m = abs(V_AO);//V_AO_m=magnitude of V_AO in volt
+V_AO_a = atan(imag(V_AO) /real(V_AO))*180/%pi;//V_AO_a=phase angle of V_AO in degrees
+
+V_BO = I_B * Z_B ; // Phase voltage V_BO in volt
+V_BO_m = abs(V_BO);//V_BO_m=magnitude of V_BO in volt
+V_BO_a = atan(imag(V_BO) /real(V_BO))*180/%pi - 180;//V_BO_a=phase angle of V_BO in degrees
+
+V_CO = I_C * Z_C ; // Phase voltage V_CO in volt
+V_CO_m = abs(V_CO);//V_CO_m=magnitude of V_CO in volt
+V_CO_a = atan(imag(V_CO) /real(V_CO))*180/%pi;//V_CO_a=phase angle of V_CO in degrees
+
+// Display the results
+disp("Solution : ");
+printf(" \n a: I_1 in A = ");disp(I_1);
+printf(" \n    I_1 = %.2f <%.2f A \n ",I_1_m, I_1_a );
+printf(" \n    I_2 in A = ");disp(I_2);
+printf(" \n    I_2 = %.2f <%.2f A\n ",I_2_m, I_2_a );
+
+printf(" \n b: I_A in A = ");disp(I_1);
+printf(" \n    I_A = %.2f <%.2f A\n",I_A_m, I_A_a );
+
+printf(" \n    I_B in A = ");disp(I_B);
+printf(" \n    I_B = %.2f <%.2f A\n",I_B_m, I_B_a );
+
+printf(" \n    I_C in A = ");disp(I_C);
+printf(" \n    I_C = %.2f <%.2f A \n",I_C_m,  I_C_a );
+
+printf(" \n c: V_AO = %.2f <%.2f V",V_AO_m, V_AO_a );
+printf(" \n    V_BO = %.2f <%.2f V",V_BO_m, V_BO_a );
+printf(" \n    V_CO = %.2f <%.2f V\n",V_CO_m, V_CO_a );
+
+printf(" \n d: The phasor diagram is shown in Fig.7-23b, with the phase voltages");
+printf(" \n    inscribed inside the (equilateral) triangle of given line voltages");
diff --git a/1092/CH7/EX7.2/Example7_2.sce b/1092/CH7/EX7.2/Example7_2.sce
new file mode 100755
index 000000000..3f4111a97
--- /dev/null
+++ b/1092/CH7/EX7.2/Example7_2.sce
@@ -0,0 +1,40 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_a = 0.1 ; // Armature resistance in ohm
+R_f = 100 ; // Field ckt resistance in ohm
+V_L_b = 120 ; // Bus voltage in volt
+V_L_a = 140 ; // Voltage of the generator in volt
+V_f = V_L_a ; // Voltage across the field in volt
+
+// Calculations
+// case a
+I_f_a = V_f / R_f ; // Field current in A
+I_a_a = I_f_a ; // Armature current in A
+E_g_a = V_L_a + I_a_a * R_a ; // Generated EMF in volt
+P_g_a = E_g_a * I_a_a ; // Generated power in W
+
+// case b
+I_a_b = ( E_g_a - V_L_b ) / R_a ; // Armature current in A
+I_f_b = V_L_b / R_f ; // Field current in A
+I_Lg = I_a_b - I_f_b ; // Generated line current in A
+P_L = V_L_b * I_Lg ; // Power generated across the lines in W
+E_g_b = V_L_a ;
+P_g_b = E_g_b * I_a_b ; // Generated power in W
+
+// Display the results 
+disp("Example 7-2 Solution : ");
+printf(" \n a: Before it is connected to the bus ");
+printf(" \n    I_a = I_f = %.1f A \n    E_g = %.2f V\n    P_g = %.1f W \n", I_a_a,E_g_a,P_g_a);
+
+printf(" \n b: After it is connected to the bus ");
+printf(" \n    I_a = %.1f A \n    I_f = %.1f A \n    I_Lg = %.1f A \n", I_a_b, I_f_b, I_Lg );
+printf(" \n    P_L = %.f W \n    P_g = %.f W ", P_L , P_g_b );
diff --git a/1092/CH7/EX7.3/Example7_3.sce b/1092/CH7/EX7.3/Example7_3.sce
new file mode 100755
index 000000000..d08f6534a
--- /dev/null
+++ b/1092/CH7/EX7.3/Example7_3.sce
@@ -0,0 +1,66 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_a = 0.1 ; // Armature resistance in ohm of 3 shunt generators
+R_a1 =R_a ; 
+R_a2 =R_a ; 
+R_a3 =R_a ; 
+R_L = 2 ; // Load resistance in ohm
+E_g1 = 127 ; // Voltage generated by generator 1 in volt
+E_g2 = 120 ; // Voltage generated by generator 2 in volt
+E_g3 = 119 ; // Voltage generated by generator 3 in volt
+// Neglect field currents
+
+// Calculations
+// case a
+// Terminal bus voltage in volt
+V_L = ( (127/0.1) + (120/0.1) + (119/0.1) ) / ( (1/0.1) + (1/0.1) + (1/0.1) + 0.5);
+
+// case b
+I_L1 = (E_g1 - V_L)/R_a1 ; // Current delivered by generator 1 in A
+I_L2 = (E_g2 - V_L)/R_a2 ; // Current delivered by generator 2 in A
+I_L3 = (E_g3 - V_L)/R_a3 ; // Current delivered by generator 3 in A
+I_L_2ohm = V_L / R_L ; // Current delivered by 2 ohm load in A
+
+// case c
+I_a1 = I_L1 ; // Armature current in A for generator 1
+I_a2 = I_L2 ; // Armature current in A for generator 2
+I_a3 = I_L3 ; // Armature current in A for generator 3
+
+P_g1 = E_g1 * I_a1 ; // Power generated by generator 1 in W
+P_g2 = E_g2 * I_a2 ; // Power generated by generator 2 in W
+P_g3 = E_g3 * I_a3 ; // Power generated by generator 3 in W
+
+// case d
+P_L1 = V_L * I_L1 ; // Power delivered to or received from generator 1 in W
+P_L2 = V_L * I_L2 ; // Power delivered to or received from generator 2 in W
+P_L3 = V_L * I_L3 ; // Power delivered to or received from generator 3 in W
+P_L = V_L * -I_L_2ohm ; // Power delivered to or received 2 ohm load in W
+
+// Display the results
+disp("Example 7-3 Solution : ");
+printf(" \n a: Converting each voltage source to a current source and applying");
+printf(" \n    Millman`s theorem yields ")
+printf(" \n    V_L = %d V \n ", V_L );
+
+printf(" \n b: I_L1 = %d A (to bus)", I_L1 );
+printf(" \n    I_L2 = %d A ", I_L2 );
+printf(" \n    I_L3 = %d A (from bus)", I_L3 );
+printf(" \n    I_L_2ohm = -%d A (from bus) \n", I_L_2ohm );
+
+printf(" \n c: P_g1 = %d W ",P_g1 );
+printf(" \n    P_g2 = %d W (floating)",P_g2 );
+printf(" \n    P_g3 = %d W \n",P_g3 );
+
+printf(" \n d: P_L1 = %d W ", P_L1);
+printf(" \n    P_L2 = %d W ", P_L2 );
+printf(" \n    P_L3 = %d W ", P_L3 );
+printf(" \n    P_L = %d W ", P_L );
diff --git a/1092/CH7/EX7.4/Example7_4.sce b/1092/CH7/EX7.4/Example7_4.sce
new file mode 100755
index 000000000..3269a85be
--- /dev/null
+++ b/1092/CH7/EX7.4/Example7_4.sce
@@ -0,0 +1,46 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+P1 = 300 ; // Power rating of generator 1 in kW
+P2 = 600 ; // Power rating of generator 2 in kW
+V = 220 ; // Voltage rating of generator 1 and 2 in volt
+V_o = 250 ; // No-load voltage applied to both the generators in volt
+// Assume linear characteristics
+V_1 = 230 ; // Terminal voltage in volt (case a)
+V_2 = 240 ; // Terminal voltage in volt (case b)
+
+// Calculations
+// case a
+kW1_a = (V_o - V_1)/(V_o - V) * P1 ; // kW carried by generator 1
+kW2_a = (V_o - V_1)/(V_o - V) * P2 ; // kW carried by generator 2
+
+// case b
+kW1_b = (V_o - V_2)/(V_o - V) * P1 ; // kW carried by generator 1
+kW2_b = (V_o - V_2)/(V_o - V) * P2 ; // kW carried by generator 2
+
+// case c
+frac_a = (V_o - V_1)/(V_o - V); // Fraction of rated kW carried by each generator
+frac_b = (V_o - V_2)/(V_o - V); // Fraction of rated kW carried by each generator
+
+// Display the results
+disp("Example 7-4 Solution : ");
+printf(" \n a: At 230 V, using Eq.(7-3) below : ");
+printf(" \n    Generator 1 carries = %d kW ", kW1_a );
+printf(" \n    Generator 2 carries = %d kW \n", kW2_a );
+
+printf(" \n b: At 240 V, using Eq.(7-3) below : ");
+printf(" \n    Generator 1 carries = %d kW ", kW1_b );
+printf(" \n    Generator 2 carries = %d kW \n", kW2_b );
+
+printf(" \n c: Both generators carry no-load at 250 V; ");
+printf(" \n    %f rated load at %d V; ", frac_b , V_2 );
+printf(" \n    %f rated load at %d V; ", frac_a , V_1 );
+printf(" \n    and rated load at %d V. ", V );
diff --git a/1092/CH7/EX7.5/Example7_5.sce b/1092/CH7/EX7.5/Example7_5.sce
new file mode 100755
index 000000000..393202959
--- /dev/null
+++ b/1092/CH7/EX7.5/Example7_5.sce
@@ -0,0 +1,38 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_1 = 220 ; // Terminal voltage of alternator 1 in volt 
+E_2 = 222 ; // Terminal voltage of alternator 2 in volt 
+f_1 = 60 ; // Frequency of alternator 1 in Hz
+f_2 = 59.5 ; // Frequency of alternator 2 in Hz
+// Switch is open
+
+// Calculations
+// case a
+E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt 
+E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt 
+
+// case b
+f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps
+
+// case c
+E_peak = E_max / 0.7071 ; // Peak value of the voltage in volt across each lamp
+
+// case d
+n = (1/2)*f_1 ; // Number of maximum light pulsations per minute
+
+// Display the results
+disp("Example 7-5 Solution : ");
+printf(" \n a: E_max/lamp = %d V (rms)\n ", E_max );
+printf(" \n    E_min/lamp = %d V \n ", E_min );
+printf(" \n b: f = %.1f Hz \n ", f );
+printf(" \n c: E_peak = %.f V \n ", E_peak );
+printf(" \n d: n = %d pulsations/min ", n );
diff --git a/1092/CH7/EX7.6/Example7_6.sce b/1092/CH7/EX7.6/Example7_6.sce
new file mode 100755
index 000000000..6c1cf6a18
--- /dev/null
+++ b/1092/CH7/EX7.6/Example7_6.sce
@@ -0,0 +1,32 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E = 220 ; // Voltage generated in volt
+E_1 = E ; // Voltage generated by alternator 1 in volt
+E_2 = E ; // Voltage generated by alternator 2 in volt
+f_1 = 60 ; // Frequency in Hz of alternator 1
+f_2 = 58 ; // Frequency in Hz of alternator 2
+// Switch is open
+
+// Calculations
+// case a
+E_max = (E_1 + E_2)/2 ; // Maximum effective voltage across each lamp in volt
+f = f_1 - f_2 ; // Frequency in Hz of the voltage across the lamps
+
+// case c
+E_min = (E_2 - E_1)/2 ; // Minimum effective voltage across each lamp in volt 
+
+// Display the results
+disp("Example 7-6 Solution : ");
+printf(" \n a: E_max/lamp = %d V \n    f = %d Hz \n ", E_max, f );
+printf(" \n b: The voltages are equal and opposite in the local circuit. \n ");
+printf(" \n c: E_min/lamp = %d V at zero frequency \n ", E_min );
+printf(" \n d: The voltages are in phase in the local circuit. ");
diff --git a/1092/CH7/EX7.7/Example7_7.sce b/1092/CH7/EX7.7/Example7_7.sce
new file mode 100755
index 000000000..7e08faf5b
--- /dev/null
+++ b/1092/CH7/EX7.7/Example7_7.sce
@@ -0,0 +1,57 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data as per Ex.(7-5)
+E1 = 220 ; // Terminal voltage of alternator 1 in volt 
+E2 = 222 ; // Terminal voltage of alternator 2 in volt 
+f1 = 60 ; // Frequency of alternator 1 in Hz
+f2 = 59.5 ; // Frequency of alternator 2 in Hz
+// Switch is open
+
+// Given data as per Ex.(7-6)
+E = 220 ; // Voltage generated in volt
+E_1 = E ; // Voltage generated by alternator 1 in volt
+E_2 = E ; // Voltage generated by alternator 2 in volt
+f_1 = 60 ; // Frequency in Hz of alternator 1
+f_2 = 58 ; // Frequency in Hz of alternator 2
+// Switch is open
+
+// Given data as per Ex.(7-7)
+R_a1 = 0.1 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.1 ; // armature resistance of alternator 2 in ohm
+X_a1 = 0.9 ; // armature reactance of alternator 1 in ohm
+X_a2 = 0.9 ; // armature reactance of alternator 2 in ohm
+
+Z_1  = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm
+Z_2  = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm
+// Switches are closed at the proper instant for paralleling.
+
+// Calculations
+// In Ex.7-5,
+E_r = E2 - E1 ; // Effective voltage generated in volt 
+I_s = E_r / (Z_1 + Z_2); // Synchronizing current in the armature in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// In Ex.7-6,
+Er = E_2 -E_1 ; // Effective voltage generated in volt 
+Is = Er / ( Z_1 + Z_2); // Synchronizing current in the armature in A
+
+// Display the results
+disp("Example 7-7 Solution : ");
+printf(" \n In Ex.7-5, ");
+printf(" \n E_r = %d V ", E_r);
+printf(" \n I_s = ");disp(I_s);
+printf(" \n I_s = %.3f <%.2f A ",I_s_m, I_s_a);
+printf(" \n where %.3f is magnitude in A and %.2f is phase angle in degrees \n",I_s_m,I_s_a);
+
+printf(" \n In Ex.7-6, ");
+printf(" \n E_r = %d V ", Er );
+printf(" \n I_s = %d A",Is);
diff --git a/1092/CH7/EX7.8/Example7_8.sce b/1092/CH7/EX7.8/Example7_8.sce
new file mode 100755
index 000000000..fc8a8afc0
--- /dev/null
+++ b/1092/CH7/EX7.8/Example7_8.sce
@@ -0,0 +1,71 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// EMF's are opposed exactly by 180 degrees
+E_gp1 = 200 ; // Terminal voltage of alternator 1 in volt 
+E_gp2 = 220 ; // Terminal voltage of alternator 2 in volt 
+R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm
+X_a1 = 2 ; // armature reactance of alternator 1 in ohm
+X_a2 = 2 ; // armature reactance of alternator 2 in ohm
+
+Z_p1  = R_a1 + %i*X_a1 ; // Effective impedance of alternator 1 in ohm
+Z_p2  = R_a1 + %i*X_a2 ; // Effective impedance of alternator 2 in ohm
+// Switches are closed at the proper instant for paralleling.
+
+// Calculations
+// case a
+E_r = (E_gp2 - E_gp1) ; // Effective voltage generated in volt 
+I_s = E_r / (Z_p1 + Z_p2); // Synchronizing current in the armature in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+P_2 = E_gp2 * I_s_m * cosd(I_s_a); // Generator action developed by alternator 2 in W
+
+// case b
+theta = I_s_a;
+// P_1 = E_gp1 * I_s_m * cosd(180 - theta)
+// P_1 = -E_gp1 * I_s_m * cosd(theta), 
+P_1 = -E_gp1 * I_s_m * cosd(theta); // Synchronizing power received by alternator 1 in W 
+
+// case c
+// but consider +ve vlaue for P_1 for finding losses, so
+P1 = abs(P_1);
+losses = P_2 - P1 ; // Power losses in both armatures in W
+check = E_r * I_s_m * cosd(I_s_a); // Verifying losses by Eq.7-7
+double_check = (I_s_m)^2 * (R_a1 + R_a2); // Verifying losses by Eq.7-7
+
+// case d
+V_p2  = E_gp2 - I_s*Z_p1 ; // Generator action
+V_p1  = E_gp1 + I_s*Z_p1 ; // Motor action
+
+// Display the results
+disp("Example 7-8 Solution : ");
+printf(" \n a: E_r = %d V ",E_r);
+printf(" \n    I_s = %.2f <%.2f A ", I_s_m, I_s_a );
+printf(" \n    P_2 = %.1f W (total power delivered by alternator 2 ) \n", P_2);
+
+printf(" \n b: P_1 = %f W (synchronizing power received by alternator 1)",P_1);
+printf(" \n    Note:Scilab considers phase angle of I_s as %f instead ",I_s_a);
+printf(" \n         of -84.3 degrees,so slight variation in the answer P_1.\n");
+
+printf(" \n c: Consider +ve value of P_1 for calculating losses"); 
+printf(" \n    Losses: P_2 - P_1 = %.1f W ",losses );
+printf(" \n    Check: E_a*I_s*cos(theta) = %.1f W ",check );
+printf(" \n    Double check : (I_s)^2*(R_a1+R_a2) = %.1f W as given in Eq.(7-1)",double_check );
+
+printf("\n\n d: From Fig.7-14, V_p2, the terminal phase voltage of ");
+printf(" \n    alternator 2, is, from Eq.(7-1)");
+printf(" \n    V_p2 = %d V (generator action)\n\n    From section 7-2.1 ",V_p2);
+printf(" \n    V_p1 = %d V ( motor action)\n",V_p1);
+
+printf(" \n e: The phasor diagram is shown in Fig.7-14.");
+
diff --git a/1092/CH7/EX7.9/Example7_9.sce b/1092/CH7/EX7.9/Example7_9.sce
new file mode 100755
index 000000000..d2aaf5290
--- /dev/null
+++ b/1092/CH7/EX7.9/Example7_9.sce
@@ -0,0 +1,73 @@
+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-9
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+E_2_mag = 230 ; // Magnitude of voltage generated by alternator 2 in volt
+E_1_mag = 230 ; // Magnitude of voltage generated by alternator 1 in volt
+
+theta_2 = 180 ; // Phase angle of generated voltage by alternator 2 in degrees
+theta_1 = 20 ; // Phase angle of generated voltage by alternator 1 in degrees
+
+R_a1 = 0.2 ; // armature resistance of alternator 1 in ohm
+R_a2 = 0.2 ; // armature resistance of alternator 2 in ohm
+
+// writing given voltage in exponential form as follows
+// %pi/180 for degrees to radians conversion
+E_2 = E_2_mag * expm(%i * theta_2*(%pi/180) ); // voltage generated by alternator 2 in volt
+E_1 = E_1_mag * expm(%i * theta_1*(%pi/180) ); // voltage generated by alternator 1 in volt
+
+// writing given impedance(in ohm)in exponential form as follows
+Z_1 = 2.01 * expm(%i * 84.3*(%pi/180) ); // %pi/180 for degrees to radians conversion
+Z_2 = Z_1 ;
+Z_1_a = atan(imag(Z_1) /real(Z_1))*180/%pi;//Z_1_a=phase angle of Z_1 in degrees
+
+// Calculations
+E_r = E_2 + E_1 ; // Total voltage generated by Alternator 1 and 2 in volt
+E_r_m = abs(E_r);//E_r_m=magnitude of E_r in volt
+E_r_a = atan(imag(E_r) /real(E_r))*180/%pi;//E_r_a=phase angle of E_r in degrees
+
+// case a
+I_s = E_r / (Z_1 + Z_2); // Synchronozing current in A
+I_s_m = abs(I_s);//I_s_m=magnitude of I_s in A
+I_s_a = atan(imag(I_s) /real(I_s))*180/%pi;//I_s_a=phase angle of I_s in degrees
+
+// case b
+E_gp1 = E_1_mag;
+P_1 = E_gp1 * I_s_m * cosd(I_s_a - theta_1); // Synchronozing power developed by alternator 1 in W
+
+// case c
+E_gp2 = E_2_mag;
+P_2 = E_gp2 * I_s_m * cosd(I_s_a - theta_2); // Synchronozing power developed by alternator 2 in W
+
+// case d
+// but consider +ve vlaue for P_2 for finding losses, so
+P2 = abs(P_2);
+losses = P_1 - P2 ; // Losses in the armature in W
+
+// E_r_a yields -80 degrees which is equivalent to 100 degrees, so
+theta = 100 - I_s_a ; // Phase difference between E_r and I_a in degrees
+
+check = E_r_m * I_s_m * cosd(theta); // Verifying losses by Eq.7-7
+R_aT = R_a1 + R_a2 ; // total armature resistance of alternator 1 and 2 in ohm
+double_check = (I_s_m)^2 * (R_aT); // Verifying losses by Eq.7-7
+
+// Display the results
+disp("Example 7-9 Solution : ");
+printf(" \n a: I_s = ");disp(I_s);
+printf(" \n    I_s = %.2f <%.2f A \n ",I_s_m, I_s_a );
+
+printf(" \n b: P_1 = %.f W (power delivered to bus)",P_1);
+printf(" \n    Slight variation in P_1 is due slight variations in ")
+printf(" \n    magnitude of I_s,& angle btw (E_gp1,I_s)\n")
+printf(" \n    P_2 = %.f W (power received from bus)\n",P_2);
+
+printf(" \n c: Losses: P_1 - P_2 = %d",losses);
+printf(" \n    Check: E_a*I_s*cos(theta) = %d W ",check );
+printf(" \n    Double check : (I_s)^2*(R_a1+R_a2) = %d W ",double_check );