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-rwxr-xr-x2642/CH5/EX5.1/Ex5_1.sce21
-rwxr-xr-x2642/CH5/EX5.2/Ex5_2.sce26
-rwxr-xr-x2642/CH5/EX5.3/Ex5_3.sce24
-rwxr-xr-x2642/CH5/EX5.4/Ex5_4.sce26
-rwxr-xr-x2642/CH5/EX5.5/Ex5_5.sce31
-rwxr-xr-x2642/CH5/EX5.6/Ex5_6.sce32
-rwxr-xr-x2642/CH5/EX5.7/Ex5_7.sce31
-rwxr-xr-x2642/CH5/EX5.8/Ex5_8.sce40
-rwxr-xr-x2642/CH5/EX5.9/Ex5_9.sce37
9 files changed, 268 insertions, 0 deletions
diff --git a/2642/CH5/EX5.1/Ex5_1.sce b/2642/CH5/EX5.1/Ex5_1.sce
new file mode 100755
index 000000000..cf855405b
--- /dev/null
+++ b/2642/CH5/EX5.1/Ex5_1.sce
@@ -0,0 +1,21 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 5.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+l = 10 // conductor length in m
+B = 0.56 // magnetic flux density in T
+I = 2 // current through conductor in A
+
+// caclulations
+F = B*I*l // magnitude of force in N
+
+// display the result
+disp("Example 5.1 solution");
+printf(" \n Magnitude of force \n F = %.1f N ", F);
diff --git a/2642/CH5/EX5.2/Ex5_2.sce b/2642/CH5/EX5.2/Ex5_2.sce
new file mode 100755
index 000000000..580dcc0de
--- /dev/null
+++ b/2642/CH5/EX5.2/Ex5_2.sce
@@ -0,0 +1,26 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.2
+
+clc;clear; // clears the console and command history
+// Given data
+I = 20 // total current in A
+V_t = 250; // supply voltage in volt
+R_sh = 200; // shunt field resistance in Ω
+R_a = 0.3; // armature resistance in Ω
+
+// caclulations
+ I_sh = V_t/R_sh; // shunt field current in A
+ I_a = I-I_sh' // armature current
+ E_b = V_t - R_a*I_a; // the back emf in V
+ P_md = E_b*I_a; // mechanical power developed in W
+
+ // display the result
+disp("Example 5.2 solution");
+printf("\n The back emf is \n E_b = %.1f V \n\n",E_b );
+printf(" \n Mechanical power developed is \n P_md = %.1f W" ,P_md );
+printf(" \n NOTE : error in calculation they has taken I_a=18.13, instead of I_a=18.75");
diff --git a/2642/CH5/EX5.3/Ex5_3.sce b/2642/CH5/EX5.3/Ex5_3.sce
new file mode 100755
index 000000000..49b0f9c72
--- /dev/null
+++ b/2642/CH5/EX5.3/Ex5_3.sce
@@ -0,0 +1,24 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.3
+
+clc;clear; // clears the console and command history
+// Given data
+R_a = 0.7; // armature circuit resistance in Ω
+V_t = 5; // applied voltage in V
+I_anl = 5; // no-load armature current in A
+I_afl = 35; // full-load armature current in A
+
+
+// caclulations
+E_bnl = V_t - R_a*I_anl; // back emf under no-load in V
+E_bfl = V_t - R_a*I_afl; // back emf under full-load in V
+E_bc = E_bnl - E_bfl; // change in back emf from no-load to full load in V
+
+ // display the result
+disp("Example 5.3 solution");
+printf("\n The change in back emf is \n E_bc = %d V ",E_bc );
diff --git a/2642/CH5/EX5.4/Ex5_4.sce b/2642/CH5/EX5.4/Ex5_4.sce
new file mode 100755
index 000000000..f01c0b067
--- /dev/null
+++ b/2642/CH5/EX5.4/Ex5_4.sce
@@ -0,0 +1,26 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.4
+
+clc;clear; // clears the console and command history
+// Given data
+I = 40; // current in A
+V_t = 230; // supply in V
+N = 1100; // speed in rpm
+R_a = 0.25; // armature resistance in Ω
+R_sh = 230; //shunt resistance in Ω
+
+
+// caclulations
+I_sh = V_t/R_sh; // shunt field current in A
+I_a = I - 1; // armature current in A
+E_b = V_t - I_a*R_a; // back emf
+T_a = 9.55*E_b*I_a/N; // amrature torque in N-m
+
+ // display the result
+disp("Example 5.4 solution");
+printf("\n The armature torque is \n T_a = %.2f N-m ",T_a );
diff --git a/2642/CH5/EX5.5/Ex5_5.sce b/2642/CH5/EX5.5/Ex5_5.sce
new file mode 100755
index 000000000..5da7290f5
--- /dev/null
+++ b/2642/CH5/EX5.5/Ex5_5.sce
@@ -0,0 +1,31 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 6 // number of poles
+V_t = 230 // supply to shunt motor in V
+Z = 450 // number of conductors
+R_a = 0.8 // armature resistance in ohm
+I = 30 // supply current in A
+P_0 = 5560 // out put power in W
+I_F = 3 // current through field winding
+phi = 25*10^-3 // flux per pole in Wb
+
+// caclulations
+A = 6 // for lap wond A=P
+I_a = I-I_F // armature current in A
+E_b = V_t-I_a*R_a // back emf in V
+N = 60*A*E_b/(P*Z*phi) // speed in rpm
+T_sh = 9.55*P_0/N // shaft torque in N-m
+
+// display the result
+disp("Example 5.5 solution");
+printf(" speed \n N = %.1f rpm \n", N );
+printf(" shaft torque \n T_sh = %.1f N-m \n", T_sh );
diff --git a/2642/CH5/EX5.6/Ex5_6.sce b/2642/CH5/EX5.6/Ex5_6.sce
new file mode 100755
index 000000000..ab90e80b9
--- /dev/null
+++ b/2642/CH5/EX5.6/Ex5_6.sce
@@ -0,0 +1,32 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+I_L1 = 5 // dc shunt motor current
+V_t = 230 // terminal voltage in V
+N_1 = 1000 // speed in rpm
+R_a = 0.2 // armature resistance in ohm
+R_F = 230 // field resistance in ohm
+I_L2 = 30 // dc shunt motor current
+
+// caclulations
+// at noload condition
+I_sh = V_t/R_F // shunt field current in A
+I_a1 = I_L1-I_sh // armature current in A
+E_b1 = V_t-I_a1*R_a // back emf in V
+// under load condition
+I_a2 = I_L2-I_sh // armature current in A
+E_b2 = V_t-I_a2*R_a // back emf in V
+N_2 = (E_b2/E_b1)*N_1 // motor speed under load condtion in rpm
+
+// display the result
+disp("Example 5.6 solution");
+printf(" \n Speed under load condition \n N_2 = %.1f rpm \n", N_2 );
+
diff --git a/2642/CH5/EX5.7/Ex5_7.sce b/2642/CH5/EX5.7/Ex5_7.sce
new file mode 100755
index 000000000..053bc52fb
--- /dev/null
+++ b/2642/CH5/EX5.7/Ex5_7.sce
@@ -0,0 +1,31 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.7
+
+clc;clear; // clears the console and command history
+
+// Given data
+I_a1 = 65 // supply current to dc series motor in A
+V_t = 230 // supply voltage in V
+N_1 = 900 // speed in rpm
+R_a = 0.2 // armature resistance in ohm
+R_sh = 0.25 // field resistance in ohm
+I_a2 = 15 // supply current to dc series motor in A
+// phi_2 = 0.4*phi_1 value of flux
+
+
+// caclulations
+E_b1 = V_t-I_a1*(R_a+R_sh) // initial back emf in V
+E_b2 = V_t-I_a2*(R_a+R_sh) // final back emf in V
+// phi_2 = 0.4*phi_1 value of flux
+N_2 = (E_b2/E_b1)*(2.5)*N_1 // motor speed when line current is 15A in rpm
+
+// display the result
+disp("Example 5.7 solution");
+printf(" \n motor speed when line current is 15A \n N_2 = %.0f rpm \n", N_2 );
+
+printf(" NOTE: in question they given I_a1=56A , but in solution they took I_a1=65A");
diff --git a/2642/CH5/EX5.8/Ex5_8.sce b/2642/CH5/EX5.8/Ex5_8.sce
new file mode 100755
index 000000000..dfaf920e3
--- /dev/null
+++ b/2642/CH5/EX5.8/Ex5_8.sce
@@ -0,0 +1,40 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.8
+
+clc;clear; // clears the console and command history
+
+// Given data
+I_L1 = 5 // dc shunt motor current in A
+V_t = 230 // supply voltage in V
+R_a = 0.25 // armature resistance in ohm
+R_sh = 115 // field resistance in ohm
+I_L2 = 40 // dc shunt motor current in A
+
+
+// caclulations
+// at noload condition
+P_in1 = V_t*I_L1 // input power in W
+I_sh = V_t/R_sh // shunt field current in A
+I_a1 = I_L1-I_sh // armature current in A
+P_acu1 = I_a1^2*R_a // armature copper loss in W
+P_shcu = I_sh^2*R_sh //shunt field copper loss in W
+P_iron_friction = P_in1-(P_acu1+P_shcu) // iron and friction losses in W
+// under load condition
+I_a2 = I_L2-I_sh // armature current in A
+P_acu2 = I_a2^2*R_a // armature copper loss in W
+P_loss = P_iron_friction+P_shcu+P_acu2 // total losses in W
+P_in2 = V_t*I_L2 // input power in W
+P_0 = P_in2-P_loss // output power in W
+n = (P_0/P_in2)*100 // efficiency in percent
+
+// display the result
+disp("Example 5.8 solution");
+printf(" \n iron and friction losses \n P_iron_friction = %.2f W \n", P_iron_friction );
+printf(" \n efficiency \n n = %.0f percent \n", n)
+
+
diff --git a/2642/CH5/EX5.9/Ex5_9.sce b/2642/CH5/EX5.9/Ex5_9.sce
new file mode 100755
index 000000000..e808f5721
--- /dev/null
+++ b/2642/CH5/EX5.9/Ex5_9.sce
@@ -0,0 +1,37 @@
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.9
+
+clc;clear; // clears the console and command history
+
+// Given data
+I_L= 80 // dc shunt motor current in A
+V_t = 220 // supply voltage in V
+N = 800 // speed in rpm
+R_a = 0.1 // armature resistance in ohm
+R_sh = 50 // shunt field resistance in ohm
+P_if = 1600 // iron and friction losses in W
+
+// caclulations
+I_sh = V_t/R_sh // shunt field current in A
+I_a = I_L-I_sh // armature current
+E_b = V_t-I_a*R_a // back emf in V
+P_in = V_t*I_L // i/p power in W
+P_md = E_b*I_a // mechanical power developed in the armature in W
+P_cu = P_in-P_md // copper loss in W
+T_a = 9.55*(E_b*I_a)/N // armature torque in N-m
+P_0 = P_md-P_if // o/p power in W
+T_sh = 9.55*(P_0/N) // shaft torque in N-m
+n= (P_0/P_in)*100 // efficiency
+
+// display the result
+disp("Example 5.9 solution");
+printf(" \n Copper loss \n P_cu = %.2f W \n", P_cu);
+printf(" \n Armature torque \n T_a = %.2f N-m \n", T_a);
+printf(" \n Shaft torque \n T_sh = %.2f N-m \n", T_sh);
+printf(" \n Efficiency \n n = %.0f percent \n", n);
+
generated by cgit (git 2.34.1) at 2025-01-08 00:39:05 +0000