diff options
Diffstat (limited to '2642/CH6')
| -rwxr-xr-x | 2642/CH6/EX6.1/Ex6_1.sce | 26 | ||||
| -rwxr-xr-x | 2642/CH6/EX6.2/Ex6_2.sce | 35 | ||||
| -rwxr-xr-x | 2642/CH6/EX6.3/Ex6_3.sce | 32 | ||||
| -rwxr-xr-x | 2642/CH6/EX6.4/Ex6_4.sce | 31 | ||||
| -rwxr-xr-x | 2642/CH6/EX6.5/Ex6_5.sce | 31 | ||||
| -rwxr-xr-x | 2642/CH6/EX6.6/Ex6_6.sce | 40 |
6 files changed, 195 insertions, 0 deletions
diff --git a/2642/CH6/EX6.1/Ex6_1.sce b/2642/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..f4b8cdee5 --- /dev/null +++ b/2642/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // supply voltage in V
+I_a1 = 10 // dc shunt motor armature current in A
+N_1 = 900 // speed in rpm
+R_a = 1 // armature resistance in ohm
+N_2 = 500 // speed in rpm
+
+// caclulations
+E_b1 = V_t-I_a1*R_a // initial back emf in V
+R = (E_b1/10)*(1-(N_2/N_1)) // additional value of resistance in ohm
+
+// display the result
+disp("Example 6.1 solution");
+printf(" \n additional value of resistance \n R = %.1f ohm \n", R);
+
+
diff --git a/2642/CH6/EX6.2/Ex6_2.sce b/2642/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..9eff18b7b --- /dev/null +++ b/2642/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,35 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 230 // supply voltage in V
+I_a1 = 15 // dc shunt motor armature current in A
+N_1 = 650 // speed in rpm
+R_a = 0.4 // armature resistance in ohm
+R = 1 // variable resistance in series with the armature
+
+// caclulations
+// at full load
+E_b1 = V_t-I_a1*R_a // initial back emf in V
+E_b2 = V_t-I_a1*(R+R_a) // final back emf in V
+N_2 = N_1*(E_b2/E_b1) // speed at full load in rpm
+
+// at half load
+I_a21 = I_a1/2 // armature current in A
+E_b21 = V_t-I_a21*(R+R_a) // back emf in V
+N_21 = N_1*(E_b21/E_b1) // speed at half load torque in rpm
+
+
+
+// display the result
+disp("Example 6.2 solution");
+printf(" \n speed at full load \n N_2 = %.1f rpm \n", N_2);
+printf(" \n speed at half load torque \n N_21 = %.1f rpm \n", N_21);
+
diff --git a/2642/CH6/EX6.3/Ex6_3.sce b/2642/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..a6fef8ef3 --- /dev/null +++ b/2642/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 230 // supply voltage series motor in V
+R_a = 0.2 // armature resistance in ohm
+R_f = 0.2 // field resistance in ohm
+I_a1 = 10 // dc series motor armature current in A
+N = 1000 // speed in rpm
+I_a2 = 8 // dc series motor armature current in A when R=2 ohm
+R_2 = 2 // series resistance in ohm
+R_1 = 0 // series resistance in ohm
+// phi2 = 0.8*phi1
+// phi1/phi2 = 1.25
+
+// caclulations
+R_a1 = R_a+R_1 // new armature resistance in ohm
+E_b1 = V_t-I_a1*R_a1 // back emf in V
+R_a2 = R_a+R_2 // new armature resistance in ohm
+E_b2 = V_t-I_a2*R_a2 // back emf in V
+N_2 = (E_b2/E_b1)*1.25*N // new speed in rpm
+
+// display the result
+disp("Example 6.3 solution");
+printf(" \n New speed \n N_2 = %.1f rpm \n", N_2);
diff --git a/2642/CH6/EX6.4/Ex6_4.sce b/2642/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..0cc5c81e7 --- /dev/null +++ b/2642/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+N = 1100 // speed of dc series motor in rpm
+P = 4 // number of poles
+I_a1 = 15 // supply current to dc series motor in A
+V_t = 220 // supply voltage in V
+R_a = 0.9 // series armature resistance in ohm
+R_se1 = 0.6 // series field resistance in ohm
+I_a2 = 25 // supply current to dc series motor in A
+// phi2 = 0.8*phi1
+// phi1/phi2 = 1.25
+
+// caclulations
+E_b1 = V_t-I_a1*(R_a+R_se1) // back emf in V
+R_se2 = 0.6/4 // value of resistance per path in ohm
+E_b2 = V_t-I_a2*(R_a+R_se2) // back emf in V
+N_2 = (E_b2/E_b1)*1.25*N // new speed in rpm
+
+// display the result
+disp("Example 6.4 solution");
+printf(" \n New speed \n N_2 = %.1f rpm \n", N_2);
+
diff --git a/2642/CH6/EX6.5/Ex6_5.sce b/2642/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..94ee69ee6 --- /dev/null +++ b/2642/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 230 // shunt motor supply voltage in V
+R_a = 0.4 // armature resistance in ohm
+I_a = 30 // armature current in A
+n = 3 // number of steps
+
+// caclulations
+R_1 = V_t/I_a // maximum value of current in ohm
+k = (R_1/R_a)^(1/3) // constant
+R_2 = R_1/k // other value of resistance in ohm
+R_3 = R_2/k // other value of resistance in ohm
+R_4 = R_3/k // other value of resistance in ohm
+R_1step = R_1-R_2 // resistance of the 1st step in ohm
+R_2step = R_2-R_3 // resistance of the 1st step in ohm
+R_3step = R_3-R_4 // resistance of the 1st step in ohm
+
+// display the result
+disp("Example 6.5 solution");
+printf(" \n resistance of the 1st step in ohm \n R_1step = %.1f ohm \n", R_1step);
+printf(" \n resistance of the 2nd step in ohm \n R_2step = %.1f ohm \n", R_2step);
+printf(" \n resistance of the 3rd step in ohm \n R_3step = %.2f ohm \n", R_3step);
diff --git a/2642/CH6/EX6.6/Ex6_6.sce b/2642/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..924048f2b --- /dev/null +++ b/2642/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,40 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS
+// Example : 6.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // shunt motor supply voltage in V
+P_0 = 3550 // o/p power in W
+n = 0.85 // efficiency
+// condition given : starting current of the motor should not exceed twice the load current
+
+// caclulations
+P_in = P_0/n // i/p power in W
+P_tloss = P_in-P_0 // total loss in W
+I_a = P_in/V_t // neglet the field current then armature current in A
+P_cu = P_tloss/2 // copper loss in W
+R_a = P_cu/I_a^2 // armature resistance in ohm
+I_1 = 2*I_a // maximum starting current in A
+R_1 = V_t/I_1 // maximum resistance in ohm
+k = (R_1/R_a)^(1/4) // constant
+R_2 = R_1/k // other value of resistance in ohm
+R_3 = R_2/k // other value of resistance in ohm
+R_4 = R_3/k // other value of resistance in ohm
+R_5 = R_4/k // other value of resistance in ohm
+R_1step = R_1-R_2 // resistance of the 1st step in ohm
+R_2step = R_2-R_3 // resistance of the 1st step in ohm
+R_3step = R_3-R_4 // resistance of the 1st step in ohm
+R_4step = R_4-R_5 // resistance of the 1st step in ohm
+
+// display the result
+disp("Example 6.6 solution");
+printf(" \n resistance of the 1st step in ohm \n R_1step = %.1f ohm \n", R_1step);
+printf(" \n resistance of the 2nd step in ohm \n R_2step = %.2f ohm \n", R_2step);
+printf(" \n resistance of the 3rd step in ohm \n R_3step = %.2f ohm \n", R_3step);
+printf(" \n resistance of the 4th step in ohm \n R_3step = %.2f ohm \n", R_4step);
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