diff options
Diffstat (limited to '2642')
104 files changed, 3107 insertions, 0 deletions
diff --git a/2642/CH1/EX1.1/Ex1_1.sce b/2642/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..ce1be8491 --- /dev/null +++ b/2642/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,21 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+Q = 4 // chagre in C
+t = 0.54 // time in sec
+
+
+// caclulations
+I = Q/t // current in A
+
+// display the result
+disp("Example 1.1 solution");
+printf(" \n Current is \n I = %.2f A ", I);
diff --git a/2642/CH1/EX1.10/Ex1_10.sce b/2642/CH1/EX1.10/Ex1_10.sce new file mode 100755 index 000000000..7fab056c6 --- /dev/null +++ b/2642/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,39 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.10
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_AN = 200 // voltage in V
+ Z = 3+%i*4 // impedance in Ω
+
+// caclulations
+
+V_AB = sqrt(3)*(V_AN*exp(%i*(30)*(%pi/180))) // phase voltage in V
+V_BC = sqrt(3)*(V_AN*exp(%i*(-90)*(%pi/180))) // phase voltage in V
+V_CA = sqrt(3)*(V_AN*exp(%i*(150)*(%pi/180))) // phase voltage in V
+
+I_ab = V_AB/Z // magnitude of load current in A
+I_bc = V_BC/Z // magnitude of loadcurrent in A
+I_ca = V_CA/Z // magnitude of load current `in A
+
+I_Aa = I_ab-I_ca // magnitude of line current in A
+I_Bb = I_bc-I_ab // magnitude of line current in A
+I_Cc = I_ca-I_bc // magnitude of line current in A
+
+// display the result
+disp("Example 1.10 solution");
+printf(" \n Current in each load \n I_ab = %.2f<%.2f A \n", abs(I_ab),atand(imag(I_ab),real(I_ab)) );
+printf(" I_bc = %.2f<% 2f A \n", abs(I_bc),atand(imag(I_bc),real(I_bc)) );
+printf(" I_ca = %.2f<% 2f A \n\n", abs(I_ca),atand(imag(I_ca),real(I_ca)) );
+
+printf(" \n Line currents \n I_Aa = %.2f<%.2f A \n", abs(I_Aa),atand(imag(I_Aa),real(I_Aa)) );
+printf(" I_Bb = %.2f<% 2f A \n", abs(I_Bb),atand(imag(I_Bb),real(I_Bb)) );
+printf(" I_Cc = %.2f<% 2f A \n\n", abs(I_Cc),atand(imag(I_Cc),real(I_Cc)) );
+
+printf(" NOTE : The line currents as given in this example in text book are wrong since ,the quesion says that both the three phase supply and load are balanced so the line should be balanced which they fail to do in text book.");
diff --git a/2642/CH1/EX1.11/Ex1_11.sce b/2642/CH1/EX1.11/Ex1_11.sce new file mode 100755 index 000000000..5c5404275 --- /dev/null +++ b/2642/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,39 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.11
+
+clc;clear; // clears the console and command history
+
+// Given data
+Z = 3+%i*4 // impedance in Ω
+V_AN = 150 // voltage in V
+R = 3 // resistance in Ω from Z
+
+// caclulations
+
+V_AN = (150*exp(%i*(0)*(%pi/180))) //source voltage in V
+V_BN = (150*exp(%i*(-120)*(%pi/180))) // source voltage in V
+V_CN = (150*exp(%i*(120)*(%pi/180))) // source voltage in V
+
+I_Aa = V_AN/Z // line current in A
+I_Bb = V_BN/Z // line current in A
+I_Cc = V_CN/Z // line current in A
+
+pf = R/abs(Z) // power factor
+I = V_AN/abs(Z) // current in A
+P = V_AN*I*pf // power supplied in W
+P_t = 3*P // total power supplied in W
+
+// display the result
+disp("Example 1.11 solution");
+printf(" \n Line currents are \n I_Aa = %.2f<%.2f A \n", abs(I_Aa),atand(imag(I_Aa),real(I_Aa)) );
+printf(" I_Bb = %.2f<% 2f A \n", abs(I_Bb),atand(imag(I_Bb),real(I_Bb)) );
+printf(" I_Cc = %.2f<% 2f A \n\n", abs(I_Cc),atand(imag(I_Cc),real(I_Cc)) );
+printf(" Power factor \n pf = %.1f \n", pf);
+printf(" \n Power supplied to each phase is \n P = %.2f W \n", P);
+printf(" \n Total power supplied \n P_t = %.2f W \n", P_t);
+
diff --git a/2642/CH1/EX1.12/Ex1_12.sce b/2642/CH1/EX1.12/Ex1_12.sce new file mode 100755 index 000000000..9d0089b02 --- /dev/null +++ b/2642/CH1/EX1.12/Ex1_12.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.12
+
+clc;clear; // clears the console and command history
+
+
+// Given data
+P = 120 // total power in kW
+cos_teta = 0.6 // power factor
+
+// caclulations
+teta = acosd(0.6) // power factor angle
+P_1 = tand(teta)*P/sqrt(3)
+P_2 = (P_1+P)/2 // second wattmeter reading in kW
+
+// display the result
+disp("Example 1.12 solution");
+printf(" \n Second wattmeter reading \n P_2= %.1f kW \n", P_2);
+
diff --git a/2642/CH1/EX1.2/Ex1_2.sce b/2642/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..d9ca65be7 --- /dev/null +++ b/2642/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,20 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = -24 // voltage in V
+I = 3 // current in A
+
+// caclulations
+ P = V*I // power for the element A in network in W
+
+// display the result
+disp("Example 1.2 solution");
+printf(" \n Power for the element A in network \n P = %.0f W ", P);
diff --git a/2642/CH1/EX1.3/Ex1_3.sce b/2642/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..efb217a2b --- /dev/null +++ b/2642/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.3
+
+clc;clear; // clears the console and command history
+// Given data
+R1 = 5 // resistance in ohm
+R2 = 4 // resistance in ohm
+R3 = 9 // resistance in ohm
+R4 = 6 // resistance in ohm
+V1 = 10 // voltage in V
+V2 = 6 // voltage in V
+
+
+// caclulations
+// Remove R3 and find R_th by short circuiting V1 and V2 than R1 and R4 wiil be in parallel
+R = (R1*R4)/(R1+R4) // equivalent resistance in ohm
+// R is connected in series with R2
+R_th = R2+R // Thevenin's resistance in ohm
+I = 4/11 // current in the figure applying KVL in A
+V_th = (6*0.36)+6 // voltage in V
+I_9 = V_th/(R_th+R3) // current through R3 in A
+
+
+// display the result
+disp("Example 1.3 solution");
+printf(" \n Current through 9Ω \n I_9Ω = %.2f A ", I_9);
diff --git a/2642/CH1/EX1.4/Ex1_4.sce b/2642/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..844830ced --- /dev/null +++ b/2642/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t1 = 30 // magnitudes of voltages in V 0 < t1 < 2
+V_t2 = -10 // magnitudes of voltages in V 2 < t2 < 4
+T = 4 // time period from figure
+
+// caclulations
+
+V1 = 30;
+V2 = -10;
+X= sqrt((1/4)*(integrate('V1^2','x',0,2) + integrate('V2^2','x',2,4)));
+
+//display the result
+disp("Example 1.4 solution");
+printf("\n RMS value of the voltage waveform : \n V_rms = %.2f V ", X);
diff --git a/2642/CH1/EX1.5/Ex1_5.sce b/2642/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..c1829c573 --- /dev/null +++ b/2642/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,23 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.5
+
+clc;clear; // clears the console and command history
+
+
+// Given data
+V_P = 200 // magnitude of phase voltage
+V_an = 200 // magnitude of phase voltage in V
+V_bn = 200*exp(%i*(-120)*(%pi/180)) // magnitude of phase voltage in V
+V_cn = 200*exp(%i*(120)*(%pi/180)) // magnitude of phase voltage in V
+
+// caclulations
+V_L = sqrt(3)*V_P // magnitude of line voltage in V
+
+// display the result
+disp("Example 1.5 solution");
+printf(" \n Magnitude of line voltage \n V_L = %.2f V ", V_L);
diff --git a/2642/CH1/EX1.6/Ex1_6.sce b/2642/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..7d2962ef2 --- /dev/null +++ b/2642/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.6
+
+clc;clear; // clears the console and command history
+
+
+// Given data
+R = 10 // resistance in Ω
+L = 15 // inductance reactance in Ω
+V_L = 420 // voltage in V
+f = 50 // frwuency in Hz
+
+// caclulations
+V_an = (V_L/sqrt(3))*exp(%i*(-30)*(%pi/180)) // phase voltage
+V_bn = (V_L/sqrt(3))*exp(%i*(-150)*(%pi/180)) // phase voltage
+V_cn = (V_L/sqrt(3))*exp(%i*(90)*(%pi/180)) // phase voltage
+Z_p = R+%i*L // phase impedance in Ω
+I_L1 = V_an/Z_p // line current
+I_L2 = V_bn/Z_p // line current
+I_L3 = V_cn/Z_p // line current
+pf = R/abs(Z_p) // lagging power factor
+
+// display the result
+disp("Example 1.6 solution");
+printf(" \n Line currents are \n I_L1 = %.2f<%.2f A \n", abs(I_L1),atand(imag(I_L1),real(I_L1)) );
+printf(" I_L2 = %.2f<% 2f A \n", abs(I_L2),atand(imag(I_L2),real(I_L2)) );
+printf(" I_L3 = %.2f<% 2f A \n", abs(I_L3),atand(imag(I_L3),real(I_L3)) );
+printf(" \n Power factor \n pf = %.1f lag \n", pf );
diff --git a/2642/CH1/EX1.7/Ex1_7.sce b/2642/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..6e3d85605 --- /dev/null +++ b/2642/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.7
+
+clc;clear; // clears the console and command history
+
+
+// Given data
+Z_p = 10+%i*15 // per phase impedance in Ω
+V_ab = 420 // magnitude of phase voltage in V
+V_bc = 420*exp(%i*(-120)*(%pi/180))// magnitude of phase voltage in V
+V_ca = 420*exp(%i*(120)*(%pi/180)) // magnitude of phase voltage in V
+
+
+
+// caclulations
+I_ab = V_ab/Z_p // phase current in A
+I_bc = V_bc/Z_p // phase current in A
+I_ca = V_ca/Z_p // phase current in A
+I_P = V_ab/abs(Z_p)
+I_L = sqrt(3)*I_P // line current in A
+
+// display the result
+disp("Example 1.7 solution");
+printf(" \n Phase currents are \n I_ab = %.2f<%.2f A \n", abs(I_ab),atand(imag(I_ab),real(I_ab)) );
+printf(" I_bc = %.2f<% 2f A \n", abs(I_bc),atand(imag(I_bc),real(I_bc)) );
+printf(" I_ca = %.2f<% 2f A \n", abs(I_ca),atand(imag(I_ca),real(I_ca)) );
+printf(" \n Line current \n I_L = %.2f A \n", I_L);
diff --git a/2642/CH1/EX1.8/Ex1_8.sce b/2642/CH1/EX1.8/Ex1_8.sce new file mode 100755 index 000000000..4c88700a1 --- /dev/null +++ b/2642/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,42 @@ +
+// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.8
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_L = 280 // generator phase voltage in V rms
+Z_1 = 2+%i*3 // line impedance
+Z_L = 4+%i*5 // load impedance
+
+// caclulations
+Z_t = Z_1+Z_L //
+V_An = (V_L*exp(%i*(0)*(%pi/180))) // phase voltage
+V_Bn = (V_L*exp(%i*(-120)*(%pi/180))) // phase voltage
+V_Cn = (V_L*exp(%i*(120)*(%pi/180))) // phase voltage
+
+I_Aa = V_An/Z_t // line current for phase A in A
+I_Bb = V_Bn/Z_t // line current for phase B in A
+I_Cc = V_Cn/Z_t // line current for phase C in A
+
+V_an = I_Aa*Z_L // voltage of load in V
+V_bn = I_Bb*Z_L // voltage of load in V
+V_cn = I_Cc*Z_L // voltage of load in V
+
+// display the result
+disp("Example 1.8 solution");
+// There is error in the textbook regarding phasor angles of
+printf(" \n Line currents are \n I_Aa = %.2f<%.2f A \n", abs(I_Aa),atand(imag(I_Aa),real(I_Aa)) );
+printf(" I_Bb = %.2f<% 2f A \n", abs(I_Bb),atand(imag(I_Bb),real(I_Bb)) );
+printf(" I_Cc = %.2f<% 2f A \n\n", abs(I_Cc),atand(imag(I_Cc),real(I_Cc)) );
+printf(" NOTE : PHASOR ANGLES CALCULATED IN TEXTBOOK FOR Van,Vbn & Vcn are wrong,\n because 4+j5 = 6.4<51.34, but in the textbook it is taken as 6.4<38.6");
+// NOTE : PHASOR ANGLES CALCULATED IN TEXTBOOK FOR Van,Vbn & Vcn are wrong,because 4+j5 = 6.4<51.34, but in the textbook it is taken as 6.4<38.6
+printf(" \n\n Phase voltages are ");
+printf("\n V_an = %.2f<%.2f V \n", abs(V_an),atand(imag(V_an),real(V_an)) );
+printf(" V_bn = %.2f<% 2f V \n", abs(V_bn),atand(imag(V_bn),real(V_bn)) );
+printf(" V_cn = %.2f<% 2f V \n", abs(V_cn),atand(imag(V_cn),real(V_cn)) );
diff --git a/2642/CH1/EX1.9/Ex1_9.sce b/2642/CH1/EX1.9/Ex1_9.sce new file mode 100755 index 000000000..820f123d3 --- /dev/null +++ b/2642/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 1 : REVIEW OF ELECRTIC CIRCUITS
+// Example : 1.9
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_P = 340 // phase voltage in V
+Z = 6+%i*8 // per phase impedance
+
+// caclulations
+
+V_AN = (V_P/exp(%i*(0)*(%pi/180))) // phase voltage
+V_BN = (V_P/exp(%i*(-120)*(%pi/180))) // phase voltage
+V_CN = (V_P/exp(%i*(120)*(%pi/180))) // phase voltage
+
+I_Aa = V_AN/Z // line current for phase A in A
+I_Bb = V_BN/Z // line current for phase B in A
+I_Cc = V_CN/Z // line current for phase C in A
+
+
+
+// display the result
+disp("Example 1.9 solution");
+printf(" \n Line currents are \n I_Aa = %.2f<%.2f A \n", abs(I_Aa),atand(imag(I_Aa),real(I_Aa)) );
+printf(" I_Bb = %.2f<% 2f A \n", abs(I_Bb),atand(imag(I_Bb),real(I_Bb)) );
+printf(" I_Cc = %.2f<% 2f A \n", abs(I_Cc),atand(imag(I_Cc),real(I_Cc)) );
+printf(" \n The load is balanced, so the value of the neutral current will be zero")
diff --git a/2642/CH10/EX10.1/Ex10_1.sce b/2642/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..f10336aec --- /dev/null +++ b/2642/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 10 : SYNCHRONOUS MOTOR
+// Example : 10.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 2.5*10^3 // supply voltage in V
+R_r = 0.12 // per phase resistance in ohm
+X_r = 3.2 // syncronous reactance in ohm
+I_a = 185 // line current in A
+pf = 0.8 // leading power factor
+
+// caclulations
+phi = acosd(pf)
+V_t = V/sqrt(3) // terminal voltage per phase in V
+Z_s = R_r+%i*X_r // impedance per phase ohm
+beta = atand(X_r/R_r)
+E_r = I_a*Z_s // resultant voltage due to impedance in V
+E_f = sqrt(V_t^2+abs(E_r)^2-2*V_t*abs(E_r)*cosd(beta+phi)) // excitation voltage per phase in V
+
+
+// display the result
+disp("Example 10.1 solution");
+printf(" \n Excitation voltage per phase \n E_f = %.2f V \n", E_f );
+
diff --git a/2642/CH10/EX10.2/Ex10_2.sce b/2642/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..e9344ff4f --- /dev/null +++ b/2642/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 10 : SYNCHRONOUS MOTOR
+// Example : 10.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 1200 // kVA ratings
+V = 14*10^3 // supply voltage in V
+R_r = 4.8 // per phase resistance in ohm
+X_r = 35 // syncronous reactance in ohm
+pf = 0.95 // leading power factor
+
+// caclulations
+phi = acosd(pf)
+Z_s = R_r+%i*X_r // impedance per phase ohm
+I_a = kVA*10^3/(sqrt(3)*V) // armature current in A
+E_r = I_a*Z_s // resultant voltage due to impedance in V
+V_t = V/sqrt(3) // terminal voltage per phase in V
+b = atand(X_r/R_r) // beta value
+E_f = sqrt(V_t^2+abs(E_r)^2-2*V_t*abs(E_r)*cosd(b-phi)) // excitation voltage per phase in V
+teta = sind(64)
+D = (E_r*teta/E_f) // torque angle
+delta = asind(abs(D))
+
+// display the result
+disp("Example 10.2 solution");
+printf(" \n Excitation voltage per phase \n E_f = %.2f V \n", E_f );
+printf(" \n Torque angle at 0.95 power factor lagging \n delta = %.2f degree \n", delta );
diff --git a/2642/CH10/EX10.3/Ex10_3.sce b/2642/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..5af584787 --- /dev/null +++ b/2642/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 10 : SYNCHRONOUS MOTOR
+// Example : 10.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 440 // supply voltage in V
+R_a = 1.5 // per phase resistance in ohm
+X_a = 8 // syncronous reactance in ohm
+P = 4 // number of poles
+f = 50 // supply frequency in Hz
+pf = 0.9 // leading power factor
+I_a = 50 // armature current in A
+
+// caclulations
+V_t = V/sqrt(3) // terminal voltage per phase in V
+phi = acosd(pf) // angle in degree
+Z_s = R_a+%i*X_a // impedance per phase ohm
+E_r = I_a*abs(Z_s) // resultant voltage due to impedance in V
+bet = atand(X_a/R_a)
+E_f = sqrt(V_t^2+E_r^2-2*V_t*E_r*cosd(bet+phi)) // excitation voltage per phase in V
+P_dm = (((E_f*V_t)/Z_s)-((E_f^2*R_a)/Z_s^2)) // maximum power per phase in W
+
+// display the result
+disp("Example 10.3 solution");
+printf(" \n Maximum power per phase \n P_dm = %.2f W \n", P_dm );
+printf(" \n In textbook solution they took E_f = 513.5V instead of 533.33V");
diff --git a/2642/CH10/EX10.4/Ex10_4.sce b/2642/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..fd29f7240 --- /dev/null +++ b/2642/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 10 : SYNCHRONOUS MOTOR
+// Example : 10.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+f = 50 // supply frequency in Hz
+V_t = 1500 // terminal voltage per phase in V
+E_f = 1000 // excitation voltage per phase in V
+Z_s = 12 // synchronnous impedance in ohm
+R_a = 1.5 //armature resistance in ohm
+
+// caclulations
+P_dm = (((E_f*V_t)/Z_s)-((E_f^2*R_a)/Z_s^2)) // maximum power per phase in W
+T_dm = 9.55*P_dm/1500 // maximum torque in N-m
+
+// display the result
+disp("Example 10.4 solution");
+printf(" \n Maximum power developed \n P_dm = %.0f W \n", P_dm );
+printf(" \n Maximum toruqe \n T_dm = %.1f N-m \n", T_dm );
diff --git a/2642/CH11/EX11.1/Ex11_1.sce b/2642/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..a20aed7cf --- /dev/null +++ b/2642/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,34 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 220 // supply voltage in V
+P = 4 // number of poles
+f = 50 // frequency in Hz
+N_l = 1450 // speed in rpm
+P_tloss = 2000 // total power loss in W
+R_2 = 10 // rotor resistance at standstill in ohm
+
+// caclulations
+N_s = 120*f/P // synchronous speed in rpm
+s_f = (N_s-N_l)/N_s // slip due to forward field
+s_b = 2-s_f // slip due to backward field
+R_f = R_2/s_f // effective rotor resistance due to forward slip in ohm
+R_b = R_2/(2-s_f) // effective rotor resistance due to backward slip in ohm
+
+// display the result
+disp("Example 11.1 solution");
+printf(" \n Slip due to forward field \n s_f = %.2f \n", s_f );
+printf(" \n Slip due to backward field \n s_b = %.2f \n", s_b );
+printf(" \n Effective rotor resistance due to forward slip \n R_f = %.2f ohm \n", R_f );
+printf(" \n Effective rotor resistance due to backward slip \n R_b = %.2f ohm \n", R_b );
+
+printf(" \n NOTE : for caclulating R_f, s_f is taken as 0.033333 so we got R_f=300");
+
diff --git a/2642/CH11/EX11.2/Ex11_2.sce b/2642/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..8756de858 --- /dev/null +++ b/2642/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,36 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // supply voltage in V
+R_1 = 6 // equivalent parameters of single phase induction motor in ohm
+R_2 = 6 // equivalent parameters of single phase induction motor in ohm
+X_1 = 10 // equivalent parameters of single phase induction motor in ohm
+X_2 = 10 // equivalent parameters of single phase induction motor in ohm
+N = 1500 // speed in rpm
+s = 0.03 // slip
+X_m = 150 // equivalent parameters of single phase induction motor in ohm
+
+// caclulations
+Z_f = 0.5*%i*X_m*((R_2/s)+%i*X_2)/((R_2/s)+%i*(X_2+X_m)) // impedance due to forward field in ohm
+R_f = real(Z_f) // from Z_f
+Z_b = 0.5*%i*X_m*((R_2/(2-s))+%i*X_2)/((R_2/(2-s))+%i*(X_2+X_m)) // impedance due to backward field in ohm
+R_b = real(Z_b) // from Z_b
+Z_t = R_1+%i*X_1+Z_f+Z_b // total impedence in ohm
+I_1 = V_t/Z_t // input current in A
+P_d = (abs(I_1))^2*(R_f-R_b)*(1-s) // power developed in W
+T_d = 9.55*P_d/N // torque in N-m
+
+// display the result
+disp("Example 11.2 solution");
+printf(" \n input current \n I_1 = %.2f<%.2f A \n", abs(I_1),atand(imag(I_1),real(I_1)));
+printf(" \n power developed \n P_d = %.2f \n", P_d );
+printf(" \n torque \n T_d = %.2f \n", T_d );
+printf("\n NOTE : ERROR : There is calculation mistake in Z_b in textbook. So there is change in answers from textbook")
diff --git a/2642/CH11/EX11.3/Ex11_3.sce b/2642/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..efe82fbac --- /dev/null +++ b/2642/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // supply voltage in V
+f = 50 // frequency in Hz
+Z_m = 3+%i*5 // main winding impedance of motor in ohm
+Z_s = 5+%i*3 // starting impedance of motor in ohm
+
+// caclulations
+alpha_s = atand(imag(Z_s),real(Z_s)) // angle in degree from Z_s
+I_s = V_t/Z_s // starting current in A
+alpha_m = atand(imag(Z_m),real(Z_m)) // angle in degree from Z_m
+I_m = V_t/(Z_m) // main winding current in A
+alpha = alpha_m-alpha_s // angle of line current
+I = sqrt((abs(I_s))^2+(abs(I_m))^2+2*abs(I_s)*abs(I_m)*cosd(alpha)) // line current in A
+
+// display the result
+disp("Example 11.3 solution");
+printf(" \n Input current \n I_s = %.2f<%.2f A \n", abs(I_s),atand(imag(I_s),real(I_s)) );
+printf(" \n Main winding current \n I_m = %.2f<%.f A \n", abs(I_m),atand(imag(I_m),real(I_m)));
+printf(" \n Line current \n I = %.2f A \n", I );
diff --git a/2642/CH11/EX11.4/Ex11_4.sce b/2642/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..3a1a61156 --- /dev/null +++ b/2642/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // supply voltage in V
+f = 50 // frequency in Hz
+Z_m = 4+%i*3.5 // main winding impedance of motor in ohm
+Z_s = 5+%i*3 // starting impedance of motor in ohm
+R_s = 5 // from Z_s
+X_s = 3 // from Z_s
+
+// caclulations
+alpha_m = 41.2 // angle in degree from Z_m
+// Let X_c be connected in series with the starting winding. Then the total impedance of starting winding is Z_s = Z_m-%i*X_c
+// The torque will be the maximum when the angle between the starting winding and main winding currents in 90 electrical degree.The value of the angle of the starting winding current is
+alpha_s = alpha_m-90
+X_c = X_s-R_s*tand(alpha_s)
+C = 1/(2*%pi*f*X_c) // starting capacitance to get maximum torque in F
+
+// display the result
+disp("Example 11.4 solution");
+printf(" \n Starting capacitance for getting maximum torque \n C = %.2e F \n", C );
diff --git a/2642/CH11/EX11.5/Ex11_5.sce b/2642/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..b502848e9 --- /dev/null +++ b/2642/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,37 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+f = 50 // supply frequency in Hz
+V_nl = 100 // no-load voltage in v
+I_nl = 2.5 // no-load current in A
+P_nl = 60 // no-load power in W
+
+V_br = 60 // Block rotor test voltage in v
+I_br = 3 // Block rotor test current in A
+P_br = 130 // Block rotor test power in W
+R_1 = 2 // main windning resistance in ohm
+
+// caclulations
+Z_br = V_br/I_br // impedance due to blocked rotor test
+R_br = P_br/I_br^2 // resistance due to blocked rotor test in ohm
+X_br = sqrt(Z_br^2-R_br^2) // reactance under blocked condition in ohm
+X_1 = X_br/2 // reactance in ohm X_1=X_2
+R_2 = R_br-R_1 // resistance in ohm
+Z_nl = V_nl/I_nl // impedance due to no-load in ohm
+R_nl = P_nl/I_nl^2 // resistance due to no-load in ohm
+X_nl = sqrt(Z_nl^2-R_nl^2) // reactance due to no-load in ohm
+X_m =2*(X_nl-X_1-0.5*X_1) // magnetizing reactance in ohm
+P_rot = P_nl-I_nl^2*(R_1+((R_2)/4)) // rotational loss in W
+
+// display the result
+disp("Example 11.5 solution");
+printf(" \n Magnetizing reactance \n X_m = %.1f ohm \n", X_m );
+printf(" \n Rotational loss \n P_rot = %.0f W \n", P_rot );
diff --git a/2642/CH11/EX11.6/Ex11_6.sce b/2642/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..1c62d307f --- /dev/null +++ b/2642/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,23 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 11 : SINGLE-PHASE MOTORS
+// Example : 11.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+r_t = 36 // rotor teeth of stepper motor
+N = 4 // stator phases
+
+// caclulations
+T_p = 360/r_t // tooth pitch
+teta = 360/(N*r_t) // step angle
+
+// display the result
+disp("Example 11.6 solution");
+printf(" \n Tooth pitch \n T_p = %.0i degree \n", T_p );
+printf(" \n Strp angle \n teta = %.1f degree \n", teta );
+
diff --git a/2642/CH2/EX2.1/Ex2_1.sce b/2642/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..92c8673af --- /dev/null +++ b/2642/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,25 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+l = 4 // lenght in m
+w = 2 // width in m
+B = 0.12 // magnetis flux density in tesla
+
+// caclulations
+A = l*w // area in m^2
+flux = B*A // magnetic flux in Wb
+
+
+// display the result
+disp("Example 2.1 solution");
+printf(" \n Magnetic flux \n flux = %.2f Wb \n", flux);
+
+
diff --git a/2642/CH2/EX2.10/Ex2_10.sce b/2642/CH2/EX2.10/Ex2_10.sce new file mode 100755 index 000000000..d14afb2ba --- /dev/null +++ b/2642/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.10
+clc;clear; // clears the console and command history
+
+// Given data
+A = 2*10^-4 // cross sectional area in m^2
+N = 200 // number of turns
+flux = 1.5*10^-3 // flux in Wb
+myu_r = 4000 // relative permiability of core
+l_1 = 0.01 // length in m
+a = 9 // length in cm
+w = 3 // width in cm
+
+// caclulations
+myu = myu_r*4*%pi*10^-7 // permiability
+l_2 = (4*(a-w-w+(1.5+1.5))-1) // mean length in cm
+R_mg = l_1/(4*%pi*10^-7*A) // reluctance of iron for air gap At/Wb
+R_mi = l_2*10^-2/(myu*A) // reluctance of iron for air gapAt/Wb
+R_mt = R_mg+R_mi // total relectance in At/Wb
+I = R_mt*flux/N // current in A
+
+// display the result
+disp("Example 2.10 solution");
+printf(" \n Current flowing through the coil \n I = %.0f A \n", I);
+
+// NOTE : In question they given flux=2.5mWb but in solution they took flux=1.5mWb
diff --git a/2642/CH2/EX2.11/Ex2_11.sce b/2642/CH2/EX2.11/Ex2_11.sce new file mode 100755 index 000000000..51e103ac0 --- /dev/null +++ b/2642/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,21 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.11
+clc;clear; // clears the console and command history
+
+// Given data
+I = 150 // current through conductor in A
+l = 2 // conductor length in m
+B = 0.35 // magnetic flux density in T
+
+// caclulations
+F = B*l*I // force in N
+
+// display the result
+disp("Example 2.11 solution");
+printf("\n Force \n F = %.0f N \n", F);
+
diff --git a/2642/CH2/EX2.12/Ex2_12.sce b/2642/CH2/EX2.12/Ex2_12.sce new file mode 100755 index 000000000..c7e19a120 --- /dev/null +++ b/2642/CH2/EX2.12/Ex2_12.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.12
+clc;clear; // clears the console and command history
+
+// Given data
+l = 25*10^-2 // length of air-core coil in m
+A = 4*10^-4 // cross sectional area in m^2
+N = 65 // number of turns
+myu_0 = 4*%pi*10^-7
+myu_r = 1
+
+// caclulations
+myu = myu_0*myu_r
+L = N^2*myu*A/l // inductance in H
+
+// display the result
+disp("Example 2.12 solution");
+printf(" \n Inductance of the coil \n L = %.1e H \n", L);
+
diff --git a/2642/CH2/EX2.13/Ex2_13.sce b/2642/CH2/EX2.13/Ex2_13.sce new file mode 100755 index 000000000..ee597f493 --- /dev/null +++ b/2642/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,23 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.13
+clc;clear; // clears the console and command history
+
+// Given data
+k_h = 110 // hysteresis co-efficient in J/m^3
+V_cvol = 0.005 // volume of the core in m^3
+B_m = 1.12 // flux density in T
+f = 60 // frequency in Hz
+n = 1.6
+
+// caclulations
+P_h = k_h*V_cvol*B_m^n*f // hysteresis loss in W
+
+// display the result
+disp("Example 2.13 solution");
+printf(" \n Hysteresis loss \n P_h = %.2f W \n", P_h);
+
diff --git a/2642/CH2/EX2.2/Ex2_2.sce b/2642/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..01dba0a73 --- /dev/null +++ b/2642/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+d_in = 3 // inside diameter of iron toroid in cm
+d_out = 6 // outside diameter of iron toroid in cm
+N =200 // number of turns
+I = 3 // current in A
+flux = 0.015 // flux in Wb
+
+// caclulations
+d = d_in+((d_out-d_in)/2) // distance in cm
+l = %pi*d // mean length in cm
+A = %pi*d^2/4 // area in cm^2
+B =flux/(A*10^-4) // flux density in mWb/m^2
+MMF = N*I // magnetomotive force in At
+H = (N*I)/(l*10^-2) // magnetic field intensity in At/m
+
+// display the result
+disp("Example 2.2 solution");
+printf("\n Flux density \n B= %.6f mWb/m^2 \n", B);
+printf(" \n Magnetomotive force \n MMF= %.2f At \n", MMF);
+printf(" \n Magnetic field intensity \n H= %.2f At/m \n", H);
+printf(" NOTE:correction in solution they took d=1.5 insted of 4.5")
diff --git a/2642/CH2/EX2.3/Ex2_3.sce b/2642/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..26fd44419 --- /dev/null +++ b/2642/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+myu_r = 625 // relative permiability of rectangular core
+N = 25 // number of turns
+I = 2 // current in A
+a = 5.5 // lenght of rectangular box in cm
+b = 1.5 // width of rectangular box in cm
+
+// caclulations
+l = 2*(a+b) // mean length of core in cm
+H = N*I/(l*10^-2) // magnetic field intensity in At/m
+myu = 4*%pi*10^-7*myu_r // permeabilty
+B = myu*H // magnetic flux density in Wb/m^2;
+
+// display the result
+disp("Example 2.3 solution");
+printf(" \n Magnetic field intensity \n H= %.0f At/m \n", H);
+printf(" \n Permeabilty \n myu= %.2e \n", myu);
+printf(" \n Magnetic flux density \n B= %.2f Wb/m^2 \n", B);
+
+
diff --git a/2642/CH2/EX2.4/Ex2_4.sce b/2642/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..14c766b82 --- /dev/null +++ b/2642/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,25 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+N = 6 // number of turns
+I =3 // current in A
+flux = 0.056 // flux in Wb
+
+// caclulations
+MMF = N*I // magnetomotive force in At
+R_m = MMF/flux // reluctance in At/Wb
+
+// display the result
+disp("Example 2.4 solution");
+printf(" \n Magnetomotive force \n MMF= %.0f At \n", MMF);
+printf( "\n Reluctance \n R_m= %.1f At/Wb \n", R_m);
+
+
diff --git a/2642/CH2/EX2.5/Ex2_5.sce b/2642/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..61e100188 --- /dev/null +++ b/2642/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.5
+clc;clear; // clears the console and command history
+
+// Given data
+I = 15 // current through conductor in A
+N = 10 // number of turns
+myu_0 = 4*%pi*10^-7 // permiability
+myu_r = 1 // relative permiability of air
+r = 0.015
+
+// caclulations
+B = myu_0*myu_r*N*I/(2*%pi*r) // magnetic flux in T
+
+// display the result
+disp("Example 2.5 solution");
+printf(" \n Magnetic flux \n B= %.0e T \n", B);
+
+
diff --git a/2642/CH2/EX2.6/Ex2_6.sce b/2642/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..510375511 --- /dev/null +++ b/2642/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+N = 200 // number of turns
+d_in = 7 // inner diameter of wooden toroidal coil in cm
+d_out = 10 // outer diameter of wooden toroidal coil in cm
+A = 0.005 // cross sectional area m^2
+I = 5 // current through coil in A
+R = d_out-d_in
+myu_0 = 4*%pi*10^-7
+
+// caclulations
+l = 2*%pi*R*10^-2 // mean circumference length in m
+H = N*I/l // magnetic field intensity in At/m
+B = myu_0*H // flux density in Wb/m^2
+flux = B*A // flux in Wb
+
+// display the result
+disp("Example 2.6 solution");
+printf(" \n Magnetic field intensity \n H= %.0f At/m \n", H);
+printf(" \n Flux density \n B= %.2e Wb/m^2 \n", B);
+printf(" \n Flux \n flux= %.1e Wb \n", flux);
diff --git a/2642/CH2/EX2.7/Ex2_7.sce b/2642/CH2/EX2.7/Ex2_7.sce new file mode 100755 index 000000000..1f307ee5c --- /dev/null +++ b/2642/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.7
+clc;clear; // clears the console and command history
+
+// Given data
+l = 0.1 // length in m
+w = 0.01 // width in m
+h =0.1 // height in m
+N = 450 // number of turns
+I = 0.2 // current in A
+myu_r = 850 // relative permiability magnetic material
+
+// caclulations
+MMF = N*I // magnetomotive force in At
+l_c = (h-w)*4 // mean length of the path in m
+A = w*w // cross sectional area in m^2
+R_m = l_c/(4*%pi*10^-7*myu_r*A) // relectance in At/Wb
+flux = MMF/R_m // flux in Wb
+B = flux/A // magnetic flux density in Wb/m^2
+H = B/(4*%pi*10^-7*myu_r) // field intensity in At/m
+
+// display the result
+disp("Example 2.7 solution");
+printf(" \n Flux \n flux= %.2e Wb \n", flux);
+printf(" \n Magnetic flux density \n B= %.4f Wb/m^2 \n", B);
+printf(" \n Field intensity \n H= %.2f At/m \n", H);
diff --git a/2642/CH2/EX2.8/Ex2_8.sce b/2642/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..bb52067bf --- /dev/null +++ b/2642/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,38 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.8
+
+clc;clear; // clears the console and command history
+
+// Given data
+N = 450 // number of turns wound on left side of limb
+A = 4 // cross sectional area in m^2
+I = 5 // current through coil in A
+fulx = 3 // flux in right limb Wb
+myu_r = 500 // relative permiability magnetic material
+l_1 = 0.12
+l_2 = 0.24
+phi2 = 3
+
+// caclulations
+// we have relation
+// phi1*rm1 = phi2*rm2
+// phi1*l_1/uA = pi2*l_2/uA
+phi1 = phi2*l_2/l_1 // flux on left side in Wb
+flux = phi1+phi2 // total flux in Wb
+B_1 = flux/A // flux density in the left limb
+H_1 = B_1/(4*%pi*10^-7*myu_r) // magnetic flux in At/m
+MMF_1 = H_1*l_2 // magnetomotive force in At
+B_2 = phi2/A // flux density in the right limb
+H_2 = B_2/(4*%pi*10^-7*myu_r) // magnetic flux in At/m
+MMF_2 = H_2*l_2 // magnetomotive force in At
+MMF_t = MMF_1+MMF_2 // total magnetomotive force in At
+I = MMF_t/N // current in A
+
+// display the result
+disp("Example 2.8 solution");
+printf(" \n Current\n I= %.2f A \n", I);
diff --git a/2642/CH2/EX2.9/Ex2_9.sce b/2642/CH2/EX2.9/Ex2_9.sce new file mode 100755 index 000000000..2b5237828 --- /dev/null +++ b/2642/CH2/EX2.9/Ex2_9.sce @@ -0,0 +1,31 @@ + // FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 2 : BESICS OF MAGNETIC CIRCUITS
+// Example : 2.9
+clc;clear; // clears the console and command history
+
+// Given data
+l = 0.45 // mean lenght in m
+A = 25*10^-4 // cross sectional area in m^2
+l_ag = 0.8*10^-3 // air gap in m
+N = 500 // number of turns
+I = 1.25 // current in A
+fulx = 2.25*10^-3 // flux in Wb
+phi = 1.5*10^-3 // flux in Wb
+
+// caclulations
+B = phi/A // magnetic flux density in Wb/m^2
+MMF = N*I // magnetomotive force in At
+H = B/(4*%pi*10^-7) // magnetomotizing force in At/m
+MMF_ag = H*l_ag // magnetomotive force in At
+MMF_i = MMF-MMF_ag // magnetomotive force for iron ring in At
+H_i = MMF_i/l // magnetic field intensity for iron part in At/m
+myu_r = B/((4*%pi*10^-7)*H_i) // relative permiability for iron
+
+// display the result
+disp("Example 2.9 solution");
+printf(" \n Relative permiability for iron \n myu_r = %.2f \n", myu_r);
+printf(" given current value in question is 2.25A, but in solution they took value of current as 1.25A ");
diff --git a/2642/CH2/EX9.2/Ex9_2.sce b/2642/CH2/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..8746869ab --- /dev/null +++ b/2642/CH2/EX9.2/Ex9_2.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 8 // number of poles
+m = 3 // number of phase
+S = 144 // number of slots
+
+// caclulations
+T_p = S/P // pole pitch interms of slots
+slots_1 = 180/T_p // pole pitch per slots
+y = 2*slots_1 // short pitch angle in degree
+k_p = cosd(y/2) // pitch factor
+
+// display the result
+disp("Example 9.2 solution");
+printf(" \n Pitch factor is \n k_p = %.2f \n", k_p );
+
+
diff --git a/2642/CH3/EX3.1/Ex3_1.sce b/2642/CH3/EX3.1/Ex3_1.sce new file mode 100755 index 000000000..f0c310134 --- /dev/null +++ b/2642/CH3/EX3.1/Ex3_1.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.1
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 2200 // primary voltage of transformer in V
+V_2 = 220 // secondary voltage of transformer in V
+N_2 = 56 // number of turns in the secondary coil of transformer
+kVA = 25 // kVA rating of transformer
+
+// caclulations
+a = V_1/V_2 // turn ratio
+N_1 = a*N_2 // number of primary turns
+I_1 = kVA*10^3/V_1 // primary full load current in A
+I_2 = kVA*10^3/V_2 // secondary full load current in A
+
+// display the result
+disp("Example 3.1 solution");
+printf(" \n Number of primary turns \n N_1 = %.0f \n", N_1);
+printf(" \n Primary full load current \n I_2 = %.2f A \n", I_1);
+printf(" \n Secondary full load current \n I_2 = %.1f A \n", I_2);
diff --git a/2642/CH3/EX3.10/Ex3_10.sce b/2642/CH3/EX3.10/Ex3_10.sce new file mode 100755 index 000000000..2f3d2ebc6 --- /dev/null +++ b/2642/CH3/EX3.10/Ex3_10.sce @@ -0,0 +1,38 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.10
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 12 // kVA ratingss of transformer
+n = 0.97 // maximum efficiency at unity power factor
+t_1 = 8 // time in hours
+P_1 = 10 // load in kW
+pf_1 = 0.8 // lagging power factor
+t_2 = 10 // time in hours
+P_2 = 15 // load in kW
+pf_2 = 0.90 // leading power factor
+t_3 = 6 // time in hours at no load
+P_3 = 0 // load in kW
+
+// caclulations
+P_01 = kVA*1 // o/p power at full load and unity factor in kW
+P_in1 = (P_01/n) // i/p power at full load
+P_tloss = P_in1-P_01 // total loss in kW
+P_cu = P_tloss/2 // copper loss at 12 kVA P_cu=P_i in kW
+P_024 = P_1*t_1+P_2*t_2+P_3*t_3 // all day o.p power in kWh
+P_i24 = 24*P_cu // iron loss for 24 hours in kWh
+P_cu24 = P_cu*t_1*((P_1/pf_1)/P_01)^2+P_cu*t_2*((P_2/pf_2)/P_01)^2 // copper loss for 24 hours
+P_in24 = P_024+P_i24+P_cu24 // all day i/p power in kWh
+n_allday = (P_024/P_in24)*100 // all day efficiency
+
+// display the result
+disp("Example 3.10 solution");
+printf(" \n All day efficiency \n n_allday = %.0f percent \n", n_allday);
+
diff --git a/2642/CH3/EX3.11/Ex3_11.sce b/2642/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..ea9938e0e --- /dev/null +++ b/2642/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.11
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 200 // voltage in V
+f = 50 // frequency in Hz
+I_0 = 0.6 // single phase current in A
+P_0 = 80 // power in W
+
+// caclulations
+cos_phi0 = P_0/(V_1*I_0) // power factor
+sin_phi0 = 0.74 // from above expression
+I_w = I_0*cos_phi0 // working component of no load current in A
+I_m = I_0*sin_phi0 // working component of no load current in A
+R_0 = V_1/I_w // no load circuit resistance in ohm
+X_0 = V_1/I_m // no load circuit reactance in ohm
+
+// display the result
+disp("Example 3.11 solution");
+printf(" \n No-load circuit resistance \n R_0 = %.2f ohm \n", R_0);
+printf(" \n No-load circuit reactance \n X_0 = %.1f ohm \n", X_0);
diff --git a/2642/CH3/EX3.12/Ex3_12.sce b/2642/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 000000000..6d743ace0 --- /dev/null +++ b/2642/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,42 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.12
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 25 // kVA ratings of transformer
+V1 = 2200 // primary side voltage in V
+V2 = 220 // secondary side voltage in V
+V_1 = 40 // voltage at high voltage side in V
+I_1 = 5 // current at high voltage side in A
+P = 150 // power at high voltage side in W
+
+// caclulations
+Z_01 = V_1/I_1 // reactance to primary sidec in ohm
+R_01 = P/I_1^2 // resistance to primary side in ohm
+phi = acosd(R_01/Z_01) // power factor angle
+X_01 = Z_01*sind(phi) // impedance to primary side in ohm
+a = V1/V2 // turn ratio
+Z_02 = Z_01/a^2 // reactance to secondary side in ohm
+R_02 = R_01/a^2 // resistance to secondary side in ohm
+X_02 = X_01/a^2 // impedance to secondary side in ohm
+I_2 = kVA*10^3/V2 // secondary side current in A
+E_2 = V2+I_2*Z_02 // secondary induced voltage in V
+VR = ((E_2-V2)/V2)*100 // voltage regulation
+
+// display the result
+disp("Example 3.12 solution");
+printf(" \n Resistance to primary side \n Z_01 = %.2f ohm \n", Z_01);
+printf(" \n Resistance to primary side \n R_01 = %.1f ohm \n", R_01);
+printf(" \n Impedance to primary side \n X_01 = %.2f ohm \n", X_01);
+printf(" \n Reactance to secondary side \n Z_02 = %.2f ohm \n", Z_02);
+printf(" \n Resistance to secondary side \n R_02 = %.2f ohm \n", R_02);
+printf(" \n Impedance to secondary side \n X_02 = %.3f ohm \n", X_02);
+printf(" \n oltage regulation \n VR = %.0f percent \n", VR);
+
diff --git a/2642/CH3/EX3.13/Ex3_13.sce b/2642/CH3/EX3.13/Ex3_13.sce new file mode 100755 index 000000000..3ce3efd85 --- /dev/null +++ b/2642/CH3/EX3.13/Ex3_13.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.13
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 120 // kVA ratings of autotransformer
+V1 = 2200 // lower part voltage of autotransformer in V
+V2 = 220 // upper part voltage of autotransformer in V
+
+// caclulations
+I_pq = kVA*10^3/V2 // currents of respective windings
+I_qr = kVA*10^3/V1 // currents of respective windings
+I_1 = I_pq+I_qr // current in primary side in A
+V_2 = V1+V2 // voltage across the secondary side in V
+kVA_1 = I_1*V1/1000 // kVA ratings of autotrnsformer
+kVA_2 = I_pq*V_2/1000 // kVA ratings of autotrnsformer
+
+
+// display the result
+disp("Example 3.13 solution");
+printf(" \n kVA ratings of autotrnsformer \n kVA_1 = %.0f kVA \n", kVA_1);
+printf(" \n kVA ratings of autotrnsformer \n kVA_2 = %.0f kVA \n", kVA_2);
+
diff --git a/2642/CH3/EX3.14/Ex3_14.sce b/2642/CH3/EX3.14/Ex3_14.sce new file mode 100755 index 000000000..b585a9279 --- /dev/null +++ b/2642/CH3/EX3.14/Ex3_14.sce @@ -0,0 +1,28 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.14
+
+clc;clear; // clears the console and command history
+
+// Given data
+E_1 = 500 // secondary induced voltages in V
+E_2 = 450 // secondary induced voltages in V
+kVA_1 = 100 // kVA ratings of transformer
+kVA_2 = 200 // kVA ratings of transformer
+Z_1 = 0.05 // impedance of transformer
+Z_2 = 0.08 // impedance of transformer
+
+// caclulations
+Z1 =Z_1*E_1/(kVA_1*10^3/E_1) // actual impedance of 1st transformer in ohm
+Z2 = Z_2*E_2/(kVA_1*10^3/E_2) // actual impedance of 2nd transformer in ohm
+Z = %i*(Z1+Z2)
+I_c = (E_1-E_2)/(Z) // value of the circulating current
+
+// display the result
+disp("Example 3.14 solution");
+printf(" \n Value of the circulating current \n I_c = %.3f<%.f A \n", abs(I_c),atand(imag(I_c),real(I_c)));
+
diff --git a/2642/CH3/EX3.15/Ex3_15.sce b/2642/CH3/EX3.15/Ex3_15.sce new file mode 100755 index 000000000..5f9a94737 --- /dev/null +++ b/2642/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,36 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.15
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_L1 = 11 // three-phase transformer supply in kV
+I_P1 = 6 // three-phase transformer current in A
+a = 11 // turns ratio
+
+// caclulations
+// delta-wye
+V_dP2 = V_L1*10^3/a // phase voltage at secondary side in V
+V_dL2 = sqrt(3)*V_dP2 // line voltage at secondary side in V
+I_dP1 = a/sqrt(3) // phase current in the primary in A
+I_dL2 = a*I_dP1 // line current in secondary in A
+//Wye delta connection
+V_wP1 = V_L1*10^3/sqrt(3) // phase voltage at primary in V
+V_wP2 = V_wP1/a // phase voltage at secondary in V, V_L2=V_P2
+I_wP2 = a*I_P1 // phase current in secondary in A
+I_wL2 = sqrt(3)*I_wP2 // line current in secondary in A
+
+// display the result
+disp("Example 3.15 solution");
+printf(" \n For delta-wye connection-");
+printf(" \n Phase voltage at secondary side \n V_dL2 = %.f V \n", V_dL2);
+printf(" \n Line voltage at secondary side \n I_dL2 = %.2f A \n", I_dL2);
+printf(" \n For wye-delta connection-")
+printf(" \n Phase voltage at secondary side \n V_wL2 = %.2f V \n", V_wP2);
+printf(" \n Line current in secondary side \n I_wL2 = %.2f A \n", I_wL2);
+
diff --git a/2642/CH3/EX3.16/Ex3_16.sce b/2642/CH3/EX3.16/Ex3_16.sce new file mode 100755 index 000000000..9ab6e7a35 --- /dev/null +++ b/2642/CH3/EX3.16/Ex3_16.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.16
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_b = 220 // voltage in V
+f = 50 // frequency in Hz
+S_b = 600 // power ratings
+R = 3 // resistance in ohm
+X_L = 5 // inducatnce in ohm
+Z = R+X_L // impedance
+
+// caclulations
+I_b = S_b/V_b // base value of current in A
+Z_b = V_b^2/S_b // base value of impedance in ohm
+R_pu = R/Z_b // per unit value of resistance in ohm
+X_Lpu = X_L/Z_b // per unit value of impedance in ohm
+Z_pu = abs(Z)/Z_b // per unit of value of impedance in ohm
+Z_pu = R_pu+%i*X_Lpu // per unit of value of impedance in ohm NOTE:alternative method
+
+// display the result
+disp("Example 3.16 solution");
+printf(" \n Per unit value of resistance \n R_pu = %.3f ohm \n", R_pu);
+printf(" \n Per unit value of impedance \n X_Lpu = %.3f ohm \n", X_Lpu);
+printf(" \n Per unit of value of impedance \n Z_pu = %.3f<%.f \n", abs(Z_pu),atand(imag(Z_pu),real(Z_pu)));
+
diff --git a/2642/CH3/EX3.17/Ex3_17.sce b/2642/CH3/EX3.17/Ex3_17.sce new file mode 100755 index 000000000..e963b79f3 --- /dev/null +++ b/2642/CH3/EX3.17/Ex3_17.sce @@ -0,0 +1,45 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.17
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+S_b1 = 100 // base apparent power
+V_bT11 = 220 // voltage of 1st transformer in kV
+V_bT12 = 132 // voltage of 1st transformer in kV
+X_T1 = 0.02 // impedance of 1st transformer in pu
+S_b2 = 50 // base apparent power
+V_bT21 = 132 // voltage of 2nd transformer in kV
+V_bT22 = 66 // voltage of 2nd transformer in kV
+X_T2 = 0.05 // impedance of 2nd transformer in pu
+X_L = 4 // line impedance in ohm
+P = 50 // power absorded in MW
+pf = 0.6 // lagging power factor from transmission line
+Z_p = 0.32*%i //Reactance of transformer in ohm
+
+// caclulations
+S_b = S_b1 //Base power(MW)
+V_b = V_bT11 //Base voltage(kV)
+a = V_bT11/V_bT12 // turn ratio for 1st transformer
+Vb_line = (V_bT11/a) // base voltage of line in kV
+Zb_line = Vb_line^2/S_b1 // base impedance of line in ohm
+Xpu_line = X_L/Zb_line // per unit reactance of line
+Xpu_T1 = X_T1*(V_bT11/V_b)^2*(S_b/S_b1) // 1st grid transformer ,the per unit reactance
+Vb_load = (V_bT12/(V_bT12/V_bT22)) // load side base voltage in kV
+Xpu_load = X_T2*(V_bT22/Vb_load)^2*(S_b/S_b2) // second load transformer ,the per unit reactance
+I_b = S_b*1000/(sqrt(3)*Vb_load) // base current
+I_L = S_b2*1000/(sqrt(3)*V_bT22*pf) // actualcurrent in load in A
+I_Lpu = I_L/I_b // per unit value of the load
+V_L = V_bT22/V_bT22 //per unit value of the voltage at the load terminal(bus4)
+V_gb = I_Lpu*exp(%i*acos(pf))*Z_p + 1 // per unit value of bus voltage
+V_gba = abs(V_gb)*V_bT11 // actual value of grid to bus voltage
+
+// display the result
+disp("Example 3.17 solution");
+printf(" \n Actual value of grid to bus voltage \n V_gba = %.2f kV \n", V_gba);
diff --git a/2642/CH3/EX3.2/Ex3_2.sce b/2642/CH3/EX3.2/Ex3_2.sce new file mode 100755 index 000000000..30cd6af41 --- /dev/null +++ b/2642/CH3/EX3.2/Ex3_2.sce @@ -0,0 +1,22 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.2
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 220 // voltage in V
+N_1 = 150 // number of turns in the primary coil of transformer
+N_2 = 300 // number of turns in the secondary coil of transformer
+f = 50 // frequency in Hz
+
+// caclulations
+a = N_1/N_2 // turn ratio
+phi_m = V_1/(4.44*f*N_1) // mutual flux in Wb
+
+// display the result
+disp("Example 3.2 solution");
+printf(" \n Mutual flux \n phi_m = %.2e Wb \n", phi_m);
diff --git a/2642/CH3/EX3.3/Ex3_3.sce b/2642/CH3/EX3.3/Ex3_3.sce new file mode 100755 index 000000000..7f8988bd5 --- /dev/null +++ b/2642/CH3/EX3.3/Ex3_3.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.3
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 2200 // primary voltage of transformer in V
+V_2 = 220 // secondary voltage of transformer in V
+I_0 = 0.5 // no-load current in A
+P_0 = 350 // absorbed power by transformer
+
+// caclulations
+phi = acos(.32)
+sin_phi = sin(phi)
+cos_phi = P_0/(V_1*I_0) // no-load power factor
+I_w = I_0*cos_phi // iron loss component of current A
+I_m = I_0*sin_phi // magnetizing component of current A
+
+
+// display the result
+disp("Example 3.3 solution");
+printf(" \n The iron loss component of current A \n I_w = %.2f A \n", I_w);
+printf(" \n The magnetizing component of current A \n I_m = %.2f A \n", I_m);
diff --git a/2642/CH3/EX3.4/Ex3_4.sce b/2642/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..295de1627 --- /dev/null +++ b/2642/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.4
+clc;clear; // clears the console and command history
+
+// Given data
+N_1 = 450 // number of turns in the primary coil of transformer
+N_2 = 45 // number of turns in the secondary coil of transformer
+Z_L = 3 // load impedance in Ω
+V_1 = 15 // primary coil voltage of transformer in V
+
+// caclulations
+a = N_1/N_2 // turn ratio
+Z_1 = a^2*Z_L // load impedance referred to primary ohm
+I_1 = V_1/Z_1 // primary current in A
+
+
+// display the result
+disp("Example 3.4 solution");
+printf(" \n Turn ratio \n a = %.0f \n", a);
+printf(" \n Load impedance referred to primary \n Z_1 = %.0f Ω \n", Z_1);
+printf(" \n Primary current \n I_1 = %.2f A \n", I_1);
diff --git a/2642/CH3/EX3.5/Ex3_5.sce b/2642/CH3/EX3.5/Ex3_5.sce new file mode 100755 index 000000000..a88344b6e --- /dev/null +++ b/2642/CH3/EX3.5/Ex3_5.sce @@ -0,0 +1,28 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.5
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 400 // primary voltage of transformer in V
+V_2 = 100 // secondary voltage of transformer in V
+I_0 = 0.4 // no-load current in A
+I_2 = 100 // load draws current in A
+cos_phi0 = 0.3 // power factor lagging from the supply
+cos_phi2 = 0.6 // power factor lagging from the secondary
+
+// caclulations
+phi0 = acosd(0.3)
+phi2 = acosd(0.6)
+phi1 = phi0-phi2
+a = V_1/V_2 // turn ratio
+I_2! = I_2/a // secondary current equivalent to the primary
+I_1 = sqrt((I_2!^2)+(I_0^2)+(2*I_2!*I_0*cosd(19.4))) // primary current in A
+
+// display the result
+disp("Example 3.5 solution");
+printf(" \n Primary current \n I_1 = %.2f A \n", I_1);
diff --git a/2642/CH3/EX3.6/Ex3_6.sce b/2642/CH3/EX3.6/Ex3_6.sce new file mode 100755 index 000000000..63857c585 --- /dev/null +++ b/2642/CH3/EX3.6/Ex3_6.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.6
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 2000 // primary voltage of transformer in V
+V_2 = 400 // secondary voltage of transformer in V
+kVA = 200 // kVA rating of transformer
+R_1 = 3 // primary resistance in Ω
+X_1 = 12 // primary reactance in Ω
+R_2 = 0.3 // secondary resistance in Ω
+X_2 = 0.1 // secondary reactance in Ω
+
+// caclulations
+a = V_1/V_2 // turn ratio
+R_01 = R_1+(a^2*R_2) // total resistance referred to primary side Ω
+X_01 = X_1+(a^2*X_2) // total reactance referred to primary side Ω
+Z_01 = sqrt((R_01^2)+(X_01^2)) // equivalent impedance reffered to primary side in Ω
+R_02 = R_2+(R_1/a^2) // total resistance referred to secondary side Ω
+X_02 = X_2+(X_1/a^2) // total reactance referred to secondary side Ω
+Z_02 = sqrt((R_02^2)+(X_02^2)) // equivalent impedance reffered to secondary side in Ω
+
+// display the result
+disp("Example 3.6 solution");
+printf(" \n Equivalent impedance reffered to primary side \n Z_01 = %.1f Ω \n", Z_01);
+printf(" \n Equivalent impedance reffered to secondary side \n Z_02 = %.2f Ω \n", Z_02);
diff --git a/2642/CH3/EX3.7/Ex3_7.sce b/2642/CH3/EX3.7/Ex3_7.sce new file mode 100755 index 000000000..876496b4f --- /dev/null +++ b/2642/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.7
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 200 // primary voltage of transformer in V
+V_2 = 400 // secondary voltage of transformer in V
+R_1 = 0.3 // primary resistance in Ω
+X_1 = 0.6 // primary reactance in Ω
+R_2 = 0.8 // secondary resistance in Ω
+X_2 = 1.6 // secondary reactance in Ω
+I_2 = 10 // secondary supply current in A
+cos_phi2 = 0.8 // power factor lagging
+
+// caclulations
+a = V_1/V_2 // turn ratio
+phi2 = acosd(0.8)
+sin_phi2 = sind(phi2)
+R_02 = R_2+(R_1/a^2) // total resistance referred to secondary side Ω
+X_02 = X_2+(X_1/a^2) // total reactance referred to secondary side Ω
+E_2 = (V_2*cos_phi2+I_2*R_02)+(%i*(V_2*sin_phi2+I_2*X_02)) // no-load voltage
+V_r = (abs(E_2)-V_2)/V_2*100 // voltage regulation
+
+// display the result
+disp("Example 3.7 solution");
+printf(" \n Voltage regulation \n V_r = %.0f percent \n", V_r);
+
+
diff --git a/2642/CH3/EX3.8/Ex3_8.sce b/2642/CH3/EX3.8/Ex3_8.sce new file mode 100755 index 000000000..eca25ca92 --- /dev/null +++ b/2642/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.8
+clc;clear; // clears the console and command history
+
+// Given data
+P_i = 1 // iron loss of transformer in kW
+P_cu = 2 // copper loss of transformer in kW
+kVA = 200 // kVA ratingss of transformer
+pf = 0.95 // power factor
+
+// caclulations
+P_cu1 = (3/4)^2*P_cu // copper loss at 1/2th of full load in kW
+P_cu2 = (1/2)^2*P_cu // copper loss at 1/2th of full load in kW
+P_01 = (3/4)*kVA*P_i // o/p power at 3/4 full load and unity power factor in kW
+P_in1 = P_01+P_i+P_cu1 // i/p power at 3/4 full load and unity power factor in kW
+n_1 = (P_01/P_in1)*100 // efficiency at 3/4 full load and unity power factor
+P_02 = (1/2)*kVA*pf // o/p power factor at1/2 full load and 0.95 power factor in kW
+P_in2 = P_02+P_i+P_cu2 // i/p power at 1/2 full load and 0.95 power factor in kW
+n_2 = (P_02/P_in2)*100 // efficiency at 1/2 full load and 0.95 power factor
+
+// display the result
+disp("Example 3.8 solution");
+printf(" \n Efficiency at 3/4 full load and unity power factor \n n_1 = %.2f percent \n", n_1);
+printf(" \n Efficiency at 1/2 full load and 0.95 power factor \n n_2 = %.2f percent \n", n_2);
+
diff --git a/2642/CH3/EX3.9/Ex3_9.sce b/2642/CH3/EX3.9/Ex3_9.sce new file mode 100755 index 000000000..5b1b69703 --- /dev/null +++ b/2642/CH3/EX3.9/Ex3_9.sce @@ -0,0 +1,34 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
+// Example : 3.9
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+P_i = 350 // iron loss of transformer in W
+P_cu = 650 // copper loss of transformer in W
+kVA = 30 // kVA ratingss of transformer
+pf = 0.6 // power factor
+
+// caclulations
+P_tloss = (P_i+P_cu)*10^-3 // total full load loss in kW
+P_out = kVA*pf // o/p power at full load in kW
+P_in = P_out+P_tloss // i/p power at full load
+n_1 = (P_out/P_in)*100 // efficiency at full load
+kVA_out = kVA*sqrt(P_i/P_cu) // o/p kVA corresponding to maximum efficiency
+P_01 = kVA_out*pf // o/p power in W
+P_tloss1 = 2*P_i // maximum efficiency iron loss=copper loss in W
+P_in1 = P_01+P_tloss1*10^-3 // i/p power in kW
+n_2 = (P_01/P_in1)*100 // efficiency
+
+// display the result
+disp("Example 3.9 solution");
+printf(" \n Efficiency at full load \n n_1 = %.2f percent \n", n_1);
+printf(" \n Out put power \n P_01 = %.1f kW \n", P_01);
+printf(" \n Efficiency \n n_2 = %.2f percent \n", n_2);
+
diff --git a/2642/CH4/EX4.10/Ex4_10.sce b/2642/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..66839a973 --- /dev/null +++ b/2642/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.10
+
+clc;clear; // clears the console and command history
+
+// Given data
+Pw = 12 // power in kW
+P = 4 // number of poles
+Z = 500 // number of conductors
+V_t = 250 // generator voltage in V
+N = 1000 // speed in rpm
+P_cu = 600 // full load copper loss in W
+brush_drop = 2 // brush drop in V
+
+// caclulations
+A = 4 // for lab wound A=P
+I_a = Pw*10^3/V_t // armature current in A
+R_a = P_cu/I_a^2 // from copper loss equestion R_a in ohm
+E_g = V_t+I_a*R_a+brush_drop // generated voltage in V
+phi = E_g*60*A/(P*Z*N) // flux per pole in Wb
+
+
+// display the result
+disp("Example 4.10 solution");
+printf(" \n Flux per pole \n phi = %.3f Wb \n", phi );
diff --git a/2642/CH4/EX4.11/Ex4_11.sce b/2642/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..57707d495 --- /dev/null +++ b/2642/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,30 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.11
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+I_L = 25 // generator delivering current in A
+V_t = 230 // generator terminal voltage in V
+R_a = 0.2 // armature resistance in ohm
+R_sh = 55 // shunt field resistance in ohm
+brush_drop = 1 // brush drop in V
+
+// caclulations
+I_sh = V_t/R_sh // shunt field current in A
+I_a = I_L+I_sh // armature current in A
+E_g = V_t+I_a*R_a+brush_drop // induced voltage in V
+P_arm = E_g*I_a // power generated in armature in W
+P_L = V_t*I_L // power absorbed by load in W
+n = (P_L/P_arm)*100 // efficiency
+
+// display the result
+disp("Example 4.11 solution");
+printf(" \n Induced voltage \n E_g = %.1f V \n", E_g );
+printf(" \n Efficiency \n n = %.1f percent \n", n );
diff --git a/2642/CH4/EX4.2/Ex4_2.sce b/2642/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..3c395ae32 --- /dev/null +++ b/2642/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,23 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.2
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+l = 0.65 // length of conductor in m
+v = 35 // speed in m/s
+B = 0.8 // magnetic flux density in T
+
+// caclulations
+e = B*l*v // induced voltage in V
+
+// display the result
+disp("Example 4.2 solution");
+printf(" \n Induced voltage \n e = %.1f V \n", e);
+
diff --git a/2642/CH4/EX4.3/Ex4_3.sce b/2642/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..4aef86cf3 --- /dev/null +++ b/2642/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.3
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+l = 1.5 // length of conductor in m
+v = 20 // speed in m/s
+B = 0.9 // magnetic flux density in Wb/m^2
+teta = 35 // angle of rotation in degree
+
+// caclulations
+e = B*l*v*sind(teta) // induced voltage in V
+
+// display the result
+disp("Example 4.3 solution");
+printf(" \n Induced voltage \n e = %.1f V \n", e);
+
diff --git a/2642/CH4/EX4.4/Ex4_4.sce b/2642/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..e18553573 --- /dev/null +++ b/2642/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,25 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.4
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+Z = 40*10 // number of conductors
+phi = 0.02 // flux per pole in Wb
+N = 1200 // speed in rpm
+
+// caclulations
+A = P/2
+E_g = (P*phi*Z*N)/(60*A) // generated voltage in V
+
+// display the result
+disp("Example 4.4 solution");
+printf("\n Generated voltage \n E_g = %.0f V \n", E_g);
+
diff --git a/2642/CH4/EX4.5/Ex4_5.sce b/2642/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..7996df6df --- /dev/null +++ b/2642/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,28 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.5
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 6 // number of poles
+Z = 600 // number of conductors
+phi = 0.05 // flux per pole in Wb
+N = 1000 // speed in rpm
+I_a = 120 // generetor supply current in A
+
+// caclulations
+A=6 // for lap-wound A=P
+E_g = (P*phi*Z*N)/(60*A) // generated voltage in V
+T_em = ((P*Z*phi)/(2*%pi*A))*I_a // electromagnetic torque in N-m
+
+
+// display the result
+disp("Example 4.5 solution");
+printf(" \n Generated voltage \n E_g = %.0f V \n", E_g);
+printf(" \n Electromagnetic torque \n T_em = %.2f N-m \n", T_em);
diff --git a/2642/CH4/EX4.6/Ex4_6.sce b/2642/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..cf9abfb43 --- /dev/null +++ b/2642/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.6
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // shunt generator voltage in V
+I = 250 // shunt generator current in A
+R_sh = 50 // shunt field resistance in ohm
+R_a = 0.02 // armature resistance in ohm
+
+// caclulations
+I_sh = V_t/R_sh // shunt field current in A
+I_a = I+I_sh // armature current in A
+E_g = V_t+I_a*R_a // generated voltage in V
+
+
+// display the result
+disp("Example 4.6 solution");
+printf(" \n Generated voltage \n E_g = %.2f V \n", E_g);
+
diff --git a/2642/CH4/EX4.7/Ex4_7.sce b/2642/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..c63d5b623 --- /dev/null +++ b/2642/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,34 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.7
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+E = 25 // power of compound generator in kW
+V_t = 220 // terminal voltage in V
+R_se = 0.05 // series resistance in ohm
+R_sh = 55 // shunt field resistance in ohm
+R_a = 0.07 // armature resistance in ohm
+brush_drop = 1 // voltage drop per brush in V
+
+// caclulations
+I_L = E*10^3/V_t // load current in A
+I_sh1 = V_t/R_sh // shunt field current in A
+I_a1 = I_sh1+I_L // armature current in A
+E_g1 = V_t+I_a1*(R_a+R_se)+2*brush_drop // generator voltage in V
+V_ab = V_t+I_L*R_se // voltage across the shunt field in V for short shunt generator
+I_sh2 = V_ab/R_sh // current in the shunt field in A for short shunt generator
+I_a2 = I_sh2+I_L // armature current in A for short shunt generator
+E_g2 = V_ab+I_a2*R_a+2*brush_drop // generator voltage in V for short shunt generator
+
+// display the result
+disp("Example 4.7 solution");
+printf(" \n Generated emf when generatar is connected in long shunt \n E_g1 = %.f V \n", E_g1);
+printf(" \n Generated emf when generatar is connected in short shunt \n E_g2 = %.1f V \n", E_g2);
+
diff --git a/2642/CH4/EX4.8/Ex4_8.sce b/2642/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..21f6aef73 --- /dev/null +++ b/2642/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.8
+
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_t = 220 // shunt generator voltage in V
+I_L = 146 // generator delivering current in A
+R_sh = 50 // shunt field resistance in ohm
+R_a = 0.012 // armature resistance in ohm
+R_s = 0.02 // series field resistance in ohm
+R_d = 0.03 // diverter field resistance in ohm
+
+// caclulations
+I_sh = V_t/R_sh // shunt field current in A
+I_a = I_L+I_sh // armature current in A
+R_com = R_s*R_d/(R_s+R_d) // combined resistance in ohm
+E_g = V_t+(I_a*(R_a+R_com)) // generated voltage in V
+P_lsd = I_a^2*R_com // power loss in series and diverter in W
+P_la = I_a^2*R_com // power loss in the armature circuit resistance in W
+P_lsh = V_t*I_sh // power loss in shunt field resistance in W
+P_dl = I_L*V_t // power delivered in W
+
+// display the result
+disp("Example 4.8 solution");
+printf(" \n Generated voltage \n E_g = %.1f V \n", E_g);
+printf(" \n Power distribution \n P_dl = %.0f W \n", P_dl);
diff --git a/2642/CH4/EX4.9/Ex4_9.sce b/2642/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..4749e112d --- /dev/null +++ b/2642/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 4 : DIRECT CURRENT GENERATORS
+// Example : 4.9
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+Z = 500 // number of conductors
+I_a = 30 // generetor supply current in A
+alpa = 6 // brushes displaced angle in degree
+
+// caclulations
+A = P/2 // for wave connected A=P/2
+I_c = I_a/A // current per conductor in A
+AT_d = Z*I_c*alpa/360 // demagnetizing ampere turns per pole in At
+AT_c = Z*I_c*((1/(2*P))-(alpa/360)) // cross magnetizing ampere turn per pole in At
+
+
+// display the result
+disp("Example 4.9 solution");
+printf(" \n Demagnetizing ampere turns per pole \n AT_d = %.1f At \n", AT_d );
+printf(" \n Cross magnetizing ampere turn per pole \n AT_c = %.1f At \n", AT_c );
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);
+
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);
diff --git a/2642/CH7/EX7.1/Ex7_1.sce b/2642/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..cd74940ef --- /dev/null +++ b/2642/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 230 // supply voltage in V
+P = 4 // number of poles
+f = 50 // frequecny Hz
+N_l = 1445 // speed in rpm
+
+// caclulations
+N_s = 120*f/P // synchronous speed in rpm
+s = (N_s-N_l)/N_s // slip
+f_r = s*f // rotor frequency in Hz
+
+// display the result
+disp("Example 7.1 solution");
+printf(" \n Synchronous speed \n N_s = %.1f rpm \n", N_s);
+printf(" \n Slip \n s = %.4f \n", s);
+printf(" \n Rotor frequency \n f_r = %.1f Hz \n", f_r);
+
diff --git a/2642/CH7/EX7.2/Ex7_2.sce b/2642/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..588142d29 --- /dev/null +++ b/2642/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+E_BR = 120 // voltage found under blocked condition in V
+P = 4 // number of poles
+f = 50 // frequecny in Hz
+N_l = 1450 // speed in rpm
+
+// caclulations
+N_s = 120*f/P // synchronous speed in rpm
+s = (N_s-N_l)/N_s // slip
+f_r = s*f // rotor frequency in Hz
+E_r = s*E_BR // Rotor voltage in V
+
+// display the result
+disp("Example 7.2 solution");
+printf(" \n Synchronous speed \n N_s = %.1f rpm \n", N_s);
+printf(" \n Slip \n s = %.3f \n", s);
+printf(" \n Rotor frequency \n f_r = %.2f Hz \n", f_r);
+printf(" \n Rotor voltage \n E_r = %.2f V \n", E_r);
+
diff --git a/2642/CH7/EX7.3/Ex7_3.sce b/2642/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..e68296e89 --- /dev/null +++ b/2642/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,25 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_0 = 230 // supply voltage in V
+P = 4 // number of poles
+T_0 = 230 // original torque in N-m
+V_s = 150 // stator voltage in V
+I_0 = 560 // starting cuurent in A
+
+// caclulations
+T_st = (V_s/V_0)^2*T_0 // starting torque in N-m
+I_st = I_0*(V_s/V_0) // starting current in A
+
+// display the result
+disp("Example 7.3 solution");
+printf(" \n Starting torque \n T_st = %.1f N-m \n", T_st);
+printf(" \n Starting current \n I_st = %.1f A \n", I_st);
diff --git a/2642/CH7/EX7.4/Ex7_4.sce b/2642/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..42025d507 --- /dev/null +++ b/2642/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+f = 50 // frequency in Hz
+P = 8 // number of poles
+a = 0.03 // full load slip
+R_2 = 0.01 // rotor resistance in ohm
+X_2 = 0.1 // standstill resistance in ohm
+
+// caclulations
+N_s = 120*f/P // synchronous speed in rpm
+s = R_2/X_2 // slip at maximum torque
+N_l = (1-s)*N_s // rotor maximum speed in rpm
+T = (2*a*s)/(s^2+a^2) // ratio of full load torque to maximum torque
+T_ratio = 1/T // ratio of maximum torque to full load torque
+
+// display the result
+disp("Example 7.4 solution");
+printf(" \n Rotor speed at maximum torque \n N_l = %.0f rpm \n", N_l);
+printf(" \n Ratio of maximum torque to full load torque \n T_ratio = %.2f \n", T_ratio);
+
diff --git a/2642/CH7/EX7.5/Ex7_5.sce b/2642/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..8ca3d77d0 --- /dev/null +++ b/2642/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 440 // supply voltage in V
+P = 4 // number of poles
+P_ag = 1500 // rotor i/p in kW
+P_rcu = 250 // copper loss in W
+f = 50 // frequency in Hz
+
+// caclulations
+s = P_rcu/P_ag // slip
+N_s = 120*f/P // synchronous speed in rpm
+N_l = (1-s)*N_s // rotor maximum speed in rpm
+P_mech = (1-s)*P_ag // mechanical power developed in W
+
+// display the result
+disp("Example 7.5 solution");
+printf(" \n Slip \n s = %.3f \n", s);
+printf(" \n Synchronous speed \n N_s = %.f rpm \n", N_s);
+printf(" \n Rotor speed \n N_l = %.0f rpm \n", N_l);
+printf(" \n Mecahnical power developed \n P_mech = %.0f W \n", P_mech);
+
+// NOTE : small change in answer instead 1245 got 1250W
diff --git a/2642/CH7/EX7.6/Ex7_6.sce b/2642/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..33417cc13 --- /dev/null +++ b/2642/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+V_1 = 150 // supply voltage in V
+P = 4 // number of poles
+f = 50 // frequency in Hz
+Z_1 = 0.12+%i*0.16 // per phase standstill stator impedance in ohm
+Z_2 = 0.22+%i*0.28 // per phase standstill rotor impedance in ohm
+R_2 = real(Z_2) // from Z_2
+
+// caclulations
+Z_eq = Z_1+Z_2 // equivalent impedance in ohm
+P_mech = 3*V_1^2/(2*(R_2+abs(Z_eq))) // maximum mechanical power developed in W
+s_mp = R_2/(abs(Z_eq)+R_2) // slip
+W_s = 2*%pi*2*f/P // since N_s = f/(P/2) and W_s = 2*%pi*N_s
+W = (1-s_mp)*W_s // speed of rotor in rad/s
+T_mxm = P_mech/W // miximum torque in N-m
+
+// display the result
+disp("Example 7.6 solution");
+printf(" \n maximum mechanical power developed \n P_mech = %.f W \n", P_mech);
+printf(" \n Maximum torque \n T_mxm = %.3f N-m \n", T_mxm);
+printf(" \n Maximum slip \n s_mp = %.2f \n", s_mp);
+printf(" \n NOTE : Error in calculation of P_mech and T_mxm ");
+
diff --git a/2642/CH7/EX7.7/Ex7_7.sce b/2642/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..8559c9eca --- /dev/null +++ b/2642/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,35 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.7
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 440 // supply voltage in V
+P = 6 // number of poles
+f = 50 // frequency in Hz
+P_a = 45000 // i/p power
+N_l = 900 // speed in rpm
+P_tloss = 2000 // total power loss in W
+
+// caclulations
+N_s = 120*f/P // synchronous speed in rpm
+s = (N_s-N_l)/N_s // slip
+P_ag = (P_a-P_tloss) // air gap power in W
+P_rcu = s*P_ag // rotor copper loss in W
+P_mech = P_ag-P_rcu // mechanical power in W
+P_0 = P_mech-3000 // o/p power in W
+n = (P_0/P_ag)*100 // efficiency since n = P_o/P_in
+
+// display the result
+disp("Example 7.7 solution");
+printf(" \n Slip \n s = %.1f \n", s );
+printf(" \n Rotor copper loss \n P_rcu = %.f W \n", P_rcu );
+printf(" \n Out put power \n P_0 = %.f W \n", P_0 );
+printf(" \n Efficiency \n n = %.f percent \n", n );
+
+
diff --git a/2642/CH7/EX7.8/Ex7_8.sce b/2642/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..09744502c --- /dev/null +++ b/2642/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 7 : THREE-PHASE INDUCTION MOTOR
+// Example : 7.8
+
+clc;clear; // clears the console and command history
+
+// Given data
+v_s = 120 // train speed in km/h
+f = 50 // frequency in Hz
+
+// caclulations
+v_s = 120*1000/(60*60) // train speed in m/s
+w = v_s/(2*f) // length of the pole-pitch in m
+
+// display the result
+disp("Example 7.8 solution");
+printf(" \n Length of the pole-pitch lenear induction motor \n w = %.2f m \n", w );
+
+
+
diff --git a/2642/CH8/EX8.2/Ex8_2.sce b/2642/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..a2c924c20 --- /dev/null +++ b/2642/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+T_ratio = 1/2 // T_st/T_f ratio of starting torque to full load torque
+s_f = 0.03 // full load slip
+I_ratio = 5 // I_sc/I_f ratio pf short circuit current to full load current
+
+// caclulations
+x = (1/I_ratio)*sqrt(T_ratio*(1/s_f)) // percentage of tapping
+
+// display the result
+disp("Example 8.2 solution");
+printf(" \n Percentage of tapping \n x = %.3f \n", x );
+
+
+
diff --git a/2642/CH8/EX8.3/Ex8_3.sce b/2642/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..f820c9f77 --- /dev/null +++ b/2642/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,24 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+T_ratio = 0.25 // T_st/T_f ratio of starting torque to full load torque
+I_ratio = 3.6 // I_sc/I_f ratio pf short circuit current to full load current
+ // since given I_sc = 3*1.2*I_f
+
+// caclulations
+s_f = T_ratio*3/(I_ratio)^2 // full load slip
+
+// display the result
+disp("Example 8.3 solution");
+printf(" \n Full load slip \n s_f = %.2f \n", s_f );
+
+
+
diff --git a/2642/CH8/EX8.4/Ex8_4.sce b/2642/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..5dea475ce --- /dev/null +++ b/2642/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+Z_icr = 0.04+%i*0.5 // inner cage impedance per phase at standstill
+Z_ocr = 0.4+%i*0.2 // outer cage impedance per phase at standstill
+V = 120 // per phase rotor induced voltage at standstill in V
+
+// caclulations
+Z_com = (Z_icr*Z_ocr)/(Z_icr+Z_ocr) // combined impedance in ohm
+I_2 = V/abs(Z_com) // rotor current per phase in A
+R_2 = real(Z_com) // combined rotor resistance in ohm from Z_com
+T = I_2^2*R_2 // torque at stand still condition in syn.watt
+s = 0.06 // slip of 6%
+Z_icr6 = 0.04/s+%i*0.5 // inner cage impedance per phase at standstill
+Z_ocr6 = 0.4/s+%i*0.2 // outer cage impedance per phase at standstill
+Z_com6 = ((Z_icr6)*Z_ocr6)/((Z_icr6)+Z_ocr6) // combined impedance in ohm at 6% slip
+I2_6 = V/abs(Z_com6) // rotor current per phase in A at 6% slip
+R2_6 = real(Z_com6) // combined rotor resistance in ohm from Z_com6
+T6 = I2_6^2*R2_6 // torque at stand still condition in syn.watt
+
+// display the result
+disp("Example 8.4 solution");
+printf(" \n Torque at stand still condition \n T = %.2f syn.watt \n", T );
+printf(" \n Torque at stand 6 percent slip \n T6 = %.2f syn.watt \n", T6 );
diff --git a/2642/CH8/EX8.5/Ex8_5.sce b/2642/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..b8f1f476f --- /dev/null +++ b/2642/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 210 // supply voltage in V
+f = 50 // supply frequency in Hz
+P = 50 // i/p power in W
+I_br = 2.5 // line current in A
+V_br = 25 // line voltgae
+R_1 = 2.4 // resistance between two resistance in ohm
+
+// caclulations
+V_br = V_br/sqrt(3) // phase voltage
+P_br = P/3 // power per phase in W
+R_eq = P_br/I_br^2 // equivalent resistance in ohm
+R_2 = R_eq-(R_1/2) // per phase rotor resistance in ohm
+Z_eq = V_br/I_br // equivalent impedance in ohm
+X_eq = sqrt(Z_eq^2-R_2^2) // equivalent reactance in ohm
+X_1 = 0.5*X_eq
+
+// display the result
+disp("Example 8.5 solution");
+printf(" \n Equivalent resistance \n R_eq = %.1f ohm \n", R_eq );
+printf(" \n Equivalent reactance \n X_eq = %.1f ohm \n", X_eq );
+printf(" \n Equivalent reactance \n X_1 = %.1f ohm \n", X_1 );
+
diff --git a/2642/CH8/EX8.6/Ex8_6.sce b/2642/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..6b283da42 --- /dev/null +++ b/2642/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,34 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 210 // supply voltage in V
+f = 50 // supply frequency in Hz
+P = 4 // number of poles
+P_0 = 400 // i/p power in W
+I_0 = 1.2 // line current in A
+V_0 = 210 // line voltage
+P_fw = 150 // total friction and windage losses in W
+R = 2.2 // stator resistance between two terminals in ohm
+
+// caclulations
+R_1 = R/2 // per phase stator resistance in ohm
+P_scu = 3*I_0^2*R_1 // copper loss in W
+P_core = P_0-P_fw-P_scu // stator core loss in W
+R_0 = (V_0/sqrt(3))^2/(P_core/3) // no-load resistance in ohm
+// alternate approach
+phi_0 = acosd(P_core/(sqrt(3)*V_0*I_0)) // power factor angle
+X_0 = (V_0/sqrt(3))/(I_0*sind(phi_0)) // magnetizing reactance per phase in ohm
+
+// display the result
+disp("Example 8.6 solution");
+printf(" \n No-load resistance \n R_0 = %.1f ohm \n", R_0 );
+printf(" \n Magnetizing reactance per phase \n X_0 = %.0f ohm \n", X_0 );
+
diff --git a/2642/CH8/EX8.7/Ex8_7.sce b/2642/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..d2e439103 --- /dev/null +++ b/2642/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR
+// Example : 8.7
+
+clc;clear; // clears the console and command history
+
+// Given data
+P_1 = 6 // number of pole
+P_2 = 4 // number of pole
+f =50 // supply frequency in Hz
+P = 60 // power in kW
+
+// caclulations
+s = P_2/(P_1+P_2) // combined slip
+N_cs = 120*f/(P_1+P_2) // combined synchronous speed in rpm
+P_0 = P*(P_2/(P_1+P_2)) // o/p of 4-pole motor in kW
+
+// display the result
+disp("Example 8.7 solution");
+printf(" \n Combined slip \n s = %.1f \n", s );
+printf(" \n Combined synchronous speed \n N_cs = %.0f rpm \n", N_cs );
+printf(" \n Out-power of 4-pole motor \n P_0 = %.f kW \n", P_0 );
+
diff --git a/2642/CH9/EX9.1/Ex9_1.sce b/2642/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..559c1b874 --- /dev/null +++ b/2642/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,22 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.1
+
+clc;clear; // clears the console and command history
+
+// Given data
+N = 300 // speed of water turbine in rpm
+f = 50 // frequency in Hz
+
+// caclulations
+P = 120*f/N // number of poles
+
+// display the result
+disp("Example 9.1 solution");
+printf(" \n Number of poles of the generator \n P = %.0f poles \n", P );
+
+
diff --git a/2642/CH9/EX9.10/Ex9_10.sce b/2642/CH9/EX9.10/Ex9_10.sce new file mode 100755 index 000000000..bd1229875 --- /dev/null +++ b/2642/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,29 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.10
+
+clc;clear; // clears the console and command history
+
+// Given data
+MVA = 30 // MVA ratings
+V = 20 // supply voltage in kV
+N = 1800 // speed in rpm
+V_t = 15 // voltage per phase in kV
+E_f = 10 // per phase terminal voltage in kV
+teta = 40 // power angle in degree
+X_s = 6 // 3 phase synchronous reactance in ohm
+
+// caclulations
+P = 3*V_t*E_f*sind(teta)/X_s // power delivered to the load in MW
+P_max = 3*V_t*E_f/X_s // three phase maximum power in MW
+
+// display the result
+disp("Example 9.10 solution");
+printf(" \n Three phase power delivered to the load \n P = %.2f MW \n", P );
+printf(" \n Three phase maximum power \n P_max = %.0f MW \n", P_max );
+
+
diff --git a/2642/CH9/EX9.11/Ex9_11.sce b/2642/CH9/EX9.11/Ex9_11.sce new file mode 100755 index 000000000..1d86c8379 --- /dev/null +++ b/2642/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,35 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.11
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 25 // kVA ratings
+V = 440 // suppy voltage in V
+f = 50 // supply frequency in Hz
+pf = 0.8 // lagging power factor
+R_a = 0.3 // resistance of machine per phase in ohm
+X_d = 5 // reactance of machine per phase in ohm
+X_q = 3 // reactance of machine per phase in ohm
+
+// caclulations
+phi = acosd(pf)
+V_t = V/sqrt(3) // terminal voltage per phase in V
+I_a = 25*10^3/(sqrt(3)*V) // armature current
+delta = atand(I_a*X_q*pf/(V_t+I_a*X_q*sind(phi))) // torque angle
+I_d = I_a*sind(delta+phi) // direct axis component of the current in A
+E_f = V_t*cosd(delta)+I_d*X_d // induced voltage per phase in V
+VR = ((E_f-V_t)/V_t)*100 // voltage regulation in V
+
+// display the result
+disp("Example 9.11 solution");
+printf(" \n Torque angle \n delta = %.2f degree \n", delta );
+printf(" \n Induced voltage per phase \n E_f = %.2f V \n", E_f );
+printf(" \n Voltage regulation \n VR = %.2f percent \n", VR );
+
+
diff --git a/2642/CH9/EX9.12/Ex9_12.sce b/2642/CH9/EX9.12/Ex9_12.sce new file mode 100755 index 000000000..ad031e126 --- /dev/null +++ b/2642/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.12
+
+clc;clear; // clears the console and command history
+
+// Given data
+E_1 = 220 // induced voltage per phase by an alternator1
+E_2 = 220*exp(%i*5*%pi/180) // induced voltage per phase by an alternator2
+Z_1 = %i*3 // impedance of an alternator1
+Z_2 = %i*4 // impedance of an alternator2
+R = 5 // resistance in ohm
+Z = 5
+
+// caclulations
+I = (E_1*Z_2+E_2*Z_1)/(Z_1*Z_2+Z*(Z_1+Z_2)) // load current in A
+V_t = I*R // terminal voltage in V
+I_a1 = ((E_1-E_2)*Z+E_1*Z_2)/(Z_1*Z_2+Z*(Z_1+Z_2)) // armature current in A
+D = atand(imag(I),real(I))// from V_t
+A = atand(imag(V_t),real(V_t)) // from I_a1
+P_1 = abs(V_t*I_a1)*cosd(D-A) // power per phase delivered by the 1st alternator in W
+
+// display the result
+disp("Example 9.12 solution");
+printf(" \n load current \n I = %.1f<%.2f degree \n", abs(I),atand(imag(I),real(I)) );
+printf(" \n terminal voltage \n V_t = %.f<%.2f V \n", abs(V_t),atand(imag(V_t),real(V_t)) );
+printf(" \n Power per phase delivered by the 1st alternator \n P_1 = %.2f W \n", P_1 );
+printf(" \n NOTE : ERROR : Calculation mistakes in textbook \n")
diff --git a/2642/CH9/EX9.2/Ex9_2.sce b/2642/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..8746869ab --- /dev/null +++ b/2642/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,26 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.2
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 8 // number of poles
+m = 3 // number of phase
+S = 144 // number of slots
+
+// caclulations
+T_p = S/P // pole pitch interms of slots
+slots_1 = 180/T_p // pole pitch per slots
+y = 2*slots_1 // short pitch angle in degree
+k_p = cosd(y/2) // pitch factor
+
+// display the result
+disp("Example 9.2 solution");
+printf(" \n Pitch factor is \n k_p = %.2f \n", k_p );
+
+
diff --git a/2642/CH9/EX9.3/Ex9_3.sce b/2642/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..8bae1ca7a --- /dev/null +++ b/2642/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,28 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+m = 3 // number of phase
+S = 40 // number of slots
+
+// caclulations
+T_p = S/P // pole pitch interms of slots
+T_c = 9-1 // coil span 1 to 9 i.e. coil pitch in terms of slots
+slots_1 = 180/T_p // pole pitch per slots
+y = T_p-T_c // short pitch angle
+y_angle = y*slots_1 // in terms of angle
+k_p = cosd(y_angle/2) // pitch factor
+
+// display the result
+disp("Example 9.3 solution");
+printf(" \n Pitch factor is \n k_p = %.2f \n", k_p );
+
+
diff --git a/2642/CH9/EX9.4/Ex9_4.sce b/2642/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..57ad56acb --- /dev/null +++ b/2642/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,25 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.4
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+S = 48 // number of slots
+
+// caclulations
+T_p = S/P // pole pitch interms of slots
+slots_1 = 180/T_p // pole pitch per slots
+n = S/(P*3) // number of slots or coils per pole per phase
+k_d = sind((n*slots_1)/2)/(n*sind(slots_1/2)) // distribution factor
+
+// display the result
+disp("Example 9.4 solution");
+printf(" \n Distribution factor is \n k_d = %.2f \n", k_d );
+
+
diff --git a/2642/CH9/EX9.5/Ex9_5.sce b/2642/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..d167e70fc --- /dev/null +++ b/2642/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,33 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.5
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 12 // number of poles
+S = 180 // number of slots
+phi_m = 0.05 // flux per pole in Wb
+N = 600 // speed of machine in rpm
+
+// caclulations
+T_p = S/P // pole pitch interms of slots
+slots_1 = 180/T_p // pole pitch per slots
+n = S/(P*3) // number of slots or coils per pole per phase
+k_d = sind((n*slots_1)/2)/(n*sind(slots_1/2)) // distribution factor
+k_p = 1 // pitch factor
+Z = (180/3)*slots_1 // number of conductor per phase
+T = Z/2 // number of turns per phase
+f = P*N/120 // frequency in Hz
+E = 4.44*k_p*k_d*f*phi_m*T // induced voltage in V
+E_L = sqrt(3)*E // line voltage in V
+
+// display the result
+disp("Example 9.5 solution");
+printf(" \n Line voltage is \n E_L = %.0f V \n", E_L );
+
+// NOTE : correction in answer
diff --git a/2642/CH9/EX9.6/Ex9_6.sce b/2642/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..a1a1ec7bc --- /dev/null +++ b/2642/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.6
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 4 // number of poles
+m = 3 // number of phase
+f = 50 // frequency in Hz
+phi_m = 0.05 // flux per pole in Wb
+
+// caclulations
+T_p = 6*3 // pole pitch interms of slots 6 slots per pole per phase hence for 3 phase
+slots_1 = 180/T_p // pole pitch per slots
+T_c = (5/6)*T_p // coil per pitch
+y = T_p-T_c // short pitch angle
+y_angle = y*10 // short pitch in terms of angle
+k_p = cosd(y_angle/2) // pitch factor
+n = 6 // number of slots
+k_d = sind((n*slots_1)/2)/(n*sind(slots_1/2)) // distribution factor
+T = (1/2)*n*P*2*5 // 2=no. of layers, 5=condctor per layer
+E = 4.44*k_p*k_d*f*phi_m*T // induced voltage in V
+
+// display the result
+disp("Example 9.6 solution");
+printf(" \n Voltage per phase is \n E = %.0f V \n", E );
+
diff --git a/2642/CH9/EX9.7/Ex9_7.sce b/2642/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..f69245998 --- /dev/null +++ b/2642/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,35 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.7
+
+clc;clear; // clears the console and command history
+
+// Given data
+P = 10 // number of poles
+m = 3 // number of phase
+f = 50 // frequency in Hz
+phi_m1 = 0.05 // flux per pole in Wb
+phi_m3 = 0.006 // flux per pole in Wb
+T_c = 150 // coil
+
+// caclulations
+T_p = 3*3 // pole pitch interms of slots 3 slots per pole per phase hence for 3 phase
+slots_1 = 180/T_p // pole pitch per slots
+y = 180-T_c // short pitch angle
+n = 3 // number of slots
+k_p1 = cosd(y/2) // pitch factor
+k_d1 = sind((n*slots_1)/2)/(n*sind(slots_1/2)) // distribution factor
+E_1 = 4.44*k_p1*k_d1*f*phi_m1*T_c // induced voltage in V
+k_p3 = cosd(y/2) // pitch factor
+k_d3 = sind((n*slots_1)/2)/(n*sind(slots_1/2)) // distribution factor
+E_3 = 4.44*k_p3*k_d3*f*phi_m3*T_c // induced voltage in V
+E = sqrt(E_1^2+E_3^2) // induced voltage per phase in V
+
+// display the result
+disp("Example 9.7 solution");
+printf(" \n Induced voltage per phase is \n E = %.0f V \n", E );
+
diff --git a/2642/CH9/EX9.8/Ex9_8.sce b/2642/CH9/EX9.8/Ex9_8.sce new file mode 100755 index 000000000..b567cefa8 --- /dev/null +++ b/2642/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,27 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.8
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 50 // kVA ratings
+V_t = 220 // terminal voltage in V
+R_a = 0.011 // effective resistance in ohm
+X_s = 0.09 // synchronous reactance in ohm
+
+// caclulations
+phi = acosd(0.85) // since power factor in 0.85
+I_a = kVA*10^3/V_t // armature current in A
+E_f = sqrt((V_t*cosd(phi)+I_a*R_a)^2+(V_t*sind(phi)+I_a*X_s)^2) // induced voltage per phase in V
+VR = ((E_f-V_t)/V_t)*100 // voltage regulation
+
+// display the result
+disp("Example 9.8 solution");
+printf(" \n No-load induced voltage per phase \n E_f = %.1f V \n", E_f );
+printf(" \n Voltage regulation is \n VR = %.1f percent \n", VR );
+
diff --git a/2642/CH9/EX9.9/Ex9_9.sce b/2642/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..920d6a9fc --- /dev/null +++ b/2642/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,36 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 9 : SYNCHRONOUS GENERATOR
+// Example : 9.9
+
+clc;clear; // clears the console and command history
+
+// Given data
+kVA = 200 // kVA ratings
+V_t = 33*10^3 // terminal voltage in V
+R_a = 0.54 // armature resistance in ohm
+V_L = 415 // voltage between lines for SC test in V
+I_sh = 25 // short circuit current in A
+
+// caclulations
+phi = acosd(0.9) // since power factor in 0.9
+V_P = V_L/sqrt(3) // phase voltage during short circuit test in V
+Z_s = V_P/I_sh // synchronous impedance in ohm
+X_s = sqrt(Z_s^2 - R_a^2) // synchronous reactance in ohm
+I_a = kVA/(sqrt(3)*V_t*10^-3) // full loa current in A
+V_ta = V_t/sqrt(3) // voltage per phase alternator
+E_f = sqrt((V_ta*cosd(phi)+I_a*R_a)^2+(V_ta*sind(phi)+I_a*X_s)^2) // no-load voltage per phase in V
+VR = ((E_f-V_ta)/V_ta)*100 // voltage regulation
+
+// display the result
+disp("Example 9.9 solution");
+printf(" \n Synchronous impedance \n Z_s = %.1f ohm \n", Z_s );
+printf(" \n Synchronous reactance is \n X_s = %.2f ohm \n", X_s );
+printf(" \n Voltage regulation is \n VR = %.2f percent \n", VR );
+printf(" \n NOTE : error in calculation, R_a is taken instead of X_s in E_f calculation \n");
+
+
+
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