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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1092/CH6 | |
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Diffstat (limited to '1092/CH6')
| -rwxr-xr-x | 1092/CH6/EX6.1/Example6_1.sce | 52 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.2/Example6_2.sce | 53 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.3/Example6_3.sce | 41 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.4/Example6_4.sce | 77 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.5/Example6_5.sce | 88 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.6/Example6_6.sce | 51 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.7/Example6_7.sce | 70 | ||||
| -rwxr-xr-x | 1092/CH6/EX6.8/Example6_8.sce | 68 |
8 files changed, 500 insertions, 0 deletions
diff --git a/1092/CH6/EX6.1/Example6_1.sce b/1092/CH6/EX6.1/Example6_1.sce new file mode 100755 index 000000000..8afa90d79 --- /dev/null +++ b/1092/CH6/EX6.1/Example6_1.sce @@ -0,0 +1,52 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 1000 ; // kVA rating of the 3-phase alternator
+V_L = 4600 ; // Rated line voltage in volt
+// 3-phase, Y-connected alternator
+R_a = 2 ; // Armature resistance in ohm per phase
+X_s = 20 ; // Synchronous armature reactance in ohm per phase
+cos_theta_a = 1 ; // Unity power factor (case a)
+cos_theta_b = 0.75 ; // 0.75 power factor lagging (case b)
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 );
+
+// Calculations
+V_P = V_L / sqrt(3) ; // Phase voltage in volt
+I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A
+I_a = I_P ; // Armature current in A
+
+// a: At unity PF
+E_g_a = ( V_P + I_a * R_a ) + %i*(I_a*X_s);
+// Full-load generated voltage per-phase (case a)
+E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt
+E_g_a_a=atan(imag(E_g_a) /real(E_g_a))*180/%pi;//E_g_a_a=phase angle of E_g_a in degrees
+
+// b: At 0.75 PF lagging
+E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b + I_a*X_s );
+// Full-load generated voltage per-phase (case b )
+E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt
+E_g_b_a=atan(imag(E_g_b) /real(E_g_b))*180/%pi;//E_g_b_a=phase angle of E_g_b in degrees
+
+
+// Display the results
+disp("Example 6-1 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.", sqrt(3));
+printf("\n\n a: At unity PF, \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_a);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_a_m , E_g_a_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_a_m,E_g_a_a);
+
+printf(" \n b: At 0.75 PF lagging , \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_b);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_b_m , E_g_b_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_b_m,E_g_b_a);
diff --git a/1092/CH6/EX6.2/Example6_2.sce b/1092/CH6/EX6.2/Example6_2.sce new file mode 100755 index 000000000..fecdf70b8 --- /dev/null +++ b/1092/CH6/EX6.2/Example6_2.sce @@ -0,0 +1,53 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-2
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 1000 ; // kVA rating of the 3-phase alternator
+V_L = 4600 ; // Rated line voltage in volt
+// 3-phase, Y-connected alternator
+R_a = 2 ; // Armature resistance in ohm per phase
+X_s = 20 ; // Synchronous armature reactance in ohm per phase
+cos_theta_a = 0.75 ; // 0.75 PF leading (case a)
+cos_theta_b = 0.40 ; // 0.40 PF leading (case b)
+sin_theta_a = sqrt( 1 - (cos_theta_a)^2 ); // (case a)
+sin_theta_b = sqrt( 1 - (cos_theta_b)^2 ); // (case b)
+
+// Calculations
+V_P = V_L / sqrt(3) ; // Phase voltage in volt
+I_P = ( kVA * 1000 ) / ( 3*V_P ) ; // Phase current in A
+I_a = I_P ; // Armature current in A
+
+// a: At 0.75 PF leading
+E_g_a = ( V_P*cos_theta_a + I_a * R_a ) + %i*( V_P*sin_theta_a - I_a*X_s);
+// Full-load generated voltage per-phase (case a)
+E_g_a_m=abs(E_g_a);//E_g_a_m=magnitude of E_g_a in volt
+E_g_a_a=atan(imag(E_g_a) /real(E_g_a))*180/%pi;//E_g_a_a=phase angle of E_g_a in degrees
+
+// b: At 0.40 PF leading
+E_g_b = ( V_P*cos_theta_b + I_a * R_a ) + %i*( V_P*sin_theta_b - I_a*X_s );
+// Full-load generated voltage per-phase (case b )
+E_g_b_m=abs(E_g_b);//E_g_b_m=magnitude of E_g_b in volt
+E_g_b_a=atan(imag(E_g_b) /real(E_g_b))*180/%pi;//E_g_b_a=phase angle of E_g_b in degrees
+
+
+// Display the results
+disp("Example 6-2 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.", sqrt(3));
+printf("\n\n a: 0.75 PF leading, \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_a);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_a_m , E_g_a_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_a_m,E_g_a_a);
+
+printf(" \n b: At 0.40 PF leading , \n ");
+printf("\n Rectangular form :\n E_g = "); disp(E_g_b);
+printf("\n Polar form :");
+printf(" \n E_g = %d <%.2f V/phase ", E_g_b_m , E_g_b_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",E_g_b_m,E_g_b_a);
diff --git a/1092/CH6/EX6.3/Example6_3.sce b/1092/CH6/EX6.3/Example6_3.sce new file mode 100755 index 000000000..ad2f089a5 --- /dev/null +++ b/1092/CH6/EX6.3/Example6_3.sce @@ -0,0 +1,41 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-3
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// From Ex.6-1 and Ex.6-2 we have V_P and E_g values as follows
+// Note : approximated values are considered when root 3 value is taken as 1.73
+// as in textbook
+V_P = 2660 ; // Phase voltage
+E_g_a1 = 3836 ; // E_g at unity PF (Ex.6-1 case a)
+E_g_b1 = 4814 ; // E_g at 0.75 PF lagging (Ex.6-1 case b)
+
+E_g_a2 = 2364 ; // E_g at 0.75 PF leading (Ex.6-2 case a)
+E_g_b2 = 1315 ; // E_g at 0.40 PF leading (Ex.6-2 case b)
+
+// Calculations
+VR_a = ( E_g_a1 - V_P )/V_P * 100 ; // voltage regulation at unity PF (Ex.6-1 case a)
+VR_b = ( E_g_b1 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF lagging (Ex.6-1 case b)
+
+VR_c = ( E_g_a2 - V_P )/V_P * 100 ; // voltage regulation at 0.75 PF leading (Ex.6-2 case a)
+VR_d = ( E_g_b2 - V_P )/V_P * 100 ; // voltage regulation at 0.40 PF leading (Ex.6-2 case b)
+
+// Display the results
+disp("Example 6-3 Solution : ");
+printf(" \n a: At unity PF : ");
+printf(" \n VR = %.1f percent \n ", VR_a );
+
+printf(" \n b: At 0.75 PF lagging : ");
+printf(" \n VR = %.2f percent \n ", VR_b );
+
+printf(" \n c: At 0.75 PF leading : ");
+printf(" \n VR = %.2f percent \n ", VR_c );
+
+printf(" \n d: At 0.40 PF leading : ");
+printf(" \n VR = %.1f percent \n ", VR_d );
diff --git a/1092/CH6/EX6.4/Example6_4.sce b/1092/CH6/EX6.4/Example6_4.sce new file mode 100755 index 000000000..1e9a42df0 --- /dev/null +++ b/1092/CH6/EX6.4/Example6_4.sce @@ -0,0 +1,77 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-4
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)
+cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)
+sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)
+sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculations
+// Assuming that the alternator is Y-connected
+// case a :
+I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A
+I_a = sqrt(3)*I_a_rated ; // Rated Line current in A
+
+R_dc = V_l/(2*I_a1); // effective dc armature resistance in ohm/winding
+R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase
+R_a = R_ac ; // effective ac armature resistance in ohm.phase from dc resistance test
+
+Z_p = E_gp2 / I_a ; // Synchronous impedance per phase
+X_s = sqrt( Z_p^2 - R_a^2 ); // Synchronous reactance per phase
+
+// case b :
+V_p = V_L / sqrt(3); // Phase voltage in volt (Y-connection)
+
+// At 0.8 PF lagging
+E_gp1 = ( V_p*cos_theta_b1 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b1 + I_a_rated * X_s);
+E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt
+E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees
+V_n1 = E_gp1_m ; // No-load voltage in volt
+V_f1 = V_p ; // Full-load voltage in volt
+VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging
+
+
+// At 0.8 PF leading
+E_gp2 = ( V_p*cos_theta_b2 + I_a_rated * R_a ) + %i*( V_p*sin_theta_b2 - I_a_rated*X_s);
+E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt
+E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees
+V_n2 = E_gp2_m ; // No-load voltage in volt
+V_f2 = V_p ; // Full-load voltage in volt
+VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading
+
+// Display the results
+disp("Example 6-4 Solution : ");
+printf(" \n Assuming that the alternator is Y-connected ");
+printf(" \n a: R_dc = %.1f ohm/winding ", R_dc );
+printf(" \n R_ac = %.2f ohm/phase ", R_ac );
+printf(" \n Z_p = %.2f ohm/phase ", Z_p );
+printf(" \n X_s = %.2f ohm/phase \n", X_s );
+
+printf(" \n b: At 0.8 PF lagging ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR1 );
+
+printf(" \n At 0.8 PF leading ");
+printf(" \n Percent voltage regulation = %.1f percent ", VR2 );
diff --git a/1092/CH6/EX6.5/Example6_5.sce b/1092/CH6/EX6.5/Example6_5.sce new file mode 100755 index 000000000..2dfac51b0 --- /dev/null +++ b/1092/CH6/EX6.5/Example6_5.sce @@ -0,0 +1,88 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta_b1 = 0.8 ; // 0.8 PF lagging (case b)
+cos_theta_b2 = 0.8 ; // 0.8 PF leading (case b)
+sin_theta_b1 = sqrt( 1 - (cos_theta_b1)^2 ); // (case b)
+sin_theta_b2 = sqrt( 1 - (cos_theta_b2)^2 ); // (case b)
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculations
+// Assuming that the alternator is delta-connected
+// case a :
+I_a_rated = (kVA*1000)/(V_L*sqrt(3)); // Rated current per phase in A
+I_L = I_a_rated ; // Line current in A
+
+V_p = E_gp2 ; // Phase voltage in volt
+V_l = V_p ; // Line voltage in volt (from short circuit data)
+
+I_p = I_L / sqrt(3) ; // Phase current in A (delta connection)
+I_a = I_p ; // Rated current in A
+
+Z_s = V_l / I_p ; // Synchronous impedance per phase
+R_dc = E_gp1/(2*I_a1); // effective dc armature resistance in ohm/winding
+R_ac = R_dc * 1.5 ; // effective ac armature resistance in ohm.phase
+
+// R_eff in delta = 3 * R_eff in Y
+R_eff = 3 * R_ac ; // Effective armature resistance in ohm
+R_a = R_eff ; // effective ac armature resistance in ohm.phase from dc resistance test
+
+X_s = sqrt( Z_s^2 - R_a^2 ); // Synchronous reactance per phase
+
+V_p = V_L ; // Phase voltage in volt (delta-connection)
+
+// At 0.8 PF lagging
+E_gp1 = ( V_p*cos_theta_b1 + I_a * R_a ) + %i*( V_p*sin_theta_b1 + I_a*X_s);
+E_gp1_m=abs(E_gp1);//E_gp1_m=magnitude of E_gp1 in volt
+E_gp1_a=atan(imag(E_gp1) /real(E_gp1))*180/%pi;//E_gp1_a=phase angle of E_gp1 in degrees
+V_n1 = E_gp1_m ; // No-load voltage in volt
+V_f1 = V_p ; // Full-load voltage in volt
+VR1 = ( V_n1 - V_f1 )/ V_f1 * 100; // percent voltage regulation at 0.8 PF lagging
+
+
+// At 0.8 PF leading
+E_gp2 = ( V_p*cos_theta_b2 + I_a * R_a ) + %i*( V_p*sin_theta_b2 - I_a*X_s);
+E_gp2_m=abs(E_gp2);//E_gp2_m=magnitude of E_gp2 in volt
+E_gp2_a=atan(imag(E_gp2) /real(E_gp2))*180/%pi;//E_gp2_a=phase angle of E_gp2 in degrees
+V_n2 = E_gp2_m ; // No-load voltage in volt
+V_f2 = V_p ; // Full-load voltage in volt
+VR2 = ( V_n2 - V_f2 )/V_f2 * 100 ; // percent voltage regulation at 0.8 PF leading
+
+// Display the results
+disp("Example 6-5 Solution : ");
+printf(" \n Assuming that the alternator is delta-connected : \n ");
+printf(" \n a: I_p = %.3f A ", I_p );
+printf(" \n Z_s = %.2f ohm/phase ", Z_s );
+printf(" \n R_eff in delta = %.2f ohm/phase ", R_eff );
+printf(" \n X_s = %.1f ohm/phase \n", X_s );
+printf(" \n R_eff, reactance and impedance per phase in delta is 3 times")
+printf(" \n the value when connected in Y. \n")
+
+printf(" \n b: At 0.8 PF lagging ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR1 );
+
+printf(" \n At 0.8 PF leading ");
+printf(" \n Percent voltage regulation = %.1f percent \n", VR2 );
+printf(" \n Percentage voltage regulation remains the same both in Y and delta connection.");
diff --git a/1092/CH6/EX6.6/Example6_6.sce b/1092/CH6/EX6.6/Example6_6.sce new file mode 100755 index 000000000..74dbed591 --- /dev/null +++ b/1092/CH6/EX6.6/Example6_6.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-6
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// 3-phase Y-connected alternator
+E_L = 11000 ; // Line voltage generated in volt
+kVA = 165000 ; // kVA rating of the alternator
+R_p = 0.1 ; // Armature resistance in ohm/per phase
+Z_p = 1.0 ; // Synchronous reactance/phase
+Z_r = 0.8 ; // Reactor reactance/phase
+
+// Calculations
+E_p = E_L / sqrt(3); // Rated phase voltage in volt
+I_p = (kVA * 1000)/(3*E_p); // Rated current per phase in A
+
+// case a
+I_max_a = E_p / R_p ; // Maximum short-circuit current in A (case a)
+overload_a = I_max_a / I_p ; // Overload (case a)
+
+// case b
+I_steady = E_p / Z_p ; // Sustained short-circuit current in A
+overload_b = I_steady / I_p ; // Overload (case b)
+
+// case c
+Z_t = R_p + %i*Z_r ; // Total reactance per phase
+I_max_c = E_p / Z_t ; // Maximum short-circuit current in A (case b)
+I_max_c_m=abs(I_max_c);//I_max_c_m=magnitude of I_max_c in A
+I_max_c_a=atan(imag(I_max_c) /real(I_max_c))*180/%pi;//I_max_c_a=phase angle of I_max_c in degrees
+overload_c = I_max_c_m / I_p ; // Overload (case a)
+
+// Display the results
+disp("Example 6-6 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.\n", sqrt(3));
+printf(" \n a: I_max = %d A ", I_max_a );
+printf(" \n overload = %.1f * rated current \n", overload_a );
+
+printf(" \n b: I_steady = %d A ", I_steady );
+printf(" \n overload = %.2f * rated current \n", overload_b );
+
+printf(" \n c: Rectangular form :\n I_max = "); disp(I_max_c);
+printf(" \n Polar form :");
+printf(" \n I_max = %d <%.2f A ", I_max_c_m , I_max_c_a );
+printf(" \n where %d is magnitude and %.2f is phase angle\n",I_max_c_m,I_max_c_a);
+printf(" \n overload = %.3f * rated current \n", overload_c );
diff --git a/1092/CH6/EX6.7/Example6_7.sce b/1092/CH6/EX6.7/Example6_7.sce new file mode 100755 index 000000000..81d6cae8d --- /dev/null +++ b/1092/CH6/EX6.7/Example6_7.sce @@ -0,0 +1,70 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-7
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta = 0.8 ; // 0.8 PF lagging
+sin_theta = sqrt( 1 - (cos_theta)^2 ); //
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculated data from Ex.6-4
+I_L = 52.5 ; // Rated line current in A
+I_a = I_L ; // Rated current per phase in A
+E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging
+X_s = 4.6 ; // Synchronous reactance per phase
+V_p = 635 ; // Phase voltage in volt
+
+// Calculations
+// case a
+P_T = sqrt(3) * V_L * I_L * cos_theta ; // Total output 3-phase power
+
+// case b
+P_p_b = P_T / 3 ; // Total output 3-phase power per phase
+
+// case c
+E_gp_m=abs(E_gp);//E_gp_m=magnitude of E_gp in volt
+E_gp_a=atan(imag(E_gp) /real(E_gp))*180/%pi;//E_gp_a=phase angle of E_gp in degrees
+
+// case d
+theta = acos(0.8)*180/%pi; // phase angle for PF in degrees
+theta_plus_deba = E_gp_a ; // phase angle of E_gp in degrees
+deba = theta_plus_deba - theta ; // Torque angle in degrees
+
+// case e
+P_p_e = (E_gp_m/X_s)*V_p*sind(deba); // Approximate output power/phase (Eq.(6-10))
+
+// case f
+P_p_f = E_gp_m * I_a * cosd(theta_plus_deba); // Approximate output power/phase (Eq.(6-9))
+
+// Display the results
+disp("Example 6-7 Solution : ");
+printf("\n root 3 value is taken as %f , so slight variations in the answer.\n", sqrt(3));
+printf(" \n a: P_T = %d W \n", P_T );
+printf(" \n b: P_p = %.2f W \n", P_p_b );
+printf(" \n c: E_gp = %d <%.2f V \n", E_gp_m, E_gp_a );
+printf(" \n where %d is magnitude in V and %.2f is phase angle in degrees.\n",E_gp_m,E_gp_a);
+printf(" \n d: Torque angle, deba = %.2f degrees \n", deba );
+printf(" \n e: P_p = %d W \n", P_p_e );
+printf(" \n f: P_p = %d W ", P_p_f );
+
+
diff --git a/1092/CH6/EX6.8/Example6_8.sce b/1092/CH6/EX6.8/Example6_8.sce new file mode 100755 index 000000000..c3afce4d2 --- /dev/null +++ b/1092/CH6/EX6.8/Example6_8.sce @@ -0,0 +1,68 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS
+// Example 6-8
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+
+kVA = 100 ; // kVA rating of the 3-phase alternator
+V_L = 1100 ; // Line voltage of the 3-phase alternator in volt
+S = 1200 ; // Synchronous speed in rpm
+
+// dc-resistance test data
+E_gp1 = 6 ; // generated phase voltage in volt
+V_l = E_gp1 ; // generated line voltage in volt
+I_a1 = 10 ; // full-load current per phase in A
+cos_theta = 0.8 ; // 0.8 PF lagging
+sin_theta = sqrt( 1 - (cos_theta)^2 ); //
+
+// open-circuit test data
+E_gp2 = 420 ; // generated phase voltage in volt
+I_f2 = 12.5 ; // Field current in A
+
+// short-circuit test data
+I_f3 = 12.5 ; // Field current in A
+// Line current I_l = rated value in A
+
+// Calculated data from Ex.6-4 & Ex.6-7
+I_L = 52.5 ; // Rated line current in A
+I_a = I_L ; // Rated current per phase in A
+E_gp = 532 + %i*623 ; // Generated voltage at 0.8 PF lagging
+E_g = 819 ; // E_g = magnitude of E_gp in volt
+X_s = 4.6 ; // Synchronous reactance per phase
+V_p = 635 ; // Phase voltage in volt
+deba = 12.63 ; // Torque angle in degrees
+
+// Calculations
+// case a
+T_p_a = ( 7.04 * E_g * V_p * sind(deba) ) / (S*X_s); // Output torque per phase in lb.ft
+T_3phase_a = 3 * T_p_a ; // Output torque for 3-phase in lb.ft
+
+// case b
+omega = S * 2*%pi *(1/60); // Angular frequency in rad/s
+T_p_b = ( E_g * V_p * sind(deba))/(omega*X_s); // Output torque per phase in lb.ft
+T_3phase_b = 3 * T_p_b ; // Output torque for 3-phase in lb.ft
+
+// case c
+T_p_c = T_p_a * 1.356 ; // Output torque per phase in N.m
+T_3phase_c = 3 * T_p_c ; // Output torque for 3-phase in N.m
+
+// Display the results
+disp("Example 6-8 Solution : ");
+pi = %pi;
+printf(" \n Slight variations in the answers are due to value of pi = %f ",pi);
+printf(" \n and omega = %f, which are slightly different as in the textbook.\n",omega);
+printf(" \n a: T_p = %d lb-ft ",T_p_a);
+printf(" \n T_3phase = %d lb-ft \n", T_3phase_a);
+
+printf(" \n b: T_p = %.1f N-m ",T_p_b);
+printf(" \n T_3phase = %.1f N-m \n", T_3phase_b);
+
+printf(" \n c: T_p = %.1f N-m ",T_p_c);
+printf(" \n T_3phase = %.1f N-m \n", T_3phase_c);
+printf(" \n Answers from cases b and c almost tally each other ");
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