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| author | priyanka | 2015-06-24 15:03:17 +0530 |
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| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1092/CH8/EX8.5 | |
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| -rwxr-xr-x | 1092/CH8/EX8.5/Example8_5.sce | 51 |
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diff --git a/1092/CH8/EX8.5/Example8_5.sce b/1092/CH8/EX8.5/Example8_5.sce new file mode 100755 index 000000000..db33e8140 --- /dev/null +++ b/1092/CH8/EX8.5/Example8_5.sce @@ -0,0 +1,51 @@ +// Electric Machinery and Transformers
+// Irving L kosow
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 8: AC DYNAMO TORQUE RELATIONS - SYNCHRONOUS MOTORS
+// Example 8-5
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+// Y-connected synchronous dynamo
+P = 2 ; // No. of poles
+hp = 1000 ; // power rating of the synchronous motor in hp
+V_L = 6000 ; // Line voltage in volt
+f = 60 ; // Frequency in Hz
+R_a = 0.52 ; // Effective armature resistance in ohm
+X_s = 4.2 ; // Synchronous reactance in ohm
+P_t = 811 ; // Input power in kW
+PF = 0.8 ; // Power factor leading
+
+// Calculated values
+E_gp = 3687 ; // Generated voltage/phase in volt
+V_p = V_L / sqrt(3); // Phase voltage in volt
+E_r = 412.8 ; // Resultant EMF across armature/phase in volt
+deba = 119.81 ; // Difference angle at 0.8 leading PF in degrees
+theta = 36.87 ; // Power factor angle in degrees
+IaXs = 409.7 ; // Voltage drop across synchronous reactance in volt
+IaRa = 50.74 ; // Voltage drop across armature resistance in volt
+
+// Calculations
+
+// Torque angle alpha in degrees calculated by different Eqns
+// case a
+alpha1 = acosd( ( E_gp^2 + V_p^2 - E_r^2 ) / ( 2*E_gp*V_p ) ); // Eq.8-12
+
+// case b
+alpha2 = asind( ( E_r * sind(deba) ) / ( E_gp ) ); // Eq.8-13
+
+// case c
+alpha3 = theta - atand( (V_p*sind(theta) + IaXs) / (V_p*cosd(theta) - IaRa) );// Eq.8-14
+
+// Display the results
+disp("Example 8-5 Solution : ");
+printf(" \n a: Using Eq.(8-12) \n alpha = %.2f degrees \n ", alpha1 );
+
+printf(" \n b: Using Eq.(8-13) \n alpha = %.2f degrees \n ", alpha2 );
+
+printf(" \n c: Using Eq.(8-14) \n alpha = %.2f degrees \n ", alpha3 );
+printf(" \n Slight variation in case c alpha is due to tan inverse value ");
+printf(" \n which was calulated to be 42.445604 degrees, instead of 42.44 degrees(textbook).")
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