From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 1092/CH8/EX8.5/Example8_5.sce | 51 +++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) create mode 100755 1092/CH8/EX8.5/Example8_5.sce (limited to '1092/CH8/EX8.5') 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).") -- cgit