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

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 1092/CH7/EX7.1/Example7_1.sce | 74 +++++++++++++++++++++++++++++++++++++++++++
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+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 7: PARALLEL OPERATION
+// Example 7-1
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+R_sh = 120 ; // Shunt field resistance in ohm
+R_a = 0.1 ; // Armature resistance in ohm
+V_L = 120 ; // Line voltage in volt
+E_g1 = 125 ; // Generated voltage by dynamo A
+E_g2 = 120 ; // Generated voltage by dynamo B
+E_g3 = 114 ; // Generated voltage by dynamo C
+
+// Calculations
+// case a
+// 1:
+I_gA = ( E_g1 - V_L ) / R_a ; // Current in the generating source A ( in  A)
+I_f = V_L / R_sh ; // Shunt field current in A
+I_a1 = I_gA + I_f ; // Armature current in A for generator A
+I_L1 = I_gA ; // Current delivered by dynamo A to the bus in A
+
+// 2:
+I_gB =  ( E_g2 - V_L ) / R_a ; // Current in the generating source B ( in  A)
+I_a2 = I_gB + I_f ; // Armature current in A for generator B
+I_L2 = I_gB ; // Current delivered by dynamo B to the bus in A
+
+// 3:
+I_gC =  ( V_L - E_g3 ) / R_a ; // Current in the generating source C ( in  A)
+I_a3 = I_gC ; // Armature current in A for generator C
+I_L3 = I_gC + I_f ; // Current delivered by dynamo C to the bus in A
+
+// case b
+// 1:
+P_LA = V_L * I_L1 ; // Power delivered to the bus by dynamo A in W
+P_gA = E_g1 * I_a1 ; // Power generated by dynamo A
+
+// 2:
+P_LB = V_L * I_L2 ; // Power delivered to the bus by dynamo B in W
+P_gB = E_g2 * I_a2 ; // Power generated by dynamo B
+
+// 3:
+P_LC = V_L * I_L3 ; // Power delivered to the bus by dynamo C in W
+P_gC = E_g3 * I_a3 ; // Power generated by dynamo C
+
+// Display the results
+disp("Example 7-1 Solution : ");
+printf(" \n a: 1. I_gA = %d A \t I_f = %d A ", I_gA,I_f );
+printf(" \n       Thus,dynamo A delivers %d A to the bus and has an armature", I_gA);
+printf(" \n       current of %d A + %d A = %d \n", I_gA,I_f,I_a1 );
+
+printf(" \n    2. I_gB = %d A ", I_gB);
+printf(" \n       Thus, dynamo B is floating and has as armature & field current of %d A \n",I_f);
+
+printf(" \n    3. I_gC = %d A ",I_gC);
+printf(" \n       Dynamo C receives %d A from the bus & has an armature current of %d A\n",I_L3,I_a3);
+
+printf(" \n b: 1. Power delivered to the bus by dynamo A is : ");
+printf(" \n       P_LA = %d W ",P_LA);
+printf(" \n       Power generated by dynamo A is \n       P_gA = %d W \n",P_gA);
+ 
+printf(" \n    2. Since dynamo B neither delivers power to nor receives power from the bus, ");
+printf(" \n       P_B = %d W ",P_LB);
+printf(" \n       Power generated by dynamo B,to excite its field, is");
+printf(" \n       P_gB = %d W \n ", P_gB); 
+
+printf(" \n    3. Power delivered by the bus to dynamo C is ");
+printf(" \n       P_LC = %d W ", P_LC);
+printf(" \n       while the internal power delivered in the direction of rotation");
+printf(" \n       of its prime mover to aid rotation is \n       P_gC = %d W", P_gC );
-- 
cgit