From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 1319/CH10/EX10.1/10_1.sce | 14 ++++++++++++++
 1 file changed, 14 insertions(+)
 create mode 100644 1319/CH10/EX10.1/10_1.sce

(limited to '1319/CH10/EX10.1/10_1.sce')

diff --git a/1319/CH10/EX10.1/10_1.sce b/1319/CH10/EX10.1/10_1.sce
new file mode 100644
index 000000000..724b42d54
--- /dev/null
+++ b/1319/CH10/EX10.1/10_1.sce
@@ -0,0 +1,14 @@
+//Determine the additional load which can be supplied
+
+clc;
+clear;
+
+printf('a) D.C two wire: \nPower transmitted by DC two wire = P and Voltage between the wires = V\n')
+printf('Copper Loss = 2 *(P/V)^2)*R ; where R is the resistance of each wire\n')
+printf('Per Unit Loss = 2*P*R/(V^2)\n\n')
+printf('b) 3 phase 3 wire: \nPower transmitted = P''\n ')
+printf('I'' = P''/(sqrt(3)*V) for unity p.f\n')
+printf('Copper Loss = 3*((P''/(sqrt(3)*V))^2)*R = ((P/V)^2)*R\n')
+printf('Per Unit Loss = P''*R/(V^2)\n\n')
+printf('Equating the per unit loss we have\n')
+printf('2*P*R/(V^2) = P''*R/(V^2) or P'' = 2P\n Which proves that 100 %% aditional pwer can be supplied by 3 phase 3 wire system\n')
-- 
cgit