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

---
 647/CH4/EX4.7/Example4_7.sce | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100755 647/CH4/EX4.7/Example4_7.sce

(limited to '647/CH4/EX4.7')

diff --git a/647/CH4/EX4.7/Example4_7.sce b/647/CH4/EX4.7/Example4_7.sce
new file mode 100755
index 000000000..fbfbde81c
--- /dev/null
+++ b/647/CH4/EX4.7/Example4_7.sce
@@ -0,0 +1,26 @@
+clear;
+clc;
+
+// Example: 4.7
+// Page: 125
+
+printf("Example: 4.7 - Page: 125\n\n");
+
+// Solution
+
+//*****Data*****//
+// HC : Heat of Combustion
+HC_C2H2 = -310600; // [cal]
+//**************//
+
+// C2H2 + (5/2)O2 = 2CO2 + H2O
+Q = -HC_C2H2;// [cal]
+// The gases present in the flame zone after combustion are carbon dioxide, water vapor and the unreacted nitrogen of the air.
+// Since (5/2) mole of oxygen were required for combustion, nitrogen required would be 10 mol.
+// Hence the composition of the resultant gas would be 2 mol CO2, 1  ol H2 & 10 mol N2.
+// Q = integrate('Cp(T)','T',T,298);
+// On integrating we get:
+// Q = 84.52*(T - 298) + 18.3*10^(-3)*(T^2 - 298^2)
+deff('[y] = f(T)','y = Q - 84.52*(T - 298) - 18.3*10^(-3)*(T^2 - 298^2)');
+T = fsolve(7,f);// [K]
+printf("The maximum attainable temperature is %.1f K",T);
\ No newline at end of file
-- 
cgit