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

---
 2090/CH4/EX4.3/Chapter4_Example3.sce | 25 +++++++++++++++++++++++++
 1 file changed, 25 insertions(+)
 create mode 100755 2090/CH4/EX4.3/Chapter4_Example3.sce

(limited to '2090/CH4/EX4.3')

diff --git a/2090/CH4/EX4.3/Chapter4_Example3.sce b/2090/CH4/EX4.3/Chapter4_Example3.sce
new file mode 100755
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+++ b/2090/CH4/EX4.3/Chapter4_Example3.sce
@@ -0,0 +1,25 @@
+clc
+clear
+//Input data
+r=8;//The compression ratio
+af=15;//Air/fuel ratio
+p1=1;//The pressure at the beginning of a compression stroke in bar
+t=60;//The temperature at the beginning of a compression stroke in degree centigrade
+cv=44000;//The calorific value of the fuel in kJ/kg
+n=1.32;//The index of the compression 
+Cv=0.717;//specific heat at constant volume in kJ/kgK
+
+//Calculations
+T1=t+273;//The temperature at the beginning of a compression stroke in K
+p2=p1*(r)^n;//The pressure at the end of a compression stroke in bar
+T2=T1*r^(n-1);//The temperature at the end of a compression stroke in K
+f=(1/(af+1));//The amount of fuel present in 1 kg of mixture in kg
+a=(af/(af+1));//The amount of air present in 1 kg of mixture in kg
+q23=cv/(af+1);//The heat transfer during process 2-3 per kg of mixture in kJ/kg
+T3=[[-10430+[(10430)^2+(4*494.8*10^5)]^(1/2)]/2];//The temperature at point 3 in K
+p3=(T3/T1)*(r)*p1;//The pressure at point 3 in bar
+T31=(q23/Cv)+T2;//The pressure at point 3 in K
+p31=(T31/T1)*r*p1;//The pressure at point 3 in bar
+
+//Output
+printf('(a) The Maximum temperature in the cylinder T3 = %3.0f K \n The Maximum pressure in the cylinder P3 = %3.0f bar \n (b)With constant value of Cv \n The Maximum temperature in the cylinder T3 = %3.0f K \n The Maximum pressure in the cylinder P3 = %3.1f bar ',T3,p3,T31,p31)
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cgit