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/EX4.10/Chapter4_Example10.sce | 23 +++++++++++++++++++++++
 1 file changed, 23 insertions(+)
 create mode 100755 2090/CH4/EX4.10/Chapter4_Example10.sce

(limited to '2090/CH4/EX4.10/Chapter4_Example10.sce')

diff --git a/2090/CH4/EX4.10/Chapter4_Example10.sce b/2090/CH4/EX4.10/Chapter4_Example10.sce
new file mode 100755
index 000000000..371862643
--- /dev/null
+++ b/2090/CH4/EX4.10/Chapter4_Example10.sce
@@ -0,0 +1,23 @@
+clc
+clear
+//Input data
+r=7.8;//Compression ratio 
+p=1;//The pressure at the start of compression in atm
+T1=335;//The temperature at the start of compression in K
+
+//Calculations
+W1=100;//Isentropic compression function for T1 in J/kg air K 
+W2=W1-(292*log(1/r));//Isentropic compression function in J/kg air K 
+T2=645;//The temperature corresponding to isentropic compression function in J/kg air K 
+V1=(292*T1)/(p*10^5);//Volume at initial in m^3/kg air 
+p2=p*(T2/T1)*r;//The pressure at the end of compression stroke in atm 
+V2=V1/r;//The volume per unit mass of air at the end of the compression stroke in m^3/kg air 
+U1=35;//Internal energy corresponding to temp T1 in kJ/kg air 
+U2=310;//Internal energy corresponding to temp T2 in kJ/kg air 
+W=U2-U1;//Work input during compression in kJ/kg air 
+E1=120;//Isentropic compression function at T1 
+E2=910;//Isentropic compression function at T2 
+p21=[exp((E2-E1)/292)];//The pressure at the end of compression stroke in atm 
+
+//Output
+printf('(a)At the end of the compression stroke, \n  The temperature is %3.0f K \n The pressure is %3.0f atm \n The volume per unit mass of air is %3.3f m^3/kg air \n The pressure is %3.0f atm \n (b)The work input during compression is %3.0f kJ/kg air ',T2,p2,V2,p21,W)
-- 
cgit