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

---
 167/CH15/EX15.6/ex6.sce | 17 +++++++++++++++++
 1 file changed, 17 insertions(+)
 create mode 100755 167/CH15/EX15.6/ex6.sce

(limited to '167/CH15/EX15.6')

diff --git a/167/CH15/EX15.6/ex6.sce b/167/CH15/EX15.6/ex6.sce
new file mode 100755
index 000000000..343de4863
--- /dev/null
+++ b/167/CH15/EX15.6/ex6.sce
@@ -0,0 +1,17 @@
+//example 6
+//First-Law Analysis of Steady-Flow Combustion
+clear
+clc
+mair=7.5*4.76*29 //mass of air in kg
+mfuel=3*12+4*2 //mass of fuel in kg
+AF=mair/mfuel //air fuel ratio
+Mfuel=0.05 //Mass flow rate of fuel in kg/min
+Mair=AF*Mfuel //mass flow rate of air in kg/min
+qout=1*(-118910)+7.5*(8150-8682)+28.2*(0+8141-8669)-2.7*(-393520+71078 -9364)-0.3*(-110530+47517-8669)-4*(-241820+57999-9904)-2.65*(0+49292-8682)-28.2*(0+47073-8669) //in kJ/kmol C3H8
+disp('This heat is transferred from the combustion chamber for each kmol (44kg) of propane.therefore qout = qout/44 kJ/kg')
+qout=qout/44 //in kJ/kg propane
+M=0.05 //mass flow rate of liquid propane in kg/min
+Qout=M*qout //rate of heat transfer in kJ/min
+Qout=Qout/60 //rate of heat reansfer in kW
+printf("\n Hence, the mass flow rate of air is = %.2f kg/min. \n",Mair);
+printf("\n and the rate of heat transfer from combustion chamber is = %.2f kW. \n",Qout);
\ No newline at end of file
-- 
cgit