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

---
 572/CH7/EX7.1/c7_1.sce | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)
 create mode 100755 572/CH7/EX7.1/c7_1.sce

(limited to '572/CH7/EX7.1')

diff --git a/572/CH7/EX7.1/c7_1.sce b/572/CH7/EX7.1/c7_1.sce
new file mode 100755
index 000000000..23015cfec
--- /dev/null
+++ b/572/CH7/EX7.1/c7_1.sce
@@ -0,0 +1,21 @@
+// (7.1)  A cylinder of an internal combustion engine contains 2450 cm3 of gaseous combustion products at a pressure of 7 bar and a temperature of 867C just before the exhaust valve opens. Determine the specific exergy of the gas, in kJ/kg. Ignore the effects of motion and gravity, and model the combustion products as air as an ideal gas. Take T0 = 300 K (27C) and p0=  1.013 bar.
+
+
+//solution
+
+
+//variable initialization
+v = 2450                                                    //volume of gaseous products in cm^3
+P = 7                                                       //pressure of gaseous product in bar
+T = 867                                                     //temperature of gaseous product in degree celcius
+T0 = 300                                                    //in kelvin
+P0 = 1.013                                                  //in bar
+
+//from table A-22
+u = 880.35                                                  //in kj/kg
+u0 = 214.07                                                 //in kj/kg
+s0(T) = 3.11883                                             //in kj/kg.k
+s0(T0) = 1.70203                                            //in kj/kg.k
+
+e = (u-u0) + (P0*(8.314/28.97)*[((T+273)/P)-(T0/P0)]) - T0*[s0(T)-s0(T0)-(8.314/28.97)*log(P/P0)]           //in kj/kg
+printf('the specific exergy of the gas, in kJ/kg is \n\t e = %f',e)
\ No newline at end of file
-- 
cgit