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 --- 172/CH8/EX8.4/ex4.sce | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) create mode 100755 172/CH8/EX8.4/ex4.sce (limited to '172/CH8/EX8.4/ex4.sce') diff --git a/172/CH8/EX8.4/ex4.sce b/172/CH8/EX8.4/ex4.sce new file mode 100755 index 000000000..9399182cc --- /dev/null +++ b/172/CH8/EX8.4/ex4.sce @@ -0,0 +1,20 @@ +//example 4 +//entropy change with different assumptions +clear +clc +T1=300 //initial temperature in kelvins +T2=1500 //final temperature in kelvins +P1=200 //initial pressure in kPa +P2=150 //final pressure in kPa +R=0.2598 // in kJ/kg-K +Cp=0.922 //specific heat in kJ/kg-K at constant pressure +dsT2=8.0649 //in kJ/kg-K +dsT1=6.4168 //in kJ/kg-K +dS1=dsT2-dsT1-R*log(P2/P1) //entropy change calculated using ideal gas tables +dS2=integrate('0.88/x-0.0001+0.54*x-0.33*x^2','x',0.3,1.5)-R*log(P2/P1) //entropy change calculated using empirical equation +dS3=Cp*log(T2/T1)-R*log(P2/P1) //entropy change assuming constant specific heat in kJ/kg-K +dS4=1.0767*log(T2/T1)+0.0747 //entropy change assuming specific heat is constant at its value at 990K +printf("\n hence,change in entropy using ideal gas tables is dS1=%.4f kJ/kg-K.\n",dS1) +printf("\n hence,change in entropy using empirical equation is dS2=%.4f kJ/kg-K.\n",dS2) +printf("\n hence,change in entropy using the value of specific heat at 300K is dS3=%.4f kJ/kg-K.\n",dS3) +printf("\n hence,change in entropy assuming specific heat is constant at its value at 900K is dS4=%.4f kJ/kg-K.\n",dS4) \ No newline at end of file -- cgit