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

---
 23/CH6/EX6.3/Example_6_3.pdf | Bin 0 -> 37310 bytes
 23/CH6/EX6.3/Example_6_3.sce |  85 +++++++++++++++++++++++++++++++++++++++++++
 23/CH6/EX6.3/Figure_6_3.jpg  | Bin 0 -> 53418 bytes
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 create mode 100755 23/CH6/EX6.3/Example_6_3.pdf
 create mode 100755 23/CH6/EX6.3/Example_6_3.sce
 create mode 100755 23/CH6/EX6.3/Figure_6_3.jpg

(limited to '23/CH6/EX6.3')

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diff --git a/23/CH6/EX6.3/Example_6_3.sce b/23/CH6/EX6.3/Example_6_3.sce
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+clear;
+clc;
+
+//To find Approx Value
+function[A]=approx(V,n)
+  A=round(V*10^n)/10^n;//V-Value  n-To what place
+  funcprot(0)
+endfunction  
+
+function[Q]=ICPS(T0,T,A,B,C,D)
+  t=T/T0;
+  Q=((A)*log(t))+(((B*T0)+(((C*T0*T0)+(D/(t*t*T0*T0)))*(t+1)/2))*(t-1))
+  funcprot(0);
+endfunction
+
+function[Q]=ICPH(T0,T,A,B,C,D)
+  t=T/T0;
+  Q=(A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t^2)+t+1))+(D/(t*T0*T0)))*(T-T0)
+  funcprot(0);
+endfunction
+
+//Example 6.3
+//Caption : Program to find Entropy and Enthalpy of Saturated isobutane Vapor 
+
+//Given Values
+
+T0=300;//[K]
+T=360;//[K]
+R=8.314;
+P=15.14;//[bar]
+A=1.7765;
+B=33.037*10^-3;
+C=0;
+D=0;
+H0=18115;  //J/mol
+S0=295.976;//J/mol/K
+
+//Graph
+X=[0,0.10,0.50,2,4,6,8,10,12,14,15.41];
+Y1=[1.780,1.700,1.514,1.293,1.290,1.395,1.560,1.777,2.073,2.432,2.720];//[(dZ/dT)p/P]
+Y2=[2.590,2.470,2.186,1.759,1.591,1.544,1.552,1.592,1.658,1.750,1.835];//[-(Z-1)/P]
+subplot(1,2,1);
+plot2d(Y1,X);
+xgrid();
+xtitle("(a)","P(bar)","[(dZ/dT)p/P]X10^4(K^-1 bar^-1)");
+subplot(1,2,2);
+plot2d(Y2,X);
+xgrid();
+xtitle("(b)","P(bar)","[-(Z-1)/P]X10^2(bar^-1)");
+
+
+//Area Under the Curve (a)
+Y1=Y1*10^-4;
+A1=0;
+for i=2:11;
+  A1=A1+((X(i-1)-X(i))*Y1(i));
+end  
+disp('(X 10^-4) K^-1',A1*10000,'Area under the graph(a)')
+//Area Under the Curve (b)
+Y2=Y2*10^-2;
+A2=0;
+for i=2:11;
+  A2=A2+((X(i-1)-X(i))*Y2(i));
+end  
+disp(approx(A2,4),'Area under the graph(b)')
+
+
+K=A1*T;//Hr/RT
+//From Eqn(6.47)
+Hr=R*T*(K);//[J/mol]
+//From Eqn(6.48)
+Sr=R*(K-(A2));//[J/mol/K]
+
+//From Eqn(6.49) and Eqn(6.50)
+H1=R*ICPH(T0,T,A,B,C,D);
+S1=R*ICPS(T0,T,A,B,C,D);
+
+H=H0+H1+Hr;
+S=approx(S0+S1+Sr-(R*log(P)),3);
+
+disp('J/mol',H,'Enthalpy')
+disp('J/mol/K',S,'Entropy')
+disp('Note: The Answer is different with that of the Book because the Method Used to find the Area under the Graph is done by finding the area of small       Rectangles')
+
+//End
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