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diff --git a/479/CH3/EX3.13/Example_3_13.sce b/479/CH3/EX3.13/Example_3_13.sce
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+//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.13
+clear;
+clc;
+
+//Given
+//Q=W=delPE=delKE=0;
+//M2=0; no exit stream
+Ti = 288;//initial temperature in K
+H = 7*Ti;//enthalpy of air in Kcal/Kgmole
+Ei = 5*Ti;// initial internal energy of air in Kcal/Kgmole
+//Ef=5*Tf;Final internal energy of air in Kcal/Kgmole
+Pi = 0.3;//initial pressure in atm
+V = 0.57;//volume of the tank in m^3
+R = 848;//gas constant in mKgf/Kg mole K
+Pf = 1;//final prssure in atm
+
+//To calculate the final weight and the final temperature of the air in the tank
+Mi = (Pi*V*1.03*10^4)/(R*Ti);//initial quantity of air in tank in Kg mole
+//Tf=(Pf*V*1.033*10^4)/(Mf*R)..(a) final temperature,Mf=final quantity of air in tank in Kg mole
+//M1=Mf-Mi..(b) M1 is mass of steam added in Kg mole
+//H*M1=(Ef*Mf)-(Ei*Mi)
+//H*M1=((5*Pf*V*1.033*10^4)/(Tf*R))*Tf-(Ei*Mi)...(c)
+A = [1 -1;0 -H];
+B = [Mi;((Ei*Mi)-((5*Pf*V*1.03*10^4)/R))];
+x = A\B;
+Mf = x(1);
+mprintf('The final weight of air in the tank is%f Kg',Mf);
+
+Tf = (Pf*V*1.03*10^4)/(Mf*R);
+mprintf('\n The final temperature of air in the tank is %f K',Tf);
+//end
+\ No newline at end of file
diff --git a/479/CH3/EX3.13/Example_3_13.txt b/479/CH3/EX3.13/Example_3_13.txt
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+For example 3.13,
+ the answers are given in the book, Mf = 0.017 Kgmole & Tf = 406K.
+ But i am getting Mf = 0.019232 Kgmole & Tf = 360K.	
+ There might be some calculation mistake in the book.