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diff --git a/926/CH5/EX5.1/Chapter5_Example1.sce b/926/CH5/EX5.1/Chapter5_Example1.sce new file mode 100644 index 000000000..d2ca16bb0 --- /dev/null +++ b/926/CH5/EX5.1/Chapter5_Example1.sce @@ -0,0 +1,52 @@ +//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504
+
+//Chapter-5, Illustration 1, Page 111
+//Title: Calculation of composition of a saturated mixture
+//=============================================================================
+clear
+clc
+
+//INPUT
+P = [442,745 760]; //Vapor pressure of ethyl ether, working pressure and standard pressure in mm Hg
+T1 = [20 0]; //Temperature of system and standard temperature in degree C
+V = 1; //cu ft of mixture(Basis of calculation for part(a) )
+MW = [74 28]; //Molecular weight of ethyl ether and nitrogen in lb/lb-mole
+
+//CALCULATIONS
+//Part(a)
+V1 = V*(P(1)/P(2)); //Pure component volume of vapor in cu ft
+v1 = V1*100; //Composition by volume of Ether vapor
+v2 = (1-V1)*100; //Composition by volume of Nitrogen
+//Part(b)
+W1 = V1*MW(1); //lb of ether vapor present
+W2 = (1-V1)*MW(2); //lb of Nitrogen present
+W = W1+W2; //Total weight of mixture in lb
+w1 = W1*100/W; //Composition by weight of Ether vapor
+w2 = W2*100/W; //Composition by weight of Nitrogen
+//Part(c)
+T = T1+273; //Converting temperature in K
+V2 = 359*(P(3)/P(2))*(T(1)/T(2)); //cu ft of mixture
+w3 = W1/V2; //lb of vapor per cu ft of mixture
+//Part(d)
+w4 = W1/W2; //lb of vapor per lb lb of nitrogen
+//Part(e)
+v3 = V1/(1-V1); //lb mole of vapor per lb moles of Nitrogen
+
+//OUTPUT
+// Console output
+mprintf('\n (a) Composition by Volume \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %% ',v1,v2);
+mprintf('\n (b) Composition by Weight \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %% ',w1,w2);
+mprintf('\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mprintf('\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mprintf('\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+
+// File output
+fd= mopen('.\Chapter5_Example1_Output.txt','w');
+mfprintf(fd,'\n (a) Composition by Volume \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %%',v1,v2);
+mfprintf(fd,'\n (b) Composition by Weight \n Ether Vapor %3.1f %% \n Nitrogen %3.1f %%',w1,w2);
+mfprintf(fd,'\n (c) Weight of ether per cu ft of mixture is %4.3f lb ',w3);
+mfprintf(fd,'\n (d) Weight of vapor per lb Nitrogen is %3.2f lb ',w4);
+mfprintf(fd,'\n (e) Moles of vapor per mole of nitrogen is %4.3f',v3);
+mclose(fd);
+
+//=============================END OF PROGRAM==================================
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