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

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 497/CH14/EX14.2/Chap14_Ex2.sce | 72 ++++++++++++++++++++++++++++++++++++++++++
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+//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491
+
+//Chapter-14, Example 2, Page 344
+//Title: Flow with Elutriation and Change in Density of Solids
+//==========================================================================================================
+
+clear
+clc
+
+//INPUT
+dt=4;//Diameter of reactor in m
+ephsilonm=0.4;//Void fraction of static bed
+rhos=2500;//Density of solid in the bed in kg/m^3
+Lm=1.2;//Height of static bed in m
+Fo=3000;//Feed rate in kg/hr
+beta1=1.2;//Increase in density of solids
+dp=[3;4;5;6;7;8;9;10;11;12;3;14;16;18;20;22;24;26;28;30]*10^-2;//Size of particles in mm
+po=[0;0.3;0.8;1.3;1.9;2.6;3.5;4.4;5.7;6.7;7.5;7.8;7.5;6.3;5.0;3.6;2.4;1.3;0.5;0];//Size distribution of solids in mm^-1
+k=[0;10;9.75;9.5;8.75;7.5;6.0;4.38;2.62;1.20;0.325;0;0;0;0;0;0;0;0;0]*10^-4;//Elutriation constant in s^-1
+pi=3.14;
+
+//CALCULATION
+W=(pi/4*dt^2)*Lm*(1-ephsilonm)*rhos;//Weight of solids in bed
+n=length(dp);
+i=1;
+F1guess=1000;//Guess value for F1
+F1c=2510:10:2700;
+while i<=n
+    function[fn]=solver_func(F1)//Function defined for solving the system
+        if k(i)==0  then    x(i)=0; break 
+                    else    x(i)=(po(i)/(W*k(i)/F1))*log(1+(W*k(i)/F1));         
+        end
+    fn=F1/(Lm*Fo)-x(i);
+    endfunction
+    [F1(i)]=fsolve(F1guess,solver_func,1E-6);//Using inbuilt function fsolve for solving Eqn.(20) for F1
+    c(i)=F1c(i)/(Lm*Fo);
+    if F1(i)==0 then a(i)=0;
+    else     a(i)=(po(i)/(W*k(i)/F1(i)))*log(1+(W*k(i)/F1(i)));
+    end    
+    i=i+1;
+end
+plot(F1,a,F1,c);
+xtitle('F1 vs a,c','F1','a,c');
+F1n=2500;//The point were both the curves meet
+F2=beta1*Fo-F1n;//Flow rate of the second leaving stream
+j=1;
+m=length(dp);
+while j<=m
+    p1(j)=(1/F1n)*((Fo*po(j))/(1+(W/F1n)*k(j)));//Size distribution of stream 1 in mm^-1 from Eqn.(16)
+    p2(j)=k(j)*W*p1(j)/F2;//Size distribution of stream 2 in mm^-1 from Eqn.(7)
+    if p1(j)==0 & p2(j)==0 then tbar(j)=0;
+    else  if  p1(j)==0 then tbar(j)=(W*p1(j))/(F2*p2(j));
+        else if  p2(j)==0 then tbar(j)=(W*p1(j))/(F1n*p1(j));
+            else tbar(j)=(W*p1(j))/(F1n*p1(j)+F2*p2(j));//Average time in hr from Eqn.(11)
+            end
+        end
+    end
+    j=j+1;
+end
+
+//OUTPUT
+printf('\nFlow rate of stream 1:%fkg/hr',F1n);
+printf('\nFlow rate of stream 2:%fkg/hr',F2);
+j=1;
+mprintf('\ntbar(hr)');
+while j<=m
+    mprintf('\n%f',tbar(j));
+    j=j+1;
+end
+
+//====================================END OF PROGRAM ======================================================
+//DISCLAIMER: The value obtained for tbar is deviating highly form the one given in textbook. However, the value obtained by manual calculation is close to the   ones obtained from the program.
\ No newline at end of file
-- 
cgit