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

---
 839/CH17/EX17.1/Example_17_1.sce | 40 ++++++++++++++++++++++++++++++++++++
 839/CH17/EX17.2/Example_17_2.sce | 44 ++++++++++++++++++++++++++++++++++++++++
 2 files changed, 84 insertions(+)
 create mode 100755 839/CH17/EX17.1/Example_17_1.sce
 create mode 100755 839/CH17/EX17.2/Example_17_2.sce

(limited to '839/CH17')

diff --git a/839/CH17/EX17.1/Example_17_1.sce b/839/CH17/EX17.1/Example_17_1.sce
new file mode 100755
index 000000000..95f209799
--- /dev/null
+++ b/839/CH17/EX17.1/Example_17_1.sce
@@ -0,0 +1,40 @@
+//clear//
+clear;
+clc;
+
+//Example 17.1
+//Given
+yb = 0.30;
+
+//Let
+Vb = 100; //[mol]
+Ace_in = yb*Vb; //[mol]
+Air_in = Vb-Ace_in; //[mol]
+//97 percent acetone aborbed, Acetone leaving is
+Ace_out = 0.03*Ace_in; //[mol]
+ya = Ace_out/(Air_in+Ace_out);
+//Acetone absorbed
+Ace_abs = Ace_in-Ace_out; //[mol]
+//10 percent acetone in the leaving solution and no acetone in the entering oil
+Lb = Ace_abs/0.1; //[mol]
+La = Lb-Ace_abs; //[mol]
+//To find out as intermediate point on the operating line, making an acetone balance
+//around the top part of the tower, assuming a particular value of yV the moles of 
+//acetone left in the gas. 
+for i=1:30  
+  y(i) = i/(i+Air_in); 
+//The moles of acetone lost by the gas in the secion, must equal to the moles gained by //the liquid 
+Ace_lost = i-Ace_out; //[mol]
+//Hence 
+x(i) = Ace_lost/(La+Ace_lost);
+end
+xe = linspace(0.001,0.15,100);
+ye = 1.9*xe;
+
+plot(x,y)
+plot(xe,ye,'r')
+xlabel('x')
+ylabel('y')
+legend('Operating line','Equilibrium line')
+title('Diagram Example 17.1')
+//The number of ideal stages determined from Fig is 4
diff --git a/839/CH17/EX17.2/Example_17_2.sce b/839/CH17/EX17.2/Example_17_2.sce
new file mode 100755
index 000000000..71a69f644
--- /dev/null
+++ b/839/CH17/EX17.2/Example_17_2.sce
@@ -0,0 +1,44 @@
+//clear//
+clear;
+clc;
+
+//Example 17.2
+//Given
+Nreal = 7;
+VbyL = 1.5;
+m = 0.8;
+yb = 0;
+xb_star = 0;
+//xb=0.1*xa;
+
+//(a)
+//Stripping Factor
+S = m*VbyL;
+//From an ammonia balance,
+//ya =0.9*xa/VbyL;
+//Also
+//xa_star = ya/m 
+//Using Eq.(17.28)
+//N = ln((xa-0.75*xa)/(0.1*xa-0))/ln(S)
+N = log(0.25/0.1)/log(S);
+disp(N,'Number of ideal trays required are')
+stage_eff = N/Nreal*100;
+disp('%',stage_eff,'Stage Efficiency is')
+
+//(b)
+VbyL = 2;
+S = m*VbyL;
+//Then,
+//Let A = (xa-xa_star)/xb
+A = exp(5.02);
+//Let 'f' be the fraction of NH3 removed. Then xb = (1-f)*xa.
+//By a material balance
+//y  = L/V*(xa-xb) = 1/2*(xa-(1-f)*xa)= 1/2*f*xa
+//xa_star = ya/m = 0.5*f*xa/0.8 = 0.625*f*xa
+//Thus, 
+//xa-xa_star = (1-0.625*f)*xa
+//Also,
+//xa-xa_star = 10.59*xb = 10.59*(1-f)*xa
+//from these
+f = 0.962
+disp('%',f,'percentage removal obtained in this case is')  
-- 
cgit