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

---
 914/CH5/EX5.12/ex5_12.sce | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)
 create mode 100755 914/CH5/EX5.12/ex5_12.sce

(limited to '914/CH5/EX5.12/ex5_12.sce')

diff --git a/914/CH5/EX5.12/ex5_12.sce b/914/CH5/EX5.12/ex5_12.sce
new file mode 100755
index 000000000..732e32ba4
--- /dev/null
+++ b/914/CH5/EX5.12/ex5_12.sce
@@ -0,0 +1,20 @@
+clc;
+warning("off");
+printf("\n\n example5.12 - pg178");
+// given
+T=0+273.15;  //[K] - temperature in Kelvins
+pa2=1.5;  //[atm] - partial presuure of a at point2
+pa1=0.5;  //[atm] - partial pressure of a at point 1
+z2=20;  //[cm] - position of point 2 from reference point
+z1=0;  //[cm] - position of point1 from reference point
+p=2;  //[atm] - total pressure
+d=1;  //[cm] - diameter
+D=0.275;  //[cm^2/sec] - diffusion coefficient
+A=(%pi*((d)^2))/4;
+R=0.082057;  //[atm*m^3*kmol^-1*K^-1] - gas constant
+k=0.75;
+// using the formula (Na/A)=-(D/(R*T*(z2-z1)))*ln((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))
+NabyA=-(D/(R*T*(z2-z1)))*(2*0.7854)*log((1-(pa2/p)*(1-k))/(1-(pa1/p)*(1-k)))/(10^6);
+printf("\n\n (Na/A)=%ekmol/sec",NabyA);
+printf("\n Note that this answer is larger than the rate for equimolar counter diffusion but smaller tahn the rate for diffusion through a stagnant film.Sometimes the rate for diffusin through a stagnant film can be considered as an upper bound, if k ties between zero and one");
+
-- 
cgit