From f35ea80659b6a49d1bb2ce1d7d002583f3f40947 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:38:01 +0530
Subject: updated the code

---
 479/CH14/EX14.12/Example_14_12.sce | 72 +++++++++++++++++++-------------------
 1 file changed, 36 insertions(+), 36 deletions(-)

(limited to '479/CH14/EX14.12')

diff --git a/479/CH14/EX14.12/Example_14_12.sce b/479/CH14/EX14.12/Example_14_12.sce
index da72bbb33..def054a07 100755
--- a/479/CH14/EX14.12/Example_14_12.sce
+++ b/479/CH14/EX14.12/Example_14_12.sce
@@ -1,37 +1,37 @@
-//Chemical Engineering Thermodynamics
-//Chapter 14
-//Thermodynamics of Chemical Reactions
-
-//Example 14.12
-clear;
-clc;
-
-//Given
-//C + 2H2 - CH4
-//Basis: 1 Kgmole of C fed
-T = 1000;//Temperature in K
-P1 = 2;//Pressure in atm
-del_F = 4580;//Standard free energy in Kcal/Kgmole
-
-
-//To Calculate the maximum CH4 concentration under the condition of 2 atm and the  quantity of methane obtained if pressure is 1 atm
-Ka = %e^(-del_F/(R*T));//Equilibrium constant
-//In relation (d) (page no 339) p_H2 = p (say)
-p = poly(0,'p');
-q = Ka*(p^2)+p-P1;
-r = roots(q);
-p_H2 = r(2);//partial pressure of H2
-p_CH4 = P1-p_H2;//partial pressure of CH4
-X_H2 = p_H2*100/P1;//mole percent of H2
-X_CH4 = p_CH4*100/P1;//mole percent of CH4
-mprintf('Under the conditions of 2 atm and 1000 K,the maximum CH4 concentration is %f percent and further increase is not pssible',X_CH4);
-//Now.pressure has become
-P2 = 1;//in atm
-q = Ka*(p^2)+p-P2;
-r = roots(q);
-p_H2 = r(2);//partial pressure of H2
-p_CH4 = P2-p_H2;//partial pressure of CH4
-X_H2 = p_H2*100/P2;//mole percent of H2
-X_CH4 = p_CH4*100/P2;//mole percent of CH4
-mprintf('\n\n Under the conditions of 1 atm and 1000 K,Methane = %f percent and Hydrogen = %f percent',X_CH4,X_H2);
+//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.12
+clear;
+clc;
+
+//Given
+//C + 2H2 - CH4
+//Basis: 1 Kgmole of C fed
+T = 1000;//Temperature in K
+P1 = 2;//Pressure in atm
+del_F = 4580;//Standard free energy in Kcal/Kgmole
+R = 1.98;
+
+//To Calculate the maximum CH4 concentration under the condition of 2 atm and the  quantity of methane obtained if pressure is 1 atm
+Ka = %e^(-del_F/(R*T));//Equilibrium constant
+//In relation (d) (page no 339) p_H2 = p (say)
+p = poly(0,'p');
+q = Ka*(p^2)+p-P1;
+r = roots(q);
+p_H2 = r(2);//partial pressure of H2
+p_CH4 = P1-p_H2;//partial pressure of CH4
+X_H2 = p_H2*100/P1;//mole percent of H2
+X_CH4 = p_CH4*100/P1;//mole percent of CH4
+mprintf('Under the conditions of 2 atm and 1000 K,the maximum CH4 concentration is %f percent and further increase is not pssible',X_CH4);
+//Now.pressure has become
+P2 = 1;//in atm
+q = Ka*(p^2)+p-P2;
+r = roots(q);
+p_H2 = r(2);//partial pressure of H2
+p_CH4 = P2-p_H2;//partial pressure of CH4
+X_H2 = p_H2*100/P2;//mole percent of H2
+X_CH4 = p_CH4*100/P2;//mole percent of CH4
+mprintf('\n\n Under the conditions of 1 atm and 1000 K,Methane = %f percent and Hydrogen = %f percent',X_CH4,X_H2);
 //end
\ No newline at end of file
-- 
cgit