initial commit / add all books

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+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");

+