SOLVE ONLY USING MATHLAB Production of propylene glycol by hydrolysis of propyle

Question

SOLVE ONLY USING MATHLAB

Production of propylene glycol by hydrolysis of propylene oxide is carried out in an adiabatic continuous-stirred tank reactor (CSTR) [Fogler, Elements of Chemical Reaction Engineering, Prentice-Hall, Englewood Cliffs, 1992]. The reaction can takes place at low temperature when sulfuric acid is used as catalyst: propylene oxide (O) + water (VV) propylene glycol The standard heat of reaction AHR at TR = 68°F) for this reac tion is -36400 Btu/lIbmol propylene oxide. The feed containing 36 mol% propylene oxide and 64 mol% methanol (M) enters the reactor at 75°FOT,n)

Explanation / Answer

fun = @ourfun;
x0 = [0,0];

x = fsolve(fun,x0)

function F=ourfun(x)

%v is volumetric flow rate (convert to ft3/sec)
v=0.09055;
%Vol is the volume of the reactor (convert gallons to ft3)
Vol=46.78;
%E is the Activation of Energy
E=32400;
%R is Universal Gas Constant
R=1.987;
%Cpin is the heat capacity of the mixture entering the reactor
Cpin=400;
%DelCp is the change in the heat capacity per mol of propylene oxide
%reacted
DelCp=-7;
%Tref is the reference temperature in R
Tref=527.67;
%Tin is the inlet temperature in R
Tin=534.67;
%delHr is the Heat of the Reaction
delHr=-36400;

F(1)=((16.96*10^12*exp(-E/(R*x(2))))*(Vol/v))/(1+(16.96*10^12*exp(-E/(R*x(2))))*(Vol/v))-x(1);
F(2)=((Cpin*(x(2)-Tin))/(-delHr-(DelCp*(x(2)-Tref))))-x(1);
end

This is the MATLAB code for which the result comes out as

x=1.00

T= 167 degree Fahrenheit

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