Problem 9.35 in Felder

Stream Compounds Reaction

1 Feed 1 C2H6 1 C2H6 -> C2H4 + H2

2 Product 2 C2H4

3 H2

n1 _____________ n2

-------->| | ------------>

n11 | reactor | n21, n22

|_____________| n23

Using the given data in the problem statement.

restart;

ns:=[[n11,0,0],[n21,n22,n23]];

sigma:=[-1,1,1];

n2:=0;

for k from 1 to 3 do

ns[2,k]:=ns[1,k]+sigma[k]*xi;

n2:=n2+ns[2,k];

od;

x:=[0,0,0];

for k from 1 to 3 do

x[k]:=ns[2,k]/n2;

od;

eq:=Kp=x[2]*x[3]*P/x[1];

s:=solve(eq,xi);

f:=s[1]/n11;

fbook:=sqrt(Kp/(P+Kp));

simplify(f-fbook,assume=positive);This proves eq 2 in the problem: part a)

Kpf:=T->7.28e6*atm*exp(-17000*K/T);Given as the Equilibrium "Constant" for the reaction

Cp2:=T->9.419*J/mol/K+.1147*J/mol/K^2*T;Cp3:=T->26.9*J/mol/K+4.167e-3*J/mol/K^2*T;Cp1:=T->11.35*J/mol/K+.1392*J/mol/K^2*T;Cps in J/mol/K with T in K

H1:=T->H1Ref+int(Cp1(T1),T1=1273*K...T);The enthalpy of ethane in kJ/mol

J:=kJ/1000.;H1(T);

H1(1273*K);

H2:=T->H2Ref+int(Cp2(T1),T1=1273*K...T);Enthalpy of C2H4

H3:=T->H3Ref+int(Cp3(T1),T1=1273*K...T)/1000.;Enthalpy of H2

dHref:=145.6*kJ/mol;eq:=H2Ref+H3Ref-H1Ref=dHref;The heat of reaction at 1273K is given as 145.6kJ/mol

H1Ref:=solve(eq,H1Ref);

q:=ns[2,1]*H1(T2)+ns[2,2]*H2(T2)+ns[2,3]*H3(T2)-ns[1,1]*H1Ref;An energy balance for the reactor with the feed at 1273 and the products at T2. If q=0, the reactor will be adiabatic.

xi:=n11*fbook;

eq1:=simplify((Kp+P)*q*mol/n11/atm/kJ,assume=positive);

Kp:=Kpf(T2);

eq1:=simplify(eq1,assume=positive);

eq2:=unapply(eq1,P,T2);

eq2(.1*atm,800*K);

eq3:=eq2(p*atm,t*K);

eq4:=unapply(simplify(eq3,assume=positive),p,t);

eq4(.1,800);

plot(eq4(0.1,t),t=600...1000);

T01:=fsolve(eq4(0.1,t),t,800...900);

T1:=fsolve(eq4(1,t),t,800...1000);

T10:=fsolve(eq4(10,t),t,900...1100);

T001:=fsolve(eq4(.01,t),t,700...900);

fract:=(P,T)->sqrt(Kpf(T)/(Kpf(T)+P));

f001:=fract(.01*atm,T001*K);

f01:=fract(.1*atm,T01*K);

f1:=fract(atm,T1*K);

f10:=fract(10*atm,T10*K);