$(2)(40$ points$)$ Successive substitution method:

Problem $6\mathrm{.}6$: A synthetic$-$ammonia plant

A continuous bleed of $25$mol$\frac{s}{s}$ of mixture of $N_2,H_2,$ and Ar is provided

to prevent Ar concentration from increasing beyond a certain value.

If the incoming feed consists of $100$mol$\frac{s}{s}$ of $N_2,300$mol$\frac{s}{s}$ of $H_2,$ and $1mo\frac{l}{s}$ of Ar $,$

simulate this system by successive substitution

As the argon flow rate increases, the $\%$ conversion decreases as

Conversion percentage $(\%)=50*exp(-0\mathrm{.}016w)$

where w is the argon flow rate in $mo\frac{l}{s}.$

Please use the following trial values:

$A_4=15$mol$\frac{s}{s}$;$H_4=200$mol$\frac{s}{s}$;$N_4=60$mol$\frac{s}{s},$TOT$4=A_4+H_4+N_4$

$6-6.($Cont$.)$

Calculation:

Insert trial values of $\}\backslash$mp@subsup$\{\backslash$textrm$\{$H$\}\}\{4\}\{\},\backslash$mp@subsup$\{$H$\}\{4\}\{\}\backslash$mathrm$\{,$ and $\}\backslash$mp@subsup$\{\backslash$textrm$\{$F$\}\}\{4\}\{\},\backslash$mathrm$\{$ TOR$4$

TOT$4=0.$

$\backslash$Delta$\_\backslash$

$\backslash$mp@subsup$\{$H$\}\{1\}\{\}\}=100+\backslash$mp@subsup$\{\backslash$textrm$\{$N$\}\}\{4\}\{$

$\backslash$mp@subsup$\{$H$\}\{1\}\{\}=300+\backslash$mp@subsup$\{$H$\}\{8\}\{*\}$

$\backslash ==\mathrm{.}6(2$ mole of amonia per nol of $\backslash$mp@subsup$\{$n$\}\{2\}\{\})(\backslash$eta$)(\backslash$mp@subsup$\{$n$\}\{2\}\{\})$

A

N

H

A

H

n

If absolute magnitude of $((\backslash$

terainate

if $>\}>0\mathrm{.}00$

set TON$4=$ A $4/4+$ H

After $400100$ps$,$ A$6=188$ mol$/$a$,$ A

Determine the flow rates of mixture through the reactor and the rate of liquid ammonia

production $(A_6)$ in mols $\frac{?}{s}.$ Note that $A_6=2^{*}{}^{*}{N}_1$;

After the first iteration:

$A_4=$mol$\frac{s}{s}$;$H_4=$mol$\frac{s}{s}$;$N_4=$mol$\frac{s}{s},A_6=$mol$\frac{s}{s}$

After the second iteration:

$A_4=$mol$\frac{s}{s}$;$H_4=$mol$\frac{s}{s}$;$N_4=$mol$\frac{s}{s},A_6=$mol$\frac{s}{s}$

After the $3$rd iteration:

$A_4=$mol$\frac{s}{s}$;$H_4=$mol$\frac{s}{s}$;$N_4=$mol$\frac{s}{s},A_6=$mol$\frac{s}{s}$