Reactor + Separator control outlet concentration

[gfd43tg]

Homework Statement

Homework Equations

The Attempt at a Solution

Hello,

I get confused with input and output and state variables. Since we want to control $C_{Ao}$, I assume that needs to be the output variable in the vector $\vec {y}$. I think the inputs should be $C_{Ai}$ and $F_{i}$, since those are not something that we can control. I think mostly the state and output variables should be the same.

I am also convinced my prof. never did this problem in class, because I have no recollection and I always come to class (and I don't fall asleep!).

So for third order kinetics, I assume the reaction should be
$$3A \xrightarrow{k} P$$

I am unsure where I should do the mass balance, around just the reactor or the whole system (reactor and separator). My balance equations are ($V_{t}$ denotes volume of reactor and separator). I wonder if $V_{t}$ should be a constant? Or do I do a separate balance with control volumes being the reactor and separator?

State $\vec {x}$, Input $\vec {u}$, and output $\vec {y}$

$$ \vec{x} = \begin{bmatrix} C_{Ao} \\ C_{Po} \\ C_{A} \\ C_{P} \\ V \end{bmatrix} $$
$$ \vec{u} = \begin{bmatrix} C_{Ai} \\ F_{i} \end{bmatrix} $$
$$ \vec{y} = \begin{bmatrix} C_{Ao} \\ C_{Po} \\ C_{A} \\ C_{P} \\ V \end{bmatrix} $$

Balance around the reactor
$$ \frac {dV}{dt} = F_{i} + F_{r} - F $$
$$ \frac {d(C_{A}V)}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3} $$
$$ \frac {d(C_{P}V)}{dt} = F_{r}C_{Pr} - FC_{P} + \frac {kC_{A}^{3}}{3} $$

Balance around reactor + separator
$$ \frac {d(C_{Ao}V_{t})}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - F_{o}C_{Ao} - kC_{A}^{3} $$
$$ \frac {d(C_{Po}V_{t})}{dt} = F_{r}C_{Pr} - F_{o}C_{Po} + \frac {kC_{A}^{3}}{3} $$

Am I on the right track here?

 

[gfd43tg]

The instructor posted a note (I was right about the problem not being done in class), so now I have some more information to move forward with the problem.

Please assume that the separator is operating at steady-state and that species A only reacts in the reactor (the reaction does not continue in the separator).

Please use CAo as your output variable (as specified in the problem statement) and classify the following as input or state variables: Fi, Fr, Fo, CA, CAi, V.

First off, I don't need to worry about the product, just species A. Second, the separator is acting at steady state. The output is $C_{Ao}$, and the states are $C_{A}$ and $V$.So my governing equations and variable vectors are
$$ \vec{x} = \begin{bmatrix} C_{A} \\ V \end{bmatrix} $$
$$ \vec{u} = \begin{bmatrix} F_{i} \\ C_{Ai} \\ F_{o} \\ F_{r} \end{bmatrix} $$
$$ \vec{y} = \begin{bmatrix} C_{Ao} \end{bmatrix} $$

Reactor
(1) $$ \frac {dV}{dt} = F_{i} + F_{r} - F $$
(2) $$ \frac {d(C_{A}V)}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
Separator (steady state)
(3) $$ 0 = F - F_{r} - F_{o} $$
(4) $$ 0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao} $$

For equation (2), I expand out
$$ V \frac {dC_{A}}{dt} + C_{A} \frac {dV}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
I substitute (1) into $\frac {dV}{dt}$
$$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} + F_{r} - F) = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
The $FC_{A}$ terms cancel
$$V \frac {dC_{A}}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} -F_{i}C_{A}-F_{r}C_{A} - kC_{A}^{3}V $$
I know I need to use the separator balance equations to substitute into this equation to cast in the form needed for part (c), but I am unsure which way to substitute.

 

[gfd43tg]

A new thought came to mind, with equation (2),
$$ V \frac {dC_{A}}{dt} + C_{A} \frac {dV}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
I will modify equation (1)
$$\frac {dV}{dt} = F_{i} + F_{r} - F$$
Using equation (3)
$$ 0 = F - F_{r} - F_{o} $$
Solving for $F_{o} = F - F_{r}$, I will substitute back into (1)
$$ \frac {dV}{dt} = F_{i} - F_{o} $$
Then substitute this into (2)
$$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V $$
I will use equation (4)
$$0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao}$$
Solve $F_{r}C_{Ar} = FC_{A}-F_{o}C_{Ao}$. I substitute this into (2) and now have
$$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} +FC_{A}-F_{o}C_{Ao} - FC_{A} - kC_{A}^{3}V $$
The $FC_{A}$ terms cancel, leaving me with
$$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} -F_{o}C_{Ao} - kC_{A}^{3}V $$
But I'm still not quite sure how to get the the form they are looking for.

 

[gfd43tg]

Okay, I have some idea now. This is what I got so far.
Using (3)
$$ 0 = F - F_{r} - F_{o} $$
I divide through by $F$, $0 = 1 - \frac {F_{r}}{F} - \frac {F_{o}}{F} = 1 - r - \frac {F_{o}}{F}$
Then I use equation (4)
$$ 0 = FC_{A} - F_{r}C_{Ar} - F_{o}C_{Ao} $$
I isolate $F_{o}C_{Ao} = FC_{A}-F_{r}C_{Ar}$, then I divide through by $FC_{Ao}$, leaving me with $\frac {F_{o}}{F}= \frac {C_{A}}{C_{Ao}}-K_{H}r$. I know that $\frac {F_{o}}{F}=1-r$, so I put into the equation $1-r = \frac {C_{A}}{C_{Ao}}-K_{H}r$.
Next, I isolate
$$\frac {C_{A}}{C_{Ao}} = 1 - r + K_{H}r = 1-r(1-K_{H}) $$
I take the inverse
$$\frac {C_{Ao}}{C_{A}} = \frac{1}{1-r(1-K_{H})}$$

We will go back to my equation (2) that I had gotten from the previous post
$$ V \frac {dC_{A}}{dt} + C_{A}(F_{i} - F_{o}) = F_{i}C_{Ai} -F_{o}C_{Ao} - kC_{A}^{3}V $$
I move the $C_{A}(F_{i} - F_{o})$ term to the RHS,
$$ V \frac {dC_{A}}{dt} = F_{i}(C_{Ai}-C_{A}) -F_{o}(C_{Ao}-C_{A}) - kC_{A}^{3}V $$
I solve for $C_{Ao}$
$$C_{Ao} = \frac {C_{A}}{1-r(1-K_{H})}$$
I substitute this into equation (2)
$$V \frac {dC_{A}}{dt} = F_{i}(C_{Ai}-C_{A}) -F_{o} \bigg (\frac {C_{A}}{1-r(1-K_{H})} -C_{A} \bigg ) - kC_{A}^{3}V $$
Pull out $C_{A}$ and divide by $V$, leaving me with
$$ \frac {dC_{A}}{dt} = \frac {F_{i}}{V}(C_{Ai}-C_{A}) - \frac {F_{o}C_{A}}{V} \bigg ( \frac {1}{1-r(1-K_{H})}+1 \bigg ) - kC_{A}^{3} $$
This looks similar to the equation in part (c) (part (c) assumed the volume in the reactor was constant)

 

[Chestermiller]

In your final equation, the +1 should be a -1. With this change, the final equation is identical to the equation in the problem statement, except for the additional term:$$C_A\frac{(F_0-F_i)}{V}$$

Chet

 

[gfd43tg]

I noticed that, but my math did not come up with that expression. I wonder if it is correct or if I made a sign error in any of my steps

 

[Chestermiller]

You made a sign error in your last step.

 

[gfd43tg]

Okay, I see it now. Thank you