# Reactor + Separator control outlet concentration

1. Sep 16, 2015

### Maylis

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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?

Last edited: Sep 16, 2015
2. Sep 16, 2015

### Maylis

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.
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) $$\frac {d(C_{A}V)}{dt} = F_{i}C_{Ai} + F_{r}C_{Ar} - FC_{A} - kC_{A}^{3}V$$
(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.

Last edited: Sep 16, 2015
3. Sep 16, 2015

### Maylis

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.

4. Sep 16, 2015

### Maylis

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)

Last edited: Sep 16, 2015
5. Sep 17, 2015

### Staff: Mentor

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

6. Sep 17, 2015

### Maylis

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

7. Sep 17, 2015