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Concentration change in CSTR

by Maylis
Tags: concentration, cstr
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Maylis
#1
Sep1-14, 06:49 AM
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P: 578
Hello,

Suppose A and B are continuously added into a tank, and exit stream consists of A,B,C, and D. I am a little confused on the following point as I follow the example in my textbook.
The reaction is
##A + B → C + D##

It says because (and I quote) ''...Because the system is operated at steady state, if we were to withdraw liquid samples at some location in the tank at various times and analyze them chemically, we would find that the concentrations of the individual species in the different samples were identical. That is, the concentration of the sample taken at 1 P.M. is the same as that of the sample taken at 3 P.M. Because the concentrations are constant and therefore do not change with time,
[itex] \frac {dC_{A}}{dt} = 0 [/itex]
which would lead to
[itex]r_{A} = 0[/itex]
which is incorrect because C and D are being formed from A and B at a finite rate. Consequently, the rate of reaction defined by equation 1-1 cannot apply to a flow system and is incorrect if it is defined in this matter''.

equation 1-1 just equates the change in concentration of a species to the rate of formation/disappearance of a species.

Where I am confused here is exactly what is meant by ##\frac {dC_{A}}{dt}##. It seems to be in the stream where A enters, it is 100% A. When it gets into the tank, it is now mixed with B, so it's concentration is less. So that means ##\frac {dC_{A}}{dt} \ne 0##.

Perhaps the wording is confusing. So it is saying, if I pick an arbitrary location in the tank, the concentration of the chemical species A should be the same at all times. In my head, I am thinking that the ##\frac {dC_{A}}{dt}## is the change in concentration of A as it moves through the CSTR. At ##t = 0##, it enters the reactor as some concentration ##C_{A_{0}}## and leaves the reactor at concentration ##C_{A_{\tau}}##, where ##t = \tau## is the residence time of the CSTR. So in this scenario there is a change in concentration of species A over a finite time, hence that derivative should not be equal to zero. This is clearly not what is meant by that equation, which is somewhat alluding me.

It appears to mean that the change in concentration in a very specific location is not changing with time, hence the derivative is zero. But then that begs the question why are we analyzing it in that way, and not what I mentioned before? What am I missing here?
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Chestermiller
#2
Sep1-14, 10:10 AM
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In the case of both a well-stirred batch reactor and a CSTR, dC/dt is equal to the rate of change of concentration C within the tank. It is not the rate of change of concentration following a parcel of fluid.

Before you can get a feel for this, it is helpful to first consider the case with no reaction in a CSTR. Suppose that the concentration of a trace species is held constant in the feed at C0, so that the concentration within the tank is also C0, as is the concentration in the exit stream. Now, at some time t = 0, we suddenly lower the concentration in the feed to 0. A mass balance on the concentration within the tank will be:
[tex]V\frac{dC}{dt}=-FC[/tex]
where V is the volume of fluid in the tank and F = feed rate = exit flow rate. What do you get for the concentration in the tank as a function of time?

Chet
Maylis
#3
Sep1-14, 06:20 PM
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P: 578
When you say it is the change in concentration within the tank, there are still some doubts in my mind. Why wouldn't the change of the concentration of a parcel of fluid entering and exiting the tank be the same as the change of concentration within the tank? The term ''change in concentration within the tank'' is ambiguous to me, so I am not really sure what is meant.

Chestermiller
#4
Sep1-14, 08:32 PM
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Concentration change in CSTR

Please be patient and bear with me. I'll be using the problem I posed to help you understand what is happening. Please provide the solution to the problem I posed so the we can start talking about Residence Time Distributions.

Chet
Maylis
#5
Sep1-14, 09:26 PM
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Yeah I will get back to that, I had my solution typed with latex then my computer crashed right when I finished
Maylis
#6
Sep1-14, 11:32 PM
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P: 578
Okay, so now for my solution to the problem posed.

To start off with the most general mole balance, since we have an unreacting system. For a species ##i##, the general mole balance is for ##t < 0##

##\frac {dn_{i}}{dt} = \dot {n_{i,in}} - \dot {n_{i,out}} + \dot {G_{i}}##

With no mole generation, and the number of moles entering the tank equal to the moles exiting the tank, we have

##\frac {dn_{i}}{dt} = 0##.

For the problem posed, at ##t = 0##, the general mole balance

##\frac {dn_{i}}{dt} = \dot {n_{i,in}} - \dot {n_{i,out}} + \dot {G_{i}}##

Since the concentration is zero, and we know that the number of moles is equal to the product of volume and concentration, the number of moles coming into the tank is zero, and the generation term is zero as well. This simplifies to

##\frac {dn_{i}}{dt} = \dot {-n_{i,out}}##

We remember that ##F## is the flow rate, Substituting ##\dot{n_{i}} = C_{i}F##, we get

## \frac {d(C_{i}V)}{dt} = -C_{i}F##
Since the volume is constant, it can be removed from the derivative,

## V \frac {dC_{i}}{dt} = -C_{i}F##

Using algebra, we get

##\frac {dC_{i}}{C_{i}} = - \frac {F}{V}dt##

Integrating this

##\int_{C_{0}}^{C_{i}} \frac {dC_{i}}{C_{i}} = \int_{0}^{t} - \frac {F}{V}dt##

which becomes

##ln(\frac {C_{i}}{C_{0}}) = - \frac {F}{V}t##

Definining ##\frac {V}{F}## to be the residence time, ##\tau##, we get the concentration in the tank as a function of time,

##C_{i} = C_{0}e^{-t/\tau}##
Maylis
#7
Sep1-14, 11:55 PM
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P: 578
How do I get the dot to be above the letter, instead of how it is right now over the subscripts?
Chestermiller
#8
Sep2-14, 01:43 PM
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Quote Quote by Maylis View Post
Okay, so now for my solution to the problem posed.

To start off with the most general mole balance, since we have an unreacting system. For a species ##i##, the general mole balance is for ##t < 0##

##\frac {dn_{i}}{dt} = \dot {n_{i,in}} - \dot {n_{i,out}} + \dot {G_{i}}##

With no mole generation, and the number of moles entering the tank equal to the moles exiting the tank, we have

##\frac {dn_{i}}{dt} = 0##.

For the problem posed, at ##t = 0##, the general mole balance

##\frac {dn_{i}}{dt} = \dot {n_{i,in}} - \dot {n_{i,out}} + \dot {G_{i}}##

Since the concentration is zero, and we know that the number of moles is equal to the product of volume and concentration, the number of moles coming into the tank is zero, and the generation term is zero as well. This simplifies to

##\frac {dn_{i}}{dt} = \dot {-n_{i,out}}##

We remember that ##F## is the flow rate, Substituting ##\dot{n_{i}} = C_{i}F##, we get

## \frac {d(C_{i}V)}{dt} = -C_{i}F##
Since the volume is constant, it can be removed from the derivative,

## V \frac {dC_{i}}{dt} = -C_{i}F##

Using algebra, we get

##\frac {dC_{i}}{C_{i}} = - \frac {F}{V}dt##

Integrating this

##\int_{C_{0}}^{C_{i}} \frac {dC_{i}}{C_{i}} = \int_{0}^{t} - \frac {F}{V}dt##

which becomes

##ln(\frac {C_{i}}{C_{0}}) = - \frac {F}{V}t##

Definining ##\frac {V}{F}## to be the residence time, ##\tau##, we get the concentration in the tank as a function of time,

##C_{i} = C_{0}e^{-t/\tau}##
Excellent. But τ is not the residence time, it is the mean residence time. Your equation shows that, because the tank is well-mixed, some of the trace species comes out right away at time t = 0, while the reaminder comes out at later times. So, there is a distribution of residence times for the trace species in the tank. At any time t, the fraction of the original amount of trace species molecules still remaining in the tank is ##e^{-t/\tau}##. So, if reaction were taking place, some of the trace species would be able to leave the tank before any significant amount of reaction occurred, while some of the trace species would have much more time to react before it left the tank.

So the parameter t in the general CSTR equation is just the amount of elapsed time since some change was made (not the reaction time for all of the trace species), and the equation takes into account everything regarding the residence time distribution that I discussed in the previous paragraph in a mathematically correct way. For example, if the trace species is dissociating by a first order reaction, the mass balance equation for the CSTR (with steady flow rate in and out) leads to:

[tex]V\frac{dC}{dt}=F(C_{in}-C)-kVC[/tex]

If the flow rate in and out, F, is equal to zero, you recover the batch reactor equation (and t is identical to the reaction time), but, if the reaction rate is zero, you recover the flow-with-mixing equation. At steady state, the time derivative is zero (assuming Cin is constant), and you obtain:

[tex]C=\frac{C_{in}}{1+kτ}[/tex]

Hope this helps.

Chet


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