Mass Transport Limited Regime

[Suspicious-Teach8788]

So we're studying for finals, and a few of us are arguing.

In a typical reaction that's governed by flow, temp, pressure, etc, like a CVD chamber (what I deal with in my research), you have a mass transport limited regime and a reaction rate limited regime.

Now if you're in the mass transport regime, and you cut the flow in half, shouldn't your deposition rate shrink in approximately half?

 

[Gibsons]

Not entirely sure I understand this, but I think this sounds similar to what happens in multi step metabolic pathways. In which case the answer is not necessarily. If your reaction rate is extremely slow, cutting the mass flow in half might not slow your deposition down - it had more than enough mass to work with before anyway.

In other words, the overall reaction rate is determined by the slowest step in the path.

Changing other steps doesn't affect the overall path unless they become slower than it. Cool thing about this is you can regulate the whole pathway just by regulating that one step.

 

[CycloWizard]

As always, it depends. How do you define flow? Is it the mass flowrate of the depositing species feed? In CVD, there are generally multiple driving forces, so changing the "flow" won't necessarily have a linear effect on the deposition rate, even if the reaction rate is arbitrarily fast. The only way to answer this question definitively is to perform the appropriate mass balance and calculate how the input affects the output.

The model should be pretty straightforward: assume an infinitely fast, heterogeneous reaction such that the boundary at the deposition site has zero concentration in the depositing species. The appropriate field equations are then dictated by the type of CVD you're working with. I can be more specific if you give a more specific case you're interested in solving.

 

[Suspicious-Teach8788]

As always, it depends. How do you define flow? Is it the mass flowrate of the depositing species feed? In CVD, there are generally multiple driving forces, so changing the "flow" won't necessarily have a linear effect on the deposition rate, even if the reaction rate is arbitrarily fast. The only way to answer this question definitively is to perform the appropriate mass balance and calculate how the input affects the output.

The model should be pretty straightforward: assume an infinitely fast, heterogeneous reaction such that the boundary at the deposition site has zero concentration in the depositing species. The appropriate field equations are then dictated by the type of CVD you're working with. I can be more specific if you give a more specific case you're interested in solving.

See, this is the thing, because we're trying to do a book problem, and this book problem significantly dumbs things down.

In my undergrad, we did a lot of detailed problem where there are a lot of factors controlling your deposition rate. Clearly the geometry, laminar flow, pressure, etc.

http://twitpic.com/3gi4s4

is the problem we've been talking about. It seems very dumbed down with just a few factors here and there. The part where I'm referring to is (c). Like all we're supposed to worry about is either the temp or the flow of silane. It seems as if the book just wants us to work with the mass transport limited regmine, so if the flow were to be cut in half, then shouldn't the dep rate be cut in half? This would definitely be an approximation. Sure we could possibly go fit the curve and do some other crazy things, but I'm not sure if this is what the problem is aiming for...

 

[McWatt]

I think the point of the question is to get you to think about boundary layer thickness scaling. Mass transport rate will scale inversely with boundary layer thickness and the BL thickness scales with the inverse square root of the free stream velocity in the Blasius solution. So if you halve the volumetric flow rate you also halve the linear flow velocity, meaning the BL grows by a factor of root 2, so the answer to c should be 0.60 um/min.

 

[CycloWizard]

See, this is the thing, because we're trying to do a book problem, and this book problem significantly dumbs things down.

In my undergrad, we did a lot of detailed problem where there are a lot of factors controlling your deposition rate. Clearly the geometry, laminar flow, pressure, etc.

http://twitpic.com/3gi4s4

is the problem we've been talking about. It seems very dumbed down with just a few factors here and there. The part where I'm referring to is (c). Like all we're supposed to worry about is either the temp or the flow of silane. It seems as if the book just wants us to work with the mass transport limited regmine, so if the flow were to be cut in half, then shouldn't the dep rate be cut in half? This would definitely be an approximation. Sure we could possibly go fit the curve and do some other crazy things, but I'm not sure if this is what the problem is aiming for...

That's a classic figure for the effectiveness factor as a function of Thiele modulus in a porous, spherical catalyst particle, so I have an analogy which works for me but maybe not for you, depending on your background. The analogy for the catalyst particle is that the effectiveness factor (y-axis) decreases as the length scale for diffusion (x-axis) increases. In other words, as you move right along the x-axis, mass transfer effects become increasingly important (as temperature increases, mass transfer resistance decreases since your x-axis is 1000/T). Thus, part C is actually asking what happens to the deposition rate in the absence of mass transfer effects.

 

[McWatt]

That's a classic figure for the effectiveness factor as a function of Thiele modulus in a porous, spherical catalyst particle, so I have an analogy which works for me but maybe not for you, depending on your background. The analogy for the catalyst particle is that the effectiveness factor (y-axis) decreases as the length scale for diffusion (x-axis) increases. In other words, as you move right along the x-axis, mass transfer effects become increasingly important (as temperature increases, mass transfer resistance decreases since your x-axis is 1000/T). Thus, part C is actually asking what happens to the deposition rate in the absence of mass transfer effects.

Part C is asking for precisely the opposite of that. It says that you're at high temperature, meaning the system is mass transfer limited because the rate constant is very high on the left side of the graph. Then it says that you slow mass transfer down even more, so of course you'll be even more mass limited. I'm fairly confident that my previous post gave the correct answer to the problem.

If part C were truly asking what happens in the absence of mass transfer effects, the answer would be "nothing" because the change described in part C only affects mass transfer, not reaction rates.

 

[CycloWizard]

Part C is asking for precisely the opposite of that. It says that you're at high temperature, meaning the system is mass transfer limited because the rate constant is very high on the left side of the graph. Then it says that you slow mass transfer down even more, so of course you'll be even more mass limited. I'm fairly confident that my previous post gave the correct answer to the problem.

If part C were truly asking what happens in the absence of mass transfer effects, the answer would be "nothing" because the change described in part C only affects mass transfer, not reaction rates.

I wrote my previous answer during a phone interview, so I thought maybe I took my analogy the wrong direction and was off track. I went back and worked out the problem for steady-state diffusion through a boundary layer of thickness t with a first-order reaction occurring at the wall. I'm pretty sure I was actually correct before, though I agree the results might seem a bit counter-intuitive (I think this is because of the semi-log axes used with an inverse x-axis, which makes interpretation a bit tricky).

The concentration profile is simply
C=(C_bulk-C_0)*(x/t)+C_0,
where C_0 is the concentration at the wall. The deposition rate is just
r=k*C_0
since I assumed a first-order reaction. Since, at steady state, the flux through the boundary layer must be equal to the rate of reaction (conservation of mass), the concentration at the wall is given by
C_0=C_bulk/(1+k*t/D).
I'm using k for the rate constant and D for the diffusivity. Since both k and D are activated processes, their temperature dependence may be modeled using an Arrhenius relationship:
k=k_0 exp(-k_a/(R*T))
and
D=D_0 exp(-D_a/(R*T))
From the data given in part C of the problem, k_0=0.85 (not worrying about units at this point), and you can choose the other parameters (D_0, k_a, and D_a) to get a figure that looks similar to the one in the problem statement. Then, if you assume no mass transfer resistance (i.e. C_0=C_bulk for all T), you will see that the upper plateau is actually the rate in absence of mass transfer effects.

edit: I think most people would tend to assume that the temperature influences the reaction rate much more heavily than the diffusivity, but this is usually not the case since the diffusivity also rises rapidly with temperature. You will achieve the same result if you model the diffusivity as a power law of temperature as suggested approximately by the kinetic theory of gases. But you also get this same qualitative result if you assume that the diffusivity is completely independent of temperature. Like I said, I think the problem is interpreting the figure due to the complex axes.

edit #2: I think another misconception people have about reaction rates is that they increase unbounded. However, from the Arrhenius form given above, it should be clear that as T goes to infinity, k goes to k_0, not infinity (since exp(anything/infinity)=exp(0)=1).

 

[McWatt]

The figure itself shows us that the transport is mass transport limited in part c. At "the highest temperature," i.e., all the way to the left in the figure, the slope isn't the same as it is on the right, at lower temperatures. The constant slope on the right shows us that in that region the process is limited by a single Arrhenius relationship. You say that the choice of nonlinear axes is counterintuitive, but it's the combination of a log vertical axis and inverse horizontal axis always used to show Arrehenius dependence because it makes the situation very obvious in a case like this.

At low temperatures we know from the real world that we're always going to be reaction rate limited. Also, part b of the question shows us indirectly that we're reaction rate limited on the right, because it's only answerable if there's a clearly measurable single slope on the graph, which is the case on the right. Yes, diffusion constants scale with temperature as well, but also from the real world we know that they scale more slowly than nearly every reaction rate constant because their activation energy is so much lower.

Anyway, the fact that we move from the constant slope seen on the right, where we know we're reaction rate limited, to a different slope on the left tells us without any ambiguity that on the left we're no longer reaction rate limited. Notice that the slope is close to zero at "the highest temperature" referenced in part C, which is indicative of the slow scaling of D with T at this temperature.

I should also point out that I've taught this class and asked this question of students, though when I asked it I made it a little clearer. Still, I don't think the problem as presented in this textbook is particularly obfuscated. As I've suggested to students in the past, sometimes it's better to think about the problem conceptually before you start writing down equations. Often, the equations then become unnecessary.

 

[CycloWizard]

The figure itself shows us that the transport is mass transport limited in part c. At "the highest temperature," i.e., all the way to the left in the figure, the slope isn't the same as it is on the right, at lower temperatures. The constant slope on the right shows us that in that region the process is limited by a single Arrhenius relationship. You say that the choice of nonlinear axes is counterintuitive, but it's the combination of a log vertical axis and inverse horizontal axis always used to show Arrehenius dependence because it makes the situation very obvious in a case like this.

At low temperatures we know from the real world that we're always going to be reaction rate limited. Also, part b of the question shows us indirectly that we're reaction rate limited on the right, because it's only answerable if there's a clearly measurable single slope on the graph, which is the case on the right. Yes, diffusion constants scale with temperature as well, but also from the real world we know that they scale more slowly than nearly every reaction rate constant because their activation energy is so much lower.

Anyway, the fact that we move from the constant slope seen on the right, where we know we're reaction rate limited, to a different slope on the left tells us without any ambiguity that on the left we're no longer reaction rate limited. Notice that the slope is close to zero at "the highest temperature" referenced in part C, which is indicative of the slow scaling of D with T at this temperature.

I should also point out that I've taught this class and asked this question of students, though when I asked it I made it a little clearer. Still, I don't think the problem as presented in this textbook is particularly obfuscated. As I've suggested to students in the past, sometimes it's better to think about the problem conceptually before you start writing down equations. Often, the equations then become unnecessary.

Surely it's not asking too much for you to show me where I'm going wrong in my analysis, especially if you've taught a course in this area before. It took me about five minutes to solve the steady state, first-order heterogeneous reaction model I presented above (which could be why there is an error in it), so you should presumably be able to develop the correct solution in about as much time.

edit: Looks like the error was in the transcription from my scrap paper to MATLAB. 😛