Cocurrent diffusion (can't be this hard....)

[KarenRei]

Hi all. I've been working on this for two days and I feel embarrassed that I haven't been able to figure it out. I'm dealing with the following cocurrent diffusion situation between an aqueous and gaseous flow moving at the same velocity:

X% of the liquid is removed and replaced with fresh (0% concentration) each runthrough, so the incoming liquid concentration C_l_in is X% of the outgoing liquid concentration C_l_out. The incoming gas concentration C_g_in is known. The diffusion constant and Henry's law constants are also known, as is the length of time the two are in contact, the total contact area, and the flow rates of both the liquid and gas phases

It seems this should be a common use case, but while I find tons of articles on Fick's Law and diffusion, all of my wading through them hasn't revealed any covering this case. And I've been trying to derive it on my own, but I suck at differential equations. :( Does anyone know how to solve for C_l_out and C_g_out?

(In case it matters: the above diagram is stylized; in practice it's droplets entrained in a gas stream)

 

[Chestermiller]

Hi all. I've been working on this for two days and I feel embarrassed that I haven't been able to figure it out. I'm dealing with the following cocurrent diffusion situation between an aqueous and gaseous flow moving at the same velocity:

View attachment 113548

X% of the liquid is removed and replaced with fresh (0% concentration) each runthrough, so the incoming liquid concentration C_l_in is X% of the outgoing liquid concentration C_l_out. The incoming gas concentration C_g_in is known. The diffusion constant and Henry's law constants are also known, as is the length of time the two are in contact, the total contact area, and the flow rates of both the liquid and gas phases

It seems this should be a common use case, but while I find tons of articles on Fick's Law and diffusion, all of my wading through them hasn't revealed any covering this case. And I've been trying to derive it on my own, but I suck at differential equations. :( Does anyone know how to solve for C_l_out and C_g_out?

(In case it matters: the above diagram is stylized; in practice it's droplets entrained in a gas stream)

Is the liquid solvent evaporating. Are you supposed to treat each phase as semi-infinite, or are you supposed to take into account the drop size? Please provide the exact statement of the problem. Your interpretation of the problem statement is not adequate to answer your question.

 

[KarenRei]

Thank you very much for your response.

Is the liquid solvent evaporating. Are you supposed to treat each phase as semi-infinite, or are you supposed to take into account the drop size?

The solvent is an ionic liquid and thus has negligible vapour pressure, and gases diffused into the droplets will not cause a relevant increase in their mass or volume; the area and volume can be taken as constant. Bulk temperature, pressure, density and velocity of the gas will also change little, enough to be negligible. However various partial pressures will change as diffusion occurs. The species of interest are primarily in ppm quantities. The bulk species will also diffuse into the liquid and quickly reach saturation, but their maximum solubility is limited.

The droplets are small enough and contact times are long enough that it would be fair to treat them as uniformly mixed. The gas however is not uniformly mixed. The flow is fairly close to laminar - fan driven, but by the time of liquid injection it has already passed through an electrostatic precipitator (hex grid of tubes), and thus has significantly reduced vorticity. The conduit is relatively straight and unobstructed (no packing - just mist collection at the end for recovery)

 

[Chestermiller]

Three questions:

1. Is this a homework problem?
2. If you knew the mass transfer coefficient, could you solve the problem?
3. Where is that exact problem statement that I asked for?

 

[KarenRei]

Three questions:

1. Is this a homework problem?
2. If you knew the mass transfer coefficient, could you solve the problem?
3. Where is that exact problem statement that I asked for?

1. No, it is not a homework problem

2. The problem I've had with trying to solve it myself is that I end up with an iterative series of three equations representing how the state evolves from one timetstep to the next (those three being liquid phase concentration, average gas phase concentration, and mass transfer rate), but I can't seem to reduce them into a single integral or summation that I could solve. I could always "solve" it by writing a python program or whatnot to iterate over tiny timesteps and evolve the concentrations until they converge, but that's clearly not the "right" way to go about it (is it not?), and I couldn't insert that into my spreadsheet for automatic updates if I did it that way.

3. I'd like to be more precise but I'm not sure what you consider to be missing from the above. I can try rewording, if that would help.

"Liquid is injected into a gas stream forming droplets in a concurrent, largely laminar, constant velocity/temperature/density stream where gas can diffuse into the droplets . The droplets are represented as fixed mass/volume/surface area with a concentration that varies with time but not depth. The gas is represented as a fixed mass/volume flow with a concentration that varies with time and distance from the solid-liquid interface, to some finite distance (halfway to the next droplet). The two phases remain in contact for a fixed timeperiod. The concentration of the species in the liquid phase at inflow equals the concentration in the liquid phase at outflow times some percentage X (aka, X percent of the liquid is replaced with fresh). The inflow gas concentration is fixed. The diffusion coefficients are known. Solve for the gas and liquid concentrations at outflow."

Is that clearer wording? Just a concurrent scrubber - and what should be an easier case than most (aka, the liquid doesn't evaporate). But I can't find anything that addresses this specific problem anywhere. The information is almost certainly out there somewhere, but mixed in with a flood of pages describing other, inapplicable problems, which is how I ended up here.

 

[Chestermiller]

Let's see what you have done so far. It seems to me the focus should be on getting the overall mass transfer coefficient between the gas and the liquid. In particular, let's see what those 3 equations look like that you are talking about under item #2.

 

[KarenRei]

Let's see what you have done so far. It seems to me the focus should be on getting the overall mass transfer coefficient between the gas and the liquid. In particular, let's see what those 3 equations look like that you are talking about under item #2.

The case where I got the furthest on was the simplified case, where diffusion through the gas was assumed to be instant (figuring once I solved that I could work on adding in diffusion through the gas as well). For it my equations were:

C_l(N+1) = C_l(N) + s * D_l(N) / F_l
C_g(N+1) = C_g(N+1) - s * K_l(N) / F_g
K_l(N + 1) = D_gl * A * (C_g(N)/H - C_l(N))

Where

C_l = concentration in liquid
C_g = concentration in gas
K_l = mass transfer rate from gas to liquid
s = timestep (infinitesimally small)
F_l = liquid flow rate
F_g = gas flow rate
D_gl = diffusion rate between the gas and liquid
A = total droplet surface area
H = Unitless Henry's law coefficient

(Hope I transcribed that right; I had actually been using cell numbers from my spreadsheet in my calculations)

I got no further than this because I was just grinding gears trying to turn those three equations that depend on each other into something I can solve. Ultimately, though, I also need to factor in diffusion through gas to the surface, and after t / s timesteps (aka, it's reached the outflow end, the gas is escaping and the liquid is being pumped back) C_g needs to be reset to its initial value and C_l needs to be multiplied by the percentage of liquid recycled, and the whole process run iteratively until it converges.

As mentioned, obviously I could do this by writing a program to iterate over tiny timesteps. But this strikes me as A) wrong, and B) slow / a pain for integrating into my spreadsheet.

 

[Chestermiller]

The equation for the mass transfer rate is not correct. Let's try something a little different. You know the ratio of the volume of gas to the volume of drops. Suppose you had a single spherical drop, immersed in a surrounding spherical volume of gas, with the ratio of the volumes equal to the actual ratio of volumes. So you are treating this combination as a unit cell. Now all you need to do is quantify the time-dependent mass transfer behavior within the unit cell. Are we together so far?

 

[KarenRei]

The equation for the mass transfer rate is not correct. Let's try something a little different. You know the ratio of the volume of gas to the volume of drops. Suppose you had a single spherical drop, immersed in a surrounding spherical volume of gas, with the ratio of the volumes equal to the actual ratio of volumes. So you are treating this combination as a unit cell. Now all you need to do is quantify the time-dependent mass transfer behavior within the unit cell. Are we together so far?

Hmm, I took it from one of the pages I found. Perhaps I transcribed it wrong.

Yes, your example is perfectly reasonable. :)

 

[Chestermiller]

OK. Here's what I get:

$$F_L\frac{dC_L}{dV}=\alpha \phi\tag{1}$$
$$F_G\frac{dC_G}{dV}=-\alpha \phi\tag{2}$$
where $\alpha$ is the mass transfer surface area per unit volume of the unit, V is the cumulative volume of the unit (up to a given location), and $\phi$ is the is the mass flux at the drop surface:
$$\phi=5\frac{D_L}{R}\left(\frac{C_{GI}}{H}-C_L\right)=\frac{D_G}{R}(C_G-C_{GI})\tag{3}$$
where $D_L$ is the diffusion coefficient in the liquid, $D_G$ is the diffusion coefficient in the gas, $C_{GI}$ is the gas concentration at the drop interface, and R is the radius of a drop. From Eqn. 3, it follows that:
$$\phi=\frac{(C_G/H-C_L)}{R\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{4}$$

I think I'll stop here for now.

 

[Chestermiller]

Continuation of previous post:

If the total volumetric flow rate of gas plus drops is F and the volume fraction of drops is f, then Eqns. 1 and 2 can be rewritten as:
$$f\frac{dC_L}{dt}=\alpha \phi\tag{5}$$
$$(1-f)\frac{dC_G}{dt}=-\alpha \phi\tag{6}$$where t is the cumulative residence time (t = V/F). If we add equation 5 and 6, we find that $$fC_L+(1-f)C_G=fC_{F0}+(1-f)C_{G0}\tag{7}$$
Combining Eqns. 4 and 5 with Eqn. 6 yields:
$$\frac{dC_L}{dt}=\frac{\alpha}{f} \frac{(C_G/H-C_L)}{R\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{8}$$
$$\frac{dC_G}{dt}=-\frac{\alpha}{(1-f)} \frac{(C_G/H-C_L)}{R\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{9}$$
If we now divide Eqn. 9 by H and subtract Eqn. 8, we obtain:
$$\frac{d(C_G/H-C_L)}{dt}=-\alpha \frac{\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}{R\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}(C_G/H-C_L)\tag{10}$$
Eqn. 10 is the same as
$$\frac{d\ln{(C_G/H-C_L)}}{dt}=-\alpha \frac{\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}{R\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{11}$$
The drop radius R in equation 11 is related to the volume fraction of drops f and the surface area per unit volume $\alpha$ by the equations:
$$4\pi R^2 n=\alpha\tag{12}$$
$$\frac{4}{3}\pi R^3 n=f\tag{13}$$where n is the number of drops per unit volume. If we divide Eqn. 13 by Eqn. 12, we obtain:
$$R=\frac{3f}{\alpha}\tag{14}$$Substitution of Eqn. 14 into Eqn. 11 then yields:
$$\frac{d\ln{(C_G/H-C_L)}}{dt}=- \frac{\alpha^2\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}{3f\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{15}$$

 

[KarenRei]

Sorry for taking a while to respond, I've been trying to make sure I understand each line.

(1 & 2) I was initially confused about the changing volume (since droplet sizes are fixed), until I realized that you're representing the whole unit as if it's growing over time. Which is interesting, I never thought about posing the problem in that manner. I was also confused by how the differentials were set up until I realized that one could convert them into the mass flow rates on the right. So no problems there.

(3) I was slightly delayed in this because "R" wasn't boldfaced as a variable, so when I didn't see it in the list of variables my mind was interpreting it as the gas constant. Got it now though. However, I'm still not seeing where that "5" is coming from. Also, on the right hand side, I don't see what the radius of the drop has to do with how quickly gas reaches the interface. Shouldn't there be two radii, one for the drop, and one for the distance across which gas travels to reach the surface? Also, part of my mind is itching about how while this makes sense in 1d, in 2d/3d the area that gas is diffusing through at any arbitrary location decreases as one nears the center of the drop. But I imagine you address that later - or it could be inherently assumed in the diffusion constants.

I'm going to take a break from staring at (3) for a bit and move onto the other equations, especially since you've posted some more.

 

[Chestermiller]

Sorry for taking a while to respond, I've been trying to make sure I understand each line.

(1 & 2) I was initially confused about the changing volume (since droplet sizes are fixed), until I realized that you're representing the whole unit as if it's growing over time. Which is interesting, I never thought about posing the problem in that manner. I was also confused by how the differentials were set up until I realized that one could convert them into the mass flow rates on the right. So no problems there.

The volume is not growing. Actually, V is the volume of the portion of the unit between the entrance and any arbitrary axial location. So dV is the volume between z and z + dz. The F's are volume rates of flow.

(3) I was slightly delayed in this because "R" wasn't boldfaced as a variable, so when I didn't see it in the list of variables my mind was interpreting it as the gas constant.

I explicitly stated that R is the radius of a drop.

Got it now though. However, I'm still not seeing where that "5" is coming from.

I'll explain that later.

Also, on the right hand side, I don't see what the radius of the drop has to do with how quickly gas reaches the interface. Shouldn't there be two radii, one for the drop, and one for the distance across which gas travels to reach the surface?

I'll explain that later. This is already the result of integrating the concentration profiles with respect to radial location.

Also, part of my mind is itching about how while this makes sense in 1d, in 2d/3d the area that gas is diffusing through at any arbitrary location decreases as one nears the center of the drop. But I imagine you address that later - or it could be inherently assumed in the diffusion constants.

That was all integrated out with respect to radius. I'll get to all this later.

 

[KarenRei]

Indeed you did - I just happened not to see the definition of R during my first readthrough of the equation because it wasn't boldfaced like the other variables, and after pondering for a whole why the gas constant would be relevant, I went back and re-read what you wrote and saw it. :)

I've now gone through all of the other equations and all of them follow logically - only #3 has been problematic. So now we have, in plain english, "the rate of change of the logarithm of (equilibrium concentration of the species in the liquid phase minus the actual liquid phase concentration), per unit time that the flow progresses through the scrubber." So let's see... if I'm thinking about this correctly, from there we need to be able to solve for C_L at an arbitrary point of time in terms of one unknown variable, evaluate at t=0 and t=t_final, so that we can set C_L_t0 / C_L_t_final = X %, and use that to solve for the unknown variable?

(By the way, I just wanted to add, thank you for taking all this time to help me with this - it's very much appreciated)

 

[Chestermiller]

Indeed you did - I just happened not to see the definition of R during my first readthrough of the equation because it wasn't boldfaced like the other variables, and after pondering for a whole why the gas constant would be relevant, I went back and re-read what you wrote and saw it. :)

I've now gone through all of the other equations and all of them follow logically - only #3 has been problematic. So now we have, in plain english, "the rate of change of the logarithm of (equilibrium concentration of the species in the liquid phase minus the actual liquid phase concentration), per unit time that the flow progresses through the scrubber."

Not exactly. $C_G/H$ is the concentration that would exist in the liquid if the liquid were in equilibrium with the local species concentration $C_G$ in the bulk of the gas. Both $C_G$ and $C_L$ are varying simultaneously with time t through the scrubber.

So let's see... if I'm thinking about this correctly, from there we need to be able to solve for C_L at an arbitrary point of time in terms of one unknown variable, evaluate at t=0 and t=t_final, so that we can set C_L_t0 / C_L_t_final = X %, and use that to solve for the unknown variable?

(By the way, I just wanted to add, thank you for taking all this time to help me with this - it's very much appreciated)

Not exactly. There are two unknowns, $C_G$ and $C_L$ that are varying through the scrubber. Eqn. 15 provides a relationship for predicting the decay of the combination $C_G/H-C_L$ (i.e., the overall concentration driving force for mass transfer) as a function of cumulative residence time through the scrubber. This solution can be combined with Eqn. 7 to provide 2 algebraic equations for the two individual unknowns $C_G$ and $C_L$.

 

[KarenRei]

Not exactly. $C_G/H$ is the concentration that would exist in the liquid if the liquid were in equilibrium with the local species concentration $C_G$ in the bulk of the gas. Both $C_G$ and $C_L$ are varying simultaneously with time t through the scrubber.

Hmm, isn't that what I wrote? "equilibrium concentration of the species in the liquid phase" = "concentration that would exist in the liquid if the liquid were in equilibrium with the local species concentration", right? And "the rate of change ... per unit time that the flow progresses through the scrubber."" = "varying simultaneously with time t through the scrubber. ", right?

I apologize if my communication here is subpar. :)

Not exactly. There are two unknowns, $C_G$ and $C_L$ that are varying through the scrubber. Eqn. 15 provides a relationship for predicting the decay of the combination $C_G/H-C_L$ (i.e., the overall concentration driving force for mass transfer) as a function of cumulative residence time through the scrubber. This solution can be combined with Eqn. 7 to provide 2 algebraic equations for the two individual unknowns $C_G$ and $C_L$.

Hmm... okay, thinking about this here. Then the next step would be to express C_F0 = C_F_tfinal * X (where tfinal is the exit time and X is the percentage recycled)? Hmm. I'd still need to work out an equation to express C_F in terms of time in order to get C_F_tfinal. Already have the change in concentration gradient with respect to time, which could be integrated.. but then that would yield total concentration gradient that's occurred over the course of the scrubber, which isn't particularly useful... hmm... (just thinking out loud here)

 

[Chestermiller]

By the way, this is probably a stupid question, but how do you know that the liquid and gas don't equilibrate with one another before reaching the exit of the scrubber?

 

[KarenRei]

By the way, this is probably a stupid question, but how do you know that the liquid and gas don't equilibrate with one another before reaching the exit of the scrubber?

I can't say that they don't, but it's important to have as low of a liquid flow rate as possible, in order to minimize the required amount of ionic liquid.

By the way, how do you post math here? I put Eq. 7 in terms of C_G:

C_G = (f*C_F0 + (1-f)*C_G0 - f*C_L) / (1-f)

I integrate both sides of Eq. 15 to get:

ln(C_G/H-C_L) = t * a² / (3 * f) * (1/f + 1/((1-f)*H)) / (1/(5*D_L)+1/(D_G*H))

I now substitute C_G into the above:

ln((f*C_F0 + (1-f)*C_G0 - f*C_L) / (1-f)/H-C_L) = t * a² / (3 * f) * (1/f + 1/((1-f)*H)) / (1/(5*D_L)+1/(D_G*H))

Now I need to solve for C_L. Since it's getting long, I punch it into Wolfram Alpha. It returns a giant mess. Combining terms into new variables it simplifies to:

C_L = (H * e^(1/3 * a² * D_G * t * (-(H² * D_G)/(f² * (H * D_G + 5 * D_L)) + H/f² - (H * D_G)/((1 - f) * (H * D_G + 5 * D_L)) - (H * D_G)/(f * (H * D_G + 5 * D_L)) + 1/(1 - f) + 1/f)) - f * H * e^(1/3 * a² * D_G * t * (-(H² * D_G)/(f² * (H * D_G + 5 * D_L)) + H/f² - (H * D_G)/((1 - f) * (H * D_G + 5 * D_L)) - (H * D_G)/(f * (H * D_G + 5 * D_L)) + 1/(1 - f) + 1/f)) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

This simplifies with the addition of temporary variables to:

r = -(H * D_G) / (H * D_G + 5 * D_L)
q = H * e^(1/3 * a² * D_G * t * (H*r/f² + H/f² + r/(1 - f) - r/f + 1/(1 - f) + 1/f))
C_L = q * (1 - f) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

Treating t as t_final and setting C_F0 equal to X% of that value, I get:

C_F0 = X * q * (1 - f) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

Which solves to:

C_F0 = X * q - f * q + f * C_G0 - C_G0) / (f*H - f - H + X*f)

From this, C_L_final and C_G_final can be solved.

Does this look right?

(By the way, C_F0 should really be C_L0, right? I'm just using your terminology. But in plain English, "concentration of the specie in the liquid phase at entrance to the scrubber)

 

[KarenRei]

If I've solved this right. then all that's left is for you to explain what you meant in post #13 :)

 

[Chestermiller]

I can't say that they don't, but it's important to have as low of a liquid flow rate as possible, in order to minimize the required amount of ionic liquid.

By the way, how do you post math here? I put Eq. 7 in terms of C_G:

C_G = (f*C_F0 + (1-f)*C_G0 - f*C_L) / (1-f)

I integrate both sides of Eq. 15 to get:

ln(C_G/H-C_L) = t * a² / (3 * f) * (1/f + 1/((1-f)*H)) / (1/(5*D_L)+1/(D_G*H))

I now substitute C_G into the above:

ln((f*C_F0 + (1-f)*C_G0 - f*C_L) / (1-f)/H-C_L) = t * a² / (3 * f) * (1/f + 1/((1-f)*H)) / (1/(5*D_L)+1/(D_G*H))

Now I need to solve for C_L. Since it's getting long, I punch it into Wolfram Alpha. It returns a giant mess. Combining terms into new variables it simplifies to:

C_L = (H * e^(1/3 * a² * D_G * t * (-(H² * D_G)/(f² * (H * D_G + 5 * D_L)) + H/f² - (H * D_G)/((1 - f) * (H * D_G + 5 * D_L)) - (H * D_G)/(f * (H * D_G + 5 * D_L)) + 1/(1 - f) + 1/f)) - f * H * e^(1/3 * a² * D_G * t * (-(H² * D_G)/(f² * (H * D_G + 5 * D_L)) + H/f² - (H * D_G)/((1 - f) * (H * D_G + 5 * D_L)) - (H * D_G)/(f * (H * D_G + 5 * D_L)) + 1/(1 - f) + 1/f)) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

This simplifies with the addition of temporary variables to:

r = -(H * D_G) / (H * D_G + 5 * D_L)
q = H * e^(1/3 * a² * D_G * t * (H*r/f² + H/f² + r/(1 - f) - r/f + 1/(1 - f) + 1/f))
C_L = q * (1 - f) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

Treating t as t_final and setting C_F0 equal to X% of that value, I get:

C_F0 = X * q * (1 - f) + f * C_G0 - f * C_F0 - C_G0)/(f * H - f - H)

Which solves to:

C_F0 = X * q - f * q + f * C_G0 - C_G0) / (f*H - f - H + X*f)

From this, C_L_final and C_G_final can be solved.

Does this look right?

(By the way, C_F0 should really be C_L0, right? I'm just using your terminology. But in plain English, "concentration of the specie in the liquid phase at entrance to the scrubber)

I'm not going to check your "arithmetic" because that part is pretty straightforward. The hard part was arriving at Eqns. 7 and 15. Incidentally, when you integrated Eqn. 15, you left of the constant of integration. Eqn. 15 should have led to:
$$C_G/H-C_L=(C_{G0}/H-C_{L0})\exp{(-t/\tau)}\tag{16}$$where
$$\tau=\frac{3f\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}{\alpha^2\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}\tag{17}$$
I would carry along $\tau$ throughout the rest of the development.
Eqns. 16 and 7 give you two linear algebraic equations in two unknowns for $C_G$ and $C_L$.

To learn how to post the math the way we do it here, go to the INFO drop down menu and click on -Help/How-To, and then choose LaTex Primer.

 

[Chestermiller]

If I've solved this right. then all that's left is for you to explain what you meant in post #13 :)

I'll provide a write-up on the derivation of Eqn. 3 in the next couple of days.

 

[KarenRei]

Hmm, you're right - I did integrate incorrectly (only with respect to time). I'll start instead with your equations 16-17:

$$C_G/H-C_L=(C_{G0}/H-C_{L0})\exp{(-t/\tau)}\tag{16}$$where
$$\tau=\frac{3f\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}{\alpha^2\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}\tag{17}$$

... and as before put Eq. 7 in terms of C_G (I'll correct "C_F0" to "C_L0" for consistency):

$$C_G=\frac{fC_{L0}+(1-f)C_{G0}-fC_L}{1-f}\tag{18}$$

I substitute equation 18 into equation 16:

$$\frac{fC_{L0}+(1-f)C_{G0}-fC_L}{(1-f)H}-C_L=(C_{G0}/H-C_{L0})\exp{(-t/\tau)}\tag{19}$$

Now I need to solve for $C_L$; it's shorter with your recommended usage of a temporary variable (τ), so I'll do this by hand. I'll represent it as a function over time.

$$C_L(t) = \frac{\frac{fC_{L0}}{1-f}+C_{G0}+(HC_{L0}-C_{G0})\exp{(-t/\tau)}}{\frac{f}{1-f} + H}\tag{20}$$

I verify the previous result by using Wolfram and then proceeding to simplify their result to mine.

Setting t to $t_{final}$ and setting $C_{F0}$ equal to X% of that value, I get:

$$C_{L0} = X\frac{\frac{fC_{L0}}{1-f}+C_{G0}+(HC_{L0}-C_{G0})\exp{(-t_{final}/\tau)}}{\frac{f}{1-f} + H}\tag{21}$$

Which solves to:

$$C_{L0} = \frac{C_{GO}X(1-\exp{(-t_{final}/\tau)}}{\frac{f}{1-f}(1-X) + H(1-X\exp{(-t_{final}/\tau)})}\tag{22}$$

Also verified by matching up my hand calcs to Wolfram.

So that all looks good. :) Take your time with your response about Eqn 3. Assuming I've done this correct, I can go ahead and start integrating this into my spreadsheet (I can change things if your comments about Eqn. 3 alter anything, such as the diffusion coefficients)

And again - my great appreciation for your help!

 

[Chestermiller]

I still think that, in this problem, you were supposed to assume that residence time is long enough for the drops and gas to equilibrate. This would happen if $t_{final}$ were more than about $4\tau$ so that $\exp{(-t_{final}/\tau)}$ approaches zero. What would that give you in Eqns. 21 and 22?

 

[KarenRei]

"was supposed to"? As mentioned, this isn't a homework problem; I'm not a student. :)

There's a wide variety of species (with varying diffusion constants and Henry's law constants), they're not all equally important to absorb. The bulk of the stream is CO2, with a few percent nitrogen; argon, sulfuric acid and carbon monoxide at several dozen ppm; helium, sulfur dioxide and neon at around 10 ppm; water vapour and sulfur particulate at around 1ppm; hydrogen chloride, sulfuryl chloride, carbonyl sulfide, phosphorus pentoxide and ferric chloride at several hundred ppb; and so on down the line (of particular note, HF at around 5 ppb).

The importance of capturing each specie varies widely; indeed, there are some that I'd rather not capture. So there comes a point that increasing fluid flow and exposure time hits diminishing returns of usefulness. By contrast, ionic liquids are very expensive, and mass in this case is even more expensive, so keeping scrubber length and liquid mass down is very important. There's a balancing point to be met; at some point, further absorption is no longer useful. There's also a question of how to allocate resources - yes, one could increase liquid mass or scrubber length to achieve greater absorption from the gas stream, but that mass could also be instead dedicated to a higher head pump to get a smaller droplet size. The biggest optimization problem is the ionic liquid itself. In the last paper I read, theoretical analysis had been done on something like 150 thousand different cases ;) They're basically infinitely tuneable, with tradeoffs over composition / manufacturing difficulty, viscosity (and thus droplet size), thermal stability, absorption for various species, and other properties.

Basically, there is no single, simple "right answer", but I'm working to constrain the problem to determine what sort of conditions would be needed to achieve a sufficient level of scrubbing while balancing all other factors.

But if you'd like, for fun I could solve the equation for an equilibrium state ;)

 

[Chestermiller]

"was supposed to"? As mentioned, this isn't a homework problem; I'm not a student. :)

There's a wide variety of species (with varying diffusion constants and Henry's law constants), they're not all equally important to absorb. The bulk of the stream is CO2, with a few percent nitrogen; argon, sulfuric acid and carbon monoxide at several dozen ppm; helium, sulfur dioxide and neon at around 10 ppm; water vapour and sulfur particulate at around 1ppm; hydrogen chloride, sulfuryl chloride, carbonyl sulfide, phosphorus pentoxide and ferric chloride at several hundred ppb; and so on down the line (of particular note, HF at around 5 ppb).

The importance of capturing each specie varies widely; indeed, there are some that I'd rather not capture. So there comes a point that increasing fluid flow and exposure time hits diminishing returns of usefulness. By contrast, ionic liquids are very expensive, and mass in this case is even more expensive, so keeping scrubber length and liquid mass down is very important. There's a balancing point to be met; at some point, further absorption is no longer useful. There's also a question of how to allocate resources - yes, one could increase liquid mass or scrubber length to achieve greater absorption from the gas stream, but that mass could also be instead dedicated to a higher head pump to get a smaller droplet size. The biggest optimization problem is the ionic liquid itself. In the last paper I read, theoretical analysis had been done on something like 150 thousand different cases ;) They're basically infinitely tuneable, with tradeoffs over composition / manufacturing difficulty, viscosity (and thus droplet size), thermal stability, absorption for various species, and other properties.

Basically, there is no single, simple "right answer", but I'm working to constrain the problem to determine what sort of conditions would be needed to achieve a sufficient level of scrubbing while balancing all other factors.

But if you'd like, for fun I could solve the equation for an equilibrium state ;)

There is no "solving" involved. Just set the exponential terms equal to zero in Eqns. 21 and 22.

 

[KarenRei]

If we assume that it's equal to precisely 0, then the equations become very short :)

$$C_L(t) = \frac{\frac{f}{1-f}C_{L0}+C_{G0}}{\frac{f}{1-f} + H}\tag{20-eq}$$
$$C_{L0} = \frac{C_{GO}X}{\frac{f}{1-f}(1-X) + H}\tag{22-eq}$$

 

[KarenRei]

Hmm, this is strange. I'm working with the numbers in the spreadsheet, and concerning tau:

$$\tau=\frac{3f\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}{\alpha^2\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}\tag{17}$$

If we break that down:

$$3f$$

... is 3 times a small number, a fraction of a percent. Starting small (since that's ideal for me) at, say 0.04%, 3f equals 1.2e-3

$$\frac{1}{5D_L}+\frac{1}{D_GH}$$

... if we just haphazardly pick, say, the diffusion rates of CO2 in air (~1,4e-5 m²/s) and in water (~1,9e-9 m²/s), with the Henry's Law coefficient for CO2 in water (8.3e-1), the result is around 1e8

So the numerator is 1.2e5. As for the denominator, the right hand side is:

$$\frac{1}{f}+\frac{1}{(1-f)H}$$

That's 2,5e3. All that's left is the area. For a 10 micron mist at 0,13 m³/s , the surface area is 7,8e4m². Which we then proceed to square (6,08e9) and put on the denominator. So we have a huge 1.52e13 on the denominator(!). This dwarfs the numerator, leading to a tiny tau, which means that in my sample 2 second exposure, $\exp(-t/\tau)$ is such a small number that my spreadsheet simply writes "0" even though it's in scientific notation mode.

Okay, let's go with a larger mist - 100 micron. Surface area is now 7,8e3 m². $\exp(-t/\tau)$ is still so small that the spreadsheet can't display it in scientific notation.

Okay, let's go with a very coarse 1mm mist. Surface area is now 7,8e2 m². $\exp(-t/\tau)$ is still too small for scientific notation.

Okay, let's go with drops a whole centimeter across.. Surface area is now 78 m². Finally the spreadsheet can display $\exp(-t/\tau)$ in scientific notation: 2.88E-128. That can't be right, can it? To get $\exp(-t/\tau)$ down to equal 0.5 I have to drop the time to under 5 milliseconds. Isn't that... an unusually small figure for centimeter-sized drops?

ED: minor edit to the above. We're using a Henry's constant that's gas over liquid, not liquid over gas like the above that I quoted for CO2. But it makes no material difference since the value is so close to 1 - I just need to make sure I handle it right in the spreadsheet.

 

[Chestermiller]

Derivation of Eqn. 3

The objective of Eqn. 3 is to estimate the mass flux at the interface between the drop and the gas in terms of (a) the difference between the average species concentration in the drop and the the species concentration in the liquid at the interface $C_L-C_{LI}$ and (b) the difference between the average species concentration in the gas and the species concentration in the gas at the interface $C_G-C_{GI}$.

Within the scope of what we are trying to do, it is too computationally intensive to obtain an exact solution for these quantities (which involves solving partial differential equations for the compositions with respect to time and spatial position). Instead, we adopt an approximate, yet very accurate, method to express the interface flux algebraically in terms of the above concentration differences. This approach gives us a lower bound for the mass transfer coefficient and species flux at the interface, and thus, a lower bound to the concentration changes for a given value of the residence time at the exit of the scrubber. However, even though they give a lower bound, they still provide a close approximation to the true changes for cases of practical interest where the amount of mass transferred is on the order of (say) 1/3 of the equilibration amount or more. The approach we use is what an experienced heat and mass transfer expert (unabashedly, like myself) would do to attack this problem.

If we suddenly took a spherical drop of fresh liquid and placed it in an infinite ocean of fresh gas (and then allowed to diffusion to occur), the concentrations and mass fluxes initially would vary very rapidly with radial position close to the interface in both the liquid and gas, while away from the interface, at short times, the concentrations would hardly be disturbed. The mass flux at the interface would be very high.

As time progressed, the region where the concentrations and mass fluxes are affected would grow larger, and the mass flux at the interface would decrease (as a result of lower concentration gradients near the interface). Eventually, the concentration would be smoothly varying everywhere within the drop, and also in an effective region on the order of one diameter away from the interface in the gas. Beyond that time, the system would approach so called "asymptotic mass transfer behavior," where the basic shape of the spatial concentration variations don't change with time, and the mass transfer coefficients at the interface approach lower bound constant values. The amount of time it takes to reach this point would be on the order of $R^2/D$ (both on the liquid side of the interface and on the gas side of the interface), where R is the drop radius and D is the diffusion coefficient. This amount of time would typically be very short compared to the residence time in the scrubber. Therefore, the "asymptotic mass transfer behavior" would be characteristic of the vast majority of the scrubber.

Our game plan then is to determine (algebraic) asymptotic mass transfer relationships between the inward mass flux $\phi$ through the interface at r = R, and the concentration driving forces on the two sides of the interface.

I think I'll stop here for now, and get back to this tomorrow.

 

[Chestermiller]

Derivation of Eqn. 3

The objective of Eqn. 3 is to estimate the mass flux at the interface between the drop and the gas in terms of (a) the difference between the average species concentration in the drop and the the species concentration in the liquid at the interface $C_L-C_{LI}$ and (b) the difference between the average species concentration in the gas and the species concentration in the gas at the interface $C_G-C_{GI}$.

Within the scope of what we are trying to do, it is too computationally intensive to obtain an exact solution for these quantities (which involves solving partial differential equations for the compositions with respect to time and spatial position). Instead, we adopt an approximate, yet very accurate, method to express the interface flux algebraically in terms of the above concentration differences. This approach gives us a lower bound for the mass transfer coefficient and species flux at the interface, and thus, a lower bound to the concentration changes for a given value of the residence time at the exit of the scrubber. However, even though they give a lower bound, they still provide a close approximation to the true changes for cases of practical interest where the amount of mass transferred is on the order of (say) 1/3 of the equilibration amount or more. The approach we use is what an experienced heat and mass transfer expert (unabashedly, like myself) would do to attack this problem.

If we suddenly took a spherical drop of fresh liquid and placed it in an infinite ocean of fresh gas (and then allowed to diffusion to occur), the concentrations and mass fluxes initially would vary very rapidly with radial position close to the interface in both the liquid and gas, while away from the interface, at short times, the concentrations would hardly be disturbed. The mass flux at the interface would be very high.

As time progressed, the region where the concentrations and mass fluxes are affected would grow larger, and the mass flux at the interface would decrease (as a result of lower concentration gradients near the interface). Eventually, the concentration would be smoothly varying everywhere within the drop, and also in an effective region on the order of one diameter away from the interface in the gas. Beyond that time, the system would approach so called "asymptotic mass transfer behavior," where the basic shape of the spatial concentration variations don't change with time, and the mass transfer coefficients at the interface approach lower bound constant values. The amount of time it takes to reach this point would be on the order of $R^2/D$ (both on the liquid side of the interface and on the gas side of the interface), where R is the drop radius and D is the diffusion coefficient. This amount of time would typically be very short compared to the residence time in the scrubber. Therefore, the "asymptotic mass transfer behavior" would be characteristic of the vast majority of the scrubber.

Our game plan then is to determine (algebraic) asymptotic mass transfer relationships between the inward mass flux $\phi$ through the interface at r = R, and the concentration driving forces on the two sides of the interface.

I think I'll stop here for now, and get back to this tomorrow.

$\alpha$ is not the amount of surface area. It is the surface per unit volume. Even so, these results don't surprise me. All it is saying is that for that small value of the drop radius, the two phases equilibrate very rapidly.

I'm very concerned about your suggesting that the volume fraction of liquid is only 0.0004. With that small a value, the liquid does not have enough capacity to remove much of the undesirable species from the gas, even if the two phases equilibrated in the scrubber. What does your Eqn. 20-eq tell you about the ratio of the final- to the initial concentration of CO2 in the gas, assuming that f = 0.0004 and CL0 = 0? I get virtually no change.

 

[Chestermiller]

When the system reaches the asymptotic regime, the inward flux to the drop is changing very slowly with time. So we solve the mass transfer equations for the asymptotic case in which the inward flux of mass is taken to be virtually constant. At constant mass flux, the average concentration rises linearly with time, and, superimposed on the average concentration is a radially varying portion that is constant with time.

On the Liquid Side of the Drop Interface:
$$\frac{4}{3}\pi R^3\frac{\partial C}{\partial t}=4\pi R^2 \phi\tag{1}$$where $\phi$ is is the inward mass flux at r = R.
Solving for the rate of change of concentration in terms of the inward mass flux yields:$$\frac{\partial C}{\partial t}=\frac{3\phi }{R}\tag{2}$$
The transient mass transfer equation in spherical coordinates is given by:$$\frac{\partial C}{\partial t}=\frac{D_L}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial C}{\partial r}\right)\tag{3}$$Substituting Eqn. 2 into Eqn. 3 then yields:
$$\frac{\partial}{\partial r}\left(r^2\frac{\partial C}{\partial r}\right)=\frac{3\phi}{D_LR}r^2\tag{4}$$
Integrating once yields:
$$\frac{\partial C}{\partial r}=\frac{\phi}{D_L}\frac{r}{R}+\frac{K}{r^2}\tag{5}$$where K is the constant of integration. Since, by symmetry, the derivative of C with respect to r must be zero at r = 0, we must have that K = 0. Therefore Eqn. 5 becomes:$$\frac{\partial C}{\partial r}=\frac{\phi}{D_L}\frac{r}{R}\tag{6}$$
If we integrate again, and apply the boundary condition that, at r = R, $C=C_{LI}$ (where $C_{LI}$ is the liquid concentration at the interface), we obtain:$$C=C_{LI}-\frac{\phi}{2D_LR}(R^2-r^2)\tag{7}$$
The average species concentration in the liquid $C_L$ is given by the equation:
$$C_L=\frac{\int_0^R{4\pi r^2Cdr}}{\left[\frac{4}{3}\pi R^3\right]}\tag{8}$$If we substitute Eqn. 7 into Eqn. 8, we obtain:
$$C_L=C_{LI}-\frac{\phi R}{5D_L}$$or equivalently,$$\phi=\frac{5D_L}{R}(C_L-C_{LI})\tag{9}$$
Eqn. 9 is our desired result for the liquid side of the interface.

On the Gaseous Side of the Drop Interface:
If the volume fraction of liquid f is very small, we can model the gas side of the interface as a drop in an infinite ocean of gas. For this situation, we have:
$$4\pi r^2D_G\frac{dC}{dr}=4\pi R^2 \phi\tag{10}$$or equivalently,$$D_G\frac{dC}{dr}=\phi \frac{R^2}{r^2}\tag{11}$$Integrating between r = R and very large r gives: $$C_G-C_{GI}=\frac{\phi}{RD_G}\tag{12}$$ where $C_G$ is the species concentration in the gas far from the interface and $C_{GI}$ is the species concentration at the interface. From this, it follows that $$\phi =\frac{D_G}{R}(C_G-C_{GI}) \tag{13}$$
Eqn. 13 is our desired result for the gaseous side of the interface.

 

[KarenRei]

Thank you for the derivation - that was very thorough. I can't imagine how long it would have taken me to come into something like that on my own.

Re α: Aha, that was the bug - I was just using total surface area, not surface area per unit volume. Now all of the numbers make sense. Excellent.

I'm very concerned about your suggesting that the volume fraction of liquid is only 0.0004. With that small a value, the liquid does not have enough capacity to remove much of the undesirable species from the gas, even if the two phases equilibrated in the scrubber. What does your Eqn. 20-eq tell you about the ratio of the final- to the initial concentration of CO2 in the gas, assuming that f = 0.0004 and CL0 = 0? I get virtually no change.

That was just a starting point. After working with the numbers all morning I'm currently looking at 0.12% (0.0012). As examples of the capture rates I'm getting:

CO2: 0.017%
N2: 7.35E-06
Ar: 2.39E-05
H2SO4: 100.00%
CO: 1.62E-05
He: 6.43E-06
SO2: 1.67%
Ne: 7.71E-06
H2O: 9.30%
HCl: 43.36%
H3PO4: 95.99%
OCS: 0.036%
...
HF: 68.61%
...

... and so forth. It might benefit to being upped some more to catch more of the H2O, HF, HCl, etc, but it's already catching all of the sulfuric and essentially all phosphoric acid. They have extremely high Henry's constants. You're absolutely right that it doesn't capture much CO2. But I don't want to capture much CO2; CO2 is the bulk gas, my interest is the minor acidic species. And the interest in their acquisition, not in purifying the exhaust stream; the exhaust doesn't need to be "clean".

I had initially had some big concerns about my HCl numbers, they were a small fraction of this. But digging through papers looking into the issue I found that Henry's Law doesn't adequately describe it well at low concentrations; it has a "reactivity" component as well as an "absorption" component that needs to be factored in, and thus greatly increases how well it's absorbed at low concentrations. Technically there's many species that would be ideally represented like this, but I'm not sure how much of the table I'll be able to fill in that way. At least it means that the data I'm working with is a pessimistic case.

I had also initially had some huge concerns about the power requirements the spreadsheet was coming up with for recycling the liquid stream (outgassing via heating; the solubilities vary greatly with temperature). But I realized I had forgotten to account for the heat exchanger; when that's factored in the numbers come out much more reasonable.

I think this will get me where I'm needing to go - thank you :)