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

#### 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.

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#### Chestermiller

Mentor
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

Mentor
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

Mentor
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 :)

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

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