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

  1. Feb 21, 2017 #1
    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:

    tmp.png

    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)
     
  2. jcsd
  3. Feb 22, 2017 #2
    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.
     
  4. Feb 22, 2017 #3
    Thank you very much for your response.

    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)
     
    Last edited: Feb 22, 2017
  5. Feb 22, 2017 #4
    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?
     
  6. Feb 22, 2017 #5
    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.
     
    Last edited: Feb 22, 2017
  7. Feb 22, 2017 #6
    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.
     
  8. Feb 22, 2017 #7
    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.
     
  9. Feb 22, 2017 #8
    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?
     
  10. Feb 22, 2017 #9
    Hmm, I took it from one of the pages I found. Perhaps I transcribed it wrong.

    Yes, your example is perfectly reasonable. :)
     
  11. Feb 22, 2017 #10
    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.
     
  12. Feb 23, 2017 #11
    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:
    $$\frac{d\ln{(C_G/H-C_L)}}{dt}=- \frac{\alpha^2\left[\frac{1}{f}+\frac{1}{(1-f)H}\right]}{3f\left[\frac{1}{5D_L}+\frac{1}{D_GH}\right]}\tag{15}$$
     
    Last edited: Feb 23, 2017
  13. Feb 23, 2017 #12
    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.
     
  14. Feb 23, 2017 #13
    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.
    I explicitly stated that R is the radius of a drop.
    I'll explain that later.
    I'll explain that later. This is already the result of integrating the concentration profiles with respect to radial location.
    That was all integrated out with respect to radius. I'll get to all this later.
     
  15. Feb 23, 2017 #14
    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)
     
  16. Feb 23, 2017 #15
    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.

    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##.
     
  17. Feb 23, 2017 #16
    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. :)

    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)
     
  18. Feb 23, 2017 #17
    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?
     
  19. Feb 23, 2017 #18
    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)
     
    Last edited: Feb 23, 2017
  20. Feb 23, 2017 #19
    If I've solved this right. then all that's left is for you to explain what you meant in post #13 :)
     
  21. Feb 23, 2017 #20
    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.
     
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