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## Hot answers tagged diffusion

13

There are two mechanisms for mixing at a liquid-liquid interface, firstly diffusion and secondly physical agitation. Diffusion is negligably slow in liquids, it takes days for solutes to travel a few centimetres, so the mixing is dominated by physical agitation e.g. wave action, convention currents, wind mixing etc. In this particular case it's hard to ...

13

Let us try to rewrite the equation in approximate form of finite differences: $$\frac{A(x,t+\Delta t)-A(x,t)}{\Delta t} = C_3\frac{A(x+h,t)+A(x-h,t)-2A(x,t)}{h^2} +$$ $$+ C_2 \frac{v(x+h,t)A(x+h,t)-v(x-h,t)A(x-h,t)}{2h} + C_1 A(x,t)+C_0$$ Where $\Delta t$ -- is a time step, and $h$ -- space step. The expression becomes your PDE, in the limit $\Delta t\to0, ... 11 As Ted Bunn said, the linear concentration profile is only a steady state if there is a steady inflow at one end and a steady outflow at the other. This net flow is what preserves the concentration gradient. With the "closed box" boundary condition instead, there is indeed an error in your reasoning because the linear profile is no longer a steady state. ... 5 Nobody has thus far touched on the probability that freshwater at a river/ocean interface is quite likely to be muddy. What does this mean? It means that the water is likely to contain a stable suspension of silicate micro- or nanoparticles, which are unable to aggregate due to short range electrostatic repulsion. This is what is called a colloid. The ... 4 The differential is used to specify that the number is for a "differential range", which is a way to remind you that the notions involved are somewhat fuzzy. Let me give a purely mathematical example. Suppose I tell you that I am going to pick an arbitrary real number between 0 and 10, with the likely hood of a number being picked being proportional to the ... 4 The previous answers do not exhaust this question, so I will add a solution to the problem as posted. The question is: what is the probability distribution for the meeting time of two diffusing spheres of radii$r_1$and$r_2$starting at initial separation R. By transforming to relative coordinates (as Peter Shor said), you reduce the problem to one ... 4 The error is in your intuition. Your calculation is correct. One thing that might help your intuition is to think about what happens at the edges of the region under consideration. There must be a steady inflow from one end and a steady outflow from the other end. Fluid is continually flowing "downhill" (from high concentration to low), but the source at ... 4 Marek suggested I post my comment (which doesn't completely answer the question) as an answer. Here it is: Suppose you have two Brownian motions with diffusion coefficients$D_1$and$D_2$, which start at the same point at$t=0$. Let$x_i$be the average displacement vector for particle$i$after time$t$. Then,$\langle x_i^2 \rangle = 2D_it$, where ... 4 As an alternative to Christian Blatter's heat interpretation,$A$might describe the concentration of particles adsorbed onto a one-dimensional substrate surface (or a two-dimensional one, where we ignore one of the dimensions). New particles are adsorbed at rate$C_0$per unit length. Adsorbed particles detach from the surface at rate$-C_1$per particle. ... 4 For Brownian motion, Langevin equation, Fokker-Planck equations, Stochastic process.. from the viewpoint of physicists, the following are standard references: Brownian Motion: Fluctuations, Dynamics, and Applications The Fokker-Planck Equation: Methods of Solutions and Applications Handbook of Stochastic Methods: for Physics, Chemistry and the Natural ... 3 I initially suspected that the picture here is one of a sand bar next to deeper water, not of two "seas" not mixing, where the light-colored water is light because it is shallow, and we are seeing the sand below, and the dense region is dark because it is too deep to see the bottom, and the light is absorbed rather than reflecting back.The foam we see at the ... 3 Diffusion is an stochastical process where a single particle can move in each direction with the same probability. Another description of the diffusion coefficient is the equation: $$D = x^2/2t$$ where$t$is the time and$x^2$is the mean squared displacement of the particles at this time. The mean squared displacement$x^2$can be seen as the statistical ... 3 There are two aspects to be considered here: Does the wall reflect or absorb the radiation? If the wall reflects, is this specular or diffuse reflection? For both of these sub-questions, the answer depends on the material that the wall is made of and on the wavelength of the infrared radiation. At wavelengths close to the visible (for example, at 0.8 ... 3 I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second. It's easier to explain if we temporarily imagine$\phi$represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of$\phi$due to the ... 3 What you are seeing are artifacts caused by the hard-wall type kernel you are using. These patterns consist of horizontal and vertical streaks with a definite periodicity, which gets more pronounced as you make the averaging box bigger. The streaks occur because the Fourier transform of a hard-wall box has zeros at certain wavenumbers. To get rid of them, ... 3 There are two phenomena present diffusion, which happens due to inhomogeneity in concentration. Particles "want to" go from areas of higher concentration to the lower ones. One can write this in the form of diffusion current $$J_{diff}(x) = - D \nabla \rho(x)$$ where$\rho(x)$is the concentration. This expression is known as Fick's law but it's actually ... 3 Here is an answer in the first case (unbounded domain). As already noted, the difference between the positions of the two particles performs a Brownian motion starting from point$(d,0,0)$with diffusion coefficient$D=D_1+D_2$and one is interested in the time$\tau$of the first hitting of the ball$B(r)$around zero with radius$r=r_1+r_2$when ... 3 If the drop is very much static (in still water) and of similar fluid properties to the water around it (so that the ink just labels some initial region), then this is the correct equation to use. If, however, you want to treat the ink as having distinct properties from the water, then you want the Navier-Stokes equations. Since you are interested in ... 3 The diffusion equation is a partial differential equation. The unknown quantity is a function$C(x,t)$. To complete the problem statement you need to specify an initial condition (at$t=0$) and boundary conditions. I'm guessing that your boundary conditions are at infinity, so we take $$C(x,t) \rightarrow 0,\ x\rightarrow \pm \infty.$$ We take a delta ... 2 Think about it physically. How much gas is in the tube? There are three sources: The gas coming in at$x=0$which was already in solution. The gas being carried away by the fluid at the other end,$x=L$Diffusion across the walls of the tube. These three terms are represented in the equation $$\frac{d}{dt} \left( A \int_{0}^{L} U(x,t) dt \right) = ... 2 It's just the usual conservation law. When something is conserved then its amount in some observed region can't disappear but must be flowing in or out. So this means that time derivative of the integral of the quantity in a given volume (the LHS) must be equal to total flux associated with the quantity through a surface of that volume (the RHS). In your ... 2 Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so? Or does it always dissipate quickly? I read this and think to myself "optimization problem". Firstly, you should know the following, which is the law of diffusion:$$\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}$$For ... 2 This is covered in the standard convection-diffusion type of equation:$$ \frac{\partial C}{\partial t} + \vec{u} \cdot \nabla C= D \nabla^2 C$$Where$C$is the smoke concentration, and$D$is the diffusion coefficient of smoke. While the air may be stagnant initially, it will come to move due to buoyancy effects, thus giving some non-zero air velocity ... 2 This was intended to be a comment, but is too long so I will post it as an answer. First of all, a disclaimer, I am a physict and all that I know about quantitative finance comes from self-learning, so please feel free of correcting me if I am mistaken (also in the physics stuff, of course!). I have been doing a little of research, and perhaps you are right ... 2 I would guess you mean self diffusion: see http://en.wikipedia.org/wiki/Self-diffusion for details. Suppose you take an aqueous solution of (for example) salt that is uniform so there are no concentration gradients. There is no net diffusion, but the sodium and chloride ions wander around due to random thermal motion, so if you watch a particular sodium ... 2 I asked the oceanologist (Nikolai Koldunov) about this photo. Here is his answer: In the ocean even if the difference of density is small (e.g., of the order$0.1\,kg/m^3\$) the process of mixing between two water masses is rather slow (without strong turbulence). The picture probably was taken close to the estuary of a big river. In this case ...

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The diffusion is determined from the broadening of the usual elastic scattering peaks. This is known as quasielastic scattering. There is no requirement for the neutron speed to be related to the diffusion speed. Have a look at this book for an introduction to the subject.

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Its worth pointing out the separation of two similar liquids is a common experiment. The diffusion of the liquids into each other is governed by Fick's Law but can also be understood in terms of Entropy of Mixing. The key to this puzzle is to really understand it in terms of entropy. Although the Black and Tan shown in the picture will eventually mix ...

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If I were to hazard a guess, I would say that it's due to the intermolecular forces. The London dispersion force should increase with density (and hence pressure) making the gas molecules more attracted to each other. Now, C2H4 is lighter than CO2 and thus has less intermolecular force. So, when you have just a trace of CO2 in C2H4 at high pressure, it ...

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