# Tag Info

## Hot answers tagged diffusion

40

Summary: I find a formula for the diameter of a bubble large enough to support one human and plug in known values to get $d=400m$. I'll have a quantitative stab at the answer to the question of how large an air bubble has to be for the carbon dioxide concentration to be in a breathable steady state, whilst a human is continuously producing carbon dioxide ...

22

Maybe I should turn the comment to an answer. The physics of the situation is the same as when one can upturn a water glass with the water not falling out. The atmospheric pressure keeps it in. There exist the diving bells with open bottoms . As they are lowered the pressure in the air goes up to balance the water pressure, because the lower in the water ...

17

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 ...

15

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 $\Delta t\to0, ... 13 This is to summarize some of the excellent comments made previously by participants of this discussion, and to emphasize a couple of important points. 1) The original question implied that gas exchange between the bubble and surrounding water may be enough to sustain indefinitely breathing organisms inside. However this does not seem to be possible ... 11 As Ted Bunn said, the linear concentration profile is only a steady state if there is a steady inflow at one end and a steady outflow at the other. This net flow is what preserves the concentration gradient. With the "closed box" boundary condition instead, there is indeed an error in your reasoning because the linear profile is no longer a steady state. ... 7 Lenses and glass bottles are transparent. As you quoted above, the different has to do with diffusion. Here is an example of an image through a transparent object: Here is an example of a translucent object: This is an example of how diffusion causes translucency: As light passes through a translucent object, it either enters or exists a rough ... 6 According to this website, the diffusion coefficient of methane (which is also produced in farts) is about 20${\rm m}^2$/s. 5 Living human bodies typically have a thin envelope of air warmed by conduction, and this warm air forms a rising plume. Classic photo: the plume of warm air above a human body, visualized by Schlieren Photography: ALWAYS SILENT, SOMETIMES DEADLY! Unless your subject is nude, any ejected gas will inflate their clothing and then leak promptly from ... 5 The previous answers do not exhaust this question, so I will add a solution to the problem as posted. The question is: what is the probability distribution for the meeting time of two diffusing spheres of radii$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 ... 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 ... 5 Diffusion is a stochastic process where a single particle can move in each direction with the same probability. Another description of the diffusion coefficient is the following 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 interpreted ... 5 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 ... 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 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 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 ... 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 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 It's not that the random motion decreases when the flow rate increases. It is only that the random motion stays the same but the coherent motion dominates. If the diffusion velocity in a gas is 1 and the convective velocity of the flow is 1000 (units don't matter), then the diffusive action can be pretty safely ignored. The important thing to remember is ... 4 It depends what you mean by diffusion. If you're considering a system at equilibrium, e.g. a sodium chloride solution, and you're trying to extract energy from the diffusional motion of the sodium and chlorine atoms then you can't extract any energy. Your friends idea of a tiny turbine is effectively the same as the Maxwell's demon idea for extracting ... 4 Since the comment answered your question I'll just go ahead and set out a more generalised version. It's straightforward to simplify things back down to your case. Consider the following continuity equation: $$\dot{N}(x,t) = -\nabla\cdot\vec{\Gamma}(x,t) + S(x,t),$$ where$\vec{\Gamma}(x,t)$is the flux (in your case$\vec{\Gamma}(x,t)= -D\nabla N(x,t)$, ... 4 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 ... 4 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 ... 3 You almost give the answer in your question: With increasing flow rates, the inertial forces become larger than the viscous forces We can state it more precisely: The inertial forces scale quadratically with the flow speed$U$:$\rho \mathbf u\nabla \mathbf u\sim \rho U^2/L$, while the viscous forces scale linearly:$\mu \nabla^2\mathbf u \sim \mu ...

3

Let's consider the expression we obtain by removing the Heaviside step function from the Green's function. It is a solution to the diffusion equation, viz., $$(\partial_t - k\nabla^{2}) \left(\frac{1}{4\pi k t}\right)^{3/2} e^{-r^2/4kt} = 0$$ Furthermore, one can show that \lim_{t\rightarrow 0^{+}} \left(\frac{1}{4\pi k t}\right)^{3/2} ...

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

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 ...

3

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 ...

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible