Understanding Diffusion: Steady State and Boundary Conditions Explained

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In summary, there is a diffusion problem where a source of particles is diffusing outwards in a box with boundary conditions set so that the density is zero at the edge of the box. The problem can be solved in steady state, but the mass flux at the boundaries is zero. This leads to a paradox, but it is resolved by considering the source term as a steady input of material. The mass flux is given by \rho AV, not the derivative (d\rho/dx), and even though \rho is zero at the boundary, it is nonzero at an infinitesimal distance away, resulting in a nonzero mass flux. This is confirmed by the fact that the outward velocity outside the box needs to be very large to maintain the density
  • #1
mordechai9
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Something has been bothering me recently regarding a diffusion problem. Consider a source of particles in a box, diffusing outwards. Set the boundary conditions so that the density is zero at the edge of the box.

We can solve this problem in "steady state" and we find essentially a parabolic (or x^2) dependence of the density inside the box, dropping away from the source. However, since the density is zero at the boundaries of the box, the mass flux there is zero (rho*v*A). What's going on here? How can we have a steady state problem with mass clearly leaving the box without any mass flow actually going out of the box?

Does this simply mean that the mass interpretation is incorrect for this example?
 
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  • #2
mordechai9 said:
Something has been bothering me recently regarding a diffusion problem. Consider a source of particles in a box, diffusing outwards. Set the boundary conditions so that the density is zero at the edge of the box.

We can solve this problem in "steady state" and we find essentially a parabolic (or x^2) dependence of the density inside the box, dropping away from the source. However, since the density is zero at the boundaries of the box, the mass flux there is zero (rho*v*A). What's going on here? How can we have a steady state problem with mass clearly leaving the box without any mass flow actually going out of the box?

Does this simply mean that the mass interpretation is incorrect for this example?

I don't see any evidence that mass is "clearly leaving the box" from what you've said here. Would you extrapolate on where this information comes from? You said the flux there is zero, so nothing is passing through the boundaries.
 
  • #3
Ah, well, you have a "source" inside the box. Thus mass is being created (or injected) into the box, and unless it's flowing out, you couldn't have a steady state problem.
 
  • #4
mordechai9 said:
Ah, well, you have a "source" inside the box. Thus mass is being created (or injected) into the box, and unless it's flowing out, you couldn't have a steady state problem.

In diffusion problems, I always took source to mean where the material is all contained, dense to a point in the box. Not that material is actually being injected into the box, just that that's the point the atoms are diffusing from.

So if I spray perfume in one corner of the room and then leave the room and close the door, the place where I sprayed the perfume is the source. No additional perfume is being injected, it's just diffusing from that point until it is evenly distributed throughout the room.
 
  • #5
Right, well, that can't be steady state though -- the perfume would just diffuse to fill the room, and then you wouldn't have a source term in the equation any more. The source term has to act like a steady input of material for this problem to be steady state.
 
  • #6
mordechai9 said:
However, since the density is zero at the boundaries of the box, the mass flux there is zero (rho*v*A).

Whoa! Just because some curve has a value of zero at some point doesn't mean its slope is also zero (and the slope corresponds to the magnitude of flux).
 
  • #7
No... this is incorrect. The mass flux is given by rho*V*A, and it is the value of the density (rho) that shows up, not the derivative (d(rho)/dx).
 
  • #8
I believe you're using Fick's First Law, which assumes steady-state:

f5b7bf8f1da6d8184514659ee8ba07b1.png


So yeah, you're saying that phi (in this equation) is zero, but that doesn't mean that d(phi)/dx is. You have to solve the differential equation for phi still and get it out of that differential form. Then you can apply your boundary conditions.
 
  • #9
mordechai9 said:
No... this is incorrect. The mass flux is given by rho*V*A, and it is the value of the density (rho) that shows up, not the derivative (d(rho)/dx).

The mass flux into any infinitesimal element is [itex]\rho AV[/itex], but the mass flux out is [itex]\rho AV+\frac{\partial}{\partial x}(\rho AV)dx[/itex]. [itex]\rho[/itex] is not uniform. Even though it is zero at the boundary, it is nonzero an infinitesimal distance away, and thus the mass flux is nonzero. Does this make sense?
 
  • #10
Pythagorean - Fick's law doesn't assume steady state... You get steady state just by assuming all time derivatives are equal to zero.

Mapes - Actually, the mass flow out is just [tex]\int \rho V \cdot da[/tex], where you integrate over the surface of the box. We're not talking about the flow into and out of a differential element, we're talking about the flow into/out of a closed surface, and that's all. If rho is zero at the boundary, you don't have flow in or out, end of story.
 
  • #11
mordechai9 said:
Pythagorean - Fick's law doesn't assume steady state... You get steady state just by assuming all time derivatives are equal to zero.

Mapes - Actually, the mass flow out is just [tex]\int \rho V \cdot da[/tex], where you integrate over the surface of the box. We're not talking about the flow into and out of a differential element, we're talking about the flow into/out of a closed surface, and that's all. If rho is zero at the boundary, you don't have flow in or out, end of story.

So when you ignore our interpretation and use your own interpretation, you find a paradox, as you pointed out in your first post. What does that tell you?

EDIT: Or, looking at it another way, what does the outward velocity need to be outside the box to maintain the density of particles at the outer edge of the box at zero? To get arbitrarily close to zero density, the velocity needs to grow very large. The product is nonzero. The flux just inside and just outside the box is equal and nonzero.
 
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Related to Understanding Diffusion: Steady State and Boundary Conditions Explained

What is diffusion interpretation?

Diffusion interpretation is a scientific method used to analyze and interpret the movement of particles in a substance over time. It involves understanding how molecules and particles move and interact with each other.

How is diffusion interpretation used in scientific research?

Diffusion interpretation is used in various fields of science, such as biology, chemistry, and physics. It helps scientists understand the behavior of substances and their properties, which can lead to the development of new materials and technologies.

What are the main principles of diffusion interpretation?

The main principles of diffusion interpretation include Brownian motion, Fick's laws of diffusion, and the random walk model. These principles explain how particles move and spread in a substance over time.

What are the practical applications of diffusion interpretation?

Diffusion interpretation has many practical applications, including drug delivery systems, air and water pollution control, and material design. It is also used in industries such as food and beverage, pharmaceuticals, and environmental engineering.

What are the limitations of diffusion interpretation?

Diffusion interpretation is limited by the assumptions and simplifications made in its models. It may not accurately represent real-world systems and can be affected by external factors such as temperature and pressure. Factors such as particle size and shape can also affect diffusion rates.

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