# Can molecules diffuse upstream when emptying a cup of tea?

I was drinking some tea, and I randomly decided to pour some into the toilet in an unbroken stream; can particles travel "upstream" quickly enough to reach the mug of tea? How fast can diffusion occur in this system?

What if I pour it into highly concentrated toxic waste? Given that even trace amounts could make it toxic, should I not drink my tea?

• @jonsca: To me it looks more like, "If I pour some of liquid X into liquid Y, can any of liquid Y diffuse back up the stream into my vessel of liquid X?" – Aesin Sep 3 '12 at 22:02
• @Aesin That makes sense. If you can edit and formulate this into more of a "chemistry" question, be my guest. – jonsca Sep 3 '12 at 22:07
• Aesin seems to have the correct idea formulated in a more mathematical point of view. Sorry.. I was describing a layman's point of view. – James Sep 3 '12 at 22:48
• (Note that diffusing concentrations of discrete classical particles cannot be continuous in a fixed volume - I removed that bit for that reason.) – Aesin Sep 5 '12 at 17:48
• A bit oddly worded, but it seems clear to me. Also, I'd say diffusion is as much chemistry as physics, since we use it all the time. I mean, that is why we stir things, right? – Canageek Sep 6 '12 at 19:04

## 2 Answers

Particles/solutes are always diffusing in all directions. The question to ask is, "are they diffusing fast enough to overcome being swept along by the flow?"

Short Answer: No. Don't try it, though.

Long Answer:

One governing equation that can be used to answer this question is the following:

$$\overline v\!\left(\vec r\right)\cdot \overline\nabla C_i\!\left(\vec r\right) = D_i\nabla^2C_i\!\left(\vec r\right)\tag{1}$$

This is the (simplifed) steady-state mass transport balance for species $i$ at a given point $\vec r$ in space. The left-hand side represents transport due to convection; the right-hand side encapsulates transport due to diffusion. $\overline v\!\left(\vec r\right)$ is the fluid flow velocity, $C_i\!\left(\vec r\right)$ is the concentration, and $D_i$ is the diffusivity. For concision, the explicit dependence on $\vec r$ will be omitted below.

Given that we're most interested in the transport in the direction of the flow of the tea, if we approximate the stream as a cylinder of uniformly flowing liquid with velocity $v_z$ and assume that concentration variations perpendicular to the flow are unimportant, Eq. $\left(1\right)$ becomes:

$$v_z{dC_i\over dz} = D_i{d^2C_i\over dz^2} \tag{2}$$

if we choose the $z$-axis as lying along the stream of tea.

The most useful way to use Eq. $\left(2\right)$ is to non-dimensionalize it. That is, to re-write it so that all of the variables and coefficients have no units. For this problem, $z$ and $C_i$ are the variables to be non-dimensionalized, whereas $v_z$ and $D_i$ are parameters to be grouped into non-dimensional groups. So, let's define:

$$\eta = {z\over L}; \Theta = {C_i\over C_i^*} \tag{3}$$

$L$ is some length scale representative of the system of interest. Here, the best choice is probably the height you're holding the cup over the surface of the toxic waste. $C_i^*$ is a representative concentration of a chemical species of interest, such as the nastiest player in that concentrated toxic waste. To substitute these definitions into Eq. $\left(2\right)$, I'll rewrite them as $z=\eta L$ and $C_i = \Theta C_i^*$:

$${v_zC_i^*\over L}{d\Theta\over d\eta} = {D_iC_i^*\over L^2}{d^2\Theta\over d\eta^2} \tag{4}$$

The terms with derivatives in Eq. $\left(4\right)$ are now dimensionless. To make the coefficients non-dimensionless, divide both sides by $D_iC_i^*/L^2$:

$${v_zL\over D_i}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{5}$$

The new grouping of parameters in Eq. $\left(5\right)$ has a special name: the Péclet number, abbreviated $\mathrm{Pe}$.

$$\mathrm{Pe}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{6}$$

The magnitude of $\mathrm{Pe}$ indicates the relative importance of convection relative to diffusion in a problem like this one. If the Péclet number is large $\left(\mathrm{Pe}\gg 1\right)$, it means that convection dominates the mass transfer behavior; if it's small $\left(\mathrm{Pe}\ll 1\right)$, then diffusion is the key phenomenon. If it's in the vicinity of unity $\left(\mathrm{Pe} \sim 1\right.$, or approximately $\left.0.3\leq \mathrm{Pe}\leq 3\right)$, then both diffusion and convection must be considered together.

So, can pouring tea into toxic waste poison your tea?

Well, setting aside the questionable wisdom of bringing a drink into a toxic waste area in the first place, what are some representative numbers for $D_i$, $v_z$, and $L$? The upper limit for the diffusivity of a solute in water is about $2\times 10^{-6}\,\mathrm{cm^2\over s}$ (see, e.g., here). Since we want to know if diffusion can outrace convection, we should give it the best chance possible and selecting this maximum value will do that. (That is to say, maximizing $D_i$, which is in the denominator of the definition of $\mathrm{Pe}$, will tend to minimize $\mathrm{Pe}$.)

Similarly, selecting the minimum practical values of $L$ and $v_z$ will also tend to minimize $\mathrm{Pe}$, giving us a worst-case scenario for whether or not you face a gruesome death from toxic waste poisoning. Let's say, optimistically(???), that one inch is the minimum height you can hope to pour your tea from without it splashing back into the cup. If we (rather loosely) approximate the velocity of the tea stream at the toxic waste surface as that of a freely falling particle, the velocity will be $v_z=\sqrt{2gL}$.

So, our estimated Péclet number is (Google Calculator result):

$$\mathrm{Pe} = {v_zL\over D_i} = {\sqrt{2gL^3}\over D_i} = 8.96\times 10^7$$

This is a huge Péclet number. There's no way any of that toxic waste will diffuse up into your tea. (Please don't test this, though. Splashing, breathing toxic fumes, falling in, etc. are very real possibilities.)

As a further note, observe that in combining terms to form Eq. $\left(5\right)$, the reference concentration $C_i^*$ cancelled out completely. So, in this case, the answer is the same regardless of how concentrated that toxic waste is.

was it those crazy tea fanatic Cuban scientists again?

The particles can supposedly travel "upstream" if you were to say, pour it into your toilet as closely as your hand can get without a hazmat team busting down your door. But I would be more concerned about the sneaky particles that instead diffuse through the air itself, and not the stream you are considering...

Also, note that the more highly concentrated solution Y is, and the more pure solution X is, the more likely it is for solute in solution Y to end up in solution X. Higher concentrations flow into lower concentrations. And the closer the two solutions are during the decanting, the easier it is for this counter-current to form. But it is assumed that at more than 1 centimeter height, a counter-current should not form according to that article.