I hope this a question that can be answered and isn't too vague. I'm also only after a very rough answer.

I'm adding a small amount (~500mL) of sugar solution into a carboy with a much larger volume of water (~20L).
I don't particularly want to stir it in very much because there is a lot of sediment that has already settled out.

I'm wondering roughly how long it would take for the sugar water to diffuse roughly evenly through the larger volume?

  • $\begingroup$ this will depend heavily on the modus of your "adding" - pour in a thin stream? place next to (above/below/at the side) the water and remove a barrier? sprinkle (very small drops over large surface)? As sugar solution and water have different densities, a gravity flow will ensue in most cases (apart from the sugar-below-remove-barrier one), that is very susceptible to the initial conditions and will transport sugar in a far greater flux than diffusion. $\endgroup$
    – bukwyrm
    Apr 20, 2018 at 11:23

1 Answer 1


You can get a rough idea from Fick's law (keeping to 1D for simplicity):

$$J = -D \frac{\partial \phi}{\partial x}$$

$J$ is the diffusion flow rate, $D$ is the diffusion coefficient (about $10^{-9}m^2s^{-1}$ for water and $\phi$ is the concentration in $moles.m^{-3}$. let's assume a 1 molar sugar solution (i.e. $10^3$ moles.$m^{-3}$). When you first add the sugar solution the concentration gradient is very high, so lets assume the mixing is partly under way and the sugar has diffused 10cm. That means $\partial \phi$ is $10^3$ and $\partial x$ = 0.1.

$$J = -10^{-9} \frac{10^3}{0.1} = -10^{-5}moles.m^2.s^{-1}$$

Suppose your blob of sugar solution is a freely floating sphere of volume 500mL containing 0.5 moles of sugar, then it's radius is about $0.05m$ and hence the surface area is $0.03m^2$. The flow rate out of the drop is therefore:

$$flow rate = 10^{-5} \times 0.03 = 3 \times 10^{-7} moles.s^{-1}$$

You can't just divide the amount of your sugar (0.5 moles) by the flow rate to get the time to disperse all the sugar, because the concentration gradient is changing all the time. I've just chosen what seems to me to be a reasonable average concentration gradient. However doing the division will give you a rough idea of the timescales, and it comes out at around 1.6 million seconds.

Unless you're very patient I would stir your carboy.

  • $\begingroup$ You're now using $J$ for two different quantities, this confuses me. Maybe it is possible to deduce the time-scale from the transport equation for $\phi$? $\endgroup$
    – Bernhard
    Feb 16, 2012 at 15:44
  • $\begingroup$ Oops, sorry, that was one copy and paste too far. I've corrected the 2nd equation. My conclusion remains valid though. The timescale remains of order two million seconds, so it's too long if for example you're adding sugar to get secondary fermentation. I still think you need to stir the carboy. $\endgroup$ Feb 16, 2012 at 16:01
  • $\begingroup$ If you really want to do this properly you need to solve the diffusion equation for spherical symmetry. In practice the blob of sugar solution will fall to the bottom and form a rough hemisphere so the sugar blob and carboy can be modelled as (half of) concentric spheres. I had a quick Google to see if I could find a ready baked solution for this without any luck, but there must be a solution out there somewhere. This might be fun but I don't think this will be of any practical use as the timescales are going to be very long. $\endgroup$ Feb 16, 2012 at 16:05

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