Sherwood number - effective and molecular diffusion

[TboneWalker]

I'm having a little trouble understanding the Sherwood number, (Sh, mass transfer Nusselt number), which gives the ratio of convective to diffusive mass transfer. My question is: given a system that has a molecular diffusion coefficient of D, the effective diffusion coefficient is measured to De because of convective forces that speed up the process. Will the Sherwood number then be proportional to De/D?

 

[Bill_K]

If you want my opinion, I'd say yes. But if you want my second opinion, I'd say no.

 

[TboneWalker]

Thanks, that clears it up

Any more thoughts anyone?

 

[SpectraCat]

How are you distinguishing the diffusion coefficient and "effective diffusion coefficient"? The diffusion coefficient as typically defined already takes temperature and Brownian motion into account, and thus it is hard to see what distinction you could be drawing there. As I understand it, the Sherwood number is used to express the ratio of transport by convection (diffusion plus advection) to that of transport by diffusion alone. So it is hard to see how you could measure an "effective diffusion coefficient" that is different from the actual one.

 

[TboneWalker]

The "effective diffusion coefficient" that I measure is the value the diffusion coefficient has to have in order to achieve the observed mass flux, assuming that diffusion alone is the driving mechanism.

For instance, I know that the diffusion of CO2 in water in reality is around 2E-9. In order to get the mass transport I'm observing, the diffusivity has to be 2E-8. Thus the "effective diffusion coefficient" is equal to 2E-8. The transport mechanism is off course not purely diffusion, but the effective diffusion coefficient would describe the amount of flux, and the ratio De/D would give you an idea of the ratio of diffusive transport compared to the total transport. I might be totally off here...

 

[Emreth]

This might help
http://www.msubbu.in/ln/mt/MT-2.doc