Stanton Number for Mass Transfer

Also known as StD · mass Stanton number · Stanton number · Sh over Re Sc · j factor group

StD=kcu\mathrm{St}_D = \frac{k_c}{u}

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The Stanton number is the most physically readable group in mass transfer, and it is under-used for it. It divides the transfer coefficient by the velocity that produced it, StD=kc/u\mathrm{St}_D = k_c/u, and the result is a plain fraction: what proportion of the approaching stream is actually delivered to the surface. A value of 0.003 says that three parts in a thousand of what flows past gets transferred. No other group in the field translates that directly into a sentence.

Its structural advantage is that it contains no length scale at all. Sherwood cannot be quoted without stating what LL was, and comparing two Sherwood numbers from different geometries means first arguing about characteristic dimensions. Stanton sidesteps the argument entirely, which lets a packed bed be compared with a wetted wall or a flat plate without either geometry being reconciled. It relates to the others as StD=Sh/(ReSc)=Sh/Pe\mathrm{St}_D = \mathrm{Sh}/(\mathrm{Re}\,\mathrm{Sc}) = \mathrm{Sh}/\mathrm{Pe}, and the length divides out of the ratio.

It is also the group the Chilton–Colburn analogy is genuinely written in. The j-factor is jD=StDSc2/3j_D = \mathrm{St}_D\,\mathrm{Sc}^{2/3}, and the analogy's claim is that this equals f/2f/2. Reading it that way makes the physical content obvious: the fraction of the stream transferred to the wall is set by the same wall friction that sets the pressure drop, once corrected for the sublayer where molecules must move on their own. It also puts a useful ceiling on the number — since the Fanning friction factor in turbulent flow rarely exceeds about 0.02, Stanton numbers much above 0.01 are physically suspicious.

When one turns up too large, the usual cause is that kck_c and uu were read from different places. A coefficient measured against a local or interstitial velocity divided by a superficial one, or the reverse, produces a number that is arithmetically fine and physically meaningless. In a packed bed the distinction is large — the interstitial velocity exceeds the superficial by the reciprocal of the voidage, commonly a factor of two or three — so a Stanton number from a packed bed should always be accompanied by a statement of which velocity it used.

Stanton Number for Mass Transfer
StD=kcu\mathrm{St}_D = \frac{k_c}{u}
ukcStD
Where
  • StD\mathrm{St}_D= Stanton number (mass)
  • kck_c= Mass transfer coefficient (m/s)
  • uu= Bulk velocity (m/s)
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