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Convective mass transfer and analogies

Short answer

Convective mass transfer coefficient h_m: N_A = h_m (C_As – C_A∞). Sc = ν/D, Sh = h_m L/D, Le = α/D. Chilton-Colburn: (f/2) = St Pr^{2/3} = St_m Sc^{2/3}, thus h/h_m = ρ c_p Le^{2/3}. Correlations: laminar flat plate Sh_L = 0.664 Re_L^{1/2} Sc^{1/3}; turbulent tube Sh = 0.023 Re^{0.83} Sc^{0.44}.

Define the convective mass transfer coefficient. Explain the Schmidt, Sherwood and Lewis numbers. State the analogy between momentum, heat and mass transfer (Chilton-Colburn) and give two correlations used for convective mass transfer.

Convective mass transfer coefficient

When a fluid flows over a surface from which species A is transferred, the flux is written in the form of Newton's law of cooling:

NA=hm (CAs−CA∞)N_A = h_m\,(C_{As}-C_{A\infty})

hmh_m (m/s) is the convective mass transfer coefficient, found from correlations just like hh.

Dimensionless numbers

NumberDefinitionMeaning
Schmidt, ScScνDAB\dfrac{\nu}{D_{AB}}Momentum diffusivity / mass diffusivity (mass analogue of PrPr)
Sherwood, ShShhmLDAB\dfrac{h_mL}{D_{AB}}Convective / diffusive mass transfer (analogue of NuNu)
Lewis, LeLeαDAB=ScPr\dfrac{\alpha}{D_{AB}} = \dfrac{Sc}{Pr}Thermal diffusivity / mass diffusivity
Stanton (mass), StmSt_mhmV=ShRe Sc\dfrac{h_m}{V}= \dfrac{Sh}{Re\,Sc}Mass transfer / mass capacity of the stream

Analogy between momentum, heat and mass transfer

Because all three processes are carried by the same molecular and eddy motion in the boundary layer, their equations are similar. Reynolds analogy (for Pr=Sc=1Pr = Sc = 1): f2=St=Stm\dfrac{f}{2}=St=St_m. For other fluids the Chilton-Colburn analogy applies (0.6<Pr<600.6 < Pr < 60, 0.6<Sc<30000.6 < Sc < 3000):

jH=St Pr2/3=f2=jM=Stm Sc2/3j_H = St\,Pr^{2/3}= \frac{f}{2}= j_M = St_m\,Sc^{2/3}

Hence, if hh is known, hmh_m follows from

hhm=ρcp Le2/3\frac{h}{h_m} = \rho c_p\,Le^{2/3}

Mass transfer can thus be predicted from heat transfer results by replacing Nu→ShNu \to Sh and Pr→ScPr \to Sc.

Correlations

  • Laminar flow over a flat plate: ShL=0.664 ReL1/2 Sc1/3Sh_L = 0.664\,Re_L^{1/2}\,Sc^{1/3}.
  • Turbulent flow in tubes (Gilliland-Sherwood): Sh=0.023 Re0.83 Sc0.44Sh = 0.023\,Re^{0.83}\,Sc^{0.44}. The Linton-Sherwood form Sh=0.023 Re0.83 Sc1/3Sh = 0.023\,Re^{0.83}\,Sc^{1/3} is also used.
  • Flow past a sphere (Frossling): Sh=2+0.552 Re1/2Sc1/3Sh = 2 + 0.552\,Re^{1/2}Sc^{1/3}.

✓ Answer checked by Claude Sonnet 5.5

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