Mass transfer coefficients and correlations

Definitions and interconversion of gas- and liquid-phase mass transfer coefficients, the Sherwood, Schmidt and Stanton numbers, standard correlations and the Chilton–Colburn analogy.

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Why it matters

In real equipment the fluid is flowing, so the concentration profile near an interface is too complicated to solve with Fick's law alone. Engineers lump the whole effect into a mass transfer coefficient, flux = coefficient × driving force, and estimate the coefficient from dimensionless correlations. Every absorber, dryer, dissolver and catalytic reactor design uses these numbers.

Key ideas

Definition. A mass transfer coefficient is the flux per unit driving force between the interface and the bulk of one phase. Because the driving force can be written in partial pressure, mole fraction or concentration, there are several coefficients for the same physical situation, and they differ only in units:

  • gas phase: k_G (driving force Δp_A), k_y (Δy_A), k_c (ΔC_A);
  • liquid phase: k_L or k_c (ΔC_A), k_x (Δx_A).

Two physical situations. Coefficients for equimolar counter-diffusion (often written with a prime, k'_G, k'_y, …) and for diffusion of A through stagnant B (k_G, k_y, …) differ by the bulk-flow factor. For dilute systems (y_BM ≈ 1) the difference vanishes. Always check which convention your textbook uses; this lesson follows the Treybal style.

Film picture. If all the resistance is imagined to lie in a stagnant film of thickness δ, then k'_c = D_AB/δ. This gives k ∝ D, whereas real data show k ∝ D^0.5 to D^0.67 – the reason for the penetration and boundary-layer theories (next topic). δ is a fictitious length, not something you can measure.

Dimensionless groups.

  • Sherwood number Sh = k_c·L/D_AB – convective mass transfer relative to pure diffusion across length L (the mass-transfer Nusselt number).
  • Schmidt number Sc = μ/(ρ·D_AB) = ν/D_AB – momentum diffusivity over mass diffusivity (≈ 0.5–2 for gases, 10²–10⁴ for liquids).
  • Reynolds number Re = ρ·u·L/μ.
  • Mass-transfer Stanton number St_m = Sh/(Re·Sc) = k_c/u.

Correlations are written Sh = a·Re^m·Sc^n. Choose the one for your geometry and flow regime, and use the characteristic length and velocity defined with it. Correlations hold only within the Re and Sc range of the data behind them.

Analogies. Because momentum, heat and mass transport obey similar equations, a known friction factor or heat transfer coefficient can be converted into a mass transfer coefficient. The Chilton–Colburn analogy j_D = j_H = f/2 works for 0.6 < Sc < 2500 when there is no form drag (pipes, flat plates); for bluff bodies only j_D = j_H holds.

What increases k. Higher velocity (thinner boundary layer), higher D, lower viscosity, smaller characteristic length, and more turbulence. Temperature raises D and lowers liquid viscosity, so liquid-side coefficients rise noticeably with temperature.

Formulas

Rate equations (A through stagnant B): N_A = k_G (p_A,b − p_A,i) = k_y (y_A,b − y_A,i) = k_c (C_A,b − C_A,i) Liquid side: N_A = k_L (C_A,i − C_A,b) = k_x (x_A,i − x_A,b)

Conversions (gas, ideal): k_y = k_G·P, k_c = k_G·R·T, k_x = k_L·C (C = total molar concentration of liquid) k_G·p_BM = k'_G·P and k_y·y_BM = k'_y

  • k_G in mol/(m²·s·Pa); k_y, k_x in mol/(m²·s); k_c, k_L in m/s; P in Pa; T in K; R = 8.314 J/(mol·K).

Film theory: k'_c = D_AB/δ

Selected correlations (take constants from your data book if different):

  • Turbulent flow in a pipe or wetted-wall column (Linton–Sherwood): Sh = 0.023·Re^0.83·Sc^(1/3), Re > 2100 (approximately 4000–60 000 in the data).
  • Laminar flat plate, average: Sh_L = 0.664·Re_L^0.5·Sc^(1/3), Re_L < 5 × 10⁵.
  • Single sphere (Frössling / Ranz–Marshall): Sh = 2 + 0.6·Re^0.5·Sc^(1/3); Sh → 2 in a stagnant fluid.
  • Fully developed laminar pipe flow, constant wall concentration: Sh = 3.66.

Chilton–Colburn analogy: j_D = (k_c/u)·Sc^(2/3), j_H = (h/(ρ·c_p·u))·Pr^(2/3), j_D = j_H = f/2 which gives k_c = (h/(ρ·c_p))·(Pr/Sc)^(2/3)

  • f – Fanning friction factor; h – W/(m²·K); c_p – J/(kg·K); Pr = c_p·μ/k.

Worked examples

Example 1 – sublimation of a naphthalene sphere (standard). Given: sphere d = 1 cm in air at 300 K, 1 atm, u = 1 m/s; ν_air = 1.6 × 10⁻⁵ m²/s; D = 6.2 × 10⁻⁶ m²/s; vapour pressure of naphthalene 11 Pa; bulk air free of naphthalene (dilute, so k_c ≈ k'_c).

  1. Sc = ν/D = 1.6 × 10⁻⁵/6.2 × 10⁻⁶ = 2.58. Re = u·d/ν = 1 × 0.01/1.6 × 10⁻⁵ = 625.
  2. Sh = 2 + 0.6·Re^0.5·Sc^(1/3) = 2 + 0.6 × 25 × 1.372 = 22.6.
  3. k_c = Sh·D/d = 22.6 × 6.2 × 10⁻⁶/0.01 = 0.0140 m/s.
  4. Surface concentration C_A,i = p/(RT) = 11/(8.314 × 300) = 4.41 × 10⁻³ mol/m³.
  5. N_A = k_c·C_A,i = 0.0140 × 4.41 × 10⁻³ = 6.17 × 10⁻⁵ mol/(m²·s).
  6. Rate = N_A·π·d²·M = 6.17 × 10⁻⁵ × 3.142 × 10⁻⁴ × 128.17 g/mol = 2.49 × 10⁻⁶ g/s.

N_A = 6.17 × 10⁻⁵ mol/(m²·s); sublimation ≈ 9.0 mg/h.

Example 2 – mass transfer coefficient from heat transfer data (GATE level). Given: for air flowing past a wet surface, h = 50 W/(m²·K); ρ = 1.16 kg/m³, c_p = 1007 J/(kg·K), Pr = 0.71, Sc (water vapour–air) = 0.60, T = 300 K. Find k_c and k_G.

  1. k_c = (h/(ρ·c_p))·(Pr/Sc)^(2/3).
  2. h/(ρ·c_p) = 50/(1.16 × 1007) = 0.0428 m/s.
  3. (Pr/Sc)^(2/3) = (0.71/0.60)^(2/3) = 1.119.
  4. k_c = 0.0428 × 1.119 = 0.0479 m/s.
  5. k_G = k_c/(R·T) = 0.0479/(8.314 × 300) = 1.92 × 10⁻⁵ mol/(m²·s·Pa).

k_c = 0.0479 m/s; k_G = 1.92 × 10⁻⁵ mol/(m²·s·Pa).

Common mistakes

  • Mixing coefficients and driving forces with different units (k_G with Δy, k_y with Δp).
  • Ignoring the p_BM or y_BM factor when the solute is not dilute.
  • Using a turbulent-pipe correlation at Re = 2000, or a flat-plate correlation for a sphere.
  • Using the wrong characteristic length (diameter vs radius, plate length vs distance x).
  • Forgetting that a sphere in still fluid has Sh = 2, not 0.
  • Writing Sc as D/ν (it is ν/D).

For GATE CH

Expect conversions between k_G, k_y and k_c, estimation of k from Sh–Re–Sc correlations given in the question, analogy problems that obtain k_c from h or f, and ratio questions (how k changes if velocity doubles, using the exponent on Re, or if D changes, using the exponent from film, penetration or boundary-layer theory). Evaporation and sublimation rates from spheres and plates are favourites.

Quick check

  1. Define the Schmidt number and give typical values for gases and liquids.
  2. Convert k_G = 2 × 10⁻⁵ mol/(m²·s·Pa) to k_y at 1 bar.
  3. What is Sh for a sphere in a stagnant fluid?
  4. In turbulent pipe flow, if velocity is doubled, by what factor does k_c change (Sh ∝ Re^0.83)?

Answers: 1. Sc = ν/D, about 0.5–2 for gases and 10²–10⁴ for liquids; 2. k_y = k_G·P = 2 mol/(m²·s); 3. 2; 4. 2^0.83 ≈ 1.78.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is a mass transfer coefficient and why is it important in chemical engineering?Concept

    A mass transfer coefficient is the flux of a species per unit driving force between an interface and the bulk of one phase, for example N_A = k_G(p_A,b − p_A,i) or N_A = k_L(C_A,i − C_A,b). It lumps the effects of molecular diffusion and fluid flow, which are too complex to solve exactly in real equipment, into one number obtained from correlations or experiments. Its units depend on the driving force (k_G in mol/(m²·s·Pa), k_y in mol/(m²·s), k_c in m/s). It is the basis for sizing absorbers, dryers, extractors and catalytic reactors.

  2. 2.How are mass transfer coefficients typically determined in practice?Concept

    Mass transfer coefficients are often determined experimentally by measuring the rate of mass transfer under controlled conditions and using correlations based on dimensionless numbers like Reynolds, Schmidt, and Sherwood numbers. These correlations help relate the experimental data to the mass transfer coefficient.

  3. 3.Why is the Sherwood number used in mass transfer correlations?Application

    The Sherwood number is a dimensionless number that represents the ratio of convective mass transfer to diffusive mass transfer. It is used in mass transfer correlations to help predict the mass transfer coefficient by relating it to other dimensionless numbers like Reynolds and Schmidt numbers.

  4. 4.What happens to the mass transfer coefficient if the flow rate of a fluid increases?Application

    If the flow rate of a fluid increases, the mass transfer coefficient typically increases as well. This is because higher flow rates enhance turbulence, which reduces the thickness of the boundary layer and increases the rate of mass transfer.

  5. 5.Calculate the mass transfer coefficient if the Sherwood number is 100, the characteristic length is 0.05 m, and the diffusion coefficient is 2 × 10⁻⁹ m²/s.Numerical

    The mass transfer coefficient (k) can be calculated using the formula: Sh = k·L/D, where Sh is the Sherwood number, L is the characteristic length, and D is the diffusion coefficient. Rearranging gives k = Sh·D/L. Substituting the given values: k = 100 × (2 × 10⁻⁹ m²/s) / 0.05 m = 4 × 10⁻⁶ m/s.

  6. 6.What is the impact of temperature on mass transfer coefficients?Application

    Temperature acts mainly through the diffusivity and the viscosity. In liquids D rises and viscosity falls with temperature, so k_L increases appreciably, often by a few per cent per kelvin. In gases D rises roughly as T^1.75, but viscosity also rises and density falls, so k_c changes only modestly; k_G = k_c/(RT) also carries an explicit 1/T. Temperature additionally shifts equilibrium (solubility), which often matters more than the change in the coefficient.

  7. 7.Explain how the Reynolds number affects mass transfer in a pipe flow.Application

    The Reynolds number indicates the flow regime in a pipe, whether it is laminar or turbulent. In turbulent flow, the mass transfer coefficient is generally higher due to increased mixing and reduced boundary layer thickness. In laminar flow, mass transfer is primarily driven by molecular diffusion, which is less efficient.

  8. 8.Gas flows turbulently through a wetted-wall column with Re = 20 000 and Sc = 0.7. Estimate the Sherwood number using the Linton–Sherwood correlation Sh = 0.023 Re^0.83 Sc^(1/3).Numerical

    Sh = 0.023 × 20 000^0.83 × 0.7^(1/3) = 0.023 × 3 714 × 0.888 ≈ 76. Then k_c = Sh·D/d once the tube diameter and diffusivity are known. The correlation applies only to turbulent flow (Re above about 2100); at Re = 2000 the flow is laminar and a laminar result such as Sh = 3.66 for fully developed flow would be used instead.

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