Mass transfer theories: film, penetration and surface renewal

Film, penetration, surface-renewal and boundary-layer models of interfacial mass transfer, their expressions for k and how each predicts the dependence of k on diffusivity and contact time.

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Why it matters

Mass transfer coefficients come from correlations, but you need a physical model to extrapolate them – to a new solute with a different diffusivity, to a different contact time, or to a reacting system. The film, penetration and surface-renewal theories give that model. They differ mainly in how k depends on D, and that single exponent is one of the most frequently tested ideas in mass transfer.

Key ideas

Film theory (Whitman, Lewis–Whitman).

  • All resistance lies in a stagnant film of thickness δ next to the interface; beyond it the fluid is perfectly mixed.
  • Diffusion through the film is steady, so the profile is linear.
  • Result: k = D/δ, so k ∝ D.
  • δ is fictitious: it is whatever thickness makes the equation match the measured k. The theory is simple and widely used (two-film theory of interphase transfer), but the D¹ dependence disagrees with most experiments, which show k ∝ D^0.5 to D^0.67.

Penetration theory (Higbie).

  • Fluid elements arrive at the interface from the bulk, stay for the same exposure (contact) time t_c, then return to the bulk.
  • During exposure the solute penetrates by unsteady diffusion into an element that behaves as a semi-infinite medium (valid when √(D·t_c) is small compared with the element size).
  • Instantaneous coefficient k(t) = √(D/(π·t)) – it falls with time as the profile spreads.
  • Averaged over the exposure: k_av = 2√(D/(π·t_c)), so k ∝ D^0.5.
  • Amount absorbed per unit area during one exposure ∝ √t_c.
  • Natural applications: gas bubbles rising through liquid (t_c ≈ bubble diameter / rise velocity), liquid flowing over packing pieces, falling films with short contact, wetted-wall columns.

Surface-renewal theory (Danckwerts).

  • Elements at the surface are replaced at random; every element has the same chance of being replaced, whatever its age. This gives an exponential surface-age distribution φ(t) = s·e^(−s·t), where s is the fractional rate of surface renewal (s⁻¹).
  • Result: k = √(D·s), again k ∝ D^0.5.
  • s is not predicted by the theory; it is fitted to data, like δ and t_c.

Boundary-layer theory. Solving the laminar boundary-layer equations gives k ∝ D^(2/3) (Sh ∝ Sc^(1/3)), which lies between the other two and is the basis of many correlations.

Combined models. Film-penetration theory (Toor–Marchello, Dobbins) gives k ∝ D for long contact or thin films and k ∝ D^0.5 for short contact, explaining why measured exponents fall between 0.5 and 1.

Equivalence. All three theories give the same k for one solute if their parameters are chosen accordingly (δ = D/k, t_c = 4D/(π·k²), s = k²/D). They differ in predictions when D changes, and in how they handle chemical reaction (penetration and renewal models handle fast reactions more realistically).

Formulas

Film theory: k = D_AB / δ

Penetration theory (Higbie): k(t) = √(D_AB / (π·t)) (instantaneous) k_av = 2·√(D_AB / (π·t_c)) (average over exposure time t_c) N_A,av = k_av (C_A,i − C_A,b); amount absorbed per area in one exposure Q = 2 (C_A,i − C_A,b)·√(D_AB·t_c/π)

Surface renewal (Danckwerts): k = √(D_AB·s); age distribution φ(t) = s·e^(−s·t)

Ratio for two solutes A and B in the same hydrodynamics: k_A / k_B = (D_A / D_B)^n, with n = 1 (film), 0.5 (penetration, renewal), 2/3 (boundary layer)

  • k in m/s; D_AB in m²/s; δ in m; t, t_c in s; s in s⁻¹; C in mol/m³; Q in mol/m².

Worked examples

Example 1 – absorption into a rising bubble (standard). Given: CO₂-free water around gas bubbles of diameter 3 mm rising at 0.25 m/s; D = 2.0 × 10⁻⁹ m²/s; interface concentration exceeds bulk by 0.2 mol/m³. Use penetration theory.

  1. Contact time t_c = d_b/u_b = 0.003/0.25 = 0.012 s.
  2. k_av = 2√(D/(π·t_c)) = 2 × √(2.0 × 10⁻⁹/(π × 0.012)) = 2 × √(5.31 × 10⁻⁸) = 4.61 × 10⁻⁴ m/s.
  3. N_A = k_av × ΔC = 4.61 × 10⁻⁴ × 0.2 = 9.21 × 10⁻⁵ mol/(m²·s).

k_L = 4.61 × 10⁻⁴ m/s, N_A = 9.21 × 10⁻⁵ mol/(m²·s).

Example 2 – extrapolating to another solute (GATE level). Given: in a stirred vessel k_L for solute A (D_A = 2.0 × 10⁻⁹ m²/s) is measured as 4.0 × 10⁻⁵ m/s. Predict k_L for solute B (D_B = 1.2 × 10⁻⁹ m²/s) under the same conditions by (a) film theory, (b) penetration theory, (c) boundary-layer theory. Also find the equivalent film thickness and renewal rate for A.

  1. D_B/D_A = 1.2/2.0 = 0.6.
  2. (a) Film: k_B = 4.0 × 10⁻⁵ × 0.6 = 2.40 × 10⁻⁵ m/s.
  3. (b) Penetration/renewal: k_B = 4.0 × 10⁻⁵ × 0.6^0.5 = 3.10 × 10⁻⁵ m/s.
  4. (c) Boundary layer: k_B = 4.0 × 10⁻⁵ × 0.6^(2/3) = 2.85 × 10⁻⁵ m/s.
  5. Equivalent δ for A: δ = D/k = 2.0 × 10⁻⁹/4.0 × 10⁻⁵ = 5.0 × 10⁻⁵ m (50 µm).
  6. Equivalent s for A: s = k²/D = (4.0 × 10⁻⁵)²/2.0 × 10⁻⁹ = 0.80 s⁻¹.

The predictions differ by about 30 %, which is why the exponent matters in design.

Common mistakes

  • Forgetting the factor 2 in the average penetration coefficient (the instantaneous value is √(D/πt)).
  • Using √(D·t_c) instead of √(D/t_c): k must have units of m/s.
  • Thinking a larger renewal rate or shorter contact time lowers k – both raise it (k ∝ t_c^(−0.5), k ∝ s^0.5).
  • Treating δ as a measurable physical thickness.
  • Assuming penetration theory works for long contact times, when the element no longer behaves as semi-infinite.

For GATE CH

Expect ratio questions (k for one solute from another, or effect of halving contact time), direct evaluation of k from Higbie or Danckwerts expressions, identifying the D-exponent of each theory, and matching theory to assumption (stagnant film, fixed exposure time, random renewal). Practise converting between δ, t_c and s for the same k.

Quick check

  1. What is the exponent of D in film, penetration and boundary-layer theories?
  2. If the exposure time in penetration theory is halved, by what factor does k_av change?
  3. Write the Danckwerts expression for k.
  4. Which theory assumes every surface element is exposed for the same time?

Answers: 1. 1, 0.5 and 2/3; 2. increases by √2 ≈ 1.41; 3. k = √(D·s); 4. Higbie's penetration theory.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is the film theory in mass transfer?Concept

    Film theory (Whitman) assumes that all resistance to transfer lies in a stagnant film of thickness δ next to the interface, with perfect mixing beyond it. Transfer through the film is by steady molecular diffusion, so the profile is linear and k = D/δ. The film thickness is fictitious – it is fitted so that the equation matches measured k – and the theory predicts k ∝ D, whereas experiments usually show k ∝ D^0.5 to D^0.67. It remains popular because it is simple and underpins the two-film model of interphase transfer.

  2. 2.Explain the penetration theory in mass transfer.Concept

    Higbie's penetration theory assumes that fluid elements from the bulk reach the interface, stay there for a fixed exposure time t_c, and then return to the bulk. During exposure the solute penetrates by unsteady diffusion into an element treated as semi-infinite. Solving Fick's second law gives an instantaneous coefficient √(D/πt) and an average k_av = 2√(D/(π t_c)), so k ∝ D^0.5 and k rises as contact time shortens. It suits bubbles, falling films and liquid flowing over packing pieces, where contact times are short.

  3. 3.Describe the surface renewal theory in mass transfer.Concept

    Danckwerts' surface renewal theory modifies the penetration theory: instead of a fixed exposure time, surface elements are replaced at random, each having the same chance of replacement regardless of its age. This gives an exponential age distribution φ(t) = s e^(−st), where s is the fractional rate of surface renewal. Averaging the penetration result over this distribution gives k = √(D·s), so k ∝ D^0.5 like the penetration theory. The renewal rate s must be found from experiment.

  4. 4.What happens to mass transfer rates if the surface renewal rate increases?Application

    If the surface renewal rate increases, the mass transfer rates generally increase as well. This is because more fresh fluid elements are brought to the interface, reducing the concentration boundary layer thickness and enhancing the diffusion process. This is particularly relevant in turbulent systems where surface renewal is frequent.

  5. 5.How does the penetration theory differ from the film theory in terms of assumptions?Concept

    Film theory assumes steady-state diffusion across a stagnant film of fixed thickness, giving a linear profile and k = D/δ. Penetration theory assumes unsteady diffusion into fluid elements that are periodically brought to the interface for a fixed short time and treated as semi-infinite, giving k = 2√(D/(π t_c)). The key testable difference is the D-dependence: k ∝ D for film theory and k ∝ D^0.5 for penetration theory. Penetration theory is also more realistic for short contact times and for fast chemical reaction near the interface.

  6. 6.Calculate the mass transfer coefficient using the film theory for a system with a diffusion coefficient of 2.5 × 10⁻⁹ m²/s and a film thickness of 0.001 m.Numerical

    The mass transfer coefficient (k) can be calculated using the formula k = D/δ, where D is the diffusion coefficient and δ is the film thickness. Substituting the given values: k = (2.5 × 10⁻⁹ m²/s) / (0.001 m) = 2.5 × 10⁻⁶ m/s.

  7. 7.If the contact time of fluid elements in the penetration theory is halved, what is the expected effect on mass transfer rates?Application

    In penetration theory k_av = 2√(D/(π t_c)), so k ∝ t_c^(−0.5). Halving the contact time increases k by a factor √2 ≈ 1.41, and the flux rises by the same factor for the same driving force. Physically, fresher elements have steeper concentration gradients at the surface because the solute has had less time to penetrate.

  8. 8.A system follows the surface renewal theory with a renewal rate of 0.2 s⁻¹. If the diffusion coefficient is 1.8 × 10⁻⁹ m²/s, estimate the mass transfer coefficient.Numerical

    Danckwerts' theory gives k = √(D·s). Here D·s = 1.8 × 10⁻⁹ m²/s × 0.2 s⁻¹ = 3.6 × 10⁻¹⁰ m²/s², so k = √(3.6 × 10⁻¹⁰) ≈ 1.9 × 10⁻⁵ m/s. This is a typical order of magnitude for a liquid-side coefficient in a gently agitated liquid.

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