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Dissolution and Solubility Modeling · Tutorial 247 of 799

Boundary-Layer Thickness and Hydrodynamic Effects

Learn to use the diffusion-layer model to quantify how fluid motion and fluid properties change dissolution, while keeping those effects distinct from changes in particle surface area.

Advanced 11 min read

What You'll Learn

  • Define diffusion-layer thickness and the mass-transfer coefficient used in dissolution models.
  • Derive the film-model expression for dissolution rate from Fick’s law.
  • Calculate how thinning the layer changes rate when area and concentration driving force are fixed.
  • Separate hydrodynamic effects from particle-area effects using rate ratios.
  • Explain how diffusivity and other fluid properties affect predictions.
  • Recognize assumptions, diagnostics, and limitations of the effective-layer model.

Why the Diffusion Layer Matters

A solid dissolves at its surface, but molecules must then move from the solid–liquid interface into the surrounding fluid. Agitation changes this transport: it continually moves fluid near a particle and can reduce the effective distance over which dissolved molecules must diffuse. That change can increase dissolution rate even when the solid’s exposed surface area and its solubility have not changed.

To interpret an observed rate increase, it is important to separate this hydrodynamic effect from a change in particle surface area. For example, stronger mixing may alter both the effective diffusion layer and the degree to which particles are dispersed or wetted. A rate change alone does not identify which mechanism changed.

Definition: The diffusion-layer thickness, denoted \(h\), is an effective distance used to represent the near-surface region through which dissolved material is transported mainly by molecular diffusion. The liquid-side mass-transfer coefficient, denoted \(k_L\), relates the dissolution flux to the concentration difference between the solid surface and the well-mixed bulk. In the simple film model, \(k_L=D/h\), where \(D\) is the solute’s diffusion coefficient in the liquid.

The Film Model and Its Assumptions

Let \(C_s\) be the dissolved concentration at the solid–liquid interface and \(C_b\) the concentration in the well-mixed bulk liquid. In the film model, the concentration falls across a near-surface layer from \(C_s\) to \(C_b\). If the concentration profile across that layer is approximated as linear, Fick’s law gives the outward mass flux per unit surface area:

$$ J = -D\frac{dC}{dx} \approx \frac{D(C_s-C_b)}{h}=k_L(C_s-C_b) $$

Here \(J\) has units of mass per area per time, \(D\) has units of area per time, and \(h\) has units of length. The total dissolution rate from a wetted surface area \(A\) is therefore:

$$ -\frac{dM_s}{dt}=A k_L(C_s-C_b)=A\frac{D}{h}(C_s-C_b) $$

In this equation, \(M_s\) is the mass remaining in the solid. The negative sign indicates that solid mass decreases during dissolution. If concentrations are expressed in \(\mathrm{kg/m^3}\), \(A\) in \(\mathrm{m^2}\), and \(k_L\) in \(\mathrm{m/s}\), the rate is in \(\mathrm{kg/s}\).

This expression is useful when the interface is close to local equilibrium, so that \(C_s\) can be specified from the relevant solubility, and transport through the liquid is the rate-limiting process being represented. It also assumes a meaningful bulk concentration, a sufficiently uniform and wetted area, and conditions that do not change substantially during the calculation. The film is an effective representation, not necessarily a sharply bounded physical slab. In a real vessel, flow may be turbulent and the local layer may vary across a particle.

How Agitation and Fluid Properties Enter

In the simple film model, increasing agitation is represented by a smaller effective \(h\), and therefore a larger \(k_L\). This is a useful directional expectation, not a universal formula connecting stir rate to layer thickness. The relationship depends on the vessel, impeller, particle motion, and flow regime. A given increase in rotational speed does not imply a fixed proportional decrease in \(h\).

A more general way to organize hydrodynamic effects is with the Sherwood number, a dimensionless mass-transfer coefficient. For a characteristic length \(L_c\), fluid density \(\rho\), characteristic fluid speed \(u\), and dynamic viscosity \(\mu\), define:

$$ Sh=\frac{k_L L_c}{D}, \qquad Re=\frac{\rho u L_c}{\mu}, \qquad Sc=\frac{\mu}{\rho D} $$

Correlations between \(Sh\), \(Re\), and \(Sc\) are specific to a geometry and flow regime; there is no single correlation that applies to every dissolution apparatus. These dimensionless quantities help explain why agitation and fluid properties cannot always be treated as independent adjustments. Agitation changes fluid speed and thus \(Re\); viscosity and density affect \(Re\) and \(Sc\); and \(D\) affects both \(Sc\) and the conversion from \(Sh\) to \(k_L\).

At fixed temperature and for a dilute solute in a Newtonian liquid, the Stokes–Einstein relation gives a useful approximation: \(D\) decreases as viscosity increases, when solute size is unchanged. This does not mean that viscosity alone determines the rate. It may also change the flow field and the effective layer thickness. If \(D\) and \(h\) both change, the relevant comparison is the resulting \(k_L=D/h\).

Worked Example: Increasing Agitation at Fixed Area

Consider a hypothetical dissolution calculation with \(D=5.00\times10^{-10}\ \mathrm{m^2/s}\), wetted area \(A=0.0200\ \mathrm{m^2}\), and concentration difference \(C_s-C_b=0.400\ \mathrm{kg/m^3}\). Suppose agitation changes the effective layer thickness from \(100\ \mathrm{\mu m}\) to \(60.0\ \mathrm{\mu m}\), while \(D\), area, and concentration difference remain fixed. Find the predicted rate in both conditions.

Convert the layer thicknesses to metres. At the lower agitation condition, \(h_1=100\times10^{-6}=1.00\times10^{-4}\ \mathrm{m}\), so:

$$ k_{L,1}=\frac{5.00\times10^{-10}}{1.00\times10^{-4}} =5.00\times10^{-6}\ \mathrm{m/s} $$

The predicted rate is:

$$ R_1=A k_{L,1}(C_s-C_b) =(0.0200)(5.00\times10^{-6})(0.400) =4.00\times10^{-8}\ \mathrm{kg/s} $$

Converting to grams per hour gives \(4.00\times10^{-8}\ \mathrm{kg/s}\times3600\ \mathrm{s/h}\times1000\ \mathrm{g/kg}=0.144\ \mathrm{g/h}\). At the higher agitation condition, \(h_2=60.0\times10^{-6}=6.00\times10^{-5}\ \mathrm{m}\), giving:

$$ k_{L,2}=\frac{5.00\times10^{-10}}{6.00\times10^{-5}} =8.33\times10^{-6}\ \mathrm{m/s} $$

Thus \(R_2=(0.0200)(8.33\times10^{-6})(0.400)=6.67\times10^{-8}\ \mathrm{kg/s}\), or \(0.240\ \mathrm{g/h}\). With area and concentration difference fixed, the predicted rate ratio is \(R_2/R_1=h_1/h_2=100/60.0=1.67\). The model predicts a 1.67-fold rate, not a twofold rate, under these stated assumptions.

Separating Hydrodynamic and Surface-Area Effects

For two conditions, the film model gives a rate ratio that makes the separate contributions explicit:

$$ \frac{R_2}{R_1} = \frac{A_2}{A_1} \frac{k_{L,2}}{k_{L,1}} \frac{(C_s-C_b)_2}{(C_s-C_b)_1} $$

This factorization is valid for positive rates under the film model. If the concentration difference is unchanged and \(D\) is unchanged, then \(k_{L,2}/k_{L,1}=h_1/h_2\). In that restricted comparison, a rate increase can be decomposed into an area ratio and a layer-thickness ratio. The decomposition is a model-based calculation: it is not evidence that the factors were independently measured.

Worked Example: Distinguishing Agitation from Increased Area

In a hypothetical system, let \(D=6.00\times10^{-10}\ \mathrm{m^2/s}\), \(A_1=0.0150\ \mathrm{m^2}\), \(h_1=120\ \mathrm{\mu m}\), and \(C_s-C_b=0.500\ \mathrm{kg/m^3}\). Under stronger agitation, suppose \(h_2=80.0\ \mathrm{\mu m}\). Separately, suppose better dispersion doubles the exposed area, from \(0.0150\) to \(0.0300\ \mathrm{m^2}\). Assume the stated changes do not alter \(D\) or the concentration difference. Compare each effect and their combination.

For the baseline, \(h_1=1.20\times10^{-4}\ \mathrm{m}\), so \(k_{L,1}=6.00\times10^{-10}/1.20\times10^{-4}=5.00\times10^{-6}\ \mathrm{m/s}\). The baseline rate is:

$$ R_1=(0.0150)(5.00\times10^{-6})(0.500) =3.75\times10^{-8}\ \mathrm{kg/s} =0.135\ \mathrm{g/h} $$

With stronger agitation alone, the area is unchanged and \(h_2=8.00\times10^{-5}\ \mathrm{m}\). Thus \(k_{L,2}=7.50\times10^{-6}\ \mathrm{m/s}\), and:

$$ R_{\mathrm{agitation}} =(0.0150)(7.50\times10^{-6})(0.500) =5.625\times10^{-8}\ \mathrm{kg/s} =0.2025\ \mathrm{g/h} $$

The rate ratio is \(0.2025/0.135=1.50\), which matches \(h_1/h_2=120/80.0=1.50\). With doubled area alone and baseline \(k_L\), the rate is \(2(0.135)=0.270\ \mathrm{g/h}\). If both changes occur, the area factor is \(2\) and the mass-transfer factor is \(1.50\), so the combined prediction is \(2(1.50)(0.135)=0.405\ \mathrm{g/h}\). The contributions are distinct in the calculation: stronger agitation raises \(k_L\), whereas increased wetted area raises \(A\).

Fluid-Property Changes and Interpretation

Fluid properties can affect the rate through more than one term. The following calculation illustrates why a change in \(D\) should not automatically be described as an agitation effect. Suppose a hypothetical fluid change reduces \(D\) from \(4.00\times10^{-10}\) to \(3.00\times10^{-10}\ \mathrm{m^2/s}\), and the effective layer changes from \(80.0\) to \(100\ \mathrm{\mu m}\). Let \(A=0.0100\ \mathrm{m^2}\) and \(C_s-C_b=0.250\ \mathrm{kg/m^3}\) in both conditions.

In fluid A, \(k_{L,A}=4.00\times10^{-10}/8.00\times10^{-5}=5.00\times10^{-6}\ \mathrm{m/s}\), and \(R_A=(0.0100)(5.00\times10^{-6})(0.250)=1.25\times10^{-8}\ \mathrm{kg/s}=0.0450\ \mathrm{g/h}\). In fluid B, \(k_{L,B}=3.00\times10^{-10}/1.00\times10^{-4}=3.00\times10^{-6}\ \mathrm{m/s}\), and \(R_B=(0.0100)(3.00\times10^{-6})(0.250)=7.50\times10^{-9}\ \mathrm{kg/s}=0.0270\ \mathrm{g/h}\). The rate ratio is \(0.0270/0.0450=0.600\), equal to the mass-transfer coefficient ratio. Both the lower diffusivity and the thicker effective layer contribute to the predicted decrease.

If only viscosity changes, Stokes–Einstein can help estimate the direction and approximate change in \(D\), subject to its assumptions. But a prediction that holds \(h\) fixed while changing viscosity isolates only one model component. In an actual apparatus, viscosity may also alter the flow and therefore \(h\). Measuring or otherwise justifying the mass-transfer coefficient is preferable to attributing the full observed rate difference to diffusivity alone.

Derivation, Diagnostics, and Alternatives

The film-model derivation is short but its assumptions matter. Across a layer of thickness \(h\), take the concentration gradient to be approximately \((C_b-C_s)/h\). Fick’s law gives \(J=-D(dC/dx)\); choosing the coordinate \(x\) to increase away from the surface, this gradient is negative when \(C_s>C_b\). Therefore the outward flux is positive and equals \(D(C_s-C_b)/h\). Multiplying flux by wetted area gives total rate. This argument also shows why the model uses the product \(A k_L\): rate data alone may not distinguish a larger area from a larger mass-transfer coefficient.

Useful diagnostics compare predictions with measurements made while changing one factor at a time. If \(A\), \(D\), and \(C_s-C_b\) are controlled, thinning \(h\) by a factor of two predicts a twofold rate increase. If the observed rate does not respond as predicted, possible explanations include an incorrect effective layer estimate, incomplete or changing wetting, an altered concentration difference, or a different rate-limiting process. A rate that stops increasing with agitation can indicate that liquid-side transport is no longer the only important resistance; it does not prove that the model is wrong in every condition.

Other approaches answer related but different questions. An empirical \(k_L\) fitted to a particular apparatus can summarize measured transfer without resolving the local flow, but its value may not transfer to another geometry or agitation setting. A hydrodynamic correlation for \(Sh\) can predict how transfer varies with flow when its geometry and regime are appropriate, but using an unrelated correlation can introduce systematic error. Computational fluid dynamics can represent spatial flow and transfer in greater detail, at the cost of additional inputs and validation. The film model is simpler and transparent, but its single \(h\) averages over local variation.

Common pitfall: “The rate doubled when stirring increased, so the exposed particle area must have doubled.” This conclusion does not follow from rate alone. Under the film model, rate depends on \(A\), \(k_L\), and \(C_s-C_b\); stronger agitation can increase \(k_L\) by reducing effective \(h\). To attribute a change to area, hold or independently characterize the hydrodynamic and concentration terms, and verify the wetted area rather than inferring it from the rate.

The effective-layer approach is most useful for organizing mechanisms and comparing controlled conditions. It is limited when area, surface concentration, diffusivity, or local flow changes substantially during dissolution, or when several transport and reaction processes contribute comparably. In those cases, the model may still be a useful approximation, but fitted values of \(h\) or \(k_L\) should be interpreted as effective parameters for the specified conditions rather than universal material constants.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

  1. If \(D\) and the concentration difference are fixed, what happens to \(k_L\) when the effective layer thickness is reduced to one-third of its original value?
  2. A rate increases by a factor of \(1.8\), while the wetted area is independently found to be unchanged. Under the film model and unchanged concentration difference, what is the implied change in \(k_L\)?
  3. With \(D=4.00\times10^{-10}\ \mathrm{m^2/s}\), \(h=80.0\ \mathrm{\mu m}\), \(A=0.0100\ \mathrm{m^2}\), and \(C_s-C_b=0.200\ \mathrm{kg/m^3}\), calculate the dissolution rate in \(\mathrm{g/h}\).
  4. Why can an observed rate increase during agitation not, by itself, establish that particle surface area increased?

Solutions. (1) Since \(k_L=D/h\), reducing \(h\) to one-third makes \(k_L\) three times larger. (2) With area and concentration difference unchanged, \(R_2/R_1=k_{L,2}/k_{L,1}\), so the implied \(k_L\) increases 1.8-fold. (3) Convert \(h\) to \(8.00\times10^{-5}\ \mathrm{m}\). Then \(k_L=4.00\times10^{-10}/8.00\times10^{-5}=5.00\times10^{-6}\ \mathrm{m/s}\), and \(R=(0.0100)(5.00\times10^{-6})(0.200)=1.00\times10^{-8}\ \mathrm{kg/s}=0.0360\ \mathrm{g/h}\). (4) Rate depends on area, mass-transfer coefficient, and concentration difference. Agitation can increase the mass-transfer coefficient by thinning the effective layer, so area must be measured or otherwise independently supported.