Workshop 11: Convective mass transfer and dissolution

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Attempt these problems before opening the worked solutions.

1 Removing a polymer coating

Solvent flowing over a flat coating

In a manufacturing process, an organic solvent (methyl ethyl ketone, MEK) is used to dissolve a thin coating of polymer film away from a nonporous flat surface (length 20 cm, width 10 cm, see Figure 1). The thickness of the polymer film is initially uniform at 0.2 mm. In the process, MEK is added at a flow rate of 30 cm3/s$\mathrm{cm^3/s}$ to an open flat pan of length 30 cm and width 10 cm. The depth of the liquid MEK in the pan is maintained at 2.0 cm. Assuming that the concentration of the dissolved polymer in the bulk solvent is essentially zero (𝑐𝐴∞ =0$c_{A\infty}=0$), determine the rate of removal of polymer coating.

Data: diffusion coefficient of dissolved polymer in MEK = 3 x 10−6$10^{-6}$ cm2/s$\mathrm{cm^2/s}$; kinematic viscosity of MEK = 6 x 10−3$10^{-3}$ cm2/s$\mathrm{cm^2/s}$; density of polymer material (solid) = 1.05 g/cm3$\mathrm{g/cm^3}$; density of MEK = 0.8 g/cm3$\mathrm{g/cm^3}$; maximum solubility of polymer in MEK = 0.04 g/cm3$\mathrm{g/cm^3}$.

Figure 1: Pan and polymer coating. The pan is 30 cm long and 10 cm wide, the coating is 20 cm long and initially 0.2 mm thick, and the liquid depth is 2 cm. Original redrawing of Figure 1 from the source question sheet.

Worked solution

Data

Quantity	Symbol	Value
Coated length and width	𝐿,𝑏$L,b$	20,10 cm$20,10\ \mathrm{cm}$
Liquid depth and initial coating thickness	𝐻,𝛿$H,\delta$	2.0,0.020 cm$2.0,0.020\ \mathrm{cm}$
Solvent volume flow	𝑄$Q$	30 cm3 s−1$30\ \mathrm{cm^{3}\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵$D_{AB}$	3.0 ×10−6 cm2 s−1$3.0\times10^{-6}\ \mathrm{cm^{2}\,s^{-1}}$
Kinematic viscosity	𝜈$\nu$	0.0060 cm2 s−1$0.0060\ \mathrm{cm^{2}\,s^{-1}}$
Surface and bulk mass concentrations	𝐶𝐴,𝑠,𝐶𝐴,𝑏$C_{A,s},C_{A,b}$	0.040,0 g cm−3$0.040,0\ \mathrm{g\,cm^{-3}}$
Polymer density	𝜌𝑝$\rho_p$	1.05 g cm−3$1.05\ \mathrm{g\,cm^{-3}}$

Assumptions

Use mass concentration 𝐶𝐴$C_A$, so 𝑘𝑐(𝐶𝐴,𝑠 −𝐶𝐴,𝑏)$k_c(C_{A,s}-C_{A,b})$ is a mass flux. In centimetre units, 𝐶𝐴$C_A$ in g cm⁻³ and 𝑘𝑐$k_c$ in cm s⁻¹ give g cm⁻² s⁻¹. Assume isothermal, dilute transport with constant properties, negligible reaction in the liquid and rapid dissolution maintaining saturation at the solid-liquid interface. The coating is thin enough to neglect changes in flow geometry during an instantaneous rate calculation.

Model the exposed upper coating face as a laminar flat plate with a fresh concentration boundary layer at its leading edge. Neglect pan end effects, sidewall transfer and the velocity-profile corrections of a shallow channel. This is the source’s external-flow approximation, not a resolved pan flow.

Calculations

Use superficial velocity

𝑈=𝑄𝑏(𝐻−𝛿),$$U=\frac{Q}{b(H-\delta)},$$

coated area 𝐴 =𝑏𝐿$A=bL$ and

Re𝐿=𝑈𝐿𝜈,Sc=𝜈𝐷𝐴𝐵.$$\mathrm{Re}_L=\frac{UL}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}.$$

The average coefficient follows

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3=¯¯¯¯𝑘𝑐𝐿𝐷𝐴𝐵.$$\overline{\mathrm{Sh}}_L =0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3} =\frac{\overline{k}_cL}{D_{AB}}.$$(1)

Use Eq. 1 for Re𝐿 <5 ×105$\mathrm{Re}_L<5\times10^5$ and Sc >0.5$\mathrm{Sc}>0.5$. The initial coating mass is 𝜌𝑝𝐴𝛿$\rho_pA\delta$.

The polymer removal rate uses the coated area rather than the total pan area.

Convert thickness to 𝛿 =0.020 cm$\delta=0.020\,\mathrm{cm}$.

The initial liquid cross-section is

10(2.0−0.020)=19.8 cm2,$$10(2.0-0.020)=19.8\,\mathrm{cm^2},$$

giving

𝑈=1.51515 cm s−1Re𝐿=5050.51Sc=2000$$U=1.51515\,\mathrm{cm\,s^{-1}} \qquad \mathrm{Re}_L=5050.51 \qquad \mathrm{Sc}=2000$$

The specified laminar correlation is within its stated group ranges.

Using Eq. 1 gives ¯¯¯¯¯¯¯Sh𝐿 =594.537$\overline{\mathrm{Sh}}_L=594.537$ and ¯¯¯¯𝑘𝑐 =8.91805 ×10−5 cm s−1$\overline{k}_c=8.91805\times10^{-5}\,\mathrm{cm\,s^{-1}}$.

The exposed coating area is

𝐴=𝑏𝐿=10(20)=200 cm2,$$A=bL=10(20)=200\ \mathrm{cm^2},$$

and the dry-bulk concentration difference is 0.040 g cm−3$0.040\ \mathrm{g\,cm^{-3}}$. Substitution gives

˙𝑚𝐴=¯¯¯¯𝑘𝑐(𝐶𝐴,𝑠−𝐶𝐴,𝑏)𝐴=7.13444×10−4 g s−1=7.13444×10−7 kg s−1.$$\dot m_A=\overline{k}_c(C_{A,s}-C_{A,b})A =7.13444\times10^{-4}\,\mathrm{g\,s^{-1}} =7.13444\times10^{-7}\,\mathrm{kg\,s^{-1}}.$$

Initial coating mass is

1.05(200)(0.020)=4.20 g.$$1.05(200)(0.020)=4.20\,\mathrm g.$$

At the assumed constant rate, removal time is

4.20˙𝑚𝐴=5886.93 s=1.63526 h.$$\frac{4.20}{\dot m_A}=5886.93\,\mathrm s=1.63526\,\mathrm h.$$

Report about 1.64 h$1.64\,\mathrm h$.

The 2019 solution reports the same rounded time.

The estimate neglects the small velocity change as the coating shrinks and any change in surface behaviour near complete removal.

The outlet pickup estimated from ˙𝑚𝐴/𝑄$\dot m_A/Q$ is 2.37815 ×10−5 g cm−3$2.37815\times10^{-5}\,\mathrm{g\,cm^{-3}}$, only 0.0595% of saturation.

This supports the approximation of zero bulk concentration.

Solvent density is unnecessary because kinematic viscosity and mass concentration are given.

Answer

The initial polymer removal rate is 7.13 ×10−4 g s−1$7.13\times10^{-4}\ \mathrm{g\,s^{-1}}$. Maintaining that rate gives an estimated coating-removal time of 1.64 h$1.64\ \mathrm h$.

2 Dissolution from a tube wall

Balance and a conditional diffusivity estimate

An experiment on the dissolution of fatty deposits in blood vessels involves the setup of a tube (0.8 cm inner diameter, 50 cm in length) that is coated on the inner surface with a thin layer (0.01 cm thick) of lipophilic material that represents the fatty deposits, called species A (𝜌$\rho$ = 1.10 g/cm3$\mathrm{g/cm^3}$). A synthetic solution (𝜈$\nu$ = 0.02 cm2/s$\mathrm{cm^2/s}$, 𝜌$\rho$ = 1.04 g/cm3$\mathrm{g/cm^3}$) is pumped through the tube at a rate of 20 cm3/s$\mathrm{cm^3/s}$. The maximum solubility of A in the synthetic solution is 20 mg A/cm3$\mathrm{mg\ A/cm^3}$ solution.

a. Develop a material balance to predict the concentration of A in the solution exiting the tube

b. Experimental measurements indicate that the concentration of A in the synthetic solution exiting the tube is 0.1 mg A/cm3$\mathrm{mg\ A/cm^3}$. Determine the diffusion coefficient of A in the solution.

Worked solution

Figure 2: Liquid flows through a tube with a saturated coated wall. The wall adds dissolved material, so the outlet bulk concentration exceeds the inlet concentration.

Data

Quantity	Symbol	Value
Liquid bore diameter	𝑑$d$	0.80 cm$0.80\ \mathrm{cm}$
Coated length	𝐿$L$	50 cm$50\ \mathrm{cm}$
Volume flow	𝑄$Q$	20 cm3 s−1$20\ \mathrm{cm^{3}\,s^{-1}}$
Kinematic viscosity	𝜈$\nu$	0.020 cm2 s−1$0.020\ \mathrm{cm^{2}\,s^{-1}}$
Saturated wall concentration	𝐶𝐴,𝑠$C_{A,s}$	20 mg cm−3$20\ \mathrm{mg\,cm^{-3}}$
Inlet and measured outlet concentrations	𝐶𝐴,𝑖,𝐶𝐴,𝑜$C_{A,i},C_{A,o}$	0,0.10 mg cm−3$0,0.10\ \mathrm{mg\,cm^{-3}}$

The supplied solution uses Sh𝑑 =𝑘𝑐𝑑/𝐷𝐴𝐵 =3.66$\mathrm{Sh}_d=k_cd/D_{AB}=3.66$ for fully developed laminar mass transfer. The calculation below checks that assumption. It treats the quoted diameter as the initial liquid bore and takes the inlet concentration to be zero.

Assumptions

Assume steady liquid transport during the measurement, negligible axial diffusion, no bulk reaction, constant flow rate and a saturated coated wall. The cross-section balance uses a mixed-mean concentration, not a uniform local radial profile. Assume an upstream uncoated length establishes the laminar velocity profile before the coated section. The source correlation 3.66 additionally requires a developed concentration profile at fixed wall concentration. Reynolds number alone does not establish this condition.

Calculations

𝑈=4𝑄𝜋𝑑2,Re𝑑=𝑈𝑑𝜈,Sc=𝜈𝐷𝐴𝐵.$$U=\frac{4Q}{\pi d^2},\qquad \mathrm{Re}_d=\frac{Ud}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}.$$

Estimate entrance lengths using

𝐿ℎ≃0.05Re𝑑𝑑,𝐿𝑐≃Sc𝐿ℎ.$$L_h\simeq0.05\mathrm{Re}_d d,\qquad L_c\simeq\mathrm{Sc}L_h.$$(2)

The tube measurement determines an effective coefficient before any diffusivity correlation is chosen.

For a slice of length 𝑑𝑧$dz$, the wall adds mass at rate

𝑘𝑐(𝐶𝐴,𝑠−𝐶𝐴)𝜋𝑑 𝑑𝑧.$$k_c(C_{A,s}-C_A)\pi d\,dz.$$

The steady balance is

𝑄𝑑𝐶𝐴𝑑𝑧=𝑘𝑐𝜋𝑑(𝐶𝐴,𝑠−𝐶𝐴).$$Q\frac{dC_A}{dz}=k_c\pi d(C_{A,s}-C_A).$$(3)

Integrate Eq. 3 with 𝐶𝐴(0) =𝐶𝐴,𝑖 =0$C_A(0)=C_{A,i}=0$.

Define

Λ=𝑘𝑐𝜋𝑑𝐿𝑄,$$\Lambda=\frac{k_c\pi dL}{Q},$$

a dimensionless transfer measure.

Then

𝐶𝐴,𝑠−𝐶𝐴,𝑜𝐶𝐴,𝑠−𝐶𝐴,𝑖=exp⁡(−Λ),𝐶𝐴,𝑜=𝐶𝐴,𝑠[1−exp⁡(−Λ)].$$\begin{aligned} \frac{C_{A,s}-C_{A,o}}{C_{A,s}-C_{A,i}} &=\exp(-\Lambda),\\ C_{A,o}&=C_{A,s}[1-\exp(-\Lambda)]. \end{aligned}$$(4)

If the local coefficient varies, the exponential instead contains

𝜋𝑑𝑄∫𝐿0𝑘𝑐(𝑧) 𝑑𝑧.$$\frac{\pi d}{Q}\int_0^L k_c(z)\,dz.$$

Thus the inferred constant coefficient is the length average under this balance model.

Using Eq. 4 with consistent mg cm⁻³ units gives

𝑘𝑐=−𝑄𝜋𝑑𝐿ln⁡(20−0.1020−0)=7.97771×10−4 cm s−1=7.97771×10−6 m s−1.$$\begin{aligned} k_c&=-\frac{Q}{\pi dL}\ln\left(\frac{20-0.10}{20-0}\right)\\ &=7.97771\times10^{-4}\,\mathrm{cm\,s^{-1}}\\ &=7.97771\times10^{-6}\,\mathrm{m\,s^{-1}}. \end{aligned}$$

The measured mass pickup is

𝑄(𝐶𝐴,𝑜−𝐶𝐴,𝑖)=2.0 mg s−1.$$Q(C_{A,o}-C_{A,i})=2.0\,\mathrm{mg\,s^{-1}}.$$

Velocity is 𝑈 =39.7887 cm s−1$U=39.7887\,\mathrm{cm\,s^{-1}}$ and Re𝑑 =1591.55$\mathrm{Re}_d=1591.55$, consistent with laminar flow.

Conditional on Sh𝑑 =3.66$\mathrm{Sh}_d=3.66$,

𝐷𝐴𝐵=𝑘𝑐𝑑3.66=1.74376×10−4 cm2 s−1=1.74376×10−8 m2 s−1.$$D_{AB}=\frac{k_cd}{3.66} =1.74376\times10^{-4}\,\mathrm{cm^{2}\,s^{-1}} =1.74376\times10^{-8}\,\mathrm{m^{2}\,s^{-1}}.$$

This reproduces the coordinator’s rounded result, but its validity needs a separate check.

The inferred Schmidt number is 114.695$114.695$.

Using Eq. 2 gives 𝐿ℎ =63.662 cm$L_h=63.662\,\mathrm{cm}$ and

𝐿𝑐=7301.68 cm=73.0168 m,$$L_c=7301.68\,\mathrm{cm}=73.0168\,\mathrm m,$$

far greater than the coated length of 0.50 m$0.50\,\mathrm m$.

An upstream uncoated tube can develop the velocity profile, but cannot develop a solute profile before dissolution starts.

The supplied 3.66 model therefore does not provide a justified physical diffusivity for this entrance-region experiment.

The data establish the effective 𝑘𝑐$k_c$ and mass removal rate under the stated balance.

Inferring a reliable diffusivity additionally needs an appropriate developing-flow correlation or a resolved transport model, with the velocity entrance condition specified.

The conditional numerical result must not be reported as a validated measured diffusivity.

Coating thickness and solid density describe the available inventory; liquid density is unnecessary when kinematic viscosity is supplied.

Answer

The measured outlet concentration gives an effective 𝑘𝑐 =7.98 ×10−6 m s−1$k_c=7.98\times10^{-6}\ \mathrm{m\,s^{-1}}$ and a mass pickup of 2.0 mg s−1$2.0\ \mathrm{mg\,s^{-1}}$. Assuming Sh =3.66$\mathrm{Sh}=3.66$ gives 𝐷𝐴𝐵 =1.74 ×10−8 m2 s−1$D_{AB}=1.74\times10^{-8}\ \mathrm{m^{2}\,s^{-1}}$, but the concentration-entry check fails. This diffusivity is conditional and is not validated by the experiment.