Dissolution of a NaCl cylinder — Chilton–Colburn analogy: 23-Chem-A3 Heat and Mass Transfer, December 2015, Q4

Question 4: Dissolution of a NaCl cylinder — Chilton–Colburn analogy (Part B — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Cross-flow of water over a soluble cylinder. The heat-transfer Nusselt correlation is converted to mass transfer through the Chilton–Colburn analogy ($Nu\to Sh$, $Pr\to Sc$). The diffusivity must first be corrected from 291 K to the operating 300 K.

Quantity	Value
Cylinder diameter $D$	1.5 cm $=0.015$ m
Water velocity $v$	10 m/s
$D_{AB}$ at 291 K	$1.26\times10^{-9}$ m²/s
$\mu$ at 291 K / 300 K	$1.073\times10^{-3}$ / $8.76\times10^{-4}$ Pa·s
$\rho_{water}$	996 kg/m³

Find. The mass-transfer coefficient $k_c$ at 300 K.

Figure 4. Soluble NaCl cylinder in cross-flow; forced-convection dissolution governed by the cross-flow-cylinder correlation via the Chilton–Colburn analogy.

Approach. Correct $D_{AB}$ to 300 K (Stokes–Einstein $D\mu/T=$ const), form $Re$ and $Sc$, evaluate the Chilton–Colburn form of the correlation for $Sh$, and convert $Sh$ to $k_c$.

Diffusivity at 300 K. Stokes–Einstein gives $D\mu/T=$ const, so $$D_{300}=D_{291}\frac{T_{300}}{T_{291}}\frac{\mu_{291}}{\mu_{300}}=1.26\times10^{-9}\frac{300}{291}\frac{1.073\times10^{-3}}{8.76\times10^{-4}}=1.59\times10^{-9}\ \text{m}^2/\text{s}.$$
Reynolds and Schmidt numbers. $$Re=\frac{Dv\rho}{\mu_{300}}=\frac{(0.015)(10)(996)}{8.76\times10^{-4}}=1.71\times10^{5},\qquad Sc=\frac{\mu_{300}}{\rho D_{300}}=\frac{8.76\times10^{-4}}{(996)(1.59\times10^{-9})}=553.$$
Chilton–Colburn analogy. Replace $Nu\to Sh$ and $Pr\to Sc$: $$Sh=(0.506\,Re^{0.5}+0.00141\,Re)\,Sc^{0.33}=(0.506\sqrt{1.71\times10^{5}}+0.00141(1.71\times10^{5}))(553)^{0.33}=\boxed{3.61\times10^{3}}.$$
Mass-transfer coefficient. The liquid-side coefficient asked for is $$k_L=\frac{Sh\,D_{300}}{D}=\frac{(3.61\times10^{3})(1.59\times10^{-9})}{0.015}=\boxed{3.83\times10^{-4}\ \text{m/s}}.$$

Quantity	Value
$D_{AB}$ at 300 K	$1.59\times10^{-9}$ m²/s
$Re$ / $Sc$	$1.71\times10^{5}$ / 553
$Sh$	$3.61\times10^{3}$
Mass-transfer coefficient $k_L$	$3.83\times10^{-4}$ m/s

Concept & Theory

Topic: Heat–mass transfer analogy (Chilton–Colburn) for forced convection over a cylinder.

Key relations: $j_D=j_H$ ⇒ the same correlation applies with $Nu\to Sh$, $Pr\to Sc$; Stokes–Einstein $D\mu/T=$ const; $Sh=k_cD/D_{AB}$.

Why this works: Because the momentum, thermal and concentration boundary layers obey the same convection–diffusion equation, a validated $Nu(Re,Pr)$ correlation maps directly onto $Sh(Re,Sc)$. The Chilton–Colburn analogy formalizes this for the $Pr^{1/3}$/$Sc^{1/3}$ dependence typical of external forced convection.

Common pitfall: Using $D_{AB}$ at the tabulated 291 K instead of correcting to 300 K — the temperature and viscosity corrections together raise $D$ by ~26 %, which propagates into both $Sc$ and $k_c$.

Source: Geankoplis, Transport Processes (4th ed.), §7.3 (mass-transfer coefficients, Chilton–Colburn $j$-factors); Welty et al., §30.

Question 4 of 7: Dissolution of a NaCl cylinder — Chilton–Colburn analogy

Question 4: Dissolution of a NaCl cylinder — Chilton–Colburn analogy (Part B — equal value)