Dimensional analysis of the spinning-disk mass-transfer coefficient: 23-Chem-A3 Heat and Mass Transfer, December 2015, Q3

Question 3: Dimensional analysis of the spinning-disk mass-transfer coefficient (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. A dimensional correlation for $k_c$ in terms of the three transport parameters $D_{AB}$, $\nu$ and $\omega$. There is no numerical data — the task is to non-dimensionalize a dimensional power law by introducing a characteristic length $L$ (the disk radius) that cancels out.

Find. Definitions of $Re$ and $Sc$, the non-dimensional group $Sh=A\,Re^{n}Sc^{m}$, and the constants $A$, $n$, $m$.

Figure 3. Rotating soluble disk; the induced swirl provides the convective velocity scale $\omega L$ used to form $Re$.

Approach. Choose the natural velocity scale $u=\omega L$ and length scale $L$ (disk radius), form $Sh$, $Re$ and $Sc$, then substitute the given power law and confirm that $L$ cancels — forcing the exponents.

(a) Dimensionless groups. With characteristic velocity $u=\omega L$, $$Re=\frac{uL}{\nu}=\frac{\omega L^2}{\nu},\qquad Sc=\frac{\nu}{D_{AB}},\qquad Sh=\frac{k_cL}{D_{AB}}.$$ The rotation supplies the velocity; $L$ is the disk radius.
(b) Substitute the correlation. Insert $k_c=0.62\,D_{AB}^{2/3}\nu^{-1/6}\omega^{1/2}$ into $Sh=k_cL/D_{AB}$: $$Sh=0.62\,D_{AB}^{-1/3}\nu^{-1/6}\omega^{1/2}L.$$ Now write $$0.62\,Re^{1/2}Sc^{1/3}=0.62\left(\dfrac{\omega L^2}{\nu}\right)^{1/2}\left(\dfrac{\nu}{D_{AB}}\right)^{1/3}=0.62\,\omega^{1/2}L\,\nu^{-1/6}D_{AB}^{-1/3},$$ which is identical — $L$ appears to the first power on both sides and every exponent matches.
(c) Constants. Therefore $$Sh=\boxed{0.62\,Re^{1/2}\,Sc^{1/3}},\qquad A=0.62,\ \ n=\tfrac12,\ \ m=\tfrac13.$$

Quantity	Result
Reynolds number	$Re=\omega L^2/\nu$
Schmidt number	$Sc=\nu/D_{AB}$
Correlation	$Sh=0.62\,Re^{1/2}Sc^{1/3}$
Constants	$A=0.62,\ n=1/2,\ m=1/3$

Concept & Theory

Topic: Non-dimensionalization of a dimensional transfer correlation (rotating-disk / Levich analysis).

Key relations: $Sh=k_cL/D_{AB}$, $Re=\omega L^2/\nu$, $Sc=\nu/D_{AB}$; general convective form $Sh=A\,Re^{n}Sc^{m}$.

Why this works: The rotating disk is “uniformly accessible” — its boundary-layer thickness is independent of radius — so $k_c$ contains no length dependence. The arbitrary length $L$ inserted to build $Sh$ and $Re$ must therefore cancel, which uniquely fixes $n=1/2$ (from $\omega^{1/2}$) and $m=1/3$ (from $D_{AB}^{2/3}$ giving $Sc^{1/3}$).

Common pitfall: Trying to identify $A$, $n$, $m$ by matching dimensions of $k_c$ alone; you must build $Sh$ (which reintroduces $L$) and require $L$ to cancel — that is what pins the exponents.

Source: Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed.), §30 (dimensional analysis of convective mass transfer); Levich, Physicochemical Hydrodynamics.

Question 3 of 7: Dimensional analysis of the spinning-disk mass-transfer coefficient

Question 3: Dimensional analysis of the spinning-disk mass-transfer coefficient (Part B — equal value)