Fraction of mass transfer in the laminar zone of a flat plate: 23-Chem-A3 Heat and Mass Transfer, December 2014, Q4

Question 4: Fraction of mass transfer in the laminar zone of a flat plate (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 single flat plate carries a laminar leading region up to transition, then a turbulent region to the trailing edge. The total mass transferred is the integral of the local coefficient over the plate; we want the laminar share.

Quantity	Value
Transition Reynolds number $Re_t$	$2.0\times10^5$
Plate-end Reynolds number $Re_L$	$3.0\times10^6$
Laminar local $Sh_x$	$0.332\,Re_x^{1/2}Sc^{1/3}$
Turbulent local $Sh_x$	$0.0292\,Re_x^{4/5}Sc^{1/3}$

Find. The percentage of total mass transfer occurring within the laminar zone ($0\le x\le x_t$).

Figure 4 — The plate is short in the laminar zone (transition at $Re_t=2\times10^5$ vs. $Re_L=3\times10^6$). Although the local coefficient is highest near the leading edge, the turbulent region is long and transfers most of the total mass.

Approach. The total transfer per unit width is $\propto\int k_c\,dx=\int (Sh_x D_{AB}/x)\,dx$; substituting $x=(\nu/v)Re_x$ turns each region’s integral into a Reynolds-number group in which the common factor $D_{AB}Sc^{1/3}$ cancels when we take the ratio.

Set up the length-integrated groups. Since $k_{c,x}=Sh_x D_{AB}/x$ and $x\propto Re_x$, the laminar and turbulent contributions reduce (after $\int x^{-1/2}dx$ and $\int x^{-1/5}dx$) to $$W_\text{lam}\propto 0.664\,Re_t^{1/2},\qquad W_\text{turb}\propto 0.0365\left(Re_L^{4/5}-Re_t^{4/5}\right),$$ where $0.664=0.332\times2$ and $0.0365=0.0292\times\tfrac54$ are the integration constants. The shared $D_{AB}Sc^{1/3}$ cancels in the ratio.
Laminar group. $$W_\text{lam}=0.664\,(2.0\times10^5)^{1/2}=0.664\times447.2=297.0.$$
Turbulent group. $$W_\text{turb}=0.0365\left[(3.0\times10^6)^{4/5}-(2.0\times10^5)^{4/5}\right]=0.0365\,(1.519\times10^5-1.741\times10^4)=4910.$$
Laminar percentage. $$\%_\text{lam}=\frac{W_\text{lam}}{W_\text{lam}+W_\text{turb}}=\frac{297.0}{297.0+4910}=\boxed{5.7\%}.$$

Quantity	Result
Laminar transfer group	297.0
Turbulent transfer group	4910
Fraction in laminar zone	5.7%

Check

Only about 6% of the transfer occurs in the laminar zone even though the local coefficient there is the highest on the plate: the laminar region spans just $Re_t/Re_L=1/15$ of the plate length, and the turbulent coefficient (falling only as $x^{-1/5}$) sustains strong transfer over the remaining 93% of the plate. The result is independent of $Sc$ and $D_{AB}$, as those cancel in the ratio.

Concept & Theory

Topic. Length-integrated mass transfer over a mixed laminar/turbulent flat plate.

Key relations. Local $k_{c,x}=Sh_x D_{AB}/x$; total transfer $\propto\int_0^L k_{c,x}\,dx$; $\int x^{-1/2}dx=2x^{1/2}$ (constant $0.332\to0.664$); $\int x^{-1/5}dx=\tfrac54 x^{4/5}$ (constant $0.0292\to0.0365$); $Re_x=vx/\nu$.

Why this works. Because $Re_x$ is linear in $x$, changing the variable of integration from $x$ to $Re_x$ leaves each zone’s contribution as a pure Reynolds-number group multiplied by the same $D_{AB}Sc^{1/3}$. Forming a ratio cancels that common factor, so the split depends only on the two Reynolds numbers and the correlation exponents — not on the fluid or the species.

Common pitfall. Comparing the two local coefficients at a point, or using the average $\overline{Sh}$ correlations directly, instead of integrating the local coefficient over each zone. Another trap is forgetting the $\tfrac54$ factor from $\int x^{-1/5}dx$, which turns 0.0292 into 0.0365.

Source. Welty et al., §30 (combined laminar–turbulent boundary-layer transfer); Geankoplis, §7.3.

Question 4 of 7: Fraction of mass transfer in the laminar zone of a flat plate

Question 4: Fraction of mass transfer in the laminar zone of a flat plate (Part B — equal value)