Question 5 of 6: C1 — Lead dissolution along a pipe section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2015 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section (four of six marked). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is supplied as Appendix A in the paper and is used throughout.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances used across every problem; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe-friction and boundary-layer mass-transfer correlations; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with variable conductivity, exchanger analysis; D. Q. Kern, Process Heat Transfer (McGraw-Hill) — the Kern casing-side method and the LMTD–correction-factor chart for multiple casing passes; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — physical properties and the Moody chart.
Question 5: C1 — Lead dissolution along a pipe section (25 marks)
Given. Laminar internal flow, thin concentration boundary layer ($\delta_c\ll R$) so the curved lead wall is treated locally as a flat plate with $x$ measured from the start of the lead section; wall (saturation) concentration $C_{As}=S$; background $C_{A0}\approx0$ (lead-free inlet, dissolved lead stays dilute in the core); supplied correlation $Sh=0.332\,Re_x^{1/2}Sc^{1/3}$ with $Re_x=v_{max}x/\nu$. For fully developed laminar (Poiseuille) pipe flow the maximum (centre-line) velocity is twice the mean: $v_{max}=2Q/\pi R^2$.
Find. (a) $N_A(x)$; (b) total dissolution rate $W$; (c) mean outlet concentration $C_{out}$; (d) the factor by which $C_{out}$ changes from 25 °C to 100 °C.
Fig. C1: A lead-lined section dissolves into laminar water flow. The concentration boundary layer grows from the leading edge of the section, so the local flux is highest at $x=0$ and decays as $x^{-1/2}$.
Approach. Convert the flat-plate Sherwood correlation into a local $k_c(x)$, multiply by the driving force $S$ for the local flux, integrate over the wetted lead area for the total rate, divide by $Q$ for the mean outlet concentration, and finally collect the temperature-dependent properties for part (d).
(a) Local flux. From $Sh=k_c x/D_{AB}=0.332\,Re_x^{1/2}Sc^{1/3}$, $$k_c(x)=0.332\,\frac{D_{AB}}{x}\left(\frac{v_{max}x}{\nu}\right)^{1/2}\!\left(\frac{\nu}{D_{AB}}\right)^{1/3}=0.332\,D_{AB}^{2/3}\,v_{max}^{1/2}\,\nu^{-1/6}\,x^{-1/2}.$$ With $C_{As}=S$, $C_{A0}=0$ and $v_{max}=2Q/\pi R^2$, $$\boxed{\;N_A(x)=0.332\,S\,D_{AB}^{2/3}\,\nu^{-1/6}\left(\frac{2Q}{\pi R^2}\right)^{1/2}x^{-1/2}\;}\qquad(\propto x^{-1/2}).$$
(b) Total dissolution rate. Integrate the flux over the wetted lead surface $dA=2\pi R\,dx$, using $\int_0^L x^{-1/2}dx=2\sqrt L$: $$W=\int_0^L N_A(x)\,2\pi R\,dx=2\pi R(0.332)S\,D_{AB}^{2/3}v_{max}^{1/2}\nu^{-1/6}\,(2\sqrt L)=1.328\,\pi R\sqrt L\,S\,D_{AB}^{2/3}\,v_{max}^{1/2}\,\nu^{-1/6}.$$ Since $\pi R\,v_{max}^{1/2}=\pi R\sqrt{2Q/\pi R^2}=\sqrt{2\pi Q}$, the radius drops out: $$\boxed{\;W=1.328\,\sqrt{2\pi QL}\;S\,D_{AB}^{2/3}\,\nu^{-1/6}\;}$$
(c) Mean outlet concentration. All dissolved lead leaves in the volumetric flow $Q$ (inlet lead-free), so the mixing-cup outlet concentration is $$\boxed{\;C_{out}=\frac{W}{Q}=1.328\,\sqrt{\frac{2\pi L}{Q}}\;S\,D_{AB}^{2/3}\,\nu^{-1/6}\;}$$ Had the mean velocity $Q/\pi R^2$ been used in $Re_x$ instead of $v_{max}$, the results of (a)–(c) would each be smaller by a factor $\sqrt2$; the temperature ratio in (d) is unaffected either way.
(d) Effect of temperature. Only $S$, $D_{AB}$ and $\nu$ change; the geometry and flow $Q$ are fixed, so $C_{out}\propto S\,D_{AB}^{2/3}\,\nu^{-1/6}$. Using Stokes–Einstein $D_{AB}\propto T/\mu$ (absolute $T$) and $\nu=\mu/\rho\propto\mu$ (water density nearly constant), $$\frac{C_{out,100}}{C_{out,25}}=\frac{S_{100}}{S_{25}}\left(\frac{T_{100}}{T_{25}}\frac{\mu_{25}}{\mu_{100}}\right)^{2/3}\!\left(\frac{\mu_{25}}{\mu_{100}}\right)^{1/6}.$$ With $S_{100}/S_{25}=2$, $T_{100}/T_{25}=373/298=1.25$, $\mu_{25}/\mu_{100}=9/2.7=3.33$: $$\frac{C_{out,100}}{C_{out,25}}=2\,(4.17)^{2/3}(3.33)^{1/6}=\boxed{6.3}$$ The outlet lead concentration rises about six-fold (≈ +530 %) on going from 25 °C to 100 °C.
