21-Mat-A2 Materials Transport Phenomena · December 2015
Question 6 of 8: Liquid-Metal Flow Over a Flat Plate — Penetration-Theory Heat Transfer
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 10-Met-A2 Metallurgical Rate Phenomena. Three-hour, open-book exam; any non-communicating calculator permitted. Candidates answer Question 1 (compulsory) plus any four of Questions 2–8; the five answered count equally (20 marks each). All eight are solved below for completeness. Candidates were told to state any interpretive assumptions where doubt exists.
Reference texts: Geiger, G. H. & Poirier, D. R., Transport Phenomena in Materials Processing (TMS) — the primary reference for the momentum-, heat- and mass-transfer analyses in Questions 3, 6, 7 and 8; Szekely, J., Evans, J. W. & Sohn, H. Y., Rate Phenomena in Process Metallurgy — reactor-network and interfacial mass-transfer modelling (Questions 2, 3); Welty, J. R. et al., Fundamentals of Momentum, Heat and Mass Transfer — dimensional analysis and pipe-friction data (Questions 4, 7); Gaskell, D. R., Introduction to the Thermodynamics of Materials — Fe–C phase relations (Question 1h).
Question 6: Liquid-Metal Flow Over a Flat Plate — Penetration-Theory Heat Transfer (20 marks)
Fig. 6 — plug-flow liquid metal over a flat plate; the thermal penetration depth grows as √x.
Given. Plug-flow liquid metal at uniform velocity $U$ and bulk temperature $\theta^B$ approaching a flat plate of length $L$ held at surface temperature $\theta^S$; thermal diffusivity $\alpha$, thermal conductivity $k$.
Find. The developing temperature profile $\theta(x,y)$, the interfacial flux $\dot q''(x)$, and the local Nusselt number $Nu_x$ in the stated form.
Approach. Because the flow is plug flow (uniform velocity across the whole liquid, i.e. no momentum boundary layer), a fluid element convects downstream at constant $U$ while conducting heat only in $y$; each element's exposure time since first contacting the plate is $t=x/U$, converting the 2-D convection–conduction problem into the classical 1-D transient (semi-infinite-solid) conduction problem with $t\to x/U$.
Governing equation. With plug flow ($u=U$ everywhere) and negligible axial conduction, the thermal energy equation reduces to
$$U\dfrac{\partial\theta}{\partial x}=\alpha\dfrac{\partial^2\theta}{\partial y^2}$$
which is identical in form to the transient conduction equation $\partial\theta/\partial t=\alpha\,\partial^2\theta/\partial y^2$ with $x/U$ playing the role of time $t$.
Semi-infinite-solid solution. With boundary/initial conditions $\theta(y=0,t)=\theta^S$, $\theta(y\to\infty,t)=\theta^B$, $\theta(y,0)=\theta^B$, the standard error-function solution applies:
$$\dfrac{\theta(y,t)-\theta^S}{\theta^B-\theta^S}=\text{erf}\!\left(\dfrac{y}{2\sqrt{\alpha t}}\right)$$
Interfacial heat flux. Differentiating and evaluating at $y=0$ (using $\dfrac{d}{dy}\text{erf}\left(\dfrac{y}{2\sqrt{\alpha t}}\right)\Big|_{y=0}=\dfrac{1}{\sqrt{\pi\alpha t}}$):
$$\dot q''=-k\left.\dfrac{\partial\theta}{\partial y}\right|_{y=0}=\boxed{-\dfrac{k(\theta^B-\theta^S)}{\sqrt{\pi\alpha t}}}$$
— exactly the given target expression, confirming the semi-infinite conduction analogy.
Substitute $t=x/U$ and form $h$.
$$h_x\equiv\dfrac{\bar{\dot q}''}{\theta^B-\theta^S}=\dfrac{k}{\sqrt{\pi\alpha x/U}}=k\sqrt{\dfrac{U}{\pi\alpha x}}$$
Local Nusselt number.
$$Nu_x=\dfrac{h_x x}{k}=x\sqrt{\dfrac{U}{\pi\alpha x}}=\sqrt{\dfrac{Ux}{\pi\alpha}}$$
Writing $Re_x=Ux/\nu$ and $Pr=\nu/\alpha$, so $Re_x\,Pr=Ux/\alpha$:
$$\boxed{Nu_x=\sqrt{\dfrac{Re_x\,Pr}{\pi}}}$$
— matching the target form exactly.
