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21-Mat-A2 Materials Transport Phenomena · December 2018

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 2018 — Met-A2 Metallurgical Rate Phenomena. Three-hour, open-book exam; any Casio or Sharp approved 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, 5, 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).

Note on the questions
Question 8 specifies a fill hole of explicit length $L$ and asks whether to increase $L$, and it is solved below from first principles. Question 5 asks for the maximum thickness of slab and adds an explicit instruction to justify the four simplifying assumptions rather than merely state them.

Question 6: Liquid-Metal Flow Over a Flat Plate — Penetration-Theory Heat Transfer (20 marks)

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.

U, θᴮδₜ(x) – thermal penetration depthx = 0x = Lθˢ (plate surface)y
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 $t\to x/U$.

  1. 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 $t$.
  2. 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: $$\dfrac{\theta(y,t)-\theta^S}{\theta^B-\theta^S}=\text{erf}\!\left(\dfrac{y}{2\sqrt{\alpha t}}\right)$$
  3. 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.
  4. Substitute $t=x/U$ and form $h$. $$h_x\equiv\dfrac{\bar{\dot q}''}{\theta^B-\theta^S}=\dfrac{k}{\sqrt{\pi\alpha x/U}}=k\sqrt{\dfrac{U}{\pi\alpha x}}$$
  5. 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.
QuantityResult
Temperature profile$(\theta-\theta^S)/(\theta^B-\theta^S)=\text{erf}\left(y/2\sqrt{\alpha t}\right)$, $t=x/U$
Interfacial flux$\dot q''=-k(\theta^B-\theta^S)/\sqrt{\pi\alpha x/U}$
Local Nusselt number$Nu_x=\sqrt{Re_x\,Pr/\pi}$