Question 4 of 9: Liquid-Metal Plug Flow Over a Flat Plate — Penetration-Theory Derivation

Question 4: Liquid-Metal Plug Flow Over a Flat Plate — Penetration-Theory Derivation (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.

Given. Liquid metal (very low Prandtl number) in plug (slug) flow at uniform velocity $U$ past a plate held at $\theta^S$, bulk temperature $\theta^B$ far from the wall. Because $Pr\ll1$ for liquid metals (Q1(n)), the momentum boundary layer is negligible compared with the thermal boundary layer — the velocity profile is essentially flat (plug flow) across the region where conduction actually matters.

Find. The transient/developing temperature profile $\theta(y,t)$, the local wall flux $q''(t)$, and its restatement as $Nu_x=f(Re_x,Pr)$.

Approach. Because the velocity is uniform (plug flow), a fluid element at distance $x$ from the leading edge has been in contact with the wall for exactly $t=x/U$ — so the 2-D steady convection problem collapses to the 1-D transient conduction problem into a semi-infinite medium suddenly exposed to a fixed surface temperature, with $t\to x/U$.

1. Governing equation. With no velocity gradient (plug flow) and no streamwise conduction (thin thermal layer, high $Pe_x$), energy conservation for the liquid reduces to the 1-D transient conduction equation in the wall-normal coordinate $y$, with time re-interpreted as residence time: $$\dfrac{\partial\theta}{\partial t}=\alpha\dfrac{\partial^2\theta}{\partial y^2},\qquad \theta(y,0)=\theta^B,\ \ \theta(0,t)=\theta^S,\ \ \theta(\infty,t)=\theta^B$$
2. Similarity solution. With $\eta=y/\sqrt{4\alpha t}$, the classical error-function solution for a semi-infinite medium with a suddenly-imposed surface temperature applies directly: $$\dfrac{\theta-\theta^S}{\theta^B-\theta^S}=\text{erf}(\eta)=\text{erf}\!\left(\dfrac{y}{\sqrt{4\alpha t}}\right)$$
3. Wall heat flux. $q''(t)=-k\left.\dfrac{\partial\theta}{\partial y}\right|_{y=0}$. Differentiating $\text{erf}(\eta)$ and evaluating at $y=0$ ($d(\text{erf}\,\eta)/d\eta|_0=2/\sqrt{\pi}$): $$q''(t)=-k(\theta^B-\theta^S)\cdot\dfrac{-2/\sqrt{\pi}}{\sqrt{4\alpha t}}=\boxed{\dfrac{k(\theta^B-\theta^S)}{\sqrt{\pi\alpha t}}}$$ — exactly the quoted result, with the sign confirming heat flows from the hot bulk into the cooler wall.
4. Substitute $t=x/U$ (plug flow, position $x$ along the plate). $$q''_x=\dfrac{k(\theta^B-\theta^S)}{\sqrt{\pi\alpha x/U}}\ \Rightarrow\ h_x=\dfrac{q''_x}{\theta^B-\theta^S}=\dfrac{k}{\sqrt{\pi\alpha x/U}}=k\sqrt{\dfrac{U}{\pi\alpha x}}$$
5. Non-dimensionalise as a local Nusselt number. $Nu_x=h_x x/k$: $$Nu_x=\dfrac{x}{\sqrt{\pi\alpha x/U}}\cdot\dfrac1x\cdot x=\sqrt{\dfrac{Ux}{\pi\alpha}}=\sqrt{\dfrac{1}{\pi}\cdot\dfrac{Ux}{\nu}\cdot\dfrac{\nu}{\alpha}}=\sqrt{\dfrac{Re_x\,Pr}{\pi}}$$ using $Re_x=Ux/\nu$ and $Pr=\nu/\alpha$: $$\boxed{Nu_x=\sqrt{\dfrac{Re_x\,Pr}{\pi}}}$$ — matching the quoted form exactly. (Averaging $Nu_x$ over $0\le x\le L$ doubles the coefficient to $Nu_{L,avg}=1.128\sqrt{Re_L\,Pr}$, the classical liquid-metal flat-plate correlation used directly in Question 7.)