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This is likely very simple, but I am not sure how to do it.

We have a tube with a liquid metal ($Pr = 0$, $Re = 10^6$). The temperature at the wall ($r_1$) is $T_1$. The temperature halfway between the wall and the centerline ($r_2$) is $T_2$.

I want to know the temperature at the centerline. How would I do that?

I can solve the temperature distribution for this cylinder with no internal heat generation

$$\frac{d}{dr}\left(r\frac{dT}{dr}\right) = 0$$

This gives me, integrating twice and using the boundary conditions $T(r_i) = T_i$:

$$T(r) = \frac{T1-T2}{\ln \left( \frac{r_1}{r_2} \right)} \ln \left( \frac{r}{r_2} \right) +T_2 $$

However, this is not valid at $r=0$, so I cannot get the temperature at the centerline from it.

I could use the boundary condition $\frac{dT}{dr}\bigg\rvert_{r=0} = 0$ by a symmetry argument. But then, I would only get:

$$r\frac{dT}{dr} = C_1 \implies \frac{dT}{dr}\bigg\rvert_{r=0} = \frac{C_1}{0} \implies C_1 = 0$$

And then, $T(r)$ would be constant.

So, what am I missing here? How do I compute the centerline temperature given the surface temperature and an intermediate temperature?

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2 Answers 2

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Prandtl number, $Pr\equiv\frac{\nu}{\alpha}\to 0$ when $\alpha\to\infty$, given that $\nu\neq 0$ (this is assured because $Re$ is finite). The limit $\alpha\to\infty$ implies that the liquid under consideration is approaching the behavior of an ideal conductor. In this limit, no matter how high $Re$ is, temperature must be uniform everywhere within the liquid, because heat is conducted through the liquid infinitely fast. Therefore $T(r)=$constant is the correct solution.

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  • $\begingroup$ Alright, so I got to more textbooks and papers, and my take on this is to use Martinelli's work (Martinelli, R. C. "Heat Transfer to Molten Metals", Trans. ASME, Vol. 69 (1947), pp. 947-959). In this work, he provides solutions for the temperature gradient in pipe flow for various Reynolds and Prandtl number (and conveniently, Re = 1,000,000 and Pr = 0 is one of these solutions). I thus obtain $\frac{T_w - T(y)}{T_w-T_c} = \frac{y}{r_0}$, and from the given solutions, I can obtain $T_c$ (Temperature at the centerline) from $T_w$ (temperature at the wall), $r_0$ (radius) and a known $T(y)$. $\endgroup$ Apr 11, 2017 at 16:21
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Alright, so I got to more textbooks and papers, and my take on this is to use Martinelli's work (Martinelli, R. C. "Heat Transfer to Molten Metals", Trans. ASME, Vol. 69 (1947), pp. 947-959).

In this work, he provides solutions for the temperature gradient in pipe flow for various Reynolds and Prandtl number (and conveniently, $Re = 10^6$ and $Pr = 0$ is one of these solutions).

I thus obtain, for $Pr=0$ and $Re = 10^6$:

$$\frac{T_w - T(y)}{T_w-T_c} = \frac{y}{r_0}$$

Knowing $T_w$ and also knowing $T(y=r_0/2)$, I can now easily solve the next equation for $T_c$ (linear temperature profile):

$$\frac{T_w - T(y=r_0/2)}{T_w-T_c} = \frac{1}{2}$$

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