3
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The heat flow (in $[\pu{J/s}]$) through the wall of a tube of length $L$ and inner radius $R_0$ is

$$Q = (2πR_0L)U(T_{1\infty} - T_{2\infty}),$$

where $T_{1\infty}$ is the bulk temperature in the inside and $T_{2\infty}$ is the bulk temperature on the outside. The overall heat transfer coefficient reads as:

$$\frac{1}{UR_0} = \frac{1}{h_1R_0} + \frac{\ln{(R_a/R_0)}}{k} + \frac{1}{h_2R_a} \tag{1}$$

where $R_0$ and $R_a$ are the inner and outer radius of the tube, respectively, $h_1$ and $h_2$ are the inner and outer heat transfer coefficients, and $k$ is the thermal conductivity of the wall.

How does Eq. (1) simplify for the $T$-profile shown on the left?

enter image description here

  1. $U = k/R_0 \cdot 1/\ln(R_a/R_0)$
  2. $U = h_1$
  3. $1/U = 1/h_1 + 1/h_2 \cdot (R_0/R_a)$
  4. $1/UR_o = (\ln(R_a/R_o)/ k ) + (1/h_2R_a)$

I had seen a similar $T$-profile in one of our exercises and there they used Eq. (1), however when selecting option 1, I get "incorrect answer". How can you tell from looking at the $T$-profle what the equation for the overall heat transfer coefficient will be?

All help and hints are appreciated!

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  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Note that a screenshot or picture of an exercise is not searchable. $\endgroup$ – andselisk Feb 12 at 16:19
2
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This is the heat equation analogue to Ohm's law, in which resistances are additive (in series), and hence the inverse conductances are additive.

The way to reason is that the full equation (1) for the heat conductivity (1) contains three terms: one for the interior of the tube, one for the tube wall, and a third for the exterior. When you look at the profile, it becomes evident that, since temperatures immediately adjacent inside and outside the tube are equal, the tube has infinite heat conductance ($k=\infty$).

The right answer is therefore 3.

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  • $\begingroup$ Why is ($k=\infty$) because the temperatures adjacent to the inside and outside of the tube are equal? And also, how does this lead to 3 being the correct answer? $\endgroup$ – Johan Feb 12 at 16:55
  • $\begingroup$ Please check choice 3. Does it match equation 1 if you remove the middle term? $\endgroup$ – Try Hard Feb 12 at 16:56
  • $\begingroup$ If there was resistance to heat flow across the tube there would be a gradient across it. The absence of that gradient suggests that $k=\infty$ $\endgroup$ – Try Hard Feb 12 at 16:57

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