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I am currently studying The Art of Electronics, third edition, by Horowitz and Hill. Exercise 1.6 says the following:

C. Power in resistors
The power dissipated by a resistor (or any other device) is \$P = IV\$. Using Ohm’s law, you can get the equivalent forms \$P = I^2 R\$ and \$P = V^2 / R\$.
Exercise 1.6. Optional exercise: New York City requires about \$10^{10}\$ watts of electrical power, at 115 volts (this is plausible: 10 million people averaging 1 kilowatt each). A heavy power cable might be an inch in diameter. Let’s calculate what will happen if we try to supply the power through a cable 1 foot in diameter made of pure copper. Its resistance is \$0.05 \ \mu \Omega\$ (\$5 \times 10^{−8}\$ ohms) per foot. Calculate (a) the power lost per foot from “\$I^2R\$ losses," (b) the length of cable over which you will lose all \$ 10^{10} \$ watts, and (c) how hot the cable will get, if you know the physics involved (\$ \omega = 6 \times 10^{-12} \text{W}/\text{K}^4 \text{cm}^2 \$). If you have done your computations correctly, the result should seem preposterous. What is the solution to this puzzle.

We managed to get the answer to (a) here as \$ 3.8 \times 10^8 \ \text{W}/\text{ft} \$. I then get \$ \dfrac{10^{10} \text{W}}{3.8 \times 10^8 \ \text{W}/\text{ft}} = 26.32 \ \text{ft} \$ for (b).

Now, I'm trying to solve (c). I'm referring to this document, which claims that, to calculate the heat dissipated by the cable, we can use the Stefan-Boltzmann equation. However, according to the Wikipedia article, this equation describes the power radiated from a black body in terms of its temperature. How can such an equation for the temperature of a black body be valid in this case?

EDIT

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The Pointer
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  • You just assume it's valid at this stage and see what sort of answer you get. –  Jan 02 '21 at 16:46
  • The document you linked ("*this document*") has nowhere in its text the phrase "heat dissipated". – Andy aka Jan 02 '21 at 16:49
  • @BrianDrummond But that doesn't answer my question. That GitHub document is **not** an official solutions manual (there is no solutions manual available for this edition of the textbook) – it's a crowdsourced solutions manual. So just because someone has claimed that that is the correct solution, does not mean that it is actually the correct solution. So I think it is a valid question as to whether the Stefan-Boltzmann equation is valid for this case. – The Pointer Jan 02 '21 at 16:51
  • @Andyaka The solution for (c) of exercise 1.6 in the document indeed has the phrase "heat dissipated": "To calculate the heat dissipated by the cable, ..." – The Pointer Jan 02 '21 at 16:53
  • Related (really!) to [this question](https://electronics.stackexchange.com/questions/333823/does-the-paint-colour-matter-on-a-heat-sink) – TimWescott Jan 02 '21 at 16:53
  • OK, ctrl-F doesn't work. And the document is many pages long so, link to the page and show it as a picture in your question. – Andy aka Jan 02 '21 at 16:55
  • @Andyaka See my edit. – The Pointer Jan 02 '21 at 16:57
  • That's why you assume first and see what answer you get. An accurate temperature would involve Newtonian cooling, convection, and require knowing the local environment (air velocity etc.) and an entire book of thermodynamics. If you get a relatively high temperature, the 4th power term dominates and you can ignore the Newtonian (linear) term. As here. You are left with the fact that unless you contain the molten (probably boiling) copper you have merely blown a 28 foot long fuse. –  Jan 02 '21 at 17:25

1 Answers1

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First, the exercise is intended to present an absurd answer (I assume they're using it to lead in to the use of high-tension wires).

Second, blackbody radiation is used for more than things that are perfectly black. The Stefan-Boltzmann equation is only valid for things with a perfectly black surface -- but it's easy to adjust it's effects based on the surface's emissivity, or to adjust the surface emissivity for the purpose at hand (Which is why some heat sinks are black. See my last paragraph).

Copper isn't very emissive (because it's shiny -- emissivity is more or less the reciprocal of reflectivity) but they're trying to show that even if the thing were perfectly black it'd still reach an absurdly high temperature. So for finding the minimum temperature that the thing would reach from radiative cooling, purely radiative cooling, they're on track.

A slightly more real-world calculation would use convective cooling (i.e., air) -- but given the temperature reached with just radiative cooling, you're really just finding that the wire will take about five times longer to explode into copper vapor with convective cooling as it would with radiative.

Radiative cooling really is a thing, and while it's contribution to cooling stuff in air is usually not the biggest factor, it's enough to motivate people to often make heat sinks (and engine cooling fins) black. This post -- Does the paint colour matter on a heat sink? -- does a rough calculation for a given heat sink to find that the radiative cooling in still air is something like 25% of the total (I suspect it's less). For a heat sink in a vacuum, all the cooling is radiative, so "black body" radiation does matter (as does emissivity -- if you see some bit of space hardware wrapped in shiny foil, it's because the designer wants it to stay cool when the sun is shining on it).

TimWescott
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    As an additional point, just because a surface is reflective in the visible band doesn't mean it is also reflective in the far IR (where most of the emission occurs for more reasonable black-body problems) or the x-ray and gamma-ray bands (which are probably what matters in this problem). – The Photon Jan 02 '21 at 17:39
  • @ThePhoton: good point. However, since the copper is predicted to hit 15,000 degrees, whatever its emissivity is at room temperature is probably not going to mean much when it's a ball of ionized gas. – TimWescott Jan 02 '21 at 20:38