How fast can heat be carried away from a small source?
Someone invents a portable, incredibly high-powered laser or similar energy weapon. The mechanism of energy production can be hand-waved away. In other words, it will heat up. Lots. Gigawatts lots, or terawatts/petawatts/etc. Handwave any production rate you like into this.
Using only substances that exist in our real world, what is the practical limit on how quickly a portable energy weapon could be cooled-down? In other words, what is the maximum sustainable energy output of such a weapon, assuming cooling rather than energy production is the limiting factor?
The sort of thing I'm thinking of: Suppose you had a larger device nearby (on a truck, say) which could cool down helium to a liquid and pump it into your energy gun via a tube. The liquid helium would have to be pumped around quickly to prevent it from boiling. For sufficiently high weapon wattage, the sheer volume of helium needed, and the size of the cooling apparatus, would exceed what you can put on an accompanying truck. I suspect substances other than helium will have a better heat capacity before boiling, or a better rate of heat transfer, or some other property, but you get the idea.
Other than the internal mechanism of the weapon, there is no significant advance in materials or other technology beyond what we have here today.
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1 answer
Let's imagine that the hottest part of the apparatus is the barrel, so to speak, which is cylindrical with length $l$, radius $r$ and temperature $T_b$. We can surround it with a fluid of temperature $T_f$. Now, the barrel has a heat transfer coefficient of $h$. The change in heat energy of the barrel over time, $\dot{Q}$, is $$\dot{Q}=hA\Delta T=h(2\pi rl)\left(T_b-T_f\right)$$ where $A$ is the surface area of the barrel. Some things immediately spring out:
- A greater heat transfer coefficient leads to quicker cooling.
- A larger area over which to transfer the heat leads to quicker cooling.
- A greater temperature difference leads to quicker cooling.
$h$ depends strongly on the properties of the materials, which can be hard. We can make some estimates for the other quantities, though.
- The source is portable (which I'm taking to imply that it could fit in a large van, for instance), so I'll estimate that $l=3\text{ m}$ and $r=0.05\text{ m}$ (perhaps the latter is a bit large). This leads to $A=0.942\text{ m}^2$.
- Let's be extremely generous and say that the laser reaches temperatures of about $T_b=1,000\text{ K}$. This is really stretching it. At any rate, even if $T_f$ is close to $0\text{ K}$, the difference $\Delta T$ cannot be greater than $1,000\text{ K}$. Therefore, let $\Delta T\sim1,000\text{ K}$.
The best transfer coefficients I can find are actually water-to-water. However, air-to-steam can yield an $h=17\text{ W m}^{-2}\text{ K}^{-1}$ through copper (which has a melting point higher than the temperatures involved here). We therefore have $$\dot{Q}\sim\left(17\text{ W m}^{-2}\text{ K}^{-1}\right)\left(0.942\text{ m}^2\right)\left(1,000\text{ K}\right)\sim16,000\text{ Watts}$$ That's pretty nice . . . if the temperatures (and temperature difference, for that matter) don't cause the entire weapon to fall apart.
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