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  3. The best answers are voted up and rise to the top heat a piece of steel so its glowing yellow (1100 C)? Assuming you had a cloudless day at a latitude of, say, San Francisco...

Basically I'm wondering if it is possible/feasible to be able to do basic metal working without a traditional forge, just using the power of the sun to heat the metal. So the diameter of the heated spot would have to be about 6" in order to heat a large enough area of the metal to work it...

I always thought you would need several huge pieces of equipment to do this, but just thought I'd ask if anyone here knew how to figure out it roughly...


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up vote 1 down vote accepted

For steel, the specific heat would be $c_p=0.5 kJ/kg K$, with a density of $ \rho=7000 kg/m^3$. Suppose you want to increase the temperature bij say $\Delta T=1100K$ of a piece of size $V=(15cm)^3$

Then you would need a total energy of.

$$E=\rho c_p V \Delta T$$ Which gives you typically $E=10^7 J$

Now, the power of the sun on a bright day, would be of the order of $p=10^3 W/m^2$.

Assuming that

  • all the energy input is converted into heat
  • the mirror is perfectly aligned
  • no heat is lost during heating,
  • no melting, e.g. no latent heat

and your mirror had diameter $D$ and you let the process run for a time $t$, then

$$E=p \frac{\pi}{4}D^2 t $$

Then you will get, in approximation

$$D = \sqrt{\frac{E}{pt}}$$

So, suppose you are willing to wait for ten minutes, then the mirror diameter would be $D\approx 4m$. Considering we assumed an ideal system, this is only an order of magnitude assumption.

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One should also take into consideration whether the supplied power is sufficient to heat up the steel more while energy is lost due to thermal radiation. Assuming that it is a black body, we have $P=\sigma T^4A$ where $T=1500\textrm{ K}$, $\sigma=5.67\times 10^{−8}\textrm{ Wm}^{-2}\textrm{K}^{-4}$ and $A\approx 0.1\textrm{ m}^2$ which gives us $P=2.9\times10^4\textrm{ W}$. In other words, in order to keep it at such a temperature, we need $D \geq 6\textrm{ m}$. This is obviously a rather close call, as this power is absolutely required, while the power supplied is more of an upper bond. – Claudius Nov 20 '12 at 21:16
@Claudius, I assumed no heat losses, so the $4m$ is probably an absolute lower bound. But including it would demand solving a differential equation in time, as you reach a asymptotic temperature, didn't think it was necessary for an order of magnitude estimation. Only time $t$ is rather arbitrarily chosen in my estimation here. – Bernhard Nov 21 '12 at 7:01
If you still want to get the time $t$, you indeed have to solve a differential equation. However, if you ignore $t$ (i. e. $t \to \infty$) you can get an upper bound on the temperature reachable with a specific diametre (as given in my calculation above). Not including heat loss would effectively mean that if one waited long enough, any mirror would do. – Claudius Nov 21 '12 at 10:41
@Claudius You are absolutely right. – Bernhard Nov 21 '12 at 13:51

protected by Qmechanic Jan 7 '13 at 8:16

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