Steam Pumpkin Launcher

I am doing a pumpkin launcher project and my group is torn between using an air cannon, trebuchet, centrifugal arm, and a steam cannon. I want to take water fill the tank x % with water and heat it to above 100 °C.

When I release the pressure, if all of the water is above 100 °C, shouldn't it all vaporize? I need to know if I am right and what math would back me because I can't for the life of me remember/find the equations to do so.

• – user7951
Commented Oct 14, 2016 at 19:47
• This somehow made me recall a certain folk song. "They found him in the wreck, with his hand on the throttle, and scalded to death by the steam." I mean, be careful. No, all of it would not vaporize, but a sizeable portion would. Commented Oct 14, 2016 at 20:58
• You're essentially suggesting making an autoclave/pressure cooker and then poking a hole in it. Take a look at OSHA reports to get an idea of what could happen--these failures happened with industrial equipment that costs thousands of dollars that are designed to remain closed during operation. Your idea is certainly possible, but if you're asking internet randoms for help, maybe think about if you have the knowledge and skills necessary to pull this off safely. Commented Oct 14, 2016 at 21:16
• If you mean superheat as in make it extremely hot gas (which I don't think you do), there is a practical problem of containing and then releasing all that pressure. If you're superheating the water in terms of it not boiling while it is above boiling point, yet the pressure is approximately atmospheric, there's practical limits to that. The further from boiling the water is, the less stable the phase is. If you want to increase it enough to propel medium sized objects, I'd say you'd probably be safer with home made nitroglyercin...
– user7652
Commented Oct 14, 2016 at 21:38
• @chipbuster has the right idea, but I'd be more dogmatic. Your question is so basic that it is obvious that you do not have the necessary knowledge and skills to safely use superheated steam.
– MaxW
Commented Oct 14, 2016 at 21:58

1 Answer

If you heat water to a temperature above $T=100\ \mathrm{^\circ C}$, you need an increased pressure to keep it liquid. This is called superheated water.

When you release the pressure, the pressure drops to the atmospheric pressure (typically $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$) and the hot liquid water immediately starts to boil. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure.

At this temperature and pressure, the specific enthalpy of steam is $h_\text{steam}=2675.5\ \mathrm{kJ\ kg^{-1}}$, and the specific enthalpy of liquid water is $h_\text{liquid}=419.06\ \mathrm{kJ\ kg^{-1}}$. Thus, the enthalpy of vaporization of water is very high: $\Delta_\text{vap}h=h_\text{steam}-h_\text{liquid}=2256.5\ \mathrm{kJ\ kg^{-1}}$. Therefore, the water does not completely flash to steam. The vaporized fraction $x$ may be estimated using the following enthalpy balance:

\begin{align} h_0&=x\cdot h_\text{steam}+(1-x) \cdot h_\text{liquid}\\[6pt] x&= \frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}} \end{align}

where $h_0$ is the initial specific enthalpy of the superheated water. This value depends on the initial temperature and pressure of the superheated water and can be obtained from so-called steam tables. (If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha.)

Example

Initial temperature: $T_0=120\ \mathrm{^\circ C}$
Initial pressure: $p_0=2\ \mathrm{bar}=200\,000\ \mathrm{Pa}$
Initial specific enthalpy: $h_0=503.81\ \mathrm{kJ\ kg^{-1}}$

New pressure: $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$
New temperature: $T=99.974\ \mathrm{^\circ C}$
New specific enthalpy of steam: $h_\text{steam}=2675.5\ \mathrm{kJ\ kg^{-1}}$
New specific enthalpy of liquid water: $h_\text{liquid}=419.06\ \mathrm{kJ\ kg^{-1}}$

Evaporated fraction: \begin{align} x&= \frac{h_0-h_\text{liquid}}{h_\text{steam}-h_\text{liquid}}\\[6pt] &= \frac{503.81\ \mathrm{kJ\ kg^{-1}}-419.06\ \mathrm{kJ\ kg^{-1}}}{2675.5\ \mathrm{kJ\ kg^{-1}}-419.06\ \mathrm{kJ\ kg^{-1}}}\\[6pt] &=0.038\\[6pt] &=3.8\ \% \end{align}

This means that approximately $3.8\ \%$ of the superheated water flashes to steam when the pressure is released. (This may not sound like much; however, it means that $1\ \mathrm{kg}$ of water would generate about $63\ \mathrm l$ of steam.)