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Well, it's a simple steam engine, The piston head strike the front wall (Yellow X); will it move the spaceship forward? enter image description here

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  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/291447/2451 $\endgroup$ – Qmechanic Nov 8 '16 at 14:21
  • $\begingroup$ How can banging a wall move you forward in free space? What will happen the rest of the ship. $\endgroup$ – user108787 Nov 8 '16 at 14:32
  • $\begingroup$ If you kick a box in space it will move forward, it doesn’t meter if you kick the backside or the inner front side. $\endgroup$ – Issack K. Nov 8 '16 at 14:55
  • $\begingroup$ You are kicking the box while at the same time connected to the box, which is an internal force. This is a closed system where momentum is conserved. $\endgroup$ – Lawrence B. Crowell Nov 8 '16 at 15:00
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No, hitting a wall of the spaceship with any object will just be a waste of your steam. Momentum is conserved, so if you start with zero forward momentum, you will end up with zero momentum. For every action, there is a equal reaction, in the opposite direction.

You require an unbalanced force, that is, you need to expel something from the ship entirely. A good reference is Newton's Laws

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Before I opened this I thought the system was different. I was thinking about a steam version of a water rocket. As a child I had those toy bulbous rockets that you filled with water and pumped air into with a little hand pump. When released the air pressure in the closed container pushed the water out the back nozzle and the whole system then lifts into the air.

Your system will not work. It will not work for much the same reason the flying machine 46 seconds into this video did not work. Momentum conservation conservation defeats you.

One could have a boiler that is heated to very high temperatures, and the steam produced in there is released out a valve or nozzle. One could use the rocket equation, which would later be modified for a varying pressure. This is momentum conservation given the rocket equation. For a mass $m$ moving at velocity $v$ in the x-direction that at a momentum time $\delta t$ splits off a units of mass $\delta m$ with velocity $v~-~V$, for $V$ the magnitude of the reaction mass out the back. This results in the equation $$ mv~=~\delta m(v~-~V)~+~(m~-~\delta m)(v~+~\delta v). $$ Ignoring he $\delta m\delta v$ term you get $$ V\delta m~=~m\delta v, $$ which after rearranging and integrating gives the final velocity $v~=~V~ln(m_i/m_f)$ for $m_i$ the initial mass and $m_f$ the final mass. If the velocity of the reaction mass out of the nozzle is $5km/s$, which corresponds to the specific impulse $s~=~v/g$, $g~=~10m/s^2$ (one Earth gravity) $s~=~500s$ and $m_i~=~10m_f$, which corresponds to a final velocity $v~=~11.5km/s$. This is about escape velocity, and this is an idealized case of "one stage to orbit."

For this to be a steam system the pressure inside of the boiler would have to be enormous. We can estimate what this would be using Bernoulli's principle. The energy $v^2 = 2p/\rho$. Assume the boiler is $20m$ in height and $5m$ in diameter. The volume is then $785m^3$ in volume. Assume the initial mass is $5\times 10^4kg$ the density is given by $90\%$ of this mass and so the density of the steam is $57kg/m^3$. The pressure is then $$ p~=~.5\times (5\times 10^3m/s)^2\times 57kg/m^3~=~7.16\times 10^8Pa. $$ This is an enormous pressure! The material of the boiler would have to be exotic in some way.

One could complicate the rocket equation as well. The pressure is going to decline as the steam is jettisoned out the back. One might use the average pressure, double the initial pressure to get an average nozzle velocity and so forth. It is however clear why this approach to rocketry was never tried.

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