Tank draining with piston

Question

I am trying to solve a problem which is close to the tank draining (as here Fuel tank draining) but with a piston a the top of the tank which means $p_mathrm A$ different of $p_mathrm B$.

I search the velocity $V_mathrm A$ at which I have to push the piston to have an out flow rate $Q$ (which is imposed).

$$begin{align}
p_mathrm A+rho g z_mathrm A+frac12rho V_mathrm A^2&=p_mathrm B+rho g z_mathrm B+frac12rho V_mathrm B^2
z_mathrm A-z_mathrm B&=h
p_mathrm B&=0
V_mathrm AS_mathrm A&=V_mathrm BS_mathrm B=Q 
p_mathrm A+rho gh+frac12rholeft(V_mathrm A^2-frac{Q^2}{S_mathrm B^2}right)&=0
end{align}$$
But I have two unknows $V_mathrm A$ and $P_mathrm A$.
Should I use something else to link the pressure to the velocity?

Arietul · Answer

Your equations are correct. 
It's just that you have a degree of freedom. You can obtain a relationship between the force you exert and the velocity you obtain by modeling the piston itself:
$M_pfrac{partial^2 V_a}{partial t^2}=S_ap_a$, 
where $M_p$ is the velocity of the pistol and assuming the friction forces are negligible.

Tank draining with piston

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