I was wondering about this:
If there is a pressurized container, like a tank of compressed air at some pressure that is greater than the ambient air pressure, and this tank of air has a hole in it, what is the velocity of the escaping air through the hole? Is there a formula for this?
Best Answer
If you neglect viscosity, Bernoulli's equation (just Navier-Stokes without frictional or stress terms) will get you into the ballpark:
$$P_g + \frac{1}{2}\rho_g v_g^2 = P_a$$
Where the $g$ subscripts pertain to the gas and the $a$ subscript to the ambient. The gas density $\rho_g \equiv M / V$ is the ratio of the mass of gas (M) in the tank to the volume of the tank. If the tank is a rigid container (like a propane tank) then the volume of the gas is constant and the pressure will vary with the mass flow and the temperature. If you assume the tank remains at a constant ambient temperature, the pressure will only vary with the mass flow rate (isothermal expansion) and you can obtain that from the ideal gas law:
$$P_g = \frac{m}{M} RT$$
where $m$ is the molecular mass of the gas in question, $T$ the temperature, $R$ the gas constant, and $M$ the total mass of the gas remaining in the tank. This is a function of time because mass is leaving the tank. The rate at which mass leaves is a function of the exit velocity (it depends on the volumetric flow rate, which is a product of the exit orifice size and the exit velocity). So you can solve for $M(v_g)$ and substitute in the above equations and solve for $v_g$ self-consistently. Note this approach also ignores any pipes that might be attached to the orifice. For that, you'd need to calculate the volumetric flow rate using Poiseuille's equation.