# Calculate flow rate of air through a pressurized hole

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?

• It has a strong dependency on the shape of the orifice. Check this out: Flow through an orifice – Mike Dunlavey Aug 15 '14 at 1:45
• The answer is right. Intuitively, if you want to double the velocity of the escaping air, you need to quadruple the pressure. The reason is that pressure (force) is momentum (of the air) per unit time, and if the velocity doubles, you're doubling the momentum of each parcel of air, and you're doubling the number of parcels of air. – Mike Dunlavey Aug 18 '14 at 15:11

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.

• Being Pa<Pg, then vg must be an imaginary number. – JoeCoolman Apr 12 '20 at 2:21

I think the answer at 'Aug 14 '14 at 22:36' is generally good, but I disagree in that the process can be regarded as isothermal.

For solving the problem at hand, the temperature inside the gas canister is required. This temperature will only be constant if the heat flow (through condution and convection) is fast enough to negate the temperature drop caused by the reduction of pressure in the vessel, and I believe this would not be a good assumption. I would rather assume adiabatic expansion inside (but not outside) the canister. This would mean that the canister becomes cold when it empties (I have used co2 powered bb-guns and seen ice forming on the co2 canister, which supports my assumption).

Assuming adiabatic expansion makes the calculation a bit more complicated. You need to enter the pressure to temperature relation

$P^{(1-\gamma)}T^\gamma = constant$

into the ideal gas law

$P = \frac{m}{M}RT$

to obtain the 'new' pressure as a function of initial temperature and pressure, and the 'new' gas mass in the canister.

• Actually an isothermal assumption is reasonable in the case of velocity of a leak, if the the hole in the tank is small and the tank isn't insulating so that it remains near ambient temperature throughout the time air is escaping. – Anachronist May 17 '17 at 22:41