Vacuum Exposure

How long will it take a spacecraft to decompress?


Is there a formula or rule-of-thumb for making a rough estimate of the rate of air loss in a space craft for a given size air leak?


The quick approximation is that the air will flow out of the hole at the speed of sound.

For a more detailed calculation, Professor Andrew Higgins of McGill University gives the following answer:

The air will leak through the hole at sonic velocity (Mach one at constriction of the leak).

So, the mass flow rate is:

dm/dt = rho V A (eqn. 1)

where rho is density, V is velocity, and A is the area of the hole. The velocity equals the speed of sound (sonic orifice), but this is slightly lower than the speed of sound in the spacecraft cabin due to expansion of gas as it flows through the hole. Density is lower also. So, it is more practical to express the mass flow rate in terms of stagnation conditions, i.e., the conditions in the cabin, which I will denote po and To:

dm/dt = A po Sqrt [(g/(R To)) (2/(g+1))^((g+1)/(g-1))) ] (eqn. 2)

here "g" is gamma, the ratio of specific heats (g = 1.4 for air) and R is the gas constant (R = 287 J/kg-K for air). You can find this derived in any compressible fluid dynamics textbook (or any fluids book with a chapter on compressible flow).

For air, this simplifies to:

dm/dt = 0.04042 A*po/Sqrt[To] (eqn. 3, "Fliegner's formula")

if you stick to MKS units (using Pascals for pressure and K for temperature), this will give you the mass flow rate of air leak in kg/s.

So far, we have assumed that the spacecraft remains at the same po, To. Of course, as the leak progresses, the pressure in the spacecraft begins to drop, and this affects the mass flow rate through the leak. Thus, dm/dt is no longer constant, and we have to integrate the above differential equation coupled to the decrease in po and To as the spacecraft leaks. You can find the details of this in Saad's Compressible Fluid Flow (2nd Ed., pp. 103-106). The answer is that, to leak from an initial pressure of pi to a final pressure of pf, the time required is:

t = 0.43 V/A [(pi/pf)^0.143 - 1]/(Sqrt[Ti]) (eqn. 4)

Again using MKS units, where V is the volume of the spacecraft (Ti = initial temperature), this gives you the time "t" to leak down to pf in seconds (assuming the cabin gas is air).

This assumed that the blow-down was isentropic. In practice, any blow-down that will last tens of seconds to minutes, the process in the spacecraft is more likely to be isothermal: mass of spacecraft has huge thermal capacity compared to the (decreasing) mass of gas inside and will keep the gas warm as it expands. With the assumption of isothermal blow-down, the time required becomes:

t = 0.086 (V/A) Ln[pi/pf]/(Sqrt[T]) (eqn. 5)

where T is the (constant) spacecraft temperature.

If the atmosphere inside the spacecraft starts out at room temperature, 293K, this simplifies to:

t = 0.005 (V/A) Ln[pi/pf] (eqn. 6)

A spacecraft with a volume V=10 m^3 is initially pressurized with air at 300 K. It has a 1 cm x 1 cm hole. V/A is (10/10^-4)= 10^5, so the time it takes the pressure to drop from 1 atm to 0.5 atm (pi/pf = 2) is (from equation 5):

t = 0.086 *(10^5) * Ln[2]/(Sqrt[300]) = 344.2 s

or about six minutes.

Andrew J. Higgins
Mechanical Engineering Deptartment
Assistant Professor, Shock Wave Physics Group
McGill University, Montreal, Quebec CANADA


  1. Demetriades, S.T., "On the Decompression of a Punctured Pressurized Cabin in Vacuum Flight," Jet Propulsion, January-February, 1954, pp. 35-36.
  2. Saad, M., Compressible Fluid Flow, 2nd Ed., Pearson Education, 1998.

This document is not a work of the U.S. government, and any opinions expressed in it are the views of the author, and not NASA or the U.S. government.


Page by Geoffrey A. Landis, copyright 2003