I am trying to develop a simple tool to size a pressurization system for a spacecraft using either a blowdown pressurization system (BD) or a pressure regulated (PR) pressurization system. Though I find the results of this process rather strange.

I see that, once I set the initial and final pressure of the pressurizing gas in both of the systems, the mass of gas is the same, even though the work done by the gas in the blowdown system is higher than the one of the pressure regulated system.

I assumed that the initial pressure of the gas for the blowdown system equals the pressure of the gas in the gas tank for the pressure regulated system, and that in the latter the pressure at which the gas expands in the propellant tank is the final pressure (the pressure regulator is set at the final pressure).

The architecture considered for both of the system is shown here. enter image description here

The equations that where used for the sizing are the following.



$p_{g,i}$ is the initial pressure of the gas.

$p_{g,f}$ is the final pressure of the gas.

$V_0$ is the volume of the ullage for the blowdown system or the volume of the gas tank in the pressure regulated system.

$V_p$ is the volume of propellant to be displaced.

$k$ is the polytropic coefficient of the expansion.


$m_{gass} = \dfrac{p_{g,i}*V_0}{R_{g}*T}$

I find it strange that, using this assumptions, the work exerted by the gas in the BD system is higher than the one in the PR system. As the pressure at which the propellant is expelled from the tank in the BD system goes from $p_{g,i}$ to $p_{g,f}$ while in the PR system the pressure is always $p_{g,f}$. All of this while the mas of gas is the same.

Is there a faulty assumption in these calculations? How would you explain this difference? Is it possible to attribute the difference in work done by the gas to the pressure regulating valve in the PR system?

I will greatly appreciate any contributions! :)

  • $\begingroup$ It makes sense to me that the work would be higher in the blowdown system. In the end you displace the same amount of liquid, but the enthalpy of the propellant in the chamber will be higher. The excess work goes with the propellant. $\endgroup$
    – A McKelvy
    Jan 11 at 15:24
  • $\begingroup$ Is the timescale you are considering for both cases a) the same and b) compatible with your k value. A short expansion in a few minutes would lead to a temperature drop whereas over a year its reasonable to expect it to be isothermal. $\endgroup$
    – Puffin
    Jan 11 at 22:22

1 Answer 1


In the BD system, the gas has access to the thermal mass of the propellant (isothermal), so at the end of expansion it would have a higher temperature than in the PR system where gas expanded in the regulator (adiabatic). Different value of k

  • $\begingroup$ Thank you for your observation, I think it's true what you say, but what happens if we assume that, instead of being in contact with the propellant, the gas in the BD system is in contact with a membrane that doesn't allow thermal exchanges (between gas and propellant)? In this case the "problem" should persist, even if than both expansions are adiabatic, what do you think? $\endgroup$ Dec 12, 2023 at 10:54
  • $\begingroup$ I think this is out of my pay grade. $\endgroup$
    – Woody
    Dec 12, 2023 at 19:21
  • 1
    $\begingroup$ @MarioPastore Unless the membrane is made of a highly insulating material, there is probably going to be a larger amount of heat transfer across a large, thin membrane, versus the pressure-regulator system where the small regulator gets chilled down. $\endgroup$
    – ikrase
    Dec 15, 2023 at 10:08
  • $\begingroup$ @ikrase yes, though I think I'd go further and say I'd expect the pressurant temperature to be dominated by the satellite environment around it, as will the propellant. Mario Pastore: If the thought experiment is actually that the gas in the BD system doesn't allow meaningful equalisation of temperature whilst the tank is emptied then what timescale are we talking about, a few minutes or a mission of >= 1 year? $\endgroup$
    – Puffin
    Jan 11 at 22:10
  • 1
    $\begingroup$ @Puffin,@ikrase, both of you are sharing interesting insights, but the "anomaly" that I noticed ( the fact that even if the work exerted by the propellant in the BD and PR system is different the equation lead to the same size of the tank) is not ideally related to the temperature. So let's suppose that in both cases the initial temperature of the gas is equal to the temperature of the spacecraft and the process is adiabatic: there is no heat exchange whatsoever with the environment. Is somebody able to explain why the work done by the BD system is higher than the one of the PR? $\endgroup$ Jan 12 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.