Imagine you have a pressure chamber which is divided by a removable wall in two equal parts.

Scenario A) Air pressure on the left side is 1 bar, on the right side it is 2 bar.
Scenario B) Air pressure on the left side is 10 bar, on the right side it is 11 bar.

The absolute pressure difference between the two sides is the same (1 bar in both scenarios). The relative pressure difference is greater in scenario A than in scenario B.

When the wall in the middle is removed air flows from the right side to the left side. Now, is the flow speed of the air the same for both scenarios or is it greater in scenario A? Intuitively, I assume the flow speed to be greater in scenario A (because of a greater relative pressure difference), but I can't back it up with a solid argument.

What is happening?

  • 2
    $\begingroup$ As long as the mean free path is much less than the distance to flow you are just fine. Now, consider 0 bar on the left, and 1 bar on the right... $\endgroup$
    – Jon Custer
    Feb 1, 2016 at 14:12

1 Answer 1


Well, let's look at the equations for a flow initially at rest:

$$ \frac{\partial \rho u}{\partial t} = -\frac{\partial p}{\partial x}$$

So, if we assume density is constant in the flow, you get:

$$ \frac{\partial u}{\partial t} = -\frac{1}{\rho}\frac{\partial p}{\partial x}$$

and so you can see that in both your cases, the gradient in pressure is the same size regardless of the magnitude of pressure. So, if your case B is hotter such that the two densities between case A and B are the same, the acceleration should be the same.

On the other hand, if your temperatures are the same, then your acceleration will be higher in case A because density is lower.

Since your system is closed and transient, it doesn't make sense to look at the total flow velocity because it is always changing. But the acceleration will change depending on how you changed your pressures.

Another way to look at it. If we assume isentropic flow, the total pressure is conserved. This means that:

$$ P_0 = P_s + \frac{1}{2}\rho u^2 \rightarrow \Delta P = \frac{1}{2}\rho u^2$$

where $\Delta P = P_0 - P_s$ somewhere along the streamline. So again we can see that the velocity depends on the change in pressure and the density and we're back at the same argument we had above.

  • $\begingroup$ So when the pressure gradient is due to a density gradient, then the initial flow speed is dependent on the relative pressure difference (or relative density difference). But when the pressure gradient is due to a difference in temperature, then the initial flow speed is only dependent on the absolute pressure difference. Do I have this right? $\endgroup$
    – Chris
    Feb 1, 2016 at 22:38
  • $\begingroup$ @Chris Not initial flow speed, but acceleration. It boils down to $F = ma$ -- if your pressure is different because you crammed more mass into one side than the other, then it makes sense it will accelerate slower than if the pressure is increased because it is hotter (but density/mass is otherwise the same). We're only looking at initial acceleration here at the moment the fluid feels the gradient kick in. $\endgroup$
    – tpg2114
    Feb 2, 2016 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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