The specific example I have in mind is a car tire with a small leak in it. As the pressure increases, does the outflow of air increase linearly, i.e. $v\propto P$, or does it have some more interesting behavior?
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asked Jan 28, 2015 at 22:00
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2 Answers
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I think the best (and simplest) way to describe something like this is Bernoulli's equation.
$$P+\rho gh+\frac12 \rho v^2=constant$$
To use this, we're looking only at instantaneous velocity, because as the air leaks, the pressure will go down. We also should assume that the "valve" is really more of a small hole than anything that fluctuates too much with pressure variations, because that complicates it a bit more.
The constant in the Bernoulli equation gets applied to any point in a continuous stream. So what we want to do is pick two points, one on either side of the hole, and relate those two points using the Bernoulli equation.
What we'll get will look something like this.
$$P_{tire}+\rho gh_0+\frac12 \rho v_0^2 =P_{atm} + \rho gh_1 +\frac12 \rho v_1^2$$
In this situation, we'll say that any vertical movement of the air is small enough to neglect. Also, the velocity of the air inside the tire is negligible as well, if not in practice, than for the case of determining the relationship between pressure and velocity. Lastly, there's an important distinction between absolute pressure (which is in the equations above) and gauge pressure (which would be what we measure with a tire pressure gauge [go figure]). Gauge pressure is defined as $P_{gage}=P_{abs}-P_{atm}$. Bringing that all together we get the following.
$$P_{tire,gauge}=\frac12 \rho v_1^2$$
The other couple of important points to make with this is that it's valid for inviscid flow (no friction) with constant velocity. The first assumption is pretty valid, if the pressure differentials are significant, the second may not be, especially since that starts to drive up the fluid velocity, and constant density goes out the window when we get to compressible flows ($Ma>0.3$). Again though, for the simple case of examining the nature of the relationship, this evaluation should be fine.
answered Jan 29, 2015 at 3:55
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Yep, the answer is a bit more interesting.
The mass flow rate ($\dot m$) will vary as $C_d A \sqrt{2 \Delta P}$ (discharge coefficient, cross sectional area, and change in pressure across the valve). Barring compression of the fluid, velocity will behave the same way.
The devil of the thing is in $C_d$. This is pretty much something that you have, measure experimentally or (try to) model numerically with computational fluid dynamics. That little number captures all of the non-ideal aspects of the flow involved (viscosity, turbulence). It's always less than one (nothing is ever ideal), so a simple prediction from Bernoulli will always over-predict (just a question of how much).
In practice, $C_d$ will change as you adjust the valve (as will area). So the manufacturer will usually just give you a few values or a curve for flow as a function of valve position and pressure or a total flow coefficient $C_V$ as a function of valve position ($\sqrt{\Delta P }$ dependence assumed).
As mentioned in the other answer, this all goes funny if the flow becomes compressible.
