# Steady state approximation for compressible fluid flow (manifolds and other pressure vessels)

I've been puzzling over a question / class of questions I'd like to solve via rough estimate before turning to CFD analysis.

Say I have a pressure vessel with one inlet and one outlet. The pressure vessel is at some initial pressure $P_0$. The outlet is at some fixed pressure $P_{out}$. The inlet has some mass flow rate, $\dot{m}$ (assume it's attached to an infinite upstream reservoir and a mass flow controller).

With a compressible fluid, a known mass flow rate, and a set volume in the vessel, what mechanism would you use to approximate the final steady state (volume-average, we'll say) pressure in the vessel? Assume that the outlet pressure is such that the flow will choke. Assume no heat transfer to the walls and that the gas is ideal.

One solution I have thought of is to perform a dumb iterative solution:

• Pick some time step, $dt$
• Add $\dot{m} * dt$ mass to the tank
• Compute the new pressure in the tank via the ideal gas law
• Use choked pipe flow equation to determine $\dot{m}$ at exit (knowing exit area)
• Subtract $\dot{m}_{exit} * dt$ mass from the tank
• Repeat until "convergence"

Possibly you could do some Bernoulli-streamline-style analysis, but I'd like to also apply this sort of approximation to a multi-inlet / multi-outlet problem. This feels like some form of the Hardy Cross method might be applicable as well. Interested in anybody's suggestions.

Graphs of what the above technique gets me for a 1 kg/s flow rate into a 5 m^3 volume vessel, initial conditions T = 290K & P = 1 Pa in the vessel:

Edit: Added a few figures from my "estimation."

• "The pressure vessel and the outlet are both at initial pressure P0." - do you mean inlet here ? as the outlet is Pout. – Solar Mike Jul 4 '17 at 18:18
• @SolarMike Should be just the pressure vessel. I'll edit it. – phyllis diller Jul 5 '17 at 6:05

We model the system simply for small mass flow rates compared to the size of the vessel simply:

At point 1, the pressure can be near infinite, or the density can be near infinite, or simply the capacity may be near infinite as far as we can tell. At this point, one thing is known for certain - no matter what the mass flow regulator needs, this system has enough energy to feel the need. However the reservoir supplies it, compared to the other points, $h_1 \approx \infty$. However, since the mass flow rate is fairly small compared to the remainder, each removal of mass will keep the temperature in the reservoir fairly constant. It's fair to state that $T_1 = T$, since $V_1 \approx 0$. After all, the tank is so massive compared to the tiny amount that leaves that the tank will warm or cool the fluid compared to the amount that has left so the outside will keep $h_1$ constant.

By the dynamics of the mass flow regulator, we know that $\dot m$ mass leaves the reservoir. We don't know much about point 2, but it's very clear that $h_2 \lll h_1$. If this wasn't the case, the reservoir in question wouldn't be able to supply a consistent energy supply and mass supply through the line in question. The pressure is still completely unknown, but $P_1 \ggg P_2 > P_3 \ggg P_4$, or the fluid wouldn't leave the system at all. However, across the regulator, the tiny mass moving of the fluid compared to the size of the piping calls into question the Stanton Number of the fluid. At a high Stanton number, it can be shown that a lot of heat moves from the pipe back into the fluid, reestablishing $T_2 = T$, and bringing $h_2$ up, but less than $h_1$ since $P_2 < P_1$

As established above, the temperature of $T_2$ should be brought back to room temperature, $T$. While the energy is unknown, we are attempting to find $P_3$. Since we enter a reservoir, $V_3 \approx 0$, and the fluid stagnates. The reservoir walls may be warming the fluid inside. Since the mass flow rate is small compared to the mass of the fluid inside the reservoir during steady state situation, it can be seen the reservoir should warm the fluid until it reaches $T$. Thus, the stagnation temperature, $T_{0_3} = T$. With this temperature known, the choked flow equation can be modified to reveal $P_{3_{final}}$, since the fluid, having stagnated:

$$\dot m = \dot m_{choked} = A_e(P_0 = P_3)\sqrt{\frac{k}{R(T_0 = T_3 = T)}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}$$

$$P_{3_{final}} = \frac{\dot m}{A_e\sqrt{\frac{k}{RT}}\left(\frac{2}{k+1}\right)^{\frac{k+1}{2(k-1)}}}$$

Indeed, with these assumptions, you've effectively turned the reservoir at point 3 into it's own mass flow regulator, at the price of keeping the pressure and temperature inside of the reservoir 3 constant.

• Hi Mark - Just wanted to say thanks. This is a great analysis. – phyllis diller Jul 6 '17 at 21:19

Interesting question, but well worth discussion of the confusion of compressible flow equations. At steady state, we will always have the mass flow rate be constant, even with choked flow. The flow rate at the exit has to be the same as the rate going in. This seems counter intuitive - choked flow states that there is a maximum flow rate which a nozzle can produce, and no more. However, choked flow states this is the case for variance of the outlet conditions, the flow rate will not increase. It states nothing about variances on the inlet conditions. Indeed, The main idea behind choked flow is that it depends on a singularity in the differential equation (which is derived on constant mass flow rate assumption):

$$\frac{\partial V}{V} = -\frac{\partial A}{A}\frac{1}{1-M^2}$$

If the Mach number is exactly 1, then the differential change in area must be 0, or the differential change in velocity would be infinite! This must be at the exit nozzle of your vessel. However, the scenario proposed is intentionally attempting to feed a vessel additional mass from an infinitely pressurized reservoir without a converging/diverging nozzle.

This situation could be resolved if more information was known about the inlet stream (such as temperature, or pressure). We couldn't use the reservoir properties as it is unknown how the mass flow meter affects the stream.

• Hi @Mark - couple comments. 1 - Unless I'm mistaken, choked flow only limits the speed of the flow to M = 1, not the actual flow rate. Upstream density can still increase. I am not assuming a constant flow rate; just that the downstream pressure is low enough to choke. I'm using the actual choked flow equation with the appropriate values of upstream pressure & density in the reservoir which do change in each step of my iterative solution to produce mdot_exit. – phyllis diller Jul 5 '17 at 23:18
• 2 - We could attempt to resolve the lack of upstream information by assuming the inlet has symmetric pressure / temperature conditions with the vessel. This is obviously not entirely true, but for "reasonable" flow rates it won't be far off. If we need to be more accurate we can back out the pressure upstream from a pipe flow equation, knowing the pressure drop that will be required to drive this rate. – phyllis diller Jul 5 '17 at 23:20
• On point 1: especially since you posted graphs, my point was mdot exit would always converge to mdot inlet. So you can always treat a vessel at steady state with continuous flow rate. Re attempt with different initial vessel pressures and you'll find this will always converge to the same final pressure with the same chokes flow rate – Mark Jul 6 '17 at 1:24
• @phyllisdiller on point 2, a real flow regulator come in several varieties. A true Coriolis meter connected to a ball valve would knock down the stagnation pressure by an unknown amount. A pressure regulator would knock down the actual pressure to a known amount, but will certainly modify the density. A metering pump can supply fluid at a fairly constant rate, but the pressure and density will definitely change. So the flow regulator definitely alters the properties of the flow. Temperature conditions aren't as likely to be a match as there will be stagnation when entering the vessel. – Mark Jul 6 '17 at 3:09
• Ok, yeah, we agree on point 1 - I thought you were saying something different. The goal of what I'm looking for is that final pressure (and how well it matches to a more realistic CFD sim / real world experiment) - and whether this model I've developed actually makes any physical sense. – phyllis diller Jul 6 '17 at 4:45