# Can air be considered incompressible as long flow velocities are less than 100 m/s (Ma = 0.3)?

Multiple sources state that steady air flow (in open systems) can be considered incompressible at velocities less than 100 m/s (Ma = 0.3). Example: Deborah A. Kaminski, Michael K. Jensen, "Introduction to Thermal and Fluids Engineering", ISBN: 0-471-26873-9, 2005

Do you have a source which gives more detailled information on this?

Specifically I am interested in the following:

1. Any fluid is compressible. But what is the extent of compression under above stated conditions?

2. How is the abovementioned statement justified? Is there a mathematical derivation? Is experimental data available?

• As a general statement this is simply wrong. One can compress air in a cylinder at any velocity. The question then becomes for what boundary conditions of the flow is it a valid statement? I am not sure there is a simple answer to that, either. – CuriousOne May 25 '16 at 8:02
• Sure, when compressing air in a cylinder the statement is definetly wrong. The question refers to an open system with air inlet and air outlet. I have added this to the question. – some_weired_user May 25 '16 at 8:06
• Even then it's not true. Imagine what happens in an very long pipe in which you keep pushing gas in until the velocity of the flow is reached. – CuriousOne May 25 '16 at 9:00
• Edited again: Steady flow is meant, density is compared against ambient air density – some_weired_user May 25 '16 at 9:15
• Even in steady flow it's not true. A trivial counterexample is Hagen Poiseuille. This sounds like one of those "I know it when I see it" rules, one can find trivial counterexamples but it's still OK in many practical applications. I just wouldn't bet my career or somebody's life on it while designing a technical system with gas flowing in it. – CuriousOne May 25 '16 at 9:19

The term "compressible flow" is rather misleading, but unfortunately, it is what the fluid-dynamics folks have chosen to use. "Compressible flow" refers to gas flow where the temperature of the gas is significantly affected by the conversion of pressure differences to kinetic energy. Bernoulli's principle states that the work done by a pressure difference is equal to the gain in kinetic energy: $$\frac12 (\rho_1 v_1^2-\rho_2 v_2^2) = p_2-p_1.$$ However, if $p_1$ and $p_2$ differ, then the gas has to expand. Not only does that mean that the densities differ ($\rho_1$ versus $\rho_2$), but the expansion without further energy exchange with the pipe walls will also lower the temperature of the gas, which will tend to increase the density again. Therefore, to describe the gas flow correctly, you need to keep track of both temperature and pressure over space and do the book keeping with the specific heat. This tends to be rather tedious. The extent to which all this plays a role depends on the square of the velocity, the specific heats $C_p$ and $C_V$, and the density. It happens that the Mach number squared scales in the same way. Roughly, at $\mathit{Ma}^2<0.1$, the impact of these compressible effect is smallish; hence, the rule of thumb, $\mathit{Ma}<0.3$. So, if someone says that the gas flow should be treated as compressible flow, it means that you need to work much harder to reach a reliable answer. Incompressible flow means that you're allowed to make certain approximations, even though the gas will undergo some volume change.

The cutoff is a bit arbitrary; if you need high precision, you might wish to set the cutoff a bit lower. In "incompressible" gas flow, you will often still need to account for density variations, but at least, you can decouple them from the temperature variations.

"Smallish" depends on the specific circumstances, but count on less than 10% error in density, flow velocity, and mass flow rate. Achieving better than 10% accuracy in nontrivial flow systems is hard anyway: heat exchange at the pipe walls, turbulence, the exact value of the viscosity, viscous dissipation, and velocity gradients perpendicular to the dominant flow direction will all affect the outcome.

• How small is smallish? Do you have any literature advice? – some_weired_user May 25 '16 at 9:18
• Answer updated. – Han-Kwang Nienhuys May 25 '16 at 10:55

As far as I catch the question (without going digged in math), pressure doesn't go instantly from a point to a point, but has certain speed, that is sound speed. No matter if your system is opened or closed, the air is the same. If you compress air in cylinder slowly, then the air well compressible, but if you try to move the piston quickly, let's say with Ma just less than 1, then you face much resistance because pressure wave doesn't have enough time to go away. It means in that case you are making too high pressure in a thin layer close to the piston while rest of cylinder still has low pressure. So I think there is some confusion with the matter. You can consider the air as well compressible at low speed, but more hard compressible at higher speed, and incompressible at all (behave like water) at Ma > 1. As is already said, there is no clear criteria for transition from "compressible" to "incompressible" speed, apparently the sources suggest Ma=0.3 as rule of thumb for practical engineering problems. Below 0.3 you can simply consider air as well compressible, while starting from 0.3 you can't.

• You have understood "compressible" backwards. The choice of using this confusing term the way they do in the fluid-mechanics community is rather unfortunate. – Han-Kwang Nienhuys May 25 '16 at 10:58
• Thank you, @Han-Kwang Nienhuys. I got it and will try to learn (if I can). Still hard logic to understand. – dmafa May 25 '16 at 11:20