# How does the compressible Bernoulli's theorem reduce to the incompressible Bernoulli's theorem?

For incompressible steady flow, such as in water, Bernoulli's theorem states that $$\frac12 \rho v^2 + P = \text{constant}. \tag{1}$$ There's also an analogue of Bernoulli's theorem for compressible flow. For an ideal gas, it states $$\frac12 v^2 + \frac{\gamma}{\gamma - 1} \frac{P}{\rho} = \text{constant} \tag{2}$$ where I used the ideal gas law to eliminate $$T$$. Both of these results are well-known and commonly used in introductory textbooks. Those books often apply the incompressible result (1) to air flow, even though air is compressible; that's a reasonable thing to do because the density of air won't change much if the fluid flows subsonically, so the air is "effectively" incompressible.

Here's the puzzle: since the compressible result (2) is more general, we ought to be able to use it too. But if we naively multiply (2) by $$\rho$$, we don't recover (1) at all; instead a numeric prefactor is totally different. I suspect the reason is that in the subsonic regime we have $$v^2 \ll P/\rho$$, so that the second term in (2) is much larger than the first; then multiplying by $$\rho$$, which is merely almost constant, could cause a significant error. But I don't know how to make that precise.

What's the proper way to derive the common result (1) from the general result (2)?

• Pressure drop is the driving force for fluid flow. Before you assume that the density of a flowing compressible fluid doesn't change appreciably, it's a good idea to consider the pressure drop involved, and the ratio of pressure drop divided by upstream pressure. Commented Jun 29, 2023 at 16:02

You need to stipulate that the incompressible flow is isentropic, i.e. $$p=K\rho^\gamma$$ If you add this condition to the compressible form you will get the familiar result. The Euler equations can be used to show that both enthalpy and entropy are constant along a streamline in steady flow. The "analogue" that you quote is enthalpy.