# Difference between compressible bernoulli and compresible momentum equation

I would like to know why the compressible bernoulli equation and the momentum equation are different.

The compressible bernoulli equation is $$\frac{1}{2} u_1^2 + \frac{k}{(k-1)} \frac{p_1}{\rho_1} = \frac{1}{2} u_2^2 + \frac{k}{(k-1)} \frac{p_2}{\rho_2}$$

while the compressible momentum equation is $$p_1 + u_1^2 \rho_1 = p_2 + u_2^2 \rho_2$$

• I added math formatting. Please chack that I did not alter your meaning. Apr 21, 2021 at 14:37
• Thank your very much, it is correct. How do I do what you did? Apr 22, 2021 at 8:52
• @ Friendly You mean the edit? Just click the "edit" option and you will see the raw text of how the mathjax works. Apr 22, 2021 at 11:50
• Ah I see it now and got it. Thank you very much for your kind help! Apr 22, 2021 at 12:20
• No problem...:) But I don't undertstand you question. Why would you expect them to be the same? One is an expression of energy conservation and the other for momentum conservation. Apr 22, 2021 at 12:24

Here is a discussion of the compatibility of the energy and momentum equations for flow through a pipe of varying cross section. It's a bit more complicated that the equations cited in the question.

Consider the flow of a compressible fluid through a pipe of slowly changing area $$A(x)$$. Here slowly varying means that we can ignore all transverse velocities. The time rate of change of the $$x$$ component of momentum of the the fluid between two surfaces at $$x_1(t)$$ and $$x_2(t)$$ that bounding a moving volume of the fluid is $$\dot P=\frac{d}{dt}\int_{x_1(t)}^{x_2(t)} \rho v A dx.$$ The $$x$$ component of the total force on the same body of fluid is
$$F=\left.pA\right|_{x_1} - \left. pA\right|_{x_2} + \int_{x_1}^{x_2} p \frac{dA}{dx}dx,$$ where the integral is the force exerted on the fluid due to the longitudinal component of the gradient of pressure $$p(x)$$ by the wall. (Because the pipe is widening, the unit normal to the wall has a non-zero component in the x direction.) We can rewrite the total force as $$F=\int_{x_1}^{x_2} \left(- \frac{\partial}{\partial x}(pA) + p \frac{dA}{dx}\right)dx= \int_{x_1}^{x_2} \left(-A \frac{\partial p}{\partial x} \right)dx.$$ We can similarly write the momentum change as $$\dot P = \left. \rho v^2 A\right|_{x_2}- \left. \rho v^2 A\right|_{x_1}+ \int_{x_1}^{x_2} \frac{\partial}{\partial t}(\rho v A) dx\nonumber\\ = \int_{x_1}^{x_2} \left(A \frac{\partial \rho v}{\partial t}+ \frac{\partial} {\partial x}(\rho v^2 A)\right)dx.\nonumber$$ As $$\dot P=F$$ and $$x_1$$ and $$x_2$$ are arbitrary, we can read off the local momentum conservation law $$A\frac{\partial \rho v}{\partial t}+\frac{\partial}{\partial x}(\rho v^2 A)=- A \frac{\partial p}{\partial x}.$$
Now we also have mass conservation, so $$0=\frac{d}{dt}\int_{x_1(t)}^{x_2(t)} \rho A dx= \left. \rho v A\right|_{x_2} - \left. \rho v A\right|_{x_1} + \int_{x_1}^{x_2} A \frac{\partial \rho}{\partial t}dx\\ =\int_{x_1(t)}^{x_2(t)} \left(A\frac{\partial \rho}{\partial t}+ \frac{\partial }{\partial x}(\rho v A)\right) dx.$$ Again, as $$x_1$$ and $$x_2$$ are arbitrary, we deduce that $$A\frac{\partial \rho}{\partial t}+ \frac{\partial }{\partial x}(\rho v A)=0.$$ When we subtract $$v$$ times the mass conservation equation from the momentum conservation equation the derivatives of $$A$$ and $$\rho$$ cancel, and obtain a pipe version of Euler's equation $$A(x)\left\{\rho\left( \frac{\partial v}{\partial t}+ v \frac{\partial v}{\partial x} \right)+\frac{\partial p}{\partial x}\right\}=0.$$ If we write $$v= \partial_x \phi(x,t)$$ and observe that
$$- \frac 1 \rho\frac{\partial p}{\partial x}= -\frac{\partial h}{\partial x}$$ where $$h$$ is the specific enthalpy, i.e $$U+PV$$ per unit mass, we can rewite $$\frac{\partial v}{\partial t}+ v \frac{\partial v}{\partial x}= -\frac 1\rho \frac{\partial p}{\partial x}$$ as $$\frac{\partial}{\partial x}\left( \frac{\partial \phi}{\partial t} +\frac 12 v^2 +h \right)=0$$ The statement that $$\frac{\partial \phi}{\partial t} +\frac 12 v^2 +h$$ is independent of $$x$$ is Bernoull's theorem for compressible flow. It is a generalization of the statement that enthalpy is conserved for throttling processes so as to include the kinetic energy of the fluid.

The fun bit we can extract from these equations ocurs for steady flow where both $$\partial_t v$$ and $$\partial_t \rho$$ are zero. The mass conservation equation then becomes $$\partial_x(\rho vA)=0$$, or equivalently $$\frac 1 \rho \frac{\partial \rho}{\partial x}+ \frac 1 v \frac{\partial v}{\partial x}+\frac 1 A \frac{\partial A}{\partial x}=0.\quad (\star)$$ The square of the local speed of sound is
$$c^2 =\frac{\partial p}{\partial \rho}$$ so the time independent Euler's equation can be rewritten as
$$\rho v \frac{\partial v}{\partial x}=- \frac{\partial p}{\partial x}=- c^2 \frac{\partial \rho}{\partial x}\quad \Rightarrow \quad \frac 1 \rho \frac{\partial \rho}{\partial x}=- \frac{v}{c^2}\frac{\partial v}{\partial x}.$$ As a consequence $$(\star)$$ becomes
$$\left( 1- \frac{v^2}{c^2}\right) \frac 1 v \frac{\partial v}{\partial x}= - \frac 1 A \frac{\partial A}{\partial x}.$$ This is de Laval's equation that says that for subsonic flow a narrowing pipe makes the fluid speed up, while for supersonic flow a widening pipe makes the flow speed up. This why the nozzle of a rocket engine first narrows to a throat at which the flow reaches Mach 1, and then expands allowing the exhaust gas to become supersonic.

• Maybe I am not clever enough, but it doesn't help me much. I know that the momentum equation I have written down is the integral of the 1D Navier-Stokes equation. The second is derived from the energy equation. But I still don't get how they both can be valid at the same time. Or is the problem, that this momentum equation is only 1D? Apr 22, 2021 at 16:51
• Yes. I think that your equations are strictly 1-d steady flow. They do not apply when a pipe changes width. You can combine them with $\rho_1v_1=\rho_2v_2$ to derive the equations relating the pressure jump to the propagation speed for planar shockwave for example. See our book goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf page 281. Apr 22, 2021 at 17:18
• Thank you very much, I will look into it! I hope it will make things clear. I have also found this: tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/… Here it seems they took the compressible bernoulli equation to calculate the mass flow rate. Apr 22, 2021 at 17:30