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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 $$ p_1 + u_1^2 \rho_1 = p_2 + u_2^2 \rho_2 $$
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.
