Conservation of momentum of ideal gas in pipe

Question

I was reading this question Pressure drop in a pipe due to cooling and have a follow up question. Please note that the question is about an ideal frictionless pipe.

In the question, it is stated $$ \rho v = C_1$$ $C_1$ being a constant due to conservation of mass, which makes sense to me, that for every cross section there must be the same mass per second, but then $$ p+\rho v^2=C_2$$ is stated with the reason being "conservation of momentum". I can't find this equation anywhere else and the reasoning behind it also doesn't make sense to me. If we combine it with https://en.wikipedia.org/wiki/Bernoulli%27s_principle $$v^2/2 + gz +p/\rho = C_3$$ set $g$ to 0, we can get $$C_2 - \rho v^2 = \rho C_3 - \rho v^2/2 $$ and $$C_2 = \rho C_3 + \rho v^2/2 $$ and using the first equation (conservation of mass) we can get $$\rho C_2 = \rho^2 C_3 + C_1^2/2 $$ which implies that $\rho$ is constant, which seems completely wrong to me. Momentum is also discussed in the comments and everyone seems fine with using temperature/pressure when talking about conservation of momentum. What doesn't make sense to me is how do temperature/pressure contribute when discussing conservation of momentum. Conservation of momentum is about the total momentum, and raising/lowering temperature/pressure, don't affect total momentum. If you have a gas in a stationary box the total momentum is zero, and heating or cooling it won't change that.

Where does this equation and reasoning come from?

Answer 1

As your system is cooling, its energy is not conserved and Bernoulli does not apply. However from 1-d Euler $$ \rho \left(\frac{\partial v}{\partial t}+ v \frac{\partial v}{\partial x}\right)= -\frac{\partial P}{\partial x} $$ and 1-d mass conservation $$ \frac {\partial \rho} {\partial t}+ \frac{\partial \rho v}{\partial x}=0 $$ we find the momentum conservation law $$ \frac{\partial \rho v}{\partial t}+ \frac{\partial}{\partial x}\left(\rho v^2+P\right)=0, $$ so in steady flow we have $\rho v^2+P=const$. This is true for strictly 1-d flow even if heat energy is being added or lost. It appears at first to in conflict with Bernoulli (which holds in the flow is isentropic) but in Bernoulli the speed changes because the pipe changes area, so the Bernoulli flow is not strictly 1-d.

In a pipe with a slowly varying area the momentum law becomes $$ A\frac{\partial \rho v}{\partial t}+ \frac{\partial\rho v^2 A}{\partial x}= -A \frac{\partial P }{\partial x}. $$