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The $\rho u^2 + p$ term corresponds to the flux of $x$-momentum through some surface. According to this equation it would give a positive momentum flux if $u$ was positive or negative. This does not make sense to me as if the velocity was reversed the momentum flux should be negative. Why is this term positive? enter image description here

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The equation is derived from momentum balance, so there is nothing wrong with. But you can resolve the apparent paradox like this. Take a square control volume $C = [a,b] \times [c,d]$ and look at x-momentum balance $$ \frac{d}{dt}\int_C \rho u dx dy = -\int_c^d (\rho u^2)(b,y) dy + \int_c^d (\rho u^2)(a,y) dy + \ldots $$ The first term on the right is the convective flux of momentum across the right face of $C$. In first term, if $u > 0$, fluid is going out of $C$ and you lose x-momentum. On the other hand, if $u < 0$, fluid is entering $C$ but you still lose x-momentum because the incoming fluid has negative x-momentum. So in either case, you lose x-momentum due to momentum convection across the right face. The situation is reversed for the left face, you always gain x-momentum due to convective flux across the left face.

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  • $\begingroup$ Thanks that makes total sense! $\endgroup$ – Frosty May 3 at 5:38

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