I am confused about a very simple thing: Let's assume you have a 1D flow (fluid), where your flow velocity v$$_x (x)$$ = -$$\frac{1}{x} \hat{x}$$, i.e. your speed (= |velocity|) increases as you approach the origin, but its direction is -$$\hat{x}$$. Now, consider your density of fluid has a relationship: $$\rho(x)$$ = $$\frac{1}{x}$$, which implies that your density also increases as you approach the origin.

What does it mean in transport of momentum ($$\rho v_x$$) if you find for a particular x = $$x_0$$, $$\left(\frac{\partial (\rho v_x)}{\partial x}\right)_{x = x_0} > 0$$ ? I think it means that we are transporting momentum inwards at location x$$_0$$, i.e. towards x < x$$_0$$, but I am not sure.

Similarly, what would mean if your,let's say $$\rho(x)$$ = $$x^3$$ and you find,$$\left(\frac{\partial (\rho v_x)}{\partial x}\right)_{x = x_0} < 0$$?

Please don't consider continuity equation as in my case I am at steady state so: $$\frac{\partial \rho}{\partial t} = 0$$. If you still persist to consider continuity equation, then please consider the RHS to be non-zero like: $$\frac{\partial \rho}{\partial t} + \frac{\partial (\rho v_x)}{\partial x} = S$$, where S can be sink or source. Also, the flow is compressible and at steady-state.

Actually $$p_x = \rho v_x$$ is the momentum density, not its flux density, and you can also view it as the mass flux density. If you're really interested in the momentum flux density, you need to rather consider $$T = \rho v_x^2$$ In higher dimensions, conservation of momentum is translated by: $$\frac{\partial \rho}{\partial t}+\nabla\cdot (\rho \vec v) = j$$ with $$\rho \vec v$$ interpreted as the mass flux density and $$j$$ the mass source density. Similarly, conservation of momentum density $$\vec p = \rho \vec v$$ is written as: $$\frac{\partial \vec p}{\partial t} + \nabla \cdot (\vec p\otimes\vec v) = \vec f$$ where you interpret the tensor $$T = \vec p\otimes\vec v$$ as the momentum flux density and $$\vec f$$ the momentum source density, ie the external volume source. Note that $$T$$ is symmetric, the index you use to contract the divergence is irrelevant, but more generally, if it isn't symmetric, you must be careful and contract with the $$\vec v$$ index.
Since a 1D tensor is always a scalar, $$T$$ is reduced to the first expression. For example, if you consider a segment with $$[a,b]$$, the momentum flux will be: $$\Phi = [\rho v_x^2]_a^b$$ which will be the force applied to the segment: $$\Phi = \int_a^bfdx$$
• Thank you! My question is basically that if you find $\nabla\cdot(\rho \vec{v})$ > 0 at steady state, which implies that your $j$ > 0, what information we can get deduce about $\rho$? Like, whether the density is high in $\nabla\cdot(\rho \vec{v})$ > 0 region or low Aug 27, 2022 at 22:44