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Consider the momentum balance equation for a steady inviscid flow

$$\int_S \rho \mathbf{v}(\mathbf{v}\cdot\mathbf{n})\ dS = -\int_S p\mathbf{n}\ dS\ +\ \int_V \rho \mathbf{g}\ dV\ +\ \mathbf{F}$$

where $\mathbf{F}$ are other forces.

In my fluid mechanics course the professor said that in a flow the hydrostatic pressure distribution balances the weight of the fluid and therefore we could cut the weight term as long as we used relative pressures.

$$p = p_H + p'$$ $$\nabla p_H = \rho \mathbf{g}\ \text{[definition of hydrostatic pressure]}$$

Applying this would yield

$$\int_S \rho \mathbf{v}(\mathbf{v}\cdot\mathbf{n})\ dS = -\int_S p'\mathbf{n}\ dS\ +\ \mathbf{F}$$

There are two things that are confusing me. The first one is that we were taught that relative pressure was relative to the atmosphere, and the professor seems to be using that same concept in here, but I do not see any reason to why the hydrostatic pressure would be equal to the atmospheric pressure.

For my other question, consider a flow along a horizontal tube, with some flow device in the middle which makes the inlet and outlet pressures differ. Applying the momentum balance in the vertical direction we have both weight and hydrostatic pressure, so they cancel out. However, in the horizontal direction we do not have weight, but we will still have hydrostatic pressure, which does not cancel with anything, right?

What are the flaws in my reasoning?

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  • $\begingroup$ If you consider two points on the same horizontal level, will not the hydrostatic pressure difference between those points be zero? $\endgroup$ – Deep Oct 29 '17 at 6:09
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The combination $$p - \rho \mathbf{g} \cdot \mathbf{x}$$ is also called the modified pressure (see Batchelor, p. 176). The usual caveats for using the modified pressure are that the density is constant and that the absolute pressure does not appear in the boundary conditions of the problem. Also, note that this simplification has nothing to do with momentum balance per se and is evident from the fundamental equation of motion of the (inviscid) fluid: $$ \rho \frac{D\mathbf{v}}{Dt} = \rho \mathbf{g} - \nabla p + \mathbf{F} \; . $$ So, to your first question about relative pressure, I am not sure I understand. Relative pressure is indeed normally meant to be the pressure relative to the ambient atmospheric pressure. This is relevant for acoustics, for example, in which sound waves create a pressure field which is a very tiny fluctuation around the atmospheric pressure. What the equation of motion above shows is that any gravitational body force is formally removable from the equation of motion by shifting the pressure. Its as simple as that.

Regarding your second question, let's assume $\mathbf{g} = g \hat{\mathbf{z}}$. Then, the modified pressure is $$p-\rho gz \; .$$ In the horizontal direction, we have $$\hat{\mathbf{x}} \cdot \nabla (p-\rho gz) = \hat{\mathbf{x}} \cdot \nabla p \; ,$$ so that the equation of motion in the $x$-direction is totally unaware of gravity and the vertical offset due to the modified pressure.

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