Given a slanted pipe Question: I have a pipeline that is tilted. I know the length of the pipeline, and the pressure (90psi) that is felt at the very bottom. I'm trying to find out how much liquid has leaked out if the pressure at the bottom decreases to 80psi.
Is the pressure felt at the very bottom directly proportional to the amount of liquid that is in the pipeline?
 A: The pressure at the bottom of a pipe, or any other column of liquid, for that matter, is directly proportional to the height of the liquid.
The pressure of the liquid is equal to:
$\rho h$, where $\rho$ represents the density of the liquid, and h represents the height.
So, assuming that the cross-sectional area of the pipe is constant, then the pressure felt at the bottom is directly proportional to the amount of liquid in the pipe.
A: Assuming:


*

*the problem is static ($\rightarrow$ no velocity gradient, so no shear, only the spheric part remains non-zero),

*the density is homogeneous ($\rightarrow$ $\rho$ is a constant),

*the cross-sectional area of the pipe is constant,

*(neglecting the variations of $g$ (!))


the pressure $p$ various linearly with the height $z$:
$$\dfrac{\mathrm{d}p}{\mathrm{d}z}=-\rho(z) g\quad\Longrightarrow \quad \Delta p=-\rho  g\Delta z \quad (\text{since }\rho(z)=\rho)$$
which is known as Pascal's law (XVIIth century).
Of course, given $\Delta p$, $\rho$ and $g$ you can calculate $\Delta z$ and deduce the volume of the leakage (which is straighforward as the pipe has a constant cross-sectional area.).
The pressure is not directly proportional to the amount of the liquid (but the variation of pressure is directly proportional to the variation of amount of the liquid), except if $p_0<<p$, which is often the case. Then, $p=p_0+\rho g h \approx \rho g h \propto h \propto V$.
