I will focus on just a little bit of one of your questions - the relationship between compressibility, density and pressure - and per my comment, recommend that you narrow down the scope of your question.
As you know, in a gas we experience "pressure" because molecules hit the walls of the containing vessel. When I double the number of molecules in the same volume at the same temperature, I double the number of collisions (each imparting on average the same momentum) and thus double the pressure - this is the familiar ideal gas law.
Now when the size of the molecules becomes a sizable fraction of the volume, the rate of collisions goes up. Imagine a pingpong ball between two walls. If the distance between the walls is large compared to the size of the ball, the time for a round trip is inversely proportional to the size of the ball; but as the distance approaches the size of the ball, the rate of collisions goes up rapidly.
I think a similar thing happens with "nearly incompressible" liquids: there is a small amount of space between the molecules, but they are permanently bumping into each other and into the walls of the vessel. As you increase the pressure, they bounce more frequently as they have less far to travel before they collide with another molecule (or the wall).
All this is still treating the liquid like a non-ideal gas. In reality, not only do you have the finite size of the molecules, but also attractive forces between them. Both these things make the picture a bit more complex than I sketched. But I would say that the above reasoning nonetheless applies (with caveats).
As for the experiment you described with stoppers on the inside or outside - there are other things going on there as you go from the static (no flow) to the dynamic (flow) situation - the water needs to accelerate before it will flow out at a certain velocity. But I think all that should be the subject of another question.
At zero flow the pressure at the shower head is simply the hydrostatic pressure given by Pascal's law:
$$p=p_0+\rho gy$$
Where $p_0$ is the atmospheric pressure, $y$ the height difference between tank meniscus and shower head, $\rho$ the density of water ($g\approx 10$$\:\mathrm{m/s^2}$).
The water pressure coming from the head is too low.
What the OP really means here is that the shower only delivers a trickle of water (low flow speed). So here I'll evaluate the factors that influence that flow speed.
When flow starts, $p$ is lowered by:
1. Viscous losses in the pipe:
Acc. Darcy Weisbach pressure loss in a straight pipe due to flow is given by:
$$\Delta p=f_D\frac{\rho}{2}\frac{v^2}{D}L$$
Where $f_D$ is a friction factor, $v$ is flow speed ($mathrm{m/s}$), $D$ pipe diameter and $L$ pipe length.
For laminar flow:
$$f_D=\frac{64\mu}{\rho D v}$$
Where $\mu$ is the viscosity of the fluid.
So for laminar flow:
$$\Delta p=\frac{32\mu v}{D^2}L$$
2. Local resistances:
Valves, bends, kinks, sudden changes in diameter etc. all cause head loss $h$, usually modelled as:
$$h_r=c\frac{v^2}{2g}$$
Where $c$ is a coefficient that depends on the type of local resistance.
In the OP's stated problem the main local resistance is almost certainly the shower head itself.
3. Bernoulli's principle:
Using Bernoulli's principle we can now write (for laminar flow):
$$y=\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L+c_{shower}\frac{v^2}{2g}$$
Or:
$$y=(c_{shower}+1)\frac{v^2}{2g}+\frac{32\mu v}{\rho gD^2}L$$
This is a simple quadratic equation in $v$ and if $c_{shower}$ and the other factors where known, then it could be solved quite easily. But in the absence of that information we can still say that $v$:
- will increase with $y$,
- will increase with $D$,
- will decrease with $L$,
- will decrease with $c_{shower}$.
4. Turbulent flow:
In the case of turbulent flow (high $v$, $Re > 4000$), $f_D$ becomes a function of $v$, $f_D=f(v)$ and the calculation becomes more complicated. But the general conclusions above still hold.
Best Answer
Nice diagrams! A more complete statement would be "The pressure exerted by a static fluid relative to atmospheric pressure depends only upon the depth of the fluid relative to the surface exposed to atmospheric pressure, the density of the fluid, and the acceleration of gravity."
Following this statement, the pressure at the bottom of tanks A and C is the same and is less than the pressure at the bottom of tank B. The pressure in the space marked "vacuum" is actually the vapor pressure of the liquid at that temperature. The dependence on temperature is strong (exponential), so C can serve as a thermometer if you add a length scale to the top.