You asked a number of questions in this question.
From a hydrostatic point of view, the pressure in a fluid should be the same at the same depth/altitude.
That "should be" assumes hydrostatic equilibrium. That is a simplifying assumption. It's a reasonable starting point, but it's not a hard and fast rule. The Earth's atmosphere, it's oceans, and even its interior are approximately in hydrostatic equilibrium.
I am guessing that the main reason is the fact that the atmosphere cannot be regarded as hydrostatic. Is this the reason?
Significant deviations from hydrostatic equilibrium do occur. This is an effect, not a cause.
How exactly can we explain these pressure differences?
Ultimately, it's because the Earth
- Is round,
- Is lit by the Sun,
- Rotates about its axis once per day,
- Has distinct rotational and orbital axes, separated by about 23 degrees,
- Has a fairly clear atmosphere, and
- Is covered by lots of liquid water.
These result in climate and weather, which in turn result in the Earth's atmosphere being only approximately in hydrostatic equilibrium.
Equatorial regions receive a lot more sunlight than do polar regions. The resulting temperature gradient is one of the key drivers of the climate. On Venus, which rotates slowly, this energy transfer occurs in a pair of Hadley cells that reach from the equator almost to the poles. On Titan, which rotates in about 16 days, the Hadley cells breaks up at about 60 degrees latitude. Jupiter and Saturn are so large and rotate so quickly that they have bands instead of Hadley-type cells.
On the Earth, which rotates once per day, the Hadley cells extend to only 30 degrees. Polar cells form around the poles, and the Ferrel cells act as intermediaries between the Hadley and Polar cells.
http://www.metoffice.gov.uk/media/image/f/s/Figure-4-Global-cells(edit)2.jpg
But how fast is this? In the order of the speed of sound? Or it has nothing to do with it?
The speed of sound has nothing to do with it. Winds do, and winds generally move much slower than the speed of sound. The fastest winds recorded are inside tornados, and even there things only move at about 40% of the speed of sound.
The pressure at D is not significantly greater than the surrounding atmosphere. The jet of fluid is surrounded by the atmosphere, and the pressure within the jet (like at point D) is virtually identical to atmospheric pressure.
What you are missing is what is happening inside the tank in the immediate vicinity of the exit hole. The hydrostatic pressure distribution inside the tank is disturbed by the fluid flow converging toward the exit hole. The effective region where this disturbance occurs is within just a few hole diameters of the exit. Thus, the point A in the figure is just about close enough to the exit hole for this disturbance to begin to be felt. In the region near the exit hole, the pressure rapidly decreases from the hydrostatic pressure variation away from the hole to atmospheric pressure at the exit hole. This localized pressure gradient provides the driving force for the flow out the hole. The surfaces of constant pressure are roughly hemispherical, with their center at the hole. The streamlines near the hole are all converging radially toward the hole. As the flow converges toward the hole, the fluid velocity increases, and the pressure decreases.
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Best Answer
The net pressure on the liquid is just the atmospheric pressure. Pressure in a fluid acts in every direction, but as the point is on the surface, $\text{P}_{water}=h\rho g=0$ as $h=0$. So only atmospheric pressure will be acting on point A.
The height of the liquid column doesn't affect the pressure on top. Pressure in a liquid is affected by the weight of the liquid above it.
EXPLANATION
Consider this diagram, :
The pressure on a cross section inside the fluid that height is given by the weight of fluid above that height, divided by horizontal cross sectional area of the fluid column. We get the pressure by $$P=\frac{F}{A}=\frac{mg}{A}$$ $$=\frac{\rho Vg}{A}=\frac{\rho (A\times h)g}{A}$$ $$\boxed{P=h\rho g}$$
When we calculate the pressure at the surface of the fluid, we take the weight of fluid above the cross sectional area. But as there is no fluid above the surface, $\text{Weight}=0,~~ \therefore \text{P}=0$