# Can someone please clarify this formula? (hydrostatic pressure)

• yuheng_wu
The total pressure is the sum of the atmospheric pressure plus the pressure due to the height of the liquid. In this situation, the atmospheric pressure is 100,000Pa and the weight of the water is 1000kg so the total pressure is:300,000 + 1000 = 3000Paf

#### yuheng_wu

Homework Statement
p hydr = ρ.g.h OR p hydro = p 0 + ρ.g.h
Relevant Equations
p hydr = ρ.g.h OR p hydro = p 0 + ρ.g.h
Is in the bottom of a liquid the pressure equal to p 0 + ρ.g.h or only ρ.g.h? please can someone tell me this correctly?

The change in hydrostatic pressure after descending a depth ##\Delta z## is ##\Delta P = \rho g \Delta z##, which means that you should really have ##P(z) = P_0 + \rho g (z - z_0)##, if ##P_0## is the pressure at ##z=z_0##. Sometimes it might be convenient to choose your coordinates such that ##P_0 = 0##, or at least approximately zero if the pressure at a certain reference position is negligible in comparison to other positions of interest.

You can deduce this by considering a column of fluid, and balancing the net vertical component of the pressure force and the weight of the fluid.

Lnewqban

The pressure at the surface is atmospheric air pressure which is about 14 lb/in^2.

The absolute pressure in the vessel is the sum of the atmospheric pressure plus the pressure due to the height of the liquid.

The atmospheric pressure also presses against the external surface of the vessel so the differential pressure is that due to the only the fluid - the atmospheric bit cancels out.

Is in the bottom of a liquid the pressure equal to p 0 + ρ.g.h or only ρ.g.h? please can someone tell me this correctly?
I’d like to add a simple example which may help.

A lake, 20m deep, contains fresh water.

Density of water: ##\rho=1000kg/m^3##
Atmospheric pressure: ##P_0= 100,000Pa## (approximate)
Acceleration due to gravity: ##g = 10m/s^2## (approximate)

Q1. What is the difference in pressure (##\Delta P##) between the water surface and the bottom of the lake?

##\Delta P = ρgh = 1000 * 10 * 20 = 200,000 Pa##

The pressure at the bottom of the lake is 200,000Pa higher than at the surface.

Q2. What is the total pressure (##P_{total}##) (sometimes called the absolute pressure) at the bottom of the lake?

The atmosphere is exerting a pressure (##P_0##) on the water surface, so this value needs to be added:

##P_{total} = P_0 + \Delta P = 100,000 + 200,000 = 300,000 Pa##

Whether you need to find ##\Delta P## or ##P_{total}## depends on the particular problem you are trying to solve. That means you need to understand the physics so you can tell which formula is applicable.

The change in hydrostatic pressure after descending a depth ##\Delta z## is ##\Delta P = \rho g \Delta z##, which means that you should really have ##P(z) = P_0 + \rho g (z - z_0)##, if ##P_0## is the pressure at ##z=z_0##. Sometimes it might be convenient to choose your coordinates such that ##P_0 = 0##, or at least approximately zero if the pressure at a certain reference position is negligible in comparison to other positions of interest.

You can deduce this by considering a column of fluid, and balancing the net vertical component of the pressure force and the weight of the fluid.
There is more to it than this. We sometimes work with absolute pressure and we sometimes work with gauge pressure. The gauge pressure is the absolute pressure at a given location minus atmospheric pressure. So, in a lake, at h meters below the surface, the absolute pressure is ##P_{atmos}+\rho g h## but the gauge pressure is just ##\rho g h##.

When we do a force balance in a block in mechanics, there are also atmospheric pressure forces acting on the faces of the block. But we don't usually include them in the force balance because they cancel out. But when we do this, we are already working with "gauge" forces.

When we do a force balance in a block in mechanics, there are also atmospheric pressure forces acting on the faces of the block. But we don't usually include them in the force balance because they cancel out. But when we do this, we are already working with "gauge" forces.

Sure, so long as the whole surface of the block is exposed, this is just a consequence of the buoyant force density being ##\mathbf{f} = -\nabla P##, in which case you only need to determine ##P## up to an additive constant, i.e. ##-\nabla(P + k) = -\nabla P##.