# How to calculate the force that a liquid creates on the wall of an underground reservoir

I would like to calculate the force exerted on the wall of an underground reservoir. The volume of rocks surrounding the reservoir creates a force equal to the product of the density of the lithosphere, the gravitational acceleration and the height above the reservoir.

What I'm trying to understand is what are the forces created by the liquid inside of the reservoir. How can I calculate these forces?

• That depends on the type of reservoir and what is stored there. Is it an actual large "hole" underground, like an abandoned mine, or is it more like a petroleum reservoir, which is in fact porous rock which is merely saturated with petroleum? In the former case, is the liquid pressurized or merely poured in until it gets "full to the brim"?
– Wasabi
Feb 28 '18 at 1:30
• Let's say for simplicity that the orange could be any type of liquid and the grey is a purely elastic material. What I would like to figure is the load that fluid is creating on the surface of the reservoir and the competition it has with the surrounding rocks that will also create their own load on the reservoir. Feb 28 '18 at 9:30
• Force due to the depth of liquid plus the force created by the mass above - unless that mass is supported by something else. Feb 28 '18 at 10:11
• but the load of the liquid is not just acting vertically right? Feb 28 '18 at 10:38
• The liquid head produces the same pressure in all directions. The driving force of well "fracking." Feb 28 '18 at 15:52

In reality there is underground water and also all variety of oil and mixture of other liquids, such as bicarbonates and other minerals dissolved in water.

Depending on the geology of soils, and the pore pressure of it, we get a wide range of situations. sometimes due to flow of underground water there is even hydraulic head pressure or negative pressure to capillary and osmosis actions.

We consider an ideally simplified case where the whole is large and forces are static and balanced and there is one liquid and no trapped air.

In a case like this if we could bore a long infinitesimally small radius hole without disturbing the layers of rock and water the pressure at any given depth is equal to

$P = \sum \rho.da.h$

So the total pressure is the weighted contribution of different material added together from grade level down to depth under consideration.

This pressure in the soil layer is transferred laterally by the factors such as passive lateral pressure, bedding angle and structural properties of the soil. When the test bore gets to the level of water/liquid the pressure is equally transferred in all directions.

The soil surrounding this reservoir will react in complex ways, again depending on its structural properties and its content of aggregates. These properties can be determined by the soils engineer using a wide selection of tools. Such as sampling the soils, doing ultrasound and satellite imaging. etc.

The finer aggregates and clayed strata would act as a container of the water and the pressure of the water would be equal to the pressure of surrounding soil. The larger aggregate will diffuse the pressure and can potentially lead to caving of the reservoir.

• Hello, thank you for your reply. Indeed I agree with your response. The idea here is to find the initial conditions at the surface of the 'reservoir' which will be also impacted by external forces the lithostastic stress increasing linearly with depth (sigma_x = sigma_y = sigma_z = -rho_lith * g * Z (where rho_lith = 2500 kg/m^3, g = -9.81 m/s^2, and Z the depth considered negative). We consider the convention that compression is negative. We don't consider porosity or pore pressure for simplification. Mar 1 '18 at 13:59

Assuming the liquid is static, or mostly static - the most simple answer: The pressure. Force exerted per unit of surface of the wall, in direction normal to the surface, is equal to pressure of the liquid in contact with that wall, and nothing else. Your liquid at the bottom is at 120kPa, it exerts 120kN per meter square of bottom surface in direction normal to the surface. This is true for any infinitesimal fragment of the wall.

In fully static liquid, that will be just density times depth times gravitational acceleration. Water, at 1000kg/m^3 times 9.8m/s^2 becomes 98kPa at 10m depth. Add ambient atmospheric pressure (~1bar) if necessary (do you consider empty tank to receive 1bar of pressure on the wall or 0?)

It gets trickier if the liquid is flowing at considerable speed, exerting dynamic pressure, drag due to laminar flow etc. But for mostly static tank, the force is just pressure times unit of surface, and always in normal direction to that surface.

Water exerts pressure on all surfaces in contact. The intensity varies, depends on the depth, the properties of the surrounding medium, and the enclosure of the cavity/reservoir. Also, the externally applied pressure, if any.