Fluid Statics

Pascal’s Law tells us that “Pressure at any point in a fluid is the same in all directions”. This means that any object submerged in a fluid is subjected to an area force distribution across their entire surface. These area force distributions are called pressures.

Fluid – Any continuous substance which, when at rest, is unable to support shear force. Liquids and gases are the most common examples of fluids.

𝑃=𝑑𝐹𝑑𝐴

This can be simplified to for a constant pressure.

For the infinitesimal area at the right it can be seen that all of the pressures are equal.

𝑑𝑦=𝑑𝑠𝑠𝑖𝑛𝜃∧𝑑𝑥=𝑑𝑠𝑐𝑜𝑠𝜃

∑ 𝐹 𝑥=𝑝1𝑑𝑦𝑑𝑧 −𝑝3𝑑𝑠𝑑𝑧 𝑠𝑖𝑛𝜃=0

∑ 𝐹 𝑦=𝑝2𝑑𝑥𝑑𝑧−𝑝3𝑑𝑠𝑑𝑧𝑐𝑜𝑠𝜃=0

→𝑝1𝑑𝑦 𝑑𝑧=𝑝3 𝑑𝑦 𝑑𝑧→𝑝1=𝑝3

→𝑝2𝑑𝑥𝑑𝑧=𝑝3𝑑𝑥𝑑𝑧→𝑝2=𝑝3

Similar results can be obtained for p4 by rotating the element 90o around the y-axis.

∴𝑝1=𝑝2=𝑝3=𝑝4

Pressure in a fluid is a result of the weight of the fluid above the region of space you are examining. The pressure at the bottom of a drinking glass is due to the weight of the fluid in the glass.

This suggests that the fluid pressure varies with vertical position (depth) in a fluid. Consider the situation to the right to verify this. Note that the pressure on the sides (x and z directions) of the parcel would be concentrated at the same depth and therefore would sum to zero.

This gives the variation of pressure with depth, where P0 is the pressure at the surface of the fluid and r is the density of the fluid. Note that h is measured from the surface of the fluid.

P0 = 1.01 x 105 Pa (Atmospheric Pressure)

Pa is the unit called Pascal

Let us now consider the hydrostatic pressure on a submerged rectangular surface.

We can consider the system as if it was a loaded beam, by looking at a cross-section. Do not forget to include the width of the area!We can examine the load profile as either a composite body or a single pressure distribution.

Composite Body:

𝐴′2=[𝑃2+

12

(𝑃1−𝑃2) ]𝐿=12

(𝑃1+𝑃2 )𝐿

𝑦 1=12𝐿

𝐴′1=𝑃2𝐿

𝑦 2=23𝐿

+ =

A’ is the area of the load distribution, not the area of the plate.

→h=𝑦 𝑐𝑜𝑠𝜃=[ 56 𝑃2𝐿+ 13𝑃1 𝐿

12𝑃1+

32𝑃2 ]𝑐𝑜𝑠 𝜃We use h to represent

the vertical depth, while y is parallel to the plane of the plate.

𝑥1=𝑥2=12𝑏

Single Pressure Distribution:

𝑑𝐴=𝑏𝑑𝑦

𝑑𝐴′=𝑃𝑑𝑦𝑑𝑅=𝑃𝑑𝐴=𝑃𝑏𝑑𝑦=𝑏𝑑 𝐴′

Area of Plate

Area of Pressure Distribution

𝑃=𝑃0+𝜌 h𝑔𝑑𝑅=𝑏 (𝑃0+𝜌𝑔𝑦   )𝑑𝑦

𝑌=∫ 𝑦𝑑𝑅

∫𝑑𝑅=𝑏∫𝑦𝑑 𝐴′

𝑏∫𝑑 𝐴′=∫ 𝑦𝑑 𝐴′

∫𝑑 𝐴′=∫ 𝑦 𝑃𝑑𝑦

∫𝑃 𝑑 𝑦

𝑌=∫ 𝑦 (𝑃0+𝜌𝑔𝑦   )𝑑𝑦

∫ (𝑃0+𝜌𝑔𝑦   )𝑑𝑦

Alternatively, we can use the average pressure and average depth to determine R.

𝑃𝑎𝑣=12

(𝑃1+𝑃2 )=12

(𝑃0+𝜌𝑔h1+𝑃0+𝜌𝑔h2 )=𝑃0+𝜌 𝑔12

(h1+h2 )

𝑃𝑎𝑣=12

(𝑃1+𝑃2 )=𝑃0

+𝜌𝑔hWhere , which is the depth of the center of the plate.

→𝑅=𝑃𝑎𝑣 𝐴=𝜌𝑔 h 𝐴 The location of R must be determined as above from the centroid expression.

Curved Plate:

A curved plate is very similar to a flat plate, except that you must look in multiple directions simultaneously. The pressure profile, for example, has both x and y components no matter your choice of reference frame.

We again have two methods we can use. We can (1) analyze by direct integration, and determine Rx and Ry as well as the centroid.

𝑅𝑥=𝑏∫ (𝑃𝑑𝐿 )𝑥=𝑏∫𝑃𝑑𝑦𝑅𝑦=𝑏∫ (𝑃𝑑𝐿 )𝑦=𝑏∫ 𝑃𝑑𝑥

You would next have to determine the location of Rx and Ry.

Or (2) look at the forces exerted on the volume of water above (or below)the plate. This second method is typically easier.

Analyze the constant pressure profile (Py) on top of the water parcel, then analyze the linear pressure profile with depth (Px). Summing all of the forces for this volume of water will allow you to determine the interaction R between the water and the plate, which is equal and opposite to the net force R on the plate.

Hydrostatic Pressure on Flat Surfaces of Any Shape

For any flat surface submerged in a fluid we can examine it using a method similar to what was done for the flat plate.

𝑅=∫𝑑𝑅=∫𝑃𝑑𝐴=𝑃0 𝐴+𝜌𝑔∫h𝑑𝐴=𝑃0+𝜌𝑔 h 𝐴h 𝐴=∫ h𝑑𝐴

Similarly we could examine the “pressure volume”.

𝑅=∫𝑑𝑅=∫𝑃𝑑𝐴=𝑃0 𝐴+𝜌𝑔∫h𝑥𝑑𝑦Note that h and x must be expressed in terms of y!

It is also necessary to determine the location of the centroid.

𝑅𝑌=∫ 𝑦𝑑𝑅 →𝑌=∫ 𝑦 (𝑃𝑥𝑑𝑦 )

∫𝑃𝑥 𝑑𝑦

Note that the center of pressure and the centroid of the plate are not at the same location!!

Buoyancy

Objects that are at least partially submerged in a fluid have a buoyant force acting on them. The buoyant force is due to the difference in pressure on opposite sides of the object. The buoyant force can also be defined as the weight of water that has been displaced by the object.

𝐹𝐵=𝑊=𝑚𝑔=𝜌𝑉𝑔

Similaly,

𝐹𝐵=∆𝑃𝐴= (𝜌𝑔h2−𝜌𝑔 h1 )=𝜌 h𝑔 𝐴=𝜌𝑔𝑉