Unit 2: Fluid Statics
This section will study the forces acting on or generated by fluids at rest.
- Introduce the concept of pressure;
- Prove it has a unique value at any particular elevation;
- Show how it varies with depth according to the hydrostatic equation and
- Show how pressure can be expressed in terms of head of fluid.
This understanding of pressure will then be used to demonstrate methods of pressure measurement that will be useful later with fluid in motion and also to analyse the forces on submerges surface/structures.
1. Fluids statics
The general rules of statics (as applied in solid mechanics) apply to fluids at rest. From earlier we know that:
- a static fluid can have no shearing force acting on it, and that
- any force between the fluid and the boundary must be acting at right angles to the boundary.
Note that this statement is also true for curved surfaces, in this case the force acting at any point is normal to the surface at that point. The statement is also true for any imaginary plane in a static fluid. We use this fact in our analysis by considering elements of fluid bounded by imaginary planes.
We also know that:
- For an element of fluid at rest, the element will be in equilibrium - the sum of the components of forces in any direction will be zero.
- The sum of the moments of forces on the element about any point must also be zero.
It is common to test equilibrium by resolving forces along three mutually perpendicular axes and also by taking moments in three mutually perpendicular planes an to equate these to zero.
As mentioned above a fluid will exert a normal force on any boundary it is in contact with. Since these boundaries may be large and the force may differ from place to place it is convenient to work in terms of pressure, p, which is the force per unit area.
Units: Newton's per square metre, ,
(The same unit is also known as a Pascal, Pa, i.e. 1Pa = 1)
(Also frequently used is the alternative SI unit
the bar, where )
(Proof that pressure acts equally in all directions.)
By considering a small element of fluid in the form of a triangular prism which contains a point P, we can establish a relationship between the three pressures px in the x direction, py in the y direction and ps in the direction normal to the sloping face.
The fluid is a rest, so we know there are no shearing forces, and we know that all force are acting at right angles to the surfaces .i.e.
acts perpendicular to surface ABCD,
acts perpendicular to surface ABFE and
acts perpendicular to surface FECD.
And, as the fluid is at rest, in equilibrium, the sum of the forces in any direction is zero.
Summing forces in the x-direction:
Force due to ,
Component of force in the x-direction due to ,
Component of force in x-direction due to ,
To be at rest (in equilibrium)
Similarly, summing forces in the y-direction. Force due to ,
Component of force due to ,
Component of force due to ,
Force due to gravity,
To be at rest (in equilibrium)
The element is small i.e., and are small, and so is very small and considered negligible, hence
Considering the prismatic element again,
is the pressure on a plane at any angle ,
the x, y and z directions could be any orientation. The element
is so small that it can be considered a point so the derived expression
. indicates that pressure at any point
is the same in all directions.
(The proof may be extended to include the z axis).
In the above figure we can see an element of fluid which is a vertical column of constant cross sectional area, A, surrounded by the same fluid of mass density . The pressure at the bottom of the cylinder is at level , and at the top is at level . The fluid is at rest and in equilibrium so all the forces in the vertical direction sum to zero. i.e. we have
Taking upward as positive, in equilibrium we have
Consider the horizontal cylindrical element of fluid in the figure below, with cross-sectional area A, in a fluid of density , pressure at the left hand end and pressure at the right hand end.
The fluid is at equilibrium so the sum of the forces acting in the x direction is zero.
This result is the same for any continuous fluid. It is still true for two connected tanks which appear not to have any direct connection, for example consider the tank in the figure below.
We have shown above that and from the equation for a vertical pressure change we have
This shows that the pressures at the two equal levels, P and Q are the same.
Here we show how the above observations for vertical and horizontal elements of fluids can be generalised for an element of any orientation.
Consider the cylindrical element of fluid in the figure above, inclined at an angle to the vertical, length , cross-sectional area A in a static fluid of mass density . The pressure at the end with height is and at the end of height is.
The forces acting on the element are
There are also forces from the surrounding fluid acting normal to these sides of the element.
For equilibrium of the element the resultant of forces in any direction is zero.
Resolving the forces in the direction along the central axis gives
Or in the differential form
If then s is in the x or y directions, (i.e. horizontal),so
Confirming that pressure on any horizontal plane is zero.
If then s is in the z direction (vertical) so
Confirming the result
In a static fluid of constant density we have the relationship , as shown above. This can be integrated to give
In a liquid with a free surface the pressure at any depth z measured from the free surface so that z = -h (see the figure below)
This gives the pressure
At the surface of fluids we are normally concerned with, the pressure is the atmospheric pressure, . So
As we live constantly under the pressure of the atmosphere, and everything else exists under this pressure, it is convenient (and often done) to take atmospheric pressure as the datum. So we quote pressure as above or below atmospheric.
Pressure quoted in this way is known as gauge pressure i.e.
Gauge pressure is
The lower limit of any pressure is zero - that is the pressure in a perfect vacuum. Pressure measured above this datum is known as absolute pressure i.e.
Absolute pressure is
As g is (approximately) constant, the gauge pressure can be given by stating the vertical height of any fluid of density which is equal to this pressure.
This vertical height is known as head of fluid.
We can quote a pressure of in terms of the height of a column of water of density, . Using ,
And in terms of Mercury with density, .