Solve problems associated with the behaviour of liquids at rest

Previous sections of this unit encompassed the behaviour, effects, and the solution of problems covering fluids in motion, i.e. thermodynamics.

In this section we are concerned with the solution of problems associated with the effects of liquids at rest. Hydrostatics is the study of the forces and pressures exerted by a fluid when the fluid system is in equilibrium.

Gravitational force

The derived SI unit of force is the Newton (N) defined as ‘that force which, when applied to a body of mass one kilogram, gives it an acceleration of one metre per second squared’. From Newton’s Second Law of Motion:

$\text{Force}=\text{Mass}\times \text{Acceleration}$ 

$F=ma(kg{m}^{-2}=N)$ 

When gravitational acceleration is considered, the above equation can be re-stated in the form:

$F=mg$ 

You will also occasionally see this referred to in terms of weight:

$W=mg$ 

Hence, the downward gravitational force exerted by a liquid column of mass 25 kg will be:

$F=25\times 9.81=245.25N$ 

Pressure (ρ)

Pressure is defined as force (F) per unit cross-sectional area (A).

i.e. Pressure $p=\frac{\text{(Force)}}{\text{Cross-sectional area}}=\frac{F}{A}$ 

The unit for pressure is the Pascal, which is one Newton per metre squared. You will also often see pressures quoted in terms of Newtons per metre squared. The unit Bar is also often seen.

1Pa=1Nm^-2

1 bar=10⁵Nm^-2

Atmospheric pressure

This is the pressure exerted on the surface of the earth by the gravitational pull on the mass of air in the Earth’s atmosphere.  Atmospheric pressure is stated as an absolute, that is relative to zero.  Atmospheric pressure changes over time depending on location and many other factors, so a standard has been adopted:

Standard atmospheric pressure=1.013 bar=1.013 10⁵ Nm^-2=1 atmosphere (ATM)

Gauge Pressure and Absolute Pressure

Various types of pressure gauges are commonly used to measure fluid pressures in vessels and pipelines and read pressures normally above or below atmospheric pressure. If a gauge displays a zero reading it means the pressure is atmospheric.

If the pressure in a vessel is increased above atmospheric to a gauge pressure pg, the true or absolute pressure p in the vessel is given by:

Absolute pressure p = gauge pressure + atmospheric pressure

i.e.  $p=p_g+p_{atm}$  

Static pressure characteristics

The concept of pressure in a fluid and the manner in which it varies throughout the fluid mass is of fundamental importance in the science of hydrostatics. The experimentally established characteristics noted below need to be acknowledged.

• The intensity of pressure in a fluid at rest is the same in all directions.
• The pressure exerted by a static fluid always acts normally (perpendicular) to any boundary surface containing the fluid.
• The pressure exerted by a static fluid is directly proportional to and increases with depth below its free surface.

Pressure variation with depth in a liquid column

Consider the forces acting on a vertical cylinder of liquid of height h and cross-sectional area A within a volume of static liquid.

If the column of liquid is at rest, it must be in equilibrium:

$\text{∑Upward forces}=\text{∑Downward forces}$ 

$p_{2}A=p_{1}A+\rho Agh$ 

Because A is a common factor it can be cancelled:

$p_2=p_1+\rho gh$ 

→ $p_2-p_1=\rho gh$ 

This means that the pressure at any depth in a liquid can be derived from the equation:

$\text{Pressure }p=\rho gh$ 

Hydrostatic Pressure and Thrust

© A.Henderson, UHI

The above figure illustrates three tanks having equal base areas A. Each vessel contains the same liquid to a common vertical depth h.

Since the tanks hold different volumes of liquid, it might be thought that container C with the largest amount and greater mass of liquid would have the greatest force exerted on its base. Tank D might be considered as having the least force on its base.

For all three tanks, however, the base pressure p = ρ gh at all points on the bases since these are horizontal. Also, since the tanks have a common base area A, it follows that the forces acting on all three bases have equal magnitude.

That means that $F=pA=pghA$ and all four quantities are constant.

The arrows inside tanks B, C and D illustrate how pressure acts normal to retaining surfaces and increases in intensity with depth below the free surface.

The total force/thrust acting on the horizontal bases of tanks B, C and D will act at the centroid of each base area.

Our studies now extend into the determination of thrust exerted on submerged and partially submerged vertical plane surfaces and to fixing the point of application of this force at the centre of pressure.

Thrust on a submerged vertical plane surface

Consider a plane surface of area A immersed vertically in a liquid of density r . The pressure on one side acts at right angles to the surface and gives rise to a force or thrust on that side.

© A.Henderson, UHI

We know that $P=\frac{F}{A}$, so we can re-arrange this for force: $F=pA$ 

If we look at the figure above we can see:

Pressure p on one side of a strip with an area  $dA=\rho gy$  

Meaning that the force $dF=\rho gydA$ 

The total force on the whole area A then can be found using $F=\sum dF$ 

Because ρ and g are constants we can modify this further to say:

The total force on the whole area A then can be found using $F=\rho g\sum ydA$ 

If you look at the diagram, you can see that ∑ydA is the total moment of area about an axis through the free surface of the liquid and is also equal to
$A\times \stackrel{-}{y}$, with $\stackrel{-}{y}$ is the vertical depth of the centroid G below the free surface of the liquid.

Thus the total thrust on an immersed plane vertical surface is proportional to the depth of the centroid of the wetted area below the free surface and can be deduced from formula:

$F=\rho gA\stackrel{-}{y}$ 

Centre of pressure

We have seen that the intensity of pressure acting over any vertically submerged plane surface area increases proportionally with depth below the free surface of a liquid. Hence, for both rectangular and circular vertical surfaces, the pressure and forces applied above their horizontal geometric centre lines will be less than those acting below their centre lines.

The combined action of all the elemental forces acting over the whole surface areas can be replaced by a single resultant force, equal in magnitude to the total thrust, and acting at a point on the surface called the centre of pressure. This point will obviously lie on a vertical centre line of area at some distance below the geometric centroid of the plane surface areas.

© A.Henderson, UHI

For both configurations of immersed plane surfaces shown above, the individual centroids (G) have a depth y below the liquid free surface. Each centre of pressure (CP) is located vertically below the centroids by a distance designated GC.

Setting aside its mathematical proof from first principles we can state the appropriate formula for calculating this distance:

$\text{DISTANCE GC }=\frac{\text{RADIUS OF GYRATION ABOUT CENTROID SQUARED}}{\text{DEPTH OF CENTROID BELOW FREE SURFACE}}$ 

OR:

$GC=\frac{k^2}{\stackrel{-}{y}}$ 

Values for k₂ can be calculated using formulae listed alongside the following figure

© A.Henderson, UHI

For the rectangle, the geometric centroid, G, lies on the intersection of the vertical and horizontal centre lines.

For the circular area the centroid, G, lies at centre of circle.

For either immersed surface, the centre of pressure, C, will lie vertically below the centroid by a distance, say GC, and can be calculated using the stated formula for each listed below.

For a rectangular surface:

$\text{Distance }GC=\frac{k^2}{\stackrel{-}{y}}=\frac{d^2/12}{\stackrel{-}{y}}$ 

For a circular surface:

$\text{Distance }GC=\frac{k^2}{\stackrel{-}{y}}=\frac{d^2/16}{\stackrel{-}{y}}$ 

Where $\stackrel{-}{y}$ = depth of centroid below the liquid free surface.

Partially submerged rectangular surfaces and wetted areas

For rectangular plane surfaces which are partially submerged vertically in a liquid, the formulae developed for total thrust and centre of pressure remain applicable when the area of the wetted surface only is considered.

The position of the geometric centroid for a partially submerged surface area can also be established for the wetted area only:

© A.Henderson, UHI