Fluid Statics

In this section, we will consider the concept of pressure, how to determine it based on the concept of fluid head, and how it can be measured.

Pressure

We are, quite literally, under pressure all the time (unless we live in a vacuum). The atmosphere (and the air within it) surrounding us exerts a force over our body. All static fluids exert a pressure on objects and their surroundings, and this pressure is defined as the ratio of the force exerted by the fluid divided by the area over which the force is being applied:

P r e s s u r e space P space equals space space fraction numerator F o r c e space F over denominator A r e a space A end fraction

The basic unit of pressure P is the N/m2, as the unit of force is the Newton N and the unit of area is the m2. But we also often use an alternative unit called the Pascal (Pa) and the two units are directly equivalent, so 1 N/m2 = 1 Pa.

1 N/m2 is a very small pressure. To get some idea of this, think about what the pressure of the atmosphere is at sea level. It is considered to be a standard of 101,325 N/m2 (over 100,000 N/m2). Therefore, for most typical industrial pressures, we will often see them referred to in multiples like kN/m2 (1,000 N/m2) or MN/m2 (1 million N/m2). A particular multiple unit is the bar (short for barometric pressure), and this is equivalent to 100,000 = 105 N/m2. Let us do a worked example to get a feel for pressure.

Example:

A carbon dioxide storage tank is at a pressure of 1,500 kPa (or 15 bar g). What would be the pressure and the force exerted by carbon dioxide on a 1 m diameter circular access hatch in the wall of the tank?

The pressure on the hatch would be the same as everywhere in the tank, so the pressure on the hatch would 1,500 kPa.

The force on the hatch would be found by considering the definition of pressure, namely:

P r e s s u r e space P space equals space space fraction numerator F o r c e space F over denominator A r e a space A end fraction

We will need to rearrange this equation to make force F, the subject, which is:

F o r c e space F space equals space P r e s s u r e space P space cross times space A r e a space A

We know that the hatch is circular with a diameter of 1 m, so the area will be given by the following formula which you may recall from your school studies:

A r e a space A space equals space fraction numerator pi d squared over denominator 4 end fraction space equals space fraction numerator pi cross times 1 squared over denominator 4 end fraction space equals space 0.785 space m squared

Substituting this value for area into the equation for force gives:

F o r c e space F space equals space P A space equals space 1 comma 500 space cross times space 10 cubed space cross times space 0.785

F space equals space 1 comma 177 comma 500 space N space equals space 1.18 space M N

Consider for a second, the size of this force – it is very large. In fact, it would be the equivalent force that a mass of over one million kilograms would exert. This very large force has to be restrained by the tank wall. For this reason, legislation exists relating to the design and operation of such pressure tanks, to ensure that they do not fail as an unintended release of such pressure can be very dangerous.

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