__Pressure in Fluids__

__Pressure in Fluids__

__Pressure__

__Pressure__

The **pressure **acting on a surface is defined as the perpendicular force per unit area of surface. The unit of pressure is the **pascal**, **Pa**, where 1 pascal is equal to 1 newton per square metre. Thus

where F is the force in newtons acting at right angles to a surface of area A square metres.

**For example**, when a force of 20 N acts uniformly over, and perpendicular to, an area of 4 m2, then the pressure on the area, p, is given by:

**Fluid Pressure**

A fluid is either a liquid or a gas and there are four basic factors governing the pressure within fluids.

(a) The pressure, at a given depth, in a fluid is equal in all directions, as shown in Figure 32.1(a).

(b) The pressure at a given depth in a fluid is independent of the shape of the container in which the fluid is held. In Figure 32.1(b), the pressure at X is the same as the pressure at Y.

(c) Pressure acts at right angles to the surface containing the fluid. In Figure 32.1(c), the pressures at points A to F all act at right angles to the container.

(d) When a pressure is applied to a fluid, this pressure is transmitted equally in all directions. In Figure 32.1(d), if the mass of the fluid is neglected, the pressures at points A to D are all the same.

The pressure, p, at any point in a fluid depends on three factors:

(i) the density of the fluid, p, in kg/m3

(ii) the gravitational acceleration, g, taken as approximately 9.81 m/s2 (or the gravitational field force in N/kg), and

(iii) the height of fluid vertically above the point, h metres.

The relationship connecting these quantities is:

**For example**, when the container shown in Figure 32.2 is filled with water of density 1000 kg/m3 , the pressure due to the water at a depth of 0.03 m below the surface is given by: ** p **= pgh = (1000 x 9.81 x 0.03) =

**294**.

**3 Pa**

#### Atmospheric Pressure

The air above the earth’s surface is a fluid, having a density, p, which varies from approximately 1.225 kg/m3 at sea level to zero in outer space. Since p = pgh, where height h is several thousands of metres, the air exerts a pressure on all points on the earth’s surface. This pressure, called **atmospheri****c pressure**, has a value of approximately 101 kilopascals. Two terms are commonly used when measuring pressures:

(a) **absolut****e pressure**, meaning the pressure above that of an absolute vacuum (which is zero pressure), and

(b) **gaug****e pressure**, meaning the pressure above that normally present due to the atmosphere.

Thus: absolute pressure = atmospheric pressure + gauge pressure

Thus, a gauge pressure of 50 kPa is equivalent to an absolute pressure of (100 C 50) kPa, i.e. 150 kPa, since the atmospheric pressure is approximately 100 kPa.

**For example**, the absolute pressure at a point on a submarine, at a depth of 30 m below the surface of the sea, when the atmospheric pressure is 101 kPa (taking the density of sea water as 1030 kg/m3 and the gravitational acceleration as 9.81 m/s2) is calculated as follows:

The pressure due to the sea, that is, the gauge pressure (pg ) is given by:

**Archimedes’ Principle**

Archimedes’ principle states that:

*I**f a solid body floats, or is submerged, in a liquid, the liquid exerts an upthrust on the body equal to the gravitational force on the liquid displaced by the body*

In other words, if a solid body is immersed in a liquid, the apparent loss of weight is equal to the weight of liquid displaced. If V is the volume of the body below the surface of the liquid, then the apparent loss of weight W is given by:

where ω is the specific weight (i.e. weight per unit volume) and p is the density. If a body floats on the surface of a liquid all of its weight appears to have been lost. The weight of liquid displaced is equal to the weight of the floating body.

**For example**, a body weighs 2.760 N in air and 1.925 N when completely immersed in water of density 1000 kg/m3 .

The apparent loss of weight is: 2.760 N - 1.925 N D 0.835 N

This is the weight of water displaced, i.e. Vpg, where V is the volume of the body and p is the density of water.

Measurement of Pressure

There are various ways of measuring pressure, and these include by:

(a) barometers,

(b) manometers,

(c) pressure gauges, and

(d) vacuum gauges.

The construction and principle of operation of each of these devices are described in Chapter 33.

Let us look briefly at just one of these instruments — the manometer. A manometer is a device used for measuring relatively small pressures, either above or below atmospheric pressure. A simple U-tube manometer is shown in Figure 32.3. Pressure p acting in, say, a gas main, pushes the liquid in the

U-tube until equilibrium is obtained. At equilibrium:

The gauge pressure of the gas is 2.94 kPa.

By filling the U-tube with a more dense liquid, say, mercury having a density of 13 600 kg/m3 , for a given height of U-tube, the pressure which can be measured is increased by a factor of 13.6.

#### More on Hydrostatic Pressure

The pressure p at the base of the tank shown in Figure 32.4(a) is:

where w is the specific weight, i.e. the weight per unit volume, its unit being N/m3.

The pressure increases to this value uniformly from zero at the free surface.

The pressure variation is shown in Figure 32.4(b).

At any intermediate depth x the pressure is: ** **It may be shown that the average pressure on any wetted plane surface is the pressure at the centroid, the centre of area. The sloping sides of the tank

For a tank with vertical sides this is the weight of liquid in the tank.

In any vessel containing homogeneous liquid at rest and in continuous contact, the pressure must be the same at all points at the same level. In a U-tube, as shown in Figure 32.5, with the liquid in the lower part at rest, the pressure must be the same on both sides for all levels up to X1X2. The pressure at X1, however, is greater than the pressure at Y1 by the amount p1 D w1 h D p1gh, where w1 , p1 are the specific weight and density respectively of the liquid, or gas, between X1 and Y1 .

Similarly, the pressure at X2 is greater than the pressure at Y2 by an amount given by p2 D w2h D p2gh, where w2 and p2 are the specific weight and density respectively of the liquid in the bottom of the U-tube.

For practical reasons, p2 must be greater than p1 and the pressure at Y1 will exceed that at Y2 by: p2 - p1 D (w2 - w1)h D (p2 - p1)gh If the upper limits of the U-tube contain air or any other gas or gas mixture, w1 and p1 can reasonably be ignored, giving: p2 - p1 D w2 h D p2 gh

If the upper limbs contain a lighter liquid, then the pressure difference may be expressed as:

A common arrangement is mercury and water, in which case d is the relative density of mercury, approximately 13.6. This gives: p2 - p1 D 12.6pgh D 12.6wh, p and w being respectively the density and specific weight of water. The pressure difference at Z1Z2 will be the same as Y1 Y2 if both limbs contain the same liquid between these levels. This follows from the fact that the pressure increase from Z1 to Y1 is the same as the increase from Z2 to Y2.

**For example**, the pressure difference between two points on a horizontal pipe carrying water and running full is to be indicated on a U-tube below the pipe. The bottom of the U-tube contains mercury; the rest is filled with water. Taking the density of water as 1000 kg/m3 and the relative density of mercury as 13.6, the pressure difference indicated when the difference in the mercury levels in the U-tube is 310 mm is determined as follows:

The pressure difference at level XX0 in Figure 32.6 is the same as the pressure difference between the two points in the pipe.

At level YY0 , however, there is no pressure difference, as below this there is a U-tube of mercury at rest. The difference in pressure at XX0 and hence between the points on the pipe is therefore the difference in pressure between that due to a column of mercury 310 mm high (i.e. X0Y0) and that due to a column of water 310 mm (i.e. XY). The pressure difference is therefore: