MECHANICAL PROPERTIES OF FLUIDS

M ECHANICAL P ROPERTIES OF F LUIDS . INTRODUCTION In this chapter, we shall study some common physical properties of liquids and gases. Liquids and gases can flow and are therefore, called fluids. It is this property that distinguishes liquids and gases from solids in a basic way.

Fluids are everywhere around us. Earth has an envelop of air and two-thirds of its surface is covered with water. Water is not only necessary for our existence; every mammalian body constitute mostly of water. All the processes occurring in living beings including plants are mediated by fluids.

Thus understanding the behaviour and properties of fluids is important. How are fluids different from solids? What is common in liquids and gases? Unlike a solid, a fluid has no definite shape of its own.

Solids and liquids have a fixed volume, whereas a gas fills the entire volume of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure.

The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words solids and liquids have much lower compressibility as compared to gases. Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress.

The shearing stress of fluids is about million times smaller than that of solids. . PRESSURE A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force.

If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose . Introduction . Pressure .

Streamline flow . Bernoulli’s principle . Viscosity . Surface tension Summary Points to ponder Exercises Additional exercises Appendix chest a large, light but strong wooden plank is placed first, is saved from this accident.

Such everyday experiences convince us that both the force and its coverage area are important. Smaller the area on which the force acts, greater is the impact. This impact is known as pressure. When an object is submerged in a fluid at rest, the fluid exerts a force on its surface.

This force is always normal to the object’s surface. This is so because if there were a component of force parallel to the surface, the object will also exert a force on the fluid parallel to it; as a consequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot happen.

Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Fig. . (a).

The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressure-measuring device is shown in Fig. . (b).

It consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston. This device is placed at a point inside the fluid. The inward force exerted by the fluid on the piston is balanced by the outward spring force and is thereby measured. If F is the magnitude of this normal force on the piston of area A then the average pressure P av is defined as the normal force acting per unit area.

av = ( . ) In principle, the piston area can be made arbitrarily small. The pressure is then defined in a limiting sense as P = lim ∆ A → ∆ ∆ ( . ) Pressure is a scalar quantity.

We remind the reader that it is the component of the force normal to the area under consideration and not the (vector) force that appears in the numerator in Eqs. ( . ) and ( . ).

Its dimensions are [ML – T – ]. The SI unit of pressure is N m – . It has been named as pascal (Pa) in honour of the French scientist Blaise Pascal ( - ) who carried out pioneering studies on fluid pressure. A common unit of pressure is the atmosphere (atm), i.e.

the pressure exerted by the atmosphere at sea level ( atm = . × Pa). Another quantity, that is indispensable in describing fluids, is the density ρ . For a fluid of mass m occupying volume V , ρ = m V ( .

) The dimensions of density are [ML – ]. Its SI unit is kg m – . It is a positive scalar quantity. A liquid is largely incompressible and its density is therefore, nearly constant at all pressures.

Gases, on the other hand exhibit a large variation in densities with pressure. The density of water at o C ( K) is . × kg m – . The relative density of a substance is the ratio of its density to the density of water at o C.

It is a dimensionless positive scalar quantity. For example the relative density of aluminium is . . Its density is .

× kg m – . The densities of some common fluids are displayed in Table . . Table .

Densities of some common fluids at STP* Fig. . (a) The force exerted by the liquid in the beaker on the submerged object or on the walls is normal (perpendicular) to the surface at all points. (b) An idealised device for measuring pressure.

* STP means standard temperature ( C) and atm pressure. Example . The two thigh bones (femurs), each of cross-sectional area10 cm support the upper part of a human body of mass kg. Estimate the average pressure sustained by the femurs.

Answer Total cross-sectional area of the femurs is A = × cm = × – m . The force acting on them is F = kg wt = N (taking g = m s – ). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is m N − × = A P av .

. Pascal’s Law The French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points if they are at the same height. This fact may be demonstrated in a simple way. Fig.

. shows an element in the interior of a fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points.

But for clarity we have enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element as discussed above. Thus, the fluid exerts pressures P a , P b and P c on this element of area corresponding to the normal forces F a , F b and F c as shown in Fig. .

on the faces BEFC, ADFC and ADEB denoted by A a , A b and A c respectively. Then F b sin θ = F c , F b cos θ = F a (by equilibrium) A b sin θ = A c , A b cos θ = A a (by geometry) Thus, ; b c a b c a b c a ( . ) Hence, pressure exerted is same in all directions in a fluid at rest. It again reminds us that like other types of stress, pressure is not a vector quantity.

No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area. Now consider a fluid element in the form of a horizontal bar of uniform cross-section. The bar is in equilibrium.

The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere in a horizontal plane.

. . Variation of Pressure with Depth Consider a fluid at rest in a container. In Fig.

. point is at height h above a point . The pressures at points and are P and P respectively. Consider a cylindrical element of fluid having area of base A and height h .

As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top ( P A ) acting downward, at the bottom ( P A ) acting upward. If mg is weight of the fluid in the cylinder we have ( P − P ) A = mg ( . ) Now, if ρ is the mass density of the fluid, we have the mass of fluid to be m = ρ V = ρ h A so that P − P = ρ gh ( .

) Fig. . Proof of Pascal’s law. ABC-DEF is an element of the interior of a fluid at rest.

This element is in the form of a right- angled prism. The element is small so that the effect of gravity can be ignored, but it has been enlarged for the sake of clarity. Fig. .

Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical cylindrical column. Pressure difference depends on the vertical distance h between the points ( and ), mass density of the fluid ρ and acceleration due to gravity g . If the point under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, P may be replaced by atmospheric pressure (P a ) and we replace P by P.

Then Eq. ( . ) gives P = P a + ρ gh ( . ) Thus, the pressure P , at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρ gh .

The excess of pressure, P − P a , at depth h is called a gauge pressure at that point. The area of the cylinder is not appearing in the expression of absolute pressure in Eq. ( . ).

Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth). The result is appreciated through the example of hydrostatic paradox . Consider three vessels A, B and C [Fig.

. ] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same, though they hold different amounts of water.

This is so because water at the bottom has the same pressure below each section of the vessel. Fig . Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height.

Example . What is the pressure on a swimmer m below the surface of a lake? Answer Here h = m and ρ = kg m - . Take g = m s – From Eq.

( . ) P = P a + ρ gh = . × Pa + kg m – × m s – × m = . × Pa ≈ atm This is a % increase in pressure from surface level.

At a depth of km, the increase in pressure is atm! Submarines are designed to withstand such enormous pressures. ⊳ . .

Atmospheric Pressure and Gauge Pressure The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. At sea level, it is . × Pa ( atm). Italian scientist Evangelista Torricelli ( – ) devised for the first time a method for measuring atmospheric pressure.

A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury as shown in Fig. . (a). This device is known as ‘mercury barometer’.

The space above the mercury column in the tube contains only mercury vapour whose pressure P is so small that it may be neglected. Thus, the pressure at Point A= . The pressure inside the coloumn at Point B must be the same as the pressure at Point C, which is atmospheric pressure, P a . P a = ρ gh ( .

) where ρ is the density of mercury and h is the height of the mercury column in the tube. In the experiment it is found that the mercury column in the barometer has a height of about cm at sea level equivalent to one atmosphere ( atm). This can also be obtained using the value of ρ in Eq. ( .

). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of mm is called a torr (after Torricelli). torr = Pa.

The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar. bar = Pa An open tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low density liquid (such as oil) for measuring small pressure differences and a high density liquid (such as mercury) for large pressure differences.

One end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure [see Fig. . (b)]. The pressure P at A is equal to pressure at point B.

What we normally measure is the gauge pressure, which is P − P a , given by Eq. ( . ) and is proportional to manometer height h . Pressure is same at the same level on both sides of the U-tube containing a fluid.

For liquids, the density varies very little over wide ranges in pressure and temperature and we can treat it safely as a constant for our present purposes. Gases on the other hand, exhibits large variations of densities with changes in pressure and temperature. Unlike gases, liquids are, therefore, largely treated as incompressible. Example .

The density of the atmosphere at sea level is . kg/m . Assume that it does not change with altitude. Then how high would the atmosphere extend?

Answer We use Eq. ( . ) ρ gh = . kg m – × .

m s × h m = . × Pa ∴ h = m ≈ km In reality the density of air decreases with height. So does the value of g . The atmospheric cover extends with decreasing pressure over km.

We should also note that the sea level atmospheric pressure is not always mm of Hg. A drop in the Hg level by mm or more is a sign of an approaching storm. Example . At a depth of m in an ocean (a) what is the absolute pressure?

(b) What is the gauge pressure? (c) Find the force acting on the window of area cm × cm of a submarine at this depth, the interior of which is maintained at sea- level atmospheric pressure. (The density of sea water is . × kg m - , g = m s – .) (b) The open tube manometer Fig .

Two pressure measuring devices. Fig . (a) The mercury barometer. Answer Here h = m and ρ = .

× kg m - . From Eq. ( . ), absolute pressure P = P a + ρ gh = .