Chapter Ten


Mechanical Properties of Fluids

10.1 Introduction

10.2 Pressure

10.3 Streamline flow

10.4 Bernoulli’s principle

10.5 Viscosity

10.6 Surface tension


Points to ponder


Additional exercises


10.1 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.

10.2 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 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.10.1(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. 10.1(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.


                                                                                                  (a)                                                         (b)

Fig. 10.1 (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.

If  F is the magnitude of this normal force on the piston of area A then the average pressure Pav is defined as the normal force acting per unit area.

2068.png (10.1)

In principle, the piston area can be made arbitrarily small. The pressure is then defined in a limiting sense as

P = 2073.png2078.png (10.2)

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. (10.1) and (10.2). Its dimensions are [ML–1T–2]. The SI unit of pressure is N m–2. It has been named as pascal (Pa) in honour of the French scientist Blaise Pascal (1623-1662) 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 (1 atm = 1.013 × 105 Pa).

Another quantity, that is indispensable in describing fluids, is the density ρ. For a fluid of mass m occupying volume V,

2083.png (10.3)

The dimensions of density are [ML–3]. Its SI unit is kg m–3. 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 4oC (277 K) is 1.0 × 103 kg m–3. The relative density of a substance is the ratio of its density to the density of water at 4oC. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is
2.7 × 103 kg m–3. The densities of some common fluids are displayed in Table 10.1.


Table 10.1 Densities of some common fluids at STP* 

*STP means standard temperature (00C) and 1 atm pressure. 

Example 10.1 The two thigh bones (femurs), each of cross-sectional area10 cm2 support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.

Answer Total cross-sectional area of the femurs is A = 2 × 10 cm2 = 20 × 10–4 m2. The force acting on them is F = 40 kg wt = 400 N (taking g = 10 m s–2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is


10.2.1 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. 10.2 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. 10.2 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 Pa, Pb and Pc on this element of area corresponding to the normal forces Fa, Fb and Fc as shown in Fig. 10.2 on the faces BEFC, ADFC and ADEB denoted by Aa, Ab and Ac respectively. Then

Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium)

Ab sinθ = Ac, Ab cosθ = Aa (by geometry)


2093.png (10.4)

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.

10.2.2 Variation of Pressure with Depth

Consider a fluid at rest in a container. In Fig. 10.3 point 1 is at height h above a point 2. The pressures at points 1 and 2 are P1 and P2 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 (P1A) acting downward, at the bottom (P2A) acting upward. If mg is weight of the fluid in the cylinder we have

(P2 P1) A = mg (10.5)

Now, if ρ is the mass density of the fluid, we have the mass of fluid to be m = ρV= ρhA so that

P2 P1= ρgh (10.6)


Fig.10.3 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 (1 and 2), mass density of the fluid ρ and acceleration due to gravity g. If the point 1 under discussion is shifted to the top of the fluid (say, water), which is open to the atmosphere, P1 may be replaced by atmospheric pressure (Pa) and we replace P2 by P. Then Eq. (10.6) gives

P = Pa + ρgh (10.7)

Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess of pressure, P Pa, 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. (10.7). 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.10.4] 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 10.4 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height. 

Example 10.2 What is the pressure on a swimmer 10 m below the surface of a lake?

Answer Here

h = 10 m and ρ = 1000 kg m-3. Take g = 10 m s–2

From Eq. (10.7)

10.2.3 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 1.013 × 105 Pa (1 atm). Italian scientist Evangelista Torricelli (16081647) 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.10.5 (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=0. The pressure inside the coloumn at Point B must be the same as the pressure at Point C, which is atmospheric pressure, Pa.

Pa = ρgh (10.8)

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 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (10.8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli).

1 torr = 133 Pa.

The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar.

1 bar = 105 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. 10.5 (b)]. The pressure P at A is equal to pressure at point B. What we normally measure is the gauge pressure, which is P Pa, given by Eq. (10.8) and is proportional to manometer height h.


Fig 10.5 (a) The mercury barometer.


(b)The open tube manometer

Fig 10.5 Two pressure measuring devices. 

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 10.3 The density of the atmosphere at sea level is 1.29 kg/m3. Assume that it does not change with altitude. Then how high would the atmosphere extend?

Answer We use Eq. (10.7)

ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa

h = 7989 m 8 km

In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

Example 10.4 At a depth of 1000 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 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3g = 10 m s–2.)

Answer Here h = 1000 m and ρ = 1.03 × 103 kg m-3.

(a) From Eq. (10.6), absolute pressure

P = Pa + ρgh

= 1.01 × 105 Pa

+ 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m

= 104.01 × 105 Pa

104 atm

(b) Gauge pressure is P Pa = ρgh = Pg

Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m

= 103 × 105 Pa

103 atm

(c) The pressure outside the submarine is P = Pa + ρgh and the pressure inside it is Pa. Hence, the net pressure acting on the window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the force acting on it is

F = Pg A = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N

Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 10.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.


Fig 10.6 (a) Whenever external pressure is applied on any part of a fluid in a vessel, it is equally transmitted in all directions.

This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law and it has many applications in daily life.

A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure. In a hydraulic lift, as shown in Fig. 10.6 (b), two pistons are separated by the space filled with a liquid. A piston of small cross-section A1 is used to exert a force F1 directly on the liquid. The pressure P = 2098.png is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A2, which results in an upward force of P × A2. Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) F2 = PA2 = 2103.png. By changing the force at A1, the platform can be moved up or down. Thus, the applied force has been increased by a factor of 2108.png and this factor is the mechanical advantage of the device. The example below clarifies it.

Archemedes’ Principle

Fluid appears to provide partial support to the objects placed in it. When a body is wholly or partially immersed in a fluid at rest, the fluid exerts pressure on the surface of the body in contact with the fluid. The pressure is greater on lower surfaces of the body than on the upper surfaces as pressure in a fluid increases with depth. The resultant of all the forces is an upward force called buoyant force. Suppose that a cylindrical body is immersed in the fluid. The upward force on the bottom of the body is more than the downward force on its top. The fluid exerts a resultant upward force or buoyant force on the body equal to (P2 – P1) × A (Fig. 10.3). We have seen in equation 10.4 that (P2-P1)A = ρghA. Now, hA is the volume of the solid and ρhA is the weight of an equivaliant volume of the fluid. (P2-P1)A = mg. Thus, the upward force exerted is equal to the weight of the displaced fluid.

The result holds true irrespective of the shape of the object and here cylindrical object is considered only for convenience. This is Archimedes’ principle. For totally immersed objects the volume of the fluid displaced by the object is equal to its own volume. If the density of the immersed object is more than that of the fluid, the object will sink as the weight of the body is more than the upward thrust. If the density of the object is less than that of the fluid, it floats in the fluid partially submerged. To calculate the volume submerged, suppose the total volume of the object is Vs and a part Vp of it is submerged in the fluid. Then, the upward force which is the weight of the displaced fluid is ρfgVp, which must equal the weight of the body; ρsgVs = ρfgVpor ρs/ρf = Vp/Vs The apparent weight of the floating body is zero.

This principle can be summarised as; ‘the loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced’.


Fig 10.6 (b) Schematic diagram illustrating the principle behind the hydraulic lift, a device used to lift heavy loads. 

Example 10.5 Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Answer (a) Since pressure is transmitted undiminished throughout the fluid,


= 90 N

(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.



= 0.67 × 10-2 m = 0.67 cm

Note, atmospheric pressure is common to both pistons and has been ignored.

Example 10.6 In a car lift compressed air exerts a force F1 on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F1. What is the pressure necessary to accomplish this task? (g = 9.8 ms-2).

Answer Since pressure is transmitted undiminished throughout the fluid,

= 1470 N

1.5 × 103 N

The air pressure that will produce this force is


This is almost double the atmospheric pressure.

Hydraulic brakes in automobiles also work on the same principle. When we apply a little force on the pedal with our foot the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way, a small force on the pedal produces a large retarding force on the wheel. An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on
all wheels.

Archimedes (287–212 B.C.)


Archimedes was a Greek philosopher, mathematician, scientist and engineer. He invented the catapult and devised a system of pulleys and levers to handle heavyloads. The king of his native city Syracuse, Hiero II, asked him to determine if his gold crown was alloyed with some cheaper metal, such as silver without damaging the crown. The partial loss of weight he experienced while lying in his bathtub suggested a solution to him. According to legend, he ran naked through the streets of Syracuse, exclaiming “Eureka, eureka!”, which means “I have found it, I have found it!”


So far we have studied fluids at rest. The study of the fluids in motion is known as fluid dynamics. When a water tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids, we focus our attention on what is happening to various fluid particles at a particular point in space at a particular time. The flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant in time. This does not mean that the velocity at different points in space is same. The velocity of a particular particle may change as it moves from one point to another. That is, at some other point the particle may have a different velocity, but every other particle which passes the second point behaves exactly as the previous particle that has just passed that point. Each particle follows a smooth path, and the paths of the particles do not cross each other.


Fig.10.7 The meaning of streamlines. (a) A typical trajectory of a fluid particle. (b) A region of streamline flow.

The path taken by a fluid particle under a steady flow is a streamline. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. Consider the path of a particle as shown in Fig.10.7 (a), the curve describes how a fluid particle moves with time. The curve PQ is like a permanent map of fluid flow, indicating how the fluid streams. No two streamlines can cross, for if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time. How do we draw closely spaced streamlines ? If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.10.7 (b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines. This means that number of fluid particles crossing the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points are AP,AR and AQ and speeds of fluid particles are vP, vR and vQ, then mass of fluid mP crossing at AP in a small interval of time t is ρPAPvP t. Similarly mass of fluid mR flowing or crossing at AR in a small interval of time t is ρRARvR t and mass of fluid mQ is ρQAQvQ t crossing at AQ. The mass of liquid flowing out equals the mass flowing in, holds in all cases. Therefore,

ρPAPvPt = ρRARvRt = ρQAQvQt (10.9)

For flow of incompressible fluids

ρP = ρR = ρQ

Equation (10.9) reduces to

APvP = ARvR = AQvQ (10.10)

which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids. In general

Av = constant (10.11)

Av gives the volume flux or flow rate and remains constant throughout the pipe of flow. Thus, at narrower portions where the streamlines are closely spaced, velocity increases and its vice versa. From (Fig.10.7b) it is clear that AR > AQ or vR < vQ, the fluid is accelerated while passing from R to Q. This is associated with a change in pressure in fluid flow in horizontal pipes.

Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent. One sees this when a fast flowing stream encounters rocks, small foamy whirlpool-like regions called ‘white water rapids are formed.

Figure 10.8 displays streamlines for some typical flows. For example, Fig. 10.8(a) describes a laminar flow where the velocities at different points in the fluid may have different magnitudes but their directions are parallel. Figure 10.8 (b) gives a sketch of turbulent flow.


Fig.10.8 (a) Some streamlines for fluid flow. (b) A jet of air striking a flat plate placed perpendicular to it. This is an example of turbulent flow. 


Fluid flow is a complex phenomenon. But we can obtain some useful properties for steady or streamline flows using the conservation of energy.

Consider a fluid moving in a pipe of varying cross-sectional area. Let the pipe be at varying heights as shown in Fig. 10.9. We now suppose that an incompressible fluid is flowing through the pipe in a steady flow. Its velocity must change as a consequence of equation of continuity. A force is required to produce this acceleration, which is caused by the fluid surrounding it, the pressure must be different in different regions. Bernoulli’s equation is a general expression that relates the pressure difference between two points in a pipe to both velocity changes (kinetic energy change) and elevation (height) changes (potential energy change). The Swiss Physicist Daniel Bernoulli developed this relationship in 1738.

Consider the flow at two regions 1 (i.e., BC) and 2 (i.e., DE). Consider the fluid initially lying between B and D. In an infinitesimal time interval t, this fluid would have moved. Suppose v1 is the speed at B and v2 at D, then fluid initially at B has moved a distance v1t to C (v1t is small enough to assume constant cross-section along BC). In the same interval t the fluid initially at D moves to E, a distance equal to v2t. Pressures P1 and P2 act as shown on the plane faces of areas A1 and A2 binding the two regions. The work done on the fluid at left end (BC) is W1 = P1A1(v1t) = P1V. Since the same volume V passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is W2 = P2A2(v2t) = P2V or, the work done on the fluid is P2V. So the total work done on the fluid is

W1 W2 = (P1 P2) V

Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing