Chapter 2: Mechanical Properties of Fluids
In the early 1640s, the pump-makers of Florence were facing a problem they could not explain. Their suction pumps, which were used to draw water out of deep wells, simply refused to lift water beyond a height of about ten metres. It did not matter how strong the pump was or how carefully it was sealed — the water always stopped rising at nearly the same height.
The puzzle reached Evangelista Torricelli, an Italian scientist who had worked with Galileo. Torricelli suspected that the pump was not “pulling” the water at all. Instead, he thought the air above the well was pushing the water up, and that this push had a limit.
To test the idea, he needed a liquid much heavier than water. In 1643 he took mercury, which is about 13.6 times denser than water, filled a glass tube nearly a metre long with it, and turned the tube upside down into a dish of mercury. Some mercury ran out — and then the column stopped, standing steadily at a height of about 76 cm, with an empty space left above it.
That short column of mercury was being held up by the weight of the entire atmosphere pressing on the dish below. Without meaning to, Torricelli had built the first barometer, and had shown that a fluid at rest can exert pressure.
Diagram
Fig. 2.0 – Torricelli’s inverted glass tube standing in a dish of mercury, with the 76 cm mercury column and the empty space above it labelled.
Each of these questions is answered somewhere in this chapter. Keep them in mind as you read, and check them again when you reach the end.
2.1 Introduction
In Class XI, you studied how solids behave when a force is applied to them. You saw that a solid can be stretched, compressed or twisted, but it tries to return to its original form once the force is removed.
Matter around us exists in three common states — solid, liquid and gas. Among these, a solid nearly maintains its fixed shape and its fixed volume even when a large force is applied to it. A steel rod pressed hard may bend very slightly, but it does not spread out or run away from your hand.
Liquids and gases behave quite differently. They do not have a shape of their own. Instead, they take the shape of the vessel that contains them. Water poured into a round bottle becomes round; the same water in a square jar becomes square. As shown in Fig. 2.0a, only a solid keeps its own shape.
Diagram
Fig. 2.0a – Three containers of different shapes holding the same solid block, the same volume of liquid, and a gas, showing that only the solid keeps its own shape.
Because of this property, liquids and gases begin to move and flow whenever an external force acts on them. This ability to flow is what gives us the word fluid.
Why Fluids Cannot Hold Their Shape
The reason lies in a property called rigidity. Rigidity is the ability of a material to resist a change in its shape. A solid has high rigidity, which is why it resists being bent or twisted.
A fluid either has no rigidity at all, or its rigidity is extremely low. So even a small sideways force is enough to make the layers of a fluid slide over one another, and the fluid simply flows away instead of resisting.
What This Chapter Will Study
In our daily life, we constantly experience the pressure exerted by a fluid, both when the fluid is at rest and when it is in motion. Pressure is the effect of a fluid pushing on a surface, and you feel it in your ears when you dive to the bottom of a pool.
Two more properties of fluids play an important role in nature. Viscosity is the internal friction that makes some fluids pour sluggishly, and surface tension is the property that makes a liquid surface behave like a stretched skin. Both will be studied carefully in the sections ahead.
In this chapter we will study these properties of fluids one by one — first for fluids at rest, and then for fluids in motion.
2.2 Fluid
Any substance that can flow is a fluid. We can now state this idea more sharply, in the way a physicist would: a fluid is a substance that deforms continually under the action of an external force.
The word continually is the important one. Push a solid block sideways and it deforms a little and then stops. Push a fluid sideways and it keeps on deforming for as long as the force acts. It never settles into a new fixed shape.
Fluid is a phase of matter that includes liquids, gases and plasmas.
In this chapter we shall discuss the mechanical properties of only liquids and gases.
Why a Fluid Cannot Resist a Shear Force
When a force is applied along a surface rather than perpendicular to it, the stress produced is called shear stress or tangential stress. In Class XI you learnt that the shear modulus of a material measures how strongly it resists such a stress.
For a fluid, the shear modulus is zero. In simpler words, fluids are substances which cannot resist any shear force applied to them. However small the sideways force, the layers of the fluid slide over one another and the fluid flows.
Air, water, flour dough, toothpaste, etc., are some common examples of fluids. Molten lava is also a fluid.
The Ideal Fluid
A fluid flows under the action of a force or a pressure gradient — that is, whenever the pressure is higher at one place than at another. The behaviour of a fluid in motion is normally complicated, so we can understand fluids by making some simple assumptions. For this purpose we introduce the concept of an ideal fluid.
An ideal fluid has the following properties:
- It is incompressible: its density is constant. Squeezing it harder does not pack more mass into the same volume.
- Its flow is irrotational: its flow is smooth, there are no turbulences in the flow.
- It is nonviscous: there is no internal friction in the flow, i.e., the fluid has no viscosity. (Viscosity is discussed in section 2.6.1.)
- Its flow is steady: its velocity at each point is constant in time. The speed may differ from place to place, but at any one chosen point it does not change with time.
Comparing Stresses on Solids and on Fluids
It is important to understand the difference between a solid and a fluid. Solids can be subjected to shear stress (tangential stress) as shown in Fig. 2.1, where the top face of a block is pushed sideways while the bottom face is held fixed.
Diagram
Fig. 2.1 – A rectangular block ABCD with its top edge pushed sideways through a small distance by a force F, showing the shear angle.
Solids can also be subjected to normal stress, in which the force acts perpendicular to the surface, as shown in Fig. 2.2. Normal stress is of two kinds — compressive stress, which squeezes the body inward, and tensile stress, which stretches it outward.
Diagram
Fig. 2.2 – Two blocks: (a) compressive, with forces pushing inward on all faces, and (b) tensile, with forces pulling outward.
Solids oppose the shear stress either by developing a restoring force, which means that the deformations are reversible, or they require a certain initial stress before they deform and start flowing. We have studied this behaviour of solids, that is, their elastic behaviour, in Class XI.
Ideal fluids, on the other hand, can only be subjected to normal, compressive stress. This compressive stress is what we call pressure. An ideal fluid cannot be sheared and cannot be stretched — it can only be squeezed.
Most fluids offer a very weak resistance to deformation. Real fluids display viscosity and so are capable of being subjected to low levels of shear stress. This is a small resistance, not a true opposition, and it never stops the flow the way a solid stops a deformation.
Forces on a Surface Inside a Fluid at Rest
Consider Fig. 2.3, which shows a small surface of area \(dA\) at rest within a fluid.
Diagram
Fig. 2.3 – A small flat surface of area dA inside a fluid at rest, with equal and opposite normal forces dF pressing on its two faces.
The surface does not accelerate, so by Newton’s first law the forces on it must balance. The surrounding fluid therefore exerts equal normal forces \(dF\) on both sides of it. Notice that these forces are normal, that is, perpendicular to the surface — a fluid at rest cannot push along the surface, because that would be a shear force.
Properties of Fluids
- They do not oppose deformation, they get permanently deformed.
- They have ability to flow.
- They have ability to take the shape of the container.
A fluid exhibits these properties because it cannot oppose a shear stress when in static equilibrium.
2.2.1 Fluids at Rest
A fluid can be studied in two situations — when it is standing still, and when it is flowing. We begin with the simpler of the two.
The branch of physics which deals with the properties of fluids at rest is called hydrostatics.
Here, “at rest” means that the fluid as a whole is in static equilibrium — no part of it is flowing. In the next few sections we will consider some of the properties of fluids at rest.
2.3 Pressure
A fluid at rest exerts a force on the surface of contact. The surface may be a wall or the bottom of an open container of the fluid. Water standing in a tank, for instance, pushes outward on the side walls and downward on the floor of the tank at the same time.
We saw in the previous section that a fluid at rest can only push perpendicular to a surface. So this force is always a normal force, whatever the tilt of the surface.
Figure 2.4 shows a fluid exerting normal forces on a vertical surface.
Diagram
Fig. 2.4 – A vertical surface of area A inside a fluid, with equal horizontal forces F pressing on its two faces.
Figure 2.5 shows fluid exerting normal forces on a horizontal surface.
Diagram
Fig. 2.5 – A horizontal surface of area A inside a fluid, with equal vertical forces F pressing on it from above and below.
The normal force (\(F\)) exerted by a fluid at rest per unit surface area (\(A\)) of contact is called the pressure (\(p\)) of the fluid.
\[P = \frac{F}{A} \qquad \text{--- (2.1)}\]
Here \(F\) is the normal force in newton (N), \(A\) is the area of contact in square metre (m\(^2\)), and \(P\) is the pressure in newton per square metre (N m\(^{-2}\)).
Why a Small Area Means a Large Pressure
Equation (2.1) tells us that pressure depends on two things — how large the force is, and how much area that force is spread over. The same force concentrated on a smaller area produces a larger pressure.
Thus, an object having small weight can exert high pressure if its weight acts on a small surface area.
For example, a force of 10 N acting on 1 cm\(^2\) results in a pressure of 10\(^5\) N m\(^{-2}\). On the other hand, the same force of 10 N while acting on an area of 1 m\(^2\), exerts a pressure of only 10 N m\(^{-2}\). The force has not changed at all — only the area has, and the pressure has fallen by a factor of ten thousand.
Units and Nature of Pressure
The SI unit of pressure is N/m\(^2\). Also, 1 N/m\(^2\) = 1 pascal (Pa). The dimension of pressure is \([L^{-1}M^{1}T^{-2}]\).
Pressure is a scalar quantity. It has magnitude but no direction of its own.
Other common units of measuring pressure of a gas are bar and torr. One torr is one mm of mercury column.
\[1 \text{ bar} = 10^{5} \text{ N m}^{-2}\]
\[1 \text{ hectapascal (hPa)} = 100 \text{ Pa}\]
The hectapascal is the unit you see in weather reports, where atmospheric pressure at a place is usually quoted in hPa.
2.3.1 Pressure Due to a Liquid Column
A vessel is filled with a liquid. Let us calculate the pressure exerted by an imaginary cylinder of cross sectional area \(A\) inside the container.
The cylinder is only imagined — no actual tube is placed in the liquid. We simply pick out a cylindrical portion of the liquid in our mind so that we can apply the definition of pressure to it.
Let the density of the fluid be \(\rho\), and the height of the imaginary cylinder be \(h\) as shown in Fig. 2.6. The top of this cylinder lies at the free surface of the liquid, that is, the open surface in contact with the air.
Diagram
Fig. 2.6 – A vessel of water with an imaginary vertical cylinder of cross-sectional area A and height h marked inside it, its top at the free surface.
The liquid column exerts a force \(F = mg\), which is its weight, on the bottom of the cylinder. This force acts in the downward direction. Therefore, the pressure \(p\) exerted by the liquid column on the bottom of the cylinder is,
\[P = \frac{F}{A}\]
\[\therefore \; P = \frac{mg}{A}\]
Now, \(m\) = (volume of cylinder) \(\times\) (density of liquid)
\[m = (Ah) \times \rho = Ah\rho\]
Substituting this value of \(m\),
\[\therefore \; P = \frac{(Ah\rho)g}{A}\]
\[p = h\rho g \qquad \text{--- (2.2)}\]
Thus, the pressure \(p\) due to a liquid of density \(\rho\) at rest, and at a depth \(h\) below the free surface is \(h\rho g\).
In this expression \(h\) is the depth below the free surface in metre (m), \(\rho\) is the density of the liquid in kilogram per cubic metre (kg m\(^{-3}\)), \(g\) is the acceleration due to gravity in m s\(^{-2}\), and \(p\) is the pressure in N m\(^{-2}\).
Note that the pressure does not depend on the area of the imaginary cylinder used to derive the expression. The area \(A\) appeared in the numerator as well as in the denominator, and cancelled out.
2.3.2 Atmospheric Pressure
Earth’s atmosphere is made up of a fluid, namely, air. Like any other fluid, air has weight, and it exerts a downward force due to this weight. The pressure due to this force is called atmospheric pressure.
Thus, at any point, the atmospheric pressure is the weight of a column of air per unit cross section starting from that point and extending to the top of the atmosphere. Imagine a tall, thin column of air of unit cross-sectional area standing on your shoulder and reaching all the way up to where the atmosphere ends, as shown in Fig. 2.6a. The weight of that entire column is what presses on you.
Diagram
Fig. 2.6a – A column of air of unit cross-sectional area standing on a point at sea level and extending upward to the top of the atmosphere, with the air becoming thinner higher up.
Why Atmospheric Pressure Falls with Height
Clearly, the atmospheric pressure is highest at the surface of the Earth, i.e., at the sea level, and decreases as we go above the surface, as the height of the column of air above decreases. Standing on a mountain top, you have less air above your head than a person at the seashore, so less weight presses down on you.
The atmospheric pressure at sea level is called normal atmospheric pressure.
There is a second reason for the fall in pressure. The density of air in the atmosphere decreases with increase in height and becomes negligible beyond a height of about 8 km. So the higher part of the column contributes very little weight, and the height of the air column producing atmospheric pressure at sea level can be taken to be 8 km.
This is why we cannot apply \(p = h\rho g\) to the atmosphere as simply as we did to water. Water has practically the same density from top to bottom, but air does not — it is thickest near the ground and thins out steadily above.
Vacuum
The region where gas pressure is less than the atmospheric pressure is called vacuum.
Perfect or absolute vacuum is when no matter, i.e., no atoms or molecules are present.
Perfect vacuum is an ideal that is never fully achieved in the laboratory. Usually, vacuum refers to conditions when the gas pressure is considerably smaller than the atmospheric pressure.
2.3.3 Absolute Pressure and Gauge Pressure
In Section 2.3.1 we found the pressure produced by a liquid column alone. But a liquid in an open vessel also has the atmosphere pressing on its surface. We now build a more complete picture that includes both.
Consider a tank filled with water as shown in Fig. 2.7. Assume an imaginary cylinder of horizontal base area \(A\) and height \(x_1 - x_2 = h\), where \(x_1\) and \(x_2\) are heights measured from a reference point, height increasing upwards, so that \(x_1 > x_2\).
Diagram
Fig. 2.7 – A tank of water with an imaginary cylinder of base area A inside it; the top face at height x₁ and the bottom face at height x₂, with forces F₁ downward on the top, F₂ upward on the bottom and weight mg at the centre.
The vertical forces acting on the cylinder are:
- Force \(F_1\) acts downwards at the top surface of the cylinder, and is due to the weight of the water column above the cylinder.
- Force \(F_2\) acts upwards at the bottom surface of the cylinder, and is due to the water below the cylinder.
- The gravitational force on the water enclosed in the cylinder is \(mg\), where \(m\) is the mass of the water in the cylinder.
As the water is in static equilibrium, the forces on the cylinder are balanced. The upward force must therefore support both the downward force from above and the weight of the water inside. The balance of these forces in magnitude is written as,
\[F_2 = F_1 + mg \qquad \text{--- (2.3)}\]
Let \(p_1\) and \(p_2\) be the pressures at the top and bottom surfaces of the cylinder respectively, due to the fluid. Using Eq. (2.1) we can write
\[F_1 = p_1 A, \quad \text{and} \quad F_2 = p_2 A \qquad \text{--- (2.4)}\]
Also, the mass \(m\) of the water in the cylinder can be written as,
\[m = \text{density} \times \text{volume} = \rho V\]
The volume of the cylinder is its base area multiplied by its height, that is, \(A(x_1 - x_2)\). Therefore,
\[\therefore \; m = \rho A (x_1 - x_2) \qquad \text{--- (2.5)}\]
Substituting Eq. (2.4) and Eq. (2.5) in Eq. (2.3) we get,
\[p_2 A = p_1 A + \rho A g (x_1 - x_2)\]
Dividing throughout by the area \(A\),
\[p_2 = p_1 + \rho g (x_1 - x_2) \qquad \text{--- (2.6)}\]
This equation can be used to find the pressure inside a liquid, as a function of depth below the liquid surface, and also the atmospheric pressure, as a function of altitude or height above the sea level.
Absolute Pressure at a Depth
To find the pressure \(p\) at a depth \(h\) below the liquid surface, let the top of an imaginary cylinder be at the surface of the liquid. Let this level be \(x_1\). Let \(x_2\) be some point at depth \(h\) below the surface as shown in Fig. 2.8.
Diagram
Fig. 2.8 – A vessel of liquid with the free surface marked as the level x₁ = 0 and a point at depth h below it marked as x₂ = −h.
Let \(p_0\) be the atmospheric pressure at the surface, i.e., at \(x_1\).
The sign convention here is exam-critical. Heights are measured upwards from the chosen reference, so a point below the surface has a negative height. Then, substituting \(x_1 = 0\), \(p_1 = p_0\), \(x_2 = -h\), and \(p_2 = p\) in Eq. (2.6) we get,
\[p = p_0 + h\rho g \qquad \text{--- (2.7)}\]
The above equation gives the total pressure, or the absolute pressure \(p\), at a depth \(h\) below the surface of the liquid.
The total pressure \(p\), at the depth \(h\) is the sum of:
- \(p_0\), the pressure due to the atmosphere, which acts on the surface of the liquid, and
- \(h\rho g\), the pressure due to the liquid at depth \(h\).
Gauge Pressure
In general, the difference between the absolute pressure and the atmospheric pressure is called the gauge pressure. Using Eq. (2.7), gauge pressure at depth \(h\) below the liquid surface can be written as,
\[p - p_0 = h\rho g \qquad \text{--- (2.8)}\]
So the gauge pressure at a depth \(h\) is exactly the pressure produced by the liquid column alone, which is the quantity we calculated in Section 2.3.1.
Pressure Above a Liquid Surface
Eq. (2.8) is also applicable to levels above the liquid surface. It gives the pressure at a given height above a liquid surface, in terms of the atmospheric pressure \(p_0\), assuming that the atmospheric density is uniform up to that height.
To find the atmospheric pressure at a distance \(d\) above the liquid surface as shown in Fig. 2.9, we substitute \(x_1 = d\), \(p_1 = p\), \(x_2 = 0\), \(p_2 = p_0\) and \(\rho = \rho_{air}\) in Eq. (2.6) and we get,
\[p = p_{o} - d\rho_{air} g \qquad \text{--- (2.9)}\]
Diagram
Fig. 2.9 – A vessel of water with its free surface at the level x₂ = 0 where the pressure is p₀, and a point in the air at height x₁ = d above it where the pressure is p.
2.3.4 Hydrostatic Paradox
Consider the interconnected vessels as shown in Fig. 2.10 (a). These vessels have very different shapes and widths, but they are all joined to one another at the bottom.
Diagram
Fig. 2.10 (a) – Four interconnected vessels A, B, C and D of different shapes and widths joined at the base, with the liquid standing at the same level in all of them.
When a liquid is poured in any one of the vessels, it is noticed that the level of the liquid in all the vessels is the same. This observation is somewhat puzzling. It was called the ‘hydrostatics paradox’ before the principle of hydrostatics were completely understood.
Why It Seems Puzzling
One can feel that the pressure at the base of the vessel C would be more than that at the base of the vessel B, because vessel C holds much more liquid and therefore much more weight. If that were true, the liquid from vessel C would rise into the vessel B until the extra pressure was relieved.
However, it is never observed. The levels simply stay equal.
The Resolution
Equation (2.2) tells us that the pressure at a point depends only on the height of the liquid column above it. It does not depend on the shape of the vessel, and it does not depend on how much liquid the vessel holds.
In this case, the height of the liquid column is the same for all the vessels. Therefore, the pressure of the liquid column in each vessel is the same and the system is in equilibrium. That means the liquid in vessel C does not rise into vessel B.
Where Does the Extra Weight Go?
The resolution above is correct, but it may still leave you uneasy. The extra liquid in a wide vessel does have weight. If that weight does not press on the base, what supports it?
Consider Fig. 2.10 (b). The arrows indicate the forces exerted against the liquid by the walls of the vessel.
Diagram
Fig. 2.10 (b) – A vessel that widens upward, with arrows showing the forces exerted by the slanting walls on the liquid, perpendicular to the walls at every point, and the central section B marked above the base.
These forces are perpendicular to the walls of the vessel at each point. These forces can be resolved into vertical and horizontal components. The vertical components act in the upward direction.
So the slanting walls themselves carry the weight of the liquid resting against them. Weight of the liquid in section B is not balanced and contributes the pressure at the base. Thus, it is no longer a paradox.
2.3.5 Pascal’s Law
So far we have studied the pressure that a fluid produces because of its own weight. Now we ask a different question: what happens if we ourselves push on a fluid that is completely enclosed?
The answer was given by the French scientist Blaise Pascal, and it is one of the most useful results in all of fluid mechanics.
Pascal’s law states that the pressure applied at any point of an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and also on the walls of the container, provided the effect of gravity is neglected.
Every condition in that statement matters. The fluid must be enclosed, it must be at rest, and the effect of gravity is being left out — otherwise the deeper parts of the fluid would already be at a higher pressure because of \(h\rho g\).
Experimental Proof of Pascal’s Law
Consider a vessel with four arms A, B, C, and D fitted with frictionless, water tight pistons and filled with an incompressible fluid, as shown in Fig. 2.10c. Let the areas of cross section of A, B, C, and D be \(a\), \(2a\), \(3a\), and \(a/2\) respectively.
Diagram
Fig. 2.10c – A closed vessel with four arms A, B, C and D fitted with pistons of cross-sectional areas a, 2a, 3a and a/2, with force F applied on piston A and forces 2F, 3F and F/2 marked on the others.
If a force \(F\) is applied on the piston A, the pressure exerted on the liquid is \(p = F/a\).
It is observed that the other three pistons B, C, and D move outward. This already tells us that the push given at A has reached all of them. To find out how much pressure reached each one, we measure the force needed to hold each piston in place.
In order to keep these three pistons B, C, and D in their original positions, forces \(2F\), \(3F\), and \(F/2\) respectively are required to be applied on the pistons.
Therefore, the pressure on the pistons B, C, and D is:
\[\text{on B,} \quad p_B = \frac{2F}{2a} = \frac{F}{a}\]
\[\text{on C,} \quad p_C = \frac{3F}{3a} = \frac{F}{a}\]
\[\text{on D,} \quad p_D = \frac{F/2}{a/2} = \frac{F}{a}\]
i.e. \(p_B = p_C = p_D = p\). This indicates that the pressure applied on piston A is transmitted equally and undiminished to all parts of the fluid and the walls of the vessel.
Notice that the forces needed at the different pistons are very different — \(3F\) at one and \(F/2\) at another. It is the pressure, not the force, that is the same everywhere. This single idea is what the machines below are built on.
Applications of Pascal’s Law
i) Hydraulic lift: A hydraulic lift is used to lift a heavy object using a small force. The working of this machine is based on Pascal’s law.
As shown in Fig. 2.11, a tank containing a fluid is fitted with two pistons S\(_1\) and S\(_2\). S\(_1\) has a smaller area of cross section, \(A_1\), while S\(_2\) has a much larger area of cross section, \(A_2\), with \(A_2 \gg A_1\).
Diagram
Fig. 2.11 – A closed tank of fluid with a small piston S₁ of area A₁ on the left carrying a downward force F₁, and a large piston S₂ of area A₂ on the right carrying an upward force F₂, with p₁ = p₂ marked in the fluid.
If we apply a force \(F_1\) on the smaller piston S\(_1\) in the downward direction, it will generate pressure \(p = (F_1/A_1)\), which will be transmitted undiminished to the bigger piston S\(_2\).
A force \(F_2 = pA_2\) will be exerted upwards on it. Substituting for \(p\),
\[F_2 = F_1\left(\frac{A_2}{A_1}\right) \qquad \text{--- (2.10)}\]
Here \(F_1\) and \(F_2\) are in newton, and \(A_1\) and \(A_2\) are in m\(^2\). Since the ratio \(A_2/A_1\) is large, thus, \(F_2\) is much larger than \(F_1\).
A heavy load can be placed on S\(_2\) and can be lifted up or moved down by applying a small force on S\(_1\). This is the principle of a hydraulic lift.
ii) Hydraulic brakes: Hydraulic brakes are used to slow down or stop vehicles in motion. They are based on the same principle as that of a hydraulic lift.
Figure 2.12 shows a schematic diagram of a hydraulic brake system.
Diagram
Fig. 2.12 – Schematic hydraulic brake system: a master cylinder of area A₁ operated by the brake pedal, connected by hydraulic tubes to slave cylinders of larger area A₂ at each of the four wheels.
By pressing the brake pedal, the piston of the master cylinder is pushed in the forward direction. As a result, the piston in the slave cylinder, which has a much larger area of cross section as compared to that of the master cylinder, also moves in the forward direction so as to maintain the volume of the oil constant.
The slave piston pushes the friction pads against the rotating disc, which is connected to the wheel. Thus, causing a moving vehicle to slow down or stop.
The master cylinder has a smaller area of cross section \(A_1\) compared to the area \(A_2\) of the slave cylinder. By applying a small force \(F_1\) to the master cylinder, we generate pressure \(p = (F_1/A_1)\). This pressure is transmitted undiminished throughout the system. The force \(F_2\) on the slave cylinder is then,
\[F_2 = pA_2 = \left(\frac{F_1}{A_1}\right)A_2 = F_1\left(\frac{A_2}{A_1}\right)\]
This is similar to the principle used in a hydraulic lift. Since area \(A_2\) is greater than \(A_1\), \(F_2\) is also greater than \(F_1\). Thus, a small force applied on the brake pedal gets converted into a large force and slows down or stops a moving vehicle.
2.3.6 Measurement of Pressure
Instruments used to measure pressure are called pressure meters or pressure gauges or vacuum gauges. Below we will describe two instruments which are commonly used to measure pressure.
Both of them work on a single idea that we have already established: in a fluid at rest, two points at the same horizontal level are at the same pressure, and going down by a height \(h\) raises the pressure by \(h\rho g\).
Mercury Barometer
An instrument that measures atmospheric pressure is called a barometer.
One of the first barometers was invented by the Italian scientist Torricelli, in the experiment described at the beginning of this chapter. The barometer is in the form of a glass tube completely filled with mercury and placed upside down in a small dish containing mercury. Its schematic diagram is shown in Fig. 2.13.
Diagram
Fig. 2.13 – A mercury barometer: an inverted glass tube standing in a dish of mercury, with point A at the top of the mercury column, Torricelli’s vacuum above it, point B inside the tube level with point C on the mercury surface in the dish, and the column height h marked.
The working of the barometer can be understood step by step.
A glass tube of about 1 metre length and a diameter of about 1 cm is filled with mercury up to its brim. It is then quickly inverted into a small dish containing mercury. The level of mercury in the glass tube lowers as some mercury spills into the dish. A gap is created between the surface of mercury in the glass tube and the closed end of the glass tube. The gap does not contain any air and it is called Torricelli’s vacuum. It does contain some mercury vapours.
Thus, the pressure at the upper end of the mercury column inside the tube is zero, i.e., the pressure at a point such as A is \(p_A\) = zero. There is no air above the mercury there to press down on it.
Let us consider a point C on the mercury surface in the dish, and another point B inside the tube at the same horizontal level as that of the point C.
The pressure at C is equal to the atmospheric pressure \(p_0\) because it is open to the atmosphere. As points B and C are at the same horizontal level, the pressure at B is also equal to the atmospheric pressure \(p_0\), i.e., \(p_B = p_0\).
Suppose the point B is at a depth \(h\) below the point A and \(\rho\) is the density of mercury, then
\[p_B = p_A + h\rho g \qquad \text{--- (2.11)}\]
Now we put in what we know about the two ends. \(p_A = 0\) (there is vacuum above point A) and \(p_B = p_0\), therefore,
\[p_0 = h\rho g\]
where \(h\) is the length of the mercury column in the mercury barometer.
This is a remarkably convenient result. To measure the pressure of the whole atmosphere, all one has to do is measure the length of a column of mercury with an ordinary scale.
Open Tube Manometer
A barometer measures the pressure of the atmosphere. A manometer measures the pressure of a gas enclosed in a container.
A manometer consists of a U-shaped tube partly filled with a low density liquid such as water or kerosene. This helps in having a larger level difference between the level of liquid in the two arms of the manometer, since a lighter liquid must rise through a greater height to balance the same pressure difference. A larger level difference can be read off the scale more accurately.
Figure 2.14 shows an open tube manometer. One arm of the manometer is open to the atmosphere and the other is connected to the container D, of which the pressure \(p\) is to be measured.
Diagram
Fig. 2.14 – An open tube U-shaped manometer with point A at the open surface, point B in the open arm at the same level as point C in the other arm, the height difference h marked, and the far end connected to a container D.
The pressure at point A is atmospheric pressure \(p_0\), because this arm is open to the atmosphere.
To find the pressure at point C, which is exposed to the pressure of the gas in the container, we consider a point B in the open arm of the manometer at the same level as point C. The pressure at the points B and C is the same, i.e.,
\[p_C = p_B \qquad \text{--- (2.12)}\]
The pressure at point B is,
\[p_B = p_o + h\rho g \qquad \text{--- (2.13)}\]
where \(\rho\) is the density of the liquid in the manometer in kg m\(^{-3}\), \(h\) is the height of the liquid column above point B in metre, and \(g\) is the acceleration due to gravity in m s\(^{-2}\).
According to Pascal’s principle, the pressure at C is the same as that at D, i.e., inside the chamber. Therefore, the pressure \(p\) in the container is,
\[p = p_C\]
Using Eq. (2.12) and Eq. (2.13) we can write,
\[p = p_0 + h\rho g \qquad \text{--- (2.14)}\]
As the manometer measures the gauge pressure of the gas in the container D, we can write the gauge pressure in the container D as
\[p - p_0 = h\rho g\]
So the difference in the levels of the liquid in the two arms, read directly off a scale, gives the gauge pressure of the gas at once.
2.4 Surface Tension
A liquid at rest shows a very interesting property called surface tension. Until now we studied what a liquid does through its bulk — its weight and the pressure it produces. We now turn to what happens at its surface, which behaves quite unlike the interior.
We have seen that a water spider walks on the surface of steady water, a greased needle floats on the steady surface of water, and rain drops and soap bubbles always take a spherical shape. Some of these are shown in Fig. 2.14a.
Diagram
Fig. 2.14a – Three everyday observations side by side: a water spider standing on a water surface, a greased needle resting on water, and a spherical raindrop.
Each of these should surprise you. A steel needle is far denser than water and ought to sink, yet it rests on the surface as though on a thin sheet. All these phenomena arise due to surface tension. Surface tension is one of the important properties of liquids.
In the sections that follow we shall find out what causes surface tension, how it is measured, and why it explains observations as different as a floating needle and a rising column of oil in a lamp wick.
2.4.1 Molecular Theory of Surface Tension
All the above observations can be explained on the basis of different types of forces coming into play in all these situations. We will try to understand the effect of these forces and their relation to the surface tension in liquids.
To understand surface tension, we need to know some terms in molecular theory that explain the behaviour of liquids at their surface.
a) Intermolecular Force
Matter is made up of molecules. Any two molecules attract each other. This force between molecules is called intermolecular force.
There are two types of intermolecular forces — i) cohesive force and ii) adhesive force. The difference between them is simply whether the two molecules belong to the same substance or to different substances.
i) Cohesive force: The force of attraction between the molecules of the same substance is called cohesive force or force of cohesion. The force of attraction between two air molecules, or that between two water molecules, is a cohesive force.
Cohesive force is strongest in solids and weakest in gases. This is the reason why solids have a definite shape and gases do not. Small droplets of liquid coalesce into one and form a drop due to this force.
ii) Adhesive force: The force of attraction between the molecules of different substances is called adhesive force or force of adhesion. The force of attraction between a glass molecule and a water molecule is a force of adhesion.
b) Range of Molecular Force
The maximum distance from a molecule up to which the molecular force is effective is called the range of molecular force.
Intermolecular forces are effective up to a distance of the order of a few nanometre (\(10^{-9}\) m) in solids and liquids. Beyond this distance the attraction is too feeble to matter. Therefore, they are short range forces.
This is a very important point. Unlike gravity, which reaches across the solar system, a molecule in a liquid simply does not feel the molecules that lie more than a few nanometre away.
c) Sphere of Influence
An imaginary sphere with a molecule at its centre and radius equal to the molecular range is called the sphere of influence of the molecule.
The spheres around molecules A, B or C are shown in Fig. 2.15 (a) and (b). The intermolecular force is effective only within the sphere of influence.
Diagram
Fig. 2.15 (a) – A single molecule at the centre of its sphere of influence, with arrows of equal length pointing outward in all directions.
So to find the net force on any molecule, we only need to count the molecules lying inside its sphere of influence. Everything outside can be ignored.
d) Surface Film
The surface layer of a liquid with thickness equal to the range of intermolecular force is called the surface film. This is the layer shown between XY and X\('\)Y\('\) in Fig. 2.15 (b).
e) Free Surface of a Liquid
It is the surface of a fluid which does not experience any shear stress. For example, the interface between liquid water and the air above. In Fig. 2.15 (b), XY is the free surface of the liquid.
f) Surface Tension on the Basis of Molecular Theory
We now have all the terms we need, and can explain where surface tension comes from.
As shown in Fig. 2.15 (b), XY is the free surface of the liquid and X\('\)Y\('\) is the inner layer parallel to XY at a distance equal to the range of molecular force. Hence, the section XX\('\)–Y\('\)Y near the surface of the liquid acts as the surface film.
Diagram
Fig. 2.15 (b) – A vessel of liquid with the free surface XY and the parallel inner layer X’Y’ marking the surface film, showing molecule A deep inside, molecule B within the surface film and molecule C on the free surface, each with its sphere of influence.
Consider three molecules A, B, and C such that molecule A is deep inside the liquid, molecule B is within the surface film, and molecule C is on the surface of the liquid.
Molecule A. As molecule A is deep inside the liquid, its sphere of influence is also completely inside the liquid. Wherever we look around A, there are equal numbers of liquid molecules pulling it. As a result, molecule A is acted upon by equal cohesive forces in all directions. Thus, the net cohesive force acting on molecule A is zero.
Molecule B. Molecule B lies within the surface layer and below the free surface of the liquid. A larger part of its sphere of influence is inside the liquid and a smaller part is in air. Due to this, a strong downward cohesive force acts on the liquid molecule.
The adhesive force acting on molecule B, due to the air molecules above it and within its sphere of influence, is weak. It points upwards. As a result, the molecule B gets attracted inside the liquid.
Molecule C. The same holds for molecule C which lies exactly on the free surface of the liquid. Half of the sphere of influence is in air and half in the liquid.
The number of air molecules within the sphere of influence of the molecule C, above the free surface of the liquid, is much less than the number of liquid molecules within the sphere of influence that lies within the liquid. This is because the density of air is less than that of a liquid.
Therefore the adhesive force trying to pull the molecule above the liquid surface is much weaker than the cohesive force that tries to pull the molecule inside the liquid surface. As a result, the molecule C also gets attracted inside the liquid.
Why the Surface Stays Under Tension
Thus, all molecules in the surface film are acted upon by an unbalanced net cohesive force directed into the liquid. Therefore, the molecules in the surface film are pulled inside the liquid.
This minimizes the total number of molecules in the surface film. As a result, the surface film remains under tension. The surface film of a liquid behaves like a stretched elastic membrane.
This tension is known as surface tension. The force due to surface tension acts tangential to the free surface of a liquid.
2.4.2 Surface Tension and Surface Energy
a) Surface Tension
As seen previously, the free surface of a liquid in a container acts as a stretched membrane and all molecules on the surface film experience a stretching force.
To turn this idea into a measurable quantity, imagine a line PQ of length \(L\) drawn tangential to the free surface of the liquid, as shown in Fig. 2.16. The line is imaginary — we simply mark it out mentally on the surface.
Diagram
Fig. 2.16 – A liquid surface seen from above with an imaginary line PQ drawn on it, and arrows showing equal and opposite forces acting tangentially on the two sides of the line.
All the molecules on this line experience equal and opposite forces tangential to the surface, as if they are tearing the surface apart, due to the cohesive forces of molecules lying on either side.
This force per unit length is the surface tension.
Surface tension \(T\) is defined as the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of a liquid.
\[T = \frac{F}{L} \qquad \text{--- (2.15)}\]
Here \(F\) is the tangential force in newton (N) and \(L\) is the length of the imaginary line in metre (m).
SI unit of surface tension is N/m. Its dimensions are \([L^{0}M^{1}T^{-2}]\).
Since length appears to the power zero in the dimensions, surface tension has the same dimensions as force per unit length, which is also energy per unit area — a fact we shall prove in part (c) of this section.
Table 2.1 – Surface tension of some liquids at 20 °C
| Sr. No. | Liquid | S.T. (N/m) | S.T. (dyne/cm) |
|---|---|---|---|
| 1 | Water | 0.0727 | 72.7 |
| 2 | Mercury | 0.4355 | 435.5 |
| 3 | Soap solution | 0.025 | 25 |
| 4 | Glycerin | 0.0632 | 63.2 |
Notice from Table 2.1 how much larger the surface tension of mercury is than that of water, and how adding soap to water lowers its surface tension to about a third of its original value. Both of these facts will be used later in this chapter.
b) Surface Energy
We have seen that a molecule inside the volume of a liquid, like molecule A in Fig. 2.15, experiences no net cohesive force, and the molecules B and C experience a net inward cohesive force.
Thus, work has to be done to bring any molecule from inside the liquid into the surface film. You are dragging that molecule outward against a force that is pulling it inward, and that requires energy.
Clearly, the surface molecules possess extra potential energy as compared to the molecules inside the liquid. The extra energy of the molecules in the surface layer is called the surface energy of the liquid.
As any system always tries to attain a state of minimum potential energy, the liquid tries to reduce the area of its surface film. Energy has to be spent in order to increase the surface area of a liquid.
c) Relation Between the Surface Energy and Surface Tension
We now show that the two quantities defined above are numerically the same thing.
Consider a C-shaped frame of wire P\('\)PSS\('\). It is fitted with a movable arm QR as shown in Fig. 2.17. This frame is dipped in a soap solution and then taken out. A film of soap solution will be formed within the boundaries PQRS of the frame.
Diagram
Fig. 2.17 – A C-shaped wire frame carrying a soap film, with a movable arm QR of length L pulled outward to a new position Q’R’ through a distance dx, with force F inward and F’ outward marked on the arm.
Each arm of the frame experiences an inward force due to the film. Under the action of this force, the movable arm QR moves towards the side PS so as to decrease the area of the film.
If the length of QR is \(L\), then this inward force \(F\) acting on it is given by
\[F = (T) \times (2L) \qquad \text{--- (2.16)}\]
Since the film has two surfaces, the upper surface and the lower surface, the total length over which surface tension acts on QR is \(2L\).
Imagine an external force \(F'\), equal and opposite to \(F\), applied isothermally — that is, gradually and at constant temperature — to the arm QR, so that it pulls the arm away and tries to increase the surface area of the film. The arm QR moves to Q\('\)R\('\) through a distance \(dx\).
The word isothermally matters. If the film were stretched suddenly, some of the energy supplied would appear as heat, and the accounting below would not be exact.
Therefore, the work done against \(F\), the force due to surface tension, is given by
\[dw = F'dx\]
Using Eq. (2.16),
\[dw = T(2Ldx)\]
But \(2Ldx = dA\), the increase in area of the two surfaces of the film. Therefore,
\[dw = T(dA)\]
This work done in stretching the film is stored in the area \(dA\) of the film as its potential energy. This energy is called surface energy.
\[\therefore \; \text{Surface energy} = T(dA) \qquad \text{--- (2.17)}\]
Thus, surface tension is also equal to the surface energy per unit area.
2.4.3 Angle of Contact
When a liquid surface comes in contact with a solid surface, it forms a meniscus, which can be either convex (mercury–glass) or concave (water–glass), as shown in Fig. 2.18. A meniscus is simply the curved surface that the liquid takes near the wall of the container.
The angle of contact, \(\theta\), between a liquid and a solid surface is defined as the angle between the tangents drawn to the free surface of the liquid and the surface of the solid at the point of contact, measured within the liquid.
When the angle of contact is acute, the liquid forms a concave meniscus, Fig. 2.18 (a), at the point of contact.
Diagram
Fig. 2.18 (a) – A narrow tube dipped in a liquid showing a concave meniscus curving upward at the walls, with the acute angle of contact θ marked inside the liquid.
When the angle of contact is obtuse, it forms a convex meniscus, Fig. 2.18 (b).
Diagram
Fig. 2.18 (b) – A narrow tube dipped in a liquid showing a convex meniscus bulging upward at the centre, with the obtuse angle of contact θ marked inside the liquid.
For example, a water–glass interface forms a concave meniscus and a mercury–glass interface forms a convex meniscus.
This difference between the shapes of the menisci is due to the net effect of the cohesive forces between liquid molecules and the adhesive forces between liquid and solid molecules, as discussed below.
a) Shape of Meniscus
To explain the shape of the meniscus, we look at a single molecule sitting on the liquid surface right next to the wall, and find the resultant of the two forces acting on it. Throughout this discussion, \(\overrightarrow{AP}\) is the net adhesive force, \(\overrightarrow{AC}\) is the net cohesive force, and \(\overrightarrow{AR}\) is their resultant.
i) Concave meniscus – acute angle of contact:
Figure 2.19 (a) shows the acute angle of contact between a liquid surface and a solid, for example, kerosene in a glass bottle.
Diagram
Fig. 2.19 (a) – A molecule A on the liquid surface at a vertical solid wall, with the horizontal adhesive force AP into the wall, the cohesive force AC at 45° into the liquid, the resultant AR directed into the solid, and the tangent AT perpendicular to AR giving an acute angle θ.
Consider a molecule such as A on the surface of the liquid near the wall of the container. The molecule experiences both cohesive as well as adhesive forces.
In this case, since the wall is vertical, the net adhesive force (\(\overrightarrow{AP}\)) acting on the molecule A is horizontal. The net cohesive force (\(\overrightarrow{AC}\)) acting on the molecule is directed at nearly 45° to either of the surfaces.
The magnitude of the adhesive force is so large that the net force (\(\overrightarrow{AR}\)) is directed inside the solid.
For equilibrium or stability of a liquid surface, the net force (\(\overrightarrow{AR}\)) acting on the molecule A must be normal to the liquid surface at all points. If it were not, there would be a component of force along the surface, and the surface would keep moving instead of settling down.
For the resultant force \(\overrightarrow{AR}\) to be normal to the tangent, the liquid near the wall should pile up against the solid boundary, so that the tangent AT to the liquid surface is perpendicular to AR. Thus, this makes the meniscus concave. Obviously, such a liquid wets that solid surface.
ii) Convex meniscus – obtuse angle of contact:
Figure 2.19 (b) shows the obtuse angle of contact between a liquid and a solid, for example, mercury in a glass bottle.
Diagram
Fig. 2.19 (b) – A molecule A on the liquid surface at a vertical solid wall, with the horizontal adhesive force AP, the larger cohesive force AC, the resultant AR directed into the liquid, and the tangent AT perpendicular to AR giving an obtuse angle θ.
Consider a molecule such as A on the surface of the liquid near the wall of the container. The molecule experiences both cohesive as well as adhesive forces. In this case also, the net adhesive force (\(\overrightarrow{AP}\)) acting on the molecule A is horizontal, since the wall is vertical.
The magnitude of the cohesive force is so large that the net force (\(\overrightarrow{AR}\)) is directed inside the liquid. This is the one change from the previous case, and it reverses the shape of the meniscus.
For equilibrium or stability of a liquid surface, the net force (\(\overrightarrow{AR}\)) acting on all molecules similar to molecule A must be normal to the liquid surface at all points.
The liquid near the wall should, therefore, creep inside against the solid boundary. This makes the meniscus convex, so that its tangent AT is normal to AR. Obviously, such a liquid does not wet that solid surface.
iii) Zero angle of contact:
Figure 2.19 (c) shows the angle of contact between a liquid, e.g. highly pure water, which completely wets a solid, e.g. clean glass, surface. The angle of contact in this case is almost zero, i.e., \(\theta \to 0^{0}\).
Diagram
Fig. 2.19 (c) – A molecule A at a vertical solid wall with a negligible cohesive force, the adhesive force AP horizontal and equal to the resultant AR, and the tangent AT lying along the wall so that θ = 0.
In this case, the liquid molecules near the contact region are so few in number that the cohesive force is negligible, i.e., \(\overrightarrow{AC} = 0\), and the net adhesive force itself is the resultant force, i.e., \(\overrightarrow{AP} = \overrightarrow{AR}\).
Therefore, the tangent AT is along the wall within the liquid and the angle of contact is zero.
iv) Angle of contact 90° and conditions for convexity and concavity:
Consider a hypothetical liquid having an angle of contact of 90° with a given solid container, as shown in Fig. 2.19 (d). This case is the dividing line between the two behaviours studied above.
Diagram
Fig. 2.19 (d) – A molecule A at a vertical solid wall with the adhesive force AP horizontal, the cohesive force AC at 45°, and the resultant AR exactly vertical along the solid surface.
In this case, the net cohesive force \(\overrightarrow{AC}\) is exactly at 45° with either of the surfaces and the resultant force \(\overrightarrow{AR}\) is exactly vertical, along the solid surface.
For this to occur,
\[\overrightarrow{AP} = \frac{AC}{\sqrt{2}}\]
where \(AC\) is the magnitude of the net cohesive force. This is because a force at 45° has a horizontal component equal to \(AC/\sqrt{2}\), and this component must be exactly cancelled by the adhesive force for the resultant to be vertical.
From this we can write the conditions for acute and obtuse angles of contact:
For acute angle of contact, \(\overrightarrow{AP} > \dfrac{AC}{\sqrt{2}}\), and for obtuse angle of contact, \(\overrightarrow{AP} < \dfrac{AC}{\sqrt{2}}\).
In words: the stronger the adhesion compared with the cohesion, the more the liquid wets the solid.
b) Shape of Liquid Drops on a Solid Surface
When a small amount of a liquid is dropped on a plane solid surface, the liquid will either spread on the surface or will form droplets on the surface.
Which phenomenon will occur depends on the surface tension of the liquid and the angle of contact between the liquid and the solid surface. The surface tension between the liquid and air, as well as that between the solid and air, will also have to be taken into account.
Notice that three different pairs of substances meet at the edge of the drop — solid and liquid, solid and air, and liquid and air. Each of these boundaries has its own surface tension.
Let \(\theta\) be the angle of contact for the given solid–liquid pair.
\(T_1\) = Force due to surface tension at the liquid–solid interface,
\(T_2\) = Force due to surface tension at the air–solid interface,
\(T_3\) = Force due to surface tension at the air–liquid interface.
As the force due to surface tension is tangential to the surfaces in contact, the directions of \(T_1\), \(T_2\) and \(T_3\) are as shown in Fig. 2.20.
For equilibrium of the drop, we balance the forces along the solid surface. Only \(T_3\) is inclined, so only its horizontal component \(T_3 \cos\theta\) enters the balance.
\[T_3\cos\theta + T_1 = T_2, \qquad \cos\theta = \frac{T_2 - T_1}{T_3} \qquad \text{--- (2.18)}\]
From this equation we get the following cases:
- If \(T_2 > T_1\) and \((T_2 - T_1) < T_3\), \(\cos\theta\) is positive and the angle of contact \(\theta\) is acute, as shown in Fig. 2.20 (a).
Diagram
Fig. 2.20 (a) – A liquid drop resting on a flat solid surface with an acute angle of contact, showing T₁ along the solid–liquid interface, T₂ along the air–solid interface and T₃ tangential to the drop surface.
- If \(T_2 < T_1\) and \((T_2 - T_1) < T_3\), \(\cos\theta\) is negative, and the angle of contact \(\theta\) is obtuse, as shown in Fig. 2.20 (b).
Diagram
Fig. 2.20 (b) – A nearly spherical liquid drop resting on a flat solid surface with an obtuse angle of contact, showing the directions of T₁, T₂ and T₃.
If \((T_2 - T_1) = T_3\), \(\cos\theta = 1\) and \(\theta\) is nearly equal to zero.
If \((T_2 - T_1) > T_3\), or \(T_2 > (T_1 + T_3)\), then \(\cos\theta > 1\), which is impossible. The liquid spreads over the solid surface and a drop will not be formed.
c) Factors Affecting the Angle of Contact
The value of the angle of contact depends on the following factors:
The nature of the liquid and the solid in contact.
Impurity: Impurities present in the liquid change the angle of contact.
Temperature of the liquid: Any increase in the temperature of a liquid decreases its angle of contact. For a given solid–liquid surface, the angle of contact is constant at a given temperature.
The last point is worth remembering for numerical problems. Whenever a problem gives you a solid–liquid pair at a fixed temperature, the angle of contact is a fixed number that you may look up, as in Table 2.2.
Table 2.2 – Angle of contact for pairs of liquid and solid in contact
| Sr. No. | Liquid – solid in contact | Angle of contact |
|---|---|---|
| 1 | Pure water and clean glass | 0° |
| 2 | Chloroform with clean glass | 0° |
| 3 | Organic liquids with clean glass | 0° |
| 4 | Ether with clean glass | 16° |
| 5 | Kerosene with clean glass | 26° |
| 6 | Water with paraffin | 107° |
| 7 | Mercury with clean glass | 140° |
Notice how the table divides into two groups. Every liquid that wets clean glass has an angle below 90°, while water on paraffin wax and mercury on glass — the two pairs that do not wet — have angles well above 90°.
2.4.4 Effect of Impurity and Temperature on Surface Tension
The surface tension of a liquid is not a fixed number for all conditions. Two factors can change it — what is mixed into the liquid, and how hot the liquid is. Since surface tension arises from cohesive forces between molecules, anything that alters those forces alters the surface tension.
a) Effect of Impurities
When a soluble substance such as common salt, i.e., sodium chloride, is dissolved in water, the surface tension of water increases.
When a sparingly soluble substance such as phenol or a detergent is mixed with water, the surface tension of water decreases.
For example, a detergent powder is mixed with water to wash clothes. Due to this, the surface tension of water decreases, and water makes good contact with the fabric and is able to remove tough stains. Plain water, with its high surface tension, tends to stay as rounded drops on the cloth instead of soaking into the fibres.
- When an insoluble impurity is added into water, the surface tension of water decreases.
When an impurity gets added to any liquid, the cohesive force of that liquid decreases, which affects the angle of contact and hence the shape of the meniscus. If mercury gathers dust, then its surface tension is reduced. It does not form spherical droplets unless the dust is completely removed.
b) Effect of Temperature
In most liquids, as temperature increases surface tension decreases.
For example, it is suggested that new cotton fabric should be washed in cold water. In this case, water does not make good contact with the fabric due to its higher surface tension, and the fabric does not lose its colour because of this.
Hot water is used to remove tough stains on fabric because of its lower surface tension. Here we want the water to make good contact with the cloth, so the lower surface tension is an advantage.
This general trend is shown in Fig. 2.20c.
Diagram
Fig. 2.20c – A graph of surface tension against temperature for a liquid, the curve falling steadily and reaching zero at the critical temperature.
There are exceptions to this rule. In the case of molten copper or molten cadmium, the surface tension increases with increase in its temperature.
The surface tension of a liquid becomes zero at the critical temperature.
2.4.5 Excess Pressure Across the Free Surface of a Liquid
Every molecule on a liquid surface experiences forces due to surface tension which are tangential to the liquid surface at rest. The direction of the resultant force of surface tension acting on a molecule on the liquid surface depends upon the shape of that liquid surface.
This is the key idea of this section. If the surface is flat, the tangential pulls on a molecule come from all sides in the same horizontal plane and cancel out. If the surface is curved, they do not.
This force also contributes in deciding the pressure at a point just below the surface of a liquid.
Figures 2.21 (a), (b) and (c) show surfaces of three liquids with different shapes and their menisci.
Let \(\vec{f_A}\) be the downward force due to the atmospheric pressure. All the three figures show two molecules A and B. The molecule A is just above, and the molecule B is just below it, inside the liquid.
The level difference between A and B is almost zero, so that it does not contribute anything to the pressure difference. In other words, the term \(h\rho g\) plays no part here, and whatever difference we find must come from surface tension alone.
In all the three figures, the pressure at the point A is the atmospheric pressure \(p\).
a) Plane Liquid Surface
Figure 2.21 (a) shows a planar free surface of the liquid.
Diagram
Fig. 2.21 (a) – A flat liquid surface with molecule A just above it and molecule B just below it, showing the downward force f_A due to atmospheric pressure and zero resultant force due to surface tension at B.
In this case, the resultant force due to surface tension, \(\vec{f_T}\), on the molecule at B is zero. The force \(\vec{f_A}\) itself decides the pressure, and the pressure at A and B is the same.
b) Convex Liquid Surface
The surface of the liquid in Fig. 2.21 (b) is upper convex — that is, convex when seen from above.
Diagram
Fig. 2.21 (b) – An upper convex liquid surface with molecule A just above and molecule B just below, showing the downward force f_A and the resultant surface tension force f_T also acting vertically downward at B.
In this case, the resultant force due to surface tension, \(\vec{f_T}\), on the molecule at B is vertically downwards and adds up to the downward force \(\vec{f_A}\).
Since the two forces now act in the same direction, the total downward push on B is larger than atmospheric. This develops a greater pressure at point B, which is inside the liquid and on the concave side of the meniscus.
Thus, the pressure on the concave side, i.e., inside the liquid, is greater than that on the convex side, i.e., outside the liquid.
c) Concave Liquid Surface
The surface of the liquid in Fig. 2.21 (c) is upper concave — concave when seen from above.
Diagram
Fig. 2.21 (c) – An upper concave liquid surface with molecule A just above and molecule B just below, showing the downward force f_A and the resultant surface tension force f_T acting vertically upward at B.
In this case, the force due to surface tension \(\vec{f_T}\) on the molecule at B is vertically upwards. The force \(\vec{f_A}\) due to atmospheric pressure acts downwards. Forces \(\vec{f_A}\) and \(\vec{f_T}\) thus act in opposite directions.
Therefore, the net downward force responsible for the pressure at B is less than \(\vec{f_A}\). This develops a lesser pressure at point B, which is inside the liquid and on the convex side of the meniscus.
Thus, the pressure on the concave side, i.e., outside the liquid, is greater than that on the convex side, i.e., inside the liquid.
The Common Rule
The three cases above look different, but the conclusion of (b) and (c) is exactly the same statement, seen from the two sides.
This single rule is used twice in the sections that follow — first to find the excess pressure inside a drop and a bubble in Section 2.4.6, and then to explain why a liquid rises or falls in a capillary tube in Section 2.4.7.
2.4.6 Explanation of Formation of Drops and Bubbles
Liquid drops and small bubbles are spherical in shape because the forces of surface tension dominate the gravitational force. These forces always try to minimize the surface area of the liquid.
Of all the shapes that can enclose a given volume, the sphere has the smallest surface area. So a small quantity of liquid left to itself, with gravity too weak to matter, settles into a sphere.
A bubble or drop does not collapse because the resultant of the force due to external pressure and the force of surface tension is smaller than the pressure inside a bubble or inside a liquid drop.
In other words, the pressure inside is a little higher than the pressure outside, and this extra pressure holds the drop open. We now calculate how much extra pressure that is.
Excess Pressure Inside a Liquid Drop
Consider a spherical drop as shown in Fig. 2.22. Let \(p_i\) be the pressure inside the drop and \(p_0\) be the pressure outside it.
Diagram
Fig. 2.22 – A spherical liquid drop of radius r with internal pressure p_i and external pressure p₀, shown with a dashed outer circle of radius r + Δr indicating a small increase in radius.
As the drop is spherical in shape, its surface is concave towards the inside, and we saw in Section 2.4.5 that the pressure on the concave side is greater. Hence the pressure \(p_i\) inside the drop is greater than \(p_0\), the pressure outside.
Therefore, the excess pressure inside the drop is \(p_i - p_0\).
To find this excess pressure, we let the drop expand very slightly and compare the work done in two different ways.
Let the radius of the drop increase from \(r\) to \(r + \Delta r\), where \(\Delta r\) is very small, so that the pressure inside the drop remains almost constant.
Let the initial surface area of the drop be \(A_1 = 4\pi r^{2}\), and the final surface area of the drop be \(A_2 = 4\pi(r + \Delta r)^{2}\).
\[\therefore \; A_2 = 4\pi(r^{2} + 2r\Delta r + \Delta r^{2})\]
\[\therefore \; A_2 = 4\pi r^{2} + 8\pi r\Delta r + 4\pi \Delta r^{2}\]
As \(\Delta r\) is very small, \(\Delta r^{2}\) can be neglected,
\[\therefore \; A_2 = 4\pi r^{2} + 8\pi r\Delta r\]
Thus, the increase in the surface area of the drop is
\[dA = A_2 - A_1 = 8\pi r\Delta r \qquad \text{--- (2.19)}\]
Work done in increasing the surface area by \(dA\) is stored as excess surface energy. Using the result of Section 2.4.2 that surface energy equals \(T\) times the increase in area,
\[\therefore \; dW = TdA = T(8\pi r\Delta r) \qquad \text{--- (2.20)}\]
This work done is also equal to the product of the force \(F\) which causes the increase in the area of the drop and the displacement \(\Delta r\), which is the increase in the radius of the bubble.
\[\therefore \; dW = F\Delta r \qquad \text{--- (2.21)}\]
The excess force is given by,
(Excess pressure) \(\times\) (Surface area)
\[\therefore \; F = (p_i - p_0)\,4\pi r^{2} \qquad \text{--- (2.22)}\]
Equating Eq. (2.20) and Eq. (2.21), we get,
\[T(8\pi r\Delta r) = (p_i - p_0)\,4\pi r^{2}\Delta r\]
\[\therefore \; (p_i - p_0) = \frac{2T}{r} \qquad \text{--- (2.23)}\]
This equation gives the excess pressure inside a drop. This is called Laplace’s law of a spherical membrane.
Here \(T\) is in N/m, \(r\) is in metre, and the excess pressure is in N/m\(^2\).
Excess Pressure Inside a Soap Bubble
In the case of a soap bubble there are two free surfaces in contact with air, the inner surface and the outer surface. Both of them have to be stretched when the bubble grows.
For a bubble, Eq. (2.19) changes to \(dA = 2(8\pi r\Delta r)\). Hence, the total increase in the surface area of a soap bubble, while increasing its radius by \(\Delta r\), is \(2(8\pi r\Delta r)\).
The work done by this excess pressure is
\[dW = (p_i - p_0)\,4\pi r^{2}\Delta r = T(16\pi r\Delta r)\]
\[\therefore \; (p_i - p_0) = \frac{4T}{r} \qquad \text{--- (2.24)}\]
So for the same radius and the same liquid, the excess pressure inside a soap bubble is exactly twice that inside a liquid drop. In numerical problems, deciding whether the object has one surface or two is the first step, and choosing the wrong formula is the commonest mistake made here.
2.4.7 Capillary Action
A tube having a very fine bore (about 1 mm) and open at both ends is called a capillary tube.
If one end of a capillary tube is dipped in a liquid which partially or completely wets the surface of the capillary, like water in glass, the level of the liquid in the capillary rises. On the other hand, if the capillary tube is dipped in a liquid which does not wet its surface, like mercury in glass, the level of the liquid in the capillary drops.
The phenomenon of rise or fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillarity.
Capillarity is in action when,
- Oil rises up the wick of a lamp.
- A cloth rag sucks water.
- Water rises up the crevices in rocks.
- Sap and water rise up to the topmost leaves in a tree.
- Blotting paper absorbs ink.
In each of these the liquid moves upward with no pump and no external push, and in the case of a tall tree it may climb tens of metres.
When a capillary is dipped in a liquid, two effects can be observed: a) the liquid level can rise in the capillary (water in a glass capillary), or b) the liquid level can fall in the capillary (mercury in a glass capillary). Here we discuss a qualitative argument to explain the capillary fall.
a) Capillary Fall
Consider a capillary tube dipped in a liquid which does not wet the surface, for example, in mercury. The shape of the mercury meniscus in the capillary is upper convex.
Consider the points A, B, C, and D such that (see Fig. 2.23 (a)):
Point A is just above the convex surface and inside the capillary.
Point B is just below the convex surface inside the capillary.
Point C is just above the plane surface outside the capillary.
Point D is just below the plane surface and outside the capillary, and below the point C.
Diagram
Fig. 2.23 (a) – A capillary tube just immersed in mercury, with the convex meniscus inside the tube and the points A, B just above and below it, and C, D just above and below the plane mercury surface outside.
Let \(p_A\), \(p_B\), \(p_C\), and \(p_D\) be the values of the pressures at the points A, B, C, and D respectively.
As discussed previously in Section 2.4.5, the pressure on the concave side is always greater than that on the convex side. For an upper convex meniscus, the concave side is the one inside the mercury, where B lies.
\[\therefore \; p_B > p_A\]
As the points A and C are at the same level, the pressure at both these points is the same, and it is the atmospheric pressure.
\[\therefore \; p_A = p_C \qquad \text{--- (2.25)}\]
Between the points C and D, the surface is plane, and across a plane surface there is no pressure difference.
\[\therefore \; p_C = p_D = p_A \qquad \text{--- (2.26)}\]
\[\therefore \; p_B > p_D\]
But the points B and D are at the same horizontal level. In a connected liquid at rest, two points at the same horizontal level must be at the same pressure, so this situation cannot last.
Thus, in order to maintain the same pressure, the mercury in the capillary rushes out of the capillary. Because of this, there is a drop in the level of mercury inside the capillary as shown in Fig. 2.23 (b).
Diagram
Fig. 2.23 (b) – The same capillary tube in mercury after equilibrium, with the mercury level inside the tube standing below the level of the mercury in the vessel.
b) Capillary Rise
Refer to Fig. 2.24 (a) and Fig. 2.24 (b), which show a capillary tube dipped in water.
Diagram
Fig. 2.24 (a) – A capillary tube just immersed in water, showing the concave meniscus inside the tube with points marked just above and below it, and points just above and below the plane water surface outside.
The argument runs exactly as before, but every conclusion is reversed. The water meniscus inside the tube is concave when seen from above, so the region just below it now lies on the convex side, and its pressure is less than atmospheric. The point at the same level outside the tube is at atmospheric pressure. Water is therefore pushed up the capillary until the weight of the raised column restores the balance, as shown in Fig. 2.24 (b).
Diagram
Fig. 2.24 (b) – The same capillary tube in water after equilibrium, with the water column inside the tube standing at a height h above the level of water in the vessel.
Expression for Capillary Rise or Fall
Method (I): Using pressure difference
The pressure due to the liquid (water) column of height \(h\) must be equal to the pressure difference \(2T/R\) due to the concavity, where \(R\) is the radius of curvature of the meniscus.
\[\therefore \; h\rho g = \frac{2T}{R} \qquad \text{--- (2.27)}\]
where \(\rho\) is the density of the liquid in kg m\(^{-3}\) and \(g\) is the acceleration due to gravity in m s\(^{-2}\).
Let \(r\) be the radius of the capillary tube and \(\theta\) be the angle of contact of the liquid, as shown in Fig. 2.25 (a).
Diagram
Fig. 2.25 (a) – A magnified view of the meniscus at the point of contact inside a capillary of radius r, showing the angle of contact θ and the radius of curvature R of the meniscus.
Then the radius of curvature \(R\) of the meniscus is given by
\[R = \frac{r}{\cos\theta}\]
Substituting this in Eq. (2.27),
\[\therefore \; h\rho g = \frac{2T\cos\theta}{r}\]
\[\therefore \; h = \frac{2T\cos\theta}{r\rho g} \qquad \text{--- (2.28)}\]
The above equation gives the expression for capillary rise (or fall) for a liquid. Narrower the tube, the greater is the height to which the liquid rises (or falls).
If the capillary tube is held vertical in a liquid that has a convex meniscus, then the angle of contact \(\theta\) is obtuse. Therefore, \(\cos\theta\) is negative and so is \(h\). This means that the liquid will suffer capillary fall or depression.
So Eq. (2.28) handles both cases by itself. A positive \(h\) means a rise, a negative \(h\) means a fall, and the sign is decided entirely by whether the angle of contact is acute or obtuse.
Method (II): Using forces
The same result can be obtained by balancing forces instead of pressures.
Rise of water inside a capillary is against gravity. Hence, the weight of the liquid column must be equal and opposite to the proper component of the force due to surface tension at the point of contact.
The length of liquid in contact inside the capillary is the circumference \(2\pi r\). Thus, the force due to surface tension is given by,
\[f_T = (\text{surface tension}) \times (\text{length in contact}) = T \times 2\pi r\]
The direction of this force is along the tangent, as shown in Fig. 2.25 (b).
Diagram
Fig. 2.25 (b) – The liquid column inside a capillary, with the surface tension force acting along the tangent to the meniscus at the wall, its vertical component marked, and the weight of the liquid column acting downward.
Since the tangent makes the angle \(\theta\) with the wall, only the part of this force along the vertical can support the weight. The vertical component of this force is
\[(f_T)_v = T \times 2\pi r\cos\theta \qquad \text{--- (2.29)}\]
Ignoring the liquid in the concave meniscus, the volume of the liquid in the capillary rise is \(V = \pi r^{2}h\).
\[\therefore \text{ Mass of the liquid in the capillary rise, } m = \pi r^{2}h\rho\]
\[\therefore \text{ Weight of the liquid in the capillary (rise or fall), } w = \pi r^{2}h\rho g \qquad \text{--- (2.30)}\]
This must be equal and opposite to the vertical component of the force due to surface tension. Thus, equating the right sides of equations (2.29) and (2.30), we get,
\[\rho\pi r^{2}hg = 2\pi rT\cos\theta\]
\[\therefore \; h = \frac{2T\cos\theta}{r\rho g}\]
which is the same as Eq. (2.28), obtained now by an entirely different route.
In terms of capillary rise, the expression for surface tension is,
\[T = \frac{r h \rho g}{2\cos\theta} \qquad \text{--- (2.31)}\]
The same expression is also valid for the capillary fall discussed earlier.
Equation (2.31) is important in the laboratory, because it turns capillarity into a way of measuring surface tension. All that has to be observed is the radius of the tube and the height of the column.
2.5 Fluids in Motion
Up to this point every fluid we studied was standing still. We now allow it to move.
We come across moving fluids in our day to day life. The flow of water through our taps, the flow of cooking gas through tubes, or the flow of water through a river or a canal can be understood using the concepts developed in this section.
The branch of Physics which deals with the study of properties of fluids in motion is called hydrodynamics.
As the study of the motion of a real fluid is very complicated, we shall limit our study to the motion of an ideal fluid. We have discussed an ideal fluid in the beginning of this chapter — it is incompressible, irrotational, nonviscous, and its flow is steady. Study of a fluid in motion is very important.
Consider Fig. 2.26, which shows a pipe whose direction and cross sectional area change arbitrarily. The direction of flow of the fluid in the pipe is as shown. We assume an ideal fluid to flow through the pipe.
Diagram
Fig. 2.26 – A curved pipe of varying cross section with an area A marked, showing a bundle of flow lines with arrows running along the pipe and forming a flow tube.
We define a few terms used to describe the flow of a fluid.
Terms Used to Describe Fluid Flow
Steady flow: A measurable property, such as pressure or velocity of the fluid at a given point, is constant over time.
Note carefully what this does and does not say. The velocity may be different at different points — faster in a narrow part of the pipe and slower in a wide part — but at any one chosen point it must not change from one moment to the next.
Flow line: It is the path of an individual particle in a moving fluid, as shown in Fig. 2.26.
Streamline: It is a curve whose tangent at any point in the flow is in the direction of the velocity of the flow at that point.
Streamlines and flow lines are identical for a steady flow.
Flow tube: It is an imaginary bundle of flow lines bound by an imaginary wall. For a steady flow, the fluid cannot cross the walls of a flow tube. Fluids in adjacent flow tubes cannot mix.
A flow tube therefore behaves like a pipe with invisible walls. This idea will be used directly in Section 2.8 to derive the equation of continuity.
Laminar flow / Streamline flow: It is a steady flow in which adjacent layers of a fluid move smoothly over each other, as shown in Fig. 2.27 (a). A steady flow of a river can be assumed to be a laminar flow.
Diagram
Fig. 2.27 (a) – Streamline flow shown as a set of straight, parallel arrows of equal spacing moving in the same direction.
Turbulent flow: It is a flow at a very high flow rate, so that there is no steady flow and the flow pattern changes continuously, as shown in Fig. 2.27 (b). A flooded river flow, or a tap running very fast, is a turbulent flow.
Diagram
Fig. 2.27 (b) – Turbulent flow shown as irregular, curling arrows crossing one another and forming small eddies.
Table 2.3 – Streamline flow and turbulent flow
| Streamline flow | Turbulent flow |
|---|---|
| 1) The smooth flow of a fluid, with velocity smaller than a certain critical velocity (limiting value of velocity), is called streamline flow or laminar flow of a fluid. | 1) The irregular and unsteady flow of a fluid when its velocity increases beyond the critical velocity is called turbulent flow. |
| 2) In a streamline flow, the velocity of a fluid at a given point is always constant. | 2) In a turbulent flow, the velocity of a fluid at any point does not remain constant. |
| 3) Two streamlines can never intersect, i.e., they are always parallel. | 3) In a turbulent flow, at some points, the fluid may have rotational motion which gives rise to eddies. |
| 4) Streamline flow over a plane surface can be assumed to be divided into a number of plane layers. In a flow of liquid through a pipe of uniform cross sectional area, all the streamlines will be parallel to the axis of the tube. | 4) A flow tube loses its order and particles move in random directions. |
2.6 Critical Velocity and Reynolds Number
The flow of a fluid, whether streamline or turbulent, is differentiated on the basis of the velocity of the flow. A tap opened slightly gives a smooth clear stream; opened fully, the same tap gives a broken, noisy one.
The velocity beyond which a streamline flow becomes turbulent is called critical velocity.
According to Osborne Reynolds (1842–1912), critical velocity is given by
\[v_c = \frac{R_n \eta}{\rho d}, \qquad \text{--- (2.32)}\]
where,
\(v_c\) = critical velocity of the fluid, in m s\(^{-1}\)
\(R_n\) = Reynolds number
\(\eta\) = coefficient of viscosity, a measure of the internal friction of the fluid, discussed fully in Section 2.6.2
\(\rho\) = density of fluid, in kg m\(^{-3}\)
\(d\) = diameter of tube, in metre
Equation (2.32) tells us what makes turbulence more likely. A wider tube, a denser fluid or a higher speed all push the flow towards turbulence, while a more viscous fluid resists it — thick oil is far harder to make turbulent than water.
From Eq. (2.32), the equation for Reynolds number can be written as,
\[R_n = \frac{v_c \rho d}{\eta} \qquad \text{--- (2.33)}\]
Reynolds number is a pure number. It has no unit and no dimensions.
It is found that for \(R_n\) less than 1000, the flow of a fluid is streamline, while for \(R_n\) greater than 2000, the flow of the fluid is turbulent. When \(R_n\) is between 1000 and 2000, the flow of the fluid becomes unsteady, i.e., it changes from a streamline flow to a turbulent flow. These ranges are shown in Fig. 2.27c.
Diagram
Fig. 2.27c – A horizontal scale of Reynolds number marked into three regions: below 1000 labelled streamline flow, 1000 to 2000 labelled unsteady, and above 2000 labelled turbulent flow.
2.6.1 Viscosity
When we pour water from a glass, it flows freely and quickly. But when we pour syrup or honey, it flows slowly and sticks to the container.
The difference is due to fluid friction. This friction is both within the fluid itself and between the fluid and its surroundings. This property of fluids is called viscosity. Water has low viscosity, whereas syrup or honey has high viscosity.
Figure 2.28 shows a schematic section of viscous flow.
Diagram
Fig. 2.28 – Arrows of different lengths representing layers of a viscous fluid in a tube, the longest arrow at the centre and the shortest arrows at the walls.
Figure 2.29 shows that of a non-viscous flow.
Diagram
Fig. 2.29 – Arrows of equal length representing layers of a non-viscous fluid, all moving with the same velocity.
Note that there is no dragging force in the non-viscous flow, and all the layers are moving with the same velocity. Viscosity of such a fluid is zero.
The only fluid that is almost non-viscous is liquid helium at about 2 K. Every ordinary fluid you will meet has some viscosity, which is why the ideal fluid of Section 2.2 remains a model rather than a real substance. In this section, we will study the viscosity of a fluid and how it affects the flow of a fluid.
Evidence from a Flowing River
If we observe the flow of river water, it is found that the water near both sides of the river bank flows slow, and as we move towards the centre of the river, the water flows faster gradually. At the centre, the flow is the fastest.
From this observation it is clear that there is some opposing force between two adjacent layers of fluid which affects their relative motion. If there were no such force, every part of the river would move at the same speed.
Viscosity and Viscous Drag
Viscosity is that property of a fluid by virtue of which the relative motion between different layers of a fluid experiences a dragging force. This force is called the viscous drag. This is shown schematically in Fig. 2.30.
Diagram
Fig. 2.30 – A stationary bottom layer with successive layers above it moving faster, marked v − dv, v and v + dv at distances x, x + dx and so on from the stationary layer.
In liquids, the viscous drag is due to short range molecular cohesive forces, and in gases it is due to collisions between fast moving molecules.
These are two quite different mechanisms, and they behave differently with temperature — a point we shall return to in the next section.
In both liquids and gases, as long as the relative velocity between the layers is small, the viscous drag is proportional to the relative velocity.
However, in a turbulent flow, the viscous drag increases rapidly and is not proportional to the relative velocity but proportional to higher powers of the relative velocity. This is one reason why turbulence is avoided wherever possible: it costs far more energy to keep the fluid moving.
Velocity Gradient
To describe how quickly the velocity changes from layer to layer, we need one more quantity.
Velocity gradient: The rate of change of velocity (\(d\mathrm{v}\)) with distance (\(dx\)), measured from a stationary layer, is called the velocity gradient (\(d\mathrm{v}/dx\)).
Its SI unit is (m s\(^{-1}\)) per metre, that is, s\(^{-1}\). A large velocity gradient means the layers are sliding past each other rapidly, and as we shall see next, that is exactly when the viscous force is large.
2.6.2 Coefficient of Viscosity
We now put a number to the viscous drag described in the previous section.
According to Newton’s law of viscosity, for a streamline flow, the viscous force (\(f\)) acting on any layer is directly proportional to the area (\(A\)) of the layer and the velocity gradient (\(d\mathrm{v}/dx\)), i.e.,
\[f \propto A\left(\frac{d\mathrm{v}}{dx}\right)\]
\[\therefore \; f = \eta A\left(\frac{d\mathrm{v}}{dx}\right) \qquad \text{--- (2.34)}\]
where \(\eta\) is a constant, called the coefficient of viscosity of the liquid.
Both factors make physical sense. A larger area means more fluid is in contact and being dragged, and a larger velocity gradient means the layers are being forced to slide past each other more rapidly.
From Eq. (2.34) we can write,
\[\eta = \frac{f}{A\left(\dfrac{d\mathrm{v}}{dx}\right)} \qquad \text{--- (2.35)}\]
Note: ‘\(A\)’ in this expression is not the cross sectional area, it is the area of the layer, parallel to the direction of the flow.
The coefficient of viscosity can be defined as the viscous force per unit area per unit velocity gradient.
The SI unit of viscosity is N s/m\(^2\). Here \(f\) is in newton, \(A\) in m\(^2\) and the velocity gradient in s\(^{-1}\).
A Microscopic View of Viscosity
Viscosity of a fluid can be explained on the basis of molecular motion as follows.
Consider the laminar flow between plates X and Y as shown in Fig. 2.30a. Plate X is stationary and plate Y moves with a velocity \(\mathrm{v}_0\). Layers a, b, and c move with velocities \(\mathrm{v} - d\mathrm{v}\), \(\mathrm{v}\), and \(\mathrm{v} + d\mathrm{v}\) respectively.
Diagram
Fig. 2.30a – Two horizontal plates, the upper plate Y moving to the right and the lower plate X stationary, with three fluid layers a, b and c between them moving with velocities v − dv, v and v + dv.
Consider two adjacent layers, b and c. The velocity of the fluid is equal to the mean velocity of the molecules contained in that layer. Thus, the mean velocity of the molecules in layer b is \(\mathrm{v}\), while the molecules in layer c have a slightly greater mean velocity \(\mathrm{v} + d\mathrm{v}\).
As you will learn in the next chapter, each molecule possesses a random velocity whose magnitude is usually larger than that of the mean velocity. As a result, molecules are continually transferred in large numbers between the two layers.
On the average, molecules passing from layer c to layer b will be moving too fast for their ‘new’ layer by an amount \(d\mathrm{v}\), and will slow down as a result of collisions with the molecules in layer b.
The result is a transfer of momentum from faster-moving layer c to its neighbouring slower-moving layer such as b, and thus eventually to plate X. Because the original source of this transfer of momentum is plate Y, the overall result is a transfer of momentum from plate Y to plate X.
If there are no external forces applied, this momentum transfer would reduce the speed of the plate Y to zero with respect to the plate X.
Reduction in the velocity of the molecules in the direction of laminar flow is due to the fact that their directions after collision are random. This randomness, to be discussed in Chapter 3, results in an increase in the thermal energy of the fluid at the cost of its macroscopic kinetic energy. That is, the process is dissipative, or frictional.
In liquids there is an additional, stronger interaction between molecules in adjacent layers, due to the intermolecular forces that distinguish liquids from gases. As a result, there is a transfer of momentum from faster-moving layers to slower-moving layers, which results in a viscous drag.
Effect of Temperature on the Coefficient of Viscosity
The two opposite behaviours follow directly from the microscopic picture above. In a liquid the drag comes mainly from intermolecular attraction, which weakens on heating. In a gas the drag comes from molecules crossing between layers and carrying momentum with them, and heating makes them cross more often.
Table 2.4 – Coefficient of viscosity at different temperatures
| Fluid | Temperature | Coefficient of Viscosity (N s/m\(^2\)) |
|---|---|---|
| Air | 0 °C | 0.017 × 10\(^{-3}\) |
| Air | 40 °C | 0.019 × 10\(^{-3}\) |
| Water | 20 °C | 1 × 10\(^{-3}\) |
| Water | 100 °C | 0.3 × 10\(^{-3}\) |
| Machine oil | 16 °C | 0.113 × 10\(^{-3}\) |
| Machine oil | 38 °C | 0.034 × 10\(^{-3}\) |
Reading Table 2.4 across the temperature pairs shows the rule at work: the value for air rises with temperature, while the values for water and machine oil fall.
2.7 Stokes’ Law
So far the fluid has been moving past a stationary surface. We now turn the situation around and let a solid object move through a fluid that is at rest.
In 1845, Sir George Gabriel Stokes (1819–1903) stated the law which gives the viscous force acting on a spherical object falling through a viscous medium (see Fig. 2.31).
Diagram
Fig. 2.31 – A small sphere falling downward through a viscous medium, with the streamlines of the fluid curving around it and the viscous force F_v marked acting upward on the sphere.
As the sphere falls, it must push the fluid out of its way, and the layers of fluid dragging along its surface oppose its motion. That opposing force is what Stokes’ law gives.
The law states that, “The viscous force (\(F_v\)) acting on a small sphere falling through a viscous medium is directly proportional to the radius of the sphere (\(r\)), its velocity (\(\mathrm{v}\)) through the fluid, and the coefficient of viscosity (\(\eta\)) of the fluid”.
\[\therefore \; F_v \propto \eta r \mathrm{v}\]
The empirically obtained constant of proportionality is \(6\pi\).
\[\therefore \; F_v = 6\pi\eta r\mathrm{v} \qquad \text{--- (2.36)}\]
Here \(F_v\) is in newton, \(\eta\) in N s m\(^{-2}\), \(r\) in metre and \(\mathrm{v}\) in m s\(^{-1}\).
This is the expression for the viscous force acting on a spherical object moving through a viscous medium. The above formula can be derived using dimensional analysis.
Notice also that \(F_v\) grows with the velocity. A sphere that falls faster meets a larger opposing force — a fact which leads directly to the idea of terminal velocity in the next section.
2.7.1 Terminal Velocity
Consider a spherical object falling through a viscous fluid. Forces experienced by it during its downward motion are,
- Viscous force (\(F_v\)), directed upwards. Its magnitude goes on increasing with increase in its velocity.
- Gravitational force, or its weight (\(F_g\)), directed downwards, and
- Buoyant force or upthrust (\(F_u\)), directed upwards.
Diagram
Fig. 2.32 – A sphere falling inside a vessel of viscous liquid, with the upthrust F_u and viscous force F_v marked upward and the weight mg marked downward.
The net downward force is given by \(f = F_g - (F_v + F_u)\), and it is responsible for the initial increase in the velocity.
Among the given forces, \(F_g\) and \(F_u\) are constant while \(F_v\) increases with increase in velocity. So as the object speeds up, the upward forces grow while the downward force stays the same, and the net force steadily shrinks.
Thus, a stage is reached when the net force \(f\) becomes zero. At this stage,
\[F_g = F_v + F_u\]
After that, the downward velocity remains constant. This constant downward velocity is called terminal velocity.
Obviously, from now onwards the viscous force \(F_v\) is also constant, since the velocity that produced it no longer changes. The entire discussion necessarily applies to streamline flow only.
Expression for Terminal Velocity
Consider a spherical object falling under gravity through a viscous medium as shown in Fig. 2.32. Let the radius of the sphere be \(r\), its mass \(m\) and density \(\rho\). Let the density of the medium be \(\sigma\) and its coefficient of viscosity be \(\eta\).
When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the sphere.
Total downward force = Total upward force
weight of sphere (\(mg\)) = viscous force + buoyant force due to the medium
The weight of the sphere is its volume multiplied by its density and \(g\), while the buoyant force is the weight of the fluid displaced, that is, the same volume multiplied by the density of the medium and \(g\).
\[\frac{4}{3}\pi r^{3}\rho g = 6\pi\eta r\mathrm{v} + \frac{4}{3}\pi r^{3}\sigma g\]
Taking the buoyant term to the other side,
\[6\pi\eta r\mathrm{v} = \left(\frac{4}{3}\pi r^{3}\rho g\right) - \left(\frac{4}{3}\pi r^{3}\sigma g\right)\]
\[6\pi\eta r\mathrm{v} = \left(\frac{4}{3}\right)\pi r^{3}g(\rho - \sigma)\]
\[\mathrm{v} = \left(\frac{4}{3}\right)\pi r^{3}g(\rho - \sigma) \times \frac{1}{6\pi\eta r}\]
\[\mathrm{v} = \left(\frac{2}{9}\right)\frac{r^{2}g(\rho - \sigma)}{\eta} \qquad \text{--- (2.37)}\]
This is the expression for the terminal velocity of the sphere.
Notice also what happens if \(\sigma\) is greater than \(\rho\) — that is, if the object is less dense than the fluid. Then \((\rho - \sigma)\) is negative and so is \(\mathrm{v}\), which simply means that the object moves upward instead of downward. An air bubble rising steadily through oil is described by the very same equation.
From Eq. (2.37) we can also write,
\[\eta = \frac{2}{9}\frac{r^{2}(\rho - \sigma)g}{\mathrm{v}} \qquad \text{--- (2.38)}\]
The above equation gives the coefficient of viscosity of a fluid. This is useful in the laboratory, because measuring how fast a small sphere settles is a practical way of finding \(\eta\) for a liquid.
2.8 Equation of Continuity
Consider a steady flow of an incompressible fluid as shown in Fig. 2.33. For a steady flow, the velocity of a particle remains constant at a given point, but it can vary from point to point.
Diagram
Fig. 2.33 – A curved tube of flow of varying cross section, with the wider section A₁ at point A where the fluid moves through a distance v₁Δt, and the narrower section A₂ at point B where it moves through v₂Δt.
For example, consider sections \(A_1\) and \(A_2\) in Fig. 2.33. Section \(A_1\) has a larger cross sectional area than the section \(A_2\). Let \(\mathrm{v}_1\) and \(\mathrm{v}_2\) be the velocities of the fluid at sections \(A_1\) and \(A_2\) respectively. We shall find that \(\mathrm{v}_2\) is necessarily larger than \(\mathrm{v}_1\).
This is because a particle has to move faster in the narrower section, where there is less space, to accommodate the particles behind it, and hence its velocity increases. When a particle enters a wider section, it slows down because there is more space.
Because the fluid is incompressible, the particles move faster through a narrow section and slow down while moving through a wider section. If the fluid does not move faster in a narrow region, it will be compressed to fit into the narrow space — and an incompressible fluid cannot be compressed.
Conservation of Mass in a Tube of Flow
Consider a tube of flow as shown in Fig. 2.33. All the fluid that passes through a tube of flow must pass through any cross section that cuts the tube of flow. We know that all the fluid is confined to the tube of flow. Fluid cannot leave the tube or enter the tube.
Consider sections \(A_1\) and \(A_2\) located at points A and B respectively, as shown in Fig. 2.33. Matter is neither created nor destroyed within the tube enclosed between section \(A_1\) and section \(A_2\). Therefore, the mass of the fluid within this region is constant over time.
That means, if a mass \(m\) of the fluid enters the section \(A_1\), then an equal mass of fluid should leave the section \(A_2\).
Let the speed of the fluid which crosses the section EFGH at point A in a time interval \(\Delta t\) be \(\mathrm{v}_1\). In this time, the fluid at A advances through a distance \(\mathrm{v}_1 \Delta t\), so the volume that enters is \(A_1 \mathrm{v}_1 \Delta t\).
Thus, the mass of the fluid entering the tube through the cross section at point A is \(\rho A_1 \mathrm{v}_1 \Delta t\).
Similarly, let the speed of the fluid be \(\mathrm{v}_2\) at point B. The fluid crosses the section PQRS of area \(A_2\) in the time interval \(\Delta t\). Thus, the mass of the fluid leaving the tube through the cross section at B is \(\rho A_2 \mathrm{v}_2 \Delta t\).
As the fluid is incompressible, its density \(\rho\) is the same at both places, and the mass of the fluid entering the tube at point A is the same as the mass leaving the tube at B.
Mass of the fluid in section EFGH = mass of fluid in section PQRS
\[\rho A_1 \mathrm{v}_1 \Delta t = \rho A_2 \mathrm{v}_2 \Delta t \qquad \text{--- (2.39)}\]
Cancelling \(\rho\) and \(\Delta t\), which are common to both sides,
\[A_1 \mathrm{v}_1 = A_2 \mathrm{v}_2 \quad \text{or,} \quad A\mathrm{v} = \text{constant} \qquad \text{--- (2.40)}\]
Here \(A\) is in m\(^2\) and \(\mathrm{v}\) in m s\(^{-1}\), so the product \(A\mathrm{v}\) is in m\(^3\) s\(^{-1}\).
Volume Flux and Mass Flux
\(A\mathrm{v}\) is the volume rate of flow of a fluid, i.e.,
\[A\mathrm{v} = \frac{dV}{dt}\]
The quantity \(\dfrac{dV}{dt}\) is the volume of fluid per unit time passing through any cross section of the tube of flow. It is called the volume flux.
Similarly, \(\rho\,dV/dt = dm/dt\) is called the mass flux, and it is measured in kg s\(^{-1}\).
Equation (2.40) is called the equation of continuity in fluid dynamics. The continuity equation says that the volume rate of flow of an incompressible fluid for a steady flow is the same throughout the flow.
2.9 Bernoulli’s Equation
On observing a river, we notice that the speed of the water decreases in the wider region, whereas the speed of the water increases in the regions where the river is narrow. This much we already know from the equation of continuity.
From this we might think that the pressure in narrower regions is more than that in the wider region. However, the pressure within the fluid in the narrower parts is less, while that in the wider parts is more. This is the opposite of what most people expect.
Swiss scientist Daniel Bernoulli (1700–1782), while experimenting with fluid inside pipes, led to the discovery of the concept mentioned above. He observed, in his experiment, that the speed of a fluid in a narrow region increases but the internal pressure of the fluid in the same narrow region decreases. This phenomenon is called Bernoulli’s principle.
Deriving Bernoulli’s Equation
Bernoulli’s equation relates the speed of a fluid at a point, the pressure at that point and the height of that point above a reference level. It is an application of the work–energy theorem for a fluid in flow.
As Bernoulli’s principle is consistent with the principle of conservation of energy, we shall derive it using pressure energy, kinetic energy and gravitational potential energy.
Figure 2.34 shows the flow of an ideal fluid through a tube of varying cross section and height. Consider an element of fluid that lies between cross sections P and R.
Diagram
Fig. 2.34 – A tube of varying cross section rising from a lower end P of area A₁ at height h₁ to an upper end R of area A₂ at height h₂, with speeds v₁ and v₂, pressures P₁ and P₂, and the small displacements d₁ and d₂ marked.
Let,
- \(\mathrm{v}_1\) and \(\mathrm{v}_2\) be the speeds of the fluid at the lower end P and the upper end R respectively.
- \(A_1\) and \(A_2\) be the cross section areas of the fluid at the lower end P and the upper end R respectively.
- \(P_1\) and \(P_2\) be the pressures of the fluid at the lower end P and the upper end R respectively.
- \(d_1\) and \(d_2\) be the distances travelled by the fluid at the lower end P and the upper end R during the time interval \(dt\), with velocities \(\mathrm{v}_1\) and \(\mathrm{v}_2\) respectively.
- Now \(P_1A_1\) and \(P_2A_2\) are the forces acting on the areas \(A_1\) at P and \(A_2\) at R respectively.
The volume \(dV\) of the fluid passing through any cross section during the time interval \(dt\) is the same, by the equation of continuity; i.e.,
\[dV = A_1 d_1 = A_2 d_2 \qquad \text{--- (2.41)}\]
Work done on the fluid element. There is no internal friction in the fluid as the fluid is ideal. In practice also, for a fluid like water, the loss in energy due to viscous force is negligible.
So the only non-gravitational force that does work on the fluid element is due to the pressure of the surrounding fluid. Therefore, the net work, \(W\), done on the element by the surrounding fluid during the flow from P to R is,
\[W = P_1 A_1 d_1 - P_2 A_2 d_2\]
The second term in the above equation has a negative sign because the force at R opposes the displacement of the fluid.
From Eq. (2.41) the above equation can be written as,
\[W = P_1 dV - P_2 dV\]
\[\therefore \; W = (P_1 - P_2)\,dV \qquad \text{--- (2.42)}\]
As the work \(W\) is due to forces other than the conservative force of gravity, it equals the change in the total mechanical energy, i.e., kinetic energy plus gravitational potential energy, associated with the fluid element.
i.e.,
\[W = \Delta K.E. + \Delta P.E. \qquad \text{--- (2.43)}\]
The mechanical energy for the fluid between sections Q and R does not change, because in a steady flow the fluid there is simply replaced by identical fluid moving in the same way. Only the two ends need to be accounted for.
Change in kinetic energy. At the beginning of the time interval \(dt\), the mass and the kinetic energy of the fluid between P and Q are \(\rho A_1 d_1\) and \(\dfrac{1}{2}\rho A_1 d_1 \mathrm{v}_1^{2}\) respectively.
At the end of the time interval \(dt\), the kinetic energy of the fluid between sections R and S is \(\dfrac{1}{2}\rho A_2 d_2 \mathrm{v}_2^{2}\).
Therefore, the net change in the kinetic energy, \(\Delta K.E.\), during the time interval \(dt\) is,
\[\Delta K.E. = \frac{1}{2}\rho A_2 d_2 \mathrm{v}_2^{2} - \frac{1}{2}\rho A_1 d_1 \mathrm{v}_1^{2}\]
Using Eq. (2.41), both \(A_1d_1\) and \(A_2d_2\) may be replaced by \(dV\).
\[\Delta K.E. = \frac{1}{2}\rho\, dV \mathrm{v}_2^{2} - \frac{1}{2}\rho\, dV \mathrm{v}_1^{2}\]
\[\Delta K.E. = \frac{1}{2}\rho\, dV\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) \qquad \text{--- (2.44)}\]
Change in potential energy. Also, at the beginning of the time interval \(dt\), the gravitational potential energy of the mass \(m\) between P and Q is \(mgh_1 = \rho\, dV g h_1\).
At the end of the interval \(dt\), the gravitational potential energy of the mass \(m\) between R and S is \(mgh_2 = \rho\, dV g h_2\).
Therefore, the net change in the gravitational potential energy, \(\Delta P.E.\), during the time interval \(dt\) is,
\[\Delta P.E. = \rho\, dV g h_2 - \rho\, dV g h_1\]
\[\Delta P.E. = \rho\, dV g (h_2 - h_1) \qquad \text{--- (2.45)}\]
Putting it together. Substituting Eq. (2.42), (2.44) and (2.45) in Eq. (2.43) we get,
\[(P_1 - P_2)\,dV = \frac{1}{2}\rho\, dV\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) + \rho\, dV g (h_2 - h_1)\]
Dividing throughout by \(dV\),
\[\therefore \; P_1 - P_2 = \frac{1}{2}\rho\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) + \rho g (h_2 - h_1) \qquad \text{--- (2.46)}\]
This is Bernoulli’s equation. It states that the work done per unit volume of a fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow.
Equation (2.46) can also be written as,
\[P_1 + \frac{1}{2}\rho \mathrm{v}_1^{2} + \rho g h_1 = P_2 + \frac{1}{2}\rho \mathrm{v}_2^{2} + \rho g h_2 \qquad \text{--- (2.47)}\]
or,
\[P + \frac{1}{2}\rho \mathrm{v}^{2} + \rho g h = \text{constant} \qquad \text{--- (2.48)}\]
Every condition in that box matters in numerical work. The fluid must be ideal, the flow steady, and the two points must lie on the same streamline.
In other words, Bernoulli’s principle is thus consistent with the principle of conservation of energy.
Applications of Bernoulli’s Equation
a) Speed of efflux:
The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body.
Consider a liquid of density \(\rho\) filled in a tank of large cross-sectional area \(A_1\) having an orifice of cross-sectional area \(A_2\) at the bottom, as shown in Fig. 2.35. Let \(A_2 \ll A_1\).
Diagram
Fig. 2.35 – A tank of liquid of large cross-sectional area A₁ with the free surface at pressure p, and a small orifice of area A₂ at the bottom at a depth h below the surface, with liquid escaping through it at pressure p₀.
The liquid flows out of the tank through the orifice. Let \(\mathrm{v}_1\) and \(\mathrm{v}_2\) be the speeds of the liquid at \(A_1\) and \(A_2\) respectively.
As both the inlet and the outlet are exposed to the atmosphere, the pressure at these positions equals the atmospheric pressure \(p_0\). If the height of the free surface above the orifice is \(h\), Bernoulli’s equation gives us,
\[P_0 + \frac{1}{2}\rho \mathrm{v}_1^{2} + \rho g h = P_0 + \frac{1}{2}\rho \mathrm{v}_2^{2} \qquad \text{--- (2.49)}\]
Using the equation of continuity we can write,
\[\mathrm{v}_1 = \frac{A_2}{A_1}\mathrm{v}_2\]
Substituting \(\mathrm{v}_1\) in Eq. (2.49) we get,
\[\frac{1}{2}\rho\left(\frac{A_2}{A_1}\right)^{2}\mathrm{v}_2^{2} + \rho g h = \frac{1}{2}\rho \mathrm{v}_2^{2}\]
\[\left(\frac{A_2}{A_1}\right)^{2}\mathrm{v}_2^{2} + 2gh = \mathrm{v}_2^{2}\]
\[2gh = \mathrm{v}_2^{2} - \left(\frac{A_2}{A_1}\right)^{2}\mathrm{v}_2^{2}\]
\[\therefore \; \left[1 - \left(\frac{A_2}{A_1}\right)^{2}\right]\mathrm{v}_2^{2} = 2gh\]
If \(A_2 \ll A_1\), the ratio \(A_2/A_1\) is so small that its square may be neglected against 1, and the above equation reduces to,
\[\mathrm{v}_2 = \sqrt{2gh} \qquad \text{--- (2.50)}\]
This is the equation of the speed of a liquid flowing out through an orifice at a depth \(h\) below the free surface. It is the same as that of a particle falling freely through the height \(h\) under gravity.
b) Ventury tube:
A ventury tube is used to measure the speed of flow of a fluid in a tube. It has a constriction in the tube. As the fluid passes through the constriction, its speed increases in accordance with the equation of continuity. The pressure thus decreases as required by the Bernoulli equation.
Diagram
Fig. 2.36 – A horizontal tube with a narrow constriction, of areas A₁ and A₂, carrying two vertical tubes in which the liquid stands at different levels with a height difference h.
The fluid of density \(\rho\) flows through the ventury tube. The area of cross section is \(A_1\) at the wider part and \(A_2\) at the constriction. Let the speeds of the fluid at \(A_1\) and \(A_2\) be \(\mathrm{v}_1\) and \(\mathrm{v}_2\), and the pressures be \(p_1\) and \(p_2\) respectively.
The tube is horizontal, so the potential head terms are equal and drop out. From Bernoulli’s equation,
\[P_1 + \frac{1}{2}\rho \mathrm{v}_1^{2} = P_2 + \frac{1}{2}\rho \mathrm{v}_2^{2}\]
\[(P_1 - P_2) = \frac{1}{2}\rho\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) \qquad \text{--- (2.51)}\]
Figure 2.36 shows two vertical tubes connected to the ventury tube at \(A_1\) and \(A_2\). If the difference in height of the liquid levels in the tubes is \(h\), we have,
\[(p_1 - p_2) = \rho g h\]
Substituting the above equation in Eq. (2.51) we get,
\[2gh = \mathrm{v}_2^{2} - \mathrm{v}_1^{2} \qquad \text{--- (2.52)}\]
From the equation of continuity, \(A_1\mathrm{v}_1 = A_2\mathrm{v}_2\). Substituting \(\mathrm{v}_1\) in terms of \(\mathrm{v}_2\), or vice versa, in Eq. (2.52), the rate of flow of liquid passing through a cross section can be calculated by knowing the areas \(A_1\) and \(A_2\).
So a simple reading of a height difference gives the rate of flow, without any moving part being placed in the stream.
c) Lifting up of an aeroplane:
The shape of the cross section of the wings of an aeroplane is as shown in Fig. 2.37.
Diagram
Fig. 2.37 – Cross section of an aerofoil with streamlines of air crowded closely above the curved upper surface and spaced more widely below the flatter lower surface.
When an aeroplane runs on a runway, due to the aerodynamic shape of its wings, the streamlines of air are crowded above the wings compared to those below the wings. Thus, the air above the wings moves faster than that below the wings.
According to Bernoulli’s principle, the pressure above the wings decreases and that below the wings does not change. Due to this pressure difference, an upward force called the dynamic lift acts on the bottom of the wings of the plane.
When this force becomes greater than the weight of the aeroplane, the aeroplane takes off. This is why an aircraft must reach a certain speed on the runway before it can leave the ground — the lift depends on how fast the air moves past the wing.
d) Working of an atomizer:
The action of the carburettor of an automobile engine, a paint-gun, a scent-spray or an insect-sprayer is based on Bernoulli’s principle.
In all these, a tube T is dipped in a liquid as shown in Fig. 2.38.
Diagram
Fig. 2.38 – An atomizer: a piston P inside a cylinder C blowing air at high speed across the open top of a vertical tube T that dips into a container of liquid, with a fine spray emerging.
Air is blown at high speed over the tip of this tube with the help of a piston P in the cylinder C. This high speed air creates low pressure over the tube, due to which the liquid rises in it and is then blown off in very small droplets with the expelled air.
Notice that nothing pushes the liquid up the tube. The atmospheric pressure on the surface of the liquid in the container is unchanged; it is the pressure at the top of the tube that has been lowered, and the liquid is pushed up into the difference.
e) Blowing off of roofs by stormy wind:
When a high speed, stormy wind blows over a roof top, it causes low pressure \(p\) above the roof in accordance with Bernoulli’s principle.
However, the air below the roof, i.e., inside the room, is still at the atmospheric pressure \(p_0\). So, due to this difference in pressure, the roof is lifted up and is then blown off by the wind, as shown in Fig. 2.39.
Diagram
Fig. 2.39 – Wind streamlines curving swiftly over the sloping roof of a house, with the reduced pressure p marked above the roof and the atmospheric pressure p₀ marked inside the room below.
2.10 Summary
Fluids and Fluids at Rest
- A fluid is a substance that can flow and that deforms continually under an external force. Liquids and gases are fluids; the shear modulus of a fluid is zero.
- An ideal fluid is incompressible, has irrotational flow, is non-viscous, and its flow is steady.
- Hydrostatics deals with fluids at rest; hydrodynamics deals with fluids in motion.
- Pressure is the normal force exerted by a fluid per unit area of contact: \[P = \frac{F}{A} \qquad \text{--- (2.1)}\] Its SI unit is N/m\(^2\) = 1 pascal (Pa), its dimensions are \([L^{-1}M^{1}T^{-2}]\), and it is a scalar quantity. Also, 1 bar = \(10^{5}\) N m\(^{-2}\), 1 hPa = 100 Pa, and 1 torr = 1 mm of mercury column.
- Pressure due to a liquid column of density \(\rho\) at a depth \(h\) below the free surface: \[p = h\rho g \qquad \text{--- (2.2)}\] This is independent of the shape and size of the container, and is valid for liquids as well as gases.
- Atmospheric pressure at a point is the weight of a column of air of unit cross section above that point. Normal atmospheric pressure is the value at sea level, \(P_{atm}\) = 76 cm of Hg = 760 mm of Hg = 760 torr.
- General relation between the pressures at two levels in a fluid at rest: \[p_2 = p_1 + \rho g (x_1 - x_2) \qquad \text{--- (2.6)}\]
- Absolute pressure at depth \(h\): \(\;p = p_0 + h\rho g\) — (2.7)
- Gauge pressure at depth \(h\): \(\;p - p_0 = h\rho g\) — (2.8)
- Atmospheric pressure at a height \(d\) above a liquid surface: \(\;p = p_o - d\rho_{air}g\) — (2.9)
- Hydrostatic paradox: liquid stands at the same level in interconnected vessels of different shapes, because pressure depends only on the height of the column above a point.
- Pascal’s law: pressure applied at any point of an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and to the walls of the container, provided the effect of gravity is neglected.
- Hydraulic lift and hydraulic brakes are applications of Pascal’s law: \[F_2 = F_1\left(\frac{A_2}{A_1}\right) \qquad \text{--- (2.10)}\]
- A mercury barometer measures atmospheric pressure, \(p_0 = h\rho g\). An open tube manometer measures the pressure of an enclosed gas, \(p = p_0 + h\rho g\) — (2.14), and its gauge pressure is \(p - p_0 = h\rho g\).
Surface Tension
- Cohesive force acts between molecules of the same substance; adhesive force acts between molecules of different substances. Both are short range, effective only up to a few nanometre (\(10^{-9}\) m), within the sphere of influence of a molecule.
- Molecules in the surface film experience a net inward cohesive force, so the surface behaves like a stretched membrane and tries to minimize its area. Gases have no free surface and so show no surface tension.
- Surface tension is the tangential force per unit length acting on both sides of an imaginary line drawn on the free surface of a liquid: \[T = \frac{F}{L} \qquad \text{--- (2.15)}\] Its SI unit is N/m and its dimensions are \([L^{0}M^{1}T^{-2}]\).
- For a film with two surfaces, the force on an arm of length \(L\) is \(F = (T)\times(2L)\) — (2.16).
- Surface energy is the extra potential energy of the molecules in the surface layer. Work done isothermally in increasing the area by \(dA\) is stored as surface energy: \[\text{Surface energy} = T(dA) \qquad \text{--- (2.17)}\] Hence surface tension is numerically equal to surface energy per unit area.
- The angle of contact \(\theta\) is measured between the tangent to the liquid surface and the solid surface at the point of contact, within the liquid. An acute angle gives a concave meniscus and wetting; an obtuse angle gives a convex meniscus and non-wetting; \(\theta = 0\) for pure water on clean glass and \(\theta = 140^{0}\) for mercury on clean glass.
- For a drop resting on a solid surface, with \(T_1\), \(T_2\) and \(T_3\) the surface tensions at the liquid–solid, air–solid and air–liquid interfaces: \[\cos\theta = \frac{T_2 - T_1}{T_3} \qquad \text{--- (2.18)}\] If \(T_2 > (T_1 + T_3)\), then \(\cos\theta > 1\), which is impossible, and the liquid spreads without forming a drop.
- The angle of contact depends on the nature of the liquid and solid, on impurities, and on temperature; it decreases as temperature rises.
- Soluble impurities such as common salt increase the surface tension of water; sparingly soluble substances such as detergents, and insoluble impurities, decrease it. In most liquids surface tension decreases with rise in temperature, and becomes zero at the critical temperature. Molten copper and molten cadmium are exceptions.
- Across a curved liquid surface, the pressure on the concave side is always greater than that on the convex side; across a plane surface the two are equal.
- Excess pressure inside a liquid drop (Laplace’s law of a spherical membrane): \[(p_i - p_0) = \frac{2T}{r} \qquad \text{--- (2.23)}\]
- Excess pressure inside a soap bubble, which has two surfaces: \[(p_i - p_0) = \frac{4T}{r} \qquad \text{--- (2.24)}\]
- Capillarity is the rise or fall of a liquid in a tube of fine bore. For a capillary of radius \(r\) and a liquid of angle of contact \(\theta\): \[h = \frac{2T\cos\theta}{r\rho g} \qquad \text{--- (2.28)}\] A positive \(h\) means capillary rise and a negative \(h\) (obtuse \(\theta\)) means capillary fall. The narrower the tube, the greater the rise or fall.
- In terms of capillary rise, the surface tension is \[T = \frac{r h \rho g}{2\cos\theta} \qquad \text{--- (2.31)}\]
Fluids in Motion
- A flow line is the path of a fluid particle; a streamline is a curve whose tangent gives the flow velocity at each point; the two coincide for steady flow. A flow tube is an imaginary bundle of flow lines that the fluid cannot cross.
- Laminar (streamline) flow is smooth and orderly; turbulent flow is irregular, unsteady and forms eddies. Two streamlines never intersect.
- Critical velocity is the speed beyond which streamline flow becomes turbulent: \[\mathrm{v}_c = \frac{R_n \eta}{\rho d} \qquad \text{--- (2.32)}\] \[R_n = \frac{\mathrm{v}_c \rho d}{\eta} \qquad \text{--- (2.33)}\] Reynolds number \(R_n\) is a pure number with no unit or dimensions. Flow is streamline for \(R_n\) less than 1000, turbulent for \(R_n\) greater than 2000, and unsteady in between.
- Viscosity is the property by which relative motion between layers of a fluid experiences a dragging force, called viscous drag. The velocity gradient is \(d\mathrm{v}/dx\).
- Newton’s law of viscosity: \[f = \eta A\left(\frac{d\mathrm{v}}{dx}\right) \qquad \text{--- (2.34)}\] \[\eta = \frac{f}{A\left(\dfrac{d\mathrm{v}}{dx}\right)} \qquad \text{--- (2.35)}\] Here \(A\) is the area of the layer parallel to the flow, not the cross section. The SI unit of \(\eta\) is N s/m\(^2\). For most liquids \(\eta\) decreases with rise in temperature, while for gases it increases.
- Stokes’ law, for a small sphere moving through a viscous medium: \[F_v = 6\pi\eta r\mathrm{v} \qquad \text{--- (2.36)}\]
- Terminal velocity is reached when weight is balanced by the viscous force and the upthrust, \(F_g = F_v + F_u\): \[\mathrm{v} = \left(\frac{2}{9}\right)\frac{r^{2}g(\rho - \sigma)}{\eta} \qquad \text{--- (2.37)}\] \[\eta = \frac{2}{9}\frac{r^{2}(\rho - \sigma)g}{\mathrm{v}} \qquad \text{--- (2.38)}\] where \(\rho\) is the density of the sphere and \(\sigma\) that of the medium.
- Equation of continuity, for the steady flow of an incompressible fluid: \[A_1\mathrm{v}_1 = A_2\mathrm{v}_2 \quad \text{or} \quad A\mathrm{v} = \text{constant} \qquad \text{--- (2.40)}\] \(A\mathrm{v} = dV/dt\) is the volume flux and \(\rho\,dV/dt = dm/dt\) is the mass flux.
- Bernoulli’s principle: where the speed of a flowing fluid is greater, the pressure within it is smaller.
- Bernoulli’s equation, for the steady streamline flow of an ideal fluid: \[P_1 - P_2 = \frac{1}{2}\rho\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) + \rho g(h_2 - h_1) \qquad \text{--- (2.46)}\] \[P_1 + \frac{1}{2}\rho \mathrm{v}_1^{2} + \rho g h_1 = P_2 + \frac{1}{2}\rho \mathrm{v}_2^{2} + \rho g h_2 \qquad \text{--- (2.47)}\] \[P + \frac{1}{2}\rho \mathrm{v}^{2} + \rho g h = \text{constant} \qquad \text{--- (2.48)}\] The three terms are called the pressure head, the velocity head and the potential head.
- Speed of efflux through an orifice at depth \(h\), for an orifice much smaller than the tank: \[\mathrm{v}_2 = \sqrt{2gh} \qquad \text{--- (2.50)}\] which is the same as the speed of a body falling freely through a height \(h\).
- Ventury tube, for a horizontal tube with a constriction: \[(P_1 - P_2) = \frac{1}{2}\rho\left(\mathrm{v}_2^{2} - \mathrm{v}_1^{2}\right) \qquad \text{--- (2.51)}\] \[2gh = \mathrm{v}_2^{2} - \mathrm{v}_1^{2} \qquad \text{--- (2.52)}\]
- Other applications of Bernoulli’s principle include the dynamic lift on an aeroplane wing, the working of an atomizer, carburettor and spray gun, and the blowing off of roofs by stormy winds.
2.11 Objective Questions
A hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg. The area of cross section of the piston carrying the load is \(2.25 \times 10^{-2}\) m\(^2\). What is the maximum pressure the piston would have to bear? (A) \(0.8711 \times 10^{6}\) N/m\(^2\) (B) \(0.5862 \times 10^{7}\) N/m\(^2\) (C) \(0.4869 \times 10^{5}\) N/m\(^2\) (D) \(0.3271 \times 10^{4}\) N/m\(^2\)
Two capillary tubes of radii 0.3 cm and 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1
The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly (A) \(0.9 \times 10^{-3}\) J (B) \(0.4 \times 10^{-3}\) J (C) \(0.7 \times 10^{-3}\) J (D) \(0.5 \times 10^{-3}\) J
Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is (A) 1:2 (B) 1:12 (C) 1:16 (D) 1:8
In Bernoulli’s theorem, which of the following is conserved? (A) linear momentum (B) angular momentum (C) mass (D) energy
2.12 Check Your Concepts
A solid block and an equal volume of water are each pushed sideways by a small tangential force. Explain why the block settles into a new shape and stops, while the water keeps moving.
Force is a vector quantity, and pressure is defined as force divided by area. Explain why pressure is nevertheless a scalar quantity.
Two vessels of very different shapes are filled with the same oil up to the same height. Their bases have equal area. Compare the pressures at the two bases and also the total forces on the two bases, and explain any difference between your two answers.
A tyre gauge reads zero when connected to a completely flat tyre, although the air inside the tyre is certainly not absent. What does the gauge actually measure, and what is the absolute pressure inside that tyre?
A hydraulic lift raises a car of weight 12000 N when a person presses the small piston with a force of 200 N. Explain why this does not violate the principle of conservation of energy.
Explain why a barometer made with water would have to be more than ten metres tall, whereas a mercury barometer stands about 76 cm high.
Why do gases not exhibit surface tension, although they are fluids and were treated along with liquids while studying pressure?
A greased steel needle, which is about eight times denser than water, can be made to rest on a water surface. Explain why this is not an example of flotation by buoyancy.
Assertion: A liquid drop always tends to take a spherical shape. Reason: For a given volume, a sphere has the smallest surface area. State whether the assertion and the reason are correct, and whether the reason correctly explains the assertion.
A small soap bubble and a large soap bubble are blown at the two ends of a tube fitted with a tap in the middle. Predict what happens when the tap is opened, and justify your prediction using the expression for excess pressure.
For the same radius and the same liquid, the excess pressure inside a soap bubble is twice that inside a liquid drop. Explain the physical reason for this factor of two.
Water rises in a clean glass capillary, but mercury falls in the same capillary. Explain both observations using the angle of contact and the shape of the meniscus.
A student performs a capillary rise experiment but obtains a much smaller rise than expected. On inspection, the inside of the capillary tube is found to be slightly greasy. Explain how this would affect the result.
Explain why a detergent is added to water for washing clothes, while cold water is recommended for washing a new coloured cotton fabric.
The coefficient of viscosity of a liquid decreases with rise in temperature, while that of a gas increases. Explain both trends in terms of the mechanism responsible for viscous drag in each case.
Two spherical raindrops of the same material fall through air, one of radius \(r\) and the other of radius \(3r\). Without calculating any numerical value, state which one falls faster at terminal velocity and by what factor, and explain your reasoning.
Why is Reynolds number defined as a combination of quantities that leaves it dimensionless? State one practical advantage of this.
The relation \(A_1\mathrm{v}_1 = A_2\mathrm{v}_2\) is used freely for water but not for a gas being compressed as it flows. Explain which more general relation holds in both cases, and what extra condition reduces it to the simple form.
Water flows steadily through a horizontal pipe that narrows and then widens again. Sketch qualitatively how the speed and the pressure vary along the pipe, and explain why the pressure is least where the speed is greatest.
A stormy wind blows over a house and the roof is lifted off while the walls remain intact. Explain why the roof is affected and the walls are not.
Explain why the speed of efflux from an orifice does not depend on the direction in which the orifice points, nor on the density of the liquid in the tank.
A ventury tube contains no moving parts and nothing is inserted into the flowing stream, yet it can measure the rate of flow. Explain the principle on which it works.
Why is the surface tension of paints and lubricating oils kept low?
How much amount of work is done in forming a soap bubble of radius r?
What is the basis of Bernoulli’s principle?
Why is a low density liquid used as a manometric liquid in a physics laboratory?
What is an incompressible fluid?
Why do two or more mercury drops form a single drop when brought in contact with each other?
Why does velocity increase when water flowing in a broader pipe enters a narrow pipe?
Why does the speed of a liquid increase and its pressure decrease when a liquid passes through a constriction in a horizontal pipe?
Derive an expression for excess pressure inside a liquid drop.
Obtain an expression for conservation of mass starting from the equation of continuity.
Explain the capillary action.
Derive an expression for capillary rise for a liquid having a concave meniscus.
2.13 Practice with Numericals
A force of 45 N acts normally on a surface of area 9.0 cm\(^2\). Calculate the pressure exerted on the surface in N m\(^{-2}\) and express it in hectapascal.
Calculate the gauge pressure and the absolute pressure at a depth of 25 m below the surface of a fresh water lake. Take the density of water as 1000 kg m\(^{-3}\), \(p_0 = 1.01 \times 10^{5}\) Pa and \(g\) = 9.8 m s\(^{-2}\).
Two immiscible liquids of densities 800 kg m\(^{-3}\) and 1200 kg m\(^{-3}\) exert the same pressure at the bottoms of two separate vessels. If the column of the lighter liquid is 30 cm tall, find the height of the column of the heavier liquid.
In a hydraulic press, the smaller piston has a radius of 3.0 cm and the larger piston a radius of 15 cm. A force of 400 N is applied to the smaller piston. Calculate the pressure produced in the liquid and the force obtained at the larger piston.
A mercury barometer reads 72 cm of mercury at a certain hill station. Taking the density of mercury as \(13.6 \times 10^{3}\) kg m\(^{-3}\) and \(g\) = 9.8 m s\(^{-2}\), calculate the atmospheric pressure there in N m\(^{-2}\).
An open tube manometer containing kerosene of density 800 kg m\(^{-3}\) shows a difference of level of 18 cm between its two arms when connected to a gas cylinder. Calculate the gauge pressure of the gas. Take \(g\) = 9.8 m s\(^{-2}\).
A circular beaker of radius 8.0 cm is filled with glycerin, whose surface tension is 0.0632 N/m. Calculate the force of surface tension acting on a diametrical line drawn on the free surface.
A soap film is formed on a rectangular frame one side of which is a movable wire of length 12 cm. Calculate the work done in moving this wire outward through 5.0 cm, if the surface tension of the soap solution is 0.025 N/m.
Calculate the excess pressure inside a water drop of radius 0.5 mm. The surface tension of water is 0.0727 N/m.
A soap bubble of radius 1.5 cm is blown from a solution of surface tension 0.030 N/m. Calculate the excess pressure inside it and the total surface energy of the bubble.
Water rises in a clean glass capillary tube of radius 0.20 mm. Taking the surface tension of water as \(7.0 \times 10^{-2}\) N/m, the angle of contact as zero, the density of water as 1000 kg m\(^{-3}\) and \(g\) = 9.8 m s\(^{-2}\), calculate the height of the rise.
A capillary tube of radius \(4.0 \times 10^{-4}\) m is dipped in mercury of surface tension 0.465 N/m and density \(13.6 \times 10^{3}\) kg m\(^{-3}\). If the angle of contact is 140°, calculate the depression of the mercury level in the tube. Take \(g\) = 9.8 m s\(^{-2}\).
A liquid of density 900 kg m\(^{-3}\) rises to a height of 2.4 cm in a capillary tube of radius 0.30 mm. If the angle of contact is zero, calculate the surface tension of the liquid. Take \(g\) = 9.8 m s\(^{-2}\).
A metal plate of area 0.20 m\(^2\) rests on a layer of oil 2.0 mm thick. A horizontal force of 0.60 N moves the plate with a uniform velocity of 0.030 m s\(^{-1}\). Calculate the velocity gradient and the coefficient of viscosity of the oil.
A glass sphere of radius 0.50 mm and density \(2.5 \times 10^{3}\) kg m\(^{-3}\) falls through a liquid of density \(1.0 \times 10^{3}\) kg m\(^{-3}\) and coefficient of viscosity 0.90 N s m\(^{-2}\). Calculate its terminal velocity. Take \(g\) = 9.8 m s\(^{-2}\).
Glycerin of density \(1.26 \times 10^{3}\) kg m\(^{-3}\) and coefficient of viscosity 0.83 N s m\(^{-2}\) flows through a tube of diameter 1.5 cm at a speed of 0.40 m s\(^{-1}\). Calculate the Reynolds number and state the nature of the flow.
Water flows through a horizontal pipe of internal diameter 6.0 cm at a speed of 1.0 m s\(^{-1}\). The pipe narrows to an internal diameter of 2.0 cm. Calculate the speed of water in the narrow section and the volume rate of flow in m\(^3\) s\(^{-1}\).
Water of density 1000 kg m\(^{-3}\) flows through a horizontal pipe. At a point where the speed is 2.0 m s\(^{-1}\) the pressure is \(1.8 \times 10^{5}\) Pa. Calculate the pressure at a second point in the same pipe where the speed is 6.0 m s\(^{-1}\).
A large tank of water has a small orifice in its side at a depth of 3.2 m below the free surface. Calculate the speed with which water emerges from the orifice. Take \(g\) = 9.8 m s\(^{-2}\).
In a ventury tube carrying water, the speed at the wider section is 1.2 m s\(^{-1}\) and that at the constriction is 4.0 m s\(^{-1}\). Calculate the difference in the water levels in the two vertical tubes attached to it. Take \(g\) = 9.8 m s\(^{-2}\).
A gale blows over the flat roof of a hut at a speed of 25 m s\(^{-1}\) while the air inside is still. The roof measures 6.0 m by 4.0 m and the density of air is 1.29 kg m\(^{-3}\). Calculate the pressure difference across the roof and the resultant upward force on it.
Find the pressure 200 m below the surface of the ocean if the pressure on the free surface of the liquid is one atmosphere. (Density of sea water = 1060 kg/m\(^3\))
In a hydraulic lift, the input piston has surface area 30 cm\(^2\) and the output piston has surface area of 1500 cm\(^2\). If a force of 25 N is applied to the input piston, calculate the weight on the output piston.
Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is \(1.8 \times 10^{-5}\) Ns/m\(^2\).
A horizontal force of 1 N is required to move a metal plate of area \(10^{-2}\) m\(^2\) with a velocity of \(2 \times 10^{-2}\) m/s, when it rests on a layer of oil \(1.5 \times 10^{-3}\) m thick. Find the coefficient of viscosity of the oil.
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m\(^2\) and specific gravity 0.9? Density of air is 1.29 kg/m\(^3\).
The speed of water is 2 m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of the nozzle of the pipe if the speed of water at the nozzle is 4 m/s?
With what velocity does water flow out of an orifice in a tank with gauge pressure \(4 \times 10^{5}\) N/m\(^2\) before the flow starts? Density of water = 1000 kg/m\(^3\).
The pressure of water inside a closed pipe is \(3 \times 10^{5}\) N/m\(^2\). This pressure reduces to \(2 \times 10^{5}\) N/m\(^2\) on opening the valve of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = 1000 kg/m\(^3\))
Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension \(7 \times 10^{-2}\) N/m. The angle of contact between water and glass is zero, density of water = 1000 kg/m\(^3\), g = 9.8 m/s\(^2\).
An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = \(7.2 \times 10^{-2}\) N/m.
Twenty seven droplets of water, each of radius 0.1 mm, coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.
A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm.
How much work is required to form a bubble of 2 cm radius from a soap solution having surface tension 0.07 N/m?
A rectangular wire frame of size 2 cm × 2 cm is dipped in a soap solution and taken out. A soap film is formed. If the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of the soap film is \(3 \times 10^{-2}\) N/m.