Course Content

Properties of Matter  Lessons

Properties of Matter  Quiz 1 and 2

Properties of Matter  Quiz 3 and 4
Mechanical Properties of Fluids
Introduction
In this chapter, let us study some common physical properties of liquids and gases. Liquids and gases can flow and are therefore fluids. It is this property that distinguishes liquids and gases from solids basically.
Fluids are everywhere around us. Earth has an envelop of air and twothirds of its surface is covered with water. Water is not necessary for our existence; every mammalian body constitute mostly of water. All the processes occurring in living beings including plants are meditated by fluids. Thus understanding the behaviour and properties of fluids is important.
How are fluids different from solids? What is common in liquids and gases? Unlike a solid, a fluid has no definite shape of its own. Solids and liquids have a fixed volume, whereas gas fills the entire volume of of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words, solids and liquids have much lower compressibility as compared to gases.
Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress. The shearing stress of fluids is about million times smaller than that of solids.
Pressure
A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose chest a large, light but strong wooden plank is placed first, is saved from this accident. Such everyday experiences convince us that both the force and its coverage area are important. Smaller the area on which the force acts, greater is the impact. This concept is known as pressure.
When an object is submerged in a fluid at rest, the fluid exerts a force on its surface. This force is always normal to the object’s surface. This is so because if there were a component of force parallel to the surface, the object will also exert a force on the fluid parallel to it; as a consequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot happen. Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Fig. 1(a).
(a) The force exerted by the liquid in the beaker on the submerged object or on the walls is normal (perpendicular) to the surface at all points
(b) An idealised device for measuring pressure
The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressuremeasuring device is shown is in Fig. 1(b). It consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston. This device is placed at a point inside the fluid. The inward force exerted by the fluid on the piston is balanced by the outward spring force and is thereby measured.
If F is the magnitude of this normal force on the piston of area A then the average pressure P_{av} is defined as the normal force acting per unit area.
P_{av} = F/A …………………………………………………………………………..(1)
In principle, the piston area can be made arbitrarily small. The pressure is then defined in limiting sense as
Pressure is a scalar quantity. We remind the reader that it is the component of the force normal to the area under consideration and not the (vector) force that appears in the numerator in Eqs. (1) and (2). Its dimensions are [ML^{1}T^{2}]. The SI unit of pressure is N m^{2}. It has been named as pascal (Pa) in honour of the French scientist Blaise Pascal (16231662) who carried out pioneering studies on the fluid pressure.
Another quantity, that is indispensable in describing fluids, is the density ρ. For a fluid of mass m occupying volume V,
ρ = m/V …………………………………………………………………………….(3)
The dimensions of density are [ML^{3}]. Its SI unit is kg m^{3}. It is a positive scalar quantity. A liquid is largely incompressible and its density is therefore, nearly constant at all pressures. Gases, on the other hand exhibit a large variation in densities with pressure.
The density of water at 4°C (277 K) is 1.0 × 10 ³ kg m^{3}. The relative density of a substance is the ratio of its density to the density of water at 4°C. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is 2.7 × 10^{3}kg m^{3}. The densities of some common fluids are displayed in Table 1.
Table 1 – Densities of some common fluids at STP(STP means standard temperature 0°C and 1 atm pressure).
Fluid  ρ ( kg m^{3} ) 

Water  1.00 × 10³ 
Sea water  1.03 × 10³ 
Mercury  13.6 × 10³ 
Ethyl alcohol  0.806 × 10³ 
Whole blood  1.06 × 10³ 
Air  1.29 
Oxygen  1.43 
Hydrogen  9.0 × 10^{2} 
Interstellar space  ≈ 10^{20} 
Example 1
The two thigh bones (femurs), each of crosssectional area 10 cm² support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer
The crosssectional area of the femurs is A = 2 × 10 cm² = 20 × 10^{4} m². The force acting on them is F = 40 kg wt = 400 N (taking g = 10 m s^{2}). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
P_{av} = F/A = 2 ×10^{5 }N m^{2}
Pascal’s Law
French scientist Blaise Pascal observed that the pressure in a fluid at rest is the same at all points if they are at the same height. This fact may be demonstrated in a simple way.
Proof of Pascal’s law, ABCDEF is an element of the interior of a fluid at rest. This element is in the form of a rightangled prism. The element is small so that the effect of gravity can be ignored, but it has been enlarged for the sake of clarity.
Fig. 2 shows an element in the interior of a fluid at rest. This element ABCDEF is in the form of a right angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points. But for clarity, we have enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element as discussed above. Thus, the fluid exerts pressures P_{a}, P_{b} and P_{c} on this element of area corresponding to the normal forces F_{a}, F_{b} and F_{c} as shown in Fig. 2 on the faces BEFC, ADFC and ADEB denoted by A_{a}, A_{b} and A_{c} respectively. Then
F_{b }sin θ = F_{c}, F_{b }cos θ = F_{a }(by equilibrium)
A_{b }sin θ = A_{c}, A_{b }cos θ = A_{a }(by geometry)
Thus,
F_{b }/A_{b }= F_{c }/A_{c}; P_{b }/P_{c }= P_{a } ………………..(4)
Hence, pressure exerted is same in all directions in a fluid at rest. It again reminds us that like other types of stress, pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.
Now consider a fluid element in the form of a horizontal bar of uniform crosssection. The bar is in equilibrium. The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere. Wind is Flow of air due to pressure differences.
Variation of Pressure with Depth
Consider a fluid at rest in a container. In Fig. 3 point 1 is at height h above a point 2. The pressure at points 1 and 2 are P_{1 }and P_{2 }respectively. Consider a cylindrical element of fluid having area of base A and height h. As the fluid is at rest the resultant horizontal forces should be zero and the resultant vertical forces should balance the weight of the element. The forces acting in the vertical direction are due to the fluid pressure at the top (P_{1 }A) acting downward, at the bottom (P_{2 }A) acting upward. If mg is the weight of the fluid in the cylinder we have,
(P_{2 }– P_{1 }) A = mg …………………………………………………(5)
Now, if ρ si the mass density of the fluid, we have the mass of fluid to be m = ρV = ρhA so that
(P_{2 }– P_{1 }) = ρgh …………………………………………………….(6)
Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical cylindrical column
Pressure difference depends on the vertical distance h between the points(1 and 2), mass density of the fluid ρ and acceleration due to gravity g. If the point 1 under discussion is shifted to the top of the fluid (say water), which is open to the atmosphere, P_{1 }may be replaced by atmospheric pressure (P_{a }) and we replace P_{2 }by P. Then Eq. (6) gives
P = P_{a }+ ρgh ………………………………………………………………………….(7)
Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh. The excess of pressure, P – P_{a }, at depth h is called a gauge pressure at that point.
The area of the cylinder is not appearing in the expression of absolute pressure in Eq. (7). Thus, the height of the fluid column is important and not crosssectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth). The result is appreciated through the example of hydrostatic paradox. Consider three vessels A, B and C [Fig. (4)] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water the level in the three vessels is the same though they hold different amounts of water. This is so, because water at the bottom has the same pressure below each section of the vessel.
Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all up to the same height
What is the pressure on a swimmer 10 m below the surface of lake?
Answer
Here,
h = 10 m and ρ = 1000 kg m^{3}. Take g = 10 m s^{2}.
From Eq. (7)
P = P_{a }+ ρgh
= 1.01 × 10^{3}Pa + 1000 kg m^{3 }× 10 ms^{2}× 10m
= 2.01 × 10^{5}Pa
≈ 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.
Atmospheric Pressure and Gauge Pressure
The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the atmosphere. At sea level it is 1.013 × 10 ^{5 }Pa (1 atm). Italian scientist Evangelista Torricelli (16081647) devised for the first time, a method for measuring atmospheric pressure. A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury as shown in Fig. 5(a). This device is known as mercury barometer. The space above the mercury column in the tube contains only mercury vapour whose pressure P is so small that it may be neglected. The pressure inside the column at point A must equal the pressure at point B, which is at the same level. Pressure at B = atmospheric pressure = P_{a}.
P_{a } = ρgh …………………………………………………….(8)
where ρ is the density of mercury and h is the height of the mercury column in the tube.
In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli).
1 torr =133 Pa.
The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar.
1 bar = 10^{5 }Pa
An opentube manometer is a useful instrument for measuring pressure differences. It consists of a Utube containing a suitable liquid. i.e., a low density liquid (such as oil) for measuring small pressure differences and a high density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure [see Fig. 5(b)]. The pressure P at A is equal to pressure at point B. What we normally measure is the gauge pressure, which is PP_{a }, given by Eq. (8) and is proportional to manometer height h.
Fig. 5 Two pressure measuring devices(a) The mercury barometer. (b) the open tube manometer
Pressure is same at the same level on both sides of the Utube containing a fluid. For liquids the density varies very little over wide ranges in pressure and temperature and we can treat it safely as a constant for our present purposes. Gases on the other hand, exhibits large variations of densities with changes in pressure and temperature. Unlike gases, liquids are therefore, largely treated as incompressible.
Example 3
The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Then how high would the atmosphere extend?
Answer
We use Eq. (7)
ρgh = 1.29 kg m^{3} × 9.8 m s^{2} × h m = 1.01 × 10^{5}Pa
∴ h = 7989 m = 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.
Example 4
At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sealevel atmospheric pressure. (The density of sea water is 1.03 × 10³ kg m^{3}, g = 10 m s^{2}.)
Answer
Here h = 1000 m and ρ = 1.03 × 10³ kg m^{3}.
(a) From Eq. (6), absolute pressure
P = P_{a }+ ρgh
=1.01 × 10^{5} Pa + 1.03 × 10³ kg m^{3 }× 10 ms^{2 }× 1000 m
= 104.01× 10^{5} Pa
≈104 atm
(b) Gauge pressure is P – P_{a }= ρgh = P_{g}
P_{g }= 1.03 × 10³ kg m^{3 }× 10 ms^{2 }× 1000 m
= 103 × 10^{5} Pa
≈ 103 atm
(c) The pressure outside the submarine is P = P_{a }+ ρgh and the pressure inside it is P_{a}. Hence,the net pressure inside it is P_{a}. Hence, the net pressure acting on the window is gauge pressure, P_{g }= ρgh. Since the area of the window is A = 0.04 m², the force acting on it is
F = P_{g }A = 103 × 10^{5} Pa × 0.04 m² = 4.12 × 10^{5} N
Hydraulic Machines
Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.
This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is the Pascal’s law for transmission of fluid pressure and has many applications in daily life.
A number of devices such as hydraulic lift and hydraulic brakes are based on the Pascal’s law. In these devices fluids are used for transmitting pressure. In a hydraulic lift as shown in Fig. 6 two pistons are separated by the space filled with a liquid. A piston of small crosssection A_{1} is used to exert a force F_{1} directly on the liquid. The pressure P = F_{1}/A_{1 }is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A_{2}, which results in an upward force of P × A_{2}. Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) F_{2 }= PA_{2 }= F_{1}A_{2}/A_{1}. By changing the force at A_{1}, the platform can be moved up or down. Thus, the applied force has been increased by a factor of A_{2}/A_{1 }and this factor is the mechanical advantage of the device. The example below clarifies it.
Archemedes’ Principle
Fluid appears to provide partial support to the objects placed in it. When a body is wholly or partially
immersed in a fluid at rest, the fluid exerts pressure on the surface of the body in contact with the fluid. The pressure is greater on lower surfaces of the body than on the upper surfaces as pressure in a fluid increases with depth. The resultant of all the forces is an upward force called buoyant force.
Suppose that a cylindrical body is immersed in the fluid. The upward force on the bottom of the body is more than the downward force on its top. The fluid exerts a resultant upward force or buoyant force on the body equal to (P_{2} – P_{1}) × A (Fig. 3). We have seen in equation 10.4 that (P_{2}P_{1})A = ρghA. Now, hA is the volume of the solid and ρhA is the weight of an equivalent volume of the fluid. (P_{2}P_{1})A = mg.
Thus, the upward force exerted is equal to the weight of the displaced fluid. The result holds true irrespective of the shape of the object and here cylindrical object is considered only for convenience. This is Archimedes’ principle. For totally immersed objects the volume of the fluid displaced by the object is equal to its own volume. If the density of the immersed object is more than that of the fluid, the object will sink as the weight of the body is more than the upward thrust. If the density of the object is less than that of the fluid, it floats in the fluid partially submerged. To calculate the volume submerged, suppose the total volume of the object is V_{s} and a part V_{p} of it is submerged in the fluid. Then, the upward force which is the weight of the displaced fluid is ρ_{f}gV_{p}, which must equal the weight of the body; ρ_{s}gV_{s} = ρ_{f}gV_{p}or ρ_{s}/ρ_{f} = V_{p}/V_{s}. The apparent weight of the floating body is zero. This principle can be summarized as; ‘the loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced’.
Example 5
Two syringes of different crosssections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?
Answer
(a) Since pressure is transmitted undiminished throughout the fluid,
F_{2 }= F_{1}A_{2}/A_{1} = 10N ×π (3/2 × 10^{2} m²)/π(1/2× 10^{2}m²) = 90 N
(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume outwards to the larger piston.
L_{1}A_{1 }= L_{2}A_{2}≈ 0.67 × 10^{2} m = 0.67 cm
Note, atmospheric pressure is common to both pistons and has been ignored.
Example 6
In a car lift compressed air exerts a force F_{1 }on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig. 7). If the mass of the car to be lifted is 1350 kg, calculate F_{1}. What is the pressure necessary to accomplish this task? (g = 9.8 ms^{2}).
Answer:
Since pressure is transmitted undiminished throughout the fluid,
F_{1 }= F_{2}A_{1}/A_{2} = 1350N × (5 × 10^{2} m²)/(15× 10^{2}m²)× 9.8 ms^{2}
= 1470 N
≈ 1.5 × 10^{3} N
The air pressure that will produce this force is
P_{ }=F_{1}/A_{1 }={ 1.5 × 10³ N/(5 × 10^{2})² m} × 1.9 × 10^{5}Pa
This is almost double the atmospheric pressure.
Hydraulic brakes in automobiles also work on the same principle. When we apply a little force on the pedal with our foot the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area. A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way a small force on the pedal produces a large retarding force on the wheel. An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.
Streamline Flow
So far we have studied fluids at rest. The study of the fluids in motion is known as fluid dynamics. When a watertap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids we focus our attention on what is happening to various fluid particles at a particular point in space at a particular time. The flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant in time. This does not mean that the velocity at different points in space is same. The velocity of a particular particle may change as it moves from one point to another. That is, at some other point the particle may have a different velocity, but every other particle which passes the second point behaves exactly as the previous particle that has just passed that point. Each particle follows a smooth path, and the paths of the particles do not cross each other.
The path taken by a fluid particle under a steady flow is a streamline. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. Consider the path of a particle as shown in Fig. 7(a), the curve describes how a fluid particle moves with time. The curve PQ is like a permanent map of fluid flow, indicating how the fluid streams. No two streamlines can cross, for if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time. How do we draw closely spaced streamlines? If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig. 7(b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines. This means that number of fluid particles crossing the surfaces as indicated at P, R and Q is the same. If area of crosssections at these points A_{P}, A_{R} and A_{Q }and speeds of fluid particles v_{P}, v_{R} and v_{Q, }then mass of fluid Am_{P }crossing at A_{P }in a small interval of time At is ρ_{P}A_{P}v_{P }At. Similarly mass of fluid Am_{R }flowing or crossing at A_{R }in a small interval of time At is ρ_{R}A_{R}v_{R }At and mass of fluid Am_{Q }is ρ_{Q}A_{Q}v_{Q }At crossing at A_{Q}. The mass of liquid flowing out equals the mass flowing in, holds in all cases.
Therefore,
ρ_{P}A_{P}v_{P }At = ρ_{R}A_{R}v_{R }At = ρ_{Q}A_{Q}v_{Q }At ……………………………………(9)
For flow of incompressible fluids
ρ_{P }= ρ_{R }= ρ_{g}
Equation (9) reduces to A_{P}v_{P }= A_{R}v_{R }= A_{Q}v_{Q } ………………………………………………(10)
which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids. In general
Av = constant ……………………………………………………….(11)
Av gives the volume flux or flow rate and remains constant throughout the pipe of flow. Thus, at narrower portions where the streamlines are closely spaced, velocity increases and its vice versa. From Fig. 7(b) it is clear that A_{R} > A_{Q }or v_{R} > v_{Q }, the fluid is accelerated while passing from R to Q. This is associated with a change in pressure in fluid flow in horizontal pipes.Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent. One sees this when a fast flowing stream encounters rocks, small foamy whirlpoollike regions called ‘white water rapids are formed.Fig. 8 displays streamlines for some typical flows. For example, Fig. 8(a) describes a laminar flow where the velocities at different magnitudes but their directions are parallel. Fig.8(b) gives a sketch of turbulent flow.
Bernoulli’s Principle
Fluid flow is a complex phenomenon. But we can obtain some useful properties for steady or streamline flows using conservation of energy.
Consider a fluid moving in a pipe of varying crosssectional area. Let the pipe be at varying heights as shown in Fig.9. We now suppose that an incompressible fluid is flowing through the pipe in a steady flow. Its velocity must change as a consequence of equation of continuity. A force is required to produce this acceleration, which is caused by the fluid surrounding it, the pressure must be different in different regions. Bernoulli’s equation is a general expression that relates the pressure difference between two points in a pipe to both velocity changes (kinetic energy change) and elevation (height) changes (potential energy change). The Swiss Physicist Daniel Bernoulli developed this relationship in 1738.Consider the flow at two regions 1 (i.e., BC) and 2 (i.e., DE). Consider the fluid initially lying between B and D. In an infinitesimal time interval Δt, this fluid would have moved. Suppose v_{1} is the speed at B and v_{2} at D, then fluid initially at B has moved a distance v_{1}Δt to C (v_{1}Δt is small enough to assume constant crosssection along BC). In the same interval Δt the fluid initially at D moves to E, a distance equal to v_{2}Δt. Pressures P_{1 }and P_{2 }act as shown on the plane faces of areas A_{1 }and A_{2 }binding the two regions. The work done on the fluid at left end (BC) is W_{1} = P_{1}A_{1} (v_{1}Δt) = P_{1}ΔV. Since the same volume ΔV passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is W_{2} = P_{2}A_{2} (v_{2}Δt) = P_{2}ΔV or, the work done on the fluid is –P_{2}ΔV. So the total work done on the fluid is
W_{1} – W_{2} = (P_{1 }– P_{2})ΔV
Part of this work goes in to changing the kinetic energy of the fluid, and part goes in to changing the gravitational potential energy. If the density of the fluid, is ρ and Δm = ρA_{1} v_{1}Δt = ρΔV is the mass passing through the pipe in time Δt, then change in gravitational potential energy is
ΔU = ρgΔV(h_{2 }– h_{1})
The change in its kinetic energy is ΔK= (1/2)ρΔV[(v_{2 })² (v_{1})²]
We can employ the workenergy theorem to this volume of fluid and this yields
(P_{1 }– P_{2})ΔV = (1/2)ρΔV[(v_{2 })² (v_{1})²] + ρgΔV(h_{2 }– h_{1})
We now divide each term by ΔV to obtain (P_{1 }– P_{2}) = (1/2)ρ[(v_{2 })² (v_{1})²] + ρg(h_{2 }– h_{1})
We can rearrange the above terms to obtain P_{1 }+ (1/2)ρ(v_{1 })² + ρgh_{1 }= P_{2 }+ (1/2)ρ(v_{2})² + ρgh_{2 }…………………………….(12)
This is Bernoulli’s equation. Since 1 and 2 refer to any two locations along the pipeline, we may write the expression in general as
P + (1/2)ρv² + ρgh = constant ……………………………………………………………..(13)
In words, the Bernoulli’s relation may be stated as follows: As we move along a streamline the sum of the pressure (P), the kinetic energy per unit volume v²/2 and the potential energy per unit volume (ρgh) remains constant.
Note that in applying the energy conservation principle, there is an assumption that no energy is lost due to friction. But in fact, when fluids flow, some energy does get lost due to internal friction. This arises due to the fact that in a fluid flow, the different layers of the fluid flow with different velocities. These layers exert frictional forces on each other resulting in a loss of energy. This property of the fluid is called viscosity and is discussed in more detail in a later section. The lost kinetic energy of the fluid gets converted in to heat energy. Thus, Bernoulli’s equation ideally applies to fluids with zero viscosity or nonviscous fluids. Another restriction on application of Bernoulli’s theorem is that the fluids must be incompressible, as the elastic energy of the fluid is also not taken in to consideration. In practice, it has a large number of useful applications and can help explain a wide variety of phenomena for low viscosity incompressible fluids. Bernoulli’s equation also does not hold for nonsteady flow or turbulent flows, because in that situation velocity and pressure are constantly fluctuating in time.
When a fluid is at rest i.e., its velocity is zero everywhere, Benoulli’s equation becomes P_{1} + ρgh_{1} = P_{2} + ρgh_{2}(P_{1} – P_{2}) = + ρg(h_{2 }– h_{1}) which is Eq. (6)
Speed of Efflux: Torricelli’s Law
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 tank containing a liquid of density ρ with a small hole in its side at a height y_{1 }from the bottom (see Fig. 10). The air above the liquid, whose surface is at height y_{2 }is at pressure P. From the equation of continuity (Eq. 10) we have
v_{1} A_{1} = v_{2} A_{2}
v_{2} = (A_{1}/A_{2})v_{1}
If the cross sectional area of the tank A_{2 }is much larger than that of the hole (A_{2} » A_{1}) then we may take the fluid to be approximately at rest at the top. i.e., v_{2 }= 0. Now applying the Bernoulli equation at points 1 and 2 and noting that at the hole P_{1 }= P_{a }, the atmospheric pressure we have from Eq. 12
P_{a }+ (1/2)(v_{1 })² + gy_{1 }= P_{ }+ gy_{2 }
Taking y_{2}y_{1}= h we have
v_{1}= √[2gh + 2(PP_{a})] …………………………………………………………(14)
When P » P_{a }and 2gh may be ignored, the speed efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand if the tank is open to the atmosphere, then P = P_{a }and
v_{1 }= √(2gh) ……………………………………………………………………..(15)
This is the speed of a freely falling body. Eq. 15 is known as Torricelli’s law.
Venturimeter
The Venturimeter is a device to measure the flow speed of incompressible fluid. It consists of a tube with a broad diameter and a small constriction at the middle as shown in Fig. 11. A manometer in the form of a Utube is also attached to it, with one arm at the broad neck point of the tube and the other at constriction as shown in Fig. 11. The manometer contains a liquid of density ρ_{m}. The speed v_{1 }of the liquid flowing through the tube at the broad neck area A is to be measured from equation of continuity Eq. (10) the speed at the constriction becomes v_{2 }= (A/a)v_{1}. Then using Bernoulli’s equation 12 for (h_{1}=h_{2}), we get
P_{1 }+ (1/2)ρ(v_{1 })² _{ }= P_{2}_{ }+ (1/2)ρ(v_{1 })²(A/a)²
So that
P_{1 }– P_{2 } _{ }= (1/2)ρ(v_{1 })²[(A/a)²1] ………………………………………….(16)
This pressure difference causes the fluid in the U tube connected at the narrow neck to rise in comparison to the other arm. The difference in height h measure the pressure difference.
P_{1 }– P_{2 } _{ }= ρ_{m}gh = (1/2)ρ(v_{1 })²[(A/a)²1]
So that the speed of fluid at wide neck is
v_{1 }= √(2ρ_{m}gh/ρ) {(A/a)² – 1}^{1/2 }………………………………………………………(17)
The principle behind this meter has many applications. The carburetor of automobile has a Venturi channel (nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck and the petrol (gasolene) is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. Filter pumps or aspirators, Bunsen burner, atomisers and sprayers (See Fig. 12) used for perfumes or to spray insecticides work on the same principle.
Example 7
Blood velocity: The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a crosssectional area equal to that of the artery. A= 8 mm². The narrower part has an area a = 4 mm². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?
Answer
We take the density of blood from Table 1 to be 1.06 × 10³ kg m^{3}. The ratio of the areas is (A/a)= 2. Using Eq. 17 we obtain
v_{1 }= √(2×24Pa/1060 kg m^{3} × (2² – 1)= 0.125 ms^{1}
Blood Flow and Heart Attack
Bernoulli’s principle helps in explaining blood flow in artery. The artery may get constricted due to the accumulation of plaque on its inner walls. In order to drive the blood through this constriction a greater demand is placed on the activity of the heart. The speed of the flow of the blood in this region is raised which lowers the pressure inside and the artery may collapse due to the external pressure. The heart exerts further pressure to open this artery and forces the blood rushes through the opening, the internal pressure once again drops due to same reasons leading to a repeat collapse. This may result heart attack.
Dynamic Lift
Dynamic lift is the force is the force that acts on a body, such as airplane wing, a hydrofoil or a spinning ball, by virtue of its motion through a fluid. In many games such as cricket, tennis, base ball, or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through air. This deviation can be partly explained on the basis of Bernoulli´s principle.
(i) Ball moving without spin:Fig. 13(a) shows the streamlines around a non spinning ball moving relative to a fluid. From the symmetry of streamlines it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball.
(ii) Ball moving with spin: A ball which is spinning drags air along with it. If the surface is rough more air will be dragged. Fig. 13(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving forward and relative to it the air is moving backwards. Therefore, the velocity of air above the ball relative to it is larger and below it is smaller. The streamlines thus get crowded above and rarified below.This difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called Magnus effect.
(iii) Aerofoil or lift on aircraft wing: Fig. 13 (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air. The crosssection of the wings of an aeroplane looks somewhat like the aerofoil shown in Fig. 13(c) with streamlines around it. When the aerofoil moves against the wind, the orientation of the wing relative to flow direction causes the streamlines to crowd together above the wing more than those below it. The flow speed on top is higher than that below it. There is an upward force resulting in a dynamic lift of the wings and this balances the weight of the plane. The following example illustrates this.
Example 8
A fully loaded Boeing aircraft has mass of 3.3 × 10^{5} kg. Its total wing area is 500 m². It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kg m^{3} ]
Answer
(a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
ΔP × A = (3.3 × 10^{5} kg × 9.8 m s^{2})
ΔP =(3.3 × 10^{5} kg × 9.8 m s^{2})/ 500 m²
= 6.5 × 10^{3} N m^{2}
(b) We ignore the small height difference between the top and bottom sides in Eq. (12). The pressure difference between them is then
ΔP_{ } = ρ/2[(v_{2})² – (v_{1 })²]
where v_{2}is the speed of air over the upper surface and v_{1 }is the speed under the bottom surface.
v_{2 }– v_{1 }= 2 ΔP/(v_{2 }+ v_{1})
Taking the average speed v_{av }= (v_{2 }+ v_{1})/2 = 960 km/h = 267 m s^{1}
we have
v_{2 }– v_{1}/v_{av }= P/(v_{av})² ≈ 0.08
The speed above the wing needs to be only 8% higher than that below.
Viscosity
Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when a solid moves on a surface. It is called viscosity. This force exists when there is relative motion between layers of the liquid. Suppose we consider a fluid like oil enclosed between two glass plates as shown in Fig. 15(a). The bottom plate is fixed while the top plate is moved with a constant velocity v relative to the fixed plate. If oil is replaced by honey, a greater force is required to move the plate with the same velocity. Hence we say that honey is more viscous than oil. The fluid in contact with a surface has the same velocity as that of the surfaces. Hence, the layer of the liquid in contact with top surface moves with a velocity v and the layer of the liquid in contact with the fixed surface is stationary. The velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity v). For any layer of liquid, its upper layer pulls it forward while lower layer pulls it backward. This results in force between the layers. This type of flow is known as laminar. The layers of liquid slide over one another as the pages of a book do when it is placed flat on a table and a horizontal force is applied to the top cover. When a fluid is flowing in a pipe or a tube, then velocity of the liquid layer along the axis of the tube is maximum and decreases gradually as we move towards the walls where it becomes zero, Fig. 14(b). The velocity on a cylindrical surface in a tube is constant.On account of this motion, a portion of liquid, which at some instant has the shape ABCD, take the shape of AEFD after short interval of time (Δt). During this time interval the liquid has undergone a shear strain of Δx/l. Since, the strain in a flowing fluid increases with time continuously, unlike a solid here the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain ‘depend on ‘strain rate’ i.e., Δx/(lΔt) or v/l instead of strain itself. The coefficient of viscosity (pronounced ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate,
(F/A)/(v/l) = Fl/vA ……………………………………………………………………….(18)
The SI unit of viscosity is poiseiulle (PI). Its other units are N s m² or Pa s. The dimensions of viscosity are [ML^{1}T^{1}]. Generally thin liquids like water, alcohol etc., are less viscous than thick liquids like coal tar, blood, blood, glycerin etc. The coefficients of viscosity for some common fluids are listed in Table 2. We point out two facts about blood and water that you may find interesting. As Table 2, indicates blood is ‘thicker'(more viscous) than water that you find interesting. Further the relative viscosity (η/η_{water}) of blood remains constant between 0°C and 37°C.
The viscosity of liquids decreases with temperature while it increases in the case of gases.
Example 9
A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 15. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m s^{1}, find the coefficient of viscosity of the liquid.
Answer
The metal block moves to the right because the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is F= T = mg = 0.010 kg × 9.8 m s^{2}= 9.8 × 10^{2 }N
Shear stress on the fluid = F/A = 9.8×10^{2}/0.10 N/m^{2}
Strain rate = v/l = 0.085/0.030×10^{3}
η = stress/strain rate = (9.8×10^{2}N) × (0.30 ×10^{3}m)/(0.085m s^{1} × 0.10m²) = 3.46 × 10^{3}Pa s
Table 2 – The viscosity of some fluids
Fluid  T°C  Viscosity (mPl) 

Water  20  1.0 
100  0.3  
Blood  37  2.7 
Machine Oil  16  113 
38  34  
Glycerine  20  830 
Honey  –  200 
Air  0  0.017 
40  0.019 
Stokes’ Law
When a body falls through a fluid it drags the layer of the fluid in contact with it. A relative motion between the different layers of the fluid is set and as a result the body experiences a retarding force. Falling of a raindrop and swinging of a pendulum bob are common examples of such motion. It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion. The other quantities on which the force F depends on viscosity η of the fluid and radius a of the sphere. Sir George G stokes (18191903), an English scientist enunciated clearly the viscous drag force F as
F = 6πηRv …………………………………………………………(19)
This is known as Stokes’ law. We shall not derive Stokes’ law.This law is an interesting example of retarding force which is proportional to velocity. We can study its consequences on an object falling through a viscous medium. We consider a rain drop in air. It accelerates initially due to gravity. As the velocity increases, the retarding force also increases. Finally when viscous force plus buoyant force becomes equal to force due to gravity, the net force becomes zero and so does the acceleration. The sphere (rain drop) then descends with a constant velocity. Thus in equilibrium, this terminal velocity v_{1}is given by6πηav_{t }= (4π/3) a³(ρ – σ)g
where ρ and σ are mass densities of sphere and the fluid respectively. We obtain
v_{t }= 2a²(ρ – σ)g/9r ………………………………………………………………….(20)
So the terminal velocity v_{t }depends on the square of the radius of the sphere and inversely on the viscosity of the medium.You may like to refer back to Example 2 in this context.
Reynolds Number
When the rate of flow of a fluid is large, the flow no longer remain laminar, but becomes turbulent. In a turbulent flow the velocity of the fluids at any point in space varies rapidly and randomly with time. Some circular motions called eddies are also generated. An obstacle placed in the path of a fast moving fluid causes turbulence [Fig. 8(b)]. The smoke rising from a burning stack of wood, oceanic currents are turbulent. Twinkling of stars is the result of atmospheric turbulence. The wakes in the water and in the air left by cars, aeroplanes and boats are also turbulent.Osborne Reynolds (18421912) observed that turbulent flow is less likely for viscous fluid flowing at low rates. He defined a dimensionless number, whose value gives one an approximate idea whether the flow would be turbulent. This number is called Reynolds R_{e}.
R_{e }= ρvd/η ………………………………………..(10.21)
where ρ is the density of the fluid flowing with a speed v, d stands for the dimension of the pipe, and η is the viscosity of the fluid. R_{e }is a dimensionless number and therefore, it remains same in any system of units. It is found that flow is streamline or laminar or R_{e }between 1000 and 2000. The critical value of R_{e }(known as critical critical Reynolds number), at which turbulence sets, is found to be the same for geometrically similar flows. For example, when oil and water with different densities and viscosities, flow in pipes of same shapes and sizes, turbulence sets in at almost the same value of R_{e}. Using this fact that a small scale laboratory model can be set up the to study the character of fluid flow. They are useful in designing of ships, submarines, racing cars and aeroplanes.
R_{e }can be written as
R_{e }= ρv²/(ηv/d) = ρAv²/(ηAv/d) ……………………………………………….(22)
= internal force /force of viscosity.
Thus R_{e }represents the ratio of internal force (force due to inertia i.e., mass of moving fluid or due to inertia of obstacle in its path) viscous force.Turbulence dissipates kinetic energy usually in the form of heat. Racing cars and planes are engineered to precision in order to minimise turbulence. The design of such vehicles involves experimentation and trial and error. On the other hand turbulence (like friction) is sometimes desirable. Turbulence promotes mixing and increases the rates of transfer of mass, momentum and energy. The blades of a kitchen mixer induce turbulent flow and provide thick milk shakes as well as beat eggs in to a uniform texture.
Example 10
The flow rate of water from a tap of diameter 1.25 cm is 0.48 lit/min. The coefficient of viscosity of water is 10^{3 }Pa s. After sometime the flow rate is increased to 3 lit/min. Characterise the flow for both the flow rates.
Answer
Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is is
Q = v × πd²/4
v= 4Q/d²
We then estimate the Reynolds number to be R_{e}= 4 ρQ/πdη
= 4 × 10³ kgm^{3 }× Q/(3.14 × 1.25 × 1.25 × 10^{2}m × 10^{3 }Pa s)
= 1.019 × 10^{8} m^{3} s Q
Since initially
Q= 0.48 lit/min = 8 cm³/s = 8 × 10^{6} m³s^{1}
We obtain,
R_{e}= 815
Since this is below 1000, the flow is steady. After some time when
Q = 3 lit/min = 50 cm³/s = 5 × 10^{5} m^{3} s^{1}
We obtain,R_{e}= 5095
The flow will be turbulent. You may carry out an experiment in your wash basin to determine the transition from laminar to turbulent flow.
Surface Tension
You must have noticed, oil and water do not mix; water wets you and me but not ducks; mercury does not wet glass but water sticks to it, oil rise up a cotton wick, inspite of gravity. Sap and water rise to the top of the leaves of the tree, hairs of a paint brush do not cling together when dry and even when dipped in water but form a fine tip when taken out of it. All these and many more such experiences are related with the free surfaces of liquids. As liquids have no definite shape but have a definite volume, they acquire a free surface when poured in a container. These surfaces possess some additional energy. This phenomenon is known as surface tension and it is concerned with only liquid as gases do not have free surfaces. Let us now understand this phenomena.
Surface Energy
A liquid stays together because of attraction between molecules. Consider a molecule well inside a liquid. The inter molecular distances are such that it is attracted to all the surrounding molecules [Fig. 16(a)]. This attraction results in a negative potential energy for the molecule, which depends on the number and distribution of molecules around the chosen one. But the average potential energy of all the molecules is the same. This is supported by the fact that to take a collection of such molecules (the liquid) and to disperse them far away from each other in order to evaporate or vaporize, the heat of evaporation required is quite large. For water it is of the order of 40 kJ/mol.
Let us consider a molecule near the surface Fig. 16(b). Only lower half side of it is surrounded by liquid molecules. There is some negative potential energy due to these, but obviously it is less than that of a molecule in bulk, i.e., the one fully inside. Approximately it is half of the latter. Thus, molecules on a liquid surface have some extra energy in comparison to molecules in the interior. A
liquid, thus, tends to have the least surface area which external conditions permit. Increasing surface area requires energy. Most surface phenomenon can be understood in terms of this fact. What is the energy required for having a molecule at the surface? As mentioned above, roughly it is half the energy required to remove it entirely from the liquid i.e., half the heat of evaporation.
Finally, what is a surface? Since a liquid consists of molecules moving about, there cannot be a perfectly sharp surface. The density of the liquid molecules drops rapidly to zero around z = 0 as we move along the direction indicated Fig 16 (c) in a distance of the order of a few molecular sizes.
Fig 16 Schematic picture of molecules in a liquid, at the surface and balance of forces (a) Molecule inside a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (A) and repulsive (R) forces.
Surface Energy and Surface Tension
As we have discussed that an extra energy is associated with surface of liquids, the creation of more surface (spreading of surface) keeping other things like volume fixed requires additional energy. To appreciate this, consider a horizontal liquid film ending in bar free to slide over parallel guides. (Fig.17)
Suppose that we move the bar by a small distance d as shown. Since the area of the surface increases, the system now has more energy, this means that some work has been done against an internal force. Let this internal force be F, the work done by the applied force is F.d = Fd. From conservation of energy, this is stored as additional energy in the film. If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between, so there area two surfaces and the extra energy is
S(2dl) = Fd ………………………………………(23)
Or, S = Fd/2dl = F/2l ………………………..(24)
This quantity S is the magnitude of surface tension. It is equal to the surface energy per unit area of the liquid interface and is also equal to the force per unit length exerted by the fluid on the movable bar.
So far we have talked about the surface of one liquid. More generally, we need to consider fluid surface in contact with other fluids or solid surfaces. The surface energy in that case depends on the materials on both sides of the surface. For example, if the molecules of the materials attract each other, surface energy is reduced while if they repel each other, the surface energy is reduced while if they repel each other the surface energy is increased. Thus, more appropriately, the surface energy is the energy of the interface between two materials and depends on both of them.
We make the following observations from above:
(i) Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of the interface between the plane of the liquid and any other substance: it also is the extra energy that the molecules at the interface have as compared to molecules in the interior.
(ii) At any point on the interface besides the boundary, we can draw a line and imagine equal and opposite surface tension forces S per unit length of the line acting perpendicular to the line, in the plane of the interface. The line is in equilibrium. To be more specific, imagine a line of atoms or molecules at the surface. The atoms to the left pull the line towards them; those to the right pull the line towards them! This line of atoms is in equilibrium under tension. If the line really marks the end of the interface, as in Fig. 16(a) and (b) there is only the force S per unit length acting inwards.
Table 3 gives the surface tension of various liquids. The value of surface tension depends on temperature. Like viscosity, the surface tension of tension of a liquid usually falls with temperature.
Table 3 – Surface tension of some liquids at the temperatures indicated with the heats of the vaporisation
Liquid  Temp°C  Surface Tension (N/m)  Heat of vaporisation (kJ/mOl) 

Helium  270  0.000239  0.115 
Oxygen  183  0.0132  7.1 
Ethanol  20  0.0227  40.6 
Water  20  0.0727  44.16 
Mercury  20  0.4355  63.2 
A fluid will stick to a solid surface if the surface energy between fluid and the solid is smaller than the sum of surface energies between solidair, and fluidair. Now there is cohesion between the solid surface and the liquid. It can be directly measured experimentally as schematically shown in Fig. 18. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side slightly till the liquid just touches the glass plate and pulls down a little because of surface tension. Weights are added till the plate just clears water.
Suppose the additional weight required is W. Then from Eq. 24 and the discussion given there, the surface tension of the liquidair interface is S_{la} = (W/2l) = (mg/2l) where m is the extra mass and l is the length of the plate edge. The subscript (la) emphasizes the fact that the liquidair interface tension is involved.
Angle of Contact
The surface of liquid near the plane of contact, with another medium is in general curved. The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact. It is denoted by θ. It is different at interfaces of of different pairs of liquids and solids. The value θ determines whether a liquid will spread on the surface of a solid or it will form droplets on it. For example, water forms droplets on lotus leaf as shown in Fig. 19(a) while spreads over a clean plastic plate as shown in Fig 19(b).
We consider the three inter facial tensions at all the three interfaces, liquidair, solidair and solidliquid denoted by S_{la}, S_{sa} & S_{sl}respectively as given in Fig. 19(a) and (b). At the line of contact, the surface forces between the three media must be in equilibrium. From the Fig. 19(b) the following relation is easily derived.
S_{la }cos θ + S_{sl }= S_{sa }……………………………………………………………………..(26)
The angle of contact is an obtuse angle if S_{sl} > S_{la }as in the case of water leaf interface while it is an acute angle if S_{sl} < S_{la} as in the case of water plastic interface. When θ is an obtuse angle then molecules of liquids are attracted strongly to themselves and weakly to those of solid, it costs a lot of energy to create a liquidsolid surface, and liquid then does not wet the solid. This is what happens with water on a waxy or oily surface, and with mercury on any surface. On the other hand, if the molecules of the liquid are strongly attracted to those of the solid, this will reduce S_{sl }and therefore cos θ may increase or θ may decrease. In this case θ is an acute angle. This is what happens for water on glass or on plastic and for kerosene oil on virtually anything(it just spreads). Soaps detergents and dying substances are wetting agents. When they are added the angle of contact becomes small so that these may penetrate well and become effective. Water proofing agents on the other hand are added to create a large angle of contact between the water and fibres.
Drops and Bubbles
One consequence of surface tension is that free liquid drops and bubbles are spherical if effects of gravity can be neglected. You must have seen this especially clearly in small drops just formed in a highspeed spray or jet, and in soap bubbles blown by most of us in childhood. Why are drops and bubbles spherical? What keeps soap bubbles stable?
As we have been saying repeatedly, a liquidair interface has energy, so far a given volume the surface with minimum energy, is the one with the least area. The sphere has this property. Though it is out of the scope of these lessons, but you can check that a sphere is better than at least a cube in this respect! So, if gravity and other forces (e.g., air resistance) were ineffective, liquid drops would be spherical.
Another interesting consequence of surface tension is that the pressure inside a spherical drop Fig. 20(a) is more than the pressure outside. Suppose a spherical drop of radius r is in equilibrium, if its radius increase by Δr. The extra surface energy is
[4π(r + Δr)² – 4πr²] S_{la }= 8πr Δr S_{la }…………………………………………(27)
If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference (P_{1 }– P_{0}) between the inside of the bubble and the inside of the bubble and the outside. The work done isW = (P_{1 }– P_{0}) 4πr²Δr …………………………………………………….(28)
so that
(P_{1 }– P_{0}) = (2 S_{la }/r) ………………………………………………………….(29)
In general, for a liquidgas interface, the convex side has a higher pressure than the concave side. For example, an air bubble in a liquid, would have higher pressure inside it. See Fig. 20(b)
A bubble Fig. 20(c) differs from a drop and a cavity; in this it has two interfaces. Applying the above argument we have for a bubble
(P_{i }– P_{0}) = (4 S_{la }/r) ………………………………………………………….(30)
This is probably why you have to blow hard, but not too hard, to form a soap bubble. A little extra air pressure is needed inside!
CapillaryRise
One consequence of the pressure difference across a curved liquidair interface is the well known effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin;if the tube were thin, the rise would be very large. To see this,
consider a vertical capillary tube of circular cross section (radius a) inserted in to an open vessel of water (Fig. 21). The contact angle between water in the capillary is concave. This means that there is a pressure difference between the two sides of the top surface. This is given by
(P_{i }– P_{0}) = (2S/r) = 2 S/(a sec θ) = (2S/a) cos θ …………………………………………….(31)
Thus the pressure of the water inside the tube, just at meniscus (airwater interface) is less than the atmospheric pressure. Consider the two points A and B in Fig. 21(a). They must be at the same pressure, namely
P_{0 }+ hρg = P_{i }=P_{A }………………………………………………………(32)
where ρ is the density of water and h is called the capillary rise. [Fig. 21(a)]. Using Eq. (31) and (32) we have
hρg = (P_{i }– P_{o}) = (2S cos θ)/a ……………………………………….(33)
The discussion here, and the Eqs. (28) and (29) make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few cm for fine capillaries. For example, if a = 0.05 cm, using the value of surface tension for water (Table 3), we find thath = 2S/(ρga) = 2 × 0.073 Nm^{1}/(10³kg m^{3 }× 9.8 m s^{2}× 5 × 10^{4}) = 2.98 × 10^{2 }m = 2.98 cmNotice that if the liquid meniscus is convex, as for mercury, i.e., if cos θ is negative then from Eq. (32) for example, it is clear that the liquid will be lower in the capillary!
Detergents and Surface Tension
We clean dirty clothes containing grease and oil stains sticking to cotton or other fabrics by adding detergents or soap to water, soaking clothes in it and shaking. Let us understand this process better.Washing with water does not remove grease stains. This is because water does not wet greasy dirt; i.e., there is very little area of contact between them. If water could wet grease, the flow of water could carry some grease away. Something of this sort is achieved through detergents. The molecules of detergents are hairpin shaped with one end attached to water and the other to molecules of grease, oil or wax, thus tending to form wateroil interfaces. The result is shown in Fig. 22 as a sequence of figures. In our language, we would say that addition of detergents, whose molecules attract at one end say oil on the other, reduces drastically the surface tension S (wateroil). It may even become energetically favorable to form such interfaces, i.e., globs of dirt surrounded detergents and then by water. This kind of process using surface active detergents or surfactants is important not only for cleaning, but also in recovering oil, mineral ores etc.
Example 11
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is 7.30 × 10^{2} Nm^{1}. 1 atmospheric pressure = 1.01 × 10^{5 }Pa. density of water = 1000 kg/m³. g = 9.80 m s^{2}. Also calculate the excess pressure.
Answer
The excess pressure in a bubble of gas in a liquid is given by 2S/r, where S is the surface tension of the liquidgas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S/r). The radius of the bubble is r. Now the pressure outside the bubble P_{0 }equals atmospheric pressure plus the pressure due 8.00 cm of water column. That is P_{0 }= (1.01 × 10^{5 }Pa + 0.08 m × 1000 kg m^{3 }× 9.80 m s^{2}) = 1.01784 × 10^{5 }Pa. Therefore, the pressure inside the bubble isP_{i }= P_{0 }+ 2S/r = 1.01784 × 10^{5 }Pa + (2 × 7.3 × 10^{2 }Pam/10^{2 }m) = 1.02× 10^{5 }Pa where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical! (The answer has been rounded off to three significant figures). The excess pressure in the bubble is 146 Pa.