10.0 Preview:
10.1 Introduction:
10.2 Pressure
10.3 Streamline Flow
- Questions for section 10.3:
10.4 Bernoulli’s Principle
10.5 Viscosity
- 10.5.1 Stokes’ Law
- Questions from section 10.5:
10.6 Reynolds Number
- Questions for sections 10.6:
10.7 Surface Tension

10.0 Preview:

This chapter revolves around properties of fluids; such as, density, pressure and flow velocity. We shall begin with describing a pressure-measuring device and thus find average pressure, $P_{av}$ as the normal force acting per unit area.

We will then describe another property of fluid, Density, ρ as the ratio of mass to volume of the fluid; given by: $ρ = \frac {m}{v} $.

We shall also learn about Pascal’s law which states that, “A change in pressure exerted at any part of a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.”

Next, we establish a relation to study variations in pressure with varying depth. We will also be dealing with differences in pressure between two points situated at different heights in a fluid. It is found that Pressure difference depends on the vertical distance, h between the points, mass density of the fluid, ρ and acceleration due to gravity, g. The liquid pressure is thus, the same at all points, at the same horizontal level (same depth). Hydrostatic paradox is used to augment this.

Next, a mercury barometer is used to find the atmospheric pressure; the mercury column in the barometer has a height of about 76 cm at sea level, which is considered equivalent to one atmosphere (1atm). We shall look at another pressure measuring device called, an open-tube manometer.

We shall then extend Pascal’s law which states that to describe how a hydraulic lift and hydraulic brakes work. The underlying principle of working of these devices is based on transmission of pressure through the fluid.

Further, we will discuss in brief—the types of flow; 1) Steady flow (or Streamline flow): A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ from point to point but do not change with time and vice versa for unsteady flow! 2) Turbulent flow: The flow in which the velocity of the fluids at any point in space varies rapidly and randomly with time.

We will then arrive at the equation of continuity and it is paramount to the statement of conservation of mass in flow of incompressible fluids. Bernoulli’s Principle or Bernoulli’s Theorem will be dealt with next. It states that, “For streamline flow of a non-viscous incompressible fluid, the sum of the pressure energy, kinetic and potential energy per unit mass remains constant”.

Further, we will look at Venturi-meter (as an application of Bernoulli’s Theorem) which is the device used to measure the speed of an incompressible fluid in a pipe. Then we shall again use Bernoulli’s theorem to explain blood flow in artery.

We shall discuss in brief, Dynamic lift as the force that acts on a body, such as airplane wing, a hydro foil or a spinning ball, by virtue of its motion through a fluid.

Next, the property of a liquid by virtue of which it opposes the relative motion between its different layers is called viscosity. 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. We will further define coefficient of viscosity (called ‘eta’) for a fluid as the ratio of shearing stress to the strain rate; Reynolds number ($R_e$) as a Dimensionless number, whose value gives one, an approximate idea whether the flow would be turbulent; The value of $R_e$ at which turbulence is just set is called Critical Reynolds Number.

Surface tension as the property of fluid by virtue of which the liquid tries to occupy least surface area which is defined as: “surface energy per unit area of the liquid interface” and is also “the force per unit length exerted by the fluid on the movable bar”.

Let us try to picture a liquid droplet, we can say that, the surface of liquid, near the plane of contact with another medium is in general, curved. This will bring us to define the Angle of contact (θ) as the angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid.

Lastly, we will learn why we have to blow hard, to start forming a soap bubble as a consequence of surface tension; another consequence of which is the capillary action. The rise of a liquid in a capillary tube is known as capillarity. The cleaning action achieved through surfactants or detergents are again based on the concept of Surface tension.

10.1 Introduction:

Matter can be broadly classified into: Solids, Liquids and Gases. Liquids and gases can flow and therefore are called fluids. Fluids play an important role in our life. The air that we breathe, the blood that circulates in our body and the water that we drink are all fluids. Fluids control our weather too.

Unlike solids, a fluid does not have a definite shape of its own. Ultimately, it assumes the shape of the container. Solids and liquids have a fixed volume, whereas, a gas occupies the entire volume of a container. Gases are easily compressed, whereas, liquids are nearly incompressible.

However, in studying mechanical properties of fluids, we shall only describe those properties which are connected with their ability to flow. The description of fluid motion is based on Newton’s laws of motion formulated in terms of measurable properties of fluids; such as, density, pressure and flow velocity

10.2 Pressure

A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force.

Thus, we can say that 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. Thisforce is always normal to the object’s surface.This is so because if there were a component offorce parallel to the surface, the object will also exert a force on the fluid parallel to it; as aconsequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface.Since the fluid is at rest, we can say that the force exerted by the fluid on the object is not parallel.
Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Figure 10.1(a).
The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressure-measuring device is shown in Figure 10.1(b). It consists of an evacuated chamber with a spring that is calibrated to measure the force acting on the piston. This device is placed at a point inside the fluid. The inward force exerted by the fluid on the piston is balanced by the outward spring force and is thereby measured.

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

$$ P_{av} = \frac {F}{A} ……(10.1) $$

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

$$ P = \lim_{\Delta A \to 0} \frac {F}{A} ……(10.2) $$

Note box:
1. Pressure is a scalar quantity. Its dimensions are [$ML^{–1}T^{–2}$]. The SI unit of pressure is N $m^{–2}$ or pascal (Pa).
2. A common unit of pressure is the atmosphere (atm), that is, the pressure exerted by the atmosphere at sea level (1 atm = 1.013 × $10^5$ Pa).

Another quantity, that is important in describing fluids, is the density, ρ. For a fluid of mass m occupying volume V,
$$ ρ = \frac {m}{v} ……(10.3) $$

Definition box:
The relative density of a substance is the ratio of its density to the density of water at 4^oC.

Note box:
1. The dimensions of density are $[ML^{–3}]$. Its SI unit is kg $m^{–3}$. It is a positive scalar quantity.
2. The density of water at 4oC (277 K) is 1.0 × $10^3$ kg $m^{–3}$.
3. Relative density is a dimensionless positive scalar quantity. For example, the relative density of aluminium is 2.7. Its density is 2.7 × $10^3$ kg $m^{–3}$.

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. Let us see how!

Q and A:

Why does not the density of a liquid vary with pressure, whereas that for a gas varies with variations in pressures?

We know that, .$ ρ= \frac {m}{v} $ Thus, density varies with variation in volume. Having the fact that gases are highly compressible and liquids are hardly compressible, the volume for a liquid does not change much on application of force as it would for a mass of gas. Thus, the density for a liquid almost remains constant while that for gas varies greatly with the variation of volume.

The densities of some common fluids are displayed in Table 10.1 below;

Example 10.1: The two thigh bones (femurs), each of cross-sectional area 10 cm$^2$ support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.

Solution: Total cross-sectional area of the femurs is A = 2 × 10 cm$^2$ = 20 × 10$^{–4}$ m$^2$. 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
$$ P_{av} = \frac {F}{A} = 2 × 10^5 \space N m^{−2} $$

Questions for section 10.2:

How is normal force exerted by a fluid at a point measured using a pressure measuring device? Also, what is the average pressure obtained from this?
Define density and relative density for a substance.
What are the dimensions of density and relative density?
What is the SI unit of density?

10.2.1 Pascal’s Law

Definition box:
Pascal’s law states that, “A change in pressure exerted at any part of a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.”

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

Figure 10.2 shows an element inside a fluid at rest.
This element ABC-DEF is in the form of a right-angled prism.
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 the prismatic element are those forces that are exerted by the fluid in contact with the prism and they must be normal to the surfaces of the element as discussed earlier.
Thus, the pressures are $P_a, P_b$ and $P_c$ on this element of area, due to the normal forces $F_a, F_b$ and $F_c$ as shown in Figure 10.2 on the faces BEFC, ADFC and ADEB denoted as areas, $A_a, A_b$ and $A_c$ respectively. Then,
$$ F_b \space sinθ = F_c, F_b \space cosθ = F_a (by \space equilibrium) $$
$$ A_b \space sinθ = A_c, A_b \space cosθ = A_a (by \space geometry) $$
Thus,
$$ \frac {F_b}{A_b} = \frac {F_c}{A_c} = \frac {F_a}{A_a} ; P_b = P_c = P_a \hspace{10mm} (10.4) $$
Hence, pressure exerted is same in all directions in a fluid at rest.

Note box: 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, let us consider a fluid element in the form of a horizontal bar of uniform cross-section. The bar is in equilibrium.
The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium, the pressure is same at all points in a horizontal plane.
Suppose that the pressure were not equal in different parts of the fluid, then there would be a flow due to the differential distribution of pressure caused by some 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.

Questions for section 10.2.1:

State and prove Pascal’s law.

10.2.2 Variation of Pressure with Depth

Let us consider a fluid at rest in a container as shown in Figure 10.3; point-1 is at height, h above point-2.

The pressures 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 (P1A) acting downward, at the bottom $(P_2A)$ acting upward (since, P = F/A or F = P x A). If mg is weight of the fluid in the cylinder we have,
$$ (P_2-P_1 ) A=mg …………….. (10.5)$$
Now, if ρ is the mass density of the fluid, we have the mass of fluid to be;
m = ρV= ρhA so that,
$$(P_2-P_1 )=ρgh ……………. (10.6)$$
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 Equation (10.6) gives, $$P = P_a + ρgh............... (10.7)$$
Thus, the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh (due to the additional pressure exerted by the liquid).
The excess of pressure, $P – P_a$, at depth h is called a gauge pressure at that point.

Note box:
The area of the cylinder is not appearing in the expression of absolute pressure in Eq. (10.7). Thus, the pressure depends on height of the fluid column and not cross sectional or base area or the shape of the container.

The liquid pressure is thus, the same at all points, at the same horizontal level (same depth).
The result is verified through the example of hydrostatic paradox.

Let us consider three vessels A, B and C [Refer figure 10.4] of different shapes. They are connected at the bottom by a horizontal pipe. On filling with water, the level in the three vessels is the same though they hold different amounts of water. This is so, because water at the bottom has the same pressure below each section of the vessel.

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

Solution: Here h = 10 m and ρ = 1000 kg m$^{-3}$. Take g = 10 m s$^{–2}$ From Eq. (10.7)
$$ P = P_a + ρgh $$
$$ = 1.01 × 10^5 Pa + 1000 \space kg m^{–3} × 10 \space m s^{–2} × 10 \space m $$
$$ = 2.01 × 10^5 \space Pa $$
$$ \approx 2 atm $$
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.

Questions for section 10.2.2:

Show that, at depth below the surface of a liquid open to the atmosphere, the pressure is greater than the atmospheric pressure. What is gauge pressure?
Describe Hydrostatic paradox.

10.2.3 Atmospheric Pressure and Gauge Pressure

The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the atmosphere.
At sea level, it is $1.013 × 10^5 \space Pa$ (1atm). Italian scientist Evangelista Torricelli devised for the first time, a method for measuring atmospheric pressure;
A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury as shown in Figure, 10.5 (a). This device is known as mercury barometer.

The space above the mercury column in the tube contains only mercury vapour whose pressure P is so small that it may be neglected.

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 = $P_a$
$$ P_a=ρgh ……………………… (10.8) $$
Where, ρ is the density of mercury and h is the height of the mercury column in the tube.
In the experiment, it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1atm).
The height can also be obtained using the value of ρ in Equation (10.8).

Note box:
1. A common way of stating pressure is in terms of cm or mm of mercury (Hg).
2. A pressure equivalent of 1 mm is called a torr (after Torricelli). 1 torr = 133 Pa.
3. 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 open-tube manometer (in figure, 10.5b) is a useful instrument for measuring pressure differences.
It consists of a U-tube containing a suitable liquid; that is, 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 other end is connected to the system whose pressure we want to measure [see Figure 10.5 (b)].
The pressure P at A is equal to pressure at point B. What we normally measure is the gauge pressure, which is $P – P_a$, given by Equation (10.8) and is proportional to manometer height, h.

Pressure is same at the same level on both sides of the U-tube containing a fluid. For liquids, the density varies very little over wide ranges in pressure and temperature, as discussed earlier. Thus, we can treat it safely as a constant for our present purposes.

Example 10.3: The density of the atmosphere at sea level is 1.29 kg/m$^3$. Assume that it does not change with altitude. Then how high would the atmosphere extend?

Solution: We use Eq. (10.7)
$$ ρgh = 1.29 \space kg m^{–3} × 9.8 \space m s^2 × h \space m = 1.01 × 10^5 \space Pa $$
$$ ∴ h = 7989 m \approx 8 \space km $$

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

Example 10.4: At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03 × 10$^3$ kg m$^{-3}$, g = 10m s$^{–2}$.) (Pg 236)

Solution: Here h = 1000 m and ρ = 1.03 × 10$^3$ kg m$^{-3}$.

(a) From Eq. (10.6), absolute pressure
$$ P = P_a + ρgh $$
$$ =1.01 × 10^5 \space P_a + 1.03 × 10^3 \space kg \space m^{–3} × 10 \space m s^{–2} × 1000 \space m $$
$$= 104.01 × 10^5 \space P_a$$
≈ 104 atm

(b) Gauge pressure is
$$ P – P_a = ρgh = P_g $$
$$ P_g = 1.03 × 10^3 \space kg m^{–3} × 10 \space ms^2 × 1000 \space m $$
$$= 103 × 10^5 \space P_a $$
≈ 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 acting on the window is gauge pressure,$ P_g = ρgh $. Since the area of the window is A = 0.04 m$^2$, the force acting on it is

$$ F = P_g A = 103 × 10^5 \space P_a × 0.04 \space m^2 = 4.12 × 10^5 \space N $$

Questions for section 10.2.3:

What is pressure of the atmosphere (atmospheric pressure)? How is it measured using a mercury barometer?
How is gauge pressure measured using open tube manometer? Explain with figure.

10.2.4 Hydraulic Machines

Let us now consider what happens when we change the pressure on a fluid contained in a vessel.

Let us 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 necessary that, they are all at the same height.
If we push the piston, the fluid level rises in all the tubes, again reaching the same levels in each one of the tubes. This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. This was stated as a law as described below by Pascal.
“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. (You may refer either definition for Pascal’s law as defined in 10.2.1 or in this, 10.2.4!)
Applications of the law: A number of devices such as hydraulic lift and hydraulic brakes are based on the Pascal’slaw. In these devices fluids are used fortransmitting pressure.
In a hydraulic lift asshown in Fig. 10.6 two pistons are separatedby the space filled with a liquid. A piston of smallcross section $A_1$ is used to exert a force F1directly on the liquid. The pressure P = $F_1/A_1$.
This pressure istransmitted throughout the liquid to the largercylinder 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 alarge force, (large weight of, say a car, or a truck,placed on the platform) $F_2 = PA_2 = F_1A_2/A_1$.

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.

Example 10.5: Two syringes of different cross sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston whena 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?

Solution: (a) Since pressure is transmitted undiminished throughout the fluid,
$$ F_2 = \frac {A_2}{A_1} F_1 = \frac { \pi (3/2×10^{-2} \space m )^2}{ \pi (1/2×10^{-2} \space m)^2} 10 N $$
(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.
$$ L_1 A_1 = L_2 A_2 $$$$ L_2 = \frac {A_1}{A_2}L_1 =\frac {A_2}{A_1} F_1 = \frac { \pi (1/2×10^{-2} \space m )^2}{ \pi (3/2×10^{-2} \space m)^2}×6×10^{-2} m $$
$$ \simeq 0.67×10^{-2} m =0.67 cm$$
Note, atmospheric pressure is common to both pistons and has been ignored.

Example 10.6: In a car lift compressed air exerts a force $F_1$ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). 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).

Solution: Since pressure is transmitted undiminished throughout the fluid,
$$ F_1 = \frac {A_1}{A_2} F_2 = \frac { \pi (5×10^{-2} \space m )^2}{ \pi (15×10^{-2} \space m)^2} (1350 \space N× 9.8 \space ms^{-2}) $$
$$ =1470 \space N $$
$$ \approx 15×10^3 \space N $$
The air pressure that will produce this force is
$$ P = \frac {F_1}{A_1} = \frac {1.5×10^3 \space N}{\pi (5×10^{-2} \space m )^2}=1.9×10^5 \space Pa $$

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 this system is that, the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.

Archimedes’ Principle

Fluids appear 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. 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 “$ρ_fgV_p$”, which must equal the weight of the body;
$$ ρ_sgV_s = ρ_fgV_p(or)$$
$$ ρ_s/ρ_f = V_p/V_s.$$
The apparent weight of the floating body is zero.
This principle can be summarised as; ‘the loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced’.

Story:

Archimedes (287 – 212 B.C.)

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

Questions for section 10.2.4:

State and explain Pascal’s law.
Explain the two important applications of Pascal’s law.
Explain Archimedes’ Principle for fluids.

10.3 Streamline Flow

So far we have studied fluids at rest. The study of the fluids in motion is known as fluid dynamics.
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.
When a water-tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. This brings us to the definition of steady flow.
Steady flow: A steady flow is one in which the conditions (velocity, pressure and cross-section) may differ from point to point but do not change with time.
In other words, during a steady flow, conditions at a particular point are always the same for any particle passing through that point. That is, at some other point the particle may have a different velocity. Every particle which passes a particular point behaves exactly as every other particle that passes through 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.

Definition box:
Streamline is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point

Let us consider the path of a particle as shown in Figure, 10.7 (a) above, the curve describes how a fluidparticle moves with time.
The curve PQ is atypical map of fluid flow, indicating how thefluid streams.
No two streamlines can cross,for if they do, an oncoming fluid particle can goeither one way or the other and the flow would not be steady.

Note box:
Unsteady flow: If at any point in the fluid, the conditions (like, velocity, pressure and cross-section) change with time, the flow is described as unsteady.

Equation of continuity:

We have seen that the fluid flowing through a pipe does not mix with the fluid outside the pipe. The total mass of fluid flowing through any cross section of the pipe should therefore be equal to the total mass coming out of the same pipe from any other cross section in the same time. This leads to the equation of continuity.

Let us consider a streamline flow of a fluid through a pipe AB of non-uniform cross section as shown in figure, 10.7b above. Let $v_1$ be the velocity of the liquid entering at A of cross-sectional area, $a_1$ normal to the surface. Let $v_2$ be the velocity with which it flows out at B where area of cross-section is$a_2$ normal to $v_2$.
Since the flow is steady, the mass of the fluid entering per second is equal to the mass of the fluid leaving per second.

Hence,
$$ ρ_1 a_1 v_1=ρ_2 a_2 v_2……………… (10.9)$$
If the fluid is incompressible,$ ρ_1=ρ_2$.Then, the equation becomes;
$$ a_1 v_1=a_2 v_2…………………………. (10.10) $$

This equation is called equation of continuity and it is the statement of conservation of mass in flow of incompressible fluids. In general,
av=constant……………………. (10.11)

The product av gives the volume flux or flow rate and remains constant throughout the pipe flow. Thus, at narrower portions where the streamlines are closely spaced, velocity increases and vice versa.

Questions for section 10.3:

1. Define:

a) Fluid dynamics

b) Steady flow

c) Streamline

d) Unsteady flow

2. Arrive at the equation for continuity.

10.4 Bernoulli’s Principle

Definition box:
Bernoulli’s Principle or Bernoulli’s Theorem states that, “For streamline flow of a non-viscous incompressible fluid, the sum of the pressure energy, kinetic and potential energy per unit mass remains constant”.

It can alternatively be stated as, “at constant height, pressure is inversely proportional to velocity of the fluid”.

Explanation:

Bernoulli’s Theorem is a statement of the law of conservation of energy as applied to a fluid. It relates the speed of a fluid at a given point, the pressure at that point and the height of that point above the reference level.
Let us consider a liquid contained between cross-sections A and B of the tube as shown in figure. The heights of A and B are $h_1$ and $h_2$ respectively above the reference level.

Let the pressure at A= $p_1$, the pressure at B= $p_2$, density of the liquid = ρ, speed of the liquid at A=$v_1$ and the speed of liquid at B=$v_2$. Then, according to Bernoulli’s Theorem,

$$ \frac {p_1}{ρ}+gh_1+\frac {1}{2} v_1^2= \frac {p_2}{ρ}+gh_2+ \frac {1}{2} v_2^2 $$
That is,$ \frac {p}{ρ}+gh+\frac {1}{2} v_2^2$=constant

Dividing throughout by “g” we get,

$$ \frac {p}{ρg}+h+ \frac {1}{2g v_2^2}=constant$$

Here, $ \frac {p}{ρg}$ is called the pressure head; $ \frac {(v_2^2)}{2g} $ is called the velocity head and “h” is called the potential head.

For a horizontal flow, h is constant.
$$ \frac {p}{ρg}+ \frac {(v_2^2)}{2g}=constant $$

It is clear from this expression that for a horizontal flow, as pressure head increases the velocity head decreases and vice-versa. This principle has many practical applications. A few of them are discussed below.

Proof of Bernoulli’s Theorem:

Let us consider an incompressible and non-viscous fluid flowing through a pipe AB of varying cross-sectional area as shown in figure, 10.9.

Let the flow be streamline.
Let $P_1$ and $P_2$ be the pressures and $a_1$ and $a_2$ be the areas of the cross sections of A and B respectively. If $v_1$ is the speed of the fluid at A and $v_2$ is that at B, then in an infinitesimally small interval of time, Δt, the fluid at A moves through a distance $v_1$ Δt to C.
At the same time, the fluid initially at D moves to B through a distance $v_1$ Δt. The work done on the fluid in the region AC is,
$$ W_1 = P_1a_1(v_1 \Delta t)=P_1 \Delta V $$ $(\because av \Delta t = \Delta V$=volume of element AC)
Work done on the fluid in the region DB against pressure is
$$W_2 = P_2a_2(v_2 \Delta t) $$
Since the flow is streamline from the equation of continuity, $a_1v_1 = a_2v_2$
$$ \therefore a_2v_2 \Delta t = a_1v_1 \Delta t = \Delta V $$
$$ \therefore W_2 = P_2 \Delta V $$
The net work done on the fluid is
$$ W_{net} = W_1-W_2 = (P_1-P_2) \Delta V $$
A part of this work done is utilised to change the kinetic energy, and a remaining is utilised in changing the gravitational potential energy.If $\rho$ is the density of the fluid passing through the pipe in time $\Delta t$ is,
$$ \Delta m = \rho a_1v_1 \Delta t = \rho a_2 v_2 \Delta t = \rho \Delta V $$
Change in kinetic energy is $ \Delta K = \frac {1}{2} \rho \Delta V (v^2_2 – v^2_1)$
Change in gravitational potential energy is
$$ \Delta U = \rho g \Delta V (h_2-h_1)$$
Applying work energy theorem for this volume of the fluid,we get
$$ W_{net} = \Delta U + \Delta K $$
$$ \implies (P_1-P_2)\Delta V = \rho g \Delta V (h_2-h_1) + \frac {1}{2} \rho \Delta V(V_2^2-v_1^1)$$
Dividing each term by $\Delta V$,we get
$$P_1-P_2 = \rho(h_2-h_1) + \frac {1}{2} \rho(v_2^2-v_1^2)$$
$$ or \space P_1-P_2 = \rho gh_2- \rho g h_1 + \frac {1}{2} \rho v^2_2 – \frac {1}{2} \rho v^2_1$$
Rearranging terms ,we get
$$ P_1 + \rho g h_1 + \frac {1}{2} \rho v^2_1 = P_2 + \rho g h_2 + \frac {1}{2} \rho v^2_2 ……(10.12)$$
Since the subscripts 1 and 2 refer to any two arbitary locations in the pipe,equation (10.12) can be written as,
$$ P+ \frac {1}{2} \rho v^2 + \rho gh = constant ….(10.13)$$
This is known as Bernoulli’s equation.

Limitations of Bernoulli’s theorem

While proving Bernoulli’s theorem, we have assumed that no energy is lost due to friction. But in practice, due to viscosity of the fluid, some energy is lost. The kinetic energy lost gets converted into hear. Thus, the Bernoulli’s equation applies only to non-viscous fluids.
Bernoulli’s theorem cannot be applied to compressible fluids, as the elastic energy of the fluid is not taken into consideration.
Bernoulli’s equation does not hold good for non-steady or turbulent flow as the viscosity and pressure may vary with time and position.

Questions for section 10.4:

State Bernoulli’s theorem and explain its proof.
State the limitations of Bernoulli’s theorem.

10.4.1 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.
Let us consider a tank containing a liquid of density, ρ with a small hole in its side at a height $y_1$ from the bottom (refer Figure, 10.10). The air above the liquid, whose surface is at height $y_2$, is at pressure P. From the equation of continuity [Equation- (10.10)] we have,
$$ v_1 A_1 = v_2 A_2(or) v_2=\frac {A_1}{A_2} v_1$$

Figure. 10.10 Torricelli’s law. The speed of efflux, $v_1$, from the side of the container is given by the application of Bernoulli’s equation. If the container is open at the top to the atmosphere, then $v_1$ = .

If the cross sectional area of the tank $A_2$ is much larger than that of the hole (that is, $A_2>>A_1$), then we may take the fluid to be approximately atrest at the top, i.e. $v_2$ = 0.
Now, applying the Bernoulli equation at points 1 and 2 and noting that at the hole, $P_1$ = Pa (the atmospheric pressure), we have from Equation (10.12).
insert equations from base 3A, page 41, ncert starting from $P_a+ 1/2ρv_1^2$+…………..label according to pencil marks
$$ P_a + \frac {1}{2} \rho v^2_1 + \rho gy_1 = P+ \rho gy_2$$
If $ y_2-y_1 = h$,we get
$$ v_1 = \sqrt { 2gh+ \frac {2(P-P_a)}{\rho}}$$
If the tank is open to atmosphere,then $P=P_a$
$$ \therefore v_1 = \sqrt {2gh}$$
This is the same as the speed of a free-falling body. This equation is known as Torricelli’s Law or Torricelli’s theorem.

Note box:
When $P >>P_a$ and 2gh may be ignored, the speed of efflux is determined by the pressure inside the container. Such a situation occurs in rocket propulsion.

10.4.2 Venturi-meter

The Venturi-meter is a device used to measure the speed of an incompressible fluid in a pipe.

It consists of a tube with a broad diameter and a small constriction at the middle as shown in Figure (10.11).
A manometer in the form of a U-tube is also attached to it, with one arm atthe broad neck of the tube and the otherat the constriction as shown in Figure (10.11).
Themanometer contains a liquid of density, $ρ_m$.
The speed, $v_1$ of the liquid flowing through the tube at the broad neck area, A will be measured from equation of continuity— Equation (10.10) and the speed at the constriction becomes: $v_2= (A/a)v_1$ . Then, using Bernoulli’s equation, we get,

$$ P_1 + \frac {1}{2}ρv^2_1 = P_2 + \frac {1}{2}ρv^2_1(A/a)^2 $$
So that
$$ P_1-P_2 = \frac {1}{2}ρv^2_1 \bigg \bigg( \frac {A}{a}\bigg)^2-1 \bigg$$

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 is the measure of the pressure difference.
$$ P_1-P_2=ρ_mgh=\frac {1}{2}ρv^2_1 \bigg[ \bigg( \frac {A}{a} \bigg)^2-1 \bigg] $$
So that the speed of fluid at wide neck is
$$ v_1 = \sqrt { \bigg( \frac {2ρ_mgh}{ρ} \bigg) } \bigg( \bigg( \frac {A}{a} \bigg)^2-1 \bigg)^{-1/2} ….(10.17)$$

Applications of Venturi meter:

The carburettor of automobile has a Venturi channel (nozzle) through which airflows with a large speed. The pressure is thenlowered at the narrow neck and the petrol(gasoline) is sucked up in the chamber to providethe correct mixture of air and fuel necessary for combustion.
Filter pumps or aspirators, Bunsen burner, atomisers and sprayers [Refer Figure 10.12]used for perfumes or to spray insecticides workon the same principle.

Example 10.7: Blood velocity: The flow of blood in a large artery of an anesthetized 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 mm$^2$. The narrower part has an area a = 4 mm$^2$. The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?

Solution: We take the density of blood from Table 10.1 to be 1.06 × 10$^3$ kg m$^{-3}$. The ratio of the areas is $ \frac {A}{a} = 2 $ .Using Eq. (10.17) we obtain
$$ v_1 = \sqrt { \frac { 2×24 \space Pa}{1060 \space kg \space m^{-3}×(2^2-1)}} =0.125 \space ms^{-1} $$

10.4.3 Blood Flow and Heart Attack

We shall recollect Bernoulli’s principle which states that, “at constant height, pressure is inversely proportional to velocity of the fluid”.
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. The speed of the flow of blood in the constricted region is raised which lowers the pressure inside the artery (since Pressure and Velocity are inversely proportional according to the principle). This results in a pressure differential with low pressure of the flowing blood and a high external pressure on the artery. Thus, resulting in a collapse of the artery due to the external pressure.
The heart exerts further pressure to open this artery and forces the blood through. As the blood rushes through the opening, the internal pressure once again drops due to same reasons.
This pattern of collapse and revival may thus result in a heart attack.

10.4.4 Dynamic Lift

Dynamic lift is the force that acts on a body, such as airplane wing, a hydro foil or a spinning ball, by virtue of its motion through a fluid.
In many games such as cricket, tennis, baseball or golf, we notice that a spinning ball deviates from its parabolic trajectory as it moves through the air. This deviation can be partly explained on the basis of Bernoulli’s principle.

Case1: Ball moving without spin:

Figure 10.13(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.

Case 2: Ball moving with spin:

A ball which is spinning drags air along with it. If the surface is rough, more air will be dragged. Figure 10.13(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 stream lines thus get crowded above and rarefied 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.

Case 3: Aerofoil or lift on aircraft wing:

Figure 10.13 (c) shows an aerofoil, which is a solid piece shaped to provide an upward dynamic lift when it moves horizontally through air.

The cross-section of the wings of an aeroplane looks somewhat like the aerofoil shown in Fig. 10.13 (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 thus, 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 10.8: A fully loaded Boeing aircraft has a mass of 3.3 × 10$^5$ kg. Its total wing area is 500 m$^2$. 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$^{-3}$]

Solution: (a) The weight of the Boeing aircraft is balanced by the upward force due to the pressure difference
$$ ΔP × A = 3.3 × 10^5 \space kg × 9.8 $$
$$ ΔP = (3.3 × 10^5 \space kg × 9.8 \space m s^{–2}) / 500 \space m^2$$
$$= 6.5 × 10^3 \space Nm^{-2} $$

(b) We ignore the small height difference between the top and bottom sides in Eq. (10.12).
The pressure difference between them is then
where $v_2$ is the speed of air over the upper surface and $v_1$ is the speed under the bottom surface.
$$ (v_2-v_1) = \frac{2 \Delta P }{ \rho (v_2-v_1)}$$

Taking the average speed
$$ v_{av} = (v_2 + v_1)/2 = 960 \space km/h = 267 \space m s^{-1} $$,

we have
$$ (v_2-v_1)/v_{av} = \frac { \Delta P}{ \rho v^2_{av}} \approx 0.08 $$
The speed above the wing needs to be only 8 % higher than that below.

Questions from sections 10.4.1 to 10.4.4:

Arrive at Torricelli’s law from the equation of continuity.
What is a Venturi meter? Describe the apparatus and arrive at the equation for speed of fluid at the wide neck of a Venturi meter.
What are the applications of Venturi meter?
Explain through Bernoulli’s principle, how heart attack occurs.
What is dynamic lift?
What is Magnus effect?
Explain how dynamic lift arises on an Aerofoil or Aircraft.

10.5 Viscosity

Definition box:
The property of a liquid by virtue of which it opposes the relative motion between its different layers is called viscosity.

Suppose we consider a fluid like oil enclosed between two glass plates as shown in Figure 10.14 (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 the surface has the same velocity as that of the surface. 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 (here, the bottom layer) 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, as shown in Figure 10.14 (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 in solids), the stress is found experimentally to depend on ‘rate of change of strain’ or ‘strain rate’ that is, Δx/(lΔt) or v/l instead of strain itself.

Definition box:
The coefficient of viscosity (called ‘eta’) for a fluid is defined as the ratio of shearing stress to the strain rate.

It can be expressed as,
(F/A)/ (v /l) = Fl/A ………………. (10.18)

Note box:
1. The SI unit of viscosity is poiseiulle (Pl). Its other units are N s $m^{-2}$ or Pa s.
2. The dimensions of viscosity are$ [ML^{-1}T^{-1}]$.
3. Generally, thin liquids like water, alcohol etc. are less viscous than thick liquids like coal tar, blood, glycerine etc.

Example 10.9: A metal block of area 0.10 m$^2$ is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 10.15. 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.

Solution: The metal block moves to the right because of 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
$$ F = T = mg = 0.010 \space kg × 9.8 \space m s^{–2} = 9.8 × 10^{-2} \space N $$
$$ Shear \space stress \space on \space the \space fluid = F/A = \frac {9.8× 10^{-2} \space N}{0.10} $$
$$ Strain \space rate = \frac {v}{l} = \frac {0.085}{0.030} $$
$$ η = \frac {stress}{strain \space rate} $$

$$ = \frac { ( 9.8 × 10^{-2} \space N )× ( 0.30 × 10^{-3} \space m)}{(0.085 \space ms^{-1})× (0.10 \space m^2)} $$
$$ = 3.45 × 10^{-3} \space Pa \space s $$

The coefficients of viscosity for some common fluids are listed in Table 10.2. We point out two facts about blood and water that you may find interesting. As Table 10.2 indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity $(η/η_{water})$ of blood remains constant between 0oC and 37oC.
The viscosity of liquids decreases with temperature while it increases in the case of gases.

10.5.1 Stokes’ Law

When an object moves relative to a fluid, the fluid exerts a friction like retarding force on the object. This force is called drag force or viscous force as it is due to the viscosity of the fluid.
Experimentally, it is seen that viscous force is proportional to the velocity of the object and is always opposite to the direction of motion.
If we consider a spherical object moving through a viscous fluid, then the viscous drag also depends on co-efficient of viscosity of the fluid, η and radius of sphere, “a”.
An English scientist, Sir George G Stokes showed that, when the fluid around a small spherical object is laminar, the viscous drag force, F is given by;
F = $6^{\pi}$ ηav………………………………… (10.19)
This equation is known as Stoke’s law.
In the case of an object falling through a fluid under gravity, the driving force is the effective weight of the object, (ρ-σ) V g
Where, ρ = density of the object, σ = density of the fluid, $ V = \frac {4}{3} πa^3 $ is the volume of the spherical object of radius, “a”.
As the object falls, its velocity increases and the viscous force also increases until it is equal in magnitude to the effective weight of the object. When the viscous force balances the effective weight of the object, the velocity does not increase further. This maximum velocity is called terminal velocity ($v_1$).

Therefore, F = $6^{\pi}$ ηav $= \frac {4}{3} πa^3 (ρ-σ)g$

This equation can be used to determine the value of co-efficient of viscosity of a fluid by measuring terminal velocity.

Solving for $v_t$ from the above equation, we get, $ v_t= \frac {(2a^2 (ρ-σ)g)}{gη} ……………. (10.20)$
Therefore, the terminal velocity, $v_t$ depends on the square of a radius of the sphere and inversely on viscosity of the medium.

Example 10.10: The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20$^o$C is 6.5 cm s$^{-1}$. Compute the viscosity of the oil at 20$^o$C. Density of oil is 1.5 × 10$^3$ kg m$^{-3}$, density of copper is 8.9 × 10$^3$ kg m$^{-3} $.

Solution: We have $v_t$ = 6.5 × 10$^{-2}$ ms$^{-1}$, a = 2 × 10$^{-3}$ m, g = 9.8 ms$^{-2}$, ρ = 8.9 × 10$^3$ kg m$^{-3}$, σ =1.5 × 10$^3$ kg m$^{-3}$. From Eq. (10.20)

$$ η = \frac {2}{9}× \frac { (2×10^{-3})m^2×9.8 \space ms^{-2}}{6.5×10^{-2} \space ms^{-1}}×7.4×10^3 \space kgm^{-3} $$

$$ = 9.9 \times 10^{-1} \space kg m^{-1} s^{-1} $$

Questions from section 10.5:

Define Viscosity.
Explain Laminar flow.
Define coefficient of viscosity.
What is the SI Unit of Viscosity? Give the dimensional formula.
What is drag force or viscous force? Give the expression for Stoke’s law pertaining to it.
Arrive at the equation for terminal velocity.

10.6 Reynolds Number

When the rate of flow of a fluid is large, the flow no longer remains 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 fast moving fluid causes turbulence [as shown in the figure below].

A jet of air striking a flat plate placed perpendicular to it. This is an example of turbulent flow.

Example box:
1. For instance, the smoke rising from a burning stack of wood and oceanic currents are turbulent.
2. Twinkling of stars is the result of atmospheric turbulence.
3. The wakes in the water and in the air left by cars, aeroplanes and boats are also turbulent.

Osborne Reynolds observed that turbulent flow is less likely to occur for a 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 the Reynolds number ($R_e$).

$$ R_e = ρvd/η …………………………… (10.21)$$
Where, ρ = density of the fluid flowing with a speed = v, d = diameter of the pipe, and η is the viscosity of the fluid.

Note box:
1. $R_e$ is a dimensionless number and therefore, it remains same in any system of units.
2. It is found that flow is streamline or laminar for $R_e$ less than 1000. The flow is turbulent for $R_e$ > 2000.
3. The flow becomes unsteady for $R_e$ between 1000 and 2000.

Concept box:
The value of $R_e$ at which turbulence is just set is called Critical Reynolds Number.

The critical value of $R_e$ at which turbulence is set, is found to be the same for the geometrically similar flows.
For example, when oil and water with their 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, a small scale laboratory model can be set up to study the character of fluid flow. They are useful in designing of ships, submarines, racing cars and aeroplanes.
$R_e$ can also be written as;

$$ R_e = ρv^2 / (ηv/d) = ρAv^2 / (ηAv/d) …………………… (10.22)$$
Therefore, $R_e$ represents the ratio of inertial to viscous force;
$R_e$ = Inertial force/ Force of Viscosity

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 with 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. For instance, the blades of a kitchen mixer induce turbulent flow and provide thick milk shakes as well as beat eggs into a uniform texture.

Example 10.11: The flow rate of water from a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is 10$^{-3}$ Pa s. After sometime the flow rate is increased to 3 L/min. Characterise the flow for both the flow rates.

Solution: 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

$$ Q = v × π d^2 / 4 $$

$$ v = 4 Q / d^2 π$$

We then estimate the Reynolds number to be
$$ R_e = 4 ρ Q / π d η $$
$$ = 4 ×10^3 \space kg m^{–3} × Q/(3.14 ×1.25 ×10^{-2} \space m ×10^{-3} \space Pa s) $$
$$ = 1.019 × 10^8 \space m^{–3} \space s \space Q $$

Since initially

$$ Q = 0.48 L / min = 8 \space cm^3 / s = 8 × 10^{-6} \space m^3 s^{-1} $$,
we obtain,

$$R_e = 815 $$

Since this is below 1000, the flow is steady. After some time when

$$ Q = 3 L / min = 50 \space cm^3 / s = 5 × 10^{-5} \space m^3 s^{-1}$$,

we obtain,

$$R_e = 5095 $$

The flow will be turbulent. You may carry out an experiment in your washbasin to determine the transition from laminar to turbulent flow.

Questions for sections 10.6:

What is turbulent flow? Why is it caused?
What is the significance of Reynold’s number? Give the necessary relation.
What is the value of Re for streamline, turbulent and unsteady flow?
What is Critical Reynolds number?

10.7 Surface Tension

A brief introduction:

You must have noticed that, 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 rises up a cotton wick despite of spite of gravity, Sap and water rise up 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 are related with the nature of 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 phenomenon.

10.7.1 Surface Energy

A liquid stay together because of attraction between molecules. Let us consider a molecule well inside a liquid. The intermolecular distances are so small that, each molecule is attracted to all the surrounding molecules [refer Figure 10.16(a)].
This attraction results in a negative potential energy for the molecule, which depends on the number and distribution of molecules around it. 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 vaporise, the heat of evaporation required is quite large. For water, it is of the order of 40 kJ/mol.
Now, let us consider a molecule near the surface [refer Figure 10.16(b)]. Only lower half side of it is surrounded by liquid molecules. There is some negative potential energy due to these. However, it is obviously less than negative potential energy for a molecule fully inside the liquid. Approximately, for these half immersed, the negative potential energy is half that 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 attain the least surface area which external conditions permit because increasing surface area requires energy. Most surface phenomenon can be understood in terms of this fact.

Concept box:
Heat of evaporation is the amount of energy that must be supplied to the liquid substance, to transform a quantity of that substance into a gas.

Q and A:

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 that is, half the heat of evaporation; while, obviously that for a molecule fully submerged inside would be the complete heat of evaporation.

Note box:
The density of the liquid molecules drops rapidly to zero around z = 0 as we move along the direction indicated Fig 10.16 (c) in a distance of the order of a few molecular sizes.

10.7.2 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) and keeping other things like volume, fixed, requires additional energy. To understand this, let us consider a horizontal liquid film in a bar such that the film is free to slide over parallel guides as depicted in Figure (10.17).

**Figure 10.17:** Stretching a film. (a) A film in equilibrium; (b) The film stretched an extra distance.

Suppose that we move the bar by a small distance, d, the area of the surface increases. This results in increased energy of the system which indicates that some work has been done against an internal force.
Let this internal force be F. Then, 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.
The film has two sides and the liquid in between, so there are two surfaces and the extra energy is,
S (2dl) = Fd …………………………… (10.23)

Or, S = Fd/2dl = F/2l ………………….. (10.24)

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. We need to consider fluid surface in contact with other fluids or solid surfaces.
The surface energy in this 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 increased. Thus, the surface energy is the energy of the interface between the two materials.

We make the following observations from above:

(i) Surface tension is the 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, let us 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 it towards themselves! This line of atoms is in equilibrium under tension. If the line really marks the end of the interface, as in Figure 10.16 (a) and (b), there is only the force S per unit length acting inwards.

Table 10.3 gives the surface tension of various liquids. The value of surface tension depends on temperature. Like viscosity, the surface tension of a liquid usually falls with temperature.

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 solid-air, and fluid-air.
Now, there is cohesion between the solid surface and the liquid. It can directly be measured experimentally as shown in schematic diagram below (Figure 10.18)

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 by placing its horizontal edge just over water.
The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension.
Weights are added till the plate just rises above water.
Suppose the additional weight required is W, then from Equation 10.24 and the discussion given there, the surface tension of the liquid-air interface is;
$$S_{la} = (W/2l) = (mg/2l) ……………………………. (10.25) $$
Where, m is the extra mass and l is the length of the plate edge. The subscript, “la” indicates the liquid-air interface tension.

Note box: The SI unit of Surface Tension is Newton per meter ($Nm^{-1}$).

10.7.3 Angle of Contact

The surface of liquid, near the plane of contact with another medium is in general, curved.

Definition box:
Angle of contact: 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 different pairs of liquids and solids. The value of θ determines whether a liquid will spread on the surface of a solid or if it will form droplets on it.
For example, water forms droplets on lotus leaf as shown in Figure 10.19 (a) while spreads over a clean plastic plate as shown in Figure 10.19(b).

We consider the three interfacial tensions for all the three interfaces: liquid-air, solid-air and solid-liquid denoted by $S_{la}, S_{sa}$ & $S_{sl}$ respectively as given in Figure 10.19 (a) and (b).
At the line of contact, the surface forces for the three sets of media must be in equilibrium.
From the Figure, 10.19(b) the following relation is easily derived;
$$ S_{la} \space cosθ + S_{sl} = S_{sa} …………………… (10.26)$$

Note box:
1. 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.

2. 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 liquid-solid surface. Thus, liquid does not wet the solid. We can conclude that for an obtuse angle of θ, the surface does not wet. Examples include, water on a waxy or oily surface and mercury on any surface.

3. 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, thus, 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). We can conclude that for an acute angle of θ, the surface gets wet.

Examples include: Soaps, detergents and dying substances (and are called 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.

10.7.4 Drops and Bubbles

A consequence of surface tension is that, free liquid drops and bubbles are spherical if effects of gravity can be neglected.
A liquid-air interface has energy and for any given volume, the surface with minimum energy is the one with the least area. The sphere has this property. (We can say that sphere is better than at least a cube in this respect!).
Thus, if gravity and other forces (like, air resistance) were ineffective, liquid drops would be spherical.

Another interesting consequence of surface tension is that the pressure inside a spherical drop Figure 10.20(a) is more than the pressure outside.
Suppose a spherical drop of radius, r is in equilibrium and increases by Δr, the extra surface energy is;
$$ [4π(r+ \Delta r)^2-4πr^2]S_{la}=8πr \space \Delta r \space S_{la} …..(10.27)$$
If the drop is in equilibrium, this energy cost is balanced by the energy gain due to expansion under the pressure difference, $“P_i – P_o”$ that is between the inside of the bubble and the outside. The work done is then,
$$ W = (P_i-P_o) 4πr^2 \Delta r ……(10.28)$$
So that
$$ (P_i-P_o)=(2 \space S_{la}/r ) ……(10.29)$$
In general, for a liquid-gas interface, the convex side has a higher pressure than the concave side. For example, an air bubble inside a liquid would have higher pressure inside it [Refer Figure 10.20 (b)].
A bubble shown in Figure 10.20 (c) differs from a drop and a cavity. Here, there are two interfaces.
A soap bubble has two liquid surfaces with air, one inside the bubble and the other outside the bubble. Therefore, the force due to surface tension = 2 x 2πr$S_{la}$. Therefore,
$$(P_i – P_o) = (4 S_{la}/ r) ……………….. (10.30)$$
Thus, excess of pressure inside a drop is inversely proportional to its radius. Therefore, the pressure needed to form a very small bubble is high. This is why we have to blow hard, to start forming a soap bubble. Once the bubble has grown, air pressure inside needed to make it expand further, becomes less. Same concept applies for blowing a balloon.

10.7.5 Capillary Rise

The property of surface tension gives rise to another interesting phenomenon called Capillarity.
When a capillary tube is dipped in water, the water rises up in the tube. The level of water in the tube is above the free surface of water in the beaker (called capillary rise).
When a capillary tube is dipped in mercury, mercury also rises in the tube. But, the level of mercury is depressed below the free surface of mercury in the beaker (called capillary fall).

Definition box:
The rise of a liquid in a capillary tube is known as capillarity.

Expression for capillary rise:

Let us consider a capillary tube of radius, “a” dipped vertically in water contained in clean beaker. Let ρ be the density of water, as shown in Figure 10.21.

In the case of water, since the force of adhesion (bond with the walls of the tube) predominates over the force of cohesion (intermolecular bonding), water molecules have a tendency to rise along the walls of the capillary and hence, water rises in the capillary tube. This is called, capillary rise.
The contact angle between water and glass is acute. Thus, the surface of water in the capillary is concave. This means that there is a pressure difference between the two sides of the top surface. By referring equations, 10.28 and 10.29, this is given by;

$$(P_i – P_o) = (2S/r) = 2S/(a \space sec θ )$$
$$= (2S/a) cos \spaceθ …………………………….. (10.31)$$

Thus, the pressure of the water inside the tube, just at the meniscus (air-water interface) is less than the atmospheric pressure. Consider the two points A and B in Fig. 10.21(a). They must be at the same pressure, namely
$$P_0 + h ρ g = P_i = PA ………………………………… (10.32)$$
Where, ρ is the density of water and h is called the capillary rise (refer figure 10.21a).
$$ h ρ g = (P_i – P_0) = (2S \space cos θ )/a ……………………. (10.33)$$

Note box:
1. The capillary rise is larger, for a smaller radius. Typically it is of the order of a few cm for fine capillaries.
2. Notice that if the liquid meniscus is convex, as for mercury, that is, if cos θ is negative then, it is clear that the liquid will be lower in the capillary!

Example box:
For example, if a = 0.05 cm, using the value of surface tension for water (Table 10.3), we find that,
h = 2S/(ρ g a)
$$ = \frac {2 \times 0.073 \space Nm^{-1}}{10^3 \space kg \space m^{-3} \space 9.8 \space ms^{-2} \times 5 \times 10^{-4}m}$$
$$ = 2.98 × 10^{–2} \space m = 2.98 \space cm $$

10.7.6 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; since there is very little area of contact between them.
If water could wet grease, the flow of water could carry some grease away. This is achieved through use of detergents.
The molecules of detergents are long, round and pin shaped with one end attracted to water and the other to molecules of grease, oil or wax, thus tending to form water-oil interfaces.

Thus, they tend to form oil-water interfaces. In other words, the addition of detergents reduces drastically, the surface tension of 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 10.12: 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 \times 10^{-2} \space Nm^{-1}$. 1 atmospheric pressure = 1.01 × $10^5$ Pa, density of water = 1000 kg/$m^3$, g = 9.80 m $s^{-2}$. Also calculate the excess pressure.

Solution:

The excess pressure in a bubble of gas in a liquid is given by 2S/r, where S is the surface tension of the liquid-gas 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 $P_o$ equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
$$ P_o = (1.01 \times 10^5 Pa + 0 .08 \space m \times 1000 \space kg \space m^{-3} \times 9.80 \space ms^{-2})$$
$$ = 1.01784 \times 10^5 Pa $$
Therefore, the pressure inside the bubble is
$$P_1 = P_o + 2S/r$$
$$ = 1.01784\times 10^5 Pa + (2 \times 7.3 \times 10^{-2} Pa-m/10^{-3}m)$$
$$=(1.01784 + 0.00146) \times 10^5 \space Pa$$
$$ = 1.02 \times 10^5 \space Pa$$
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.

Questions for section 10.7.1 to 10.7.6:

What is Surface Tension? Give the SI unit.
Define Angle of contact for a liquid in contact with a solid.
What does the value of θ denote?
What happens when θ is acute and when it is acute?
Give the expression for work done in expanding a bubble.
Define capillarity.
Give the expression for capillary rise.
Explain with necessary figures, the cleaning action of detergents.

Chapter 10 – Mechanical Properties of Fluids – 11th Physics

Table of Contents