## Mechanical Properties of Fluids

### 9.1 INTRODUCTION

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

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

How are fluids different from solids? What is common in liquids and gases? Unlike a solid, a fluid has no definite shape of its own. Solids and liquids have a fixed volume, whereas a gas fills the entire volume of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words solids and liquids have much lower compressibility as compared to gases.

Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress. The shearing stress of fluids is about million times smaller than that of solids.

### 9.2 PRESSURE

A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose chest a large, light but strong wooden plank is placed first, is saved from this accident. Such everyday experiences convince us that both the force and its coverage area are important. Smaller the area on which the force acts, greater is the impact. This impact is known as pressure.

When an object is submerged in a fluid at rest, the fluid exerts a force on its surface. This force is always normal to the object’s surface. This is so because if there were a component of force parallel to the surface, the object will also exert a force on the fluid parallel to it; as a consequence of Newton’s third law. This force will cause the fluid to flow parallel to the surface. Since the fluid is at rest, this cannot happen. Hence, the force exerted by the fluid at rest has to be perpendicular to the surface in contact with it. This is shown in Fig.9.1(a).

The normal force exerted by the fluid at a point may be measured. An idealised form of one such pressure-measuring device is shown in Fig. 9.1(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 $P_{a v}$ is defined as the normal force acting per unit area.

$$ \begin{equation*} P_{a v}=\frac{F}{A} \tag{9.1} \end{equation*} $$

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

$$ \begin{equation*} P=\lim _{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A} \tag{9.2} \end{equation*} $$

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. (9.1) and (9.2). Its dimensions are $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$. The SI unit of pressure is $\mathrm{N} \mathrm{m}^{-2}$. It has been named as pascal $(\mathrm{Pa})$ in honour of the French scientist Blaise Pascal (1623-1662) who carried out pioneering studies on fluid pressure. A common unit of pressure is the atmosphere (atm), i.e. the pressure exerted by the atmosphere at sea level $\left(1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{~Pa}\right)$.

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

$$ \begin{equation*} \rho=\frac{m}{V} \tag{9.3} \end{equation*} $$

The dimensions of density are $\left[\mathrm{ML}^{-3}\right]$. Its SI unit is $\mathrm{kg} \mathrm{m}^{-3}$. 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 $4^{\circ} \mathrm{C}(277 \mathrm{~K})$ is $1.0 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. The relative density of a substance is the ratio of its density to the density of water at $4^{\circ} \mathrm{C}$. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7 . Its density is $2.7 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$. The densities of some common fluids are displayed in Table 9.1.

Table 9.1 Densities of some common fluids at STP*

Fluid | $\rho\left(\mathbf{k g} \mathbf{~ m}^{-3}\right)$ |
---|---|

Water | $1.00 \times 10^{3}$ |

Sea water | $1.03 \times 10^{3}$ |

Mercury | $13.6 \times 10^{3}$ |

Ethyl alcohol | $0.806 \times 10^{3}$ |

Whole blood | $1.06 \times 10^{3}$ |

Air | 1.29 |

Oxygen | 1.43 |

Hydrogen | $9.0 \times 10^{-2}$ |

Interstellar space | $\approx 10^{-20}$ |

#### 9.2.1 Pascal’s Law

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

Fig. 9.2 shows an element in the interior of a fluid at rest. This element $\mathrm{ABC}-\mathrm{DEF}$ is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points But for clarity we have enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element as discussed above. Thus, the fluid exerts pressures $P_{\mathrm{a}}, P_{\mathrm{b}}$ and $P_{\mathrm{c}}$ on this element of area corresponding to the normal forces $F_{\mathrm{a}}, F_{\mathrm{b}}$ and $F_{\mathrm{c}}$ as shown in Fig. 9.2 on the faces BEFC, ADFC and ADEB denoted by $A_{a}, A_{b}$ and $A_{c}$ respectively. Then

$F_{\mathrm{b}} \sin \theta=F_{\mathrm{c}}, \quad F_{\mathrm{b}} \cos \theta=F_{\mathrm{a}} \quad$ (by equilibrium)

$A_{\mathrm{b}} \sin \theta=A_{\mathrm{c}}, \quad A_{\mathrm{b}} \cos \theta=A_{\mathrm{a}}^{\mathrm{a}}$ (by geometry) Thus,

$$ \begin{equation*} \frac{F_{b}}{A_{b}}=\frac{F_{c}}{A_{c}}=\frac{F_{a}}{A_{a}} ; \quad P_{b}=P_{c}=P_{a} \tag{9.4} \end{equation*} $$

Hence, pressure exerted is same in all directions in a fluid at rest. It again reminds us that like other types of stress, pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.

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

#### 9.2.2 Variation of Pressure with Depth

Consider a fluid at rest in a container. In Fig. 9.3 point 1 is at height $h$ above a 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 $\left(P_{1} A\right)$ acting downward, at the bottom $\left(P_{2} A\right)$ acting upward. If $m g$ is weight of the fluid in the cylinder we have

$$ \begin{equation*} \left(P_{2}-P_{1}\right) A=m g \tag{9.5} \end{equation*} $$

Now, if $\rho$ is the mass density of the fluid, we have the mass of fluid to be $m=\rho V=\rho h A$ so that

$$ \begin{equation*} P_{2}-P_{1}=\rho g h \tag{9.6} \end{equation*} $$

Pressure difference depends on the vertical distance $h$ between the points ( 1 and 2 ), mass density of the fluid $\rho$ 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, $\mathrm{P}_1$ may be replaced by atmospheric pressure $\left(\mathrm{P}_a\right)$ and we replace $\mathrm{P}_2$ by P. Then Eq. (9.6) gives

$$ \begin{equation*} P=P_{\mathrm{a}}+\rho g h \tag{9.7} \end{equation*} $$

Thus, the pressure $P$, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount $\rho g h$. The excess of pressure, $P-P_{\mathrm{a}}$, 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. (9.7). Thus, the height of the fluid column is important and not cross-sectional or base area or the shape of the container. The liquid pressure is the same at all points at the same horizontal level (same depth). The result is appreciated through the example of hydrostatic paradox. Consider three vessels A, B and C [Fig.9.4] 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.

#### 9.2.3 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 $1.013 \times 10^{5} \mathrm{~Pa} \mathrm{(1} \mathrm{atm).} \mathrm{Italian} \mathrm{scientist}$ Evangelista Torricelli (1608-1647) 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.9.5 (a). This device is known as ‘mercury barometer’. The space above the mercury column in the tube contains only mercury vapour whose pressure $P$ is so small that it may be neglected. Thus, the pressure at Point $\mathrm{A}=0$. The pressure inside the coloumn at Point B must be the same as the pressure at Point $\mathrm{C}$, which is atmospheric pressure, $\mathrm{P}_{a}$.

$$ \begin{equation*} P_{\mathrm{a}}=\rho g h \tag{9.8} \end{equation*} $$

where $\rho$ 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 $76 \mathrm{~cm}$ at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of $\rho$ in Eq. (9.8). A common way of stating pressure is in terms of $\mathrm{cm}$ or $\mathrm{mm}$ of mercury $(\mathrm{Hg})$. A pressure equivalent of $1 \mathrm{~mm}$ is called a torr (after Torricelli).

1 torr $=133 \mathrm{~Pa}$.

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

1 bar $=10^{5} \mathrm{~Pa}$

An open tube manometer is a useful instrument for measuring pressure differences. It consists of a U-tube containing a suitable liquid i.e., a low density liquid (such as oil) for measuring small pressure differences and a high density liquid (such as mercury) for large pressure differences. One end of the tube is open to the atmosphere and the other end is connected to the system whose pressure we want to measure [see Fig. 9.5 (b)]. The pressure $P$ at A is equal to pressure at point $B$. What we normally measure is the gauge pressure, which is $P-P_{\mathrm{a}}$, given by Eq. (9.8) 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 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.

#### 9.2.4 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 [Fig. 9.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.

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

A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure. In a hydraulic lift, as shown in Fig. 9.6 (b), two pistons are separated by the space filled with a liquid. A piston of small cross-section $A_{1}$ is used to exert a force $F_{1}$ directly on the liquid. The pressure $P=\frac{F_{1}}{A_{1}}$ 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 $P \times A_{2}$. Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) $F_{2}=P A_{2}=\frac{F_{1} A_{2}}{A_{1}}$. 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 $\frac{A_{2}}{A_{1}}$ and this factor is the mechanical advantage of the device. The example below clarifies it.

### 9.3 STREAMLINE FLOW

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

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.9.7 (a), the curve describes how a fluid particle moves with time. The curve $P Q$ 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 $\mathrm{Q}$ in Fig. 9.7 (b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines. This means that number of fluid particles crossing the surfaces as indicated at $P, R$ and $Q$ is the same. If area of cross-sections at these points are $A_{\mathrm{P}}, A_{\mathrm{R}}$ and $A_{Q}$ and speeds of fluid particles are $v_{\mathrm{P}}, v_{\mathrm{R}}$ and $v_{Q}$, then mass of fluid $\Delta m_{\mathrm{P}}$ crossing at $A_{\mathrm{P}}$ in a small interval of time $\Delta t$ is $\rho_{\mathrm{P}} A_{\mathrm{P}} V_{\mathrm{P}} \Delta t$. Similarly mass of fluid $\Delta m_{\mathrm{R}}$ flowing or crossing at $A_{\mathrm{R}}$ in a small interval of time $\Delta t$ is $\rho_{\mathrm{R}} A_{\mathrm{R}} V_{\mathrm{R}} \Delta t$ and mass of fluid $\Delta m_{Q}$ is $\rho_{Q} A_{Q} V_{Q} \Delta t$ crossing at $A_{\mathrm{Q}}$. The mass of liquid flowing out equals the mass flowing in, holds in all cases. Therefore,

$$\rho_{\mathrm{P}} A_{\mathrm{P}} V_{\mathrm{P}} \Delta t=\rho_{\mathrm{R}} A_{\mathrm{R}} V_{\mathrm{R}} \Delta t=\rho_{\mathrm{Q}} A_{\mathrm{Q}} V_{\mathrm{Q}} \Delta t \tag{9.9}$$

For flow of incompressible fluids

$\rho_{\mathrm{P}}=\rho_{\mathrm{R}}=\rho_{\mathrm{Q}}$

Equation (9.9) reduces to

$$A_{\mathrm{P}} V_{\mathrm{P}}=A_{\mathrm{R}} V_{\mathrm{R}}=A_{\mathrm{Q}} V_{\mathrm{Q}} \tag{9.10}$$

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

$$ Av= \text{constant} \tag{9.11} $$

$A v$ 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 9.7b) it is clear that $A_{\mathrm{R}}>A_{\mathrm{Q}}$ or $v_{\mathrm{R}}<v_{\mathrm{Q}}$, the fluid is accelerated while passing from $\mathrm{R}$ to $\mathrm{Q}$. This is associated with a change in pressure in fluid flow in horizontal pipes.

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

Figure 9.8 displays streamlines for some typical flows. For example, Fig. 9.8(a) describes a laminar flow where the velocities at different points in the fluid may have different magnitudes but their directions are parallel. Figure 9.8 (b) gives a sketch of turbulent flow.

### 9.4 BERNOULLI’S PRINCIPLE

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

Consider a fluid moving in a pipe of varying cross-sectional area. Let the pipe be at varying heights as shown in Fig. 9.9. 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 $\mathrm{B}$ and $\mathrm{D}$. In an infinitesimal time interval $\Delta t$, this fluid would have moved. Suppose $v_{1}$ is the speed at $\mathrm{B}$ and $v_{2}$ at $\mathrm{D}$, then fluid initially at B has moved a distance $v_{1} \Delta t$ to $\mathrm{C}$ ( $v_{1} \Delta t$ is small enough to assume constant cross-section along $\mathrm{BC})$. In the same interval $\Delta t$ the fluid initially at $\mathrm{D}$ moves to $\mathrm{E}$, a distance equal to $v_{2} \Delta t$. 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 $W_{1}=$ $P_{1} A_{1}\left(V_{1} \Delta t\right)=P_{1} \Delta V$. Since the same volume $\Delta V$ passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is $W_{2}=P_{2} A_{2}\left(V_{2} \Delta t\right)=P_{2} \Delta V$ or, the work done on the fluid is $-P_{2} \Delta V$. So the total work done on the fluid is

$$ W_{1}-W_{2}=\left(P_{1}-P_{2}\right) \Delta V $$

Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If the density of the fluid is $\rho$ and $\Delta m=\rho A_{1} v_{1} \Delta t=\rho \Delta V$ is the mass passing through the pipe in time $\Delta t$, then change in gravitational potential energy is

$$ \Delta U=\rho g \Delta V\left(h_{2}-h_{1}\right) $$

The change in its kinetic energy is

$$ \Delta K=\frac{1}{2} \quad \rho \Delta V\left(V_{2}^{2}-V_{1}^{2}\right) $$

We can employ the work - energy theorem (Chapter 6) to this volume of the fluid and this yields

$$ \left(P_{1}-P_{2}\right) \Delta V=\frac{1}{2} \rho \Delta V\left(v_{2}^{2}-V_{1}^{2}\right)+\rho g \Delta \mathrm{V}\left(h_{2}-h_{1}\right) $$

We now divide each term by $\Delta V$ to obtain

$$ \left(P_{1}-P_{2}\right)=\frac{1}{2} \rho\left(V_{2}^{2}-V_{1}^{2}\right)+\rho g\left(h_{2}-h_{1}\right) $$

We can rearrange the above terms to obtain

$$ \begin{equation*} P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2} \tag{9.12} \end{equation*} $$

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

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 $\frac{\rho v^{2}}{2}$ and the potential energy per unit volume ( $\rho g h$ )remains a 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 into heat energy. Thus, Bernoulli’s equation ideally applies to fluids with zero viscosity or nonviscous fluids. Another restriction on application of Bernoulli theorem is that the fluids must be incompressible, as the elastic energy of the fluid is also not taken into 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 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, Bernoulli’s equation becomes

$$ \begin{aligned} & P_{1}+\rho g h_{1}=P_{2}+\rho g h_{2} \\ & \left(P_{1}-P_{2}\right)=\rho g\left(h_{2}-h_{1}\right) \end{aligned} $$

which is same as Eq. (9.6).

#### 9.4.1 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. Consider a tank containing a liquid of density $\rho$ with a small hole in its side at a height $y_1$ from the bottom (see Fig. 9.10). The air above the liquid, whose surface is at height $y_2$, is at pressure $P$. From the equation of continuity [Eq. (9.10)] we have

$$ \begin{aligned} & V_{1} A_{1}=V_{2} A_{2} \\ & v_{2}=\frac{A_{1}}{A_{2}} v_{1} \end{aligned} $$

If the cross-sectional area of the tank $A_{2}$ is much larger than that of the hole $\left(A_{2} > > A_{1}\right)$, 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. (9.12)

$$ P_{a}+\frac{1}{2} \rho v_{1}^{2}+\rho g y_{1}=P+\rho g y_{2} $$

Taking $y_{2}-y_{1}=h$ we have

$$ \begin{equation*} v_{1}=\sqrt{2 g h+\frac{2\left(P-P_{a}\right)}{\rho}} \tag{9.14} \end{equation*} $$

When $P > > P_{a}$ and $2 g h$ may be ignored, the speed of 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

$$ \begin{equation*} v_{1}=\sqrt{2 g h} \tag{9.15} \end{equation*} $$

This is also the speed of a freely falling body. Equation (9.15) represents Torricelli’s law.

#### 9.4.2 Dynamic Lift

Dynamic lift 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, baseball, 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. 9.11(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 9.11(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 the ball is larger and below it is smaller (see Section 9.3). The stream lines, 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 spining is called Magnus effect.

Aerofoil or lift on aircraft wing: Figure 9.11 (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. 9.11 (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.

### 9.5 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. 9.12 (a). The bottom plate is fixed while the top plate is moved with a constant velocity $\mathbf{v}$ 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 $\mathbf{v}$ 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. 9.12 (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 $A B C D$, take the shape of AEFD after short interval of time $(\Delta t)$. During this time interval the liquid has undergone a shear strain of $\Delta x / 1$. 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 rate’ i.e. $\Delta x /(1 \Delta t)$ or $v / 1$ 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.

$$ \begin{equation*} \eta=\frac{F / A}{v / l}=\frac{F l}{v A} \tag{9.16} \end{equation*} $$

The SI unit of viscosity is poiseiulle (Pl). Its other units are $\mathrm{N} \mathrm{s} \mathrm{m}^{-2}$ or $\mathrm{Pa} \mathrm{s}$. The dimensions of viscosity are $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$. Generally, thin liquids, like water, alcohol, etc., are less viscous than thick liquids, like coal tar, blood, glycerine, etc. The coefficients of viscosity for some common fluids are listed in Table 9.2. We point out two facts about blood and water that you may find interesting. As Table 9.2 indicates, blood is ’thicker’ (more viscous) than water. Further, the relative viscosity $\left(\eta / \eta_{\text {water }}\right)$ of blood remains constant between $0{ }^{\circ} \mathrm{C}$ and $37{ }^{\circ} \mathrm{C}$.

The viscosity of liquids decreases with temperature, while it increases in the case of gases.

#### 9.5.1 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 some 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 are viscosity $\eta$ 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

$$ \begin{equation*} F=6 \pi \eta a v \tag{9.17} \end{equation*} $$

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 raindrop 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 the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus, in equilibrium, this terminal velocity $v_{t}$ is given by

$$ 6 \pi \eta a v_{\mathrm{t}}=(4 \pi / 3) a^{3}(\rho-\sigma) g $$

where $\rho$ and $\sigma$ are mass densities of sphere and the fluid, respectively. We obtain

$$ \begin{equation*} v_{\mathrm{t}}=2 a^{2}(\rho-\sigma) g /(9 \eta) \tag{9.18} \end{equation*} $$

So the terminal velocity $v_{\mathrm{t}}$ 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 6.2 in this context.

### 9.6 SURFACE TENSION

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, inspite of gravity,

Sap and water rise up to the top of the leaves of the tree, hair 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.

#### 9.6.1 Surface Energy

A liquid stays together because of attraction between molecules. Consider a molecule well inside a liquid. The intermolecular distances are such that it is attracted to all the surrounding molecules [Fig. 9.14(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 vaporise, the heat of evaporation required is quite large. For water it is of the order of $40 \mathrm{~kJ} / \mathrm{mol}$.

Let us consider a molecule near the surface Fig. 9.14(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 9.14 (c) in a distance of the order of a few molecular sizes.

#### 9.6.2 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) keeping other things like volume fixed requires a horizontal liquid film ending in bar free to slide over parallel guides Fig (9.15).

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 $\mathbf{F}$, the work done by the applied force is F.d $=F d$. 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 $2 d l$. A film has two sides and the liquid in between, so there are two surfaces and the extra energy is

$$ \begin{align*} & S(2 d)=F d \tag{9.19}\\ & \text { Or, } S=F d / 2 d l=F / 21 \tag{9.20} \end{align*} $$

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 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 it towards them! This line of atoms is in equilibrium under tension. If the line really marks the end of the interface, as in Figure 9.14 (a) and (b) there is only the force $S$ per unit length acting inwards.

Table 9.3 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.

Table 9.3 Surface tension of some liquids at the temperatures indicated with the heats of the vaporisation

Liquid | Temp $\left({ }^{\circ} \mathbf{C}\right)$ | Surface Tension $(\mathbf{N} / \mathbf{m})$ |
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 solid-air, and fluid-air. Now there is attraction between the solid surface and the liquid. It can be directly measured experimentaly as schematically shown in Fig. 9.16. 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, with 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 clears water.

Suppose the additional weight required is $W$. Then from Eq. 9.20 and the discussion given there, the surface tension of the liquid-air interface is

$$ \begin{equation*} S_{1 \mathrm{la}}=(\mathrm{W} / 2 \mathrm{l})=(\mathrm{mg} / 2 \mathrm{l}) \tag{9.21} \end{equation*} $$

where $\mathrm{m}$ is the extra mass and $l$ is the length of the plate edge. The subscript (la) emphasises the fact that the liquid-air interface tension is involved.

#### 9.6.3 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 $\theta$. It is different at interfaces of different pairs of liquids and solids. The value of $\theta$ 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. 9.17 (a) while spreads over a clean plastic plate as shown in Fig. 9.17(b).

We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air and solid-liquid denoted by $S_{\mathrm{la}}$, $S_{\mathrm{sa}}$ and $S_{\mathrm{sl}}$, respectively as given in Fig. 9.17 (a) and (b). At the line of contact, the surface forces between the three media must be in equilibrium. From the Fig. 9.17(b) the following relation is easily derived.

$$ \begin{equation*} S_{\mathrm{la}} \cos \theta+S_{\mathrm{sl}}=S_{\mathrm{sa}} \tag{9.22} \end{equation*} $$

The angle of contact is an obtuse angle if $S_{\mathrm{sl}}>S_{\mathrm{la}}$ as in the case of water-leaf interface while it is an acute angle if $S_{\mathrm{sl}}<S_{\mathrm{la}}$ as in the case of water-plastic interface. When $\theta$ 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, 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 $S_{\mathrm{sl}}$ and therefore, $\cos \theta$ may increase or $\theta$ may decrease. In this case $\theta$ 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.

#### 9.6.4 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 high-speed 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 for 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 this book, 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. 9.18(a) is more than the pressure outside. Suppose a spherical drop of radius $r$ is in equilibrium. If its radius increase by $\Delta r$. The extra surface energy is

$$ \begin{equation*} \left[4 \pi(r+\Delta r)^{2}-4 \pi r^{2}\right] S_{1 \mathrm{a}}=8 \pi r \Delta r S_{\mathrm{la}} \tag{9.23} \end{equation*} $$

If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference $\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)$ between the inside of the bubble and the outside. The work done is

$$ \begin{equation*} W=\left(P_{\mathrm{i}}-P_{\mathrm{o}}\right) 4 \pi r^{2} \Delta r \tag{9.24} \end{equation*} $$

so that

$$ \begin{equation*} \left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)=\left(2 S_{\mathrm{la}} / r\right) \tag{9.25} \end{equation*} $$

In general, for a liquid-gas 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 9.18 (b)

A bubble Fig 9.18 (c) differs from a drop and a cavity; in this it has two interfaces. Applying the above argument we have for a bubble

$$ \begin{equation*} \left(P_{\mathrm{i}}-P_{\mathrm{o}}\right)=\left(4 S_{\mathrm{la}} / r\right) \tag{9.26} \end{equation*} $$

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!

#### 9.6.5 Capillary Rise

One consequence of the pressure difference across a curved liquid-air interface is the wellknown effect that water rises up in a narrow tube in spite of gravity. The word capilla means hair in Latin; if the tube were hair thin, the rise would be very large. To see this, consider a vertical capillary tube of circular cross section (radius a) inserted into an open vessel of water (Fig. 9.19). 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. This is given by

$$ \begin{align*} & \left(P_{i}-P_{o}\right)=(2 S / r)=2 S /(a \sec \theta) \\ & =(2 S / a) \cos \theta \tag{9.27} \end{align*} $$

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. 9.19(a). They must be at the same pressure, namely

$$ \begin{equation*} P_{O}+h \rho g=P_{i}=P_{A} \tag{9.28} \end{equation*} $$

where $\rho$ is the density of water and $h$ is called the capillary rise [Fig. 9.19(a)]. Using Eq. (9.27) and (9.28) we have

$$ \begin{equation*} h \rho g=\left(P_{i}-P_{o}\right)=(2 S \cos \theta) / a \tag{9.29} \end{equation*} $$

The discussion here, and the Eqs. (9.24) and (9.25) 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 $\mathrm{cm}$ for fine capillaries. For example, if $a=0.05 \mathrm{~cm}$, using the value of surface tension for water (Table 9.3), we find that

$$ \begin{aligned} h & =2 S /(\rho g a) \\ & =\frac{2 \times\left(0.073 \mathrm{~N} \mathrm{~m}^{-1}\right)}{\left(10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(5 \times 10^{-4} \mathrm{~m}\right)} \\ & =2.98 \times 10^{-2} \mathrm{~m}=2.98 \mathrm{~cm} \end{aligned} $$

Notice that if the liquid meniscus is convex, as for mercury, i.e., if $\cos \theta$ is negative then from Eq. (9.28) for example, it is clear that the liquid will be lower in the capillary!

### Summary

**1.** The basic property of a fluid is that it can flow. The fluid does not have any resistance to change of its shape. Thus, the shape of a fluid is governed by the shape of its container.

**2.** A liquid is incompressible and has a free surface of its own. A gas is compressible and it expands to occupy all the space available to it.

**3.** If $F$ is the normal force exerted by a fluid on an area $A$ then the average pressure $P_{\text {av }}$ is defined as the ratio of the force to area

$$ P_{a v}=\frac{F}{A} $$

**4.** The unit of the pressure is the pascal (Pa). It is the same as $\mathrm{N} \mathrm{m}^{-2}$. Other common units of pressure are

$1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{~Pa}$

$1 \mathrm{bar}=10^{5} \mathrm{~Pa}$

1 torr $=133 \mathrm{~Pa}=0.133 \mathrm{kPa}$

$1 \mathrm{~mm}$ of $\mathrm{Hg}=1$ torr $=133 \mathrm{~Pa}$

**5.** Pascal’s law states that: Pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.

**6.** The pressure in a fluid varies with depth $\mathrm{h}$ according to the expression $P=P_{\mathrm{a}}+\rho g h$ where $\rho$ is the density of the fluid, assumed uniform.

**7.** The volume of an incompressible fluid passing any point every second in a pipe of non uniform crossection is the same in the steady flow.

$v A=$ constant ( $v$ is the velocity and $A$ is the area of crossection)

The equation is due to mass conservation in incompressible fluid flow.

**8.** Bernoulli’s principle states that as we move along a streamline, the sum of the pressure $(P)$, the kinetic energy per unit volume $\left(\rho v^{2} / 2\right)$ and the potential energy per unit volume ( $\rho g y$) remains a constant.

$P+\rho v^{2} / 2+\rho g y=$ constant

The equation is basically the conservation of energy applied to non viscuss fluid motion in steady state. There is no fluid which have zero viscosity, so the above statement is true only approximately. The viscosity is like friction and converts the kinetic energy to heat energy.

**9.** Though shear strain in a fluid does not require shear stress, when a shear stress is applied to a fluid, the motion is generated which causes a shear strain growing with time. The ratio of the shear stress to the time rate of shearing strain is known as coefficient of viscosity, $\eta$.

where symbols have their usual meaning and are defined in the text.

**10.** Stokes’ law states that the viscous drag force $\mathbf{F}$ on a sphere of radius a moving with velocity $\mathbf{v}$ through a fluid of viscosity is, $\mathbf{F}=6 \pi \eta \mathbf{v}$.
11. Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of interface between the liquid and the bounding surface. It is the extra energy that the molecules at the interface have as compared to the interior.

### POINTS TO PONDER

**1.** Pressure is a scalar quantity. The definition of the pressure as “force per unit area” may give one false impression that pressure is a vector. The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed. While describing fluids as a concept, shift from particle and rigid body mechanics is required. We are concerned with properties that vary from point to point in the fluid.

**2.** One should not think of pressure of a fluid as being exerted only on a solid like the walls of a container or a piece of solid matter immersed in the fluid. Pressure exists at all points in a fluid. An element of a fluid (such as the one shown in Fig. 9.4) is in equilibrium because the pressures exerted on the various faces are equal.

**3.** The expression for pressure

$P=P_{\mathrm{a}}+\rho g h$

holds true if fluid is incompressible. Practically speaking it holds for liquids, which are largely incompressible and hence is a constant with height.

**4.** The gauge pressure is the difference of the actual pressure and the atmospheric pressure. $P-P_{\mathrm{a}}=P_{\mathrm{g}}$

Many pressure-measuring devices measure the gauge pressure. These include the tyre pressure gauge and the blood pressure gauge (sphygmomanometer).

**5.** A streamline is a map of fluid flow. In a steady flow two streamlines do not intersect as it means that the fluid particle will have two possible velocities at the point.

**6.** Bernoulli’s principle does not hold in presence of viscous drag on the fluid. The work done by this dissipative viscous force must be taken into account in this case, and $P_{2}$ [Fig. 9.9] will be lower than the value given by Eq. (9.12).

**7.** As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity, $\eta$ falls. In a gas the temperature rise increases the random motion of atoms and $\eta$ increases.

**8.** Surface tension arises due to excess potential energy of the molecules on the surface in comparison to their potential energy in the interior. Such a surface energy is present at the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid alone.

Physical Quantity | Symbol | Dimensions | Unit | Remarks |
---|---|---|---|---|

Pressure | $P$ | $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$ | pascal $(\mathrm{Pa})$ | $1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{~Pa}$, Scalar |

Density | $\rho$ | $\left[\mathrm{M} \mathrm{L}^{-3}\right]$ | $\mathrm{kg} \mathrm{m}^{-3}$ | Scalar |

Specific Gravity | No | No | $\frac{\rho_{\text {substance }}}{\text { Pwater }}$ Scalar | |

Co-efficient of viscosity | $\eta$ | $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-1}\right]$ | Pa s or poiseiulles (Pl) |
Scalar |

Surface Tension | $S$ | $\left[\mathrm{M} \mathrm{T}^{-2}\right]$ | $\mathrm{N} \mathrm{m}^{-1}$ | Scalar |