Applied Physics Past Paper PDF (Beni-Suef University) 2023-2024

Beni-Suef University College of Technology and Education Applied Physics First year students first semester Basic Science Department 2023-2024  Contents Unit one Mechanical properties 1-36...

Beni-Suef University College of Technology and Education Applied Physics First year students first semester Basic Science Department 2023-2024  Contents Unit one Mechanical properties 1-36 Unit two Heat and Thermodynamics 37-72 Unit Three MCQ 73- 95 Unit Four Laboratory experiments 96-110  MECHANICAL P ROPERTIES 1.1 INTRODUCTION In this chapter, we shall study some common physical properties of liquids and gases. Liquids and gases can flow Intr oduction and are ther efore, called fluids. It is this property that Pressure distinguishes liquids and gases fr om solids in a basic way. Str eamline flow Fluids are everywhere around us. Earth has an envelop of Ber noulli’s principle air and two-thirds of its sur face is covered with water. Water Viscosity is not only necessary for our existence; every mammalian body constitute mostly of water. All the processes occurring Reynolds number in living beings including plants are mediated by fluids. Thus Sur face tension understanding the behaviour and properties of fluids is Summary important. Points to ponder How are fluids dif ferent from solids? What is common in Exer cises liquids and gases? Unlike a solid, a fluid has no definite Additional exercises shape of its own. Solids and liquids have a fixed volume, Appendix wher eas a gas fills the entire volume of its container. We have lear nt in the previous chapter that the volume of solids can be changed by str ess. The volume of solid, liquid or gas depends on the str ess or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The dif ference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pr essure is rather small. In other wor ds solids and liquids have much lower compressibility as compar ed to gases. Shear stress can change the shape of a solid keeping its volume fixed. The key pr operty of fluids is that they offer very little r esistance to shear stress; their shape changes by application of very small shear str ess. The shearing stress of fluids is about million times smaller than that of solids. 1.2 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 per former across whose 1  MECHANICAL PROPERTIES OF FLUIDS 247 chest a large, light but strong wooden plank is In principle, the piston area can be made placed first, is saved from this accident. Such arbitrarily small. The pressure is then defined everyday experiences convince us that both the in a limiting sense as force and its coverage area are important. Smaller the area on which the force acts, greater is the lim ∆F P = ∆A →0 (1.2) impact. This concept is known as pressure. ∆A Pressure is a scalar quantity. We remind the When an object is submerged in a fluid at reader that it is the component of the force rest, the fluid exerts a force on its sur face. This normal to the area under consideration and not force is always normal to the object’s surface. the (vector) force that appears in the numerator This is so because if there wer e a component of in Eqs. (1.1) and (1.2). Its dimensions are force parallel to the surface, the object will also [ML– 1T– 2]. The SI unit of pr essure is N m– 2. It has exert a force on the fluid parallel to it; as a been named as pascal (Pa) in honour of the consequence of Newton’s third law. This for ce French scientist Blaise Pascal (1623-1662) who will cause the fluid to flow parallel to the surface. carried out pioneering studies on fluid pressure. Since the fluid is at rest, this cannot happen. A common unit of pressure is the atmosphere Hence, the force exerted by the fluid at rest has (atm), i.e. the pr essur e exerted by the to be perpendicular to the sur face in contact atmosphere at sea level (1 atm = 1.013 × 105 Pa). with it. This is shown in Fig.1.1(a). Another quantity, that is indispensable in The normal force exerted by the fluid at a point describing fluids, is the density ρ. For a fluid of may be measured. An idealised form of one such mass m occupying volume V, pressure-measuring device is shown in Fig. m 1.1(b). It consists of an evacuated chamber with ρ = (1.3) V a spring that is calibrated to measure the force The dimensions of density are [ML – 3]. Its SI acting on the piston. This device is placed at a unit is kg m– 3. It is a positive scalar quantity. A point inside the fluid. The inward for ce exerted liquid is largely incompressible and its density by the fluid on the piston is balanced by the is therefore, nearly constant at all pressures. outward spring force and is ther eby measured. Gases, on the other hand exhibit a lar ge variation in densities with pressure. The density of water at 4 oC (277 K) is 1.0 × 10 3 kg m–3. The relative density of a substance is the ratio of its density to the density of water at 4oC. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is 2.7 × 103 kg m–3. The densities of some common fluids are displayed in Table 1.1. Table 1.1 Densities of some common fluids (a) (b) at STP* Fig. 1.1 (a) The force exerted by the liquid in the beaker on the submerged object or on the walls is nor mal (perpendicular) to the surface at all points. (b) An idealised device for measuring pressure. If F is the magnitude of this nor mal force on the piston of area A then the average pressure Pav is defined as the normal force acting per unit area. F Pa v = (1.1) A * STP means standard temperature (00C) and 1 atm pressure. 2 t this element of area corresponding to the normal Example 1.1 The two thigh bones forces Fa, Fb and Fc as shown in Fig. 1.2 on the (femurs), each of cross-sectional area10 cm2 faces BEFC, ADFC and ADEB denoted by Aa , Ab support the upper part of a human body of and Ac respectively. Then mass 40 kg. Estimate the average pressure Fb sinθ = F c, Fb cosθ = F a (by equilibrium) sustained by the femurs. Ab sinθ = Ac, Ab cos θ = Aa (by geometry) Thus, Answer Total cr oss-sectional area of the femurs is A = 2 × 10 cm 2 = 20 × 10 –4 m 2. The Fb F F = c = a ; Pb = Pc = Pa (1.4) force acting on them is F = 40 kg wt = 400 N Ab Ac Aa (taking g = 10 m s–2). This force is acting Hence, pr essure exerted is same in all vertically down and hence, normally on the directions in a fluid at rest. It again reminds us femurs. Thus, the average pressure is that like other types of stress, pr essure is not a F vector quantity. No direction can be assigned Pav = = 2 × 105 N m − 2 t A to it. The force against any area within (or bounding) a fluid at rest and under pr essure is 1.2.1 Pascal’s Law normal to the area, regardless of the orientation of the area. The French scientist Blaise Pascal observed that Now consider a fluid element in the for m of a the pressure in a fluid at rest is the same at all horizontal bar of uniform cr oss-section. The bar points if they are at the same height. This fact is in equilibrium. The horizontal forces exerted may be demonstrated in a simple way. 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. Wind is flow of air due to pr essure differences. 1.2.2 Variation of Pressure with Depth Fig. 1.2 Proof of Pascal’s law. ABC-DEF is an Consider a fluid at r est in a container. In element of the interior of a fluid at rest. Fig. 1.3 point 1 is at height h above a point 2. This element is in the form of a right- The pressur es at points 1 and 2 are P1 and P2 angled prism. The element is small so that respectively. Consider a cylindrical element of the effect of gravity can be ignored, but it has been enlarged for the sake of clarity. fluid having area of base A and height h. As the fluid is at rest the resultant horizontal forces Fig. 1.2 shows an element in the interior of should be zero and the resultant vertical forces a fluid at rest. This element ABC-DEF is in the should balance the weight of the element. The for m of a right-angled prism. In principle, this forces acting in the vertical direction are due to prismatic element is very small so that every the fluid pressur e at the top (P1 A ) acting part of it can be considered at the same depth downward, at the bottom (P2A) acting upwar d. from the liquid surface and therefore, the effect If mg is weight of the fluid in the cylinder we of the gravity is the same at all these points. have But for clarity we have enlarged this element. (P2 − P1 ) A = mg (1.5) The forces on this element ar e those exerted by Now, if ρ is the mass density of the fluid, we the rest of the fluid and they must be normal to have the mass of fluid to be m = ρV= ρhA so the surfaces of the element as discussed above. that Thus, the fluid exerts pressures Pa , Pb and Pc on P2 − P1 = ρgh (1.6) 3  MECHANICAL PROPER TIES OF FLUIDS 249 Fig 1.4 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height. Example 1.2 What is the pressure on a t swimmer 10 m below the surface of a lake? Fig.1.3 Fluid under gravity. The effect of gravity Answer Here is illustrated through pressure on a vertical h = 10 m and ρ = 1000 kg m-3. T ake g = 10 m s–2 cylindrical column. From Eq. (1.7) P = Pa + ρgh Pressure difference depends on the vertical = 1.01 × 105 Pa + 1000 kg m–3 × 10 m s–2 × 10 m distance h between the points (1 and 2), mass = 2.01 × 105 Pa density of the fluid ρ and acceleration due to ≈ 2 atm gravity g. If the point 1 under discussion is This is a 100% increase in pressur e from shifted to the top of the fluid (say water), which surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed is open to the atmosphere, P1 may be replaced to withstand such enormous pressures. t by atmospheric pressure (Pa ) and we replace P2 by P. Then Eq. (1.6) gives 1.2.3 Atmospheric Pressure and Gauge P = Pa + ρgh (1.7) Pressure Thus, the pressure P, at depth below the The pressure of the atmosphere at any point is surface of a liquid open to the atmosphere is equal to the weight of a column of air of unit gr eater than atmospheric pr essure by an cross sectional area extending from that point amount ρgh. The excess of pressure, P − Pa , at to the top of the atmosphere. At sea level it is depth h is called a gauge pressure at that point. 1.013 × 1 05 Pa (1 atm). Italian scientist Evangelista T orricelli (1608-1647) devised for The area of the cylinder is not appearing in the first time, a method for measuring the expression of absolute pr essure in Eq. (1.7). atmospheric pressure. A long glass tube closed Thus, the height of the fluid column is important at one end and filled with mer cury is inverted and not cross sectional or base area or the shape into a trough of mercury as shown in Fig.1.5 (a). of the container. The liquid pressure is the same This device is known as mercury barometer. The at all points at the same horizontal level (same space above the mercury column in the tube depth). The result is appreciated through the contains only mer cury vapour whose pressure example of hydrostatic paradox. Consider three P is so small that it may be neglected. The vessels A, B and C [Fig.1.4] of different shapes. pressure inside the column at point A must They are connected at the bottom by a horizontal equal the pressure at point B, which is at the pipe. On filling with water the level in the three same level. Pr essur e at B = atmospheric vessels is the same though they hold different pressure = P a amounts of water. This is so, because water at Pa = ρgh (1.8) the bottom has the same pressure below each where ρ is the density of mercury and h is the section of the vessel. height of the mercury column in the tube. 4  In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (1.8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressur e equivalent of 1 mm is called a torr (after Torricelli). 1 torr = 133 Pa. The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar. 1 bar = 105 Pa (b) the open tube manometer An open-tube manometer is a useful instrument for measuring pressure dif ferences. Fig 1.5 Two pressure measuring devices. It consists of a U-tube containing a suitable liquid i.e. a low density liquid (such as oil) for Pressure is same at the same level on both measuring small pressure dif ferences and a sides of the U-tube containing a fluid. For high density liquid (such as mercury) for large liquids the density varies very little over wide pressur e differences. One end of the tube is open ranges in pressure and temperature and we can to the atmosphere and other end is connected to the system whose pressure we want to treat it safely as a constant for our present measure [see Fig. 1.5 (b)]. The pressure P at A purposes. Gases on the other hand, exhibits is equal to pressur e at point B. What we large variations of densities with changes in normally measure is the gauge pressure, which pressure and temperature. Unlike gases, liquids is P − Pa, given by Eq. (1.8) and is proportional are therefore, largely treated as incompressible. to manometer height h. E x a m p l e 1. 3 The density of the t atmosphere at sea level is 1.29 kg/m3. Assume that it does not change with altitude. Then how high would the atmosphere extend? Answer We use Eq. (1.7) ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa ∴ h = 7989 m ≈ 8 km In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pr essure over 100 km. We should also note that the sea level atmospheric pressur e is not always 760 mm of Hg. A drop in the Hg level by 10 mm or mor e is a sign of an approaching storm. t t Example 1.4 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 Fig 1.5 (a) The mercury barometer. 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressur e. (The density of sea water is 1.03 × 10 3 kg m-3, g = 10m s–2.) 5  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. Answer Here h = 1000 m and ρ = 1.03 × 10 3 kg m-3. This indicates that when the pressure on (a) From Eq. (1.6), absolute pressure the cylinder was increased, it was distributed uniformly throughout. We can say whenever P = Pa + ρgh external pressure is applied on any part of a = 1.01 × 105 Pa fluid contained in a vessel, it is transmitted + 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m undiminished and equally in all directions. = 104.01 × 105 Pa This is the Pascal’s law for transmission of ≈ 104 atm fluid pressure and has many applications in (b) Gauge pressure is P − Pa = ρgh = Pg daily life. Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m A number of devices such as hydraulic lift = 103 × 105 Pa and hydraulic brakes are based on the Pascal’s ≈ 103 atm law. In these devices fluids are used for (c) The pressure outside the submarine is transmitting pressure. In a hydraulic lift as P = Pa + ρgh and the pressure inside it is shown in Fig. 1.6 two pistons are separated Pa. Hence, the net pressure acting on the by the space filled with a liquid. A piston of small window is gauge pressure, P g = ρgh. Since cross section A1 is used to exert a force F1 the area of the window is A = 0.04 m 2, the F1 force acting on it is directly on the liquid. The pressure P = A is F = Pg A = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N 1 t transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A2 , 1.2.4 Hydraulic Machines which results in an upward force of P × A2. Therefore, the piston is capable of supporting a Let us now consider what happens when we large force (large weight of, say a car, or a truck, change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a F1A 2 placed on the platform) F2 = PA2 = A. By piston and three vertical tubes at different 1 points. The pressure in the horizontal cylinder changing the force at A1, the platform can be Archemedes’ Principle Fluid appears to provide partial support to the objects placed in it. When a body is wholly or partially immersed in a fluid at rest, the fluid exerts pressur e on the surface of the body in contact with the fluid. The pressur e is greater on lower surfaces of the body than on the upper sur faces as pressure in a fluid increases with depth. The r esultant of all the forces is an upwar d 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 for ce on its top. The fluid exerts a resultant upwar d force or buoyant force on the body equal to (P2-P1 ) A. We have seen in equation 1.4 that ( P2-P1 )A = ρghA. Now hA is the volume of the solid and ρhA is the weight of an equivaliant volume of the fluid. (P2 -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 Vs and a part Vp of it is submerged in the fluid. Then the upward for ce which is the weight of the displaced fluid is ρf gVp , which must equal the weight of the body; ρs gVs = ρf gVpor ρs /ρf = Vp /Vs The apparent weight of the floating body is zero. This principle can be summarised as; ‘the loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced’. 6 moved up or down. Thus, the applied force has (b) Water is considered to be perfectly incompressible. Volume covered by the A2 been increased by a factor of A and this factor movement of smaller piston inwards is equal to 1 volume moved outwards due to the larger piston. is the mechanical advantage of the device. The L1 A1 = L2 A2 example below clarifies it. 2 A L2 = 1 L1 = ( π 1/2 × 10 –2 m ) × 6 × 10 –2 m 2 A2 ( π 3 /2 × 10 –2 m ) j 0.67 × 10-2 m = 0.67 cm Note, atmospheric pressure is common to both pistons and has been ignored. t Example 1.6 In a car lift compressed air t 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 1.7). If the mass of the car to Fig 1.6 Schematic diagram illustrating the be lifted is 1350 kg, calculate F1. What is principle behind the hydraulic lift, a device the pressure necessary to accomplish this used to lift heavy loads. task? (g = 9.8 ms-2). Answer Since pressure is transmitted t Example 1.5 Two syringes of different cross sections (without needles) filled with undiminished throughout the fluid, water are connected with a tightly fitted rubber tube filled with water. Diameters 2 of the smaller piston and larger piston are A ( π 5 × 10 –2 m ) 1.0 cm and 3.0 cm respectively. (a) Find F1 = 1 F2 = A2 2 (1350 Kg× 9.8 m s ) –2 the force exerted on the larger piston when ( π 15 × 10–2 m ) a force of 10 N is applied to the smaller = 1470 N piston. (b) If the smaller piston is pushed ≈ 1.5 × 103 N in through 6.0 cm, how much does the The air pressure that will produce this larger piston move out? force is Answer (a) Since pressure is transmitted F1 1.5 × 10 3 N P = = = 1.9 × 105 Pa undiminished throughout the fluid, A1 π 5 × 10 –2 2 m ( ) 2 A2 ( π 3 /2 × 10 –2 m) This is almost double the atmospheric 2 × 10N F2 = F1 = pressure. t A1 ( π 1/2 × 10 –2 m ) Hydraulic brakes in automobiles also work on the same principle. When we apply a little = 90 N Archimedes (287 – 212 B.C.) Ar chimedes 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 deter mine 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!” 7 force on the pedal with our foot the master The path taken by a fluid particle under a piston moves inside the master cylinder, and steady flow is a streamline. It is defined as a the pressure caused is transmitted through the curve whose tangent at any point is in the brake oil to act on a piston of larger area. A direction of the fluid velocity at that point. large force acts on the piston and is pushed Consider the path of a particle as shown in down expanding the brake shoes against brake Fig.1.7 (a), the curve describes how a fluid lining. In this way a small force on the pedal particle moves with time. The curve PQ is like a produces a large retarding force on the wheel. permanent map of fluid flow, indicating how the An important advantage of the system is that fluid streams. No two streamlines can cross, the pressure set up by pressing pedal is for if they do, an oncoming fluid particle can go transmitted equally to all cylinders attached to either one way or the other and the flow would the four wheels so that the braking effort is not be steady. Hence, in steady flow, the map equal on all wheels. of flow is stationary in time. How do we draw closely spaced streamlines ? If we intend to show 1.3 STREAMLINE FLOW streamline of every flowing particle, we would end up with a continuum of lines. Consider So far we have studied fluids at rest. The study planes perpendicular to the direction of fluid flow of the fluids in motion is known as fluid e.g., at three points P, R and Q in Fig.1.7 (b). dynamics. When a water-tap is turned on The plane pieces are so chosen that their slowly, the water flow is smooth initially, but boundaries be determined by the same set of loses its smoothness when the speed of the streamlines. This means that number of fluid outflow is increased. In studying the motion of particles crossing the surfaces as indicated at fluids we focus our attention on what is P, R and Q is the same. If area of cross-sections happening to various fluid particles at a at these points are AP ,AR and AQ and speeds of particular point in space at a particular time. fluid particles are v P , vR and v Q, then mass of The flow of the fluid is said to be steady if at fluid ÄmP crossing at AP in a small interval of any given point, the velocity of each passing time Ät is ρP AP vP Ät. Similarly mass of fluid ÄmR fluid particle remains constant in time. This flowing or crossing at AR in a small interval of does not mean that the velocity at different time Ät is ρR ARvR Ät and mass of fluid ÄmQ is points in space is same. The velocity of a ρ QA Qv Q Ät crossing at AQ. The mass of liquid particular particle may change as it moves from flowing out equals the mass flowing in, holds one point to another. That is, at some other point in all cases. Therefore, the particle may have a different velocity, but ρ PAP vP Ät = ρ RARv RÄt = ρ QAQvQÄt (1.9) every other particle which passes the second For flow of incompressible fluids point behaves exactly as the previous particle ρ P = ρR = ρ Q Equation (1.9) reduces to that has just passed that point. Each particle AP vP = AR vR = AQvQ (1.1 ) follows a smooth path, and the paths of the which is called the equation of continuity and particles do not cross each other. it is a statement of conservation of mass in flow of incompressible fluids. In general Av = constant (1.11) Av gives the volume flux or flow rate and remains constant throughout the pipe of flow. Thus, at narrower portions wher e the streamlines are closely spaced, velocity increases and its vice versa. From (Fig 1.7b) it is clear that AR > AQ or vR < vQ, the fluid is accelerated while passing from R to Q. This is associated with a change in pressure in fluid flow in horizontal pipes. Steady flow is achieved at low flow speeds. Fig. 1.7 The meaning of streamlines. (a) A typical Beyond a limiting value, called critical speed, trajectory of a fluid particle. this flow loses steadiness and becomes (b) A region of streamline flow. turbulent. One sees this when a fast flowing 8 254 PHYSICS stream encounters rocks, small foamy change). The Swiss Physicist Daniel Bernoulli whirlpool-like regions called ‘white water rapids developed this relationship in 1738. are formed. Consider the flow at two regions 1 (i.e. BC) Figure 1.8 displays streamlines for some and 2 (i.e. DE). Consider the fluid initially lying typical flows. For example, Fig. 1.8(a) describes between B and D. In an infinitesimal time a laminar flow where the velocities at different interval ∆t, this fluid would have moved. points in the fluid may have diff e rent Suppose v 1 is the speed at B and v2 at D, then magnitudes but their directions are parallel. fluid initially at B has moved a distance v1 ∆t to Figure 1.8 (b) gives a sketch of turbulent flow. C (v1 ∆t is small enough to assume constant cross-section along BC). In the same interval ∆t the fluid initially at D moves to E, a distance equal to v2 ∆t. Pressures P1 and P2 act as shown on the plane faces of areas A1 and A2 binding the two regions. The work done on the fluid at left end (BC) is W 1 = P1 A1(v 1∆t) = P1 ∆V. Since the same volume ∆V passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is Fig. 1.8 (a) Some str eamlines for fluid flow. W 2 = P2 A2 (v2 ∆t) = P2∆V or, the work done on the (b) A jet of air striking a flat plate placed perpendicular to it. This is an example of fluid is –P2∆V. So the total work done on the turbulent flow. fluid is W 1 – W 2 = (P1− P2 ) ∆V 1.4 BERNOULLI’S PRINCIPLE Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing Fluid flow is a complex phenomenon. But we the gravitational potential energy. If the density can obtain some useful properties for steady or of the fluid is ρ and ∆m = ρA1 v1∆t = ρ∆V is the streamline flows using the conservation of mass passing through the pipe in time ∆t, then energy. change in gravitational potential energy is Consider a fluid moving in a pipe of varying ∆U = ρg∆V (h 2 − h1 ) cross-sectional area. Let the pipe be at varying The change in its kinetic energy is heights as shown in Fig. 1.9. We now suppose that an incompressible fluid is flowing through 1  ∆K =   ρ ∆V (v2 2 − v 12 ) the pipe in a steady flow. Its velocity must 2  change as a consequence of equation of We can employ the work – energy theorem continuity. A force is required to produce this (Chapter 6) to this volume of the fluid and this acceleration, which is caused by the fluid yields surrounding it, the pressure must be different 1  in different regions. Bernoulli’s equation is a (P1 − P2) ∆V =  2  ρ ∆V (v22 − v 12 ) + ρg∆V (h2 − h1)   general expression that relates the pressure We now divide each term by ∆V to obtain difference between two points in a pipe to both velocity changes (kinetic energy change) and 1  (P1 − P2) =   ρ (v 22 − v1 2) + ρg (h2 − h1 ) elevation (height) changes (potential energy 2  Daniel Bernoulli (1700-1782) Daniel Ber noulli was a Swiss scientist and mathematician who along with Leonard Euler had the distinction of winning the Fr ench Academy prize for mathematics ten times. He also studied medicine and served as a professor of anatomy and botany for a while at Basle, Switzerland. His most well known work was in hydrodynamics, a subject he developed from a single principle: the conservation of ener gy. His work included calculus, probability, the theory of vibrating strings, and applied mathematics. He has been called the founder of mathematical physics. 9  We can rearrange the above terms to obtain restriction on application of Bernoulli theorem is that the fluids must be incompressible, as 1  1  P1 +   ρv1 2 + ρgh1 = P2 +   ρv2 2 + ρgh2 the elastic energy of the fluid is also not taken   2 2  into consideration. In practice, it has a large (1.12) number of useful applications and can help This is Bernoulli’s equation. Since 1 and 2 explain a wide variety of phenomena for low refer to any two locations along the pipeline, viscosity incompressible fluids. Bernoulli’s we may write the expression in general as equation also does not hold for non-steady or 1  turbulent flows, because in that situation P +   ρv 2 + ρgh = constant (1.13) velocity and pressure are constantly fluctuating 2  in time. When a fluid is at rest i.e. its velocity is zero everywhere, Bernoulli’s equation becomes P1 + ρgh1 = P2 + ρgh2 (P1 − P2) = ρg (h 2 − h1 ) which is same as Eq. (1.6). 1.4.1 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 ρ with a small hole in its side at a height y1 from the bottom (see Fig. 1.10). Fig. 1.9 The flow of an ideal fluid in a pipe of The air above the liquid, whose surface is at varying cross section. The fluid in a height y2, is at pressure P. From the equation section of length v1∆ t moves to the section of continuity [Eq. (1.10)] we have of length v2∆ t in time ∆t. v 1 A1 = v2 A2 A1 In words, the Bernoulli’s relation may be v2 = v stated as follows: As we move along a streamline A2 1 the sum of the pressure (P), the kinetic energy ρv2 per unit volume and the potential energy 2 per unit volume (ρgh) 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 Fig. 1.10 Torricelli’s law. The speed of efflux, v1, later section. The lost kinetic energy of the fluid from the side of the container is given by gets converted into heat energy. Thus, the application of Bernoulli’s equation. Bernoulli’s equation ideally applies to fluids with If the container is open at the top to the zero viscosity or non-viscous fluids. Another atmosphere then v1 = 2 g h. 10 256 PHYSICS If the cross sectional area of the tank A2 is A much larger than that of the hole (A2 >>A1), then a 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 = Pa , the atmospheric 2 pressure, we have from Eq. (1.12) 1 h 1 Pa + ρ v12 + ρ g y1 = P + ρ g y 2 2 Taking y2 – y1 = h we have 2 ( P − Pa ) v1 = 2 g h + (1.14) ρ Fig. 1.11 A schematic diagram of Venturi-meter. When P >>Pa and 2 g h may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket 1  A 2  propulsion. On the other hand if the tank is P1 – P2 = ρ mgh = ρv1  a  – 1 2 2    open to the atmosphere, then P = Pa and So that the speed of fluid at wide neck is v1 = 2g h (1.15) –½ This is the speed of a freely falling body.  2 ρm gh    A 2      –1  Equation (1.15) is known as Torricelli’s law. v1 =  ρ   a   (1.17) 1.4.2 Venturi-meter The principle behind this meter has many applications. The carburetor of automobile has The Venturi-meter is a device to measure the a Venturi channel (nozzle) through which air flow speed of incompressible fluid. It consists flows with a large speed. The pressure is then of a tube with a broad diameter and a small lowered at the narrow neck and the petrol constriction at the middle as shown in (gasoline) is sucked up in the chamber to provide Fig. (1.11). A manometer in the form of a the correct mixture of air to fuel necessary for U-tube is also attached to it, with one arm at combustion. Filter pumps or aspirators, Bunsen the broad neck point of the tube and the other burner, atomisers and sprayers [See Fig. 1.12] at constriction as shown in Fig. (1.11). The used for perfumes or to spray insecticides work manometer contains a liquid of density ρm. The on the same principle. speed v1 of the liquid flowing through the tube at the broad neck area A is to be measured from equation of continuity Eq. (1.10) the A speed at the constriction becomes v 2 = v1. a Then using Bernoulli’s equation, we get 1 1 P1 + ρv 12 = P2 + ρv 12 (A/a)2 2 2 So that 2 1 A P1 - P2 = ρv 1 2 [   – 1 ] (1.16) 2 a  This pressure difference causes the fluid in the U tube connected at the narrow neck to rise Fig. 1.12 The spray gun. Piston forces air at high in comparison to the other arm. The difference speeds causing a lowering of pressure in height h measure the pressure difference. at the neck of the container. 11 MECHANICAL PROPER TIES OF FLUIDS 257 t 1.4.4 Dynamic Lift Example 1.7 Blood velocity: The flow of blood in a large artery of an anesthetised Dynamic lift is the force that acts on a body, dog is diverted through a Venturi meter. such as airplane wing, a hydrofoil or a spinning The wider part of the meter has a cross- ball, by virtue of its motion through a fluid. In sectional area equal to that of the artery. many games such as cricket, tennis, baseball, A = 8 mm2. The narrower part has an area or golf, we notice that a spinning ball deviates a = 4 mm2. The pressure drop in the artery from its parabolic trajectory as it moves through is 24 Pa. What is the speed of the blood in air. This deviation can be partly explained on the artery? the basis of Bernoulli’s principle. (i) Ball moving without spin: Fig. 1.13(a) Answer We take the density of blood from Table shows the streamlines around a non- 1.1 to be 1.06 × 103 kg m -3. The ratio of the spinning ball moving relative to a fluid. A  From the symmetry of streamlines it is clear areas is   = 2. Using Eq. (1.17) we obtain a  that the velocity of fluid (air) above and below the ball at corresponding points is the same 2 × 24 Pa resulting in zero pressure difference. The v1 = = 0.125 m s –1 ( 1060 kg m –3 × 22 – 1 ) t air therefore, exerts no upward or downward force on the ball. (ii) Ball moving with spin: A ball which is 1.4.3 Blood Flow and Heart Attack spinning drags air along with it. If the surface is rough more air will be dragged. Bernoulli’s principle helps in explaining blood Fig 1.13(b) shows the streamlines of air flow in artery. The artery may get constricted due to the accumulation of plaque on its inner for a ball which is moving and spinning at walls. In order to drive the blood through this the same time. The ball is moving forward constriction a greater demand is placed on the and relative to it the air is moving activity of the heart. The speed of the flow of backwards. Therefore, the velocity of air the blood in this region is raised which lowers above the ball relative to it is larger and the pressure inside and the artery may collapse below it is smaller. The stream lines thus due to the external pressure. The heart exerts get crowded above and rarified below. further pressure to open this artery and forces This difference in the velocities of air results the blood through. As the blood rushes through in the pressure difference between the lower the opening, the internal pressure once again and upper faces and there is a net upward force drops due to same reasons leading to a repeat on the ball. This dynamic lift due to spining is collapse. This may result in heart attack. called Magnus effect. (a) (b) (c) Fig 1.13 (a) Fluid streaming past a static sphere. (b) Streamlines for a fluid around a sphere spinning clockwise. (c) Air flowing past an aerofoil. 12 258 PHYSICS Aerofoil or lift on aircraft wing: Figure 1.13 1.5 VISCOSITY (c) shows an aerofoil, which is a solid piece Most of the fluids are not ideal ones and offer some shaped to provide an upward dynamic lift when resistance to motion. This resistance to fluid motion it moves horizontally through air. The cross- is like an internal friction analogous to friction when section of the wings of an aeroplane looks a solid moves on a surface. It is called viscosity. somewhat like the aerofoil shown in Fig. 1.13 (c) This force exists when there is relative motion with streamlines around it. When the aerofoil between layers of the liquid. Suppose we consider moves against the wind, the orientation of the a fluid like oil enclosed between two glass plates wing relative to flow direction causes the as shown in Fig. 1.14 (a). The bottom plate is fixed streamlines to crowd together above the wing while the top plate is moved with a constant more than those below it. The flow speed on velocity v relative to the fixed plate. If oil is top is higher than that below it. There is an replaced by honey, a greater force is required upward force resulting in a dynamic lift of the to move the plate with the same velocity. Hence wings and this balances the weight of the plane. we say that honey is more viscous than oil. The The following example illustrates this. fluid in contact with a surface has the same t velocity as that of the surfaces. Hence, the layer Example 1.8 A fully loaded Boeing of the liquid in contact with top surface moves aircraft has a mass of 3.3 × 105 kg. Its total with a velocity v and the layer of the liquid in wing area is 500 m2. It is in level flight contact with the fixed surface is stationary. The with a speed of 960 km/h. (a) Estimate velocities of layers increase uniformly from the pressure difference between the lower bottom (zero velocity) to the top layer (velocity and upper surfaces of the wings (b) v). For any layer of liquid, its upper layer pulls Estimate the fractional increase in the it forward while lower layer pulls it backward. speed of the air on the upper surface of This results in force between the layers. This the wing relative to the lower surface. [The type of flow is known as laminar. The layers of density of air is ρ = 1.2 kg m-3] liquid slide over one another as the pages of a book do when it is placed flat on a table and a Answer (a) The weight of the Boeing aircraft is horizontal force is applied to the top cover. When balanced by the upward force due to the a fluid is flowing in a pipe or a tube, then pressure difference velocity of the liquid layer along the axis of the ∆P × A = 3.3 × 105 kg × 9.8 tube is maximum and decreases gradually as ∆P = (3.3 × 105 kg × 9.8 m s–2) / 500 m2 we move towards the walls where it becomes = 6.5 ×10 3 Nm -2 zero, Fig. 1.14 (b). The velocity on a cylindrical (b) We ignore the small height difference surface in a tube is constant. between the top and bottom sides in Eq. (1.12). On account of this motion, a portion of liquid, The pressure difference between them is which at some instant has the shape ABCD, then take the shape of AEFD after short interval of time (∆t). During this time interval the liquid has undergone a shear strain of where v 2 is the speed of air over the upper ∆x/l. Since, the strain in a flowing fluid surface and v1 is the speed under the bottom increases with time continuously. Unlike a solid, surface. here the stress is found experimentally to 2 ∆P depend on ‘rate of change of strain’ or ‘strain ( v2 – v1 ) = rate’ i.e. ∆x/(l ∆t) or v/l instead of strain itself. ρ (v 2 + v1 ) The coefficient of viscosity (pronounced ‘eta’) for Taking the average speed a fluid is defined as the ratio of shearing stress vav = (v 2 + v 1 )/2 = 960 km/h = 267 m s-1, to the strain rate. we have ∆P F /A F l ( v2 – v1 ) / vav = 2 ≈ 0.08 η= = (1.18) ρv av v /l v A The speed above the wing needs to be only 8 The SI unit of viscosity is poiseiulle (Pl). Its % higher than that below. t other units are N s m-2 or Pa s. The dimensions 13  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. (a) Fig. 1.15 Measur ement of the coef ficient of viscosity of a liquid. Answer 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 kg × 9.8 m s –2 = 9.8 × 10 -2 N (b) 9.8 × 1 0–2 Shear stress on the fluid = F/A = 0.10 Fig 1.14 (a) A layer of liquid sandwiched between two parallel glass plates in which the v 0.085 Strain rate = = lower plate is fixed and the upper one l 0.030 is moving to the right with velocity v (b) velocity distribution for viscous flow stress η= in a pipe. strain rate of viscosity are [ML-1T-1]. Generally thin liquids (9.8 × 10 N )(0.30 × 10 m ) –2 –3 like water, alcohol etc. are less viscous than = thick liquids like coal tar, blood, glycerin etc. (0.085 m s )( 0.10 m ) –1 2 The coefficients of viscosity for some common = 3.45 ×10 -3 Pa s t fluids are listed in Table 1.2. We point out two facts about blood and water that you may find Table1.2 The viscosities of some fluids interesting. As Table 1.2 indicates, blood is ‘thicker’ (more viscous) than water. Further the Fluid T(oC) Viscosity (mPl) relative viscosity (η/ηwater ) of blood remains Water 20 1.0 constant between 0oC and 37oC. 100 0.3 The viscosity of liquids decreases with Blood 37 2.7 temperature while it increases in the case of Machine Oil 16 113 gases. 38 34 Glycerine 20 830 t 2 Honey 200 Example 1.9 A metal block of area 0.10 m Air 0 0.017 is connected to a 0.010 kg mass via a string 40 0.019 that passes over an ideal pulley (considered massless and frictionless), as in Fig. 1.15. 1.5.1 Stokes’ Law A liquid with a film thickness of 0.30 mm When a body falls through a fluid it drags the is placed between the block and the table. layer of the fluid in contact with it. A relative 14 motion between the different layers of the fluid 1.6 REYNOLDS NUMBER is set and as a result the body experiences a When the rate of flow of a fluid is large, the flow retarding force. Falling of a raindrop and no longer remain laminar, but becomes swinging of a pendulum bob are some common turbulent. In a turbulent flow the velocity of examples of such motion. It is seen that the the fluids at any point in space varies rapidly viscous force is proportional to the velocity of and randomly with time. Some circular motions the object and is opposite to the direction of motion. The other quantities on which the force called eddies are also generated. An obstacle F depends are viscosity η of the fluid and radius placed in the path of a fast moving fluid causes a of the sphere. Sir George G. Stokes (1819- turbulence [Fig. 1.8 (b)]. The smoke rising from 1903), an English scientist enunciated clearly a burning stack of wood, oceanic currents are the viscous drag force F as turbulent. Twinkling of stars is the result of atmospheric turbulence. The wakes in the water F = 6 π η av (1.19) and in the air left by cars, aeroplanes and boats This is known as Stokes’ law.We shall not are also turbulent. derive Stokes’ law. Osborne Reynolds (1842-1912) observed that This law is an interesting example of retarding turbulent flow is less likely for viscous fluid force which is proportional to velocity. We can flowing at low rates. He defined a dimensionless study its consequences on an object falling number, whose value gives one an approximate through a viscous medium. We consider a idea whether the flow would be turbulent. This raindrop in air. It accelerates initially due to number is called the Reynolds Re. gravity. As the velocity increases, the retarding Re = ρvd/η (1.21) force also increases. Finally when viscous force where ρ is the density of the fluid flowing with plus buoyant force becomes equal to force due a speed v, d stands for the dimension of the to gravity, the net force becomes zero and so pipe, and η is the viscosity of the fluid. Re is a does the acceleration. The sphere (raindrop) dimensionless number and therefore, it remains then descends with a constant velocity. Thus same in any system of units. It is found that in equilibrium, this terminal velocity vt is given flow is streamline or laminar for Re less than by 1000. The flow is turbulent for Re > 2000. The 6π ηavt = (4π/3) a3 (ρ-σ)g flow becomes unsteady for Re between 1000 and where ρ and σ are mass densities of sphere and 2000. The critical value of Re (known as critical the fluid respectively. We obtain Reynolds number), at which turbulence sets, is vt = 2a2 (ρ-σ)g / (9η) (1.20) found to be the same for the geometrically So the terminal velocity v t depends on the similar flows. For example when oil and water square of the radius of the sphere and inversely with their different densities and viscosities, flow on the viscosity of the medium. in pipes of same shapes and sizes, turbulence You may like to refer back to Example 6.2 in sets in at almost the same value of Re. Using this context. this fact a small scale laboratory model can be Example 1.10 The terminal velocity of a t set up to study the character of fluid flow. They copper ball of radius 2.0 mm falling are useful in designing of ships, submarines, through a tank of oil at 20o C is 6.5 cm s-1. racing cars and aeroplanes. Compute the viscosity of the oil at 20oC. Re can also be written as Density of oil is 1.5 ×103 kg m-3, density of Re = ρv 2 / (ηv/d) = ρAv 2 / (ηAv/d) (1.22) copper is 8.9 × 103 kg m -3. = inertial force/force of viscosity. Thus Re represents the ratio of inertial force Answer We have vt = 6.5 × 10-2 ms-1, a = 2 × 10-3 m, (force due to inertia i.e. mass of moving fluid or g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3, due to inertia of obstacle in its path) to viscous σ =1.5 ×103 kg m-3. From Eq. (1.20) force. 2 Turbulence dissipates kinetic energy usually ( –3 ) 2 2 × 10 m × 9.8 m s η= × 2 –2 × 7.4 × 103 kg m –3 in the form of heat. Racing cars and planes are 9 6.5 × 10–2 m s –1 engineered to precision in order to minimise = 9.9 × 10-1 kg m–1 s–1 t turbulence. The design of such vehicles involves 15 MECHANICAL PROPER TIES OF FLUIDS 261 experimentation and trial and error. On the liquids have no definite shape but have a other hand turbulence (like friction) is definite volume, they acquire a free surface when sometimes desirable. Turbulence promotes poured in a container. These surfaces possess mixing and increases the rates of transfer of some additional energy. This phenomenon is mass, momentum and energy. The blades of a known as surface tension and it is concerned kitchen mixer induce turbulent flow and provide with only liquid as gases do not have free thick milk shakes as well as beat eggs into a surfaces. Let us now understand this uniform texture. phenomena. t Example 1.11 The flow rate of water from 1.7.1 Surface Energy a tap of diameter 1.25 cm is 0.48 L/min. The coefficient of viscosity of water is A liquid stays together because of attraction 10-3 Pa s. After sometime the flow rate is between molecules. Consider a molecule well increased to 3 L/min. Characterise the flow inside a liquid. The intermolecular distances are for both the flow rates. such that it is attracted to all the surrounding molecules [Fig. 1.16(a)]. This attraction results Answer Let the speed of the flow be v and the in a negative potential energy for the molecule, diameter of the tap be d = 1.25 cm. The which depends on the number and distribution volume of the water flowing out per second is of molecules around the chosen one. But the Q = v × π d2 / 4 average potential energy of all the molecules is v = 4 Q / d2 π the same. This is supported by the fact that to We then estimate the Reynolds number to be take a collection of such molecules (the liquid) Re = 4 ρ Q / π d η = 4 ×103 kg m–3 × Q/(3.14 ×1.25 ×10-2 m ×10-3 Pa s) and to disperse them far away from each other = 1.019 × 108 m–3 s Q in order to evaporate or vaporise, the heat of Since initially evaporation required is quite large. For water it Q = 0.48 L / min = 8 cm3 / s = 8 × 10-6 m3 s-1, is of the order of 40 kJ/mol. we obtain, Let us consider a molecule near the surface Re = 815 Fig. 1.16(b). Only lower half side of it is Since this is below 1000, the flow is steady. surrounded by liquid molecules. There is some After some time when negative potential energy due to these, but Q = 3 L / min = 50 cm3 / s = 5 × 10-5 m 3 s-1, obviously it is less than that of a molecule in we obtain, bulk, i.e., the one fully inside. Approximately it Re = 5095 is half of the latter. Thus, molecules on a liquid The flow will be turbulent. You may carry out surface have some extra energy in comparison an experiment in your washbasin to determine to molecules in the interior. A liquid thus tends the transition from laminar to turbulent to have the least surface area which external flow. t conditions permit. Increasing surface area requires energy. Most surface phenomenon can 1.7 SURFACE TENSION be understood in terms of this fact. What is the You must have noticed that, oil and water do energy required for having a molecule at the not mix; water wets you and me but not ducks; surface? As mentioned above, roughly it is half mercury does not wet glass but water sticks to the energy required to remove it entirely from it, oil rises up a cotton wick, inspite of gravity, the liquid i.e., half the heat of evaporation. Sap and water rise up to the top of the leaves of Finally, what is a surface? Since a liquid consists the tree, hairs of a paint brush do not cling of molecules moving about, there cannot be a together when dry and even when dipped in perfectly sharp surface. The density of the liquid water but form a fine tip when taken out of it. molecules drops rapidly to zero around z = 0 as we All these and many more such experiences are move along the direction indicated Fig 1.16 (c) in related with the free surfaces of liquids. As a distance of the order of a few molecular sizes. 16 Fig. 1.16 Schematic picture of molecules in a liquid, at the surface and balance of for ces. (a) Molecule inside a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (A) and repulsive (R) for ces. 1.7.2 Surface Energy and Surface Tension unit area of the liquid interface and is also equal to the force per unit length exerted by the fluid As we have discussed that an extra energy is on the movable bar. associated with surface of liquids, the creation So far we have talked about the surface of of more surface (spreading of surface) keeping one liquid. More generally, we need to consider other things like volume fixed requires fluid surface in contact with other fluids or solid additional energy. To appreciate this, consider surfaces. The surface energy in that case a horizontal liquid film ending in bar free to depends on the materials on both sides of the slide over parallel guides Fig (1.17). 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: Fig. 1.17 Stretching a film. (a) A film in equilibrium; (i) Surface tension is a force per unit length (b) The film stretched an extra distance. (or surface energy per unit area) acting in the plane of the interface between the plane Suppose that we move the bar by a small of the liquid and any other substance; it also distance d as shown. Since the area of the is the extra energy that the molecules at surface increases, the system now has more the interface have as compared to molecules energy, this means that some work has been in the interior. done against an internal force. Let this internal (ii) At any point on the interface besides the force be F, the work done by the applied force is boundary, we can draw a line and imagine F.d = Fd. From conservation of energy, this is equal and opposite surface tension forces S stored as additional energy in the film. If the per unit length of the line acting surface energy of the film is S per unit area, perpendicular to the line, in the plane of the extra area is 2dl. A film has two sides and the interface. The line is in equilibrium. To the liquid in between, so there are two surfaces be more specific, imagine a line of atoms or and the extra energy is molecules at the surface. The atoms to the S (2dl) = Fd (1.23) left pull the line towards them; those to the Or, S=Fd/2dl = F/2l (1.24) right pull it towards them! This line of atoms This quantity S is the magnitude of surface is in equilibrium under tension. If the line tension. It is equal to the surface energy per really marks the end of the interface, as in 17  Figure 1.16 (a) and (b) there is only the Sla = (W/2l) = (mg/2l) (1.25) force S per unit length acting inwards. where m is the extra mass and l is the length of Table 1.3 gives the surface tension of various the plate edge. The subscript (la) emphasises liquids. The value of surface tension depends the fact that the liquid-air interface tension is on temperature. Like viscosity, the surface involved. tension of a liquid usually falls with temperature. 1.7.3 Angle of Contact The surface of liquid near the plane of contact, Table 1.3 Sur face tension of some liquids at the temperatures indicated with the with another medium is in general curved. The heats of the vaporisation angle between tangent to the liquid surface at the point of contact and solid surface inside the Liquid Temp (oC) Surface Heat of liquid is termed as angle of contact. It is denoted Tension vaporisation by θ. It is different at interfaces of different pairs (N/m) (kJ/mol) of liquids and solids. The value of θ determines whether a liquid will spread on the surface of a Helium –270 0.000239 0.115 solid or it will form droplets on it. For example, Oxygen –183 0.0132 7.1 water forms droplets on lotus leaf as shown in Ethanol 20 0.0227 40.6 Water 20 0.0727 44.16 Fig. 1.19 (a) while spreads over a clean plastic Mercury 20 0.4355 63.2 plate as shown in Fig. 1.19(b). 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 be directly measured experimentaly as schematically shown in Fig. 1.18. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of (a) 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. (b) Fig. 1.19 Different shapes of water drops with interfacial tensions (a) on a lotus leaf (b) on a clean plastic plate. We consider the three interfacial tensions at Fig. 1.18 Measuring Surface Tension. all the three interfaces, liquid-air, solid-air and solid-liquid denoted by S la, Ssa & Ssl respectively Suppose the additional weight required is W. as given in Fig. 1.19 (a) and (b). At the line of Then from Eq. 1.24 and the discussion given contact, the surface forces between the three media there, the surface tension of the liquid-air must be in equilibrium. From the Fig. 1.19(b) the interface is following relation is easily derived. 18 264 PHYSICS Sla cos θ + Ssl = Ssa (1.26) If the drop is in equilibrium this energy cost is The angle of contact is an obtuse angle if balanced by the energy gain due to Ssl > Sla as in the case of water-leaf interface expansion under the pressure difference (Pi – Po) while it is an acute angle if Ssl < Sla as in the between the inside of the bubble and the case of water-plastic interface. When θ is an outside. The work done is obtuse angle then molecules of liquids are W = (Pi – Po) 4πr2 ∆r (1.28) attracted strongly to themselves and weakly to so that those of solid, it costs a lot of energy to create a (Pi – Po ) = (2 Sla / r) (1.29) liquid-solid surface, and liquid then does not In general, for a liquid-gas interface, the wet the solid. This is what happens with water convex side has a higher pressure than the on a waxy or oily surface, and with mercury on concave side. For example, an air bubble in a any surface. On the other hand, if the molecules liquid, would have higher pressure inside it. of the liquid are strongly attracted to those of See Fig 1.20 (b). the solid, this will reduce S sl and therefore, cos θ may increase or θ may decrease. In this case θ is an acute angle. This is what happens for water on glass or on plastic and for kerosene oil on virtually anything (it just spreads). Soaps, detergents and dying substances are wetting agents. When they are added the angle of contact becomes small so that these may penetrate well and become effective. Water proofing agents on the other hand are added to Fig. 1.20 Dr op, cavity and bubble of radius r. create a large angle of contact between the water and fibres. A bubble Fig 1.20 (c) differs from a drop and a cavity; in this it has two interfaces. 1.7.4 Drops and Bubbles Applying the above argument we have for a One consequence of surface tension is that free bubble liquid drops and bubbles are spherical if effects (Pi – Po) = (4 S la/ r) (1.30) of gravity can be neglected. You must have seen This is probably why you have to blow hard, this especially clearly in small drops just formed but not too hard, to form a soap bubble. A little in a high-speed spray or jet, and in soap bubbles extra air pressure is needed inside! blown by most of us in childhood. Why are drops and bubbles spherical? What keeps soap 1.7.5 Capillary Rise bubbles stable? One consequence of the pressure difference As we have been saying repeatedly, a liquid- across a curved liquid-air interface is the well- air interface has energy, so for a given volume known effect that water rises up in a narrow the surface with minimum energy is the one tube in spite of gravity. The word capilla means 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. 1.20(a) is more than the pressure outside. Suppose a spherical drop of radius r is in equilibrium. If its radius increase by ∆r. The Fig. 1.21 Capillary rise, (a) Schematic picture of a extra surface energy is narrow tube immersed water. [4π(r + ∆r) 2 - 4πr2 ] S la = 8πr ∆r Sla (1.27) (b) Enlarged picture near interface. 19 MECHANICAL PROPER TIES OF FLUIDS 265 hair in Latin; if the tube were hair thin, the rise hairpin shaped, with one end attracted to water would be very large. To see this, consider a and the other to molecules of grease, oil or wax, vertical capillary tube of circular cross section thus tending to form water-oil interfaces. The result (radius a) inserted into an open vessel of water is shown in Fig. 1.22 as a sequence of figures. (Fig. 1.21). The contact angle between water In our language, we would say that addition and glass is acute. Thus the surface of water in of detergents, whose molecules attract at one the capillary is concave. This means that end and say, oil on the other, reduces drastically there is a pressure difference between the the surface tension S (water-oil). It may even two sides of the top surface. This is given by become energetically favourable to form such (Pi – Po) =(2S/r) = 2S/(a sec θ ) interfaces, i.e., globs of dirt surrounded by = (2S/a) cos θ (1.31) detergents and then by water. This kind of Thus the pressure of the water inside the process using surface active detergents or tube, just at the meniscus (air-water interface) surfactants is important not only for cleaning, is less than the atmospheric pressure. Consider but also in recovering oil, mineral ores etc. the two points A and B in Fig. 1.21(a). They must be at the same pressure, namely P0 + h ρ g = P i = PA (1.32) where ρ is the density of water and h is called the capillary rise [Fig. 1.21(a)]. Using Eq. (1.31) and (1.32) we have h ρ g = (Pi – P 0) = (2S cos θ )/a (1.33). The discussion here, and the Eqs. (1.28) and (1.29) make it clear that the capillary rise is due to surface tension. It is larger, for a smaller a. Typically it is of the order of a few cm for fine capillaries. For example, if a = 0.05 cm, using the value of surface tension for water (Table 1.3), we find that h = 2S/(ρ g a) = ( 2 × 0.073 Nm – 1 ) (10 kg m 3 –3 ) (9.8 m s ) (5 × 10 –2 –4 m ) –2 = 2.98 × 10 m = 2.98 cm Notice that if the liquid meniscus is convex, as for mercury, i.e., if cos θ is negative then from Eq. (1.32) for example, it is clear that the liquid will be lower in the capillary ! 1.7.6 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; i.e., there is very little area of contact between them. If water could wet grease, the flow of water could carry some grease away. Something of this sort is achieved through Fig. 1.22 Deter gent action in terms of what detergents. The molecules of detergents are detergent molecules do. 20 266 PHYSICS t are two liquid surfaces, so the formula for excess Example 1.12 The lower end of a capillary pressure in that case is 4S/r.) The radius of the tube of diameter 2.00 mm is dipped 8.00 bubble is r. Now the pressure outside the bubble cm below the surface of water in a beaker. P o equals atmospheric pressure plus the What is the pressure required in the tube pressure due to 8.00 cm of water column. That is in order to blow a hemispherical bubble at Po = (1.01 × 105 Pa + 0.08 m × 1000 kg m–3 its end in water? The surface tension of × 9.80 m s–2) water at temperature of the experiments is 5 = 1.01784 × 10 Pa 7.30 ×10 -2 N m -1. Therefore, the pressure inside the bubble is 1 atmospheric pressure = 1.01 × 105 Pa, Pi = Po + 2S/r density of water = 1000 kg/m3 , g = 9.80 m s-2. = 1.01784 × 105 Pa+ (2 × 7.3 × 10-2 Pa m/10-3 m) Also calculate the excess pressure. = (1.01784 + 0.00146) × 105 Pa = 1.02 × 105 Pa Answer The excess pressure in a bubble of gas where the radius of the bubble is taken to be in a liquid is given by 2S/r, where S is the equal to the radius of the capillary tube, since surface tension of the liquid-gas interface. You the bubble is hemispherical ! (The answer has should note there is only one liquid surface in been rounded off to three significant figures.) this case. (For a bubble of liquid in a gas, there The excess pressure in the bubble is 146 Pa.t 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 pr essure Pav is defined as the ratio of the force to area F Pav = A 4. The unit of the pressur e is the pascal (Pa). It is the same as N m-2. Other common units of pressur e are 1 atm = 1.01×105 Pa 1 bar = 105 Pa 1 torr = 133 Pa = 0.133 kPa 1 mm of Hg = 1 torr = 133 Pa 5. Pascal’s law states that: Pressur e in a fluid at rest is same at all points which ar e 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 pressur e in a fluid varies with depth h according to the expr ession P = Pa + ρgh where ρ 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 cr ossection is the same in the steady flow. v A = constant ( v is the velocity and A is the ar ea of crossection) The equation is due to mass conservation in incompre