12-ELASTICITY AND FLUID MECHANICS-01-THEORY

ELASTICITY AND FLUID MECHANICS So far, we have dealt with all the solids that have been modeled as rigid bodies, that is objects do not change their shape. Real objects, however, deform to some extent when an external force is applied to them. In this chapter we formulate some systematic ways to describe qualitatively the deformation of solids that are subjected to applied forces. INTERATOMIC AND INTERMOLECULAR FORCE: What holds the atoms in a solid or liquid? This is the question we now consider. Evidently there must exist some interatomic or intermolecular forces that keep them bound: Detailed calculations as well as deductions from experiment show that the interaction between any isolated pairs of atoms or molecules may be represented by a curve that shows how the potential energy varies with separation between them as shown in the figure. This curve describes the interatomic potential. The force between the atoms can be found from the potential energy by using the relation. The resulting interatomic force curve is shown in figure. Force is along the line joining the atoms or molecules, and is shown negative for attraction & positive for repulsion. We see that as the distance decreases, the attractive force first increases and then decreases to zero at a separation where the potential energy is minimum. For smaller distance force is repulsive, because at these distance the negative charge distribution associated with one atom begins to over lap with that associated with the neighboring atom. The above picture of interatomic or intermolecular force is an over simplification of the actual situation. However, it provides a reasonable approximation. CLASSIFICATION OF MATTER: Matter can be classified into three states: solids, liquids and Gases. Solids: A solid is that state of matter in which its atoms and molecules are strongly bound so as to preserve their original shape and volume. Solids are of two types – crystalline & amorphous. (a) Crystalline Solid: A crystalline solid is one which has regular & periodic arrangement of atoms or molecules in the three dimensions. Examples of crystalline solids are diamond, rocksalt, mica sugar etc. (b) Amorphous Solids or Glassy Solids: The word ‘amorphous’ literally means ‘without any form’. There is no ‘order’ in arrangement of atoms in such a solid. An amorphous solid is one which does not have a periodic arrangement of atoms. The best example of an amorphous solid is glass In solids, the intermolecular forces are so strong that there is no change in shape and size easily. Liquids: The intermolecular forces are comparably less than that in solids, so the shape can easily be changed. But volume of a given mass of a liquid is not easy to change. It needs quite an effort to change the density of liquids. Liquids are not able to produce reaction forces to applied forces in arbitrary directions. They can not support sharing force. (stress) Gases: This is the third state of matter which can not support compressive, tensile, or sharing forces. Densities of gases change very rapidly with the increase in temperature. Together Liquids and Gases are Classified as Fluids: The word “fluid” comes from a latin word meaning. ‘to flow”. • On the average, the atoms or molecules in a gas are far apart, typically about ten atomic diameters at room temperature and pressure. They collide much less frequently than those in a liquid. Gases in general are compressible. ELASTICITY: The property of material of a body by virtue of which the body regains its original length, volume and shape after the deforming forces have been removed is called elasticity Stress, Strain and Elastic Modulii: The precise definition of stress depends on the particular situation being considered, but in general it is force per unit area. The strain indicates some fractional change in dimension or volume. The unit of stress is , whereas strain is a dimensionless number. A force applied to an object can change its dimensions and shape. In general, the response of a material to a given type of deforming force is characterized by an elastic modulus, which is defined as Elastic modulus ….(i) Hooke’s Law: Within elastic limit, the extension of an elastic body is directly proportional to the force. Stress is proportional to strain, within the elastic limit. = constant This constant is known as modulus of elasticity or coefficient of elasticity. It depends upon the nature of the materials. STRESS – STRAIN GRAPHS: The stress-strain graph of a ductile metal is shown in figure. Initially, the stress strain graph is linear and it obeys the Hooke’s Law upto the point called the proportional limit. After the proportional limit the - graph is non-linear but it still remains elastic upto the yield point where the slope of the curve is zero. At the yield point the material starts deforming under constant stress and it behaves like a viscous liquid. The yield point is the beginning of the plastic zone. After the yield point, the material starts gaining strength due to excessive deformation and this phenomenon is called strain hardening. The point shows the ultimate strength of the material. It is the maximum stress that the material can sustain without failure. After the point the curve goes down toward the breaking point because the calculation of the stress is based on the original cross-sectional area whereas the cross-sectional areas of the sample actually decreases. The stress-strain relationship for a metal. Young’s Modulus: Young’s modulus is a measure of the resistance of a solid to a change in its length when a force is applied perpendicular to its surface. Consider a rod with an unstressed length and cross-sectional area , as shown in the figure. When it is subjected to equal and opposite forces along it axis and perpendicular to the end faces its length changes by . These forces tend to stretch the rod. The tensile stress on the rod is defined as …(ii) Forces acting in the opposite direction, as shown in figure, would produce a compressive stress. The resulting strain is defined as the dimensionless ratio. …(iii) Young’s modulus Y for the material of the rod is defined as the ratio of tensile stress to tensile strain. So Young’s Modulus …(iv) A force applied normal to the end face of a rod cause a change in length Illustration 1. A solid cylindrical steel column is 4.0 m long and 9.0 cm in diameter. What will be decrease in length when carrying a load of 80000 kg? . Solution: Let us first calculate the cross-sectional area of column Then, from we have Shear Modulus: The shear modulus of a solid measures its resistance to a shearing force, which is a force applied tangentially to a surface, as shown in the figure. (Since the bottom of the solid is assumed to be at rest, there is an equal and opposite force on the lower surface). The top surface is displaced by relative to the bottom surface. The shear stress is defined as Shear stress = …(v) where is the area of the surface. The shear strain is defined as Shear strain …(vi) where y is the separation between the top and the bottom surfaces. The shear modulus G is defined as Shear modulus = …(vii) A solid block deforms under the action of shearing forces. Illustration 2. A box shaped piece of gelatin dessert has a top area of 15 cm2 and a height of 3 cm. When a shearing force of 0.50 N is applied to the upper surface, the upper surface displaces 4 mm relative to the bottom surface. What are the shearing stress, the shearing strain, and the shear modulus for the gelatin? Solution: Shear stress Pa Shear stress = Shear modulus Bulk Modulus: The bulk modulus of a solid or a fluid indicates its resistance to a change in volume. Consider a cube of some material, solid or fluid, as shown in the figure. We assume that all faces experience the same force normal to each face. (One way to accomplish this is to immerse the body in a fluid – as long as the change in pressure over the vertical height of the cube is negligible). The pressure on the cube is defined as the normal force per unit area …(viii) The SI unit of pressure is N/m2 and is given the name pascal (Pa). The change in pressure is called the volume stress and the fractional change in volume is called the volume strain. The bulk modulus of the material is defined as [A cube of some material is subject to equal forces normal to each face. This condition may be achieved by immersing the body in a fluid. Bulk modulus = or …(ix) The negative sign is included to make a positive number since an increase in pressure leads to decrease in volume . The inverse of is called the compressibility, factor …(x) Elastic Properties of Matter State Shear Modulus Bulk Modulus Solid Large Large Liquid Zero Large Gas Zero Small Energy Stored in a Wire: Consider an elastic wire of length . Suppose it is stretched by a length when a force F is applied at one end. If the elastic limit is not exceeded, then the extension is directly proportional to the applied load. Consequently, the force in the wire increases in magnitude from zero to F. So, the average force in the wire while stretching was , Now, work done = average force extension . This is the amount of energy stored in the wire. It is the gain in molecular potential energy of the molecules due to their displacement from their mean positions Again, or Work done A Note on Thermal Stress : Consider a bar which is heated & then prevented from contracting as it cools. A considerable force will be exerted at the ends of the bar, if the bar were free to contract, then decrease in length, … (i) Where is the original length, is the coefficient of linear expansion & is the fall of temperature. Again, or …(ii) Where, F is the force developed in the bar & is the cross-sectional area of the bar. From (i) & (ii), or Again, thermal stress= or Thermal stress = . FLUID STATICS: It refers to the state when there is no relative velocity between fluid elements. In this section we will learn some of the properties of the fluid statics. Density: The density of a substance is defined as the mass per unit volume of a sample of the substance. If a small mass element occupies a volume , the density is given by In general, the density of an object depends on position, so that If the object is homogeneous, its physical parameters do not change with position throughout its volume. Thus, for a homogeneous object of mass and volume , the density is defined as …(xi) The units of density are kg m-3. Specific Gravity: The specific gravity of a substance is the ratio of its density to that of water at 4º C, which is 1000 kg/m3. Specific gravity is a dimensionless quantity numerically equal to the density quoted in g/cm3. For example, the specific gravity of mercury is 13.6, and the specific gravity of water at 100 ºC is 0.998. Illustration 3. Find the density and specific gravity of gasoline if 51 g occupies 75 cm3? Solution: Density = Sp. gr = or Sp. gravity = Illustration 4. The mass of a liter of milk is 1.032 kg. The butterfat that it contains has a density of 865 kg/m3 when pure, and it constitutes 4 percent of the milk by volume. What is the density of the fat-free skimmed mill? Solution: Volume of fat in 1000 cm3 of milk = 4% 1000 cm3 = 40 cm3 Mass of 40 cm3 fat = Density of skimmed milk = = Pressure: The pressure exerted by a fluid is defined as the force per unit area at a point within the fluid. Consider an element of area as shown in figure and an external force is acting normal to the surface. The average pressure in the fluid at the position of the element is given by As , the element reduces to a point, and thus, pressure at a point is defined as …(xii) A normal force acts on a small cylindrical element of cross-section area . When the force is constant over the surface , the above equation reduces to …(xiii) The unit of pressure is Nm-2 and is also called pascal (Pa). The other common pressure units are atmosphere and bar. 1 atm = Pa 1 bar = Pa 1 atm = 1.01325 bar Illustration 5. Atmospheric pressure is about Pa. How large a force does the atmosphere exert on a area on the top of your head? Solution: Because , where is perpendicular to , we have . Assuming that 2 cm2 of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), we have Pressure is Isotropic: Imagine a static fluid and consider a small cubic element of the fluid deep within the fluid as shown in the figure. Since this fluid element is in equilibrium therefore, forces acting on each lateral face of this element must also be equal in magnitude. Because the areas of each face are equal, therefore, the pressure on each face are equal, therefore the pressure on each of the lateral faces must also be the same. In the limit as the cube element reduces to a point, the forces on top and bottom surfaces also become equal. Thus, the pressure exerted by a fluid at a point is the same in all directions – the pressure is isotropic. Since the fluid cannot support a shear stress, the force exerted by a fluid pressure must also be perpendicular to the surface of the container that holds it. A small cubical element is m equilibrium inside a fluid. Variation of Pressure with Depth: Weight of a fluid element of mass The pressure force acting on the lower face of the element is and that on the upper face is . The figure (b) shown the free body diagram of the element. Applying the condition of equilibrium we get, if be the density of the fluid at the position of the element, then and (a) (b) (a) A cylindrical fluid element in a fluid (b) Free body diagram of the element. or In the limit approaches to zero, becomes …(xiv) The above equation indicates that the slope of versus is negative. That is, the pressure decreases with height from the bottom of the fluid. In otherwords, the pressure increases with depth , i.e., …(xv) The Incompressible Fluid Model: For an incompressible fluid, the density of the fluid remains constant throughout its volume. It is a good assumption for liquids. To find pressure at the point in a fluid column as shown in the figure is obtained by integrating equation (xv). or or or …(xvi) A point A is located in a fluid at a height y from the bottom and at a depth h from the free surface. where is the density of the fluid, and is the atmospheric pressure at the free surface of the liquid. Absolute Pressure and Gauge Pressure: Absolute pressure is the total pressure at a point while gauge pressure is relative to the local atmospheric pressure. Gauge pressure may be positive or negative depending upon the fact whether the pressure is more or less than the atmospheric pressure. Pascal Law: According to equation (xvi) pressure at any depth in a fluid may be increased by increasing the pressure at the surface. Pascal recognized a consequence of this fact that we now call Pascal’s Law. A pressure applied to a confined fluid at rest is transmitted equally undiminished to every part of the fluid and the walls of the container. This principle is used in a hydraulic jack or lift, as shown in the figure. The pressure due to a small force applied to a piston of area is transmitted to the larger piston of area . The pressure at the two pistons is the same because they are at the same level. or …(xviii) Consequently, the force on the larger piston is large. Thus, a small force acting on a small area results in a larger force acting on a larger area . A hydraulic jack Brain Teaser: 1. Water is poured to the same level in each of the vessels shown, all having the same base area. If the pressure is the same at the bottom of each vessel, the force experienced by the base of each vessel is the same. Do the three vessels then have different weights when put on a scale. This apparently contradictory result is commonly known as the hydrostatic paradox Illustration 6. Find the absolute pressure and gauge pressure at point and as shown in the figure (1 atm = 105 Pa) Solution: Pa Points Gauge Pressure Absolute Pressure kPa = 108 kPa = (800)(10)(2) + (103) (10) (1.5) = 31 kPa = 131 kPa = (800)(10)(2)+(103)(10)(2)+(13.6 103)(10)(0.5) = 104 kPa = 204 kPa Illustration 7. Find the pressure in the air column at which the piston remains in equilibrium. Assume the piston to be massless and frictionless. Solution: Let be the air pressure above the piston. Applying Pascal’s law at point and . (1.73) = 138 kPa Illustration 8. A weighted piston confines a fluid of density in a closed container, as shown in the figure. The combined weight of piston and weight is N, and the cross-sectional area of the piston is cm2. Find the total pressure at point if the fluid is mercury and = 25 cm ( = 13600 kg/m3). What would an ordinary pressure gauge read at ? Solution: Pascal’s principle tells us about the pressure applied to the fluid by the piston and atmosphere. This added pressure is applied at all points within the fluid. Therefore the total pressure at is composed of three parts: Pressure of atmosphere Pa Pressure due to piston and weight Pa Pressure due to height of fluid = Pa In this case, the pressure of the fluid itself is relatively small. We have Total pressure at Pa = 380 kPa The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at kPa Illustration 9. For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of 800 cm2. The piston on the right at , has cross-sectional area 25 cm2 and negligible weight. If the apparatus is filled with oil ( g/cm3), find the force required to hold the system in equilibrium as shown in figure. Solution: The pressures at point and are equal because they are at the same level in the single connected fluid. Therefore, Pressure at = Pressure at (Pressure due to left piston) = (Pressure due to and right piston) +(8m)(780 kg/m-3)(9.8) After solving, we get, N. The Compressible – Fluid Model: For gases, the constant density assumed in the compressible model is often not adequate. However, an alternative simplifying assumption can be made that the density is proportional to the pressure, i.e., Let be the density of air at the earth’s surface where the pressure is atmospheric , then After eliminating k, we get Substituting the value of in equation (xiv) or On rearranging, we get where is the pressure at a height above the earth’s surface. After integrating, we get or …(xix) Note: That instead of a linear decrease in pressure with increasing height as in the case of an incompressible fluid, in this case pressure decreases exponentially. PRESSURE MEASURING DEVICES: Manometer: A manometer is a tube open at both the ends and bent into the shape of a “ ” and is partially filled with mercury. When one end of the tube is subjected to an unknown pressure , the mercury level drops on that side of the tube and rises on the other so that the difference in mercury level is as shown in the figure. According to Pascal’s Law, when we move down in a fluid pressure increases with depth and when we move up the pressure decreases with height. When we move horizontally in a fluid pressure remains constant. Therefore, where the atmospheric pressure, and is the density of the fluid inside the vessel An U – shaped manometer tube connected to a vessel. Illustration 10. As shown in the figure, as column of water 40 cm high supports a 31 cm of an unknown fluid. What is the density of the unknown fluid? Solution: The pressure at point due to the two fluids must be equal (or the one with the higher pressure would push the lower pressure fluid away). Therefore, Pressure due to water = pressure due to known fluid from which (1000 kg/m2) = 1290 kg/m3. Illustration 11. For the arrangement shown in the figure, determine if the pressure difference between the vessels and is 3 kN/m2. Solution: Let pressure in the horizontal tube is So in left vertical tube here, N/m2 kg/m3 kg/m3 Thus, m = 50 cm. The Mercury Barometer: It is a straight glass tube (closed at one end) completely filled with mercury and inserted into a dish which is also filled with mercury as shown in the figure. Atmospheric pressure supports the column of mercury in the tube to a height . The pressure between the closed end of the tube and the column of mercury is zero, . Therefore, pressure at points and are equal and thus At the sea level, can support a column of mercury about 76 cm in height Hence, Nm-2 for Pa A mercury barometer Illustration 12. What must be the length of a barometer tube used to measure atmospheric pressure if we are to use water instead of mercury. Solution: We know that where and are the density and height of the water column supporting the atmospheric pressure . Since and m m. Illustration 13. In the figure shown, find (a) the total force on the bottom of the tank due to the water pressure. (b) the total weight of water. Solution: (a) Pressure at the base due to water is N/m2 Force N (b) Weight of water N. Buoyancy: If a body is partially or wholly immersed in a fluid, it experiences an upward force due to the fluid surrounding it. The phenomenon of force exerted by fluid on the body called buoyancy and the force is called buoyant force. A body experiences buoyant force whether it floats or sinks, under its own weight or due to other forces applied on it. Archimedes Principle: A body immersed in a fluid experiences an upward buoyant force equivalent to the weight of the fluid displaced by it. The proof of this principle is very simple. Imagine a body of arbitrary shape completely immersed in a liquid of density as shown in the figure. A body is being acted upon by the forces from all directions. Let us consider a vertical element of height and crosssectional area as shown in the figure(b). The force acting on the upper surface of the element is (downward) and that on the lower surface is (upward). Since , therefore, the net upward force acting on the element is It can be easily seen from the figure(b), that and so Also, and The net upward force is Hence, for the entire body, the buoyant force is the weight of the volume of the fluid displaced. The buoyant force acts through the centre of gravity of the displaced fluid. (a) The fluid exerts force on the immersed body from all directions. (b) The net force experienced by every vertical element of the body is in the upward Important It buoys because the pressure in the fluid is not uniform: it increases with depth. An object floats on water if it can displace a volume of water whose weight is greater than that of the object. If the density of the material is less than that of the liquid, it will float even if the material is a uniform solid, such as a block of wood floats on water surface. If the density of the material is greater than that of water, such as iron, the object can be made to float provided it is not a uniform solid. An iron built ship is an example to this case. Brain Teaser: 2. Does Archimedes principle hold in a vessel in free fall? In a satellite moving in a circular orbit? Illustration 14. An iceberg with a density of 920 kgm-3 floats on an ocean of density 1025 kgm-3. What fraction of the iceberg is visible. Solution: Let be the volume of the iceberg above the water surface, then the volume under water will be . Under floating conditions, the weight of the iceberg is balanced by the buoyant force . Thus, or or or Since kg m-3 and kg m-3, therefore, Hence 10% of the total volume is visible. Illustration 15. When a 2.5 kg crown is immersed in water, it has an apparent weight of 22 N. What is the density of the crown? Solution: Let actual weight of the crown = apparent weight of the crown = density of crown = density of water The buoyant force is given by or Since , therefore, Eliminating from the above two equations, we get Here N; N; kg m-3 kg m-3. Stability of a Floating Body: The stability of a floating body depends on the effective point of application of the buoyant force. The weight of the body acts at its center of gravity. The buoyant force acts at the center of gravity of the displaced liquid. This is called the center of buoyancy. Under equilibrium condition the center of gravity and the center of buoyancy lies along the vertical axis of the body as shown in the figure(a). (a) The buoyant force acts at the center of gravity of the displaced fluid. (b) When the boat tilts, the line of action of the buoyant force intersects the axis of the boat at the metacenter . In a stable boat, is above the center of gravity of the boat. When the body tilts to one side, the center of buoyancy shifts relative to the center of gravity as shown in the figure(b). The two forces act along different vertical lines. As a result the buoyant force exerts a torque about the center of gravity. The line of action of the buoyant force crosses the axis of the body at the point , called the metacentre. If is below , the torque will tend to restore the body to its equilibrium position. If is above , the torque will tend to rotate the body away from its equilibrium position and the body will be unstable. Illustration 16. An ice cube of side 1 cm is floating at the interface of kerosene and water in a beaker of base area 10 cm2. The level of kerosene is just covering the top surface of the ice cube. (a) Find the depth of submergence in the kerosene and that in the water. (b) Find the change in the total level of the liquid when the whole ice melts into water. Solution: (a) Condition of floating or …(i) where and be the submerged depth of the ice in the kerosence and water, respectively. Also, …(ii) Solving equation (i) and (ii), we get cm, cm (b) Fall in the level of kerosence Rise in the level of water Net fall in the overall level cm = 0.1 mm. FLUID SUBJECTED TO CONSTANT ACCELERATION: A fluid, contained in a vessel, may be subjected to a constant linear acceleration without any relative movement being created between different element of the fluid in the vessel. The fluid orient itself to attain a new equilibrium position under the action of the acceleration. Although the fluid is undergoing an acceleration, it is moving with the appearance of a rigid body. The fluid is thus said to be in a state of relative rest. In the absence of relative motion between different fluid elements, the law of fluid statics is applicable. Consider a liquid contained in a vessel. While at rest, the free surface maintains horizontal level but in the state of relative rest under a constant acceleration , the liquid orients itself to maintain the free surface inclined at an angle with the horizontal as shown in the figure (b). The inclination may be related to the magnitude and direction of the acceleration vector by considering the dynamics of an element of the liquid. (a) Liquid at rest (b) Liquid under constant acceleration (c) Free Body Diagram of the element w.r.t. vessel Consider a small element of size and at the position coordinates as shown in the figure(b). From the free body diagram we can write the equation of dynamics along the -axis = 0 Since or …(xx) along the -axis Since , and therefore or …(xxi) Note that if and , then and That is, pressure does not vary along the -axis if , thus pressure at all the points on plane are small when pressure increases in the opposite direction of acceleration. The angle of inclination of the free surface is obtained by …(xxii) Illustration 17. An open rectangular tank 5 m 4 m 3 m high containing water upto a height of 2 m is accelerated horizontally along the longer side. (a) Determine the maximum acceleration that can be given without spilling the water. (b) Calculate the percentage of water spilt over, if this acceleration is increased by 20% (c) If initially, the tank is closed at the top and is accelerated horizontally by 9 m/s2, find the gauge pressure at the bottom of the front and rear walls of the tank. (Take m/s2) Solution: (a) Volume of water inside the tank remains constant or Since, , therefore m/s2 (b) When acceleration is increased by 20% Now, m Fraction of water spilt over Percentage of water spilt over =10% (c) g volume of air remains constant Since or ; m Gauge pressure at the bottom of the i) Front wall zero ii) Rear wall Pa Illustration 18. A vertical - tube with the two limbs 0.75 m apart is filled with water and rotated about a vertical axis 0.5 m from the left limb, as shown in the figure. Determine the difference in elevation of the water levels in the two limbs, when the speed of rotation is 60 rpm. Solution: Consider a small element of length at a distance from the axis of rotation. Considering the equilibrium of this element. or On integrating between 1 and 2 or m. FORCES ON FLUID BOUNDARIES: Whenever a fluid comes in contact with solid boundaries it exerts a force on it. Consider a rectangular vessel of base size filled with water to a height as shown in the figure. The force acting at the base of the container is given by (area of the base) because pressure is same everywhere at the base and is equal to . Therefore, Since, (volume of the liquid) Thus, weight of the liquid inside the vessel A fluid contained in a vessel exerts forces in the boundaries. Unlike the base, the pressure on the vertical wall of the vessel is not uniform but increases linearly with depth from the free surface. Therefore, we have to perform the integration to calculate the total force on the wall. Consider a small rectangular element of width and thickness at a depth from the free surface. The liquid pressure at this position is given by The force at the element is The total force is The total force acting per unit width of the vertical wall is …(xxiii) The point application (the centre of force) of the total force from the free surface is given by …(xxiv) where is the moment of force about the free surface. Here Since , therefore, ….(xxv) Illustration19. Find the force acting per unit width on a plane wall inclined at an angle with the horizontal as shown in the figure. Solution: Consider a small element of thickness at a distance measured along the wall from the free surface. The pressure at the position of the element is The force is given by The total force per unit width is given by or …(xxvi) Note: That the above formula reduces to for a vertical wall . FLUID DYNAMICS: The Equation of Continuity: In order to describe the motion of a fluid, in principle one might apply Newton’s laws to a particle (a small volume element of fluid) and follow its progress in time. This is difficult approach. Instead, we consider the properties of the fluid, such as velocity pressure, at fixed points in space. In order to simplify the discussion we make several assumptions: (i) The fluid is non viscous: There is no dissipation of energy due to internal friction between adjacent layer in the fluid. (ii) The flow is steady: (iii) The flow is irrotational: A tiny paddle wheel placed in the liquid will not rotate. In rotational flow, for example, in eddies, the fluid has angular momentum about a given point. In general the velocity of a particle will not be constant along a streamline. The density and the cross-sectional area of a tube of flow will also change. Consider two sections of a tube of flow, as shown in the figure. The mass of fluid contained in a small cylinder of length and area is . Since fluid does not leave the tube of flow, this mass will later pass through a cylinder of length and area . The mass in this cylinder is . The lengths and are related to the speeds at the respective locations: and . Since no mass is lost or gained. A “tube of flow” The fluid contained in the left cylinder of length is later contained in the right cylinder of length . and …(xxviii) This is called the equation of continuity. It is statement of the conservation of mass. If the fluid is incompressible, its density remains unchanged. This is a good approximation for liquid, but not for gases. If , the equation (xviii) becomes, …(xxix) The product is the volume rate of flow . Figure shows a pipe whose cross section narrows. From equation (xxix) we conclude that the speed of a fluid is greatest where the cross-sectional area is the least. Notice that the streamlines are close together where the speed is higher. A fluid flowing through a pipe whose cross section changes. Notice that the streamlines are closer together in the narrower section. This indicates that the fluid is moving faster. Bernoulli’s Theorem Statement When an incompressible and non-viscous liquid (or gas) flows in streamlined motion from one place to another, then at every point of its path the total energy per unit volume (Pressure energy + Kinetic energy + Potential energy) is constant. That is = constant. Thus, Bernoulli’s theorem is in one way the principle of conservation of energy for a flowing liquid (or gas). Bernoulli’s Equation: Let us focus our attention on the motion of the shaded region. This its our “system”. The lower cylindrical element of fluid of length and area is at height , which moves with speed . After some time, the leading section of our system fills the upper cylinder of fluid of length and area at height , and is then moving with speed . A pressure force acts on the lower part of the cylindrical tube towards right and pressure force acts on the upper part of the cylindrical tube towards left. The net work done on the system and is The motion of a fluid in a tube of flow. The work done by the pressure forces-equals the change in energy of the shaded volume of fluid. where we have used the relations and . The net effect of the motion of the system is to raise the height of the lower cylinder of mass and to change its speed. The changes in the potential and kinetic energies are These changes are brought about by the net work done on the system, Since the density is , we have Since the points 1 and 2 can be chosen arbitrarily, we can express this result as Bernoulli’s Equation = constant …(xxx) It is applied to all points along a streamline in a nonviscous, incompressible fluid. A fluid flowing through a tube whose cross section decreases. The pressure in the narrower tube, where the fluid is moving faster, is lower. Brain Teaser: 3. The height of the liquid in the standpipes of figure indicates that the pressure drops along the channel, even through the channel has a uniform cross section and the flowing liquid is incompressible. Explain. Illustration 20. A tank, initially at rest, is filled with water to a height m. A small orifice is made at the bottom of the wall. Find the velocity attained by the tank when it becomes completely empty. Assume mass of the tank to be negligible. Friction is negligible. Solution: Let be the instantaneous velocity of the tank, and be the instantaneous velocity of efflux with respect to the tank. Thrust exerted on the tank. where is the cross-sectional area of the orifice. where is the instantaneous height of water in the tank. Mass of the tank at any time is = cross-sectional area of the tank, Using Newton’s second law or …(i) In a time if the water level falls by dh, then according to the conservation of mass. or Equation (i) can be written as or or On integrating Since m, therefore m/s APPLICATIONS OF BERNOULLI’S PRINCIPLE: (a) Working of Aeroplane: This is also based on Bernoulli’s principle. The wings of the aeroplane are having tapering as shown in figure. Due to this specific shape of wings when the aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli’s principle, creates a pressure difference, due to which an upward force called ‘dynamic lift’ (= pressure difference area of wing) acts on the plane. If this force becomes greater than the weight of the plane, the plane will rise up. (b) Velocity of Efflux: If a liquid is filled in a vessel up to height and a hole is made at a depth below the free surface of the liquid as shown in figure, then taking the level of hole as reference level (i.e., zero point of potential energy) and applying Bernoulli’s principle to the liquid just inside and outside the hole (assuming the liquid to be at rest inside) we get or which is the same speed that an object would acquire in falling from rest through a distance and is called ‘velocity of efflux’ or velocity of flux. From this expression it is clear that: (1) The speed of the liquid coming out of the orifice is independent of the nature and quantity of liquid in the container or the area of the orifice. (2) Greater is the distance of the hole from the free surface of liquid greater will be the velocity of efflux (i.e., ). This is why liquid gush-out with maximum velocity from the orifice which is at maximum vertical distance from the free surface of the liquid. (3) As the vertical velocity of liquid at the orifice is zero and it is at a height from the base, the time taken by the liquid to reach the base-level Now during this time liquid is moving horizontally with constant velocity , so it will hit the base level at a horizontal distance (called range) as shown in figure such that From this expression it is clear that will be maximum when is maximum i.e., or or , i.e., So that i.e., range will be maximum when . (4) If the level of free surface in a container is at height from the base and there are two holes at depth and below the free surface, then and Now if , i.e., i.e., or , i.e., or i.e, the range will be same if the orifice is at a depth or below the free surface. Now as the distance from top means from the bottom, so the range is same for liquid coming out of holes at same distance below the top and above the bottom. (5) If is the area of orifice at a depth below the free surface and that of container, the volume of liquid coming out of the orifice per second will be [as ] Due to this, the level of liquid in the container will decrease and so if the level of liquid in the container above the hole changes from to in time to then , i.e., So the time taken for the level to fall from to If the hole is at the bottom of the tank, time taken to emptied the tank: [as here ]. (c) Ventury Tube: A ventury tube is used to measure the flow speed of fluid in a tube. It consists of a constriction or a throat in the tube. As the fluid passes through the constriction, its speed increases in accordance with the equation of continuity. The pressure thus decreases as required by Bernoulli’s equation. Figure shows a ventury tube through which a liquid of density is flowing. The area of cross-section is at the wider part and at the constriction. Let the speeds of the liquid at and be and and the pressures at and be and respectively. By the equation of continuity …(i) and by Bernoulli’s equation, or, …(ii) Figure also shown two vertical tubes connected to the ventury tube at and .If the difference in heights of the liquid levels in these tubes is , we have . Putting in (ii), …(iii) Knowing and , one can solve equation (i) and (iii) so as to get and . This allows one to know the rate of flow of liquid pass a cross-section. SURFACE TENSION: The properties of a surface are quite often markedly different from the properties of the bulk material. A molecule well inside a body is surrounded by similar particles from all sides. But a molecule on the surface has particles of one type on one side and of a different type on the other side. Figure shown an example. A molecule of water well inside the bulk experiences force from water molecules from all sides but a molecule at the surface interacts with air molecules from above and water molecules from below. This asymmetric force distribution is responsible for surface tension. By a surface we shall mean a layer approximately 10-15 molecular diameters. The force between two molecules decreases as the separation between them increases. The force becomes negligible if the separation exceeds 10-15 molecular diameters. Thus, if we go 10-15 molecular diameters deep, a molecule finds equal forces from all directions. Imagine a line drawn on the surface of a liquid (figure). The line divides the surface in two parts, surface on one side and the surface on the other side of the line. Let us call them surface to the left of the line and surface to the right of the line. It is found that the two parts of the surface pull each other with a force proportional to the length of the line . These forces of pull are perpendicular to the line separating the two parts and are tangential to the surface. In this respect the surface of the liquid behaves like a stretched rubber sheet. The rubber sheet which is stretched from all sides is in the state of tension. Any part of the sheet pulls the adjacent part towards itself. Let F be the common magnitude of the forces exerted on each other by the two parts of the surface across a line of length . We define the surface tension of the liquid as The unit of surface tension is N/m. Illustration 21. Calculate the force required to take away a flat circular plate of radius 4 cm from the surface of water, surface tension of water being 75 dyne cm-1. Solution: Length of the surface = circumference of the circular plate cm Required force dyne. The fact that a liquid surface has the property of surface tension can be demonstrated by a number of simple experiments. Take a ring of wire and dip it in soap solution. When the ring is taken out, a soap film bounded by the ring is formed. Now take a loop of thread, and place it gently on the soap film. The loop stays on the film in an irregular fashion as it is placed. Now prick a hole in the film inside the loop with a needle. The thread is readily pulled by the film surface outside and it takes a circular shape (figure). Before the picking, there were surfaces both inside and outside the thread loop. Taking any small part of the thread, surfaces on both sides pulled it and the net force was zero. The thread could remain in any shape. Once the surface inside was punctured, the outside surface pulled the thread to take the circular shape. Surface Energy: We have seen that a molecule well within the volume of a liquid is surrounded by the similar liquid molecules for all sides and hence there is no resultant force on it (figure). On the other hand, a molecule in the surface is surrounded by similar liquid molecules only on one side of the surface while on the other side it may be surrounded by air molecules or the molecules of the vapour of the liquid etc. These vapours having much less density exert only a small force. Thus, there is a resultant inward force on a molecule in the surface. This force tries to pull the molecule into the liquid. Thus, the surface layer remains in microscopic turbulence. Molecules are pulled back from the surface layer to the bulk and new molecules from the bulk go to the surface to fill the empty space. When a molecule is taken from the inside to the surface layer, work is done against the inward resultant force while moving up in the layer. The potential energy is increased due to this work. A molecule in the surface has greater potential energy than a molecule well inside the liquid. The extra energy that a surface layer has is called the surface energy. The surface energy is related to the surface tension as discussed below. Relation between Surface Tension and Surface Energy: Consider a U-shaped frame with a sliding wire on its arm. Suppose it is dipped in a soap solution, taken out and placed in a horizontal position (figure). The soap film that is formed may look quite thin, but on the molecular scale its thickness is not small. It may have several hundred thousands molecular layers. So it has two surfaces enclosing a bulk of soap solution. But the surfaces are in contact with the sliding wire and hence exert forces of surface tension on it. If be the surface tension of the solution and be the length of the sliding wire, each surface will pull the wire parallel to itself with a force . The net force of pull on the wire due to both the surfaces is . One has to apply an external force equal and opposite of so as to keep the wire in equilibrium. Now suppose the wire is slowly pulled out by the external force through a distance so that the area of the frame is increased by .As there are two surfaces of the solution, a new surfaces area is created. The liquid from the inside is brought to create the new surface. The work done by the external force in the displacement is As there is no change in kinetic energy, the work done by the external force is stored as the potential energy of the new surface. The increase in surface energy is Thus, or, . We see that the surface tension of a liquid is equal to the surface energy per unit surface area. In this interpretation, the unit of surface tension may be written as J/m2. It may be verified that N/m is equivalent to . Illustration 22. Calculate the work done in blowing a soap bubble of radius 10 cm, surface tension being 0.06 Nm-1. What additional work will be done in further blowing it so that its radius is doubled? Solution: In case of a soap bubble, there are two free surfaces. work done at blowing a soap bubble of radius is given by. , where is the surface tension of the soap solution. where m, 0.06 Nm-1 J = 1.51 J Similarly, work done is forming a bubble of radius 0.2 m is, Additional work done in doubling the radius of the bubble is given by, J – 1.51J = 4.52J. Angle of contact: When a solid body in the form of a tube or plate is immersed in a liquid, the surface of the liquid near the solid is, in general curved. It is defined as the angle between the tangents to the liquid surface and the solid surface at the point of contact, for that pair of solid and liquid is called angle of contact. For example when glass strip is dipped in water and mercury as shown in figure (a) and (b) respectively, the angle is angle of contact, which is acute in case of water and obtuse in case of mercury. Excess Pressure: The pressure inside a liquid drop or a soap bubble must be in excess of the pressure outside the bubble drop because without such pressure difference a drop or a bubble cannot be in stable equilibrium. Due to surface tension the drop or bubble has got the tendency to contract and disappear altogether. To balance this, there must be excess of pressure inside the bubble. To obtain a relation between the excess of pressure and the surface tension, consider a water drop of radius and surface tension . Divide the drop into two halves by a horizontal passing through its centre as shown in figure and consider the equilibrium of one-half, say, the upper half. The forces acing on it are: (i) forces due to surface tension distributed along the circumference of the section. (ii) outward thus on elementary areas of it due to excess pressure. Obviously, both the types of forces are distributed. The first type of distributed forces combine into a force of magnitude . To find the resultant of the other type of distributed forces, consider an elementary area of the surface. The outward thrust on where is the excess of the pressure inside the bubble. If this thrust makes an angle with the vertical, then it is equivalent to along the vertical and along the horizontal. The resolved component is infective as it is perpendicular to the resultant force due to surface tension. The resolved component contributes to balancing the force due to surface tension. The resultant outward thrust where = area of the projection of on the horizontal dividing plane For equilibrium of the bubble we have or, …(iv) If it is a soap bubble, the resultant force due to surface tension is , because a bubble has two surfaces. Hence for the equilibrium of a bubble we have or ...(v) Illustration 23. If a number of little droplets of water, all of the same radius coalesce to form a single drop of radius , show that the rise in temperature of the water is given by. where is the mechanical equivalent of heat, and is the surface tension of water. Solution: Let be the number of droplets, each of radius cm that coalesce to form a single drop of radius Decrease in the surface area Decrease in the surface energy Heat energy produced in the drop Suppose the whole of heat energy is used to raise the temperature of the resultant drop by , therefore, where m is the of the drop having specific heat . (density = 1 gm/cc) cal/gºC for water …(i) Since volume remains the same, or Putting this value in (i), we have Thus the rise in temperature is given by, . Capillary Action: When a glass tube of very fine bore called a capillary tube is dipped in a liquid (like water), the liquid immediately rises up into it due to the surface tension. This phenomenon of rise of a liquid in a narrow tube is known as capillarity. Suppose that a capillary tube of radius is dipped vertically in a liquid. The liquid surface meets the wall of the tube at some inclination called the angle of contact. Due to surface tension a force, act on an element of the circle of contact along which the liquid surface meets the solid surface and it is tangential to the liquid surface at inclination to the wall of the tube. (The liquid on the wall of the tube exerts this force. By the third law of motion, the tube exerts the same force on the liquid in the opposite direction.) Resolving this latter force along and perpendicular to the wall of the tube, we have along the tube vertically upward and perpendicular to the wall. The latter component is ineffective. It simple compresses the liquid against the wall of the tube. The vertical component pulls the liquid up the tube. The total vertical upward force = Due to this upward pull liquid rises up in the capillary tube till it is balanced by the downward gravitational pull. If is the height of the liquid column in the tube up to the bottom, the gravitational pull, i.e., weight of the liquid inside the tube is , where is the volume of the liquid in meniscus. For equilibrium of the liquid column in the tube If volume of the liquid in meniscus is negligible then, ….(vi) The small volume of the liquid above the horizontal plane through the lowest point of the meniscus can be calculated if is given or known. For pure water and glass and hence the meniscus is hemispherical. = volume of the cylinder of height - volume of hemisphere. For water and glass For a given liquid and solid at a given place as and are constant, constant i.e., lesser the radius of capillary greater will be the rise and vice-versa. Illustration 24. Water rises to a height of 10 cm in a certain capillary tube. The level of mercury in the same tube is depressed by 3.42 cm. Compare the surface tension of water and mercury for the contacts angles zero and 135º respectively. Solution: Using the capillarity relation, (for water) rg (for mercury) = = 32.9 rg . VISCOSITY: If water in a tube is whirled and then left to itself, the motion of the water stops after some time. This is a very common observation. What stops the motion? There is no external force to stop it. A natural conclusion is, therefore, that whenever there is relative motion between parts of a fluid, internal forces are set up in the fluid, which oppose the relative motion between the parts in the same way as forces of friction operate when a block of wood is dragged along the ground. To maintain relative motion between layers of a fluid an external force is needed. “This property of a fluid by virtue of which it oppose the relative motion between its different layers is known as viscosity and the force that is into play is called the viscous force”. Consider the slow and steady flow of a fluid over a fixed horizontal surface. Let be the velocity of a thin layer of the fluid at a distance from the fixed solid surface. Then according to Newton, the viscous force acting tangentially to the layer is proportional to the area of the layer and the velocity gradient at the layer. If is the viscous force on the layer, then where is the area of the layer The negative sign is put to account for the fact that the viscous force is opposite to the direction of motion. where is a constant depending upon the nature of the liquid and is called the coefficient of viscosity and velocity gradient If and , we have Thus the coefficient of viscosity of a liquid may be defined as the viscous force per unit area of the layer where velocity gradient is unity. The coefficient of viscosity has the dimension and its unit is Newton second per square metre ( ) or kilograme per meter per second . In the unit of viscosity is poise. Illustration 25. A metal plate 100 cm2 in area rests on a layer of castor oil ( poise) 0.2 cm thick. Calculate the horizontal force required to move the plate with a speed of 3 cm/s. Solution: where poise cm2 dyne =- 0.233 N So force required = 0.233 N Poiseuille’s Equation: When a liquid flows slowly and steadily through a capillary tube, the flow is streamline. the rate of this streamline flow through the tube is given by an equation deduced by Poiseuille and is known as Poiseuille’s equation. Consider a laminar flow through a horizontal round pipe of length as shown in figure. The equilibrium of cylinder of radius r contained in a lamina of radius is considered, For equilibrium, the net pressure force on cylindrical element due to pressure difference ‘ ’ across its ends must equal to the shear force on its periphery due to the presence of a cylindrical lamina over it. ….(viii) Total discharge through the pipe. ( ) …(ix) Equation (ix) can also be expressed as where is known as fluid resistance. Capillaries in Series and Parallel: Capillaries in series: When two capillaries are connected in series across constant pressure difference , then …(x) The volume of fluid flowing per second through both the tubes is the same i.e., …(xi) If and be the pressure differences across individual capillaries respectively, …(xii) Capillaries in parallel: When two capillaries are connected to parallel across constant pressure difference , then or …(xiii) The volume of fluid flowing through the first capillary …(xiv) Similarly, the volume of fluid flowing through second capillary Reynolds Number: The stability of laminar flow is maintained by viscous forces. It is observed, however that laminar or steady flow is disrupted if the rate of flow is large. Irregular, unsteady motion, turbulence, sets in at high flow rates. Reynolds defined a dimensionless number whose value given one a approximate idea, whether the flow rate would be turbulent. This number, called the Reynolds number is defined as, Where the density of the fluid flowing with a speed .The parameter stand for the typical dimension of the obstacle or boundary to fluid flow. Illustration 26. The diameter of the tap is 1.25 cm and the flow rate through it is . Is the flow turbulent? Given coefficient of viscosity of water is poise. Solution: Volume of water flowing out per second is Reynold’s number The flow will be turbulent. Stoke’s Law: When a solid moves through a viscous medium, its motion is opposed by a viscous force depending on the velocity and shape and size of the body. The energy of the body is continually deceases in overcoming the viscous resistance of the medium. This is why cars, aeroplanes etc. are shaped streamline to minimize the viscous resistance on them. The viscous drag on a spherical body of radius , moving with velocity , in a viscous medium of visocsity is given by . This relation is called Stokes’ law Illustration 27. An air bubble of diameter 2 cm rises through a long cylindrical column of a viscous liquid, and travels at 0.21 cms-1. If the density of the liquid is 1.47 g cm-1 find its coefficient of viscosity. Ignore the density of the air. Solution: Weight of the bubble is equal to the viscous force. or …(i) Given ; kg/m3 Substituting these values in (i) we have, . Terminal Velocity: Let the body be driven by a constant force. In the beginning the viscous drag on the body is small because the velocity is small and so the body is accelerated through the medium by the driving force with the increases of velocity of the body the viscous drag on it will also increase and eventually when it becomes equal to the driving force, the body will acquire a constant velocity. This velocity is called the terminal velocity of the body: Consider the downward motion of a spherical body through a viscous medium such as a ball falling through liquid. If is the radius of the body, the density of the material of the body and is the density of the liquid then, the weight of the body , downwards and the buoyancy of the body , upwards. The net downward driving force . If is the terminal velocity of the body, then the viscous force on the body is .For no acceleration of the body we have or, …(xvi)