Fluid Mechanics Theory PDF

9. FLUID MECHANICS 1. INTRODUCTION Fluid is a collective term for liquid and gas. A fluid cannot sustain shear stress when at rest. We will study the dynamics of non-viscous, incompressible fluid. We will be learning about pressure variation, Archemides principle, equatio...

9. FLUID MECHANICS 1. INTRODUCTION Fluid is a collective term for liquid and gas. A fluid cannot sustain shear stress when at rest. We will study the dynamics of non-viscous, incompressible fluid. We will be learning about pressure variation, Archemides principle, equation of continity, Bernoulli’s Theorem and its applications and surface tension, Stoke’s Law and Terminal velocity of a spherical body. 2. DEFINITION OF A FLUID A fluid is a substance that deforms continuously under the application of a shear (tangential) stress no matter how small the shear stress may be. F F (a) Solid (b) Fluid Figure 9.1: Behavior of a solid and a fluid, under the action of a constant shear force. 3. 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 fluid statics. 3.1 Density The density ρ of a substance is defined as the mass per unit volume of a sample of the substance. If a small mass ∆m element ∆m occupies a volume ∆V, the density is given by ρ = ∆V In general, the density of an object depends on position, so that ρ =f(x, y, z) 9. 2 | Fluid Mechanics If the object is homogeneous, its physical parameters do not change with position throughout its volume. Thus for M a homogeneous object of mass M and volume V, the density is defined as ρ = V Thus SI units of density are kg m–3. MASTERJEE CONCEPTS Note: As pressure is increased, volume decreases and hence density will increase. As the temperature of a liquid is increased, mass remains the same while the volume is increased. Vaibhav Krishnan (JEE 2009, AIR 22) 3.2 Specific Gravity The specific gravity of a substance is the ratio of its density to that of water at 4ºC, which is 1000 kg/m3. Specific gravity is a dimensionless quantity numerically equal to the density quoted in g/cm3. For example, the specific Density of substance gravity of mercury is 13.6, and the specific gravity of water at 100ºC is 0.999. RD= Density of water at 4º C Illustration 1: Find the density and specific gravity of gasoline if 51 g occupies 75 cm3?  (JEE MAIN) Sol: Density is mass per unit volume, and specific gravity is the ratio of density of substance and density of water. mass 0.051kg Density = = = 680 kg/m3 volume 75 × 10−6 m3 density of gasoline 680kg / m3 mass of 75 cm3gasoline Sp. gr = = =0.68 or Sp. gravity = density of water 1000kg / m3 mass of 75 cm3 water 51g = =0.68 75g Illustration 2: 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 milk?  (JEE MAIN) Sol: Find the mass of butterfat present in the milk. Subtract this from total mass to get mass of fat-free milk. The density of fat-free milk is equal to its mass divided by its volume. Volume of fat in 1000 cm3 of milk = 4% × 1000 cm3 = 40 cm3 Mass of 40 cm3 fat = Vρ = (40 × 10–6 m3)(865 kg/m3) = 0.0346 kg mass (1.032 − 0.0346)kg Density of skimmed milk = = volume (1000 − 40) × 10−6 m3 3.3 Pressure The pressure exerted by a fluid is defined as the force per unit area at a point within the F fluid. Consider an element of area ∆A as shown in the figure and an external force ∆F is acting normal to the surface. The average pressure in the fluid at the position of the ∆F element is given by Pav = [A normal force ∆F acts on a small cylindrical element of A ∆A cross-section area ∆A.] Figure 9.2  P hysics | 9.3 As ∆A → 0, the element reduces to a point, and thus, pressure at a point is defined as ∆F dF p = lim = ∆A →0 ∆A dA F When the force is constant over the surface A, the above equation reduces to p = A The SI unit of pressure is Nm-2 and is also called Pascal (Pa). The other common pressure units are atmosphere and bar. 1 atm = 1.01325 × 105 Pa; 1 bar = 1.00000 × 105 Pa; 1 atm = 1.01325 bar 3.3.1 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 is equal in magnitude. Therefore the pressure on each of the lateral faces must also be the same. In the limit as the cube element 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 – pressure is isotropic. Note: Since the fluid cannot support a shear stress, the force exerted by a fluid pressure Figure 9.3: A small cubical must also be perpendicular to the surface of the container that holds it. element is in equilibrium inside a fluid 3.3.2 Atmospheric Pressure (P0) It is pressure of the earth’s atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level (an average value) is 1.013 × 105 Pa. Thus, 1 atm = 1.013 × 105 Pa=1.013 Bar 3.3.3 Absolute Pressure and Gauge Pressure The excess pressure above atmospheric pressure is usually called gauge pressure and the total pressure is called absolute pressure. Thus, Gauge pressure = absolute pressure – atmospheric pressure. Aboslute pressure is always greater than or equal to zero. While gauge pressure can be negative also. Illustration 3: Atmospheric pressure is about 1.01 × 105 Pa. How large a force does the atmosphere exert on a 2 cm2 area on the top of your head? (JEE MAIN) Sol: Force = Pressure × Area Because p = F/A, where F is perpendicular to A, we have F = pA. 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 F = pA = (1.01 × 105 N/m2) (2 × 10–4 m2) ≈ 20N 3.3.4 Variation of Pressure with Depth Weight of a fluid element of mass Dm, DW = (Dm)g. The force acting on the lower face of the element is pA and that on the upper face is (p + Dp)A. The figure (b) shows the free body diagram of the element. Applying the condition of equilibrium, we get, pA – (p + Dp) A – (Dm)g = 0 if ρ is the density of the fluid at the position of the element, then Dm = ρA(Dy) 9. 4 | Fluid Mechanics and pA – (p + Dp) A – rgA(Dy) = 0 (p+ p)A ∆p or = - rg m ( m)g ∆y y pA ∆p In the limit ∆y approaches to zero, becomes y ∆y dp = −ρg. The above equation indicates that the dy (a) (b) slope of p versus y is negative. That is, the pressure p Figure 9.4 decreases with height y from the bottom of the fluid. In dp other words, the pressure p increases with depth h, i.e., = ρg dh 3.4 The Incompressible Fluid Model For an incompressible fluid, the density ρ of the fluid remains constant throughout its volume. It is a good assumption for liquids. To find pressure at the point A in a fluid column as shown in the figure, is obtained by integrating p the following equation: h A p h dp = rgdh or ∫ dp = ρg∫ dh or p – p0 = rgh or p = p0 + rgh  …(xvi) y p0 0 where ρ is the density of the fluid, and p0 is the atmospheric pressure at the free surface of the liquid. Figure 9.5: A point A is located in Note: Further, the pressure is the same at any two points at the same level in a fluid at a height from the bottom the fluid. The shape of the container does not matter. and at a deth h from the free surface P0 P0 h A B PA= PB= P0+pgh Illustration 4: Find the absolute pressure and gauge pressure at point A, B and C as shown in the Fig. 9.6 (1 atm = 105 Pa) (JEE MAIN) Sol: Gauge Pressure = rgh, Absolute Pressure is sum of gauge pressure and atmospheric pressure. 1m Kerosene A Patm = 10 Pa. 5 2m Absolute Pressure A -> PA + Patm= r1ghA = (800)(10)1 = 8 kPa 3 p1=800 kg/m p′A =pA + patm =108 kPa Water 1.5m Gauge Pressure = 8 kPa. 2m B 3 p2=1000 kg/m B -> pB = ρ1g(2) + ρ2g(1.5) C Mercury 0.5m ′ pB + patm = 131 kPa = (800)(10)(2) + (103)(10)(1.5) = 131 kPa p= B p3=13.6x10 3 kg/m3 Gauge Pressure = 31 kPa. Figure 9.6  P hysics | 9.5 C-> = pc p1g(2) + ρ2g(2) + ρ3g(0.5) ρC′ = pC + patm = 204kPa = (800)(10)(2) + (10)3(10)(2) + 1(13.6 × 103)(10)(0.5) = 204 kPa Gauge Pressure = 104 kPa. Illustration 5: A glass full of water of a height of 10 cm has a bottom of area 10 cm2, top of area 30 cm2 and volume 1 litre. (JEE ADVANCED) (a) Find the force exerted by the water on the bottom. (b) Find the resultant force exerted by the side of the glass on the water. (c) If the glass is covered by a jar and the air inside the jar is completely pumped out, what will be the answer to parts (a) and (b). (d) If a glass of different shape is used provided the height, the bottom area and the volume are unchanged, will the answers to parts (a) and (b) change. Take g = 10m/s2, density of water = 103 kg/m3 and atmospheric pressure = 1.01 × 105 N/m2. Sol: Pressure at the bottom depends on the height of water in the container. Force = Pressure × Area. The force on water surface due to atmospheric pressure plus the weight of water are balanced by the force on water by the container bottom and its walls. (a) Force exerted by the water on the bottom F1 = (P0 + rgh)A1  … (i) Here, P0 = atmospheric pressure = 1.01 × 105 N/m2; ρ = density of water = 103 kg/m3 g = 10 m/s2, h = 10 cm = 0.1 m and A1 = area of base 10 cm2 = 10–3 m2. Substituting in Eq. (i), we get F1= (1.01 × 10 + 10 × 10 × 0.1) ×10 or F1 = 102 N (downwards) 5 3 –3 (b) Force exerted by atmosphere on water F2 = (P0)A2 Here, A2 = area of top = 30 cm2 = 3 × 10–3 m2 ; F2 = (1.01 × 105)(3 × 10–3) = 303 N (downwards) Force exerted by bottom on the water F3 = – F1 or F3= 102 N (upwards) Weight of water W = (volume)(density)(g) = (10–3)(103)(10) = 10 N (downwards) Let F be the force exerted by side walls on the water (upwards). Then, from equilibrium of water Net upward force = net downward force or F + F3 = F2 + W F – F2 + W – F3 = 303 + 10 – 102 or F = 211 N (upwards) (c) If the air inside of the Jar is completely pumped out, F1 = (rgh)A1 (as P0 = 0) = (103)(10)(0.1)(10–3) = 1 N (downwards). In this case F2 = 0 and F3 = 1 N (upwards) ∴ F = F2 + W – F3 = 0 + 10 – 1 = 9 N (upwards) (d) No, the answer will remain the same. Because the answers depend upon P0, ρ, g, h , A1 and A2. Illustration 6: Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill up to a particular common height. Is the force exerted by water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to the same height give different reading on a weighing scale? (JEE MAIN) Sol: Force on the base of the vessel depends on the pressure on it, and pressure depends on the height of the liquid in the vessel. On the other hand the normal reaction from the surface on which the vessel is kept, depends on both the pressure at the base as well as the weight of the liquid in the vessel. 9. 6 | Fluid Mechanics Pressure (and therefore force) on the two equal base areas are identical. But force is exerted by water on the sides of the vessels also, which has non-zero vertical component when the sides of the vessel are not perfectly normal to the base. This net vertical component of force by water on the side of the vessel is greater for the first vessel than the second. Hence, the vessels weigh different when the force on the base is the same in the two cases. 3.4.1 Pascal’s Laws F2 According to the equation p = p0 + rgh. Pressure at any depth h in a F1  fluid may be increased by increasing the pressure p0 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 A1 A2 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 F1 applied to a piston of area A1 is transmitted to the large piston of area A2. The pressure at the two pistons is the same because they are at the same level. A hydraulic jack Figure 9.7 F1 F2 A  p = = Or F2 =  2  F1. Consequently, the force on the larger piston is large. A1 A2  A1  Thus, a small force F1 acting on a small area A1 results in a larger force F2 acting on a larger area A2. MASTERJEE CONCEPTS Since energy is always conserved, F1x1 = F2x2 where x1 and x2 are the distances moved by the pistons. Nitin Chandrol (JEE 2012, AIR 134) Illustration 7: Find the pressure in the air column at which the piston remains in equilibrium. Assume Air the pistons to be massless and frictionless. Piston  (JEE MAIN) 5m 1.73m Kerosene Sol: Apply Pascal’s law at two points at equal 60° S=0.8 height from a common datum. Datum A B Water Let pa be the air pressure above the piston. Applying Pascal’s law at point A and B. Figure 9.8 3 Patm + r wg(5) =pa + rkg(1.73) ; Pa = 138 kPa 2 Illustration 8: A weighted piston confines a fluid of density ρ in a closed container, as shown in the figure. The combined weight of piston and h container is W = 200 N, and the cross-sectional area of the piston is B A = 8 cm2. Find the total pressure at point B if the fluid is mercury and h = 25 cm (pm = 13600 kgm-3). What would be an ordinary pressure gauge reading at B? (JEE ADVANCED) Sol: Pressure difference between two points at different heights is equal to ρgh, where h is difference in heights of two points. Apply Pascal’s law at two points at different heights from a common datum. Figure 9.9  P hysics | 9.7 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 B is composed of three parts: Pressure of atmosphere = 1.0 × 105 Pa W 200N Pressure due to piston and weight = = = 2.5 × 105 Pa A −4 2 8 × 10 m Pressure due to height h of fluid = hrg = 0.33 × 105 Pa In this case, the pressure of the fluid itself is relatively small. We have Total pressure at B = 3.8 × 105 Pa = 383 kPa. The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at B = 280 kPa Illustration 9: For the system shown in the 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 F right at S, has cross-sectional area 25 cm2 and negligible weight. If the apparatus is filled with oil (ρ=0.78 g/cm3), find the force F required to hold the system in S equilibrium as shown in figure. (JEE ADVANCED) 600 kg 8m Sol: Apply Pascal’s law at two points at different heights from a common datum. L H1 H2 The pressures at point H1 and H2 are equal because they are at the same level in the single connected fluid. Therefore, Pressure at H1 = pressure at H2 = (pressure due to F plus pressure due to liquid column above H2) (600)(9.8)N F Figure 9.10 = + (8m)(780 kg/m–3)(9.8) 2 0.08m 25 × 10−4 m2 After solving, we get, F = 31 N Illustration 10: As shown in the figure, as column of water 40 cm high supports 31 cm of an unknown fluid. What is the density of the unknown fluid? (JEE MAIN) Sol: Find the hydrostatic pressure at the bottom most point A due to both the water column and the unknown fluid column. The pressure at point A due to the two fluids must be equal (or the one with the higher pressure would push lower pressure fluid away). Therefore, pressure due to water = pressure due to known 40 cm h 40 fluid; h1r1g = h2r2g, from which r2 = 1 p1 = (1000 kg/m2) = 1290 31 cm h 31 kg/m3 2 For gases, the constant density assumed in the compressible model is often not adequate. However, an alternative simplifying A assumption can be made that the density is proportional to the Figure 9.11 pressure, i.e., ρ = kp. Let r0 be the density of air at the earth’s surface ρ0 where the pressure is atmospheric po, then r0 = kp0 ; After eliminating k, we get ρ = p p0 ρ  Putting the value of ρ in equation dp = –rgdy or dp = −  00 p  gdy p    p dp ρ h On rearranging, we get ∫ p = − 0 g∫ dy where p is the pressure at a height y = h above the earth’s surface. p0 0 p0 9. 8 | Fluid Mechanics −p0 ρ gh p p0 After integrating, we get ln = – 0 gh or p = p0 p0 p0 Note: Instead of a linear decrease in pressure with increasing height as in the case of an incompressible fluid, in this case pressure decreases exponentially. 4. PRESSURE MEASURING DEVICES 4.1 Manometer P0 A manometer is a tube open at both ends and bent into the shape of a “U” and is partially filled with mercury. When one end of the tube is subjected to an unknown pressure p, the mercury level drops on that side of the tube and rises on the h other so that the difference in mercury level is h as shown in P =? the figure. P0 h0 When we move down in a fluid, pressure increases with B A depth and when we move up the pressure decreases with height. When we move horizontally in a fluid, pressure remains constant. Therefore, p + r0gh0 – rmgh = p0 where p0 is atmospheric pressure, and rm is the density of the fluid inside the vessel. Figure 9.12: An U-shaped manometer tube connected to a vessel Po= 4.2 The Mercury Barometer O Pm 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 Po Po h figure. Atmospheric pressure supports the column of mercury in the tube to A B a height h. The pressure between the closed end of the tube and the column of mercury is zero, p = 0. Therefore, pressure at points A and B are equal and thus p0 = 0 + rmgh. Hence, p0 = (13.6 × 103)(9.8)(0.76) = 1.01 × 105 Nm-2 for Pa. A mercury barometer Figure 9.13 Illustration 11: What must be the length of a barometer tube used to measure atmospheric pressure if we are to use water instead of mercury? (JEE MAIN) Sol: The length of the barometer tube will be inversely proportional to the density of fluid used in it. We know that p0 = rmghm = r wghw where r w and hw are the density and height of the water column supporting the atmospheric pressure p0. ρm ρm \ hw = hw ; Since = 13.6 ; hw = 0.76 m = (13.6)(0.76) = 10.33 m. ρw ρw 5. PRESSURE DIFFERENCE IN ACCELERATING FLUIDS Consider a beaker filled with some liquid of density p accelerating upwards with an acceleration ay along positive y-direction. Let us draw the free body diagram of a small element of fluid of area A and length dy as shown in figure. Equation of motion for this fluid element is, PA – W – (P + dP)A = (mass)(ay) or –W – (dP) A =(Aρ dy)(ay) dP or (Arg dy) – (dP)A = (Aρ dy)(ay) or = −ρ(g + ay ) dy  P hysics | 9.9 y P + dP (P + dP)A A dy A ay P x PA Figure 9.14 Similarly, if the beaker moves along positive x-direction with acceleration ax, the equation of motion for the fluid element shown in the figure is, PA – (P + dP) A = (mass)(ax) y dP or (dP)A = (Aρ dx)ax Or = −ρax dx But suppose the beaker is accelerated and it (P + dP)A has components of acceleration ax and ay in x ax PA P + dP and y directions respectively, then the pressure P A decreases along both x and y directions. The A above equation ax dx in that case reduces to, x dP dP Figure 9.15 = −ρax and = −ρ(g + ay )  ….. (i) dx dy For surface of a Liquid Accelerated in Horizontal Direction. Consider a liquid placed in a beaker which is accelerating horizontally with an y acceleration ‘a’. Let A and B be two points in the liquid at a separation x in the same horizontal line. As we have seen in this case. dP = −ρa or dP = -ra dx. Integrating this with proper limits, we get dx h1 a PA – PB = pax ….. (ii) h2 A B Further, PA = P0 + rgh1 And PB = P0 + rgh2 h1 − h2 Substituting in Eq. (ii), we get pg(h1 – h2) = pax \ x x x a a Figure 9.16 = = tan θ \ tan θ = g g Note: When ay is not equal to zero then the angle of inclination is given by     dy  (dp)  ax tan= θ = = dx   dp   g + ay       dy   9. 1 0 | Fluid Mechanics Illustration 12: A liquid of density ρ is in a bucket that spins with angular velocity ω as shown is the figure. Show that the pressure at a radial distance r from the axis is ρω2r 2 P = P0 + where P0 is the atmospheric pressure. (JEE ADVANCED) 2 Sol: The net force on the liquid surface in equilibrium is always perpendicular to it as the liquid surface cannot sustain shear stress. Consider a fluid particle P of mass m at coordinates (x, y). From a non-inertial rotating frame of reference, two forces are acting on it. Figure 9.17 (i) Pseudo force (mxω2 ) y (ii) Weight (mg) in the direction shown in figure. Net force on it should be perpendicular to the free P mx² surface (in equilibrium). Hence, P mxω2 xω2 dy xω2 x tan θ = = or = mg F net mg g dx g y x xω2 x2 ω2 ∫ dy ∴= ∫ g ⋅ dx ∴ y = 2g P(x,y) 0 0 Figure 9.18 This is the equation of the free surface of the liquid, which is a parabola. P y r 2 ω2 ρω2r 2 As x = r, y = ∴ P(r)= P0 + ρgy or P(r) = P0 + 2g 2 x=r P(r) Figure 9.19 Hence proved. a Front Rear Illustration 13: An open rectangular tank 5 m × 4 m × 3 m high containing water up to a height of 2 m is accelerated horizontally along the longer side. 3m Water 2m (a) Determine the maximum acceleration that can be given without spilling the water. (b) Calculate the percentage of water split over, if this acceleration is 5m increased by 20%. Figure 9.20 (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 0 walls of the tank. (Take g = 10 m/s2) (JEE MAIN) 3m a0 Sol: As the water column is accelerated towards right in horizontal direction, 2m the free surface will not be horizontal but will be inclined at an angle with y0 the θ horizontal, such that the left edge of the surface is at a higher level than the right edge. This is because the pressure at the left of water column 5m will be more than the pressure at the right of it. Figure 9.21 (a) Volume of water inside the tank remains constant  3 + y0  3 −1   5 × 4 = 5 × 2 × 4 or y0 = 1m \tan q0 = 5 = 0.4 3m  2  a v Since, tan q0 = 0 , therefore a0 = 0.4 g = 4 m/s2 g 5m (b) When acceleration is increased by 20% Figure 9.22  P hysi cs | 9.11 a x a = 1.2 a0 = 0.48 g ∴ tan θ = = 0.48 g Air Now, y = 3 – 5 tan θ = 3 – 5 (0.48) = 0.6 m y (3 + 0.6) 3m 4 × 2×5 − ×5× 4 Fraction of water split over = 2 = 0.1 Water W 2×5× 4 Percentage of water split over = 10% 5m a' (c) a’ = 0.9 g; tan θ’ = = 0.9 Figure 9.23 g 1 Volume of air remains constant -> 4 × yx = (5)(1) × 4 ⇒ Pressure does not change in the air. 2 1 2 Since y = x tan θ’∴ x tan θ ' =5 or x = 3.33 m; y = 3.0m 2 Gauge pressure at the bottom of the (i) Front wall pf = zero (ii) Rear wall pr = (5 tan θ’)rwg = 5(0.9)(103)(10) = 4.5 × 104 Pa 60 rpm Illustration 14: A vertical U-tube with the two limbs 0.75 m apart 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.  (JEE MAIN) 0.5m Sol: Each element of water in the tube is accelerated towards the axis. Along the horizontal 0.75m part of the tube, the pressure will increase gradually as one moves radially away Figure 9.24 from the axis. The extra pressure provides the required centripetal acceleation. Consider a small element of length dr at a distance r from the axis of rotation. Considering the equilibrium of this element. (p + dp) – p = rw2 r dr or dp = rw2 r dr On integrating between 1 and 2 r1 2 ρω2 2 2 p1 – p2 = ρω ∫ r dr = (r − r ) r2 2 1 2 ω2 2 2 (2π)2 or h1 – h2 = [r1 − r2 ] = [(0.5)2 – (0.25)2] = 0.37 m. 2g 2(10) Figure 9.25 6. 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 is 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. Note: The buoyant force is due to the fact that the hydrostatic pressure at different depths is not the same. Buoyant force is independent of: (a) Total volume and shape of the body. (b) Density of the body. 9. 1 2 | Fluid Mechanics 6.1 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 ρ. A body is being acted upon by the forces from all directions. Let us consider a vertical element of height h and cross-sectional area dA.The force acting on the upper surface of the element is F1 (downward) and that on the lower surface is F2 (upward). Since F2> F1, therefore, the net upward force acting on the element is dF = F2 – F1. It can be easily seen that F1 = (rgh1)dA and F2= (rgh2)dA. So dF = rg(h) dA Also, h2 – h1 = h and h(dA) = dV \ The net upward force is F = ∫ ρgdV = ρVg Hence, for the entire body, the buoyant force is the weight of the volume of the fluid displaced. Note: Buoyant force acts on the centre of gravity of the displacement liquid. This point is called centre of Buoyancy. MASTERJEE CONCEPTS The fluid exerts force on the immersed part of the body from all directions. The net force experienced by every vertical element of the body is in the upward direction. A uniform body floats in a liquid if density of the body is less than or equal to the density of the liquid and sinks if density of the uniform body is greater than that of the liquid. B Rajiv Reddy (JEE 2012, AIR 11) 6.1.1 Detailed Explanation 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 Apparent weight of a body immersed in a liquid = w – w0, where ‘w’ is the true weight of the body and w0 is the apparent loss in weight of the body, when immersed in the liquid. 6.1.2 Buoyant Force in Accelerating Fluids  Suppose a body is dipped inside a liquid of density ρL placed in an elevator moving with acceleration a. The buoyant force F in this case becomes, F = VρL geff ;   Here, geff = | g − a | Illustration 15: An iceberg with a density of 920 kgm-3 floats on an ocean V of density 1025 kgm-3. What fraction of the iceberg is visible?(JEE MAIN) Above water Sol: The buoyant force on the iceberg will be equal to its weight. The w(V0-V) buoyant force is equal to the weight of water displaced by the iceberg, i.e. the weight of volume of water equal to the volume of iceberg immersed. wV0 Let V be the volume of the iceberg above the water surface, then the volume under inside is V0 – V. Under floating conditions, the weight (ρIV0g) V0-V of the iceberg is balanced by the buoyant force rw(V0 – V)g. Under water Thus, ρIV0g = rw(V0 – V)g Figure 9.26 or rwV = (rw – ρI)V0  P hysi cs | 9.13 V ρ w − ρI or = V0 ρw V 1025 − 920 Since, r w = 1025 kg m-3 and ri = 920 kg m3, therefore, = = 0.10 V0 1025 Hence 10% of the total volume is visible. Illustration 16: 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? (JEE MAIN) Sol: Apply Archemides principle. Let W = actual weight of the crown and W’ = apparent weight of the crown ρ = density of crown, r0 = density of water. The buoyant force is given by FE = W – W’ or W r0Vg = W – W’. Since W = rVg, therefore, V =. Eliminating V from the above equation, we get ρg ρ0 W (10)3 (25) ρ=. Here W = 25 N; W’ = 22 N; r0 = 103 kg m-3 ; ρ = = 9.3 × 103 kg m-3. W − W' 25 − 22 Illustration 17: The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in figure is T0 when the system is at rest. What will be the tension in the string if the system has an upward acceleration a? (JEE MAIN) Sol: The weight and tension force on the block are balanced by the buoyant force on it. When the system is accelerated upwards, the effective value of g is increased. Let m be the mass of block. Initially for the equilibrium of block, F = T0 + mg ….(i) Figure 9.27 Here, F is the up thrust on the block. When the lift is accelerated upwards, geff becomes g + a instead of g. g+a Hence F' = F  ...(ii)  g  From Newton’s second law, F’ – T – mg = ma ...(iii)  a Solving equations (i), (ii) and (iii), we get = T T0  1 +   g Figure 9.28 Illustration 18: An ice cube of side 1 cm is floating at the interface of kerosene and water in beaker of base area 10 cm2. The level of kerosene is just covering Kerosine the top surface of the ice cube. S=0.8 (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. (JEE ADVANCED) Figure 9.29 Sol: Apply Archemedes principle. Sum of the buoyant forces by kerosene and water will be equal to the weight of the ice cube. (a) Condition of floating 0.8 rwghk + rwghw = 0.9 rwgh or 0.8 hk + hw = (0.9)h  … (i) Where hk and hw are the submerged depths of the ice in the kerosene and water, respectively. 9. 1 4 | Fluid Mechanics Also hk + hw = h... (ii) Here it is given that h = 1 cm Solving equations (i) and (ii), we get hk = 0.5 cm, hw = 0.5 cm m heat → 0.9 cm3 (b) 1 cm3  Ice (water) 0.5 0.9 − 0.5 0.4 Fall in the level of kerosene Dhk = ; Rise in the level of water Dhw = = A A A 0.1 0.1 Net fall in the overall level Dh = = = 0.01 cm = 0.1 mm. A 10 6.2 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 centre of gravity. The buoyant force acts at the centre of gravity of the displaced liquid. This is called the centre of buoyancy. Under equilibrium condition, the centre of gravity G and the centre of buoyancy B lie along the vertical axis of the body as shown in the figure(s). Fa M Fb W W W (a) (b) Figure 9.30 (a) The buoyant force acts at the centre 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 metacentre M. In a stable boat, M is above the centre of gravity of the boat. When the body tilts to one side, the centre of buoyancy shifts relative to the centre 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 centre of gravity. The line of action of the buoyant force crosses the axis of the body at the point M, called metacentre. If G is below M, the torque will tend to restore the body to its equilibrium position. If G is above M, the torque will tend to rotate the body away from its equilibrium position and the body will be unstable. Illustration 19: A wooden plank of length 1 m and uniform cross section is hinged at one end to the bottom of a tank as shown in the figure. The tank is filled with water up to a height of 0.5 m. The specific gravity of the plank is 0.5. Find the angle θ that mg the plank makes with the vertical in the equilibrium position. (Exclude the case θ = 0)  (JEE ADVANCED) Figure 9.31 Sol: The net torque about the hinge due the weight of the plank and due to the buoyant force acting on the plank should be zero. The forces acting on the plank are shown in the figure. The height of water level is 0.5m. The length of the plank is 1.0 = 2 . We have OB = . The buoyant force F acts through the mid-point of the dipped part OC of the plank. OC  We have OA = = ; Let the mass per unit length of the plank be ρ. 2 2cos θ  P hysi cs | 9.15    Its weight mg = 2  rg; The mass of the part OC of the plank =  ρ.  cos θ  1 1 2ρ 2ρg The mass of water displaced = ρ= ; The buoyant force F is, therefore, F =. 0.5 cos θ cos θ cos θ Now, for equilibrium, the torque of mg about O should balance the torque of F about O.  2ρ     1 1 So, mg (OB) sin θ = F(OA) sin θ or (2  ρ)  =    or cos θ = or cos θ = 2 , or θ = 45º  cos θ  2cos θ  2 2 6.3 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 l × b filled with water to a height H as shown in figure The force acting at the base of the container is given by Fb = p × (area of the base) Pressure is same everywhere at the base and is equal to rgH. Therefore, Fb Fb = rgH(lb) = ρ glb H Since, lbH = V (volume of the liquid).Thus, Fb = rgV = weight of the liquid inside the vessel. A fluid contained in a vessel exerts forces on the boundaries. Unlike the base, l the pressure on the vertical wall of the vessel is not uniform but increases Figure 9.32 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 b and thickness dh at depth h from the free surface. The liquid pressure at this position is given by p = rgh. The force at the element is dF = p(dbh) = rgbh dh; H 1 F 1 The total force is F = rgb ∫ h dh= ρgbH2. The total force acting per unit width of the critical walls is = ρgH2 O 2 b 2 H 1 F ∫0 The point of application (the centre of force) of the total force from the free surface is given by hc = h dF H Where ∫ h dF is the moment of force about the free surface. 0 H H H 1 Here ∫ h dF = ρgb ∫ h2dh = ∫ h(ρ gbh dh) = ρgH3 ; 0 0 0 3 1 2 Since F = rgbH2 , therefore, hc = H 2 3 Illustration 20: Find the force acting per unit width on a plane wall inclined at an angle θ with the horizontal as shown in the figure.  (JEE MAIN) h=y sin dF y Sol: The pressure at each point on the wall will be different, H depending on the height. Find pressure on a small element, and use dy the method of integration. Consider a small element of thickness dy at a distance y measured Figure 9.33 along the wall from the free surface. There pressure at the position of the element is p = rgh = rgy sin θ. The force given by dF = p(b dy) = rgb(y dy) sin q H/sin θ H/sin θ F  y2  The total force per unit width b is given by =ρgsin θ. ∫ y dy = ρg sin θ   b 0  2  0 9. 1 6 | Fluid Mechanics F 1 H2 Or = ρg b 2 sin θ 1 Note: That the above formula reduces to rgH2 for a vertical wall (θ = 90º) 2 6.4 Oscillations of a Fluid Column The initial level of liquid in both the columns is the same. The area of cross-section of the tube is uniform. If the liquid is depressed by x in one limb, it will rise by x along x the length of the tube is the other limb. Here, the restoring force is provided by the x hydrostatic pressure difference. ∴ F = − ( ∆P ) A = − (h1 + h2 ) ρgA = − ρgA ( sin θ1 + sin θ2 ) x 1 2 suppose, m is the mass of the liquid in the tube. Then, ma = −ρgA ( sin θ1 + sin θ2 ) x Figure 9.34 Since, F or a is proportional to –x, the motion of the liquid column is simple harmonic in nature, time period of which is given by, x m Τ = 2π or Τ = 2π a ρgA ( sin θ1 + sin θ2 ) 6.5 Oscillations of a Floating Cylinder Consider a wooden cylinder of mass m and cross-sectional area A floating in a liquid of density ρ. At equilibrium, the cylinder is floating with a depth h submerged [See Fig. 8.35]. If the cylinder is pushed downwards by a small distance y and then released, it will move up and down with SHM. It is desired to find the time period h h+y and the frequency of oscillations. According to the principle of flotation, the weight of the liquid displaced by the immersed part of the body is equal to the weight of the body. Therefore, at equilibrium, Figure 9.35 Weight of cylinder = Weight of liquid displaced by the immersed part of cylinder or mg= ( ρ Ah) g ∴ Mass of cylinder, m= ρ Ah When the cylinder is pushed down to an additional distance y, the restoring force F (upward) equal to the weight of additional liquid displaced acts on the cylinder. ∴ Restoring force, F= - (weight of additional liquid displaced) or F =− ( ρ A y ) g The negative sign indicates that the restoring force acts opposite to the direction of the displacement. F − (ρ A y ) g g Acceleration a of the cylinder is given by a = = = −  y...(i) … (i) m ρ Ah h  Since g/h is constant, a α − y Thus the acceleration a of the body (wooden cylinder) is directly proportional to the displacement y and its direction is opposite to the displacement. Therefore, motion of the cylinder is simple harmonic. h ∴ Time period T = 2π … (ii) g  1 1 g ∴ Frequency = f =  … (iii) T 2π h These very interesting results show that time period and frequency have the same form as that of simple pendulum. The submerged depth at equilibrium takes the place of the length of the pendulum.  P hysi cs | 9.17 7. FLUID DYNAMICS In the 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 a 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 take several assumptions: (i) The fluid is non viscous (ii) The flow is steady (iii) The flow is non rotational (iv) The fluid is incompressible. 7.1 Equation of Continuity B It states that for streamlined motion of the liquid, the volume of liquid flowing per unit time is constant through different cross-sections of A the container of the liquid. Thus, if v1 and v2 are velocities of fluid at respective points A and B of areas of cross-sections a1 and a2 and r1 and r2 be the densities respectively. Then the equation of continuity is a1 given by r1a1v1 = r2a2v2 ... (i) v1  a2 v2  If the same liquid is flowing, then ρ1 =ρ2 ; then the equation (i) can be written Figure 9.36 As a1v1 = a2v2 ...(ii) ⇒ av = constant ⇒ v ∝ 1/a MASTERJEE CONCEPTS Equation of continunity repersents the law of conservation of mass of moving fluids. a1v1r1 = a2v2r2 (General equation of continuity) This equation is applicable to actual liquids or to other fluids which are not incompressible. Yashwanth Sandupatla (JEE 2012, AIR 821) Illustration 21: Water is flowing through a horizontal tube of non-uniform cross-section. At a place, the radius of the tube is 1.0 cm and the velocity of water is 2 m/s. What will be the velocity of water, where the radius of the pipe is 2.0 cm? (JEE MAIN) Sol: Apply the equation of continuity. Where area of cross-section is larger, the velocity of water is lesser and vice- versa. 2 A   πr 2  r  Using equation of continuity, A1v1 = A2v2 ; v 2 =  1  v1= or v 2 = 1  v1  1  v1  A2   πr 2   r2   2   1.0 × 10−2  Substituting the value, we get v 2 =   or v2 = 0.5 m/s  2.0 × 10−2    Illustration 22: Figure shows a liquid being pushed out of a tube by pressing a piston. The area of cross-section of the piston is 1.0 cm2 and that of the tube at the outlet is 20 mm2. If the piston is pushed at a speed of 2 cm-s-1, what is the Figure 9.37 speed of the outgoing liquid? 9. 1 8 | Fluid Mechanics Sol: Apply the equation of continuity. Where area of cross-section is larger, the velocity of liquid is lesser and vice-versa. From the equation of continuity A1v1 = A2v2 or (1.0 cm2) (2 cm s-1) = (20 mm2) v2 1.0 cm2 or v2 = × 2 cm s−1 20 mm2 100 mm2 = × 2 cm s−1 = 10 cm s−1 20 mm2 SHM of fluids in tubes: Tubes form angles θ1 and θ2 with the horizontal. x m T = 2π x ρgA ( sin θ1 + sin θ2 ) 1 2 m is total mass of fluid in tubes, A is area of cross – section ρ is density of fluid. Figure 9.38 8. BERNOULLI’S THEOREM When a non-viscous and an incompressible fluid flows in a streamlined motion from one place to another in a container, then the total energy of the fluid per unit volume is constant at every point of its path. Total energy = pressure energy + Kinetic energy + Potential energy 1 = PV + Mv2 + Mgh 2 Where P is pressure, V is volume, M is mass and h is height from a reference level. 1 ∴ The total energy per unit volume = P + rv2 + rgh 2 Where ρ is density. Thus if a liquid of density ρ, pressure P1 at a height h1 which flows with velocity v1 to another point in streamline motion Figure 9.39 where the liquid has pressure P2, at height h2 which flows with velocity v 2, 1 1 then P1 + ρv12 + ρgh1= P2 + ρv 22 + ρgh2 2 2 8.1 Derivations 8.1.1 Pressure Energy If P is the pressure on the area A of a fluid, and the liquid moves through a distance due to this pressure, then Pressure energy of liquid = work done = force × displacement = PAl The volume of the liquid is Al. PAl ∴ Pressure energy per unit volume of liquid = =P Al 8.1.2 Kinetic Energy 1 If a liquid of mass m and volume V is flowing with velocity v, then the kinetic energy is mv2. 2  P hysi cs | 9.19 1m 2 1 2 ∴ Kinetic energy per unit volume of liquid. =  v = ρv. Here, ρ is the density of liquid. 2 V  2 8.1.3 Potential energy If a liquid of mass m is at a height h from the reference line (h = 0), then its potential energy is mgh. ∴ Potential m energy per unit volume of the liquid =   gh = rgh v 1 Thus, the Bernoulli’s equation P + rv2 + rgh = constant 2 This can also be written as: Sum of total energy per unit volume (pressure + kinetic + potential) is constant for an ideal fluid. MASTERJEE CONCEPTS P v2 is called the ‘pressure head’, the velocity head and h the gravitational head. ρg 2g GV Abhinav JEE 2012, AIR 329 Intresting takeaway is the SI unit of each of these is meter (m). Illustration 23: Calculate the rate of flow of glycerin of density 1.25×103 kg/m3 through the conical section of a pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 N/m2.  (JEE MAIN) Sol: Apply the equation of continuity. Where area of cross-section is larger, the velocity of fluid is lesser and vice-versa. From continuity equation, A1v1 = A2v2 2 v1 A2 πr22  r2   0.04  4 or = = =  =   =  ... (i) v 2 A1 πr 2  r1   0.1  25 1 Figure 9.40 1 1 From Bernoulli’s equation , P1 + ρv12 = P2 + ρv 22 2 2 2 × 10 or v 22 − v12 = = 1.6 × 10−2 m2 / s2 ... (ii) 3 1.25 × 10 Solving equations (i) and (ii), we get v2 = 0.128 m/s \ Rate of volume flow through the tube Q = A2v2 = (pr22) v2= π (0.04)2(0.128) = 6.43 × 10–4 m3/s Illustration 24: Figure shows a liquid of density 1200 kg m–3 flowing steadily in a tube of varying cross section. The cross section at a point A is 1.0 cm2 and that at B is 20 mm2, the points A and B are in the same horizontal plane. The speed of the liquid at A is 10 cm s-1. Calculate the difference in pressure at A and B.   (JEE ADVANCED) Figure 9.41 Sol: Apply the equation of continuity. Where area of cross-section is larger, the velocity of fluid is lesser and vice- versa. 9. 2 0 | Fluid Mechanics From equation of continuity. The speed v2 at B is given by, A1v1 = A2v2 1.0cm2 or (1.0 cm2) (10 cm s-1) = (20 mm2)v2 or v2 = ×10cm s−1 =50 cm s−1 20mm2 1 1 By Bernoulli equation, P1 + ρgh1 + ρv12= P2 + ρgh2 + ρv 22 2 2 1 2 1 2 1 Here h1 = h2. Thus P1 – P2 = ρv − ρv = × (1200 kg m−2 )(2500 cm2 s−2 − 100 cm2 s−2 ) 2 2 2 1 2 = 600 kg m-3 × 2400 cm2 s-2 = 144 Pa 8.2 Application Based on Bernoulli’s Equation 8.2.1 Venturimeter Figure shows a venturimeter used to measure flow speed in a pipe of non-uniform cross- section. We apply Bernoulli’s equation to the h wide (point 1) and narrow (point 2) parts of the pipe, with h1 = h2 1 1 P1 + ρv12 = P2 + ρv 22 p1 2 2 p2 A1 v1 From the continuity equation v 2 = v1 v2 A2 A2 Substituting and rearranging, H A1 Venturimeter 1  A2  we get P1 − P2 = ρv12  1 − 1  …(i) Figure 9.42 2  A2   2  The pressure difference is also equal to rgh, where h is the difference in liquid level in the two tubes. 2gh Substituting in equation (i), we get v1 = 2  A1    − 1  A2  MASTERJEE CONCEPTS Because A1 is greater than A2, v2 is greater than v1 and hence the pressure P2 is less than P1. dV 2gh The discharge or volume flow rate can be obtained as, = A= 1 v1 A1 dt  A1  2  A  − 1  2 Anurag Saraf (JEE 2011, AIR 226)  P hysi cs | 9.21 9. TORRICELLI’S THEOREM It states that the velocity of efflux of a liquid through an orifice is equal to that velocity which a body would attain in falling from a height from the free surface of a liquid to the orifice. If h is the height of the orifice below the free surface of a liquid and g is acceleration due to gravity, the velocity of efflux of liquid = v= 2gh. Total energy per unit volume of the liquid at the surface = KE + PE + Pressure energy = 0 + rgh + P0...(i) and total energy per unit volume at the orifice = KE + PE + Pressure 1 Figure 9.43 energy = ρv 2 + 0 + P0 2 Since total energy of the liquid must remain constant in steady flow, in accordance with Bernoulli’s equation, 1 2 we have rgh + P0 = ρv + P0 or v= 2gh 2 Range = velocity × time ; R = Vx × time = 2gh × 1 1 2 2(H − h) Now, H–h= gt ⇒ t =. From equation (i), 2 g 2(H − h) R= 2gh × = 2h × 2(H − h) × h(H − h)2 g \=R 2 h(H − h) dR H − 2h H Range is max. if = 0 ⇒2× 0 ⇒ H – 2h = 0 ⇒ h = = dh 2 h(H − h) 2 MASTERJEE CONCEPTS R h = RH – h Rh = 2 h(H − h) RH–h = 2 h(H − h) i.e. Range would be the same when the hole is at a height h or at a height H – h from the ground or from the top of the beaker. H Figure 9.44 R is maximum at h = and Rmax =H. 2 Vijay Senapathi (JEE 2011, AIR 71)  9. 2 2 | Fluid Mechanics 9.1 An Expression for the Force Experienced by the Vessel The force experienced by the vessel from which liquid is coming out. dp d d F= (Rate of change of momentum) = (mv) = (ρAvtv) dt dt dt F = ρAv 2 Where ρ = It is the density of the liquid. A = It is the area of hole through which liquid is coming out. 9.2 Time taken to Empty a Tank Consider a tank filled with a liquid of density ρ up to a height H. A small hole of area of cross section a is made at the bottom of the tank. The area of cross-section of the tank is A. Let at some instant of time the level of liquid in the tank be y. Velocity of efflux at this instant of time would be, v = 2gy.  dV  At this instant volume of liquid coming out of the hole per second is  1 .  dt   dV  Volume of liquid coming down in the tank per second is  2 .  dt  t 0 dV1 dV2  dy   dy  A −1/2 dt = dt ; \ av  dt  = A  −  Or = A  −  ∴ a 2gy  dt  ∫ dt = − ∫y a 2g H dy 0 2A A 2H ∴t = [ y ]H0 = a 2g a g Illustration 25: A tank is filled with a liquid up to a height H. A small hole is made at the bottom of this tank. Let t1 be the time taken to empty first half of the tank and t2 the time taken to empty rest half of the tank. t1 Then find. (JEE MAIN) t2 Sol: This problem needs to be solved by method of integration. Substituting the proper limit in equation (i), derived in the theory, we have t1 A H/2 −1/2 2A 2A  H ∫ dt = − a ∫ 2g H y dy Or t1 = a 2g y ]H [= H/2 Or  H− a 2g   2  0 A H Or t1 = ( 2 − 1)  …(ii) a g t2 A 0 A H Similarly ∫0 dt = − ∫ y −1/2 dy Or t2 = ... (iii) a 2g H/2 a g t1 t1 From equations (ii) and (iii), we get = 2 − 1 Or = 0.414 t2 t2  P hysi cs | 9.23 MASTERJEE CONCEPTS From here we see that t1< t2. This is because inititally the pressure is high and the liquid comes out with greater speed. Ankit Rathore (JEE Advanced 2013, AIR 158) 10. VISCOSITY When a liquid moves slowly and steadily on a horizontal surface, its layer in contact with the fixed surface is stationary and the velocity of the layers increase with the distance from the fixed surface. Consider two layers CD and MN of a liquid at distances x and x + dx from the fixed surface AB having velocities v  dv  and v + dv respectively as shown in the figure. Here   denotes the rate of change of velocity with distance and  dx  is known as velocity gradient. The tendency of the upper layer is to accelerate the motion and the lower layer tries to retard the motion of upper layer. The two layers together tend to destroy their relative motion as if there is some backward dragging force acting tangentially on the layers. To maintain the motion, an external force is applied to overcome this backward drag. Hence the property of a liquid virtue of which it opposes the relative motion between its different layers is known as viscosity. dv The viscous force is given by F = −ηA dx Where η is a constant, called the coefficient of viscosity. The SI unit of η is N-s/m2. It is also called decapoise or Pascal second. Thus, 1 decapoise = N-s/m2 = 1 Pa-s = 10 poise. Dimensions of h are [ML-1T-1] Figure 9.45 MASTERJEE CONCEPTS The negative sign in the above equation shows that the direction of viscous force F is opposite to the direction of relative velocity of the layer. Viscous force depends upon the velocity gradient whereas the mechanical frictional force is independent of the velocity gradient. Vaibhav Gupta (JEE 2009, AIR 54) 10.1 Effect of Temperature In case of liquids, coefficient of viscosity decreases with increase of temperature as the cohesive forces decrease with increase of temperature. Illustration 26: A plate of area 2 m2 is made to move horizontally with a speed of 2 m/s by applying a horizontal tangential force over the free surface of a liquid. The depth of the liquid is 1 m and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid is 0.01 poise. Find the tangential force needed to move the plate.  (JEE MAIN) 9. 2 4 | Fluid Mechanics Sol: Apply the Newton’s formula for the frictional force between two layers v=2 m/s of a liquid. F ∆v 2−0 m/s Velocity gradient = = =2 1m ∆y 1−0 m From Newton’s law of viscous force, ∆v |F| = ηA = (0.01 × 10-1)(2)(2) = 4 × 10-3 N. Figure 9.46 ∆y So, to keep the plate moving, a force of 4 × 10-3 N must be applied. 10.2 Stokes’ Law and Terminal Velocity Stokes established that the resistive force or F, due to the viscous drag, for a spherical body of radius r, moving with velocity V, in a medium of coefficient of viscosity η is given by F = 6 pη rV Mg Figure 9.47 10.3.1 An Experiment for Terminal Velocity Consider an established spherical body of radius r and density ρ falling freely from rest under gravity through a fluid of density σ and coefficient of viscosity η. When the body acquires the terminal velocity V W = Ft+ 6πηrV ; 4 2 r 2 (ρ − σ)g 6πηrV= πpr3 (ρ − σ)g ⇒ V = 3 9 η Note: From the above expression we can see that terminal velocity of a spherical body is directly proportional to the densities of the body and the fluid (ρ – σ). If the density of the fluid is greater than that of the body (.i.e. σ>ρ), the terminal velocity is negative. This means that the body instead of falling, moves upward. This is why air bubbles rise up in water. Illustration 28: Two spherical raindrops of equal size are falling vertically through air with a terminal velocity of 1 m/s. What would be the terminal speed if these two drops were to coalesce to form a large spherical drop?   (JEE MAIN) Sol: Use the formula for terminal velocity for spherical body. vT ∝ r2. Let r be the radius of small rain drops and R the radius of large drop. 4 2 4  Equating the volume, we have πR = 2  πr3  3  3  2 R vT 'R  \ R = (2). r 1/3 or = (2)1/3 \ =   = (2)2/3 r vT  r  \ vT’ = (2)2/3 vT = (2)2/3 (1.0) m/s = 1.587 m/s.  P hysi cs | 9.25 Illustration 29: An air bubble of diameter 2 mm rises steadily through a solution of density 1750 kg m-3 at the rate of 0.35 cm s-1. Calculate the coefficient of viscosity of the solution. The density of air is negligible. (JEE MAIN) Sol: As the air bubble rises with constant velocity, the net force on it is zero. 4 3 The force of buoyancy B is equal to the weight of the displaced liquid. Thus B = pr sg. 3 This force is upward. The viscous force acting downward is F = 6 π hrv. The weight of the air bubble may be neglected as the density of air is small. For uniform velocity 4 3 2r 2 σg 2 × (1 × 10−3 m)2 × (1750 kg m−3 )(9.8 ms−2 ) F = B or, 6 phrv = pr σg or, η = = ≈ 11 poise. 3 9v 9 × (0.35 × 10−2 ms−1 ) This appears to be a highly viscous liquid. 10.3 Stream Line Flow When liquid flows in such a way that the velocity at a particular point is the same in magnitude as well as in direction. As shown in figure every molecule should have the same velocity at A, if it crossed from that point. Notice that the velocity at the point B will be different from that of A. But every molecule which reaches at the point B, gets the velocity of the point B. Figure 9.48 10.4 Turbulent Flow When the motion of a particle at any point varies rapidly in magnitude and direction, the flow is said to be turbulent or beyond critical velocity. If the paths and velocities of particles change continuously and haphazardly, then the flow is called turbulent flow. 10.5 Critical Velocity and Reynolds Number When a fluid flows in a tube with small velocity, the flow is steady. As the velocity is gradually increased, at one stage the flow becomes turbulent. The largest velocity which allows a steady flow is called the critical velocity. Whether the flow will be steady or turbulent mainly depends on the density, velocity and the coefficient of viscosity ρvD of the fluid as well as the diameter of the tube through which the fluid is flowing. The quantity N = is called η the Reynolds number and plays a key role in determining the nature of flow. It is found that if the Reynolds number is less than 2000, the flow is steady. If it is greater than 3000, the flow is turbulent. If it is between 2000 and 3000, the flow is unstable. 11. SURFACE TENSION The properties of a surface are quite often marked 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 shows 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. A surface layer is approximately 10-15 molecular diameters. The force between two molecules decreases as the separation between them increases. The force becomes Figure 9.49 9. 2 6 | Fluid Mechanics 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 AB 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 AB. 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 behave 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. Figure 9.50 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 T of the liquid as T = F/  The SI unit of surface tension is N/m. Note: The surface tension of a particular liquid usually decreases as temperature increases. To wash clothing thoroughly, water must be forced through the tiny spaces between the fibers. This requires increasing the surface area of the water, which is difficult to do because of surface tension. Hence, hot water and soapy water is better for washing. MASTERJEE CONCEPTS Surface tension acts over the free surface of a liquid only and not within the interior of the liquid. Due to surface tension the insects can walk on liquid surface. Vaibhav Krishnan (JEE 2009, AIR 22) Illustration 30: 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.  (JEE MAIN) Sol: Force = Surface tension×length of the surface Length of the surface = circumference of the circular plate = 2pr = (8π) cm Required force = T × L = 72 × 8π = 1810 dyne. 12. SURFACE ENERGY When the surface area of a liquid is increased, the molecules from the interior rise to the surface. This requires work against force of attraction of the molecules just below the surface. This work is stored in the form of potential energy. Thus, the molecules in the surface have some additional energy due to their position. This additional energy per unit area of the surface is called ‘surface energy’. The surface energy is related to the surface tension as discussed below: Let a liquid film be formed on a wire frame and a straight wire of length  can slide on this wire frame as shown in figure. The film has two surfaces and both the surfaces are in Figure 9.51 contact with the sliding wire and hence, exert forces of surface tension on it. If T be the surface tension of the solution, each surface will pull the wire parallel to itself with a force T . Thus, net force on the wire due to both the surfaces is 2T . One has to apply an external force F equal and opposite to it to keep the wire in equilibrium. Thus, F = 2T   P hysi cs | 9.27 Now, suppose the wire is moved through a small distance dx, the work done by the force is, dW = F dx = (2T  )dx But (2  )(dx) is the total increase in the area of both the surfaces of the film. Let it be dA. Then, dW dW = T da or T = dA Thus, the surface tension T can also be defined as the work done in increasing the surface area by unity. Further, since there is no change in kinetic energy, the work done by the external force is stored as the potential energy of the new surface. dU ∴T= (as dW = dU) dA Thus, the surface tension of a liquid is equal to the surface energy per unit surface area. Illustration 31: How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm? Surface tension of soap solution is 3.0 × 10-2 N/m. (JEE MAIN) Sol: Work done will be equal to the increase in the surface porential energy, which is surface tension multiplied by increase in area of surface of liquid. Soap bubble has two surfaces. Hence, W = T ∆A Here, ∆A = 2[4p{(2.5×10–2)2 – (1.0×10–2)2}] = 1.32 × 10-2 m2 W = (3.0×10–2)(1.32×10–2)J = 3.96×10–4J Illustration 32: Calculate the energy released when 1000 small water drops each of same radius 10–7m coalesce to form one large drop. The surface tension of water is 7.0×10-2 N/m.  (JEE MAIN) Sol: Energy released will be equal to the loss in surface potential energy. Let r be the radius of smaller drops and R of bigger one. 4 3 4  Equating the initial and final volumes, we have πR = (1000)  πr 3  3 3  R = 10r = (10)(10–7) m = 10-6 m. Further, the water drops have only one free surface. Therefore, ∆A = 4pR2 – (1000)(4pr2) = 4p[(10–6)2 – (103)(10–7)2] = –36π(10–12)m2 Here, negative sign implies that surface area is decreasing. Hence, energy is released in the process. U = T[∆A] = (7×10–2)(36p×10–12)J = 7.9×10–12J 13. 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 the 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. 9. 2 8 | Fluid Mechanics 13.1 Excess Pressure Inside a Drop Figure 9.52 To obtain a relation between the excess of pressure and the surface tension, consider a water drop of radius r and surface tension T. 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 force acting on it are: (a) Force due to surface tension distributed along the circumference of the section. (b) Outward thrust 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 2pr×T. To find the resultant of the other type of distributed forces, consider an elementary area DS of the surface. The outward thrust on DS = pDS where p is the excess of the pressure inside the bubble. If this thrust makes an angle θ with the vertical, then it is equivalent to DSp cos θ along the vertical and DSp sin θ along the horizontal. The resolved component DSp sin θ is infective as it is perpendicular to the resultant force due to surface tension. The resolved component DSp cos θ is equal to balancing the force due to surface tension The resultant outward thrust = ΣDSp cos θ = pΣDS cos θ = pΣDS cos q = pΣDS’ where DS’ = DS cos θ = area of the projection of DS on the horizontal dividing plane = p × pr2 (  DS’ = pr2) 2T For equilibrium of the bubble we have pr2 p = 2pr T or p = r MASTERJEE CONCEPTS If we have an air bubble inside a liquid, a single surface is formed. There is air on the concave side and liquid on the convex side. The pressure in the concave side (that is in the air) is greater than P2 P1 2T the pressure in the convex side (that is in the liquid) by an amount. R 2T ∴ P2 − P1 = Figure 9.53 R Nivvedan (JEE 2009, AIR 113) 13.2 Excess Pressure Inside Soap Bubble A soap bubble consists of two spherical surface films with a thin layer of liquid between them. P'− P1 = 2S / R where R is the radius of the bubble.  P hysi cs | 9.29 As the thickness of the bubble is small on a macroscopic scale, the difference in the radii P1 of the two surfaces will be negligible. P’ Similarly, looking at the inner surface, the air is on the concave side of the surface, hence P2 − P' = 2S / R. Adding the two equations, P2 − P1 = 4S / R P2 Illustration 33: What should be the pressure ins