Properties of Fluids

CHAPTER Properties of Fluids 1 1.1 INTRODUCTION A fluid can be defined as a substance which deforms or yields continuously when shear stress is applied to it, no matter how small it is. Fluids can be subdivided into liquids and gases. Liquids occupy a certain volume and have a free surface. Gases have a tendency to expand and fill the container in which they are kept; they do not have a free surface. Gases, when subjected to normal stress change their volume considerably. Liquids can be compressed to a small extent. Solids are least compressible. Solids when subjected to shear stress deform until internal resistance to deformation equals the externally applied stress. Some of the examples of fluids are water, air, hydrogen gas, oils, paint, blood, glycerine, brine, honey, etc. In this book SI system (Systeme Internationale d unites) of units is adopted, in which the following are the units of basic quantities that are used in fluid mechanics: Length metre (m) Mass kilogramme (kg) Time second (s) Thermodynamic temperature kelvin (K) Temperature celsius ( C) A unit of force is newton (N) which is the force that produces 1 m/s acceleration is a mass of 1 kg. Unit of work is joule (J) which is the work done when 1.0 N force acts through a distance of 1.0 m. Unit of power is watt (W) which is the power necessary for 1 J of work in 1 s. 1J 1Nm 1 W = = 1s 1s Unit of frequency is hertz (H) having dimension of s 1. 1. MASS DENSITY, SPECIFIC WEIGHT, SPECIFIC VOLUME, RELATIVE DENSITY, PRESSURE Mass density, ρ (Rho) is the mass per unit volume;

Fluid Mechanics through Problems r= lim D V Æ0 DM D V and r= f ( x, y, z, T)...(1.1) where x, y, z are co-ordinates of the point in flow field and T is temperature. At 0 C and atmospheric pressure, for water: ρ = 998 kg/m air: ρ = 1.08 kg/m Specific weight γ (Gamma) is weight per unit volume; γ = ρg N/m...(1.) Specific volume is volume per unit weight and hence 1 v = m N...(1.) g Relative density is the ratio of mass density to mass density of pure water at standard pressure of 101.5 N/m and temperature of 4 C. Pressure is force acting on unit area normal to it and has unit of N/m ; 1. VISCOSITY p = lim D A Æ0 DF D A...(1.4) Viscosity is that property of fluid by which it offers resistance to shear acting on it. According to Newton s law of viscosity the shear F acting between two layers of fluid is proportional to difference in their velocities Δu and area A, and inversely proportional to the distance Δy between them (see Fig. 1.1). F Area A U y u + u u b a b a Fluid Therefore or the shear stress τ is Fig. 1.1 Shear and velocity distribution Du F =ma Dy t= F du A =m dy...(1.5)

where μ (Mu) is the constant of proportionality with dimensions of [ ] [ F A] ie [ du dy] m m m N sm Ns kg m = =,.., or ms Properties of Fluids One gm/cms dynamic viscosity is known as poise (P). du/dy gives the angular velocity of line ab or it is the rate of angular deformation. m Coefficient of kinematic viscosity, n=...(1.6) r ν (Nu) has dimensions of m /s. One cm /s kinematic viscosity is known as stoke (S). μ or ν = f (p, T) Variation of viscosity of liquids with pressure is very small and can be neglected. The variation with temperature is given by μ T = Ae B/T...(1.7) where μ T is dynamic viscosity at absolute temperature T, A and B are constants. Viscosity of liquids decreases with increase in temperature. In the presence of suspended matter the viscosity of liquids increases according to the law μ m /μ = 1 + ac ν...(1.8) where μ m is the viscosity of liquid when suspended matter concentration in absolute volume is C ν and μ is the viscosity of clear liquid; a is.5 according to Einstein for C ν upto 0.0 and 4.5 for higher concentration as recommended by Ward. Viscosity of gases increases with increase in temperature and can be given by the formula proposed by Sutherland bt m T =...(1.9) 1 + at where a and b are constants for a given gas. Fluids are classified according to the relation between shear τ and rate of angular deformation: τ = 0 Ideal fluids du t=m dy 1 Newtonian fluids Êduˆ t= const +m Á Ëdy Êduˆ t= const +m Á Ëdy n Ideal plastics of Bingham plastics Thyxotropic fluids Êduˆ t=m Ë Á dy n Non Newtonian fluids For non Newtonian fluids, if n is less than unity, they are called pseudo-plastics while fluids in which n is greater than unity are known as dilatants. These are shown in Fig. 1..

4 Fluid Mechanics through Problems Thyxotropic Ideal plastic Non Newtonian (Dilatants) Yield stress 1 Newtonian Non Newtonian (Pseudoplastics) Ideal du/dy Fig. 1. Rheological classification of fluids 1.4 BULK MODULUS OF ELASTICITY Bulk modulus of elasticity of fluid, E, is defined as follows: dp E =- v...(1.10) dv where v is the specific volume and dv is the change in v due to increase in pressure dp. E is expressed in N/m or kn/m (kilo newtons per m ). 1.5 GAS LAWS Compression or expansion of gases can occur either under isothermal condition (i.e. at constant temperature) or under adiabatic condition (i.e. at constant heat content). Isothermal condition is governed by Boyle s law according to which, for a given weight of gas at constant temperature, pv = const....(1.11) where p is the pressure in N/m or kn/m and V is the volume of given weight of gas. According to Charle s law, for a constant volume of a given weight of gas p/t = const....(1.1) where T is the absolute temperature of gas in kelvins. T = t C + 7.16 Equation of State of gas is pv/t = R...(1.1) where R is known as the universal gas constant having the units of Nm or m K. NK In adiabatic process pv k = const....(1.14) If adiabatic process is reversible, it is known as isentropic. Value of k varies with different gases (see Appendix D).

1.6 SURFACE TENSION AND CAPILLARITY Properties of Fluids 5 Coefficient of surface tension σ is force per unit length (N/m) which is caused when liquid-gas interface meets a solid wall. scosq Capillary rise or depression, h =...(1.15) gr where θ is the angle of contact between solid surface and liquid. If θ is greater than 90, there will be depression of surface in the tube (e.g. mercury in glass tube). If θ is less than 90 there will be a capillary rise (e.g. water in glass tube) (Fig. 1.). For clean glass and water, θ is nearly equal to zero. R R h h Mercury Water Fig. 1. Capillary rise and depression Difference of pressure Δp between inside and outside of a droplet of radius R s D p =...(1.16) R Appendix A gives the basic properties of common fluids at 0 C, whereas Appendix B gives properties of water at different temperatures. Appendix C lists the properties of air at different temperatures and at atmospheric pressure. Appendix D lists the properties of common gases while Appendix E gives properties of standard atmosphere. 1.7 FLOW REGIMES Depending on relative effects of viscosity and inertia, flows can be classified into laminar and turbulent flows. In pipes: Ê UDrˆ Reynolds No. Re Á= < 100 Ë m laminar flow Re > 000 turbulent flow Here U is average velocity of flow in a pipe of diameter D. Ê URrˆ In open channels: Re Á= > 500 Ë m laminar flow Re > 000 turbulent flow

6 Fluid Mechanics through Problems Depending on the relative importance of gravity and inertial forces, flow in open channels can be classified into subcritical and supercritical flows. Ê U ˆ Froude No. Fr Á = < 1.0 Ë gd subcritical flow Fr = 1.0 critical flow Fr > 1.0 supercritical flow Here R is the hydraulic radius which equals area of flow divided by wetted perimeter; D is hydraulic mean depth which equals area of flow divided by water surface width. Depending on relative importance of compressibility and inertia, flows of compressible fluids are classified as follows: ( ) Mach No. M = U E r < 1.0 subsonic M slightly less than unity to slightly greater than unity: transonic 1.0 < M < 6.0 supersonic M > 6.0 hypersonic It may be noted that E r is the velocity of sound or compression wave in the medium. Depending on distance between different molecules of the fluid relative to characteristic length of flow, one can have either continuum flow or free molecular flow. Mean free path of molecules Knudsen Number N = Characteristic length of flow If N < 0.01 continuum flow N > 10 free molecular flow In all discussions in this book, the flow is treated as continuum. ILLUSTRATIVE EXAMPLES 1.1 If.5 m of oil weighs.95 kn, calculate its specific weight, specific volume, mass density and relative density. weight.95 Specific weight g= = volume.5 = 9.414 kn/m 1 1 Specific volume v = = = 0.106 m kn g 9.414 g 9.414 10 Mass density r= = = 960.0 kg m g 9.806 960 Relative density = = 0.960 1000 1. If the kinematic viscosity of benzene is 7.4 10 stokes and its mass density is 860 kg/m, determine its dynamic viscosity in kg/m s.

Properties of Fluids 7 v = 7.4 10 S -4-7 = 7.4 10 10 m s = 7.4 10 m s -7 m=r v = 860 7.4 10-4 = 6.81 10 kg m s 1. If the mass density of a fluid is 789 kg/m, determine its specific weight and specific volume. g=r g = 789 9.806 = 776.94 N m or 7.77 kn m 1 1 Specific volume v = = = 0.19 m kn g 7.77 1.4 Determine the kinematic viscosity of air at 0 C if its dynamic viscosity is 1.85 10 4 poise and its mass density is 1.08 kg/m. -4 m= 1.85 10 poise -5 = 1.85 10 kg ms -5 m 1.85 10-5 v = = = 1.51 10 m s r 1.08 1.5 The space between two parallel horizontal plates kept 5 mm apart is filled with crude oil of dynamic viscosity.5 kg/ms. If the lower plate is stationary and upper plate is pulled with a velocity of 1.75 m/s, determine the shear stress on the lower plate. du 1.75-1 = = 0.5 10 s dy 5 10 du t=m =.5 0.5 10 dy = 875.0 N/m 1.6 A 50 mm diameter and 0.10 m long cylindrical body slides vertically down in a 5 mm diameter cylindrical tube. The space between the cylindrical body and tube wall is filled with oil of dynamic viscosity 1.9 N s/m. Determine its velocity of fall if its weight is 16 N (Fig. 1.4). Let U be its terminal velocity of fall. Shear stress τ will be du U t=m = 1.9 1 10 dy Oil 5 mm 50 mm 16 N 0.10 m = 1.9 10 U N m The shear stress will act on the surface of the cylinder. Hence Fig. 1.4

8 Fluid Mechanics through Problems Total force F = τ A = 1.9 10 U.14 50 10 0.10 = 9.849 U Under equilibrium condition, the weight will be balanced by the total shear force. Hence 16.0 = 9.849 U or U = 0.56 m/s Here it is assumed that velocity variation across the gap of 1 mm is linear. 1.7 A rectangular plate of 0.50 m 0.50 m dimensions weighing 500 N slides down an inclined plane making 0 angle with horizontal, at a velocity of 1.75 m/s. If the mm gap between the plate and inclined surface is filled with a lubricating oil, find its viscosity and express it in poise as well as in Ns/m (Fig. 1.5). 1.75 F = 0.5 0.5 m 10 = 0.19 10 μ mm For equilibrium, this must be equal to component of weight along the inclined surface. 500 sin 0 = 0.19 10 0 500 N μ 500 0.50 10 m= Fig. 1.5 0.19 = 1.14 Ns/m The viscosity in poise units will be 11.4 poise. 1.8 The velocity distribution, for small values of y, in laminar boundary layer on a flat plate is given by the equation u = 5y y in which u is the velocity in m/s at a distance y m above the plate. Determine shear stress at y = 0 and y = 0.5 m if μ = 1.85 10 5 kg/m s. du 5 6y dy = - du du = 5 s and = 5.0 0.75 = 4.65 s dy dy 1 1 y= 0 y= 0.5 τ at the boundary = 5 1.85 10 5 = 9.5 10 5 N/m At y = 0.5 m, τ = 4.65 1.85 10 5 = 8.556 10 5 N/m 1.75 m/s 1.9 If the velocity distribution in laminar boundary layer over a flat plate is given by u = y y + y 4 determine the shear stress on the plate and at y = 0.10 m if dynamic viscosity of the fluid is 1 10 kg/ms. du/dy = 6y + 4y Oil

Properties of Fluids 9 Êduˆ at y = 0, t =m Á Ëdy y= 0 = 1 10 = 10 N/m At y = 0.10 m, Êduˆ t=m Á Ëdy y= 0.10 = 1 10 ( 0.06 + 0.004) = 1.944 10 N/m 1.10 A closed vessel of volume 80 litres contains 0.50 N of gas at a pressure of 150 kn/m. If the gas is compressed isothermally to half its volume, determine the resulting pressure. For isothermal process pv = pv 1 1 p pv 1 1 150 80 = = = 00 kn m v 40 1.11 When the pressure of an enclosed gas is doubled, its new volume is 0.591 times the initial volume. Determine the value of k assuming the process to be adiabatic. For adiabatic process p 1 v 1 k = p v k k 1 1 p Ê v ˆ Ê v ˆ = 1.691 p Á \ = = Ëv Á Ë0.591v 1 1 k = 1.19 1.1 What will be the change in pressure of a gas enclosed in a container at 00 kn/m pressure if its temperature is changed from 0 C to 65 C. For constant volume p p p T 1 = or p = p1 1 1 T T T ( 7.16 + 65.00) = 00 ( 7.16 + 0.00) = 4.65 kn/m Change in pressure = (4.65 00.000) = 4.65 kn/m 1.1 Determine the specific weight of hydrogen at 40 C at a pressure of two atmospheres absolute. One atmospheric pressure = 101.5 kn/m p = 101.5 = 0.650 kn m From Appendix D, = 0.65 10 N/m R = 40. and T = 7.16 + 40 = 1.16 K k k

10 Fluid Mechanics through Problems pv 1 p = R or g = = T v RT 0.65 10 g= = 1.540 N m 40. 1.16 1.14 What should be the internal diameter of a glass tube if capillary rise of water in it is not to exceed.0 mm? Since scosq h = and q = 0 gr 0.075 R = = 7.5 10 m 10 9787 = 7.5 mm diameter = 7.5 = 15 mm 1.15 Assuming that sap in trees has the same characteristics as water and that it rises purely due to capillary phenomenon, what will be the average diameter of capillary tubes in a tree if the sap is carried to a height of 10 m? Taking θ = 0, h = s gr 0.075-6 R = = 1.50 10 m 10 9787 diameter =.004 10 6 m = 10 mm 1.16 What will be the difference of pressure between inside and outside of a droplet of water 1 mm in diameter. Since p = σ/r = 4σ/d 4 0.075 = = 0.940 10 N m 1 10 - = 94 N/m 1.17 A glass tube of mm internal diameter is immersed in an oil of mass density 960 kg/m to a depth of 10 mm. If a pressure of 17 N/m is needed to form a bubble which is just released, determine the surface tension of the oil (Fig. 1.6). (p i p 0 ) = σ/r; but p 0 = 960 9.806 10 10 = 94.18 N/m Inside pressure p i = 17.00 N/m (p i p 0 ) = (17.000 94.18) = 77.86 N/m mm 77.86 1 10 s= = 0.089 N m Fig. 1.6 Oil Bubble 10 mm

1.18 If velocity distribution for laminar flow in a pipe is given by { ( )} Properties of Fluids 11 u umax = 1 - r R where u is the velocity at a distance r from the centre line, u max is the centre line velocity and R is the pipe radius, determine expression for shear stress τ. dy Since r = R y, 1 =- or dy = dr dr du du t=m =-m dy dr However, du Ê r ˆ -ru = umax Á- = dr Ë R R max mrumax t= R 1.19 Calculate the force required to lift a thin ring of wire 0 mm in diameter from water surface. Neglect the weight of the ring. Perimeter of ring = πd =.14 0 10 m = 0.0684 m Surface tension force will be due to surface tension on both sides of wire F = 0.0684 0.075 = 9.8 10 N 1.0 In the flow conditions given in Fig. 1.7 determine the velocity at which the central plate of area 5.0 m will move if a force of 150 N is applied to it. The dynamic viscosities of the two oils are in the ratio of 1 : and viscocity of top oil 0.10 Ns/m. 10 mm 5 mm 150 N 5 mm Fig. 1.7 F = F 1 + F Êduˆ Êduˆ = Am 1Á + Am Ëdy Á Ëdy. U U 150 = 5 0.1 + 5 0. 5 10 5 10 = U(100 + 00) = 400 U U = 150/400 = 0.75 m/s

1 Fluid Mechanics through Problems 1.1 Determine whether the flow in a 1.5 m wide rectangular channel with 0.75 m depth of flow and 1.5 m/s velocity is (i) subcritical or supercritical, (ii) laminar or turbulent. By ( 1.5 0.75) R = = = 0.75 m B + y ( 1.5 + 0.75 ) Assuming water temperature to be 0 C, μ = 1 10 kg/ms, and ρ = 998 kg/m. URr 1.5 0.75 998 Re = = m 1 10 = 4.678 10 5 Hence the flow is turbulent. A 1.5 0.75 Hydraulic mean depth D = = = 0.75 m T 1.5 U 1.5 Fr = = = 0.461 gd 9.806 0.75 Hence flow is subcritical. 1. Obtain an expression for volume modulus of elasticity of atmosphere assuming the process to be isothermal. dp E =-v dv but pv = constant for isothermal process. pdv + vdp = 0 vdp or - = p dv E = p 1. What will be the velocity of sound in water and in air at atmospheric pressure and temperature of 0 C? For water.075 10 E r= = 1441.9 m s 998 For air at 0 C and assuming isothermal process 9 101.5 10 E r= p r= = 89.61 m s 1.08 If the process is assumed to be adiabatic, E = kp (see Problem 1. for solution). Taking k = 1.4, 1.4 101.5 10 E r= kp r= = 4.68 m s 1.08 1.4 Air is introduced through a nozzle into a tank of water to form a stream of bubbles. If the process requires.5 mm diameter bubbles to be formed, by how much the air pressure at the nozzle must exceed that of the surrounding water?

Properties of Fluids 1 s D p = R Assuming water temperature to be 0, σ = 0.075 N/m Δp = 0.075/1.5 10 = 117.6 N/m 1.5 Discuss how properties of the atmosphere change with altitude. The atmosphere can be broadly divided into the following four regions: Troposphere up to 11 km above mean sea level. Stratosphere 11 km to km above msl. Mesosphere km to 76 km above msl. Ionosphere Beyond 76 km above msl. The pressure of the atmosphere always decreases with the increase in altitude. The temperature in troposphere decreases at the rate of 6.5 C per km; this is known as the lapse rate. The temperature in stratosphere remains constant at 16.5 K i.e. 56.5 C. In mesosphere the temperature first rises linearly to 50 K at 50 km, remains constant at 50 K up to 60 km, and then decreases linearly to 40 K at 76 km. It remains constant at 40 K up to 84 km and then rises with increase in elevation in the ionosphere. In troposphere, which is of interest to us, the pressure, density and temperature vary as follows ( p p ) = ( T T ) 1 0 0 ar ( ) ( ) Ê Á -1 ˆ Ë a rr = TT R 0 0 T = T0 -az where p 0, ρ 0 and T 0 are the pressure, density and temperature (in K) at sea level and p, ρ, and T are the corresponding quantities at elevation Z; α is the lapse rate. The viscosity is related to the temperature by the empirical relation ( mm ) = ( T T) 0.75 0 0 one can take α = 0.0065 C/m and R = 9.57 m/k. 1.6 Calculate the capillary depression for mercury in a.5 mm diameter glass tube. Take the angle of contact θ = 140. See Appendix A; for mercury γ = 1.880 N/m and σ = 0.510 N/m. scosq 0.51cos( 140 ) h = = g 1.880 1.5 10 R = 4.7 10 m or 4.7 mm The negative sign indicates that it is a capillary depression. 1.7 A 0 mm diameter soap bubble has an internal pressure 7.576 N/m greater than the outside atmospheric pressure. Compute the surface tension of soap-air interface. The soap bubble has two surfaces with the air the inner and the outer, and almost the same radius since the soap film is very thin. Δp = 4σ/R 1

14 Fluid Mechanics through Problems D pr 7.576 10 10 s= = 4 4 = 0.06894 N/m PROBLEMS 1.1 If the relative density of a fluid is 1.59, calculate its mass density, specific weight and specific volume. (1590 kg/m, 15.591 kn/m, 0.064 m /kn) 1. If relative density of a liquid is 1.6, find its specific volume. (0.00750 m /kn) 1. If the dynamic viscosity of liquid is 0.01 poise and its relative density is 0.79, obtain its kinematic viscosity in m /s. (1.5 10 6 m /s) 1.4 Two horizontal plates are kept 1.5 mm apart and the space between them is filled with oil of dynamic viscosity of 14 poise. If the top plate is moved at a constant velocity of.5 m/s, determine the shear stress on the lower plate. (80.0 N/m ) 1.5 When a real fluid flows past a plate held parallel to flow, the velocity distribution near the plate is given by uu= ( y d) - 1 ( y d) where u = U when y = δ. Determine the shear stress at y = 0 and when y/δ = 0.50. (μu/δ, μu/δ) 1.6 A cylindrical body of 75 mm diameter and 0.15 m length falls freely in a 80 mm diameter circular tube kept vertically. If the space between the cylindrical body and tube is filled with oil of dynamic viscosity 0.9 poise, determine the weight of the body when it falls at a speed of 1.5 m/s. (1.909 N) 1.7 A circular disc of 0.0 m diameter and weighing 50 N is kept on an inclined surface with a slope of 45. The space of mm between disc and inclined surface is filled with oil of dynamic viscosity 1.0 Ns/m. What force will be required to pull the disc up the inclined plane at velocity of 0.50 m/s? (5.09 N) 1.8 What will be the dynamic viscosity of muddy water at 0 C if it contains 0 per cent fine sediment by volume? (.5 10 kg/ms) 1.9 Express the bulk modulus of elasticity in terms of mass density ρ of the fluid and pressure. Ê dp ˆ ÁE =r Ë dr 1.10 Find the increase in the pressure required to reduce the volume of water by 0.8 per cent if its bulk modulus of elasticity is.075 10 9 N/m. (1.66 10 7 N/m ) 1.11 Assuming depth of sea to be 10.000 m, determine the mass density of sea water at this depth if its relative density at sea level is 1.06. Take a constant value of E as.11 10 9 N/m. (107.60 kg/m ) 1.1 Determine the bulk modulus of elasticity of a fluid that has a density increase of 0.00 per cent for a pressure increase of 44.540 kn/m. (.7 10 9 N/m ) 1.1 What change in pressure is required to compress a given mass of gas to one third its volume under isothermal condition? (p 1 )

Properties of Fluids 15 1.14 If 1.0 m of gas is compressed to 0.0 m volume when its initial pressure was 150 kn/m determine the final pressure if the compression takes place under (i) isothermal condition, (ii) adiabatic condition. Take k = 1.0. (750 kn/m, 115.49 kn/m ) 1.15 In an adiabatic compression when final pressure was four times the initial pressure, the ratio of final to initial volume was 0.15. Determine the value of adiabatic constant. (1.0) 1.16 Determine the temperature and pressure of oxygen if from its original p, v and t of 1 bar, 1.5 m and 0 C it is compressed to 0.0 m volume under adiabatic condition. Take k = 1.9. (194.7 C, 7.974 bars) 1.17 A cylinder of 0.0 m length and 0.10 m diameter rotates about a vertical axis inside a fixed cylindrical tube of 105 mm diameter and 0.0 m length. If the space between the tube and the cylinder is filled with liquid of dynamic viscosity 0.15 N s/m, determine the speed of rotation of the cylinder which will be obtained if an external torque of 1.0 N m is applied to it. (81.0 rpm) 1.18 A circular disc of radius R is held parallel to a large plane and stationary surface at a small distance from it. If the space t between the two is filled with oil of dynamic viscosity μ and the disc is rotated at N rpm, determine the torque required to maintain this rotation. Êp Á Ë 4 NmR 60t 1.19 A circular cylinder of radius R 1 and height h rotates at N rpm in a cylindrical container of radius R with their axes vertical and coinciding. If the spacing between the bottoms of the cylindrical container and cylinder is t which is small and if the space between the cylinder and container is filled with oil of dynamic viscosity μ, obtain an expression for the total torque T required to maintain the motion. Assume R to be slightly larger than R 1. Ê p m p m ÁT = + Ë 15 4 N R1h N R1 1 60t ( R - R ) 1.0 What will be the temperature of oxygen in C if at 4.0 kn/m pressure its unit weight is 0.50 N/m? (Use value of R = 6.8 m/ K). (5.5 C) 1.1 What per cent error will be made in the viscosity of blood if clear water viscosity at 0 C is assumed when it contains 50 per cent by volume of red blood cells? Use Ward s equation. (5 per cent) 1. Determine the capillary rise in a clean glass tube of internal diameter.5 mm if the liquid is carbon tetrachloride. (.71 mm) 1. Two vertical parallel glass plates distance t apart are partially submerged in a liquid of specific weight γ and surface tension σ. Show that the capillary rise is given by scosq h = tg 1.4 Determine the diameter of a droplet of water in mm if the pressure inside is to be greater than the outside pressure by 10 N/m. (.6 mm) 1.5 A soap bubble of 50 mm diameter has a pressure difference of 0 N/m between its inside and outside. Determine the coefficient of surface tension of the solution. (.5 10 1 N/m) 1.6 For flow in a pipe take ˆ ˆ

16 Fluid Mechanics through Problems u t=m y and shear stress at the wall τ 0 to be related to shear stress at distance y by the relation Ê 0 1 y ˆ t=t - ËÁ R. Determine the velocity distribution by integration of the equation for shear assuming u = u max at the centre and u = 0 at the wall. Here R is the radius of pipe. Ê u Á Ëu max Ê r = Á1 - Ë R 1.7 Three cylindrical tubes of 0.50 m length are place co-axially and the central tube is rotated at 5 rpm applying a torque of 6.0 N m. Determine the viscosity of oil which fills the space between the tubes. Take R 1, R and R as 0.150, 0.15 and 0.154 m. (1.084 Ns/m ) 1.8 Determine whether the flow in the following cases is laminar or turbulent: (i) 0.0 m diameter pipe carrying water at 0 C at.0 m/s velocity. (ii) Flow of oil at 0.50 m/s velocity in 5 mm diameter tube. The dynamic viscosity of oil is 1.0 Ns/m and its mass density is 970 kg/m. (iii) Water flowing at 1.5 m/s velocity in a rectangular channel of 4.0 m width and 1.0 m depth. Water temperature is 0 C. Also note the Reynolds numbers in each case. (8.98 10 5, turbulent; 1.15, laminar; 9.985 10 5, turbulent) 1.9 Determine whether the flow in the following cases is subcritical, critical or supercritical (list the Fr number in each case). (i) Water flowing at 0.75 m/s velocity in a rectangular channel of width 5.0 m and depth 1.0 m. (ii) Water flowing in a triangular channel of central angle 90 when the velocity is 0.99 m/s and depth of 0.0 m. (iii) Water flowing at.0 m/s velocity in a trapezoidal channel of bottom width 1.0 m and side slope of 1 : 1, at a depth of 0.8 m. (0.9, subcritical; 1.0, critical; 1.87, supercritical) 1.0 If an aeroplane is moving through air at 800 km/hr, determine the flow regime if velocity of sound in air is 40 m/s. (Note the Mach number). (0.654, subsonic) 1.1 Obtain expression for E when the flow is adiabatic. (E = kp) 1. A long rigid pipe of 0.0 m diameter is used for pumping oil across the country. The pipe becomes plugged at some unknown point so that no fluid can flow. A piston inserted from one end of the pipe slides without leakage through 0.50 m causing an increase in pressure of 00 kn/m. Assume E for oil to be 1.8 10 9 N/m, determine the approximate location of obstruction. (4.5 km) 1. A cubical block of side 0.0 m and weighing 1.0 kn slides down an inclined plane of an angle of inclination of 0 with the horizontal. If oil of dynamic viscosity 7 10 kg/ms forms a thickness of 0.005 mm between the block surface and the inclined surface, what will be its terminal velocity? (6.107 m/s) ˆˆ

Properties of Fluids 17 1.4 Show that the capillary rise between two concentric glass tubes of radii R 1 and R (R being greater than R 1 ) and contact angle θ is given by scosq h = g - ( R R ) 1 1.5 A small circular jet of water of mm diameter issues from an opening. What is the pressure difference between inside and outside of the jet? (7.6 N/m ) 1.6 Take air properties at the sea level as p 0 = 101.5 kn/m, ρ 0 = 1.5 kg/m, T 0 = 88. K and μ 0 = 1.789 10 5 kg/ms, and determine the corresponding quantities at an elevation of 10.000 m. Take R = 9.57 m/k, α = 0.0065 C/m. (6.80 kn/m, 0.418 kg/m,. K, 1.477 10 5 kg/ms) DESCRIPTIVE QUESTIONS 1.1 Does definition of fluid include only substances in liquid phase? Explain. 1. List the differences between liquids, solids and granular material. 1. List the differences between liquids and gases. 1.4 Give five examples of fluid flow phenomena encountered in every day life. 1.5 What is continuum? Is air a continuum? Does it always remain so? 1.6 Classify the fluids A and B for which following values of deformation rate and shear stress are obtained experimentally. τ N/m 0 100 00 00 400 Fluid A du dy s -1 0 0.0 0.60 0.90 1. Fluid B du dy s -1 0 54.8 77.5 94.9 109.5 DM 1.7 How small should be the value of D V in the definition of mass density r= lim D V Æ 0 D V? 1.8 Give one example each of Newtonian fluid, pseudoplastic, dilatant, ideal plastic and thyxotropic fluid. 1.9 Classify the following fluids: Water, sugar solution, printer s ink, air, glycerine, and molten metal. 1.10 Under what conditions is the elasticity of the fluid important? 1.11 Differentiate between adiabatic and isentropic processes. 1.1 Even though needle is heavier than water, it can float on it if it is placed lengthwise on the water surface. Why? 1.1 Why does oil spread when it is poured on water surface? 1.14 Why does the viscosity of a liquid decrease with increase in temperature whereas it increases with increase in temperature in the case of a gas? 1.15 Give two examples each of (i) laminar flow, (ii) turbulent flow, and (iii) supersonic flow. 1.16 Show that the velocity gradient can be interpreted as rate of angular deformation. 1.17 Which fluid property/properties are important in the following physical phenomena?

18 Fluid Mechanics through Problems Lubrication Rise of sap in trees Ground water flow Settling of a sediment particle in water Force acting on the bottom of water tank Formation of droplets Energy loss is pipelines Water hammer phenomenon in pipes Echo sounders Formation of waves on water surface Cavitation Pumping of blood through arteries to capillaries. 1.18 Arrange the following fluids according to increasing values of dynamic viscosity and also according to decreasing values of mass density: air, water, alcohol, glycerine, castor oil. 1.19 Give one example each where air can be treated as an incompressible fluid and water has to be treated as compessible fluid. 1.0 Mention different ways in which the fluid can be set in motion; give one example each. 1.1 Draw a sketch of smoke issuing from the cigarette and sketch its path. Indicate to what regimes of flow the portion near the cigarette, away from it, and in between the two belong. 1. A fluid is a substance that (i) has to be kept in a closed container (ii) is almost incompressible (iii) has zero shear stress (iv) flows when even a small shear is applied to it. 1. A Newtonian fluid is that (i) which follows Newton s laws of motion (ii) which needs a minimum shear before it starts deforming (iii) for which shear and deformation are related as τ = μ u/ y. 1.4 Dimensions of dynamic viscosity are (i) L /T (ii) M/LT (iii) MT/L (iv) T/L 1.5 Dynamic viscosity of a gas (i) increases as temperature decreases (ii) increases as temperature increases (iii) is independent of temperature (iv) may increase or decrease with increase in temperature, depending on nature of gas. 1.6 An isentropic process is one in which (i) pv = constant (ii) pv k = constant (iii) pv k = constant, and process is reversible, (iv) none of the above. 1.7 Bulk modulus of elasticity of liquids has the dimensions (i) MT /L (ii) ML /T (iii) M/LT 1.8 If the mass of the body at sea level is.0 kg, what will be its mass on a planet where the gravitational acceleration is 1.4 m/s? 1.9 If the diameter of a capillary tube is doubled, the capillary rise will be (i) unaffected (ii) doubled (iii) halved (iv) none of the above

Properties of Fluids 19 1.0 If the depth of flow and average velocity in a wide rectangular channel are.0 m and.5 m/s respectively, the flow regime will be (i) laminar and supercritical (ii) turbulent and supercritical (iii) laminar and subcritical (iv) turbulent and subcritical 1.1 Ten poise of viscosity is equal to (i) 1.0 kg/ms, (ii) 10 kg/ms, (iii) 100 kg/ms, (iv) 1000 kg/ms. 1. Mass density of air at 0 C and atmospheric pressure of 101.5 kn/m is (i) 10.8 kg/m (ii) 1.08 kg/m (iii) 1.08 kg/m (iv) 0.108 kg/m 1. Assertion A: Viscosity of a gas increases as its temperature decreases. Reason R: With decrease in temperature the molecular activity of the gas is reduced. Choose the correct answer from the following (a) Both A and R are true and R is the correct explanation of A. (b) A is true but R is false. (c) A is false but R is true. (d) Both A and R are true but R is not the correct explanation of A. 1.4 If kerosene has a bulk modulus of elasticity E = 1.49 10 0 N/m and its relative density is 0.806, the speed of sound in kerosene in m/s will be (i) 184 (ii) 11 (iii) 1078 (iv) 100 1.5 If the rate of angular deformation du is plotted on x axis and the shear stress τ on y axis, dy Newtonian fluid will be represented by (i) a curved line passing through origin (ii) a curved line passing through yield stress and zero (iii) a straight line passing through yield stress and zero (iv) a straight line passing through origin and inclined to x axis 1.6 If a capillary rise of water in a glass tube of.5 mm diameter is.7 mm, the capillary rise in 1.5 mm diameter tube will be (i).7 mm (ii) 1.5 mm (iii) 5.4 mm (iv) none of the above