Lecture notes of Fluid Mechanics pdf

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MECHANICS OF FLUIDS (Regulation-2008 & 2013) UNIT – I DEFINITIONS AND FLUID PROPERTIES Definitions – Fluid and fluid mechanics – Dimensions and units – Fluid properties – Continuum Concept of system and control volume. UNIT – II FLUID STATICS & KINEMATICS Pascal‘s Law and Hydrostatic equation – Forces on plane and curved surfaces – Buoyancy – Meta centre – Pressure measurement – Fluid mass under relative equilibrium Fluid Kinematics Stream, streak and path lines – Classification of flows – Continuity equation (one, two and three dimensional forms) – Stream and potential functions – flow nets – Velocity measurement (Pilot tube, current meter, Hot wire and hot film anemometer, float technique, Laser Doppler velocimetry). UNIT – III FLUID DYNAMICS Euler and Bernoulli‘s equations – Application of Bernoulli‘s equation – Discharge measurement – Laminar flows through pipes and between plates – Hagen Poiseuille equation – Turbulent flow – Darcy-Weisbach formula – Moody diagram – Momentum Principle. UNIT – IV BOUNDARY LAYER AND FLOW THROUGH PIPES Definition of boundary layer – Thickness and classification – Displacement and momentum thickness – Development of laminar and turbulent flows in circular pipes – Major and minor losses of flow in pipes – Pipes in series and in parallel – Pipe network UNIT – V SIMILITUDE AND MODEL STUDY Dimensional Analysis – Rayleigh‘s method, Buckingham‘s Pi-theorem – Similitude and models – Scale effect and distorted models UNIT I DEFINITION AND FLUID PROPERTIES FLUID OR FLUID MECHANICS Fluid mechanics is that of branch of science which deals with the behaviour of the fluids at rest as well as in motion. Thus this branch of science deals with the static, kinematics and dynamics aspects of fluids. STATICS The study of fluid at rest is called fluid statics. It deals with the study under static. KINEMATICS The study of fluids in motion, where pressure forces not considered, is called fluid kinematics.It deals with velocity, acceleration and pattern of flow. DYNAMICS The pressure forces are not considered, is called fluid kinematics and if the pressure forces are also considered for the fluid in motion, that branch of science is called fluiddynamics.It deals with the relation between velocity, acceleration with various forces. Classification of fluid : Fluids are two types (i) Ideal fluids (ii) Real fluids DIMENSION AND UNITS DIMENSIONS: Fundamental dimensions are M,L,T. M for mass L for length T for time Example: Force Force = mass x acceleration = m x a And now, a = V/T = Velocity/Time And V = L/T ; Length/Time -1 = LT 2 -2 Now a = L/T or LT -2 F = M L T -2 Force dimension = MLT -2 Velocity dimension = LT -1 Acceleration dimension = LT UNITS: Basically it has two system (i) Absolute system (MFT) (ii) gravitational system ( FMT) In absolute system unit of mass is taken as unit of force In gravitational system unit of force is taken as unit of mass EXAMPLE: Force Now force = 1 Newton 2 Newton = 1 newton is the force which imparts and acceleration of one meter/second to a mass of I kg. 2 1N = 1 kg x 1m/sec 1 kg =9.81 N 5 1N = 10 mdynes. FLUID PROPERTIES Density or Mass density It is defined as the ratio of the mass of a fluid to its volume. Thus mass per unit volume of a fluid is called density. It is denoted by the symbol ρ (rho).The unit of mass Prepared By, 3 Shivram Pachaiyappan (B.E.,) 3 density in SI unit is kg per cubic meter, i.e., kg/m . The density of liquids may be considered as constant while that of gases changes with the variation of pressure and temperature. Mathematically, mass density written as ρ = mass of fluid volume of fluid 3 3 The value of density of water is 1 gm/cm or 1000 kg/m . Specific weight or weight density It is defined as the ratio of the weight of a fluid to its volume.Thus weight per unit volume of a fluid is called weight density and it is denoted by the symbol w. Thus mathematically, w= Weight of fluid Volume of fluid = ( Mass of fluid) x Acceleration due to gravity Volume of fluid = Mass of fluid x g / Volume of fluid = ρ x g w = ρg 3 the value of specific weight or weight density (w) for water is 9.81 x 1000 Newton/m in SI units. Specific volume It is defined as the ratio of the weight density of a fluid to the weight density of a standard fluid. For liquids, the standard fluid is taken water and for gases, the standard fluid is taken air. Specific gravity is also called relative density. It is dimensionless quantity and is denoted by the symbol S. Mathematically, S (for liquid ) = Weight density of liquid Weight density of water S (for gas) = Weight density of gas Weight density of air Thus weight density of a liquid = S x Weight density of water Prepared By, 4 Shivram Pachaiyappan (B.E.,) 3 = S x 1000 x 9.81 N/m The density of a liquid = S x Density of water 3 = S x 1000 kg x m If the specific gravity of a fluid is known, then the density of the fluid will be equal to specific gravity of fluid multiplied by the density of water. For example , the specific 3 gravity of mercury is 13.6, hence density of mercury = 13.6 x 1000 = 13600 kg/m . Problem: Calculate the specific weight, density and specific gravity of one liter of a liquid which weighs 7 N. Solution: Volume = 1 liter 3 = 1m 1000 Weight = 7 N (i) Specific weight (w) = Weight / Volume 3 = 7N / (1/1000) m 3 = 7000 N /m . (ii) Density (ρ) = w/g 3 = 7000/9.81 kg/m 3 = 713.5 kg/m . (iii) Specific gravity = Density of liquid / Density of water 3 Density of water = 1000 kg/m = 713.5 / 1000 = 0.7135. Viscosity: Viscosity is defined as the property of fluid which offers resistance to the movement of one layer of fluid over another adjacent layer of fluid. When two layers move one over the other at different velocities , say U and V+ du, the viscosity together with relative velocity causes a shear stress acting between the fluid layer. Prepared By, 5 Shivram Pachaiyappan (B.E.,) The top layer causes a shear stress on the adjacent lower layer while the lower layer causes a shear stress on the adjacent top layer. Shear stress du  dy This shear stress is proportional to the rate of change of velocity constant of proportionality. du  (or) dy  Coefficient of dynamic viscosity (or) only viscosity Specific Volume: Volume per unit mass of a fluid is called sp volume 3 Unit : m /kg.  Commonly applied for gases. Volume of a fluid 1 1  Sp. volume  Mass of fluid p mass of fluid    volume  Specific Gravity: (or) Relative density: Specific gravity is the ratio between the weight of a body to the weight of equal volume of water. Unit: Dimension less. Denoted as: ‗S‘ Weight density of liquid S pr.liq Weight density of water Weight density of gas S for gases Weight density of air Weight density of a liquid = S wt density of water 9.81 3  S1000 N / m g Density of a liquid = S x Density of water. 3 S x 1000 Kg/m Calculate the spwt, density and sp gr of 1 litre of liquid which weighs 7 N. Prepared By, 6 Shivram Pachaiyappan (B.E.,) Solution: 1  3 1lt m  1000  3  1lt 1000 cm  1 3 Given V 1ltre m 1000 W = 7 N weight 7N 3 i. Sp. Weight (w)  7000 N / m volume 1  3 m 1000  7000 N w 3 3 ii Density (p)  kg / m 713..5 Kg / m 3 g 9.81m Density of liquid 713.5 3) iii. Sp. Gravity  (Density of water = 1000 kg / m Density of water 1000 = 0.7135 2 g = 9.81 m / s Newtonian fluid  when obeys above relation. Non – neutron fluid  doesn‘t obey the above relation. Neutron’s Law of Viscosity: It states that the shear stress ( ) on a fluid element layer is directly proportional to the rate of shear strain. The constant of proportionality is called the co-efficient of viscosity du  dy Fig.(1): Relationship between shear stressvelocity gradient of Newtonian fluids Prepared By, 7 Shivram Pachaiyappan (B.E.,) Fig.(2): Relationship between shear stress and shear strain rate of different fluids. Types of fluids: 1. Ideal fluid 2. Real fluid 3. Newtonian fluid 4. Non-Newtonian fluid. 5. Ideal plastic fluid Kinematic Viscosity: Defined as the ratio between the dynamic viscosity and density of fluid. Vis cos ity Represented as nu. Density p 2 SI and MKS unit: m / sec. 2 CGS: cm / S. (Kinematic also known STOKE) 2 2 2 Cm 1 m  4 2 1 Stoke 10 m / s.  S 100 S  1 Centistoke means  stoke. 100 Problem: Find the kinematic viscosity of an oil having density 981 kg/m. The shear stress at a 2 point in oil is 0.2452 N/m and velocity gradient at that point is 0.2 /sec. 2 3 Mass density p = 981 kg/m , Shear stress  0.2452N / m du Velocity gradient  0.2 dy du 0.2452  0.24520.21.226Ns. dy 0.2  1.226 2 22 4 2 2  0.12510 m / s. 0.1251010 cm / Scm / s stoke p 981 2 2 2  0.12510 cm / s12.5cm / s12.5stoke. Prepared By, 8 Shivram Pachaiyappan (B.E.,) Problem: Determine the sp gr of a fluid having viscosity 0.05 poise and Kinematic viscosity 0.035 stokes. Density of liquid 1428.5 Specific gravity of liquid = = = 1.428 = 1.43 Density of water 1000 0.05 2  0.05poise Ns / m 10   p 0.05 1 4 3 0.03510 p 1428.5kg / m 10 p Compressibility: (K) Defined as the ratio of compressive stress to volumetric strain. (reciprocal of bulk modulus) Consider a cylinder filled with a piston as shown V  Volume of gas enclosed in the cylinder p  Pressure of gas when volume is  2 Increase in pressure = dp kgf / m d Decrease of volume =  d  Volumetric strain   - Ve sign  Volume with  p d p Increase of Pr essure d p   Bulk modulus K   d Volumetric strain d  Prepared By, 9 Shivram Pachaiyappan (B.E.,) 1 Compressibility K Surface Tension. Surface tension is defined as the tensile force acting on the surface of a liquid in contact with a gas or on the surface between two immiscible liquid such that the contact surface behaves like a membrane under tension. In MKS unit, kgf/m, SI  N / m. Capillarity: Capillary is defined as a phenomenon of rise of a liquid surface is a small tube relative to adjacent general level of liquid when the tube is held vertically in the liquid. The resistance of liquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression. Capillary rise in the glass tube is not to exceed 0.2 mm of water. Determine its minimum size, given that surface tension of water in contact with air = 0.0725 N/m 3 Capillary rise  h = 0.2 mm = 0.210 m Surface tension  0.0725 N / m 2 Dia of tube = d Angel  for water = 0 Density p for water = 1000 kg / m 4 40.0725 3 h 0.210 p g d 10009.81 d Prepared By, 10 Shivram Pachaiyappan (B.E.,) 40.0725 d 0.148m 14.8cm 3 10009.810.210 Minimum  of the tube = 14.8 cm. Find out the minimum size of glass tube that can be used to measure water level o if the capillary rise in the tube is to be restricted to 2mm. Consider surface tension of water in contact with air as 0.073575 N/m. 3 3 h = 2.0 mm = 2.010 m dia = d density of water = 1000 kg / m angle  0.073575 N / m 0 4 4 0.073575 3 h 2.010 p g d 1000 9.81 d d = 0.015 m = 1.5 cm. Difference between Real fluid and Ideal fluid. Real Fluid: A fluid, which possesses viscosity, is known as real fluid. All fluids, in actual practice, are real fluids. Ideal Fluid: A fluid which is incompressible and is having no viscosity is known as an ideal fluid. Ideal fluid is only an imaginary fluid as all the fluids, which exist have some viscosity. Expression for capillary fall. 4Cos h p g d where, h = height of depression in tube. d = diameter of the σ = surface tension ρ = density of the liquid. θ = Angle of contact between liquid and gas. Prepared By, 11 Shivram Pachaiyappan (B.E.,) Two horizontal plates are placed 1.25 cm apart. The space between them being filled with oil of viscosity 14 poises. Calculate the shear stress in oil if upper plate is moved with a velocity of 2.5 m/s. Solution: Given: Distance between the plates, dy = 1.25 cm = 0.0125m. Viscosity μ = 14 poise = 1.4 2 - Ns / m . 10 Velocity of upper plate, u = 2.5 m/Sec. Shear stress is given by equation as τ = μ (du / dy). Where du = change of velocity between the plates = u – 0 = u = 2.5 m/sec. dy = 0.0125m. 2 τ = (14 /10) X (2.5 / 0.0125) = 280 N/m . Calculate the capillary effect in millimeters a glasstubeof4mmdiameter, when 0 immersed in (a) water (b) mercury. The temperature of the liquid is 20 C and the 0 values of the surface tension of water and mercury at 20 + C in contact with air are 0.073575 and 0.51 N/m respectively. The angel of contact for water is zero that for 0 0 mercury 130 . Take density of water 20 C as equate 998 kg Given: 3 410 m Dia of tube  d = 4 mm = 4 cos h Capillary effect (rise or depression)  p g d  Surface tension in kg f/m  angle of contact p = density Prepared By, 12 Shivram Pachaiyappan (B.E.,) i. Capillary effect for water 0  0.073575 N / m, 0 3 0 p 998 kg / m 20 c 0 4 0.73575Cos0 3 h 7.5110 m 3 9989.81 410 = 7.51 mm. Capillary effect for mercury: 0  0.51N / m,  130 3 p sp gr100013.6100013600kg / m 0 40.51 Cos130 h 3 136009.81 410 3 2.4610 m = - 2.46 mm. -Ve capillary depression. 3 0 2 A cylinder of 0.6 m in volume contains air at 50 C and 0.3 N/ mm absolute 3 pressure. The air is compressed to 0.3 m . Find (i) pressure inside the cylinder assuming isothermal process (ii) pressure and temperature assuming adiabatic process. Take K = 1.4 Given: 3 2 Initial volume  0.36 m Pressure P = 0.3 N/mm 1 1 Prepared By, 13 Shivram Pachaiyappan (B.E.,) 6 2 0 t = 50 C  0.310 N / m 1 4 2 0 T = 273 + 50 = 323 K  3010 N / m 1 3  0.3m K = 1.4 2 i. Isothermal Process: P  Cons tan t (or) p Cons tan t p p p 1 1 2 2 4 p 30100.6 1 1 6 2 p 0.610 N / m 2  0.3 2 2 = 0.6 N / mm ii. Adiabatic Process: p  Cons tan t or K K p p cons tant K K p . p 1 1 2 2 1.4  R 0.6  1 4 4 1.4 p p 3010 3010 2  2 1  K 0.3  2 6 2 2  0.79110 N / m 0.791N / mm k p   cons tant For temperature , p RT , RT RT k  cons tan t p and   k1 RT Cons tant k1 T Cons tant R is also cons tant k1 k1 T V T V 1 1 2 2 k1 1.41.0  V 0.6  1 T T 323  2 1  V 0.3   2 Prepared By, 14 Shivram Pachaiyappan (B.E.,) 0.4 0  3232 426.2 K 0 t 426.2 273153.2 C 2 If the velocity profile of a fluid over a plate is a parabolic with the vertex 202 cm from the plate, where the velocity is 120 cm/sec. Calculate the velocity gradients and shear stress at a distance of 0, 10 and 20 cm from the plate, if the viscosity of the fluid is 8.5 poise. Given, Distance of vertex from plate = 20 cm. Velocity at vertex, u = 120 cm / sec. 8.5 Ns  8.5poise 0.85 Viscosity, 2 10 m Parabolic velocity profile equation 2 u ay by C (1) a, b and c constants values determined from boundary condition. a. at y = 0, u = 0 b. at y = 20cm, u = 120 cm/se. du c. at y = 20 cm,  0 dy Substituting (a) in equation (1), C = 0 2 120 a20 b2 400a 20' s Substituting (b) in equation (1), (ii) du Substituting (C) in equation (1),  2ay b dy 0 2a20b 40ab (iii) 400 a 20 b 0 40 a + b = 0 () 800 a 20b 0 b = - 40 a 120 400 a 20b 40 a 400 a800 a400 a Prepared By, 15 Shivram Pachaiyappan (B.E.,) 120 3 a 0.3  400 10 b40 0.31.2 2 u0.3y12y Substituting a, b and c in equation (i) du 0.3 2y120.6y12 dy Velocity gradient  du 0.6012 12 / s. at y = 0, Velocity gradient,  dy  y0  du  0.61012612 6 / s. at y =10 cm, Velocity gradient,  dy  y10  du  0.6 20121212 0 at y = 20 cm, Velocity gradient,  dy  y20 Shear Stresses: du  Shear stresses is given by, dy  du 2   0.8512.010.2N / m i. Shear stress at y = 0 ,  dy  y0  du 2   0.856.0 5.1N / m ii. Shear stress at y = 10,  dy  y10  du   0.850 0  iii. Shear stress at y = 20, dy  y20 A 15 cm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 15.10 cm. Both cylinders are 25 cm high. The space between the cylinders is filled with a liquid whose viscosity is unknown. If a torque of 12.0 Nm is required to rotate the inner cylinder at 100 rpm determine the viscosity of the fluid. Dia of cylinder = 15 cm = 0.15 m Prepared By, 16 Shivram Pachaiyappan (B.E.,) Dia of outer cylinder = 15.10 cm = 0.151 m Length of cylinder  L = 25 cm = 0.25 m Torque T= 12 Nm N = 100 rpm. Viscosity =   DN0.15100 Tangential velocity of cylinder u 0.7854 m / s 60 60 Surface area of cylinder AD L0.150.25 2 = 0.1178 m du du u0 u 0.7854 m / s  dy 0.151 0.150 dy 0.0005 m 2 0.7854  0.0005 0.7854 Shear force, F Shear Stress Area0.1178 0.0005 D Torque T F 2 0.7854 0.15 12.00.1178 0.0005 2 12.00.0005 2 2  0.864Ns / m 0.78540.11780.15  0.86410 8.64 poise. The dynamic viscosity of an oil, used for lubrication between a shaft and sleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at 190 rpm. Calculate the power last in the bearing for a sleeve length of 90 mm. The thickness of the oil film is 1.5 mm. Given, 6 Ns Ns  6poise 0.6 2 2 10 m m 3 L 90mm 9010 m D = 0.4 m Prepared By, 17 Shivram Pachaiyappan (B.E.,) 3 N = 190 rpm. t1.5mm1.510 m 2 NT Power W F Shear stress Area N 60 D DL T force Nm. 2  DN du 2  N / m u M / s. dy 60  DN0.4190 Tangential Velocity of shaft, u 3.98 m / s. 60 60 3 du = change of velocity = u – 0 = u = 3.98 m/s. dy t1.510 m. du 3.98 2  10 1592N / m 3 dy 1.510 Shear force on the shaft F = Shear stress x Area 3 F1592D L15920.49010180.05N D 0.4 Torque on the shaft, T Force 180.05 36.01 Ns. 2 2 2NT 219036.01 Power lost  716.48W 60 60 2 2 If the velocity distribution over a plate is given by u y y in which U is the 3 velocity in m/s at a distance y meter above the plate, determine the shear stress at y = 0 and y = 0.15 m. Take dynamic viscosity of fluid as 8.63 poise. Given: 2 2 u y y 3 du 2  2y dy 3 Prepared By, 18 Shivram Pachaiyappan (B.E.,)  du 2 2   20  dy 3 3  y0  du 2  20.17 0.667 0.30  dy 3  y0.15 8.63 2  8.63poise SI units = 0.863 Ns / m 10 du  dy i. Shear stress at y = 0 is given by  du 2   0.8630.667 0.5756 N / m 0  dy  y0 ii. Shear stress at y = 0.15 m is given by  du 2  0.863 0.367 0.3167 N / m y0.15  dy  y0.15 The diameters of a small piston and a large piston of a hydraulic jack at3cm and 10 cm respectively. A force of 80 N is applied on the small piston Find the load lifted by the large piston when: a. The pistons are at the same level b. Small piston in 40 cm above the large piston. 3 The density of the liquid in the jack in given as 1000 kg/m Given: Dia of small piston d = 3 cm.  2 2 2 a d3 7.068cm  Area of small piston , 4 4 Dia of large piston, D = 10 cm P 2 2 A10 78.54cm  Area of larger piston, 4 Force on small piston, F = 80 N Let the load lifted = W a. When the pistons are at the same level Pressure intensity on small piston Prepared By, 19 Shivram Pachaiyappan (B.E.,) F 80 2 P N / cm a 7.068 This is transmitted equally on the large piston. 80  Pressure intensity on the large piston  7.068  Force on the large piston = Pressure x area 80  x 78.54 N = 888.96 N. 7.068 b. when the small piston is 40 cm above the large piston Pressure intensity on the small piston F 80 2  N / cm a 7.068  Pressure intensity of section A – A F  Pressure intensity due of height of 40 cm of liquid. P = pgh. a But pressure intensity due to 40cm. of liquid 2  p gh10009.810.4N / m 10009.810.4 2 2  N / cm 0.3924N / cm 4 10  Pressure intensity at section 80 A A 0.3924 7.068 2 = 11.32 + 0.3924 = 11.71 N/cm 2 Pressure intensity transmitted to the large piston = 11.71 N/cm Force on the large piston = Pressure x Area of the large piston 11.71 A11.7178.54 = 919. 7 N. Vapour Pressure Prepared By, 20 Shivram Pachaiyappan (B.E.,)

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