Lecture notes on Fluid Mechanics pdf

what is fluid mechanics and it application and hydraulics, fluid mechanics and hydraulics lecture notes, advanced fluid mechanics lecture notes pdf free download
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Published Date:21-07-2017
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CE6303 - MECHANICS OF FLUIDS FLUID PROPERTIES TWO MARK QUESTIONS AND ANSWERS 1. Define fluid mechanics. It is the branch of science, which deals with the behavior of the fluids (liquids or gases) at rest as well as in motion. 2. Define Mass Density. Mass Density or Density is defined as ratio of mass of the fluid to its volume (V) 3 3 Density of water = 1 gm/cm or 1000 kg / m . Mass of fluid p= Volume of fluid 3. Define Specific Weight. It is the ratio between weight of a fluid to its volume. Weight of fluid æMass of fluid ö w= =ç ÷ g´= p´ g ç ÷ Volume of fluid Volume of fluid è ø w = p´g 3 Unit: N / m 4. Define 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 u+ du, the viscosity together with relative velocity causes a shear stress acting between the fluid layers. 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.This shear stress is proportional to the rate of change of velocity. du  = dy  Þ Coefficient of dynamic viscosity (or) only viscosity du / dy = rate of shear strain 5. Define Specific Volume. Volume per unit mass of a fluid is called specific volume æ Volume of a fluidö 1 1 ç ÷ Sp. volume= == ç ÷ Mass of fluid p mass of fluid æ ö è ø ç ÷ volume è ø 3 Unit: m / kg. 6. Define Specific Gravity. Specific gravity is the ratio of the weight density or density of a fluid to the weight density or density of standard fluid. It is also called as relative density. Unit : Dimension less. Denoted as: ‘S’ Weight density of liquid S( forliquid )= Weight density of water Weight density of gas S( for gases =) Weight density of air 7. Calculate the specific weight, density and specific gravity of 1 litre of liquid which weighs 7 N. Solution: 1 3 Given V = 1ltre= m 1000 W = 7 Nweight 7N 3 = = =7000 N / m i. Sp. Weight (w) volume 1 æ ö 3 ç ÷ m 1000 è ø 7000 N w 3 3 == kg / m =713..5 Kg / m ii Density (p) 3 g 9.81m Density of liquid 713.5 3) = = iii. Sp. Gravity (S) (Density of water = 1000 kg / m Density of water 1000 S = 0.7135 8. State Newton’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 9. Name the Types of fluids. 1. Ideal fluid 2. Real fluid 3. Newtonian fluid 4. Non-Newtonian fluid. 5. Ideal plastic fluid 10. Define Kinematic Viscosity. It is defined as the ratio between the dynamic viscosity and density of fluid. Vis cos ity  Represented as ;  = = Density p 2 Unit: m / sec.2 2 2 Cm 1 m æ ö - 4 2 1 Stoke = = = 10 m / s. ç ÷ S 100 S è ø 1 Centistoke means = stoke 100 11. Find the Kinematic viscosity of an oil having density 981 kg/m. The shear stress 2 at a 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  = dy 0.2452 2 0.2452  =0´ .2 Þ  = = 1.226Ns / m . 0.2  1.226 kinematicvis cosity( ) = = p 981 - 2 2 = 0.125´ 10 m / s. - 2 4 2 0= .125 ´ 10 ´ 10 cm / S = 12.5stoke. 12. Determine the specific gravity of a fluid having viscosity 0.05 poise and Kinematic viscosity 0.035 stokes. 2 Given: Viscosity, μ = 0.05 poise = (0.05 / 10) Ns / m . 2 Kinematic viscosity ν = 0.035 stokes = 0.035 cm / s -4 2 = 0.035 x 10 m / s   = p 0.05 1 - 4 3 0.035 10´ = ´ Þp= 1428.5kg / m 10 p Density of liquid 1428.5Specific gravity of liquid = = = 1.428 = 1.43 Density of water 1000 13. Define Compressibility. Compressibility is the reciprocal of the bulk modulus of elasticity, K which is defined as the ratio of compressive stress to volumetric strain. 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 Decrease of volume = d" - d" \ Volumetric strain= " - Ve sign® Volume decreases with increase in pressure d p Increase of Pr essure d p = = \ Bulk modulus K= - " - d" Volumetric strain d" " 1 Compressibility= K 14. Define 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 liquids such that the contact surface behaves like a membrance under tension. Unit: N / m.15. Define 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. It is expressed in terms of cm or mm of liquid. 16. The 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 Solution: 3 Capillary rise, h = 0.2 mm = 0.2 x 10 m Surface tension  =0.0725 N / m Let, Diameter of tube = d Angelqfor water = 0 2 Density for water = 1000 kg / m 4 4´ 0.0725 - 3 h= Þ 0.2 ´ 10 = p ´ g´ d 1000 ´ 9.81´ d 4´ 0.0725 d= = 0.148m= 14.8cm - 3 1000 9´ .81´ 0.2´ 10 Minimum  of the tube = 14.8 cm. 17. Find out the minimum size of glass tube that can be used to measure water level 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. Solution: - 3 Capillary rise h = 2.0 mm = 2.0´ 10 mLet, diameter = d 3 Density of water = 1000 kg / m  =0.073575 N / m Angle for water  =0 4 4´ 0.073575 - 3 h= Þ 2.0 ´ 10 = p ´ g´ d 1000 ´ 9.81´ d d = 0.015 m = 1.5 cm. Thus the minimum diameter of the tube should be 1.5 cm. 18. Define 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. 19. Write down the expression for capillary fall. 4 Cos h= Height of depression in tube 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.20. 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. 2 Viscosity μ = 14 poise = 14 / 10 Ns / m 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 . 16 MARKS QUESTIONS AND ANSWERS 1.Calculate the capillary effect in millimeters a glass tube of 4mm diameter, 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 angle of contact for water is zero that for 0 3 mercury 130 . Take specific weight of water as 9790 N / m Given: - 3 4´ 10 m Diameter of tube Þ d = 4 mm = 4 cos h= Capillary effect (rise or depression)Þ p ´ g´ d  =Surface tension in kg f/m  =Angle of contact and p = density i. Capillary effect for water0  =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  =0.51 N / 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 indicates capillary depression. 3 0 2 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 Initial volume "=0.36 m 1 2 Pressure P = 0.3 N/mm 1 6 2 = 0.3´ 10 N / m 0 Temperature, t = 50 C 1 0 T = 273 + 50 = 323 K 13 Final volume, "=0.3m 2 K = 1.4 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 tan t K K p . "= p" 1 1 2 2 1.4 "K 0.6 æ ö 4 4 1.4 1 p =p 30=´ 10 ´ 30=´ 10 ´ 2 ç ÷ 2 1 " K 0.3 è ø 2 6 2 2 = 0.791´ 10 N / m =0.791N / mm k p"=cons tan t For temperature, p"=RT , RT RT k ´"=cons tan t p= and " " k- 1 RT" =Cons tan t k- 1 T" =Cons tan t (QR is also cons tan t)k- 1 k- 1 T V = T V 1 1 2 2 k- 1 1.4- 1.0 æ V ö 0.6 æ ö 1 ç ÷ T = T =323 ç ÷ 2 1 ç ÷ V 0.3 è ø è 2ø 0.4 0 323 = ´ 2 =426.2 K 0 t 426 = .2 - 273= 153.2 C 2 3. 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  =8.5 poise = =0.85 Viscosity, 2 10 m 2 u ay= + by+ C Parabolic velocity profile equation, (1) Where, a, b and c constants. Their values are determined from boundary conditions. i) At y = 0, u = 0 ii) At y = 20cm, u = 120 cm/se. du iii) At y = 20 cm, =0 dy Substituting (i) in equation (1), C = 0 2 120 a= ( 20 ) b+ 2 (= 400)a+20b Substituting (ii) in equation (1), -(2)du Substituting (iii) in equation (1), = 2ay+ b dy 0 2 a=´ 20´ b+= 40a+ b -(3) solving 1 and 2, we get, 400 a + 20 b=0 40 a + b = 0 (- ) 800 a + 20 b=0 b = - 40 a 120 400 a=20b+(40- a ) 400=a 800 - a =- 400 a 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 æ duö ç ÷ 0= .6- 0´+ 12= 12 / s. at y = 0, Velocity gradient, ç ÷ dy è ø y= 0 æ duö ç ÷ 0.6 = 10-´ 12+= 6-+ 12=6 / s. at y =10 cm, Velocity gradient, ç ÷ dy è ø y= 10 æ ö du ç ÷ 0.6 = 20-´ 12+ 12=-+ 12=0 at y = 20 cm, Velocity gradient, ç ÷ dy è ø y= 20 Shear Stresses: du  = Shear stresses is given by, dyæ duö 2  = ç ÷ 0= .85 ´ 12.0= 10.2N / m i. Shear stress at y = 0 , ç ÷ dy è ø y= 0 æ duö 2  = ç ÷ 0= .85 ´ 6.0=5.1N / m ii. Shear stress at y = 10 , ç ÷ dy è ø y= 10 æ ö du  = ç ÷ 0= .85 ´ 0=0 iii. Shear stress at y = 20 , ç ÷ dy è ø y= 20 4. 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. Solution: Diameter of cylinder = 15 cm = 0.15 m Diameter of outer cylinder = 15.10 cm = 0.151 m Length of cylinderÞ L = 25 cm = 0.25 m Torque T= 12 Nm ; N = 100 rpm. Viscosity =m  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. 5. The dynamic viscosity of 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 lost in the bearing for a sleeve length of 90 mm. The thickness of the oil film is 1.5 mm. 6 Ns Ns  6= poise = =0.6 Given, 2 2 10 m m - 3 L 90=mm = 90´ 10 m D = 0.4 m - 3 N = 190 rpm. t 1= .5mm = 1.5´ 10 m 2 NT Power= W 60 D T = force´ Nm. 2 F=Shear stress´Area =´ DL du 2  = N / m dy DN u= m / s. 60  DN  ´ 0.4´ 190 Tangential Velocity of shaft, u = = =3.98 m / s. 60 60 du = change of velocity = u – 0 = u = 3.98 m/s. - 3 dy t== 1.5´ 10 m. du 3.98 2   = Þ = 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 6.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 3æ duö 2 2 ç ÷ =- 2( 0 =) ç ÷ dy 3 3 è ø y= 0 æ ö du 2 ç ÷ = 2-(´ 0.17 )= 0.667- 0.30 ç ÷ dy 3 è ø 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 æ duö 2  = ç ÷ 0= .863´ 0.667=0.5756 N / m 0 ç ÷ dy è ø y= 0 ii. Shear stress at y = 0.15 m is given by æ duö 2 () = ç ÷ 0= .863 ´ 0.367=0.3167 N / m y= 0.15 ç ÷ dy è ø y= 0.15 7.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 cmP 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 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 / m1000 ´ 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.CE6303 - MECHANICS OF FLUIDS FLUID STATICS & KINEMATICS 1. Define “Pascal’s Law”: It stats that the pressure or intensity of pressure at a point in a static fluid is equal in all directions. 2. What is mean by Absolute pressure and Gauge pressure? Absolute Pressure: It is defined as the pressure which is measured with the reference to absolute vacuum pressure. Gauge Pressure: It is defined as the pressure which is measured with the help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum. The atmospheric pressure on the scale is marked as zero. 3. Define Manometers. Manometers are defined as the devices used for measuring the pressure at a point in a fluid by balancing measuring the column of fluid by the same or another column of fluid. 1. Simple M 2. Differential M 4. A differential manometer is connected at the two points A and B . At B pr is 2 9.81 N/cm (abs), find the absolute pr at A. 1000 9´ .81´ 0.6+p Pr above X – X in right limb = B Pr above X – X in left limb = 13.6 1000´ 9.81´0.1´900+ 9´ .81´ 0.2+P A Equating the two pr head 2 Absolute pr at P = 8.887 N/cm A 5. Define Buoyancy.When a body is immersed in a fluid, an upward force is exerted by the fluid on the body. This upward force is equal to the weight of the fluid displaced by the body and is called the force of buoyancy or simply buoyancy. 6. Define META – CENTRE It is defined as the point about which a body starts oscillating when the body is fitted by a small angle. The meta – centre may also be defined as the poit at which the line of action of the force of buoyancy wil meet the normal axis of the body when the body is given a small angular displacement. 7. Write a short notes on “ Differential Manometers”. Differential manometers are the devices used for measuring the difference of pressures between two points in a pipe or in two different pipes/ a differential manometer consists of a U – tube containing a heavy liquid, whose two ends are connected to the points, whose difference of pressure is to be measured. Most commonly types of differential manometers are: 1. U – tube differential manometer. 2. Inverted U – tube differential manometers. 8. Define Centre of pressure. Is defined as the point of application of the total pressure on the surface. The submerged surfaces may be: 1. Vertical plane surface 2. Horizontal plane surface 3. Inclined plane surface 4. Curved surface 9. Write down the types of fluid flow. The fluid flow is classified as : 1. Steady and Unsteady flows. 2. Uniform and Non – uniform flows. 3. Laminar and turbulent flows. 4. Compressible and incompressible flows. 5. Rotational and irrotational flows 6. One, two and three dimensional flows. 10. Write a short notes on “Laminar flow”. Laminar flow is defined as that type of flow in which the fluid particles move along well – defined paths or stream line and all the stream lines are straight and parallel.

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