Mechanics of Fluids - Ebook & Notes.pdf

Studynama.com powers Engineers, Doctors, Managers & Lawyers in India by providing 'free' resources for aspiring students of these courses as well as students in colleges. You get FREE Lecture notes, Seminar presentations, guides, major and minor projects. Also, discuss your career prospects and other queries with an ever-growing community. Visit us for more FREE downloads: www.studynama.com ALL FILES ON STUDYNAMA.COM ARE UPLOADED BY RESPECTIVE USERS WHO MAY OR MAY NOT BE THE OWNERS OF THESE FILES. FOR ANY SUGGESTIONS OR FEEDBACK, EMAIL US AT [email protected] CE6303 - MECHANICS OF FLUIDS (FOR III – SEMESTER) UNIT – I PREPARED BY R.SURYA, M.E Assistant Professor DEPARTMENT OF CIVIL ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SRI VIDYA COLLEGE OF ENGINEERING AND TECHNOLOGY-626001 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) Density of water = 1 gm/cm3 or 1000 kg / m3. 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  Mass of fluid  w   g  p g Volume of fluid  Volume of fluid  w  p g Unit: N / m3 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   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  Volume of a fluid  1 1 Sp. volume       Mass of fluid  p  mass of fluid   volume  Unit: m3 / 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 Given V  1ltre  m3 1000 W=7N weight 7N i. Sp. Weight (w)    7000 N / m 3 volume  1  3  m  1000  w 7000 N ii Density (p)   3 kg / m 3  713. Define Kinematic Viscosity.. Ideal plastic fluid 10.   Density p Unit: m2 / sec. Ideal fluid 2. S  100  S . It states that the shear stress (  ) on a fluid element layer is directly proportional to the rate of shear strain.81 m Density of liquid 713. Gravity (S)   (Density of water = 1000 kg / m3) Density of water 1000 S = 0. 1. 2 Cm 2  1  m 2 1 Stoke     10  4 m 2 / s. State Newton’s Law of Viscosity. Sp. Vis cos ity  Represented as  . Non-Newtonian fluid. Newtonian fluid 4. Name the Types of fluids.5 iii.7135 8. Real fluid 3. The constant of proportionality is called the co-efficient of viscosity du   dy 9.5 Kg / m 3 g 9. It is defined as the ratio between the dynamic viscosity and density of fluid. 5. 2452 N/m2 and velocity gradient at that point is 0. 0.2 dy du   dy 0. 12.05 1 0. Find the Kinematic viscosity of an oil having density 981 kg/m.2  1.125  10 m / s.05 poise = (0.2452 0.035 x 10-4 m2 / s    p 0.428 = 1.035 stokes.2 /sec.43 Density of water 1000 .226 kinematicvis cos ity( )   p 981 2  0.5kg / m 3 10 p Density of liquid 1428.2  0.035 stokes = 0. μ = 0.226 Ns / m 2 . Shear stress   0.05 poise and Kinematic viscosity 0. The shear stress at a point in oil is 0. 1 Centistoke means  stoke 100 11. Determine the specific gravity of a fluid having viscosity 0.5stoke.2452    0.5 Specific gravity of liquid = = = 1.2     1. Given: Viscosity.05 / 10) Ns / m2.2452 N / m 2 du Velocity gradient  0 .125  10 2  10 4 cm 2 / S  12 . Kinematic viscosity ν = 0. Mass density p = 981 kg/m3.035  10  4    p  1428 .035 cm2 / s = 0. 13. . Define Surface Tension. Unit: N / m. K which is defined as the ratio of compressive stress to volumetric strain.Ve sign  Volume decreases with increase in pressure Increase of Pr essure  d p  dp  Bulk modulus K     d Volumetric strain d  1 Compressibility  K 14. 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. Consider a cylinder filled with a piston as shown V  Volume of gas enclosed in the cylinder P  Pressure of gas when volume is  Increase in pressure = dp kgf / m2 Decrease of volume = d  d  Volumetric strain   . Define Compressibility. Compressibility is the reciprocal of the bulk modulus of elasticity. 2 mm of water. 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.8cm 1000  9.2 mm = 0.15.2  10 3  p g d 1000  9. Solution: 3 Capillary rise h = 2.0725 h  0. 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.0 10 m Let.2 10 3 Minimum  of the tube = 14. 17. Consider surface tension of water in contact with air as 0. Diameter of tube = d Angel  for water = 0 Density for water = 1000 kg / m2 4 4  0.0725 N / m Let.81  d 4  0.073575 N/m.0725 N/m Solution: Capillary rise.0 mm = 2. 16.0725 d  0. h = 0.81  0. The Capillary rise in the glass tube is not to exceed 0.8 cm. The resistance of liquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression.148 m  14. given that surface tension of water in contact with air = 0. Determine its minimum size. It is expressed in terms of cm or mm of liquid. diameter = d Density of water = 1000 kg / m3 .2 x 103 m Surface tension   0. 5 cm. θ = Angle of contact between liquid and gas.81  d d = 0.5 cm. . Thus the minimum diameter of the tube should be 1.0  10 3  p gd 1000  9. Define Real fluid and Ideal fluid. have some viscosity. h = height of depression in tube. 19. 4Cos Height of depression in tube h  p g d Where. is known as real fluid. Ideal Fluid: A fluid. are real fluids. which possesses viscosity.073575 N / m Angle for water   0 4 4  0. Write down the expression for capillary fall. Real Fluid: A fluid. is known as an ideal fluid. 18. which exist. in actual practice.   0.015 m = 1.073575 h  2. d = diameter of the σ = surface tension ρ = density of the liquid. All fluids. Ideal fluid is only an imaginary fluid as all the fluids. which is incompressible and is having no viscosity. when immersed in (a) water (b) mercury. dy = 1.5 m/Sec. Two horizontal plates are placed 1.25 cm = 0.5 m/s.51 N/m respectively. Calculate the shear stress in oil if upper plate is moved with a velocity of 2. dy = 0. Take specific weight of water as 9790 N / m3 Given: Diameter of tube  d = 4 mm = 4  10 3 m 4 cos Capillary effect (rise or depression)  h p g d   Surface tension in kg f/m   Angle of contact and p = density . The space between them being filled with oil of viscosity 14 poises.5 m/sec. τ = (14 /10) X (2.Calculate the capillary effect in millimeters a glass tube of 4mm diameter. 16 MARKS QUESTIONS AND ANSWERS 1.0125m.20. Viscosity μ = 14 poise = 14 / 10 Ns / m2 Velocity of upper plate. The angle of contact for water is zero that for mercury 1300. Where du = change of velocity between the plates = u – 0 = u = 2.25 cm apart. The temperature of the liquid is 20 0 C and the values of the surface tension of water and mercury at 200 C in contact with air are 0.0125m.073575 and 0. Solution: Given: Distance between the plates.5 / 0. Shear stress is given by equation as τ = μ (du / dy). u = 2.0125) = 280 N/m2. 6 m3 in volume contains air at 500C and 0.46 10 3 m = .36 m 3 Pressure P1 = 0.073575 N / m. Take K = 1.4 Given: Initial volume 1  0.46 mm.3 N/ mm2 absolute pressure. 2. -Ve indicates capillary depression.   0 0 p  998 kg / m 3 @ 20 0 c 4  0. Capillary effect for mercury:   0.51  10 3 m 998  9. Find (i) pressure inside the cylinder assuming isothermal process (ii) pressure and temperature assuming adiabatic process.51 N / m. Capillary effect for water   0.3 m3. The air is compressed to 0.2.81  4  10 3  2. t1 = 500 C T1 = 273 + 50 = 323 0 K .81  4  10 = 7. A cylinder of 0.73575  Cos0 0 h 3  7.51  Cos130 0 h 13600  9.51 mm.3 N/mm2  0.3 10 6 N / m 2 Temperature.6 1000  13600 kg / m 3 4  0.i.   130 0 p  sp gr 1000  13. 6  p 2  p1  30  10 4     30  10 4  21.1  p 2  2 K K 1. p  RT .791N / mm 2 For temperature.  2  0.4 i.Final volume.6  10 6 N / m 2 2 0.6 p2    0.6 N / mm2 ii. p  cons tan t k RT RT p and   k  cons tan t   RT k 1  Cons tan t Tk 1  Cons tan t  R is also cons tan t  .4 1 K  0.3m 3 K = 1.3   0. Isothermal Process: P  Cons tan t (or ) p  Cons tan t p p11  p 2  2 p1 1 30  10 4  0. Adiabatic Process: p  Cons tan t or pK p K  cons tan t p1 .4 2 K  0.3 = 0.791 10 6 N / m 2  0. du iii) At y = 20 cm. Calculate the velocity gradients and shear stress at a distance of 0.4 1. (1) Where. T1V1k 1  T2V2k 1 k 1 1. a.   8. u = 120 cm / sec. Their values are determined from boundary conditions.4  426. 0 dy Substituting (i) in equation (1). where the velocity is 120 cm/sec. b and c constants.0 V   0.5 poise. Distance of vertex from plate = 20 cm. If the velocity profile of a fluid over a plate is a parabolic with the vertex 202 cm from the plate.2  273  153. Velocity at vertex.3   323  2 0. Given. i) At y = 0.5 poise   0. 120  a20  b2  400a  20b -----------(2) 2 . C = 0 Substituting (ii) in equation (1).85 10 m 2 Parabolic velocity profile equation.2 0 C 3. u  ay 2  by  C -------------. if the viscosity of the fluid is 8. u = 0 ii) At y = 20cm.2 0 K t 2  426.10 and 20 cm from the plate. u = 120 cm/se. 8.6  T2  T1  1   323    V2   0.5 Ns Viscosity. 3  2 y  12  0. b and c in equation (i) u  0. Velocity gradient.    0.    0.    0.   y 0 dy  du  at y =10 cm.6 y  12 dy Velocity gradient  du  at y = 0. du Substituting (iii) in equation (1).   dy .   y 10 dy  du  at y = 20 cm.6  20  12  12  12  0   y 20 dy Shear Stresses: du Shear stresses is given by.3  1.  2ay  b dy 0  2  a  20  b  40a  b -----------(3) solving 1 and 2.3 y 2  12 y du  0.6  10  12  6  12  6 / s.2 Substituting a.40 a 120  400 a  20 b 40 a  400 a  800 a  400 a 120 3 a     0. 400 a  20 b  0 ( ) 40 a + b = 0 800 a  20 b  0 b = .3  400 10 b  40   0. Velocity gradient.6  0  12  12 / s. we get. Velocity gradient. 0005 m 2   0.0 Nm is required to rotate the inner cylinder at 100 rpm determine the viscosity of the fluid .    0.  du  Shear stress at y = 0 . Both cylinders are 25 cm high.     i.2 N / m 2  dy  y 0  du  10 .151  0. The space between the cylinders is filled with a liquid whose viscosity is unknown.151 m Length of cylinder  L = 25 cm = 0.25 m Torque T= 12 Nm .15  0.     ii.0  10.10 cm.1178 m2   du du  u  0  u  0.85  6.     iii.    0.150 dy   0.15 m Diameter of outer cylinder = 15.15 100 Tangential velocity of cylinder u    0. Solution: Diameter of cylinder = 15 cm = 0. A 15 cm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 15. If a torque of 12.10 cm = 0.85  0  0   y 20 dy 4. Shear stress at y =    0.25 = 0. Viscosity =   DN   0.7854  0.0  5. N = 100 rpm.1N / m 2  dy  y 10  du  Shear stress at y = 20 .7854 m / s dy 0.7854 m/s 60 60 Surface area of cylinder A  D  L    0.0005 .85  12.   6 poise  2  0. 5.15 12.4 m L  90mm  90 10 3 m N = 190 rpm. The thickness of the oil film is 1. Calculate the power lost in the bearing for a sleeve length of 90 mm.1178  0.4 m and rotates at 190 rpm. The dynamic viscosity of oil.64 poise. used for lubrication between a shaft and sleeve is 6 poise.5 10 3 m 2 NT Power  W 60 D T  force  Nm. 2 F  Shear stress  Area    DL du   N / m2 dy . The shaft is of diameter 0.1178 0.7854  0. F  Shear Stress  Area   0.0   0.5 mm.0005 D Torque T  F  2   0.864 10  8.0  0.5mm  1.7854 0. 6 Ns Ns Given.   0.0005  2   0. t  1.15   0.6 2 10 m m D = 0.0005 2 12.7854 Shear force.1178  0.864 Ns / m 2 0. T  Force   180.98 m/s.4 190 Tangential Velocity of shaft. du 3.5 10 3 m. 60 60 du = change of velocity = u – 0 = u = 3.05 N D 0.98      10  3  1592 N / m 2 dy 1. 2 2 2NT 2 190  36.15 m. u    3. 60  DN   0.63 poise. dy  t  1.98 m / s.48 W 60 60 2 6.  DN u m / s. determine the shear stress at y = 0 and y = 0.If the velocity distribution over a plate is given by u  y  y 2 in which U is the 3 velocity in m/s at a distance y meter above the plate.5  10 Shear force on the shaft F = Shear stress x Area F  1592  D  L  1592    0.01 Power lost    716.4  90 10 3  180.4 Torque on the shaft.05   36. Given: 2 u y  y2 3 du 2   2y dy 3 . Take dynamic viscosity of fluid as 8.01 Ns. 3167 N / m 2  dy  y 0.   Area of small piston .15 m is given by   y 0.  du      20  2 2  dy  y 0 3 3  du    2  0. Shear stress at y = 0.15    du   0.15 3 8.367  0.863 Ns / m2 10 du   dy i.863  0.667  0. D = 10 cm .667  0.17   0.863  0.068cm 2 2  4 4 Dia of large piston. A force of 80 N is applied on the small piston Find the load lifted by the large piston when: a.63   8.5756 N / m 2  dy  y 0 ii.30 2    dy  y 0.The diameters of a small piston and a large piston of a hydraulic jack at3cm and 10 cm respectively.15 7. The pistons are at the same level b. Shear stress at y = 0 is given by  du   0      0. a  d2   3  7. The density of the liquid in the jack in given as 1000 kg/m3 Given: Dia of small piston d = 3 cm.63 poise  SI units = 0. Small piston in 40 cm above the large piston. when the small piston is 40 cm above the large piston Pressure intensity on the small piston F 80   N / cm 2 a 7.4 N / m 2 . a But pressure intensity due to 40cm.  10   78.068 This is transmitted equally on the large piston. P = pgh. 80  Pressure intensity on the large piston  7. When the pistons are at the same level Pressure intensity on small piston F 80 P  N / cm 2 a 7.96 N. F = 80 N Let the load lifted = W a. 4 Force on small piston.81  0.068  Pressure intensity of section A – A F   pressure intensity due of height of 40 cm of liquid.068  Force on the large piston = Pressure x area 80 = x 78. of liquid  p  g  h  1000  9. 7.54 N = 888.068 b.54cm 2 P A 2  Area of larger piston. 71 A  11.81  0.54 = 919.068 = 11.3924 7.4  4 N / cm 2  0. . 7 N.71 N/cm2 Pressure intensity transmitted to the large piston = 11.3924 N / cm 2 10  Pressure intensity at section 80 A A   0.71 78.71 N/cm2 Force on the large piston = Pressure x Area of the large piston  11. 1000  9.32 + 0.3924 = 11. M.E Assistant Professor DEPARTMENT OF CIVIL ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SRI VIDYA COLLEGE OF ENGINEERING AND TECHNOLOGY-626001 .SURYA. CE6303 .MECHANICS OF FLUIDS (FOR III – SEMESTER) UNIT – II FLUID STATICS & KINEMATICS PREPARED BY R. Hot wire and hot film anemometer. find the absolute pr at A 5 Define Buoyancy 7 6 Define META – CENTRE 8 7 Write a short notes on “ Differential Manometers” 8 8 Define Centre of pressure 8 9 Write down the types of fluid flow 8 10 Write a short notes on “Laminar flow”. find the pressure in the . 10 Water is flowing through two different pipes.81 N/cm2 (abs). 8 11 Define “ Turbulent flow”. to which an inverted 22 differential manometer having an oil of sp. 7 At B pr is 9. float technique.8 is connected the 10 pressure head in the pipe A is 2 m of water. 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. Gr 0. 9 17 Define Stream function 9 18 What is mean by Flow net? 10 19 Write the properties of stream function.NO 2 MARKS PAGE NO 1 Define “Pascal’s Law” 7 2 What is mean by Absolute pressure and Gauge pressure? 7 3 Define Manometers 7 4 A differential manometer is connected at the two points A and B . 10 20 What are the types of Motion? 10 21 Define “Vortex flow”. 8 12 What is mean by Rate flow or Discharge? 9 13 What do you understand by Continuity Equation? 9 14 What is mean by Local acceleration? 9 15 What is mean by Convective acceleration? 9 16 Define Velocity potential function. current meter. 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. Laser Doppler velocimetry) S. A differential manometer is connected at the two points A and B . The diameters of a pipe at the sections 1 and 2 are 10 cm and 15 cm respectively.81 N/cm2 (abs). 23 11 At B pr is 9. Find the weight lifted by the hydraulic pressure 11 when the force applied at the plunger is 500 N. determine also the velocity at section 2 . Find the discharge through the pipe if the velocity of 25 12 water flowing through the pipe section 1 is 5 m/s. pipe B for the manometer readings. find the absolute pr at A A hydraulic pressure has a ram of 30 cm diameter and a plunger of 24 4.5 cm diameter. 6 and circular in cross – section 6 is required to float in oil (sp. Find also the force acting horizontally at the top of the gate and posiotn of centre of pressure. the pipe branches.6 A wooden cylinder of sp. Sketch the arrangement in both cases A differential manometer is connected at the two points A and B of two pipes a shown in figure. The flow velocity in branch CE is 2. the liquid of sp.5 m/s. Determine the pressure of water in the main line. Find the L/D ratio for 21 the cylinder to float with its longitudinal axis vertical in oil. whereas on the downstream side the water is 16 3 available upto a height touching the top of the gate. 7 Branch CD is 0. If the pressure of water in pipe line is reduced to 9810 N/m2. 25 show that velocity potential exists and determine its form. Find also the force acting horizontally at the top of the gate which is capable of opening it.NO 16 MARKS PAGE NO A U-Tube manometer is used to measure the pressure of water in a pipe line.45.y – 4x. velocity components are given by u = x – 4y and v= . Gr 13. The 15 pressures at A and B are 1 kgf / cm2 respectively.8m in diameter and carries one third of the flow 23 in AB. Find the difference in mercury level in the differential manometer A vertical sluice gate is used to cover an opening in a dam.90) . Find the resultant force acting on the gate and position of centre of pressure. On the upstream of the gate. find the volume rate of flow in AB. where L is the height of cylinder and D is its diameters Water flows through a pipe AB 1. The sp gr of the cylinder – 0.9 .0 m has a height of 3m. Assume the gate is hinged at the bottom Find the density of a metallic body which floats at the interface 4 of mercury of sp. The opening is 2m wide and 1. lies upto a height of 1. The pipe A contains a liquid of 2 sp. Gr = 0.6 and water such that 40% of its volume 19 is sub-merged in mercury and 60% in water A solid cylinder of diameter 4. the velocity in CD and the diameter of CE In a two – two dimensional incompressible flow. Find the 20 5 meta – centric height of the cylinder when it is floating in water with its axis vertical. Gr 1. Calculate the new difference in the level of mercury. Gr = 1. The right limb of the manometer contains water and mercury is in the left limb. U tube is 10 13 cm and the free surface of mercury is in level with over the centre of the pipe. gr = 0. if 1 the difference in level of mercury in the limbs U.gr = 0.2m diameter at 3 m/s and then passes through a pipe BC 1. Find .S. the velocity in BC.5 m diameter at C.2m high. which is in excess of atmospheric pressure.5 while pipe b contains a liquid of sp .5m above the top of the gate. the fluid 8. also the stream function.. 2 MARKS QUESTIONS AND ANSWERS 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 9.81 N/cm2 (abs), find the absolute pr at A. Pr above X – X in right limb = 1000  9.81 0.6  p B Pr above X – X in left limb = 13.6 1000  9.81 0.1  900  9.81 0.2  PA Equating the two pr head Absolute pr at PA = 8.887 N/cm2 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. Thus the particles move in laminas or layers gliding over the adjacent layer. This type of flow is also called stream – line flow or viscous flow/ 11. Define “ Turbulent flow”. Turbulent flow is that type of flow in which the fluid particles move in a zig –zag way. Due to the movement of fluid particles in a zig – zag way. 12. What is mean by Rate flow or Discharge? It is defined as the quantity of a fluid flowing per second through a section of a pipe or channel. For an incompressible fluid( or liquid) the rate of flow or discharge is expressed as volume of fluid flowing across the section per section. For compressible fluids, the rate of flow is usually expressed as the weight of fluid flowing across the section. The discharge (Q) = A X V Where, A = Cross – sectional area of pipe. V = Average velocity of fluid across the section. 13. What do you understand by Continuity Equation? The equation based on the principle of conservation of mass is called continuity equation. Thus for a fluid flowing through the pipe at all the cross-section, the quantity of fluid per second is constant. A1V1 = A2V2.. 14. What is mean by Local acceleration? Local acceleration is defined as the rate of increase of velocity with respect to time at a given point in a flow field. In equation is given by the expression (Ә u / Әt ) , ( Әv / Әt) or ( Әw / Әt) is known as local acceleration. 15. What is mean by Convective acceleration? It is defined as the rate of change of velocity due to the change of position of fluid particles in a fluid flow. The expressions other than (Ә u / Әt ) , ( Әv / Әt) and ( Әw / Әt) in the equation are known as convective acceleration. 16. Define Velocity potential function. It is defined as a scalar function of space and time such that its negative derivative with respect to any direction gives the fluid velocity in that direction. It is defined by Φ (Phi). Mathematically, the velocity, potential is defined as Φ = f ( x,y,z) for steady flow such that. u = - (Ә Φ/Әx) v = - ( Ә Φ/Әy) w = - ( Ә Φ/Әz) where, u,v and w are the components of velocity in x.y and z directions respectively. 17. Define Stream function. It is defined as the scalar function of space and time, such that its partial derivative with respect to any direction gives the velocity component at right angles to that direction. It is denoted by ψ (Psi) and only for two dimensional flow. Mathematically. For steady flow is defined as ψ = f(x,y) such that , (Ә ψ / Әx) = v (Ә ψ / Әy) = - u. 18. What is mean by Flow net? A grid obtained by drawing a series of equipotential lines and stream lines is called a flow net. The flow net is an important tool in analyzing the two – dimensional irrotational flow problems. 19. Write the properties of stream function. The properties of stream function (ψ) are: 1. If stream function (ψ) exists, it is possible case of fluid flow which may be rotational or irrotational. 2. If stream function (ψ) satisfies the Laplace equation, it is a possible case of irrotational flow. 20. What are the types of Motion? 1. Linear Translation or Pure Translation. 2. Linear Deformation. 3. Angular deformation. 4. Rotation. 21. Define “Vortex flow”. Vortex flow is defined as the flow of a fluid along a curved path or the flow of a rotating mass of a fluid is known ‘ Vortex Flow’. The vortex flow is of two types namely: 1. Forced vortex flow, and 2. Free vortex flow. 22. Water is flowing through two different pipes, to which an inverted differential manometer having an oil of sp. Gr 0.8 is connected the pressure head in the pipe A is 2 m of water, find the pressure in the pipe B for the manometer readings. pA Pr heat at A   2m of water. pg p A  p  g  2  1000  9.81 2 = 19620 N/ m2 81 0. A hydraulic pressure has a ram of 30 cm diameter and a plunger of 4.6 1000  9.07068 m 2 4 4   Area of plunger.81 0. D = 30 cm = 0. Pr below X – X in left limb = PA – p1 gh1 = 19620 .76 N / m 2  1. F = 500 N To find : Weight lifted = W   Area of ram. we get.76  18599 . Pr above X – X in right limb = 1000  9. PB  16677  1922 .32  0. a d2  0.2  PA Equating the two pr head Absolute pr at PA = 8. 24.12  PB 1922.81 0. A D2  0.887 N/cm2. find the absolute pr at A.1  900  9.0452  0.5 cm = 0.1000 x 781 x 0.000159 m 2 4 4 .3 m Dia of plunger.8599 N / cm 2 23.3 = 16677 N/m2 Pr below X – X in right limb p B 1000  9.5 cm diameter. Given: Dia of ram.045 m Force on plunger.81 N/cm2 (abs). A differential manometer is connected at the two points A and B .1  800  9. d = 4. At B pr is 9.6  p B Pr above X – X in left limb = 13.81 0.81 0.76 Equating two pressure. Find the weight lifted by the hydraulic pressure when the force applied at the plunger is 500 N. Discharge through pipe is given by equation Q = A1 X V1 = 0. Force on plunger N Pressure intensity due to plunger   Area of plunger m 2 F 500   N / m2 a 0. At section 2. Given: At section 1.1)2 = 0.07068 Weight  314465 .01767 m2. The diameters of a pipe at the sections 1 and 2 are 10 cm and 15 cm respectively.007854 m2. We have A1V1 = A2V2. D2 = 15 cm = 0. Find the discharge through the pipe if the velocity of water flowing through the pipe section 1 is 5 m/s. determine also the velocity at section 2. Hence the pressure intensity at 500 ram   314465 .07068  22222 N  22.007854 /0.15)2 = 0. V1 = 5 m/s.01767) X 5 = 2. . A2 = (π / 4) X(0. the intensity of pressure will be equally transmitted in all distance . A1 = (π / 4) X D12 = (π / 4) X (0.007544 X 5 = 0. V2 = ( A1V1 / A1) = ( 0. 25. D1 = 10 cm = 0.1 m.03927 m3/ s.00159 weight W W Pressure intensity at ram    Area of ram A 0..00159 Due to Pascal law.4 0.07068 W  314465 .4 N / m2 0.222 KN.4  0.22 m/s. using equation. 1. Solution.15 m. Hence pressure at B should be equal to pressure at C. U tube is 10 cm and the free surface of mercury is in level with over the centre of the pipe. If the pressure of water in pipe line is reduced to 9810 N/m2.1 m. at point A) The point B and C lie on the same horizontal line. A U-Tube manometer is used to measure the pressure of water in a pipe line. Let PA = pr of water in pipe line (ie. which is in excess of atmospheric pressure. Given.  PA  p  g  h where . if the difference in level of mercury in the limbs U. But pressure at B = Pressure at A and Pressure due to 10 cm (or) 0.1 m  PA  1000  9. The right limb of the manometer contains water and mercury is in the left limb. Sketch the arrangement in both cases. Difference of mercury = 10 cm = 0.1m of water. Calculate the new difference in the level of mercury.81 0. 16 MARKS QUESTIONS AND ANSWERS 1. P = 1000kg/m3 and h = 0. Determine the pressure of water in the main line.1  PA  981N / m 2 (i) . and D* show the final conditions.6 N / m 2 Given p A  9810 N / m 2 II part: In this case the pressure at A is 9810 N/m2 which is less than the 12360.6 N (ii) But pressure at B is = to pr @ c. (or) .81 0. Hence.6 p A  13341 . C and D show the initial condition.6 1000  9.6  981  12360 . x = Ries of mercury in left limb in cm Then fall of mercury in right limb = x cm.6 1000 kg / m 3 h0 = 10 cm = 0.Pressure at C = Pressure at D + pressure due to 10 cm of mercury 0  P0  g  h0 where po for mercury  13.1 m Pressure at C  0  13. Hence the mercury in left limb will rise. Let. = pressure at D* + pressure due to (10-2x) cm of mercury. Whereas points B*. The pressure at B * = pressure at C* Pressure at A + pressure due to (10-x) cm of water.1 = 13341.6 N/m2. The rise of mercury in left limb will be equal to the fall of mercury in right limb as the total volume of mercury remains same. C*. equating (i) and (ii) PA  981  13341 . The points B. Find the difference in mercury level in the differential manometer. S1 = 1.016 cm. Gr = 1.992cm 262  New difference of mercury = 10 . The pressures at A and B are 1 kgf / cm2 respectively.81   10  2 x    9.5 while pipe b contains a liquid of sp . 2.81 . Gr of liquid at A. A differential manometer is connected at the two points A and B of two pipes a shown in figure. gr = 0.2 x cm  10  2  0. we get. Sp.81  100   0 13. Gr of liquid at B.992 = 8. The pipe A contains a liquid of sp.5  p1 = 1500 Sp. S2 = 0.9 . PA = 1kgf / cm2 = 110 kgf / m 4 2 .81  10  x     9.9  p2 = 900 Pr at A.81  100  Dividing by 9. PA  p1 g  h1  pD *  p 2  g  h2 (or) 9810 1000  9. Given.6 1000  9. 1000  100  10 x  1360  272 x 272 x  10 x  1360  1100 262 x = 260 260 x  0.8 9. 81  h  2  1.6  1000  9.81 h  1500  9. lies upto a height of 1.9h  19.5 or 12.0 0.8  18  0.7 3.5  0. Find the resultant force acting on the gate and position of centre of pressure.81h  7500  93. we get.81 N) Pr at B. Gr 1.7 h = 2.6 1000  9. we get 13.81  h  7500  9.9h  1.81.8 17. A vertical sluice gate is used to cover an opening in a dam.81 N / m 2 Density of mercury  13.1cm 12.8 kgf / cm2  1. On the upstream of the gate.81 2  3  PA  13. PB = 1.6  0.8 13.6h  17.2m high.8 10 4  9.81 Dividing by 1000  9.81N / m 2 (1 kgf= 9.5m above the top of the gate.81 h  2  PB  900  9.181m  18.8 10 4  9.45.8110 4  900  9.81  h  2  1. 13. the liquid of sp.3 2. Find also the force acting horizontally at the top of the gate and posiotn of centre of pressure. whereas on the downstream side the water is available upto a height touching the top of the gate.9  18 13.81 Equating two pressure. The opening is 2m wide and 1.3 h  0.6 1000 kg / m 3 Pr above X – X in left limb = 13.8 10 4  9.5  10  h  2. Find also the force acting horizontally at the .81  9.81  10 4 Pr above X – X in the right limb  900  9.9h  19.  10 4  9.6h  7.6 1000  9. 1.2  1.4  2.6m 2 F2  1000  9.2  2.6  14126 N . Assume the gate is hinged at the bottom.G of gate from free surface of liquid.2  0. F2  p 2 gAh2 p2 = 1000 kg/m3 h2  Depth of C.45 1000  1450 kg / m 3 F1 = force exerted by water on the gate F1  p1 g Ah1 p1  1. 1  1.5   2.4  0. b = 2m A  b  d  2 1.45 1000  1450 kg / m 2 h  depth of C.2 m p1  1.81 2. Given.1 = 71691 N F2  p 2 gAh2 Centre of Pressure (h*): Centre of pressure is calculated by using the “Principle of Moments” which states that the moment of the resultant force about an axis is equal to the sum of moments of the components about the same axis.4m 2 d = 1.1m 2 F1  1450  9.81 2.G of gate from free surface of water. top of the gate which is capable of opening it. 2 .288m 4 12 12 0.4 m2.8 = 0.288 h2*   0.1571m 2.6 Distance of F2 from hinge = 1. Resultant force on the gate  F1  F2  71691 14126 = 57565 N.i.1  2.4  0. 57565 .4  57565 38921  5650 . Position of centre of pressure of resultant force: IG h1 *   h1 Ah1 bd 3 2 1.8m 2. ii.4  m above hinge. 0.7 . h2  0. A = 2.1571 = 0. The force F2 will be acting at a depth of h 2* from free surface of water nd is given by IG h2*   h2 A h2 where IG = 0.6  0.1   Distance of F1 from hinge  1.6  0.5429  14126  0.2  h1  * = 2.4  2.6m.1  0.288 h1*   2.4 m.5429 m.0571  2.2. The resultant force 57565 N will be acting at a distance given by 71691  0.2  0.2 3 IG    0.288 m4.0.5  1. 2 = 27725.578 m above the hinge.5429  14126  0. = 0. Force of buoyancy due to water = Weight of water displaced by body. 4.4  F1  0. . Find the density of a metallic body which floats at the interface of mercury of sp.5429 F1  0.4 F 1.5 N.6Vm 3 100 For the equilibrium of the body Total buoyant force (upward force) = Weight of the body. But total buoyant force = Force of buoyancy due to water + Force of buyance due to mercury.2 71691  0. iii.5429  F2  0. F 1.2  F2  0. F1 and F2 about the hing. Solution: Let the volume of the body = V m3 Then volume of body sub-merged in mercury 40  V  0.4Vm 3 100 Volume of body sub-merged I water 60   V  0. Gr 13.6 and water such that 40% of its volume is sub-merged in mercury and 60% in water.4  1. Force at the top of gate which is capable of opening the gate: Taking moment of F. h = 3.  g  Density of mercury  Volume of mercury displaced.6 1000  g  0.0 m has a height of 3m.00 kg / m3 5. gr of cylinder = 0. The sp gr of the cylinder – 0.6 1000  0.6 1000  Volume of body in mercury  g 13.6 .6 Depth of immersion of cylinder = 0.4  600  54400 = 6040. A solid cylinder of diameter 4. Find the meta – centric height of the cylinder when it is floating in water with its axis vertical. Given Dia of cylinder .4V  p  g  V p  600  13600  0.00 kg / m3  Density of the body = 6040. Weight of the body = Density x g x volume of body  p  g V where p is the densityof the body For equilibrium.0 m Sp.6  V  13.6  V N Force of buoyancy due to mercury = Weight of mercury displaced by body.0 m Height of cyinder. Total buoyant force = Weigt of the body 1000  g  0. = Density of water x g x Volume of water displaced.  1000  g  0.6 x 3.4V N.  g 13. D = 4. = 1000  g  volume of body in water.0 . 05 m.AB = 1.centre (M) below the centre of gravity (G) .6 m.0  1.6  0.8 AB   0.0.6   0.5 – 0.55  0.   D 4  Depth of immersion 4   42  1.8m 3 4   4. Ve sign indicates meta.0 4 4 64 V = volume of cylinder in water.6 16 1.0 4 GM  64  0.8 4 4 1 4.6   4. 1 Now the meta – centric height GM is given by equation GM   BG V I = Moment of Inertia about Y – Y axis of plan of the body    D4   4.9 = 0.9m 2 3 AG   1.8 m 1. = 1.0 2 1    0.8 18 = .5m 2 BG = AG . . 6 Sp gr of oil.81  D 2  h  0. where L is the height of cylinder and D is its diameters. Find the L/D ratio for the cylinder to float with its longitudinal axis vertical in oil.6. AB   L 2 2  3  L L 3L  2L L BG  AG  AB     2 3 6 3  D4 GM  1  BG I 4 V = Volume of cylinder in oil.9 3 L Dts of centre of gravity C from A.81 4 4 L  0. Solution: Dia of cylinder = D Height of cylinder = L Sp gr of cylinder.6  L 2 h  L 0. depth of cylinder immersed in oil = h Buoyancy principle Weight of cylinder = Weight of oil dispersed. A wooden cylinder of sp.9 1000  9.6  h  0.90) . S1 = 0.6 1000  9.9 Let.gr = 0.6 and circular in cross – section is required to float in oil (sp. Gr = 0. V 64 . AG  2 h 1 2  The distance of centre of buoyancy B from A.9 0.   D 2  L  0. S2 = 0. 8m in diameter and carries one third of the flow in AB. the velocity in CD and the diameter of CE. . the velocity in BC.2m diameter at 3 m/s and then passes through a pipe BC 1. GM should be +Ve or 3D 2 L GM  0 (or)  0 32 L 6 3D 2 L 3  6 L2  (or)  2 32 L 6 32 D L2 18 9  (or) D 2 32 16 L 9 3 L 3     D 16 7 D 4 Experimental Method of Determination of Metacentric Height W = Weight of vessel including w1 G = Centre of gravity of the vessel B = Centre of buoyancy of the vessel 1 x GM  Tan 7.5 m diameter at C.5 m/s. Branch CD is 0.   D2h 4   4  D  2 2 1  64  1 D  D 3D 2   V   2  16 h 2 32 L  D h 16  L 4  3 3D 3 L GM   32 L 6 For stable equilibrium. find the volume rate of flow in AB. The flow velocity in branch CE is 2. Water flows through a pipe AB 1. the pipe branches. VCE = 2. Given: Diameter of Pipe AB. Flow through CE. DCD = 0.0 m/s. VAB X Area of pipe AB = VBC X Area of Pipe BC 3 X (π / 4) X (DAB )2 = VBC X (π / 4) X (DBC )2 3 X (1.2)2 = 3.131 / 0. Velocity of flow in pipe BC = VBC m3/s.2) 2 = VBC X (1.393 m3/s.22)/ 1. Now the flow rate through AB = Q = VAB X Area of AB = 3 X (π / 4) X (DAB )2 = 3 X (π / 4) X (1. DAB = 1.25 m/s.Solution.2 m. . Velocity of flow through AB VAB = 3.5 m/s.52 = 1. of Pipe BC. Applying the continuity equation to pipe AB and pipe BC. Diameter of pipe CE = DCE Then flow rate through CD = Q / 3 And flow rate through CE = Q – Q/3 = 2Q/3 (i). of Branched pipe CD. The flow rate through pipe CD = Q1 = Q/3 = 3. (iv).131 = VCD X (π / 4) X (0.8m.92 m/s. DBC = 1. (ii). Dia.5)2 VBC = ( 3X1.5026 = 2.393 /3 = 1.131 m3/s. Velocity of flow in pipe CE.8)2 VCD = 1. Q1 = VCD X Area of pipe CD X (π / 4) (CCD)2 1. Dia. Velocity of flow in pipe CD = VCD m3/s.5m. (iii). Let the rate of flow in pipe AB = Q m3/s. we get C = (y2 / 2) + C1.y – 4x (Әu / Әx) = 1 & (Әv/ Әy) = -1.(1) Ә Φ / Әy = .( x2 / 2 ) + 4xy + C---.131 = 2.(3) Where C is a constant of Integration. show that velocity potential exists and determine its form. which is independent of ‘x’.(2) Integrating equation(i)..( -y – 4x) = y + 4x. 8. This constant can be a function of ‘y’. Q2 = Q – Q1 = 3. In a two – two dimensional incompressible flow. which is independent of ‘x’ and ‘y’. Differentiating the above equation. Solution. Where C1 is a constant of integration. Let the velocity components in terms of velocity potential is given by Ә Φ / Әx = . Let Φ = Velocity potential.0735 m Diameter of pipe CE = 1.262 m3/s. we get Φ = . (Әu / Әx) +( Әv/ Әy) = 0 hence flow is continuous and velocity potential exists.0735m.v = . i.---------------.5 X π) = 1.263 = 2. equation (3) with respect to ‘y’.y – 4x. Q2 = VCE X Area of pipe CE = VCE X (π / 4) (DCE)2 2.u = .393 – 1. . we get Ә Φ / Әy = 0 + 4x + Ә C / Әy But from equation (3).263 X4)/ (2. Find also the stream function. the fluid velocity components are given by u = x – 4y and v= .e. we have Ә Φ / Әy = y + 4x Equating the two values of Ә Φ / Әy .x + 4y ---------------.5 X (π / 4) (DCE)2 DCE = (2. Given: u = x – 4y and v = . we get 4x + Ә C / Әy = y + 4x or Ә C / Әy = y Integrating the above equation.( x – 4y) = . Ψ = . Differentiating equation (6) w. we have Әψ/Әy = -x + 4y Equating the values of Әψ/Әy. Әψ/Әx = . Value of stream functions Let Әψ/Әx = v = -y .r.x + Әk/Әy or Әk/Әy = 4y.u = . Integrating the above equation. ‘x’ we get Ψ = .to.yx – (4x2/2) + k -----------------(6) Where k is a constant of integration which is independent of ‘x’ but can be a function ‘y’. Let Әψ/Әy = . ----------------------.( x – 4y) = x + 4y ------(5) Integrating equation (4) w. we get C = y2/ 2.yx – 2 x2 + 2 y2 . Substituting the value of k in equation (6).r. ‘y’ we get.x – 0 + Әk/Әy But from equation (5). we get .(x2/2) + 4xy + y2/2.t. we get k = 4y2 /2 = 2y2. Φ = . we get. Substituting the value of C in equation (3).(4). we get.Taking it equal to zero.4x. The total energy consists of pressure energy.204 + 5 = 35.. The flow is steady (iii).43 N/cm2 = 29. Water is flowing through a pipe of 5 cm diameter under a pressure of 29. It states that in a steady.204m .81)) = 0. The fluid is ideal. Pressure ρ = 29. datum head z =5m total head = Pressure head + Velocity head + Datum head pressure head = (p/ ρg) = ( 29. kinetic energy and potential energy or datum energy.0 m/s.23 N/m2 ` velocity.43X104/ (2X9.0 m/s. (ii). Te flow is incompressible. These energies per unit weight of the fluid are: Pressure Energy = p / ρg Kinetic energy = v2 / 2g Datum Energy =z The mathematically.204 m Total head = (p/ (ρg)) + (v2 / 2g) + z = 30 + 0. v = 2. (iv). Bernoulli’s theorem for steady flow of an incompressible fluid. which is 5 cm above the datum line.e.43 N/cm2 (gauge) and with mean velocity of 2.5 m. Viscosity is zero. 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 Assumptions made in the derivation of Bernoulli’s equation: (i). ideal flow of an incompressible fluid.81)) = 30 m kinetic head = (v2/ 2g) = ( 2X2/(2X9. Given: Diameter of the pipe 5 cm = 0. the total energy at any point of the fluid is constant. i. find the total head or total energy per unit weight of the water at cross – section. Bernoulli’s theorem is written as (p/w) + (v2 / 2g) + z = Constant. The flow is irrotational. 76 N / m 2  1.81  0.007854 /0. D2 = 15 cm = 0.007854 m2. pA Pr head at A   2m of water. V2 = ( A1V1 / A1) = ( 0.1)2 = 0.15 m. find the pressure in the pipe B for the manometer readings. Discharge through pipe is given by equation Q = A1 X V1 = 0. to which an inverted differential manometer having an oil of sp.8 is connected the pressure head in the pipe A is 2 m of water.03927 m3/ s.01767 m2. At section 2.76  18599 . we get.15)2 = 0. PB  16677  1922 .. V1 = 5 m/s.01767) X 5 = 2. We have A1V1 = A2V2.81  0. D1 = 10 cm = 0. A2 = (π / 4) X(0.007544 X 5 = 0. 1.1  800  9. A1 = (π / 4) X D12 = (π / 4) X (0. .81  2 = 19620 N/ m2 Pr below X – X in left limb = PA – p1 gh1 = 19620 .8599 N / cm 2 The diameters of a pipe at the sections 1 and 2 are 10 cm and 15 cm respectively. Gr 0. Solution.Water is flowing through two different pipes.76 Equating two pressures. Find the discharge through the pipe if the velocity of water flowing through the pipe section 1 is 5 m/s.22 m/s. determine also the velocity at section 2.3 = 16677 N/m2 Pr below X – X in right limb p B  1000  9. Using equation.1000 x 781 x 0. pg p A  p  g  2  1000  9. Given: At section 1.1 m.12  PB  1922 . gr. The stagnation pressure head is 6mm and static pressure head is 5m.6 / 1. hs = 6mm.026  h = x Sg / So  1 = 0.81 X 20 .17 m Sp. of mercury level x = 170 mm = 0.393 m/s. = ( 6. Of sea-water So = 1.34 m/s. Calculate the velocity of flow assuming the co-efficient of tube equal to 0. find the speed of the sub-marine knowing that the sp. A sub-marine moves horizontally in a sea and has its axis 15 m below the surface of water.834 = 6. the resulting net force.81 X 1 = 4.01 km/hr. Given: Stagnation Pressure head. Static pressure ht = 5 mm. Of mercury Sg = 13. Write the equations of motion. gr.6 Sp.A pitot – static tube is used to measure the velocity of water in a pipe. of mercury is 13. The difference of mercury level is found to be 170 mm. A pitot tube properly placed just in front of the sub-marine and along its axis connected to the two limbs of a U – tube containing mercury. Given : Diff.026 with respect fresh water.0834 m V= 2 gh = 2 X 9.026 )  1 = 2. Fx = (Fg)x + (Fp)x+(Fv)x+(Ft)x Where Fg = gravity force Fp = Pressure force Fv = force due to viscosity Ft = force due to turbulence Fc = force due to compressibility . h =6–5=1m Velocity of flow V = Cv 2 gh = 0. gr.6 and that of sea-water is 1.393X60X60 / 1000) km/hr = 23.98. Fx = (Fg)x + (Fp)x+(Fv)x+(Ft)x+(Fc)x If the force due to compressibility.17 (3. Fc is negligible.98 2 X 9. S = maximum vertical height attained by the particle. It consists of three pats (i). . Free liquid jet Free of liquid jet is defined as the jet of water coming out from the nozzle in atmosphere. Formula to find the maximum height attained by the jet U 2 sin 2  S= 2g Where. It is based on the principle that if the velocity of flow at a point becomes zero.Venturi meter. h = difference of pressure head in terms of fluid head flowing through venture meter Dynamics of fluid flow The study of fluid motion with the forces causing flow is called dynamics of fluid flow. θ = angle with horizontal direction. Venturimeter is a device used for measuring the rate of flow of a fluid flowing through a pipe. The path traveled by the free jet is parabolic. a2 = area at the throat Cd = co-efficient of venture meter. a1 a 2 Q = Cd X 2 gh a12  a 22 Where a1 = area of the inlet Venturi meter. which relates the acceleration with the forces. A short converging part (ii) Throat and (iii) Diverging part. Pitot – tube. U = velocity of jet of water. Write down the formulae for finding the discharge in Venturimeter. The dynamic behavior of the fluid flow is analyzed by the Newton’s second law of motion. g = acceleration due to gravity. Pitot tube is a device used for measuring the velocity of flow at any point in a pipe or channel. we get .Euler’s Equation of motion. The forces acting on the cylindrical element are: a. as = . Pressure force pdA In the direction of flow. This is equation of motion in which the forces due to gravity and pressure are taken into consideration. p b. Weight of element gdAds Let  is the angle between the direction of flow and the line of action of the weight of element.  p  pdA   p  dA  gdAds cos  s   dAdsXas        (1) Where as is the acceleration in the direction of ‘s’ dv Now. =0 dt v v as = s Substituting the value of ’as’ in equation (1) and simplifying the equation. The resultant force on the fluid element of ‘s’ must be equal to the mass on the fluid element X acceleration in the direction ‘s’. This is derived by considering the motion of a fluid element along a streamline in which flow is taking place in s-direction as shown in fig. where v is a function of ‘s’ and ‘t’. dt v dv v vv v  ds  =      v s dt t s t  dt  dv if the flow is steady. Consider a cylindrical element of cross-section dA and length dS.  s  c. Pressure force  p  ds dA opposite to the direction of flow. If the pressure at section 1 is 39. If the pressure at section 2 is 4m above the datum. we get .24 N/cm2. 4 P2 = Z1 = 4.456 m/s.10m  2 A2 = (0.314m2. we have cos  = ds 1 p dz vv g  o  s ds s p (or )  gdz  vdv  0 -------------------------------(2).035 / 0.2m  2 A1 = (0. Applying Bernoulli’s Equations at sections at 1 and 2.  Equation 2 is known as Euler’s equation of motion The water is flowing through a pipe having diameters 20 cm and 10 cm at sections 1 and 2 respectively. D2 = 0.1) = 0. find the intensity of pressure at section 2. D1 = 20 cm = 0. Z1 = 6. Given: At section 1.24 N/cm2 = 39. 4 P1 = 39. p vv  dsdA  gdAds cos  dAdsX s s p vv DividingbydsdA.0314 = 1.035 / 0.0785 = 4.24 X104 N/m2. the section 1 is 6m above datum.0785m2.035m3/s Q = A1V1 = A2V2 V1 = Q / A1 = 0.2) = 0.  g cos  s s p vv or  g cos  v 0 s s dz But from fig.0m Rate of flow Q = 35 lit/sec = 35/1000 = 0.114 m/s V2 = Q / A2 = 0.0m At section 2. The rate of flow through pipe is 35 lit/sec. 4 Dia.981 X 104 N/m2.8. calculate the flow rate.0 40+0. 2 2 p V p V 1  1  z1 = 2  2  z2 g 2g g 2g Or (39.8 X1000 = 800 kg/m3. Difference of pressure head (pB – pA)/ ρg = (9810 / (800X9. A is 2 m above B. of oil. If the gauges at A and B are replaced by tubes filled with the same liquid and connected to a U – tube containing mercury.81) + ((1.81) + 6.051X9810/104) = 40.gr.063 – 5. = 9810 N/m2.81)+((4.012+4. At B DB = 8 cm = 0.16) 2  0.81)+4. = 0. the pressure gauge readings have shown that the pressure at B is greater than at A by 0.063 = (p2/9810) + 5. So = 0.24 X104 / 1000X9.012 (p2/9810) = 46.81)) = 1.0 46.. two pressure gauges have been installed at A and B where the diameters are 16 cm and 8 cm respectively.08) 2  0. Dia at A.063+6.25 Applying Bernoulli’s theorem at A and B and taking reference line passing through section B.981 N/cm2. Difference of pressures. we get p A V A2 p V2   ZA  B  B  ZB g 2 g g 2 g .114)2/2X9. Given: Sp.005026 m2 4 (i). AB = = X (0. calculate the difference of level of mercury in the two limbs of the U-tube.0 = (p2/9810)+1. DA = 16 cm = 0. ρ = 0.012 = 41.16m  2 Area at A.456)2/2X9. pB – pA = 0.981 N/cm2.8 Density.08m  Area at B. A1 = X (0.0201m . Neglecting all losses.0 = (P2 /1000X9.051 p2 = (41.27 N/cm2 In a vertical pipe conveying oil of specific gravity 0. Difference of mercury in the U –tube.1.8  x = (0.99 X 0.75 / 16 ) = 0.75 X 2 X 9.75 =  --------------------. p A pB V2 V2   ZA  ZB  B  A g g 2g 2g p A  pB V2 V2  2.6   0.0 =  = 1.0-0 = 0.25 + 2.01989 m3/s.16) 2 V XA 4 VB = A 1   4V A A2  (.81 VA =  0. we get 16V A2 V A2 15V A2 0.75  x   1 = x X 16  0. we get VAXA1 = VBXA2  VA X (. 15 Rate of flow. 13. Q = VAXA1 = 0.08) 2 4 Substituting the Value of VB in equation (i). Expression for loss of head due to friction in pipes or Darcy – Weisbach Equation.25+2.(i) 2g 2g Now applying continuity equation at A and B. (ii).04687 cm.25 2g 2g g VB2 V A2 0.0  0.0  B  A g 2g 2g VB2 V A2 pB  p A .75 =   2g 2g 2g 0.0201 = 0.  Sg  Then h = x  1  So   pA   pB  p A  pB Where h =   Z A     Z B    ZA  ZB  g   g  g = -1. Let h = difference of mercury level. .99m / s.75. f1 = Frictional resistance per unit wetted area per unit velocity. Frictional force F1 Resolving all forces in the horizontal direction. F = f1 x π d l x V2 [ Wetted area = π d x L.V2 = are the values of pressure intensity and velocity at section 2-2. and Velocity V = V1 = V2] F1 = f1xPxLxV2 ----------. (P1/ ρg) = (P2/ρg)+hf hf = (P1/ ρg) . And P2. Applying Bernoulli’s equation between sections 1-1 & 2-2 Total head 1-1 = total head at 2-2 + loss of head due to friction between 1-1&2-2 (P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+hf ------------(1) but Z1 = Z1 [ pipe is horizontal ] V1= V2 [ diameter of pipe is same at 1-1 & 2-2] (1) becomes. But from (1) we get P1 – P2 = ρg hf . L = length of the pipe between the section 1-1 and 2-2 d = diameter off pipe. having steady flow as shown figure.(P2/ρg) frictional resistance = frictional resistance per unit wetted area per unit velocity X wetted area X velocity 2.(2).. 1) Pressure force at section 1-1 = P1X A 2) Pressure force at section 2-2 = P2 X A 3). Let P2 = Velocity of flow at section 1-1.Consider a uniform horizontal pipe. [π d = wetted perimeter = p] The forces acting on the fluid between section 1-1 and 2-2 are. Let 1 -1 and 2-2 is two sections of pipe. hf = loss of head due to friction. Let P1 = pressure intensity at section 1-1. P1 A – P2A – F1 = 0 (P1-P2)A = F1 = f1xPxLxV2 (P1-P2) = (f1xPxLxV2 / A ). Equating the values of (P1 – P2) we get ρg hf = (f1xPxLxV2 / A ). hf = ( f1 / ρg) x ( 4/d) x LxV2. It consists of three pats (i). Expression for rate of flow through Venturimeter. hf = (f1 / ρg) X (P/A) X LX V2 (P/A) = (π d / (π d2/4)) = (4/d) Hence. continuity equation at 1 & 2 a1V1 = a2V2 or V1 = ( a2V2/ a1) --------------. This equation is commonly used to find loss of head due to friction in pipes. g V22 V12 h=  ------------. where f is the co – efficient of friction 4 fLV 2 hf = 2 gd This equation is known as Darcy – Weisbach equation.(2). 2g 2g Now applying. Venturi meter is a device used for measuring the rate of flow of a fluid flowing through a pipe. A short converging part (ii) Throat and (iii). P2. V2. a2 are the corresponding values at section 2 Applying the Bernoulli’s equation at section 1 & 2 (P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2 since the pipe is horizontal Z1 = Z2 (P1/ρg) + (V12 / 2g) = (P2/ρg) + (V22 / 2g) P1  P2 V 22 V12   g 2g 2g P1  P2 We know that is the difference or pressure head and is equal to h. . Diverging part Let d1 = diameter at inlet or at section 1 Let P1 = pressure at section 1 Let V1 = velocity of fluid at section 1  Let a1 = area of section 1 = d 12 4 And d2. Putting (f1 / ρ) = (f / 2).(1). 0 m/s. Velocity of flow in pipe CD = VCD m3/s. Velocity of flow in pipe CE. The flow velocity in branch CE is 2. Q = a2V2 a 2 a1 2 gh Q= theoretical discharge a12  a 22 Actual discharge a 2 a1 2 gh Q act = Cd X a12  a 22 Where Cd = co – efficient of venturi meter. the pipe branches.5m. of Branched pipe CD.5 m diameter at C. DCD = 0. Velocity of flow in pipe BC = VBC m3/s.5 m/s.2 m. Branch CD is 0. Solution. Dia.5 m/s. DAB = 1.2m diameter at 3 m/s and then passes through a pipe BC 1. a12  a 22 Discharge. Given: Diameter of Pipe AB. of Pipe BC.8m in diameter and carries one third of the flow in AB. Water flows through a pipe AB 1. the velocity in BC. DBC = 1. Velocity of flow through AB VAB = 3. Find the volume rate of flow in AB. VCE = 2. the velocity in CD and the diameter of CE. Dia.8m. Sub (2) in equation (1) we get 2  a 2V   2  V 22  a1  V 22  a 22    h=  = 1   2g 2g 2 g  a12  V22 = 2gh ( a12 / ( a12 – a 22)) a1 V2 = 2 gh . . Let the rate of flow in pipe AB = Q m3/s. 393 /3 = 1.2)2 = 3.22)/ 1. Determine the rate of flow.5)2 VBC = ( 3x1.263 X4)/ (2.5 X π) = 1.8) 2 VCD = 1.393 – 1. Flow through CE. A horizontal Venturimeter with inlet and throat diameters 30 cm and 15 cm respectively is used to measure the flow of water.131 / 0.131 = 2.131 = VCD x (π / 4) x (0. (iii). The reading of differential manometer connected to the inlet and the throat is 20 cm of mercury.25 m/s. Q2 = VCE X Area of pipe CE = VCE X (π / 4) (DCE) 2 2. Q2 = Q – Q1 = 3.131 m3/s.98.92 m/s.263 = 2.262 m3/s. Q1 = VCD x Area of pipe CD x (π / 4) (CCD) 2 1.5026 = 2. (iv).5 X (π / 4) (DCE) 2 DCE = (2. Now the flow rate through AB = Q = VAB X Area of AB = 3 X (π / 4) X (DAB )2 = 3 X (π / 4) X (1.0735 m Diameter of pipe CE = 1. Given: d1 = 30 cm .393 m3/s. VAB X Area of pipe AB = VBC x Area of Pipe BC 3 X (π / 4) x (DAB )2 = VBC x (π / 4) x (DBC )2 3 x (1.0735m.2) 2 = VBC x (1. Take Cd = 0. Applying the continuity equation to pipe AB and pipe BC. Diameter of pipe CE = DCE Then flow rate through CD = Q / 3 And flow rate through CE = Q – Q/3 = 2Q/3 (i). (ii).52 = 1. The flow rate through pipe CD = Q1 = Q/3 = 3. 6 / 1) – 1] = 252. a 2 a1 2 gh Q act = Cd x a12  a 22 706.81x252 = 0.7 cm2 Cd = 0.  Sh  Difference of pressure head.98 x 2 706.85x176. .0 cm of mercury.85  176.98 Reading of differential manometer = x = 20 cm of mercury.7 2 = 125756 cm3 / s = 125.   a1 = d 12 = (30) 2 4 4 = 706.7 2 x9.756 lit / s. h = x  1  So  = 20 [ (13.85 cm2 d2 = 15 cm  2  a2 = d = (15) 2 4 2 4 = 176. L Major energy loss and minor energy loss in pipe The loss of head or energy due to friction in pipe is known as major loss while the loss of energy due to change of velocity of the flowing fluid in magnitude or direction is called minor loss of energy. . This phenomenon of rise in pressure is known as water hammer or hammer blow. Equivalent pipeline An Equivalent pipe is defined as the pipe of uniform diameter having loss of head and discharge of a compound pipe consisting of several pipes of different lengths and diameters. It is defined as the line which gives the sum of pressure head ( P/ρg) and datum head (z) of a flowing fluid in a pipe with respect to some reference line or is the line which is obtained by joining the top of all vertical ordinates. A sudden rise in pressure has the effect of hammering action on the walls of the pipe. Water Hammer in pipes. A pressure wave is transmitted along the pipe. there will be a sudden rise in pressure due to the momentum of water being destroyed. It is briefly written as H. Total Energy line It is defined as the line. which gives sum of pressure head. 1 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 Hydraulic gradient line. when the flowing water is suddenly brought to rest by closing the valve or by any similar cause. datum head and kinetic head of a flowing fluid in a pipe with respect to some reference line. In a long pipe.G. showing the pressure head ( P/ρg) of a pipe from the center of the pipe. by which the boundary should be displaced to compensate for the reduction in momentum of the flowing fluid of boundary δ ∫ θ = u / v(1 − u / v)dy 0 . when a main pipe divides into two or more parallel pipes. It is defined as the distance. adjacent to the solid boundary velocity for a small thickness variation in influenced by various effect. measured perpendicular to the boundary of the solid body. laminar sub layer In turbulent boundary layer region. 2 Pipes in series: Pipes in series or compound pipes is defined as the pipes of different lengths and different diameters connected end to end (in series) to form a pipe line. Pipes in parallel: The pipes are said to be parallel.99 times the free stream (v) velocity of the fluid. When a solid body is immersed in a flowing fluid. This layer is called as laminar sub layer. momentum thickness. The pipes are connected in parallel in order to increase the discharge passing through the main. there is a narrow region of the fluid in neighbourhood of the solid body. where the velocity of fluid varies from zero to free stream velocity. Boundary layer. This narrow region of fluid is called boundary layer. which again join together downstream and continues as a mainline. It is defined as the distance from the boundary of the solid body measured in the y – direction to the point where the velocity of the fluid is approximately equal to 0. Boundary layer thickness. (v) Rise of water in plants through their roots etc. It is define as the type of flow in which the density is constant for the fluid flow. (vi) No mixing between different fluid layers (except by molecular motion. (iv) Due to viscous shear. 3 Incompressible flow. Providing guide – blades in a bend. Examples: Subsonic. there is continuous dissipation of energy and for maintaining the flow must be supplied externally. Characteristics of laminar flow (i) No slip at the boundary (ii) Due to viscosity. (ii) Underground flow (iii) Movement of blood in the arteries of human body. Different methods of preventing the separation of boundary layers 1. Suction of slow moving fluid by suction slot 2. 5.. there is a shear between fluid layers. Mathematically ρ = constant.direction (iii) The flow is rotational. Examples laminar flow / viscous flow (i) Flow past tiny bodies. (v) Loss of energy is proportional to first power of velocity and first power of viscosity. which is given by τ = µ(du / dy) for flow in x. which is very small) Differentiate between laminar boundary layer and turbulent boundary layer . Providing a bypass in the slotted wring 4. Providing small divergence in diffuser 6. Supplying additional energy from a blower 3. (iv) Flow of oil in measuring instruments. Rotating boundary in the direction of flow. aerodynamics. 18 X 104 As Re lies between 4000 and 100.4X10-4 = 3.24 X 0. the boundary layer is called turbulent boundary.3 X 2 X 9.079 0.4 stoke is flowing through a pipe of diameter 300mm at the rate of 300 litres/sec.3 m3/s Length of pipe L = 50m π Velocity V = Q/ Area = 0.3m Discharge Q = 300 Lit/S = 0. Solution : .30) / 0.24 m/s 4 Renold number Re = ( V X d )/ v = (4. m = hydraulic mean depth and i = hf/ L. Of pipe d = 300mm = 0. if the Renolds number of the flow is defined as Re = U x X / v is less than 3X105 If the Renolds number is more than 5X105.4 cm2/s = 0.3 / ( ( 0. Chezy’s formula is generally used for the flow through open channel.000.4 X 10-4 m2/s Dia. C = chezy’s constant. flowing through a pipe of diameter 20 cm and giving a discharge of 10 lps. V = C mi Where .81) = 3.61 m Find the type of flow of an oil of relative density 0. Given : Kinematic viscosity v = 0. Where. A crude of oil of kinematic viscosity of 0.00591 X 50X 4.00591 ( Re ) (3.3)2) = 4. the value of “f” is given by 0. 4 The boundary layer is called laminar.9 and dynamic viscosity 20 poise.18 X 10 4 )1 / 4 Head lost due to friction hf = 4 f L V2/ 2 g d = (4 X 0.4 stoke = 0. U = Free stream velocity of flow X = Distance from leading edge v = Kinematic viscosity of fluid Chezy’s formula.079 f= 1/ 4 = = 0.242) /( 0. find the head lost due to friction for a length of 50m of the pipe. 221)/ (Re)0. Formula for finding the loss of head due to entrance of pipe hi hi = 0.647) < 2000. L = 100m Value of friction factor.222X10-3 m2 / s.222X10-3] = 28. H = total head at inlet of pipe. 5 s = relative density = Specific gravity = 0. Dia of pipe D = 0. Discharge Q = 10 lps = (10 / 1000) m3/s.221 / (Re)0.0032 + (0.3183 X 0.9X1000)] = 2.237 Renolds number Re = (ρVD/µ) = (VXD) / ν (µ/ρ = ν) VXD/ 10-6 = VXDX106 Find diameter of pipe. . hf = head lost due to friction Hydro dynamically smooth pipe carries water at the rate of 300 lit/s at 20oC (ρ = 1000 kg/m3.5 ( V2 / 2g) Formula to find the Efficiency of power transmission through pipes n = ( H – hf) / H where. 4 Kinematic viscosity = v = µ / ρ = [ 2 / (0. Determine the pipe diameter. It is Laminar flow. ν = 10-6 m2/s) with a head loss of 3m in 100m length of pipe.3183 m/s. Reynolds number Re = VD / v Re = [0. Use f = 0.647.3m3/s Density ρ = 1000 kg/m3 Kinematic viscosity ν = 10-6m2/s Head loss hf = 3m Length of pipe.2 m.2 / 2.9 µ = Dynamic viscosity = 20 poise = 2 Ns/m2. Q = AV. Q = 300 lit/sec= 0.0032 + 0. π So V = Q / A = [10 / (1000 X ( ( 0.237 equation for f where hf = ( fXLXV2)/ 2gd and Re = (ρVD/µ) Given: Discharge. f = 0.2)2) )] = 0. Since Re ( 28. H. f = 0.221 5 2 0. will go on increasing.0032 = .H.3 than L.033 – 0.0105 X 10.5886D / ( 0.0105 D 0.0032= 0 ---------------(iii) the above equation (iii) will be solved hit trial method (i).382 / D2 --------------------------.S.382   XDX 10 6  2  D  { from Equation (ii). becomes as L.237 – 0.0032 + (0. the L.033 X 1 5 – 0.0032 +  0.033 X 0.S of the equation (iii). V = 0.5886D/V2 and Re = VXDX106} 0.382/D ) = 0.0193 by increasing the value of D more than 1m .0333 D5 – 0.H.382 X 10 6 ) 0. (ii) Assume D = 0.0032 + 0.0105 – 0.237 – 0.81)/ 100V2 f = 0.3 / π ) = 0. 6 Let D = diameter of pipe Head loss in terms of friction factor is given as hf = ( fXLXV2)/ 2gXD 3 = (fX100XV2)/ 2X9.0032 = 0.237    4.237 4.237 0.30.5886D / V2 ------------------------------( i) now Q = AXV π 0.81XD f = ( 3XDX2X9.3 = ( D)2 X V or D2 X V = ( 4 X 0.0098 – 0.382 = 0.0. Assume D = 1m.0032 = 4.S = 4. then L.237 – 0.0015 X D0.S of equation (iii) becomes as L.221)/ (VXDX106)0.5886 x D / 0.0032 + (0.221)/ (Re)0.H.H.S = 4.5886/ D2 = 0. Hence decrease the value of D.00129 .(ii) f = 0.00789 – 0.237 { from equation (i).0032 +  (0.0032 = 4.382 / D2} 0.221 2 2 0.237     D 0.0333 D5 = 0.237 0.382 4 V2 = 0. 237 – 0. Consider a uniform horizontal pipe. having steady flow as shown figure.00793 – 0.308m.S = 4.3060.V2 = are the values of pressure intensity and velocity at section 2-2.0. L = length of the pipe between the section 1-1 and 2-2 d = diameter off pipe. Actually the value of D will be slightly more than 0.0108 – 0. Or Darcy – Weisbach Equation.S is approximately equal to equal to zero.(P2/ρg) frictional resistance = frictional resistance per unit wetted area per unit velocity X wetted area X velocity 2.0032 = 0. (P1/ ρg) = (P2/ρg)+hf hf = (P1/ ρg) .S of equation (iii) becomes as L.033 X 0. Let P1 = pressure intensity at section 1-1.H.3060.H.3 (iii) Assume D = 0.0032 = . hf = loss of head due to friction. the values of D will be slightly more than 0. 7 as this value of negative.00033 This value of L.0105X0. f1 = Frictional resistance per unit wetted area per unit velocity. Expression for loss of head due to friction in pipes. Let 1 -1 and 2-2 are two sections of pipe. Applying Bernoulli’s equation between sections 1-1 & 2-2 Total head 1-1 = total head at 2-2 + loss of head due to friction between 1-1&2-2 (P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+hf ------------(1) but Z1 = Z1 [ pipe is horizontal ] V1= V2 [ diameter of pipe is same at 1-1 & 2-2] (1) becomes.306m say 0. And P2.H. F = f1 X π d l X V2 [ Wetted area = π d X L. and Velocity V = V 1 = V2] . Let P2 = Velocity of flow at section 1-1.237 – 0.306 then L. where f is the co – efficient of friction 4 fLV 2 hf = 2 gd This equation is known as Darcy – Weisbach equation. Given: Discharge Q = 0. 1) Pressure force at section 1-1 = P1X A 2) Pressure force at section 2-2 = P2 X A 3). Pressure intensity in large pipe. Putting (f1 / ρ) = ( f / 2) . the diameter of the pipe which is 200mm is suddenly enlarged to 400mm. hf = (f1 / ρg) X (P/A) X LX V2 (P/A) = (π d / (π d2/4)) = (4/d) hence. 8 F1 = f1XPXLXV2 ----------.(2).25 m3/s. P1 A – P2A – F1 = 0 (P1-P2)A = F1 = f1XPXLXV2 (P1-P2) = (f1XPXLXV2 / A ). But from (1) we get P1 – P2 = ρg hf Equating the values of (P1 – P2) we get ρg hf = (f1XPXLXV2 / A ). Power lost due to enlargement.25 m3/s.772 N/cm2.. the pressure intensity in the smaller pipe is 11. This equation is commonly used to find loss of head due to friction in pipes The rate of flow through a horizontal pipe is 0. hf = ( f1 / ρg) X ( 4/d) X LXV2. Frictional force F1 Resolving all forces in the horizontal direction. [π d = wetted perimeter = p] The forces acting on the fluid between section 1-1 and 2-2 are. (iii). Determine (i). . Loss of head due to s udden enlargement (ii). 21 X ρg = 13.25 / 0.(V22 / 2g) . = 12.81 N/m2 = 13.453kW. Let the pressure intensity in large pipe = p2.81 = 1.99 2 = + − − 1.12566 m2.229 – 20.229-0.96 2 1.178 = 13.816 1000 X 9.2018-1. Of smaller pipe D1 = 200mm = 0.25X1. Then applying Bernoulli’s equation before and after the sudden enlargement. P = (ρg Q he) / 1000 = (1000X9.816 m. (P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+he But Z1 = Z2 (P1/ρg) + (V12 / 2g) = (P2/ρg) + (V22 / 2g) +he Or (P1/ρg) + (V12 / 2g) = (P2/ρg) + (V22 / 2g) + Z2+hf (P2/ρg) = (P1/ρg) + (V12 / 2g) .12566 = 1. (iii).96N/cm2.4) 2 = 0. 4 Dia of large pipe D2 = 400mm = 0.96 m/s. (i). 4 Pressure in smaller pipe p1 = 11. 9 Dia.21m of water p2 = 13.0+3.81X10-4 N/cm2.21 X 100X 9.99 m/s.99) 2 / 2X 9.772 X104 N/m2. (ii).816)/1000 = 4. .21X1000X9.03141 m2.4m π Area A2 = (0. Loss of head due to sudden enlargement.96 – 1.he 11.8160 = 15.772 N/cm2 = 11.772 X 10 4 7.81 2 X 9.81X0.2m π Area A1 = (0.81 2 X 9. Power lost due to sudden enlargement. he = ( V1 – V2)2/ 2g = (7.03414 = 7.2) 2 = 0.25 / 0.81 = 12. Velocity V2 = Q / A2 = 0. Now velocity V1 = Q / A1 = 0. For the first 25m of its length from the tank. and taking reference line passing through the center of the pipe. Considering all losses of head. 0+0+8 = (P2/ρg) + (V22 / 2g) +0+all losses 8.3m Height of water H = 8m Co-effi.0 = 0+(V22 / 2g)+hi+ hf1+ he+ hf2 . Of friction f = 0. Given: Total length of pipe. the height of water level in the tank is 8m above the centre of the pipe. L1 = 25m Dia of 1st pipe d1 = 150mm = 0.01 Applying the Bernoulli’s theorem to the surface of water in the tank and outlet of pipe as shown in fig. L = 40m Length of 1 st pipe. the pipe is 150mm diameter is suddenly enlarged to 300mm. Take f = 0. Determine the rate of flow.15m Length of 2nd pipe L2 = 40 – 25 = 15m Dia of 2nd pipe d2 = 300mm = 0. 10 A horizontal pipeline 40m long is connected to a water tank at one end and discharges freely into the atmosphere at the other end. which occur.01 for both sections of the pipe. 5 X (4V2)2 )/ 2g = 8V22/2g 4 X 0.67 126.0 x 2 xg 8.0 = + + 106.15 X 2 g 0. we have A1V1 = A2V2 π 2 d 2 XV 2  2 2   XV 2 =  4 d 0.0 0. 11 Where.01X 25 X (4V22 ) 4 X 0.67 2 0.5 V12/ 2g 4 XfXL1 XV12 hf1 = head lost due to friction in pipe 1 = d1 X 2 g he = loss of head due to sudden enlargement = (V1.67 2g 2g 8.15  4 Substituting the value of V1 in different head losses.67+9+2] = 126.15 2g 2g he = (V1 – V2)2/ 2g = (4V2 – V2)2/2g = 9V22/2g 4 X 0.01X 25 X 16 V22 V2 hf1 = = X = 106.01X 15 X (V22 ) 4 X 0.5 Vi2/ 2g = (0.3 X 2 g 0.67 2 + 2 + 2 X 2 2g 2g 2g 2g 2g V22 V22 = [ 1+8+106. we have hi = 0.V2)2/2g 4 XfXL2 XV22 hf2 = head lost due to friction in pipe 2 = d 2 X 2g But from continuity equation. hi = loss of head at entrance = 0.67 .3 2g 2g Substituting the values of these losses in equation (i). we get V 22 8V 22 V 2 9V 2 V2 8.01X 15 V22 V22 hf2 = = X = 2.81 V2 = = = 1.113 m/s 126.3  V1 = (A2V2/A1) = π =  2  XV 2 = 4V 2 d 12  d1   0.0 X 2 X 9. 1 cm2/s = 2.744X0.0526 m3/s Density ρ = 950 kg/m3 Kinematic viscosity ν = 2.and whose Kinematic viscosity is 2.81 .0526/ ( (0.07867 m3/s = 78.1 X10-4 m2/s Height of upper end = 40m Pressure at upper end = atmospheric = 0 Renolds number.67 litres/sec.hf 4 XfXL XV 2 4 X 0.744 m/s 4 Re = (0. Re = VXD/ν.8 = 0.015 X 3200 X (0.113 = 0.05m of oil. 300mm in diameter amd 3200m long is used to pump up 50kg per second of an oil whose density is 950n kg/m3.3)2 X 1.015 Head lost due to friction. The center of the pipe at upper end is 40m above than at the lower end. Given: Dia of pipe d = 300mm = 0. where V = Discharge / Area π = 0.30) / (2. 12 π Rate of flow Q = A2XV2 = ( 0.1X10-4) = 1062.3) 2) = 0.1 stokes. 4 A pipe line. Q Discharge Q = 50/ ρ = 50/950 = 0. f = 16/ Re = 16 / 1062. d X 2g 0.3 X 2 X 9. Find the pressure at the lower end and draw the hydraulic gradient and the total energy line.744) 2 = = = 18.1 stokes = 2.3m Length of pipe L = 3200m Mass M = 50kg/s = ρ. The discharge at the upper end is atmospheric.8 Co – efficient of friction. E.099 N/cm2. Then DE is the total energy line.0282 m p1/ ρg = 58. 13 Applying the Bernoulli’s equation at the lower and upper end of the pipe and taking datum line passing through the lower end. hf = 18. P2 = 0. Draw a vertical line AB through A such that AB = 58.0m respectively.. the co-efficient of friction for each parallel pipe is same and equal to 0. From A draw the centerline of the pipe in such way that point C is a distance of 40m above the horizontal line. A main pipe divides into two parallel pipes. AND T. Draw a line DE parallel to BC at a height of 0.L. which again forms one pipe as shown. V2/2g = (0. we have (P1/ρg) + (V12 / 2g) + Z1 = (P2/ρg) + (V22 / 2g) + Z2+hf but Z1 = 0. Z2 = 40m. Find the rate of flow in each parallel pipe.L.0282m above the hydraulic gradient line. H.05 m of oil p2 / ρg = 0 Draw a horizontal line AX as shown in fig. we have = 5400997 N/m2 = 54. if total flow in main is 3. while the length and diameter of 2nd parallel pipe are 2000m and 0.8m.005. Join B with C.05m. Given: . V1 = V2 as diameter is same.744)2/2X9.81 = 0. then BC is the hydraulic gradient line.G.0 m3/s.05m Substituting these values. The length and diameter for the first parallel pipe are 2000m and 1. 5026] = 3. Q1 = d12XV1 = ( 1)2X(V2 / 0.8785+0.64) X (V2) 4 4 4 Substituting the value of Q1 and Q2 in equation (i) we get π π (1)2X(V2 / 0. Q = Q1+ Q2 = 3. Substituting this value in equation (ii).8785 V2 + 0.0m Length of pipe 2 L2 = 2000m Dia of pipe 2 d2 = 0.894 π π Now.005 X 2000 XV12 4 X 0.005 let Q1 = discharge in pipe 1 let Q2 = discharge in pipe 2 from equation.8) 2X(V2) = (0.894) 4 4 π π π And Q2 = d22XV2 = (0.0 0.8 0.8 V2 V2 V1 = = 0.0 ---------------(i) using the equation we have 4 Xf 1 XL1 XV12 4 Xf 2 XL 2 XV 22 = d1 X 2 g d 2 X 2g 4 X 0.0 or 0.005 X 2000 XV 22 = 1.0 / 1.0 or V = 3.5026 V2 = 3.64)X(V2) = 3.894)+ ( 0.81 0.8 X 2 X 9.0 4 4 V 2[0.8m Total flow Q = 3.81 V12 V 22 V2 = orV12 = 2 1. .0m3/s f1 = f2 = f = 0.3811 = 2.0 X 2 X 9.8 0.17 m/s. 14 Length of Pipe 1 L1= 2000m Dia of pipe1 d1 = 1. 0 – 1.096 m3/s 4 4 Q2 = Q – Q1 = 3. where V1 = Q1 / Area = 0.906 = 1. Applying Bernoulli’s equation to point B and D .904 m3/s.006 / ( ( 0. Dia. 0. Q1 = 60lit/s = 0. ZA = ZD+ ρg +hf 4 Xf 1 XL1 XV12 π Where hf = . d1 = 30cm = 0. C are connected by a pipe system shown in fig.518 m.0 For pipe DC. ZA= 40m For pipe DB.81 p1 {ZD+ ρg } = 40. Given: Length of pipe AD.3 X 2 X 9. L1 = 1200m Dia of pipe AD.3)2) = 0. B. find the height of water level in the reservoir C.482 m Hence piezometric head at D = 36.. d1 X 2 g 4 4 X 0. d3= 30cm = 0. Hence water flows from B to D. Three reservoirs A.848 m/s.518 = 36..30m. Dia.3m Discharge through AD.0 – 3.006 for all pipes. ZB= 38.894 m/s π π 2 Hence Q1 = d12XV1 = 1 X2.848 2 hf= = 3.482m. length L3 = 800mm.20m. length L2 = 600mm.06 m3/s Height of water level in A from reference line .894 = 2. pD Applying the Bernoulli’s equations to point E and. d2 = 20cm = 0. take f = 0.427 = 1. 15 V1 = V2 / 0. Find the discharge into or from the reservoirs B and C if the rate of flow from reservoirs A is 60 litres / s.17 / 0.006 X 1200 X 0. 006 + 0. X 2 X 9.2 X 2 X 9.006 X 800 X 1.194 = 32.81 ZC = 36.0202 = 0.643 X X(0.482 = 1.006 X 600 XV 22 But hf2 = = d 2 X 2g 0.518 X 0.518m 4 XfXL2 XV 22 4 X 0.2 X 2 X 9.006 X 600 π π Discharge Q2 = V2X ( d2)2 = 0.288m A Pipe line of length 2000 m is used for power transmission.643m / s 4 X 0.518 = 0. 4 2 Applying Bernoulli’s equation to D and C pD {ZD+ ρg }= ZC+hf3 4 XfXL3 XV32 Q3 where.482 + hf2 hf2 = 38 – 36.2)2 = 0.0202m3/s = 20.0802 m3/s Q3 Q3 V3 = = = 1.81 V2 = = 0.482 = ZC + = ZC+ 4. V3 = 36. 16 pD ZB = {ZD+ ρg }+hf2 or 38 = 36.134m / s π 2 π d3 (0.81 1.9) 2 4 4 4 X 0.482 = ZC + d 3 X 2 g π 2 d3 4 but from continuity Q1 + Q2 = Q3 Q3 = Q1+Q2 = 0.482 – 4.81 4 X 0. If 110.134 2 36.2 X 2 X 9. Find the diameter of the pipe and efficiency .2lit/s.006 X 600 XV22 1.194 0.365 kW power is to be transmitted through the pipe in which water having pressure of 490.5 N/cm2 at inlet is flowing. P = [ρg X Q X ( H – hf)] / 1000 kW. Take f = 0.02812m3/s π 2 But discharge Q = AXV = d XV 4 π 2 d X V = 0.65 / d) X (0.3625 = [ 1000X9. V = 0. Pressure head at inlet.P transmitted = 150 Pressure at inlet.5 N/cm2 = 490. 4 fXLXV 2 hf = dX 2 g but.14 X d2 = 0.0065.1X104/m2 Loss of head hf = 98.81 d = (2.02812 4 V = (0.81X400) ] = 0. 110.0065 Head available at the end of the pipe = H – hf = 500 – 100 = 400m Let the diameter of the pipe = d Now power transmitted is given by.0358 / d2----------(1) Total head lost due to friction.1 X104 / ρg = 98.02812 X 4) / 3.5 X104 N/m2.81XQX400] / 1000 Q = [ 110. hf = 100m 4 XfXL XV 2 4 X 0.1 N/cm2.81) = 100m Co – efficient of friction f = 0.1 N/cm2 = 98.358/d2)2 = 0.0065 X 2000 XV 2 2.0358 / d2 . p = 490. H. Given: Length of pipe L = 2000m.65 XV 2 100 = hf = = = d X 2g dX 2 X 9.3625 X 1000 / ( 1000X9. 17 of transmission if the pressure drop over the length of pipe is 98. H = p / ρg Pressure drop = 98.1X104/ ( 1000 X 9.003396 / d5 from equation (1). Efficiency of power transmission is given by equation H −hf 500 − 100 η= = = 0.003396 / d5 d = (0.7mm.00396 / 100)1/5 = 0. 18 100 = 0.1277m = 127.80 = 80% H 500 ************************************** . area. Buckingham’s Pi-theorem – Similitude and models – Scale effect and distorted models Dimensional analysis. Fundamental dimensions The fundamental units quantities such as length (L). The advantages of model analysis are: 1. Derive the dimensions for velocity. . The merits of alternative design can be predicted with the help of model testing and the most economical and safe design may be finally adopted. List out the advantages of model analysis. Velocity is the distance (L) travelled per unit time (T) Velocity = Distance/ Time = [L/T] = LT-1. 2. Model Model is nothing but small-scale repetition of the actual structure or machine. mass (M). 1 UNIT – V SIMILITUDE AND MODEL STUDY Dimensional Analysis – Rayleigh’s method. Unit is defined as a yardstick to measure physical quantities like distance. The performance of hydraulic structure or machine can be easily predicted in advance from its model. mass etc. volume. and time (T) are fixed dimensions known as fundamental dimensions. Units. Dimensional analysis is defined as a mathematical technique used in research work for design and conducting model tests. They are called as non-dimensional parameters. Surface tension force is defined as the product of surface tension and length of surface of flowing fluid. Kinematic similarity 3. surface tension (Fs) 6. Lr = Lp/Lm = bp/bm = Dp/Dm Dynamic similarity. . Dynamic similarity Scale ratio. Gravity force (Fg) 4. Geometric similarity 2. Inertia force (fi) 2. 1. Scale ratio is the ratio of linear dimension in the model and prototype which are equal in a geometric similarity. Pressure force (Fp) 5. 1. It is denoted by Lr. Types of forces in a moving fluid. Viscous force (fv) 3. It means the similarity of forces at corresponding points in the model and prototype is equal. The three types of similarities are. Surface tension. The types of forces in a moving fluid are. Dimensionless numbers are the numbers obtained by dividing inertia force or gravity force or pressure force or elastic force or surface tension. 2 Similitude. Elastic force (Fe) Dimensionless numbers. The classifications of models are i. The types of dimensionless numbers are: 1. Euler’s number 4. Froude’s number 3. Reynold’s number 2. Elastic force. 3 Pressure force. It is denoted by (Re) Re = V x d / v = ρ V d / μ Froude’s number Froude’s number is defined as square root of ratio of inertia force of flowing fluid Fi V to gravity. It is denoted as Fe = = fg Lg Classification of models. Weber’s number 5. Undistorted models ii. Types of dimensionless numbers. Distorted models . Elastic force is defined as the product of elastic stress and the area of flowing fluid. Pressure force is the product of pressure intensity and cross-sectional area of the flowing fluid in case of pipe flow. Mach’s number Reynold’s number. Reynold’s number is defined as the ratio of inertia force of flowing fluid and viscous force of the fluid. 4 Undistorted model If the scale ratio for the linear dimensions of the model and prototype is same. Distorted model A distorted model is said to be a distorted model only when it is not geometrically similar to prototype. Weber’s model laws 5. 1. The vertical dimensions of the model can be measured accurately. then the model is said to be undistorted model. Mach’s model laws . The types of model laws are 1. 3. Euler’s model laws 4. The cost of model can be reduced. 2. List out the types of model laws. Reynolds’s model laws 2. Froude’s model laws 3. Turbulent flow in the model can be maintained. Advantages of distorted models. The advantages of distorted models are. qn ) = 0 where the qi are the n physical variables and they are expressed in terms of k independent physical units. (We) model = (We) prototype Buckingham’s ‘ π ’ theorem.. Where fluids of different densities flow over one anther Weber’s model laws When surface tensile forces alone are predominant a model may be taken to be dynamically similar to the prototype when ratio of inertial to the surface tensile forces is the same in the model and prototype. different systems which share the same description in terms of these dimensionless numbers are equivalent. Flow of jet from an orifice or nozzle 3. π2………. 5 List out the application of Froude’s model laws Froude’s model laws is applied in 1. channels etc 2. if we have a physically meaningful equation such as f ( q1.……. q2. Free surface flow such as flow over spillways. The theorem states that if we have a physically meaningful equation involving a certain number. For the purpose of the experimenter. In mathematical terms. The Buckingham’s ‘ π ’ theorem is a key theorem in dimensional analysis. πp) = 0 . Where waves are likely to be formed on surface 4. weirs. then the original expression is equivalent to an equation involving a set of p = n – k dimensionless variables constructed from the original variables. n of physical variables and these variables are expressible in terms of k independent fundamental physical qualities. sluices. the above equation can be restated as F (π1. Furthermore and most importantly it provides a method for computing these dimensionless parameters from the given variables. with the fundamental units as basis vectors and with multiplication of physical units as the “Vector addition” operation and raising to powers as the “scalar multiplication” operation. they are equivalent for the purposes of the experimental list who wants to determine the form of the equation can choose the most convenient one. The theorem describes how every physical meaningful equation involving n variables can be equivalently rewritten as an equation of n – m dimensionless parameters. Most importantly. Proofs of the π theorem often begin by considering the space of fundamental and derived physical unit’s vector space. The π theorem uses linear algebra: the space of all possible physical units can be seen as a vector space over the rational numbers if we represent a unit as the set of exponents needed for the fundamental units ( with a power of zero if the particular fundamental unit is not present) Multiplication of physical units is then represented by vector addition within this vector space. Two systems for which these parameters coincide are called similar. the Buckingham π theorem provides a method for computing sets of dimensionless parameters is not unique: Buckingham’s theorem only provides a way of generating sets of dimensionless parameters and will not choose the most ‘physically meaningful’. Making the physical units match across sets of physical equations can then the regarded as imposing linear constraints in the physical units vector space.m…….qnm Where the exponents mi are the rational numbers. The algorithm of the π . q2.. The use of the πi the dimensionless parameters was introduced by Edge Buckingham in his original 1914 paper on the subject from which the theorem draws its name. even if the form of the equation is still unknown. where m is the number of fundamental units used. the πi are the dimensionless parameters constructed from the qi by p = n–k equations of the form πi = q1m. 6 Where. Express the functional relationship between these variables and the resisting force.1 = 0 y2 = -1. m = 3 Number of π terms = n – m = 6 – 3 = 3 f (π1. (Nov 2005. x2 = -2. l. x2+y2 -3z2 . The resisting force of (R) of a supersonic flight can be considered as dependent upon the length of aircraft “l”. k) = 0 Number of variables.1+3 . Assume l. μ. Vy. air viscosity ‘μ’. μz.) Solution: Step 1. air density ‘ρ’. R π1 = l-2V-2ρ-1R = l V 2ρ 2 Step 3: π2 = l x Vy ρz μ MOLOTO = [ L]X [ LT-1] y [ML-3]z [ML-1T-1] z1+1 = 0. V. and bulk modulus of air ‘k’. π2.R MOLOTO = [ L]X [ LT-1] y [ML-3]z [MLT-2] z1+1 = 0. velocity ‘V’. V and ρ to be the repeating variables. k ) Φ ( R. ρ. μ. ρ. 7 theorem is essentially a Gauss – Jordan elimination carried out in this vector space.2. n = 6 Number of primary variable. x1+y1-3z1+1 = 0 -y1-2 = 0 z1 = -1 x1-2+3+1 = 0 y1 = . .1 = 0 -y2 – 1 = 0 z1 = -1 x2 . π1 = lx. V. π3) = 0 Step 2. R = f ( l. Take the Kinematic viscosity of seawater and air as 0. The velocity of air in the wind tunnel around the model is 30m/s and the resistance of the model is 60N.1 = 0 y2 = -2. µ π 2 = l-1V-1ρ-1 μ = l V ρ Step 4. Determine the velocity of ship in seawater and also the resistance of the ship in sea water.012 stokes and 0. . π2 = l x Vy ρz K MOLOTO = [ L]X [ LT-1] y [ML-3]z [ML-1T-2] Z3+1 = 0.  R µ K  Φ (π1. Length Lp = 300m Fluid = Sea water Density of water = 1030 kg/m3 . π2. x3+y3 -3z3 . =0  lVρ ρV 2    A Ship is 300m long moves in seawater. =0  l 2V 2 ρ lVρ ρV 2     µ K  R = l2 V2 ρ Φ  . A 1:100 model of this to be tested in a wind tunnel.2+3 . The density of air is given as 1.018 stokes respectively.1 = 0 -y3 – 2 = 0 Z3 = -1 x3 . Given: For prototype. π3) = 0 Φ  . whose density is 1030 kg/m3. K π 2 = l-0V-2ρ-1 K = ρV 2 Step 5. 8 x2 = -1.24 kg/m3. x3 = 0. 53Kgf).018 X10-4 m2/s Let velocity of ship = Vp Resitance = Fp For Model Length Lm = (1/100)X300 = 3m Velocity Vm = 30m/s Resistance Fm = 60N Density of air ρm = 1. If model experience a force of 7500N (764. If a 1:9 scale model of this spillway is to be constructed. Hp = 2.24) X (300/3)2X (0. determine force on the prototype.2 m/s.24 kg/m3 Kinematic viscosity of air vm = 0. Given: For prototype: height hp = 7. Resistance = Mass X Acceleration = ρ L3X (V/t) = ρ L2X (V/1)X (L / t ) = ρ L2V2 Then Fp / Fm = (ρ L2V2) p / (ρ L2V2)m = (ρp/ ρm) X (Lp/Lm)2X (Vp/Vm) (ρp/ ρm) = 1030 / 1.018X10-4) X (3/300) X30 = 0.2m Length.012X10-4 / 0.018 stokes = 0.2 N.018 stokes = 0. A 7. determine model dimensions.2 m height and 15m long spill way discharges 94 m3/s discharge under a head of 2.24 Fp / Fm = (1030 /1.17 Fp = 369. 9 Kineamtic viscosity vp = 0.0m. the Reynolds’s number for both of them should be equal.17 X Fm = 369.0m .018 X10-4 m2/s For dynamic similarity between the prototype and its model.2/30) = 369. Lp = 15m Discharge Qp = 94 m3/s Head. head over spillway model and the model discharge. (VpXLp / vp ) = (VmXLm /vm) or (vp / vm ) X (Lm/Lp) X Vm = (0.17 X 60 = 22150. 2 / 9 = 0.e. w.67 m. Lr = 9 Force experienced by model Fp = 7500N Find : (i) Model dimensions i. Head over model (Hm) hp/ Hm = Lr = 9 Hm = Hp / 9 = 2/9 = 0.5 Qm = (Qp / Lr2. (iv).e.e. of the size of prototype Linear scale ratio.t.222 m.5 = 0.. Discharge through model (Qm) Using equation we get.. (ii). Qm (iv). Head over model i. Qp / Qm = Lr2. Hm = 30 m.5) = 94 / 92.. Np = 428 rpm. what power will be obtained from full scale turbine assuming its n is 3% better than that of model. 10 Size of model = 1/9.8 m Lm = Lp / 9 = 15 / 9 = 1. A quarter scale turbine model is tested under a head of 12m. Hm (iii). Solution : Dm / Dp = 1/4 . (iii). If model develops 100 kW and uses 1100 l/s at this speed. Force on prototype (i. Discharge through model i. Force on the prototype(Fp) Using Fr = Fp / Fm = Lr3 Fp = Fm X Lr3 7500X93 = 5467500N.e. The full-scale turbine is to work under a head of 30m and to run at 428 rpm.387 m3/s. Model dimensions ( hm and Lm ) hp/ hm = Lp/ Lm = Lr = 9 hm = hp / 9 = 7.k. Find N for model. height and length of model (hm and Lm) (ii). Fp) (i).. (DN/ H )m = (DN/ H )p (D2N2)m/ (D2N2)P = Hm / Hp) . 22%. (Q / D2 H )m = (Q / D2 H )p Qp = ( Q / H )m ( H )p ( Dp/Dm)2. . Pm = 100 kW.1 / 12 ) X ( 30 ) X 16 = 27. (ii).1X12) = 0. It is given in this problem that the efficiency of the actual turbine is 3% better than the model.1 m3/ s. Qm = 1.82 m3/s Pp = (η 0 ) p X (ρg Q H)p = 0. Qp = (1.82 X 30 = 6514.7638 rpm.54159% We know that.2917 kW.772248 or 77.7954159 X 9810 X 27. (η 0 ) p = 79. (η 0 ) m = ( P/ ρg Q H)m = 100X103 / (9810X1. 11 2  Dp  H Nm =   X m X (N p )2 D  Hp  m  12 = 16 X X 428 2 30 = 1082.