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FLUID MECHANICS

For MECHANICAL ENGINEERING CIVIL ENGINEERING

FLUID MECHANICS SYLLABUS Fluid properties; fluid statics, manometry, buoyancy, forces on submerged bodies, stability of floating bodies; control-volume analysis of mass, momentum and energy; fluid acceleration; differential equations of continuity and momentum; Bernoulli’s equation; dimensional analysis; viscous flow of incompressible fluids, boundary layer, elementary turbulent flow, flow through pipes, head losses in pipes, bends and fittings. Turbomachinery: Impulse and reaction principles, velocity diagrams, Pelton-wheel, Francis and Kaplan turbines.

ANALYSIS OF GATE PAPERS Exam Year 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 Set-1 2014 Set-2 2014 Set-3 2014 Set-4 2015 Set-1 2015 Set-2 2015 Set-3 2016 Set-1 2016 Set-2 2016 Set-3 2017 Set-1 2017 Set-2 2018 Set-1 2018 Set-2

MECHANICAL 1 Mark 2 Mark Ques. Ques. 1 6 2 9 1 3 3 7 3 7 1 5 7 4 3 1 3 3 2 2 2 1 5 2 3 3 3 2 2 1 4 2 2 2 2 3 3 2 3 3 3 4 3 4 3 3 4 2 1

Total 13 20 7 17 17 11 14 10 7 7 6 11 8 9 6 9 6 6 9 8 9 10 10 11 4

CIVIL Exam 1 Mark 2 Mark Year Ques. Ques. 2003 3 8 2004 6 10 2005 3 7 2006 3 7 2007 2 7 2008 1 7 2009 1 3 2010 3 2 2011 3 2 2012 3 2013 3 2 2014 Set-1 2 7 2014 Set-2 2 4 2015 Set-1 3 4 2015 Set-2 3 3 2016 Set-1 1 3 2016 Set-2 1 3 2017 Set-1 1 2 2017 Set-2 1 2 2018 Set-1 4 3 2018 Set-2 1 3

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Total 19 26 17 17 16 15 7 7 7 3 7 16 10 11 9 7 7 5 5 10 7

CONTENTS Topics 1.

2.

3.

4.

5.

Page No

BASICS OF FLUID MECHANICS 1.1 Definition of Fluid 1.2 Basic Equations 1.3 System and Control Volume

1 1 1

PRESSURE & FLUID STATICS 3.1 Pressure 3.2 The Barometer and Atmospheric Pressure 3.3 Principles of Fluid Statics 3.4 Pressure Measurement 3.5 Hydrostatic Forces on Surfaces 3.6 Buoyancy & Floatation Gate Questions

12 13 14 14 21 33 40

PROPERTIES OF FLUIDS 2.1 Density 2.2 Specific Gravity/Relative Density 2.3 Viscosity 2.4 Surface Tension 2.5 Capillarity 2.6 Thermo Dynamic Properties 2.7 Compressibility and Bulk Modulus 2.8 Vapor Pressure 2.9 Cavitations Gate Questions

KINEMATICS 4.1 Introduction 4.2 Methods of Describing Fluid Motion 4.3 Types of Fluid Flow 4.4 Continuity Equation in Three-Dimensions 4.5 Continuity Equation in One Dimension 4.6 Motion of Fluid Element 4.7 Flow Patterns 4.8 Stream Function 4.9 Velocity Potential Function 4.10 Equipotential Line Gate Questions

BERNOULLI’S EQUATION & ITS APPLICATIONS 5.1 Introduction

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2 2 2 5 6 7 7 8 8 9

48 48 48 50 51 51 52 53 53 54 59 69

6.

7.

8.

5.2 5.3 5.4 5.5 5.6

Euler’s Equation Bernoulli’s Equation Application Of Bernoulli’s Equation Bernoulli’s Equation For Real Fluids Free Liquid Jets

DYNAMICS OF FLUID FLOW 6.1 Introduction 6.2 Impulse Momentum Principle 6.3 Force Exerted By a Fluid on a Pipe Bend 6.4 Force Exerted By a Fluid on Vertical Stationary Plate 6.5 Force Exerted By a Fluid on Moving Blade Gate Questions

77 77 77 77 78 85

FLOW THROUGH CONDUITS/PIPES 7.1 Internal Flow 93 7.2 Laminar Flow/Viscous Flow 93 7.3 Turbulent Flow in Pipes 95 7.4 Loss of Energy in Fluid Flow 98 7.5 Flow through Pipes in Series or Flow through Compound Pipes 99 7.6 Flows through Nozzles 100 Gate Questions 108

10.

EXTERNAL FLOW 8.1 Boundary Layer Formation 8.2 Regions of Boundary Layer 8.3 Boundary Layer Thickness 8.4 Drag Force on a Flat Plate Due To Boundary Layer 8.5 Boundary Condition for the Velocity Profile 8.6 Analysis of Turbulent Boundary Layer 8.7 Boundary Layer Separation Gate Questions HYDRAULLIC TURBINES 9.1 Introduction 9.2 Classification of Turbines 9.3 Turbine Efficiencies 9.4 Power Developed by Turbine (Euler’s Equation) 9.5 Pelton Turbine 9.6 Reaction Turbine 9.7 Francis Turbine 9.8 Kaplan Turbine 9.9 Specific Speed Significance 9.10 Model Testing (Dimensionless Turbine Parameters) Gate Questions

11.

CIVIL GATE QUESTIONS

9.

69 69 70 72 72

ASSIGNMENT QUESTIONS

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114 114 115 115 116 117 117 122 136 136 136 137 137 138 140 140 141 141 149 159

189

1

BASICS OF FLUID MECHANICS

1.1 DEFINITION OF FLUID

1.3 SYSTEM AND CONTROL VOLUME

Fluid mechanics deals with the behaviour of fluids at rest and in motion. A fluid is a substance that deforms continuously under the application of a shear (tangential) stress no matter how small the shear stress may be. Fluids comprise the liquid and gas (or vapour) phases of the physical forms in which matter exists. The distinction between a fluid and the solid state of matter is clear if you compare fluid and solid behaviour. A solid deforms when a shear stress is applied, but its deformation does not continue to increase with time.

A system is defined as a fixed, identifiable quantity of mass. The system boundaries separate the system from the surroundings.

1.2 BASIC EQUATIONS

Analysis of any problem in fluid mechanics necessarily begins, either directly or indirectly, with statements of the basic laws governing the fluid motion. The basic laws, which are applicable to any fluid, are: 1. 2. 3. 4. 5.

The boundaries of the system may be fixed or movable; however, there is no mass transfer across the system boundaries. In the familiar piston–cylinder assembly the gas in the cylinder is the system. Heat and work may cross the boundaries of the system, but the quantity of the matter within the system boundaries remains fixed. There is no mass transfer across the system boundaries. A control volume is an arbitrary volume in the space through which fluid flows. The geometric boundary of the control volume is called the control surface. The control surface may be real or imaginary; it may be at rest or in motion.

The conservation of mass Newton’s second law of motion The principle of angular momentum The first law of thermodynamics The second law of thermodynamics

Clearly, not all basic laws always are required to solve any one problem. On the other hand, in many problems it is necessary to bring into the analysis additional relations, in the form of equations of state or conservation equations, that describe the behaviour of physical properties of fluid under given conditions.

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2

PROPERTIES OF FLUIDS

2.1 DENSITY Density is defined as mass per unit volume and denoted by ρ. SI unit of density is Kg/m3 mass of fluid Density(ρ) = volume of fluid The reciprocal of density is the specific volume ( ), which is defined as volume per unit mass. SI unit of density is m3/Kg 1 v= ρ The density of a substance, in general, depends on temperature and pressure. The density of most gases is proportional to pressure and inversely proportional to temperature. Liquids and solids, on the other hand, are essentially incompressible substances, and the variation of their density with pressure is usually negligible.

2.2 SPECIFIC GRAVITY/RELATIVE DENSITY

Specific gravity or relative density of a substance is defined as the ratio of the density of a substance to the density of standard substance at a specified temperature (usually water at 4°C, for which water is 1000 kg/m3). Density of liquid Specific Gravity = Density of water 2.3 VISCOSITY When two solid bodies in contact move relative to each other, a friction force develops at the contact surface in the direction opposite to motion. Fluid is a substance that deforms continuously under the action of shear stress. The situation is similar when a fluid moves relative to a

solid or when two fluids move relative to each other. A property that represents the internal resistance of a fluid to motion is called viscosity. The force, a flowing fluid exerts on a body in the flow direction is called the drag force, and the magnitude of this force depends, in part, on viscosity.

2.3.1 VISCOUS FORCE IN LIQUIDS AND GASES 1) Molecular momentum transfer: In the flow of liquids and gases, molecules are free to move from one layer to another. When the velocity in the layers are different as in viscous flow, the molecules moving from the layer at lower speed to the layer at higher speed have to be accelerated. Similarly, the molecules moving from the layer at higher velocity to a layer at lower velocity, carry with them a higher value of momentum and these are to be slowed down. Thus, the molecules diffusing across layers transport a net momentum, introducing a shear stress between the layers. The force will be zero if both layers move at the same speed or if the fluid is at rest. 2) Cohesive force: When cohesive forces exist between atoms or molecules these forces have to be overcome, for relative motion between layers. A shear force is to be exerted to cause fluids to flow. 2.3.2 TYPES OF FLUID

Ideal fluid: Consider a fluid layer between two very large parallel plates (or equivalently, two parallel plates immersed in a large body of a fluid) separated by a

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small distance (dy). Now a constant tangential force F is applied to the upper plate while the lower plate is held fixed. After some time upper plate moves continuously under the influence of this force at a constant velocity V. The fluid in contact with the upper plate sticks to the plate surface and moves with same velocity as that of the surface. This condition is known as no slip condition.

The shear stress τ yx acting on this fluid layer is τ yx = F / A

Deformation or shear stain is denoted by dα Rate of shear stain or deformation is given dα by dt dl tan(dα) = dα = dy

 dl  dα  dt  du = = dt dy dy Where, u is the velocity is x direction. du is known as velocity gradient. dy

1) Newtonian fluid: Fluids for which the rate of deformation is proportional to the shear stress are called Newtonian fluids. τ yx ∝ rate of deformation or rate of shear strain. d τ =μ dt du τ =μ dy

Where, μ is constant of proportionality and is known as Dynamic viscosity. SI unit is kg/m-s, N. s/m2 or Pa / s

Note: 1) Pa is the pressure unit Pascal. 2) A common viscosity unit is poise 10 Poise = 1Ns / m 2 −2 1centipoise = 10 = Poise 10−3 Ns / m 2 3) The viscosity of water at 20°C is 1 centipoise, and thus the unit centipoise serves as a useful reference. 2) Non Newtonian Fluids: Fluids in which shear stress is not directly proportional to deformation rate are known as nonNewtonian fluids. Non Newtonian fluids commonly are classified as having time independent or time dependent behavior. Numerous equations have been proposed to model the observed relations between τ and du/dy for timen

 du  independent fluids. τ =K    dy  Where, n is called the flow behavior index k the consistency index.

 du  τ =k    dy 

n −1

du du = η dy dy

n −1

 du  The term η = k   is referred to as  dy  the apparent viscosity. Fluids in which the apparent viscosity decreases with increasing deformation rate (n<1) are called pseudo plastic (or shear thinning) fluids. Most nonNewtonian fluids fall into this group; Examples include polymer solutions, colloidal suspensions, and paper pulp in water. If the apparent viscosity increases with increasing deformation rate (n>1) the fluid is termed dilatants (or shear thickening).Suspension of starch and of sand are examples of dilatants fluids.

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A “fluid” that behaves as a solid until a minimum yield stress,τ, is exceeded and subsequently exhibits a linear relation between stress and rte of deformation is referred to as an ideal or Bingham plastic. Thixotropic fluids show a decrease in η with time under a constant applied shear stress. Rheopectic fluids show an increase in η with time. After deformation some fluids partially return to their original shape when the applied stress is released; such fluids are called viscoelastic.

1) Liquids: In case of liquids the viscosity force is mainly due to cohesive force. The cohesive force decreases. So, viscosity of liquids decreases when temperature increases. The relation of viscosity with temperature is given by   1 µ = µ0    1 + αT + 7βT  2) Gases: In the case of gases, the contribution to viscosity is more due to momentum transfer. As temperature increases, more molecules cross over with higher momentum differences. Hence, in the case of gases, viscosity increases with temperature. µ = µ 0 + αT + β T 2 where, μ = Viscosity atT 0in poise μ 0 = Viscosity at00 in poise α, β = are constants for liquid and gas SOLVED EXAMPLES

2.3.3 KINEMATIC VISCOSITY Kinematic viscosity is defined as the ratio between dynamic viscosity and density denoted by ‘ν’ ν = μ/ρ The unit in SI system is m2/s. Stoke is CGS unit of kinematic viscosity given by 1 (cm2/s) = 10–4 m2/s. 1 centistoke = 10–6 m2/s.

Example: ‘An infinite plate is moved with a velocity of 0.3m/s over a second plate on a layer of liquid for small gap width d=0.3mm, assume a linear velocity distribution the liquid viscosity is 0.65 ×10−3 kg / ms and S.G is 0.88. a) Calculate kinematic viscosity b) The shear stress on the lower plate Solution: a) µ V= ρ

Kinematic viscosity gives the rate of m3 −3 kg 0.65 ×10 = × momentum flux or momentum diffusivity. m.s ( 0.88 ×1000 ) kg For liquids and gases absolute (dynamic) viscosity is not influenced significantly by = 7.39 ×10−7 m 2 / s pressure. But kinematic viscosity of gases is b) influenced by pressure due to change in u τlower = µ  density. d 2.3.4 EFFECT OF TEMPERATURE ON VISCOSITY

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=0.65 ×10−3

kg m 1 × 0.3 × m.s s 0.3 ×10−3 m

=0.65Pa Direction of shear stress on lower plate  is τlower

Example: Calculate the dynamic viscosity of oil, which is used for lubrication b/w a square of size 0.8 × 0.8 m2 and an inclined plane with angle of inclination30° , wt = 300 N, slides down the inclined plane with a uniform velocity of 0.3 m/s. the thickness of oil film is 1.5 mm. Solution: W sin θ =Foil Foil ∆u = µ Fcontact ∆y Foil u =µ A y F u µ = oil . A y W sin θ.y µ= A.u 300.sin 30 ×1.5 ×10−3 = µ = 0.117 ×10Poise 0.64 × 0.3 =1.17Poise 2.4

SURFACE TENSION

Consider two liquid molecules, one at the surface and one deep within the liquid body. The attractive forces applied on the interior molecule by the surrounding molecules balance each other because of symmetry. But the attractive forces acting on the surface molecule are not symmetric, and the attractive forces applied by the gas molecules above are usually very small. Therefore, there is net attractive force acting on the molecule at the surface of the liquid, which tends to pull the molecules on the surface toward the interior of the liquid. This force is balanced by the repulsive forces from the molecules below the surface that are being compressed. The

resulting compression effect causes the liquid to minimize its surface area. This is the reason for the tendency of the liquid droplets to attain a spherical shape, which has the minimum surface area for a given volume. The surface of the liquid acts like a stretched elastic membrane under tension. The pulling force that causes this tension acts parallel to the surface and is due to the attractive forces between the molecules of the liquid. The magnitude of this force per unit length is called surface tension and is usually expressed in the unit N/m. Surface tension is also defined as the surface energy per unit surface area or work that needs to be done to increase the surface area of the liquid by a unit amount. Surface tension is a binary property of the liquid & gas or two liquids. The surface tension of air and water at 20° c is about 0.73 N/m.

If F is the tensile force on the surface, L is the length of the surface. Surface tension is given by F or σ= L E surface σ= surface Area 2.4.1 SURFACE TENSION ON LIQUID DROPLET

Let σ is surface tension R is radius of droplet

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Ap is area of projection

1) Force due to difference in pressure inside & outside the liquid drop = ∆PA p … (i) =∆PπR 2 2) Tensile force due to surface tension = σ × circumference … (ii) = σ × 2πR Under equilibrium condition, these 2 forces will be equal



2σ ∆P = R

2.4.2 SURFACE TENSION ON A SOAP BUBBLE

1) Force due to pressure inside & outside the liquid jet = ∆p.L 2R … (i)

( )

2) Tensile force due to surface tension … (ii) = 2L.σ Equating the forces, we get σ ∆P = R 2.5 CAPILLARITY

1) Force due to pressure inside the liquid drop = ∆PA p … (i) =∆PπR 2 2) Tensile force due to surface tension Fsurface = 2πR.σ Fsurface2 =2π ( R + t ) σ

Equating the forces, we have ∆Pπr2 =σ2π R + t + σ2πR

(

)

… (ii)

Assuming thickness is very small T << R 4σ ∆P = R

2.4.3SURFACE TENSION ON A LIGHT JET

Another consequence of surface tension is the capillary effect, which is the rise or fall of a liquid in a small-diameter tube inserted into the liquid. Such narrow tubes or confined flow channels are called capillaries. The curved free surface of a liquid in a capillary tube is called the meniscus. It is commonly observed that water in a glass container curves up slightly at the edges where it touches the glass surface; but the opposite occurs for mercury: it curves down at the edges.. The strength of the capillary effect is quantified by the contact (or wetting) angleβ, defined as the angle that the tangent to the liquid surface makes with the solid surface at the point of contact. The surface tension force acts along this tangent line toward the solid surface.

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1) Force due to surface tension =σ.2πr Fy = σπr.cos β

Fx = 0 2) Vertical force is responsible for lifting the liquid in capillary Fy = mg = ρπr2h.g

Equating the vertical forces, we get 2πrσ cos β = ρπr2h.g

2σ cos β ⇒h= ρgr Note: Same expression is used for capillary fall. Angle, β = 0 for glass tube & water. β = 128o for glass tube & mercury.

2.6

THERMO DYNAMIC PROPERTIES

2.6.1 IDEAL GAS EQUATION PV = nRT Where, P is pressure in Pa V is volume in m3 n is moles of gas R is universal gas constant (8.314 KJ/mole K)  M  PV =   RT  Mw   R  m  ⇒ PV =  T  Mw 

Where, R' is characteristic gas constant & for air J R ' = 287 kg − K Mw is molecular wt of gas ρ is density of gas 2.6.2 THERMODYNAMIC PROCESS

a) Isothermal: Constant Temperature PV=mRT T = const PV = const b) Adiabatic Process: No heat transfer takes place P = const ργ γ is ratio of specific heat Cp γ= Cv γ = 1.4 for air c) Isobaric process: Constant pressure process P = const V = const T d) Isochoric process: Constant volume process V = const P = const T 2.7 COMPRESSIBILITY MODULUS

AND

BULK

It is the measure of volume change under the action of external force.

( )

⇒ PV = m R' T

m ⇒P=   R''T V ⇒P= ρR''T

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Let the volume of gas decrease from V to (v-dv), the pressure is increased from P to P+dP Increase in pressure = dp kgf / m2 Decrease in volume = dv −dv Volumetric strain = v dp Bulk Modulus (k) = −dv / v Compressibility = 1/k Relationship b/w bulk modulus and pressure (p) for a gas undergoing compression process a) For Isothermal process: P = const ρ for closed system. Pv = const by taking log ln P+ ln V = ln(const) By differentiating dp dV + = 0 p V dp =P −dv / v k=p b) Adiabatic Process: PV y = const by taking log both side In P + γ ln v = ln const by differentiating dp ydV + = 0 p V dp = γP −dv / v k = γP

Likewise, at a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure Psat . The vapour pressure Pv of a

pure substance is defined as the pressure exerted by its vapour in phase equilibrium with its liquid at a given temperature. Pv is a property of the pure substance, and turns out to be identical to the saturation pressure Psat of the liquid ( Pv ” Psat ). 2.9 CAVITATIONS

The liquid pressure in liquid-flow systems drops below the vapour pressure at some locations, results in vaporization of liquid. For example, water at10°C will convert into vapour and form bubbles at locations (such as the tip regions of impellers or suction sides of pumps) where the pressure drops below 1.23 kPa. The vapour bubbles (called cavitation bubbles since they form “cavities” in the liquid) collapse as they are swept away from the low pressure regions, generating highly destructive, extremely high-pressure waves. This phenomenon, which is a common cause for drop in performance and even the erosion of impeller blades, is called cavitation, and it is an important consideration in the design of hydraulic turbines and pumps.

2.8 VAPOR PRESSURE At a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature Tsat . © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

GATE QUESTIONS Q.1

Q.2

Q.3

The SI unit of kinematic viscosity (v) is a) m2/s b) kg/m-s 2 c) m/s d) m3/ s2 [GATE–2001] For a Newtonian fluid a) Shear stress is proportional to shear strain b) Rate of shear stress is proportional to shear strain c) Shear stress is proportional to rate of shear strain d) Rate of shear stress is proportional to rate of shear strain [GATE–2006] Assuming constant temperature condition and air to be an ideal gas, the variation in atmospheric pressure with height calculated from fluid statics is a) linear b)exponential b) quadratic d) cubic [GATE–2016(2)]

Q.4 A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as shown in the figure below. A thin oil film of thickness h = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is

0.04 m2 . The velocity distribution within the oil film gap is linear as shown in the figure. Take dynamic viscosity of oil as 7 × 10 −3 P-s and acceleration due to gravity as 10 m / s2 . Neglect weight of the oil. The terminal velocity V (in m/s) of the block is _________ (correct to one decimal place).

[GATE–2018(2)]

ANSWER KEY: 1 (a)

2 (c)

3 (b)

4 10.71

EXPLANATIONS Q.1 Q.2

(a) (c)

Q.4 (10.71)

From the Newton’s law of Viscosity, the shear stress (τ) is directly proportional to the rate of shear strain (du/dy).

Terminal velocity is reached when weight force of the block is balanced by the du du So, τ ∝ or τ =μ viscous drag forces due to shear stress dy dy generated in the fluid-solid interface. Where, µ = constant of proportionality and it ∴τ A = W is known as coefficient of viscosity Using Newton's law of viscosity du V = τ μ= μ (Q linear profile) dy h V Wh mgh ⇒ μ A= W⇒ V= = Q.3 (b) h μA μA

dP = ρ gdh P ⇒ dP = ⋅ g ⋅ dh RT on Integnating, P

∫ Patm

2 10  0.1510−3 = = 10.714 m/s 7 10−3  0.04

h

dP g dh = P RT ∫0

 P  gh ln  = + RT  Patm  ⇒ P = Patm e

 gh  +   RT 

Hence, exponential

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3

PRESSURE & FLUID STATICS

3.1 PRESSURE Pressure is defined as the normal force exerted by fluid per unit area. We speak of pressure only when we deal with a gas or a liquid. The counterpart of pressure in solids is normal stress. F P= A Since pressure is defined as force per unit area, it has the unit of Newton’s per square meter (N/m2), which is called Pascal (Pa) 1bar = 105 Pa = 0.1 MPa = 100 kPa

3.1.1 ABSOLUTE, GAUGE, ATMOSPHERIC & VACUUM PRESSURE

1) The pressure values must be stated with respect to a reference level. If the reference level is vacuum (i.e. absolute zero pressure), pressures are termed absolute pressure. 2) Most pressure gauges indicate a pressure difference—the difference between the measured pressure and the ambient level (usually atmospheric pressure). Pressure levels measured with respect to atmospheric pressure are termed gauge pressures 3) Pressures below atmospheric pressure are called vacuum pressures. Absolute, gauge, and vacuum pressures are all positive quantities and are related to each other by = Pabsolute Patmospheric + Pgauge = Pvacuum Patmospheric + Pabsolute

3.1.2 PRESSURE AT A POINT Pressure is the compressive force per unit area, and it gives the impression of being a vector. However, in fluids under static conditions, pressure is found to be independent of the orientation of the area. This concept is explained by Pascal’s law which states that the pressure at a point in a fluid at rest is equal in magnitude in all directions. Pressure has magnitude but not a specific direction, and thus it is a scalar quantity. P= P= Pz x y 3.1.3 PRESSURE VARIATION IN A STATIC FLUID (HYDROSTATIC LAW):

For fluids at rest or moving on a straight path at constant velocity, all components of acceleration are zero. In fluids at rest, the pressure remains constant in any horizontal direction (P is independent of x and y) and varies only in the vertical direction. As a result of gravity, these relations are applicable for both compressible and incompressible fluids. dp dp dp = −ρ(g), = 0, = 0 dz dx dy

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The negative sign is taken because dz is taken positive in upward direction and pressure decrease in upward direction. For incompressible fluid ρ is constant. p

z

po

zo

Case 2: Acceleration in negative x direction dp dp dp = −ρ ( g ) , = −ρ ( −a x ) , = 0 dz dx dy

∫ dp = −ρg ∫ dz

3.2 THE BAROMETER & ATMOSPHERIC

P − Po = −ρg(z − z o )

PRESSURE

P − Po = ρgh For compressible fluid, ρ varies with pressure, i.e. ρ =f (P) For gases, variation of density with pressure can be expressed by ideal gas equation P ρ= RT Where , P is pressure T is temperature

Atmospheric pressure is measured by a device called barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. The pressure at point B is equal to the atmospheric pressure, and the pressure at C can be taken to be zero since there is only mercury vapour above point C and the pressure is very low relative to Patm and can be

3.1.4 FLUIDS IN RIGID-BODY MOTION

When fluid is in stationary container, the pressure remains constant along horizontal direction. The pressure varies only along vertical direction. When a fluid is placed in an accelerated container, initially fluid splashes and there is a relative motion between fluid & container boundary. After some time, the liquid comes to rest and attains fixed shape relative to container. The pressure varies in the direction of acceleration. 1) When container accelerates in vertical direction Case1: Downward acceleration of az dp dp dp = −ρ ( g − a z ) , = 0, = 0 dz dx dy Case 2: upward acceleration of az dp dp dp = −ρ ( g + a z ) , = 0, = 0 dz dx dy 2) When container accelerates in horizontal direction Case1: Acceleration in positive x direction dp dp dp = -ρ ( g ) , = -ρa x , =0 dz dx dy

neglected for an excellent approximation. Writing a force balance in the vertical direction gives Patm = ρgh Where, ρ is the density of mercury, g is the local gravitational acceleration, h is the height of the mercury column above the free surface. Note that the length and the cross-sectional area of the tube have no effect on the height of the fluid column of a barometer. A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C ( ρHg is13,595 kg / m3 ) . If water instead of mercury were used to measure the standard atmospheric pressure, a water column of about 10.3m would be needed.

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3.3 PRINCIPLES OF FLUID STATICS 1) When fluid is at rest, in a continuous fluid, fluid at the same elevation has the same pressure. 2) The pressure at the bottom of a column of fluid is equal to the pressure at the top, plus density multiplied by gravity multiplied by the height of the column of fluid. A consequence of the second principle is that when different columns of fluid stack on top of one another, the pressures due to each column simply add up. 3.3.1 HYDRAULIC LIFT

A consequence of the pressure in a fluid remaining constant in the horizontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is called Pascal’s law. The application of Pascal’s law in hydraulic lift is shown in fig P1= P2 F1 / A1 = F2 / A 2

Fig.: Schematic of a hydraulic lift

3.4 PRESSURE MEASUREMENT Hydrostatic law indicates the pressure difference b/w two points in a static fluid. A device based on this principle is called manometer, and it is commonly used to measure small and moderate pressure differences. A manometer mainly consists of a glass or plastic U tube containing one or more fluids such as mercury, water,

alcohol, or oil. The pressure of a fluid is measured by the following. 1) Manometers a. Simple • Piezometer • U-tube manometer • Single column manometer b. Differential • U-tube differential manometer • Inverted U-tube manometer 2) Mechanical gauge 3.4.1 SIMPLE MANOMETER

1) Piezometer: It is the simplest kind of manometer. It does not have any high density liquid. The tube is connected to the point where pressure is to be measured. The liquid rises in tube to balance the pressure at ‘A’. The gauge pressure PA is given by PA = ρgh Where ρ is the density of liquid inside vessel/pipe

Piezometer is used to measure low pressures. The height of column increases if pressure is high. E.g. height of water column is 10.3 m if gauge pressure at ‘A’ is 1atm (105N/m2).

2) U-tube manometer: It has a glass Utube with liquid having density higher than the density of fluid in the container.

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The atmospheric pressure exists and should be taken into account for evaluation of absolute pressure, but while evaluating gauge pressure it is not accounted in equations. The gauge pressure at ‘A’ is given by P= PA + ρc gh1 1 P2 = ρm gh 2 Where, ρC is the density of fluid in container, it

can be water or oil ρm is the density of manometric fluid. Usually mercury is chosen as manometric fluid h2 is difference in mercury level PA is pressure in vessel/pipe in gauge From principle of fluid statics, when fluid is at rest, fluid at the same elevation has the same pressure. PA = ρm gh 2 − ρw gh1 3) Single Column Manometers: In this manometer a large cross-sectional area reservoir is placed in one of the limbs. When pressure is applied, the fluid lowers slightly in the reservoir as compared to the fluid rise in the other limb.

ii) large cross-sectional area (100 times) iii) due to large area of cross section and small change of pressure, the change in level of reservoir will be very small and can be neglected.

4) Inclined Single Column Manometer: This manometer is more sensitive than straight column. The liquid rises more in the column due to inclination.

= h 2 L sin θ ah 2 = PA . ( ρ2 g − ρ1g ) + ( ρ2 gh 2 − ρ1gh1 ) A PA = ρ2 g 2 h 2 − ρ1g1h1 PA = ρ2 g 2 L sin θ − ρ1gh1 3.4.2 DIFFERENTIAL MANOMETERS Differential Manometers are devices used for measuring the difference of pressure between two points in a pipe or two different pipes. It contains of a U-tube with manometric liquid. The manometric liquid can be of higher density or lower density than pipe liquid. 1) U-tube differential manometer:

Gauge Pressure at point A is given as ah 2 = PA . ( ρm g − ρc g ) + ( ρm gh 2 − ρc gh1 ) A Since A>>α PA = ρm gh 2 − ρc gh1 Salient features of Single column manometer: i) modified form of U-tube manometer

Pressure above a-a

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• •

In the left limb = ρ1g ( h + x ) + PA

In the right limb= PB + ρ2 gy + ρHg gh By equating PA + ρ1g(h + x)= PB + ρ2 gy + ρHg g.H

PA − PB = ρ2 gy + ρHg .g.H − ρ1g ( h + x )

A & B are at same level :

PA − PB = ρ2 gx + ρHg gh − ρ1g ( h + x )

If liquid is same, then ρ1gx = ρ2 gx ∴ PA − P= gh ( ρHg − ρ1 ) B

2) Inverted Manometer:

U-tube

differential

PA − ρ1gh= PB − ρ2 gh 2 − ρs gh 1 PA − PB = ρ1gh1 − ρ2 gh 2 − ρs gh It is used for measuring difference of low pressures

respectively. If the depth of water in the tank is 1m and tank is open at the top then, Calculate: I. angle of the water surface with the horizontal II. the maximum and minimum pressure intensities at the bottom

Solution: Given: Constant acceleration a = 2.4m/ s2 Length = 6m; Width = 2.5m and depth = 2m, Depth of water in tank, h = 1m

i) The angle of the water surface to the horizontal Let θ = the angle of water surface to the horizontal Using equation, we get a 2.4 tan θ = − = − = −0.2446 g 9.81 (the –ve sign shows that the free surface of water is sloping downward as shown in Fig) ∴ tan θ = −0.2446 (slope downward)

−1 = ∴ θ tan = 0.2446 13.7446o or 13o 44.6 '

ii) The maximum and minimum pressure intensities at the bottom of the tank SOLVED EXAMPLES From the figure, depth of water at the front end, Example: = h1 1 - 3tanθ = 1 - 3 × 0.2446 = 0.2662m A rectangular tank is moving horizontally Depth of water at the rear end : in the direction of its length with a constant = h2 1 + 3tanθ = 1 + 3 × 0.2446 = 1.7338m acceleration of 2.4m/ s2 . The length, width and depth of the tank are 6m, 2.5m and 2m © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

The pressure intensity will be maximum at the bottom, where depth of water is maximum. Now, the maximum pressure intensity at the bottom will be at point A and it is given by, Pmax = ρ × g× h2 =1000×9.81×1.7338N/ m2 =17008.5N/ m2

The minimum pressure intensity at the bottom will be at point B and it is given by pmin = ρ × g× h1

=1000×9.81×0.2662= 2611.4N/ m

2

Example: A U-tube as shown in figure, filled with water to mid level is used to measure the acceleration when fixed on moving equipment. Determine the acceleration ax as a function of the angle θ and the distance A between legs.

Solution: This is similar to the formation of free surface with angle θ tan = θ –a x / g + a y As= a y 0, tan = θ –a x / g

(

)

The acute angle θ will be given by, θ =tan–1 ( a x / g ) ax = g × tan θ As tan θ =2h / A h = A a x / 2g

Example: A hydraulic press has a ram of 30cm diameter and a plunger of 4.5cm diameter. Find the weight lifted by the hydraulic press when the force applied at the plunger is 500N.

Solution: Given: Dia. of ram D=30cm = 0.3m Dia. Of plunger, d = 4.5cm = 0.045m Force on plunger, F =500N Let the weight lifted = W π 4

Area of ram, A = D2 =

2 π 0.3) = 0.07068m2 ( 4

Area of plunger, 2 π π a = d2 = 0.045 = .00159m2 4 4 Pressure intensity due to plunger Force on plunger F 500 = = = N / m2 Area of plunger a .00159

(

)

Due to Pascal’s law, the intensity of pressure will be equally transmitted in all directions. Hence the pressure intensity at the ram 500 = =314465.4N/ m2 .00159 But pressure intensity at ram Weight W W = = = N/ m2 Area of ram A .07068 W =314465.4 .07068 ∴ Weigth = 314465.4×0.7068 = 22222N = 22.222kN

Example: The diameters of a small piston and a large piston of hydraulic jack are 3 cm and 10cm 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 is 40cm above the large piston. The density of the liquid in the jack is given as 1000kg/ m3

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Solution: Given: Dia. of small piston, d =3cm ∴ Area of small piston, π π × d2 = × 32 = 7.068cm2 A= 4 4 Dia. of large piston, D=10cm ∴ Area of larger piston, π A= × 102 = 78.54cm2 4 Force on small piston, F = 80N Let the load lifted = W

a) When the pistons are at the same level Pressure intensity on small piston F 80 = N/ m2 a 7.068

This is transmitted equally to the large piston. ∴ Pressure intensity on the large piston 80 = 7.068 ∴ Force on the large piston = pressure x area 80 = × 78.54N = 888.96N 7.068

b) When the small piston is 40cm above the large piston Pressure intensity on the small piston F 80 = N/ m2 a 7.068

∴ Pressure intensity at section A-A F = + Pressure intensity due to height a of 40cm of liquid. Pressure intensity due to 40cm of liquid = ρ × g× h = 1000×9.81×0.4N/ m2

=

1000×9.81×.40 N/ cm2 = 0.3924N/ cm2 104

∴ Pressure intensity at section A-A 80 = +0.3924 7.068 11.32 + 0.3924 = 11.71 N/cm2

∴ Pressure intensity transmitted to the large piston = 11.71N/ cm2 ∴ Force on the large piston = Pressure × Area of the large piston = 11.71× A = 11.71×78.54 = 919.7N

Example: A U-tube manometer is used to measure the pressure of water in a pipe line, which is in excess of atmospheric pressure. The right limb of the manometer contains mercury and is open to water in the main line, if the difference in level of mercury is in the left limb. Determine the pressure of water in the main line, if the difference in level of mercury in the limbs of U-tube is 10 cm and the free surface of mercury is in level with the centre of the pipe. If the pressure of water in pipe line is reduced to 9810N/m2, calculate the new difference in the level of mercury. Sketch the arrangements in both cases. Solution: Given: Difference in mercury level =10 cm =0.1m The arrangement is shown in fig (a)

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1st Part Let PA = (pressure of water in pipe line (i.e., at point A) The points B and C lie on the same horizontal line. Hence pressure at B should be equal to pressure at C. But pressure at B = Pressure at A + Pressure due to 10cm (or 0.1 m) of water = pA + ρ × g× h

Where ρ = 1000kg/ m3 and h = 0.1m = pA +1000×9.81×0.1

…………(i) = pA + 981N/ m2 Pressure at C = pressure at D+ Pressure due to 10 cm of mercury = 0+ ρ0 × g× h0 Where 𝜌𝜌0 for mercury = 13.6×1000kg/ m3 And h0 =10cm = 0.1m ∴ Pressure at C

= 0 + (13.6 × 1000) × 9.81 × .01 = 13341.6N ………(ii)

But pressure at B is equal to pressure at C. Hence equating the equations (i) and (ii), we get, pA + 981 = 13341.6

∴ pA = 13341.6 - 981 = 12360.6

N

m

2

2nd Part Given, pA = 9810N/ m2 Find new difference of mercury level. The arrangement is shown in following figure. In this case, pressure at A is 9810𝑁𝑁/𝑚𝑚2 which is less than the 12360.6𝑁𝑁/𝑚𝑚2 . Hence

mercury in left limb will rise. 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. Let x =Rise of mercury in left limb in cm. Then fall of mercury in right limb = x cm The points B, C and D shows the initial conditions whereas points B*, C* and D* show the final conditions. pressure at B* = Pressure at C* Or Pressure at A + Pressure due to (10-x)cm of water = Pressure at D* + Pressure due to (10-2x) cm of mercury Or pA + ρ1 × g× h1 = pD *+ρ2 × g× h2 Or

 10 - x  1910+1000×9.81×    100 

 10 - 2x  = 0+ (13.6×1000) ×9.81×    100  Dividing by 9.81, we get 1000 + 100 - 10x = 1360 - 272x Or 272x - 10x = 1360 - 1100 Or 262x = 260 260 x= = 0.992cm ∴ 262 ∴ New difference of mercury =10 − 2xcm =10 − 2×0.992 = 8.016cm

Example: Fig. shows a conical vessel having its outlet at A to which a U-tube monometer is connected. The reading of the manometer given in the figure shows when the vessel is empty. Find the reading of the manometer when the vessel is completely filled with water.

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= 1000×9.81× (3+ h1 + y/ 100 )

13.6 × ( 0.2 + 2y / 100 ) = ( 3 + 2.72 + y / 100 )

Or 2.72+27.2y/ 100 = 3+2.72+ y/ 100 Or 27.2y- y / 100 = 3.0

(

)

Or 26.2y =3×100 =300 Solution: Vessel is empty: Given: Difference of mercury level h2 = 20cm

Let h1 = Height of water above X-X S.G. of mercury, S2 =13.6 S.G. of water, S1 = 1.0 Density of mercury, ρ2 = 13.6×1000

Density of water, ρ1 = 1000

Equating the pressure above datum line XX, we have, ρ2 × g× h2 = ρ1 × g× h1

Or 13.6×1000×9.81×0.2=1000×9.81×h1 h1 = 2.72m of water.

Vessel is full of water: When vessel is full of water, the pressure in the right limb will increase and mercury level in the right limb will go down. Let the distance through which mercury goes down in the right limb be, y cm as shown in following figure. The mercury will rise in the left by a distance of y cm. Now the datum line is Z-Z. Equating the pressure above the datum line Z-Z, Pressure in left limb =Pressure in right limb  2y  13.6×1000×9.81×  0.2+  100  

300 =11.45cm 26.2 The difference of mercury level in two limbs ∴

y=

= ( 20+2y ) cm of mercury

= 20+2×11.45= 20+22.90

= 42.90cm of mecury

∴ Reading of monometer = 42.90 cm Example: A single column manometer is connected to a pipe containing a liquid of S.G. 0.9 as shown in Fig. Find the pressure in the pipe if the area of the reservoir is 100 times the area of the tube for the manometer reading shown in Fig. The specific gravity of mercury is 13.6

Solution: Given: Sp. gr. of liquid in pipe, S1 = 0.9 ∴ Density ρ1 = 900kg/ m3 Sp. gr. of heavy liquid, S2 =13.6 ∴ Density, ρ2 = 13.6×1000

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Area of reservoir

=

A

= 100

Area of right limb a Height of liquid, h1 = 20cm = 0.2m

Rise of mercury in right limb, h2 = 40cm = 0.4m

Let pA = Pressure in pipe Using equation, a pA = h2 ρ2 g − ρ1 g  + h2 ρ2 g − h1 ρ1 g, A we get, 1 = ×0.4 13.6×1000×9.81- 900×9.81 100 +0.4×13.6×1000×9.81 − 0.2×900×9.81 0.4 133416 - 8829 +53366.4 -1765.8 = 100  = 533.664 + 53366.4 − 1765.8N / m 2 = 52134N / m 2 = 5.21N / cm 2 Example: A differential manometer is connected at the two points A and B of two pipes as shown in fig. The pipe A contains a liquid of S.G. = 1.5 while pipe B contains a liquid of S.G. = 0.9. The pressures at A and B are 1kgf/cm2and 1.80kgf/cm2 respectively. Find the difference in mercury level in the differential manometer.

PA =1kgf/ cm2 =1×104 kgf/ m2

= 104 × 9.81N / m2 (Q 1kgf = 9.81N) Pressure at B,

PB =1.8kgf/ cm2

= 1.8×104 kgf/ m2

=1.8 × 104 × 9.81N / m2(Q 1kgf =9.81N)

Density of mercury = 13.6×1000kg/ m3 Taking X-X as datum line Pressure above X-X in the left limb =13.6×1000×9.81×h+1500×9.81× ( 2+3) + pA

=13.6×1000×9.81× h+7500×9.81+9.81×104

Pressure above X-X in the right limb = 900×9.81× ( h+2) + pB = 900×9.81× ( h+2) +1.8×104 ×9.81

Equating the two pressures, we get 13.6 × 1000 × 9.81h + 7500 × 9.81 + 9.81 × 104 = 900×9.81× ( h+2) +1.8×104 ×9.81

Dividing by 1000×9.81 , we get 13.6h+7.5+10 = ( h+2.0 ) ×.9+18

Or 13.6h+17.5 = 0.9h+1.8 +18 = 0.9h+19.8 Or 13.6 - 0.9 h =19.8 -17.5or12.5h = 2.3

(

∴h=

)

2.3 = 0.181m =18.1cm 12.7

3.5 HYDROSTATIC FORCES ON SURFACES

Solutions: Given: S.G. of liquid at A, = S1 1.5 ∴ = ρ1 1500 S.G. of liquid at B, = S2 1.5 ∴ = ρ2 1500 Pressure at A,

In fluid statics, there is no relative motion between adjacent fluid layers, and thus there are no shear (tangential) stresses in the fluid trying to deform it. The only stress we deal with in fluid statics is the normal stress, which is the pressure, and the variation of pressure is only due to the weight of the fluid. The force exerted on a surface by a fluid at rest is normal to the surface at the point of contact since there is no relative motion between the fluid and the solid surface, and thus no shear forces can act parallel to the surface. Fluid statics is used to determine the forces acting on floating or submerged bodies and

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the forces developed by devices like hydraulic presses and car jacks. The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on the surfaces using fluid statics. 3.5.1 TOTAL PRESSURE

Force is exerted by a static fluid on a surface, either plane or curved when fluid comes in contact with the surfaces. This force always acts normal to the surface. 3.5.2 CENTRE OF PRESSURE

It is defined as the point of application of the total pressure on the surface. The submerged surfaces may be 1) Vertical plane submerged 2) Horizontal plane surface 3) Inclined plane 4) Curved surface 3.5.3 VERTICAL SUBMERGED IN

PLANE

SURFACE

LIQUID Consider a plane vertical surface of arbitrary shape immersed in a liquid as shown A = Total area of surface h = Distance of C. G. of the area from free surface of liquid G = Centre of Gravity of plane surface P = Centre of Pressure h∗ = Distance of centre of pressure from free surface of liquid.

a) Total Pressure Pressure Intensity at strip = ρgh Area of strip dA = b. dh

Force on strip dF = ρ. g. h. b. dh Total pressure force on the whole surface is ∫ dF = ∫ρghbdh s

∫ dF = ρg ∫ h.dA s

F = ρ.g.h.A

∫h.dA is moment of surface area about

free surface of liquid is equal moment of C.G. about free surface. ∫h.dA = A.h

b) Centre of Pressure: (𝐡𝐡∗ ) Principle of Moments: Moment of the resultant force about an axis is equal to the sum of moments of the components about the same axis. Ft . h∗ = ∑moments about free surface of liquid. …(1) = ∫ dA.h.ρgh ∑ moments

= ρg ∫ bh2dh = ρgb∫ h2dh

Where, 2 ∫dA.h = Io is the moment of Inertia of surface about free surface of liquid. …(2) ∑ moments = ρgI0

∴ Ft .h* = ρgIo h* =

h* =

ρgIo

ρgh.A

I0

hA Where, h is the distance of C.G. from free surface A is the area. From II axis theorem

= Io IC.G. + Ah2

IC.G. +Ah2 h= h.A 1) h∗ lies below the C.G. of the surface *

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2) It is independent of the density of liquid & depends only on surface area.

3.5.4 HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID As every point of the surface is at the same depth from free surface of the liquid, the pressure intensity will be equal on the entire surface and equal to P = ρgh where h is depth of surface

= F1 ρg h × Area

h= h= h*

3.5.5 INCLINED PLANE SUBMERGED IN LIQUID

∴ F = ∫ρg y sinθdA

But ∫y dA = Ay

is

∴ F = ρg A h

(Q h = y sin θ)

surface at distance ‘y’ ∴ F = ρg sinθAy

the

moment

of

Note: the above expression of force is for fluid with no pressure acting on the surface. If pressure acts on the surface = F P0 A + ρg Ah

b) Centre of Pressure Pressure force on the strip, dF = ρgh dA = ρgy sin θdA Moment of the force, dF, about axis O-O = dF × y = ρgy sinθdA.y Sum of moments of all such forces about O − O = ∫ρgsinθy 2dA

= ρgsinθ∫y 2dA

Where, 2 ∫y dA = Moment of Inertia of the Let A = Total area of surface h = Distance of C. G. of the area from free surface of liquid G = Centre of Gravity of plane surface P = Centre of Pressure h∗ = Distance of centre of pressure from free surface of liquid.

surface about O − O = Io ∴ Sum of moments of all force = ρg sinθIo F × y* = ρg sinθIo

ρgsinθIo ρg Ah I sinθ y* = o Ah I sin 2 θ h* = 0 Ah a) Total Pressure sin 2 θ * = h (IG + Ay 2 ) Pressure intensity on the strip P = ρgh Ah Pressure force dF on the strip sin2 θ  Ah2  dF = P × dA = ρghdA  IG +  = h* 2   Ah sin θ Total pressure force on the whole area,   F= ∫ dF= ∫ ρghdA a) Rectangle * h h h From fig. sinθ= = = * y y y ∴ h = y sinθ y* =

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A=ab, I xx,C =

b) Circle

A= πR , I xx,C 2

c) Triangle

ab3 12

πR 2 = , I xx,C 0.109757R 4 2 f) Semi ellipse = A

πR 4 = 4

πab 2 = A = , I xx,C 0.109757ab3 2 3.5.6 CURVED SURFACE SUB-MERGED IN LIQUID

d) Triangle

πab3 A= πab, I xx,C = 4

= A e) Semicircle

ab ab3 = , I xx,C 2 36

For a submerged curved surface, the determination of the resultant hydrostatic force is more involved since it typically requires the integration of the pressure forces that change direction along the curved surface. The way to determine the resultant hydrostatic force FR acting on a two-dimensional curved surface is to determine the horizontal and vertical components Fx and Fy separately. This is done by considering the free-body diagram of the liquid block enclosed by the curved surface and the two plane surfaces (one horizontal and one vertical) passing through the two ends of the curved surface. Note that the vertical surface of the liquid block considered is simply the projection of the curved surface on a vertical plane, and the horizontal surface is the projection of the curved surface on a horizontal plane.

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The resultant force acting on the curved Where, surface is given by = ρ 1000kg = / m3 , g 9.81m / s 2 2 2 = FR Fx + Fy 1 A = 3 × 2 = 6m 2 , h = × 3 = 1.5m Inclination of resultant with horizontal is 2 given by ∴= F 1000 × 9.81× 6 ×1.5 Fy = 88290N tan θ = Fx Depth of centre of pressure is given by 1) The horizontal component of the equation as hydrostatic force acting on a curved IG surface is equal (in both magnitude and = h* +h the line of action) to the hydrostatic Ah force acting on the vertical projection of Where, the curved surface. IG = M.O.I. about C.G. of the area of 2) The vertical component of the surface hydrostatic force acting on a curved bd 3 2 × 33 surface is equal to the weight of liquid = = = 4.5m 4 12 12 supported by the curved surface. Example: A rectangular plane surface is 2m wide and 3m deep. It lies in vertical plane in water. Determine the total pressure and position of centre of pressure on the plane surface when it’s upper edge is horizontal and (a) coincides with water surface, (b) 2.5m below the free water surface.

∴= h*

4.5 + 1.5 = 6 ×1.5

0.5+1.5 = 2.0m

b) Upper edge is 2.5, below water surface

Solution: Given: Width of plane surface, b=2m Depth of plane surface, d=3m

a) Upper edge coincides with water surface Total pressure is given by equation as F = ρ gAh

Total pressure (F) is given by F = ρ gAh

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Where, h =Distance of C.G. from free surface of water 3 = 2.5+ = 4.0m 2 ∴ F =1000×9.81×6×4.0

= 235440N

Center of pressure is given by I h* = G + h Ah Where IG = 4.5, A = 6.0,h = 4.0

4.5 + 4.0 6.0×4.0 = 0.1875+ 4.0 = 4.1875= 4.1875m.

π 4 d 64

IG +h= + 4.0 π 2 Ah d × 4.0 4 π 4  Q IG = d  64   h* =

d2 32 = +4.0= +4.0=0.14+4.0= 4.14m 16×4.0 16×4.0

h* =

Example: A circular opening, 3m diameter, in a vertical side of a tank is closed by disc of 3m diameter which can rotate a horizontal diameter. Calculate: i) The force on the disc, and ii) The torque required to maintain the disc in equilibrium in the vertical position when the head of water above the horizontal diameter is 4m. Solution: Given: Dia. of opening d =3m π 2 2 Area, A = ×3 =7.0685m 4 Depth of C.G. h = 4m i) Force on the disc is given by equation as F = ρ gAh =1000×9.81×7.0685×4.0

= 277368N = 277.368kN

ii) To find the torque required to maintain the disc in equilibrium, first calculate the point application of force acting on the disc, i.e., center of pressure of the force F. The depth of centre of pressure (h*) is given by equation as

The force F is acting at a distance of 4.14 m from free surface. Moment of this force about horizontal diameter X-X

= F ( h* -h ) = 277368 ( 4.14 - 4.0) =38831Nm

Hence a torque of 38831 Nm must be applied on the disc in the clockwise direction.

Example: A pipe line which is 4m in diameter contains a gate valve. The pressure at the centre of the pipe is 19.6N/ cm2 . If the pipe is filled with oil of S.G. 0.87; find the force exerted by the oil upon the gate and position of centre of pressure. Solution:

Given: Dia. of pipe, d = 4m ∴ Area, π A = ×42 = 4pm2 4 ∴Density of oil ρ= 0.87 × 1000 = 870kg / m3 0

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∴ Weight density of oil, w 0 = ρ0 × g = 870×9.81N/ m3

Pressure at the centre of pipe,

p = 19.6N/ cm2 = 19.6×104 N/ m3

∴ Pressure head at the centre p 19.6×104 = = = 22.988m w 0 870×9.81 ∴ The height of equivalent free oil surface from the centre of pipe =22.988m The depth of C.G. of the gate valve from free oil surface h = 22.988m F = ρ gAh

Solution: Given: Width of gate, Depth of gate, ∴ Area,

b = 2m d =1.2m

A = b×d = 2×1.2= 2.4m2 Sp. gr. of liquid =1.45 ∴ Density of liquid,

ρ1 =1.45×1000 =1450kg/ m3

Let F1 = Force exerted by the fluid of sp. gr 1.45 on gate F2 = Force exerted by water on the gate. The force F1 = is given by F1 = ρ1 g× A× h1 Where

Where ρ = density of oil = 870kg/ m3 F = 870×9.81×4π ×22.988 = 2465500N = 2.465MN (ii)Position of centre of pressure (h*) is given as ρ1 =1.45×1000 =1450kg/ m2 π 4 h1 = Depth of C.G. of gate from free surface d 2 2 IG d 4 of liquid h*= +h = 64 +h = +h = +22.988 π 2 Ah 16h 16×22.988 1.2 d ×h =1.5+ = 2.1m 4 2 = 0.043+22.988 = 23.031m ∴ F1 = 1450×9.81×2.4×2.1 = 71691N Or, centre of pressure is below the centre of Similarly, F2 = ρ2 g .Ah2 the pipe by a distance of 0.043m Example: A vertical sluice gate is used to cover an opening in a dam. The opening is 2m wide and 1.2 m high. On the upstream of the gate, the liquid of S.G. 1.45 lies upto a height of 1.5m above the top of the gate. Find the resultant force acting on the gate and position of centre of pressure. Find also the force acting horizontally at the top of the gate which is capable of opening it. Assume that the gate is hinged at the bottom.

Where ρ2 =1000kg/ m3 h2 = Depth of C.G. of gate from free surface of water 1 = ×1.2= 0.6m 2

∴ F2 = 1000×9.81×2.4×0.6 = 14126N

(i)Resultant force on the gate

= F1 − F2 = 71691 − 14126 = 57565N

(ii) Position of centre of pressure of resultant force. The force F1 will be acting at a depth of h1 * from free surface of liquid, given by the relation I h* = G + h1 Ah where bd3 2×1.23 IG = = = 0.288m4 12 12

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h1 * =

.288 +2.1= 0.0571+2.1= 2.1571m 2.4×2.1

∴ Distance of F1 from hinge

= (1.5+1.2) − h1 * = 2.7 − 2.1571= 0.5429m

The force F2 will be acting at a depth of h2 * from free surface of water and is given by I h2 * = G + h2 Ah2 Where IG = 0.288m4 ,h2 = 0.6m, A = 2.4m2

.288 +0.6 = 0.2+0.6 = 0.8m 2.4×0.6 Distance of F2 from hinge =1.2 − 0.8 = 0.4m The resultant force 57565N will be acting at a distance given by 71691×.5429 − 14126×0.4 = 57565 38921-5650.4 = m above hinge 57565 h2 * =

= 0.578m above the hinge

(iii)Force at the top of gate which is capable of opening the gate. Let F is the force required on the top of the gate to open it as shown in fig. Taking the moments of F, F1andF2 about the hinge, we get F×1.2+ F2 × 0.4 =F1 × .5429

Or

F=

=

F1 ×.5429 − F2×0.4 1.2

71691×.5429-14126×0.4 38921-5650.4 = 1.2 1.2

= 27725.5N.

Example: A tank contains water up to a height of 0.5m above the base. An immiscible liquid of sp. gr. 0.8 is filled on the top of water up to 1m height. Calculate: i) total pressure on one side of the tank,

ii) the position of centre of pressure from one side of the tank, which is 2m wide Solution: Given: Depth of water = 0.5m Depth of liquid = 1m Sp. gr of liquid = 0.8 Density of liquid, ρ1 = 0.8×1000 = 800kg/ m3 Density of water, ρ2 =1000kg/ m3 Width of tank = 2m (i) Total pressure on one side is calculated by drawing pressure diagram, which is shown in fig . Intensity of pressure on top, pA = 0 Intensity of pressure on D (or DE), pD = ρ1 g .h1

= 800×9.81×1.0 = 7848N/ m2

Intensity of pressure on base (or BC), pB = ρ1 gh1 + ρ2 g×0.5

=7848+1000×9.81×0.5=7848+ 4905=

Now Force,

F1 = Area of ∆ ADE × Width of tank

12753N m2

1 1 = × AD×DE×2.0= ×1×7848×2.0=7848N 2 2 Force, F2 = Areaof rectangle DBFE× Width of tank

= 0.5×7848×2=7848N

F3 = Area of ∆ EFC × Width of tank

1 1 = ×EF×FC×2.0= ×0.5×4905×2.0= 2452.5N 2 2

∴ Total force F = F1 + F2 + F3 =7848+7848+2452.5=18148.5N

(ii) Centre of pressure (h*). Taking the moments of all forces about A, we get

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  2 1  2  F×h* = F1 × AD+F2  AD+ BD  +F3  AD+ BD 3 2  3   

2 * 18148.5 × h= 7848 × × 1 3  0.5  +7848  1.0 +  2    2 +2452.5 1.0 + × 0.5  3  

=5232+9810+3270=18312

∴ h* =

18312 = 1.009m from top 18148.5

Example: A circular plate 3.0m diameter is immersed in water in such a way that their greatest and least depths below the free surface are 4m and 1.5m respectively. Determine the total pressure on one face of the plate and position of the centre of pressure. Solution: Given Dia. of plate, d =3.0m ∴ Area, 2 π π A = d2 = 3.0 = 7.0685m2 4 4 Distance DC =1.5m,BE = 4m

(

)

Distance of C.G. from free surface = h = CD+GC sin θ = 1.5+1.5sin θ

But

AB BE- AE 4.0- DC 4.0-1.5 sin θ = = = = BC BC 3.0 3.0 2.5 = = 0.8333 3.0 ∴ h =1.5+1.5×.8333=1.5+1.249= 2.749m

i) Total Pressure (F) F = ρ gAh =1000×9.81×7.0685×2.749=190621N ii) Centre of pressure (h*) Using equation, we have

IGsin2 θ +h Ah Where 4 4 π π IG = ( d ) = (3) =3.976m4 64 64 h* =

h* =

3.976× ( .8333) ×.8333 7.0685×2.749

= 2.891m.

+2.749=0.1420+2.749

Example: If in the above problem, the given circular plate is having a concentric circular hole of diameter 1.5m, then calculate the total pressure and position of the centre of pressure on one face of the plate. Solution: Given: [referring to given figure] Dia. of plate, d =3.0m ∴ Area of solid plate π π 2 = d2 = 3 =7.0685m2 4 4

()

Dia. of hole in the plate,

d0 = 1.5m

2 π π ∴ Area of hole = d20 = (1.5) =1.7671m2 4 4 ∴ Area of the given plate A

= Area of solid plate − Area of hole

=7.0685-1.7671=5.3014m2 Distance of CD =1.5,BE = 4m

Distance of C.G. from the free surface,

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h = CD+GCsin θ

=1.5+1.5sinθ

But

AB BE- AE 4 -1.5 2.5 = = = BC BC 3 3 2.5 =1.5+1.25= 2.75m ∴ h =1.5+1.5× 3 i) Total pressure force (F) F = ρ gAh =1000×9.81×5.3014×2.75 =143018N =143.018kN ii) Position of centre of pressure (h*) Using equation, we have sin θ =

I sin2 θ h* = G +h Ah where π π IG = d4 - d04  = 34 -1.54  m4 64 64 π π A = d2 - d20  = 32 -1.52  m2 4 4 2.5 sin θ = and h = 2.75 3 2

  π 4 4  2.5 3 − 1.5  ×  3  64   +2.75 ∴ h* = π 2 3 − 1.52  ×2.75 4 = 0.177+2.75= 2.927m

Example: An inclined rectangular sluice gate AB, 1.2m x 5m size as shown in fig is installed to control the discharge of water. The end A is hinged. Determine the force normal to the gate applied at B to open it.

A = Area of gate =1.2×5.0 = 6.0m2 Depth of C.G. of the gate from free surface of the water = h = DG = BC- BE =5.0 − BG sin45° 1 5.0 − 0.6 × = 4.576m 2 The total pressure force (F) acting on the gate,

F = ρgAh = 1000 × 9.81 × 6.0 × 4.576 = 269343N

This force is acting at H, where the depth of h from free surface is given by I sin2 θ h* = G +h Ah Where, IG = M.O.I. of gate

bd3 5.0×1.23 = = 0.72m 12 12 ∴ Depth of centre of pressure 0.72×sin2 45° h* = +4.576 = .013+4.576 = 4.589m 6×4.576 h* = sin45° But from fig, OH ∴ h* 4.589 OH = = = 4.589× 2 = 6.489m 1 sin45° =

2 5 Distance, BO = =5× 2 =7.071m sin45° Distance, BH = BO− OH = 0.071 − 6.489= 0.582m

∴ Distance,

AH = AB− BH =1.2 − 0.582= 0.618m

Taking the moments about the hinge A P× AB = F× ( AH )

Solution: Given:

Where P is the force normal to the gate applied at B ∴ P×1.2= 269343×0.618 269343×0.618 ∴ P= =138708N 1.2 Example:

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Fig shows a quadrant shaped gate of radius 2m. Find the resultant force due to water per meter length of the gate. Find also the angle at which the total force will act.

Solution: Given: Radius of gate =2m Width of gate =1m Horizontal Force, Fx = Force on the projected area of the curved surface on vertical plane = Force on BO = ρ gAh Where, 1 A = Area of BO = 2×1= 2m2 ,h = ×2=1m 2 Fx = 1000×9.81×2×1 = 19620N

This will act at depth of

2 4 ×2= m from 3 3

free surface of liquid, Vertical Force, Fy = Weight of water (imagined) supported by AB

= ρ g× Areaof AOB×1.0

π 4 =1000×9.81× ( 2) ×1.0 =30819N 4 This will act a distance of 4R 4 × 0.2 = = 0.848m from OB. 3π 3π ∴ Resultant force, F is given by F = Fx2 +Fy2

=

196202 + 308192

=36534.4N.

The angle made by the resultant with horizontal is given by F 30819 tan θ = y = =1.5708 Fx 19620

∴ θ = tan-1 1.5708 =57°31'

Example: Find the horizontal and vertical component of water pressure acting on the face of a sector gate of 90° with radius 4m as shown in fig. Take width of gate as unity.

Solution: Given: Radius of gate, R=4m Horizontal component of force acting on the gate is Fx = Force on area of gate projected on vertical plane = Force on area ADB = ρ gAh Where A = AB× Width of gate = 2× AD×1

(Q AB = 2AD)

= 2×4×sin45° = 8×.707 =5.656m2

{Q AD = 4sin 45°} h=

AB 5.656 = = 2.828m 2 2

∴ Fx =1000×9.81×5.656×2.828N =156911N

Vertical component Fy = Weight of water

supported

or

enclosed by the curved surface = Weight of water in portion ACBDA = ρ g× Area of ACBDA× Width of gate = 1000 × 9.81 ×  Area of sector ACBOA − Area of ∆ABO  × 1 π AO×BO  = 9810×  R 2 −  2  4

Q ∆ AOB is a right angled 

π 4×4  = 9810×  42 −  = 44796N 2  4

Example: A cylindrical gate of 4m diameter & 2m long has water on its both sides as shown in Fig. Determine the magnitude, location and direction of the resultant force exerted

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by the water on the gate. Find also the least weight of the cylinder so that it may not be lifted away from the floor.

i) Resultant force, F is given as F = Fx2 +Fy2 =

1

Where Fx = Force of water on area 1

projected on vertical plane = Force on area AOC =ρgA h Where A = AC × width = 4×2= 8cm2 1 h = × 4 = 2m 2 Fx = 1000×9.81×8×2 1

=156960N Fy = Weight of water enclosed by ABCOA 1

π  π =1000×9.81×  R 2  ×2.0= 9810× ×22 ×2.0=123276N 2 2 

Right Side of the Cylinder

Fx = ρ gA 2h2 = Force on vertical area CO 2

= 1000 × 9.81× (2 × 2) ×

2 =39240N 2

Fy = Weight of water enclosed by DOCD 2

π  = ρ g×  R 2  × Width of gate 4 

π =1000×9.81× ×22 ×2= 61638N 4

∴ Resultant force in the direction of x, Fx = Fx − Fx =156960 − 39240 =117720N 1

2

1

2

Resultant force in the direction of y, Fy = Fy +Fy =123276+61638 =184914N

2

2

= 219206N

ii) Direction of resultant force is given by F 184914 tan θ = y = =1.5707 Fx 117720



Solution: Given: Dia. of gate =4m Radius =2m (i) The force acting on the left sides of the cylinder are The horizontal component, Fx

(117720) + (184914)

θ =57°31'

iii) Location of the resultant force Force, Fx acts at a distance of 1

2×4 = 2.67m 3

from the top surface of

water on left side, while Fx acts at a distance of

2 ×2=1.33m 3

2

from free

surface on the right side of the cylinder .The resultant force Fx in the direction of x will act at a distance of y from the bottom as Fx × y = Fx 4 − 2.67  − Fx 2 − 1.33 Or

1

2

117720= × y 156960 ×1.33 − 39240 × .67

= 208756.8 − 26290.8 = 182466

182466 =1.55m from the bottom 117720 Force Fy acts at a distance 4R from 1 3π 4 × 2.0 AOC or at a distance = 0.8488m 3π from AOC towards left of AOC. Also Fy acts at a distance 4R = 0.8488m 2 3π from AOC towards the right of AOC. The resultant force Fy will act at a distance

∴ y=

x from AOC which is given by Fy × x = Fy ×.8488 − Fy ×.8488 1

2

Or 184914× x =123276×.8488 − 61638×.8488  52318.4 = .8488 123276 − 61638 =

∴ x = 52318.4 = 0.2829m from AOC 184914

iv) Least weight of cylinder. The resultant force in the upward direction is

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Fy = 184914N

Thus the weight of cylinder should not be less than the upward force Fy . Hence, weight of cylinder should be at least 184914N

ii) Vertical thrust exerted by water Fy = Weight of water supported by

curved surface OA upto free surface of water = Weight of water in the portion ABO = ρ g × Area of OAB × Width of dam 9  =1000×9.81×  ∫ x×dy  ×1.0  0  9   =1000×9.81×  ∫ 2y 1/2×dy  ×1.0  0 

Example:

2

 x  A dam has a parabolic shape y = y 0    x0  as shown in fig. below having x 0 = 6m and y 0 = 9m. The fluid is water with density

(Q x = 2y ) 1/2

= 1000kg/ m . Compute the horizontal, 3

vertical and the resultant thrust exerted by water per meter length of the dam.

9

 y 3/2  2  =19620× 93/2  =19620×   3  (3/ 2)  0 2 =19620× ×27 =353160N 3

iii) Resultant thrust exerted by water F=

Solution: Given: Equation of the curve OA is 2

2

x x x2 x2 y = y 0   = 9   = 9× = 36 4 6  x0  Or x 2 = 4y



3973052 + 3531602 = 531574N

iv) Direction of resultant is given by Fy 353160 tan= θ = = 0.888 Fx 397305

−1 = θ tan = 0.888 41.63o

x = 4y = 2y 1/2

Width of dam, b =1m

FX 2 + F Y2 =

i) Horizontal thrust exerted by water Fx = Force exerted by water on vertical surface OB, i.e., the surface obtained by projecting the curved surface on vertical plane = ρ gAh

3.6 BUOYANCY & FLOATATION When a body is immersed in a fluid, an upward force is exerted on the body; this upward force is known as the buoyant force. This force is because of difference in pressure.

9 =1000×9.81× ( 9×1) × =397305N 2

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Force acting on the element because of difference in pressure on the top and bottom. dF = ρg ( h2 − h1 ) dA

= FB

∫ρg ( h

FB = ρgV

2

)

− h1 dA

Where, ρ is the density of fluid Meta centric height = GM V is the volume of body immersed in fluid I ( Moment of Inertia about yy of the plan ) or volume of fluid displaced by that body. = − BG The relation ρgV is simply the weight of the Volume of fluid displaced liquid whose volume is equal to the B.G is the distance between CG & CB points. immersed volume of the body. Thus the buoyant force acting on the body is equal to 3.6.3 OSCILLATION OF A FLOATING the weight of the liquid displaced by the BODY body. Note that the buoyant force is independent of the distance of the body When body floats in the fluid and it is given from the free surface. It is also independent a disturbance in clockwise direction or anti of the density of the solid body clock wise direction. The body oscillates about its metacenter. The time period of 3.6.1 CENTRE OF BUOYANCY oscillation is given by The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume. The centroid of displaced fluid is known as centre of buoyancy.

3.6.2 META CENTRE It is defined as the point about which a body starts oscillating when the body is tilled by a small angle. The Meta centre may also be defined as the intersection point of line of action of buoyant force and normal to the body when the body is tilted by an angle.

T = 2π

k2 GM.g

Where, GM is Meta centric ht K is radius of gyration

3.6.4 CONDITIONS OF EQUILIBRIUM OF SUBMERGED & FLOATING BODIES There are 3 types of equilibrium conditions i) Stable Equilibrium ii) Neutral equilibrium iii) Unstable equilibrium

i) Stable equilibrium: Any small disturbance (someone moves the ball to the right or left) generates a restoring force (due to gravity) that returns it to its initial position.

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ii) Neutral equilibrium: If someone moves the ball to the right or left, it will stay at its new location. It has no tendency to move back to its original location, nor does it continue to move. iii) Unstable Equilibrium: It is a situation, in which the ball may be at rest at the moment, but any disturbance, even an infinitesimal one, causes the ball to roll off the hill—it does not return to its original position; rather it moves away from it. 3.6.4.1 STABILITY IN SUBMERGED BODIES

1) Stable Equilibrium: When W = FB and point B is above G. A small displacement in clockwise direction, gives couple due to FB & weight in anticlockwise direction. Thus, the body will return to its original position. Hence, equilibrium is stable. 2) Unstable Equilibrium: If W = FB and point B is below point ‘G’. A small displacement to the body, in the clock wise direction, gives couple due to W & FB also in the clockwise direction. Thus, body will move away from its original position. Hence, equilibrium is unstable 3) Neutral Equilibrium: If FB = W and B & G are at the same point, the displacement of body does not result in any couple of Wt & FB. Body remains at its displaced position

Stable equilibrium

Unstable equilibrium

3.6.42 STABILITY IN FLOATING BODY The stability of floating body is determined from position of metacentre(M). In case of

floating body, the weight of body is equal to the buoyant force. 1) Stable Equilibrium: When M is above G, because of a small displacement to the body in the clock wise direction, the couple between Wt & FB causes rotation in anti-clockwise direction. 2) Unstable Equilibrium: When M is below G, because of small displacement to the body in the clock wise direction, the couple between Wt & FB causes rotation in clockwise direction. 3) Neutral: If M lies at the C.G. of body, the displacement of body does not result in any couple of Wt & FB. Body remains at its displaced position.

Stable equilibrium

Unstable equilibrium Example: Find the volume of the water displaced and position of centre of buoyancy for a wooden block of width 2.5 m and of depth 1.5m, when it floats horizontally in water. The density of wooden block is 650kg/ m3 and its length is 6.0m. Solution: Given: Width = 2.5m Depth = 1.5m Length = 6.0m

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Volume of the block

3

= 2.5×1.5×6.0= 22.50m

Density of wood,

ρ = 65kg/ m3

∴ Weight of block = ρ ×g× Volume

= 650×9.81×22.50N = 143471N For equilibrium, the weight of water displaced = Weight of wooden block = 143471N ∴ Volume of water displaced =

Weight of water displaced 143471 = =14.625m3 Weight density of water 1000×9.81

Position of center of Buoyancy: Volume of wooden block in water = Volume of water displaced Or 2.5×h×6.0=14.625m3 , Where, h is depth of wooden block in water 14.625 = 0.975m ∴ h= 2.5×6.0 ∴ Centre of Buoyancy 0.975 = = 0.4875m from base 2

Example: Find the density of a metallic body which floats at the interface of mercury of S.G. 13.6 and water such that 40% of its volume is sub-merged in mercury and 60% in water. Solution: Let the volume of the body = Vm3 Then volume of body sub-merged in mercury 40 = V = 0.4Vm3 100 Volume of body sub-merged in water

=

60 V = 0.6Vm3 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 buoyancy due to mercury Force of buoyancy due to water = Weight of water displaced by body = Density of water × Volume of mercury displaced = 1000 × g ×Volume of body in water = 1000×g ×0.6× VN And, force of buoyancy due to mercury = Weight of mercury displaced by body = g × Density of water× Volume of mercury displaced =g ×13.6×1000× volume of body in mercury = g ×13.6×1000×0.4VN Weight of the body =Density x g x Volume of body ∴ For equilibrium, we have Total buoyant force =Weight of the body

1000×g×0.6×V+13.6×1000×g×.4V = ρ ×g×V

Or ρ =600+13600×.4 =600+54400=6040.00kg/ m3 ∴ Density of the body = 6040.00kg/ m3 Example: A float valve regulates the flow of oil of S.G. 0.8 into a cistern. The spherical float is 15 cm in diameter. AOB is a weightless link carrying the float at one end, and a valve at the other end which closes the pipe through which oil flows into the cistern. The link is mounted on a frictionless hinge at O and the angle AOB is 135°. The length of OA is 20cm, and the distance between the centre of the float and the hinge 50 cm. When of the flow is stopped AO will be

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vertical. The valve is to be pressed on to the seat with a force of 9.81 N to completely stop the flow of oil into the cistern. It was observed that the flow of oil is stopped when the free surface of oil in the cistern is 35 cm below the hinge. Determine the weight of the float.

= 800×9.81×0.00945=7.416N The buoyant force and weight of the float passes through the same vertical line, passing through B. Let the weight of float is W. Then net vertical force on float =Buoyant force –Weight of float = (7.416W) Taking moments about the hinges O, we get

P×20= (7.416 − W ) ×BD= (7.416 − W ) ×50×cos45°

(

Solution: Given: Sp. gr. of oil =0.8 ∴ Density of oil = ρ0 = 0.8×1000 = 800kg/ m3 Dia. of float, D=15cm

∠ AOB=135° OA = 20cm

Force, P=9.81N OB=50 cm Let the weight be W.

When the flow of oil is stopped, the centre of float is shown in Fig. The level of oil is also shown. The centre of float is below the level of oil, by a depth ‘h’ From ∆ BOD,

∴ W =7.416 −

20×9.81 =7.416 − 5.55=1.866N 35.355

Example: A rectangular pontoon is 5m long, 3m wide and 1.20m high. The depth of immersion of the pontoon is 0.80m in sea water. If the centre of gravity is 0.6m above the bottom of the pontoon, determine the Meta – centric height. The density for sea water = 1025kg/ m3

Solution: Given: Dimension of pontoon = 5m × 3m × 1.20m Depth of immersion =0.8m Distance AG=0.6m

OD OC+CD 35+ h = = OB OB 50 Or 50×sin45° =35+ h 1 h =50× − 35=35.355 − 35= 0.355cm = .00355m 2 The weight of float is acting through B, but the upward buoyant force is acting through the centre of weight of oil displaced Volume of oil displaced :   D 15 2 =7.5cm  = π r3 + h× π r2 r = = 2 2 3   sin45° =

3 2 2 = × π × ( 0.75) +.00355× π × ( 0.75) =0.000945m3 3 =Weight of oil displaced = ρ0 × g× Volume of oil

)

Or 9.81×20 = 7.416 − W ×35.355

1 Distance AB = ×Depth of immersion 2 1 = ×.8 = 0.4m 2

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Density for sea water = 1025kg/ m3 Meta-centre height GM, given by equation as I GM = − BG ∀ Where I = Moment of Inertia of the plan of the pontoon about Y-Y axis 1 45 = ×5×33 m4 = m4 12 4 ∀ = Volume of the body sub- merged in water

=3×0.8×5.0=12.0m3

BG = AG = AB = 0.6 - 0.4 = 0.2m

GM =

45 4

×

1

12.0

− 0.2=

45

48

− 0.2= 0.9375 − 0.2= 0.7375m

Example: A solid cylinder of diameter 4.0 m has a height of 4.0m. Find the meta-centric height of the cylinder if the specific gravity of the material of cylinder = 0.6 and it is floating in water with its axis vertical. State whether the equilibrium is stable or unstable. Solution: Given: D=4m Height, h=4m S.G. = 0.6 Depth of cylinder in water = S.G x h

= 0.6×4.0= 2.4m ∴ Distance of centre of buoyancy (B) from

A

2.4 =1.2m 2 Distance of centre of gravity (G) from A h 4.0 AG = = = 2.0m 2 2 ∴ BG = AG− AB = 2.0 − 1.2= 0.8m Now the meta–centric height GM is given by I GM = − BG ∀ Where I=M. O. I. of the plan of the body about Y-Y axis

4 π 4 π D = × ( 4.0) 64 64 ∀ = Volume of cylinder in water

=

π π ×D2×Depth of cylinder in water = ×42 ×2.4m3 4.0 4 π ×44 1 1 42 1 ∴ = 64 = = 0.4167m = × ∀ π 2 16 2.4 2.4 ×4 ×2.4 4 1 GM = − BG = 0.4167 − 0.8 = −0.3833m ∀ =

-ve sign means that the meta–centre (M) is below the centre of gravity (G). Thus the cylinder is in unstable equilibrium.

Example: A wooden cylinder of S.G. 0.6 and circular cross–section is required to float in oil (S.G. 0.90). Find the L/D ratio for the cylinder to float with its longitudinal axis vertical in oil, where L is the height of cylinder and D is its diameter. Solution: Given: Dia of cylinder =D Height of cylinder =L Sp. Gr. Of cylinder S1 = 0.6 Sp. Gr of oil S2 = 0.9 Let the depth of cylinder immersed in oil=h

AB =

For the principle of buoyancy Weight of cylinder = wt. of oil displaced

π 2 π ×D ×L×0.6×1000×9.81= ×D2×h×0.9×1000×9.81 4 4

Or

L×0.6 = h×0.9

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0.6×L 2 = L 0.9 3 The distance of centre of gravity G from A, L AG = 2 The distance of centre of buoyancy B from A h 1 2  L AB = =  L  = 2 2 3  3



h=

L L 3L− 2L L = ∴ BG = AG− AB = − = 2 3 6 6

The meta–centric height GM is given by I GM = − BG ∀ π 4 Where I = D and ∀ = Volume of 64 π 2 cylinder in oil = D ×h 4 I  π 4 π 2  1 D2 D2 3D2 ∴ = D / D h= = = ∀  64 4  16 h 16× 2 L 32L 3

 2  Q h = L  3  

3D2 L − 32L 6 For stable equilibrium, GM should be +ve or, 3D2 L GM > 0 or − >0 32L 6 Or 3D2 L 3×6 L2 > or > 32L 6 32 D2 Or L2 18 9 < or 2 16 D 32

∴ GM =





The time period of rolling of a ship of weight 29430kN in sea water is 10seconds. The centre of buoyancy of the ship is 1.5 m below the centre of gravity. Find the radius of gyration of the ship if the moment of inertia of the ship at the water line about fore and aft axis is1000m4 . The specific weight of sea water as 10100N/ m3 Solution: Given: Time period T=10sec Distance between centre of buoyancy and centre of gravity, BG=1.5m Moment of Inertia, I =10000m4 Weight W = 29430kN = 29430×1000N Let the radius of gyration =K First calculating the meta–centric height, which is given as I GM = BM− BG = − BG ∀ Where I= Moment of Inertia And ∀ = Volume of water displaced =

Weight of ship 29430×1000 = = 2912.6m3 Sp.weight of sea water 10104

∴GM =

10000 − 1.5=3.433 − 1.5=1.933m 2912.6

Using equation, T = 2π We get

10 = 2π

Or

K=

K2 GM× g

K2 2π K = 1.933×9.81 1.933×9.81

10× 1.933×9.81 = 6.93m 2π

L 9 3 < = D 16 4 L <3/ 4 D

Example:

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GATE QUESTIONS Q.1

Q.2

A static fluid can have a) non-zero normal and shear stress b) negative normal stress and zero shear stress c) positive normal stress and zero shear stress d) zero normal stress and non-zero shear stress [GATE–2001] The horizontal and vertical hydrostatic forces Fx and Fy on the semi-circular GATE, having a width w into the plane of figure, are

Q.4

Q.5

ρgwr 2 2 πρgwr 2 F = d) Fx = 2ρghrw and y 2 [GATE–2001] c) Fx = ρghrw and Fy =

A cylindrical body of cross-sectional area A, height H and density ρs , is immersed to a depth h in a liquid of density ρ, and tied to the bottom with a string. The tension in the string is

a)ρghA

b)(ρs−ρ)ghA

d)(ρh−ρsH)gA [GATE–2003] The pressure gauges G1and G2 installed on the system show pressure of PG1=5.00 bar and PG2= 1.00 bar. The value of unknown pressure P is [Atmospheric Pressure 1.01 bar]

a) 1.01 bar c) 5.00 bar

a) Fx = ρghrw and Fy = 0 b) Fx = 2ρghrw and Fy = 0

Q.3

c)(ρ−ps)ghA

b) 2.01 bar d) 7.01 bar [GATE–2004]

A closed cylinder having a radius R and height H is filled with oil of density ρ. If the cylinder is rotated about its axis at an angular velocity of ω, the thrust at the bottom of the cylinder is a) πR2 ρgH ρω2 R 2 b) πR 2 4 c) πR 2 (ρω2 R 2 + ρgH)  ρω2 R 2  d) πR 2  + ρgH   4 

[GATE–2004]

Q.6 Oil in a hydraulic cylinder is compressed from an initial volume 2 m3 to 1.96 m3. If the pressure of oil in the cylinder changes from 40 MPa to 80 MPa during compression the bulk modulus of elasticity of oil is a) 1000 MPa b) 2000 MPa c) 4000 MPa d) 8000 MPa [GATE–2007] Q.7

For the stability of a floating body,

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under the influence of gravity alone, which of the following is TRUE? a) Metacenter should be below centre of gravity. b) Metacenter should be above centre of gravity. c) Metacenter and centre of gravity must lie on the same horizontal line. d) Metacenter and centre of gravity must lie on the same vertical line. [GATE–2010] Q.8

A large tank with a nozzle attached contains three immiscible, inviscid fluids as shown. Assuming that the changes in h1, h2 and h3 are negligible, the instantaneous discharge velocity is

 ρ h ρ h  a) 2gh 3  1 + 1 1  + 2 2   ρ3 h 3 ρ3 h 3 

b) 2g ( h1 + h 2 + h 3 )

 ρ h +  ρ 2 h 2 + ρ3 h 3  c) 2g  1 1  ρ1  + ρ 2  + ρ3  

 ρ h h +  ρ 2 h 3 h1 + ρ3 h1h 2  d) 2g  1 2 3  ρ1h1 + ρ 2 h 2 + ρ3 h 3   [GATE–2012]

Q.9

A hinged GATE of length 5 m, inclined at 30° with the horizontal and with water mass on its left, as shown in the figure below. The minimum mass of the GATE in kg per unit width (perpendicular to the plane of paper), required to keep it closed is

a) 5000 c) 7546

b) 6600 d) 9623 [GATE–2013]

Q.10 For a completely submerged body with centre of gravity 'G' and centre of buoyancy ‘B', the condition of stability will be a) G is located below B b) G is located above B c) G and B are coincident d) independent of the locations of G and B [GATE–2014] Q.11 The difference in pressure (in N/m2) across an air bubble of diameter 0.001 m immersed in water (surface tension = 0.072 N/m) is______. [GATE–2014] Q.12 An aluminium alloy (density 2600 kg/m3) casting is to be produced. A cylindrical hole of 100 mm diameter and 100 mm length is made in the casting using sand core (density 1600 kg/m3). The net buoyancy force (in newton) acting on the core is ______. [GATE–2014] Q.13 A spherical balloon with a diameter of 10 m, shown in the figure below is used for advertisements. The balloon is filled with helium (RHe = 2.08 kJ/kgK) at ambient conditions of 15°C and 100 kPa. Assuming no disturbances due to wind, the maximum allowable weight (in newton) of balloon material and rope required to avoid the fall of the balloon (Rair = 0.289 kJ/kgK) is _____.

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volume is in oil while the rest is in water. The density of the body is _________ kg/m3. The specific gravity of oil is 0.7 and density of water is 1000 kg/m3. Acceleration due to gravity g = 10 m/s2. [GATE–2014] Q.14 For a floating body, buoyant force acts at the a) centroid of the floating body b) center of gravity of the body c) centroid of the fluid vertically below the body d) centroid of the displaced fluid [GATE–2016] Q.15 An inverted U-tube manometer is used to measure the pressure difference between two pipes A and B, as shown in the figure. Pipe A is carrying oil (specific gravity = 0.8) and pipe B is carrying water. The densities of air and water are 1.16 kg/m3 and 1000 kg/m3, respectively. The pressure difference between pipes A and B is __________ kPa. Acceleration due to gravity g = 10 m/s2.

[GATE–2016] Q.16 The large vessel shown in the figure contains oil and water. A body is submerged at the interface of oil and water such that 45 percent of its

[GATE–2016] Q.17 Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2𝑅𝑅 × 𝐿𝐿 at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semicylinder of radius R. The tank is filled with a liquid of density ρ up to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.

The force F required to hold the plug in its position is  π a) 2ρR2gL 1 −   4  π b) 2ρR2gL 1 +   4 2 c) πR ρgL π d) ρR2gL 2 [GATE–2016(2)]

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Q.18 For the stability of a floating body the

c) centre of gravity must be above the centre of buoyancy d) metacentre must be above the centre of gravity

a) centre of buoyancy must coincide with the centre of gravity b) centre of buoyancy must be above the centre of gravity

[GATE–2017(2)]

ANSWER KEY: 1 (c) 15

-2.19

2 (d) 16

865

3 (d) 17 (a)

4 (d) 18 (d)

5 (d)

6 (b)

7 (b)

8 (a)

9 (d)

10 (a)

11 288

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12 7.7

13 14 5303.7 (d)

EXPLANATIONS Q.1

Q.2

(c) Fluid static deals with problems associated with fluids at rest. In static fluid, there is no relative motion between adjacent fluid layers and thus there are no shear (tangential) stresses in the fluid trying to deform it. The only stress in static fluid is the normal stress, which is the pressure and the variation of pressure is only due to the weight of the fluid and it is always positive. Therefore, the topic of fluid statics has significance only in gravity field.

Q.3

(d) Given: Cross section area of body = A Height of body = H Density of body = ρs Density of liquid = ρ Tension in the string = T We have to make the FBD of the block. B = Buoyancy force

(d)

Here F1 = weight of water column above the top surface. F2 = weight of water column above the bottom surface. At the depth, pressure is given by p = ρgh then horizontal force, Fx = A×p = (2r×w) ρgh where A = Normal area, when viewed in the direction of Fx Fx = 2ρghrw Fy = F2− F1 = weight of water contained in volume of semi circular π  GATE. = Fy mg =  r 2 × w  ρg 2  m = ρV and V = A × w πρgwr 2 Fy = 2

Q.4

T + mg = ρhAg T + ρs Vg  = ρhAg T + ρsAHg = ρhAg T = ρhAg − ρsAHg = Ag(ρh−ρsH)

Q.5

(d)

(d) Given: PG1 = 5.00 bar, PG2 = 1.01 bar Absolute pressure of G2 = Atmospheric pressure + Gauge pressure =1.01 + 1.00= 2.01 bar Absolute pressure of G1 = PG1 + Pabs(G2) = 5.0 + 2.01 = 7.01 bar

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Total thrust at the bottom of cylinder = Weight of water in cylinder + Pressure force on the cylinder For rotating motion, ∂p ρV 2 ρr 2 ω2 = = = ρω2 r ∂r r r ∂p = ρω2rdr Integrating both the sides P

R

0

0

Given data : V1 = 2 m3 V2 = 1.96 m3 ∴ dV = V2 − V1 =1.96 − 2 =− 0.04 m3 p1 = 40 MPa p 2 = 80 MPa ∴ dp = p 2 − p1 =80 − 40 = 40 MPa

2 ∫∂p =∫ρω rdr

Bulk modulus of elasticity. K=−

R

 r2  [p] = ρω    2 0 ρω2 r 2 P= 2 Dividing whole area of cylinder in the infinite small rings with thickness dr, Force on elementary ring = Pressure intensity × Area of ring ρω2 r 2 = × 2πrdr 2 R ρω2 r 2 Total F   = ∫0 2 × 2πrdr P 0

2

= 2000 MPa

Q.7

(b)

Q.8

As shown in figure above. If point B’ is sufficiently far from B, these two forces (Gravity force and Buoyant force) create a restoring moment and return the body to the original position. A measure of stability for floating bodies is the metacentric height GM, which is the distance between the centre of gravity G and the metacenter M (the intersection point of the lines of action of the buoyant force through the body before and after rotation.) A floating body is stable if point M is above the point G, and thus GM is positive, and unstable if point M is below point G, and thus GM is negative. Stable equilibrium occurs when M is above G

R

Q.6

dp dp −2 40 = − V1 = −0.04 dV/V1 dV

R4 = πρω2 ∫r 3dr = πρω2 4 0 Weight = mg = 𝜌𝜌Vg = ρπR2Hg ω2 R 4 So, Net force = ρπ + ρπR 2 Hg 4 2 2  ρω R  πR 2 ρgH + 4  

(b)

(a)

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Converting all the height corresponding to liquid (3) x 3ρ3 = h1ρ1 ρ X 3 = h1 1 ρ3 y3ρ3 = h 2ρ 2 hρ y3 = 2 2 ρ3 ρ ρ H= h 3 + h1 1 + h 2 2 ∴ ρ3 ρ3 Instantaneous discharge velocity = 2gH

= V = Q.9

 ρ ρ  2g  h 3 + h1 1 + h 2 2  ρ3 ρ3  

 ρ h ρ h  2gh 3 1 + 1 1 + 2 2   ρ3 h 3 ρ3 h 3 

(d)

(A submerged equilibrium)

body

in

stable

Q.11 (288)

2σ r (this is a case of droplet) where, ∆P is pressure difference σ is surface tension = 0.072 N/m 0.001 m r is bubble radius = 2 = 0.0005m 2 × 0.072 ∴ ∆P= 0.0005 = 288 N/m2 For bubble, ∆P =

Q.12 (7.7)

FR = �ρgh� × A = 103 × 9.81× 2.5sin sin 30° × 5 ×1 =61.31kN l sin 2 θ h cp= h + G Ah 1× 53 0.25 = 1.25 + × 12 5 ×1×1.25 = 1.667m h cp = 3.34m sin 30° Taking moment about the hinge 61.31×103 × 3.34 = m × 9.81× 2.5cos 30° m = 9622.8kg

Q.10 (a)

π 2 DH 4 = 0.785×0.12 × 0.1 = 7.85×10−4 m3 Net buoyancy force =(ρAl – ρcore)Vcoreg = 1000×9.81×7.85×10−4 = 7.7 N Volume of core =

Q.13 (5303.688) Mass of Balloon:4 Volume, V = πr 3 3 3 4 × 3.14 × 5 = 3 = 523.33 m3 P 100 ×103 ρ= = He RT 2080 × 288 = 016693 kg/m3 Total mass = 0.16693 × 523.33 = 87.359 kg

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Mass of Air:P 100 ×103 = ρ = RT 289 × 288 = 1.2 kg/m3 Mass of displaced volume of air = 1.2 × 523.33 = 628 kg Net Buoyant Mass = 628 − 87.359 = 540.64 kg ∴Maximum allowable weight = 540.64 × 9.81 = 5303.688 N

Q.14 (d)

Q.15 (−2.19)

Q.16 (865) Weight of the body will be balanced by the buoyant force ∴ FB = mg ρs Vg ( 0.7ρω )( 0.45V ) g + ρω ( 0.55V ) g = ⇒ 700×0.45 + 1000 ×0.55 = ρs ρs = 865kg/m3 Q.17 (a) Force required  πR 2 =  2R × R × L − 2 

 L  ρg 

 π = 2ρR 2 gL 1 −   4 Q.18(𝒅𝒅)

For stable equilibrium

Given data Specific gravity of oil Soil=0.8 ∴ ρoil =0.8× 1000 = 800kg/m3 Density of air ρoil = 1.16 kg /m3 Density of water ρoil =1000 kg/m3 Acceleration due to gravity g = 10 kg /m2 h1 = 80 mm = 0.08 m h2 = 200 mm = 0.2 m h3 = 100 mm = 0.1 m Pressure at section1 on left limb = Pressure at section 1 on right limb pa - ρoil gh2 – ρair gh1= pb – ρwater g(h1+ h2+ h3) pa - 800×10×0.2-1.16×10×0.08= pb – 1000×10(0.08+ 0.2+ 0.1) pa – 1600-0.928= pb – 3800 pa –pb= -2199.07Pa = -2.199KpA © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

4

KINEMATICS

4.1 INTRODUCTION Fluid kinematics deals with describing the motion of fluids without necessarily considering the forces and moments that cause the motion. The velocity at any point in a flow field at any time is studied in this branch of fluid mechanics. Once the velocity is known, then the pressure distribution & hence forces acting on the fluid can be determined. 4.2 METHODS OF DESCRIBING FLUID MOTION

1) Lagrangian Method: A single fluid particle is followed during its motion and its velocity, acceleration, density etc. are described. 2) Eulerian Method: The velocity, acc, density etc. are described as a point in flow field. It is commonly used in fluid mechanics. 4.3 TYPES OF FLUID FLOW

4.3.1 STEADY & UNSTEADY FLOW Steady flow is defined as that type of flow in which the fluid characteristics like velocity etc. at a point do not change with time.  dP  =0    dt ( xo ,y o ,zo )

(

P = f x,y,z

)

Where, P is flow parameter (velocity, acceleration, density etc)

Example: Fluid flowing in pipe of uniform cross section or varying cross section with

constant mass flow rate. The velocity of fluid may vary with position but remains constant w.r.t time.

Unsteady flow is that type of flow in which the velocity, pressure, density at a point changes with respect to time.  dP  ≠0    dt ( x ,y ,z )

(

o

o

o

P = f x,y,z,t

)

Example: Fluid flowing in pipe of uniform cross section or varying cross section with variable mass flow rate 4.3.2 UNIFORM FLOW & NON UNIFORM FLOW

Uniform flow is defined as that type of flow in which the velocity at any given time does not change w.r.t. space.  dP    =0  ds t

Example: Fluid flowing through uniform cross section. The velocity remains constant w.r.t. space. The mass flow rate can vary w.r.t. time.

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i) Fluid flowing between two parallel plates V1 = V2

Velocity is along ‘x’ direction U = f (y)

Non Uniform flow is that type of flow in which velocity changes with respect to space at given instant.

ii) Velocity flow field given by

V1 ≠ V2

2) Two dimensional: Two dimensional flows is that type of flow in which the flow parameter such as velocity is a function of time and two space coordinates. Velocity flow field given by = V ax ˆi − byjˆ

Example: Fluid flowing through non-uniform cross section. The velocity varies w.r.t space. The mass flow rate can vary

1) Steady Uniform flow: Flow at constant rate through a duct of uniform crosssection. 2) Steady non-uniform flow: Flow at constant rate through a duct of nonuniform cross-section (tapering pipe). 3) Unsteady Uniform flow: Flow at varying rates through a long straight pipe of uniform cross-section. (Again the region close to the walls is ignored. 4) Unsteady non-uniform flow: Flow at varying rates through a duct of nonuniform cross-section 4.3.3 ONE-, TWODIMENSIONAL FLOWS

AND

THREE-

1) One dimensional: One dimensional flow is that type of flow in which the flow parameter such as velocity is a function of time and one space coordinate only. Example:

= V ax ˆi + bxjˆ

Here velocity has two components but depends only on one dimension. Therefore one dimensional flow

3) Three dimensional: Three dimensional flows is that type of flow in which the flow parameter such as velocity is a function of time and three spaces coordinate. Velocity flow field given by V = ax ˆi + byˆj + czkˆ = V axyˆi − byzjˆ 4.3.4 LAMINAR AND TURBULENT FLOWS

Viscous flow regimes are classified as laminar or turbulent on the basis of flow structure. In the laminar regime, flow structure is characterized by smooth motion in lamina or layers. Flow structure in the turbulent regime is characterized by random, three-dimensional motions of fluid particles in addition to the mean motion. In laminar flow there is no mixing between adjacent fluid layers. A thin filament of dye injected into a laminar flow appears as a single line; there is no dispersion of dye throughout the flow. A dye filament

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injected into turbulent flow disperses quickly throughout the flow field; the line of dye breaks up into entangled threads of dye. This behaviour of turbulent flow is due to the velocity fluctuations present; the mixing of fluid particles from adjacent layers of fluid results in rapid dispersion of the dye. If one measures the x component of velocity at a fixed locations in a pipe for both laminar and turbulent steady flow, the traces of velocity versus time appears. In the turbulent flow the flow velocity trace indicates random fluctuations of the instantaneous velocity, u= u + u′ In one dimensional laminar flow, the shear stress is related to the velocity gradient by the simple relation Tyx = µdu / dy In turbulent flow there is no universal relationship between the stress field and the mean velocity field. Thus in turbulent flows we must rely heavily on semi empirical theories and on experimental data. The type of flow is determined by Reynold’s No. ρVL Re = µ Where, ρ =Density L = Characterstic length V = Velocity μ = Dynamic Viscosity

4.3.5 COMPRESSIBLE & INCOMPRESSIBLE FLOWS

Flows in which variations in density are negligible are termed incompressible; when density variations within a flow are not negligible, the flow is called compressible. Gas flowing with negligible heat transfer may also be considered incompressible provided that the flow speed is small relative to the speed of sound.

The ratio of the flow speed V, to the local speed of sound C, in the gas is defined as the Mach number. V M= C Thus gas flows with M<0.3 can be treated as incompressible. 4.3.6 INTERNAL AND EXTERNAL FLOWS

Flows completely bounded by solid surfaces are called internal or duct flows. Flows over bodies immersed in an unbounded fluid are termed external flows. Both internal and external flows may be laminar or turbulent, compressible or incompressible. 4.4 CONTINUITY EQUATION IN THREEDIMENSIONS Continuity equation is based on mass conservation principle. (mass flow rate in) - (mass flow rate out) = (rate of change of mass in control volume) dm min − mout = dt In Cartesian coordinate system ∂ ∂ ∂ ∂ρ ρu + ρv + ρw + = 0 ∂x ∂y ∂z ∂t Where, u = f1 ( x,y,z )

( )

( )

( )

( ) w = f ( x,y,z )

v = f2 x,y,z 3

This equation is applicable to

1. Steady & unsteady 2. Compressible, incompressible. 3. Uniform, non uniform ∂ρ For steady flow =0 ∂t ∂ ∂ ∂ ρu + ρv + ρw = 0 ∂x ∂y ∂z

( )

( )

( )

For incompressible & steady flow

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∂ ∂ ∂ u + v + w = 0 ∂x ∂y ∂z For two dimensional, incompressible & steady flow ∂ ∂ u + v = 0 ∂x ∂y

( )

( )

( )

( )

( )

The magnitude of resultant velocity is given by

V=

u2 + v 2 + w 2

The acceleration is given by

a = a xi + a yj + a z  k

Where, 4.5 CONTINUITY EQUATION IN ONE du dv dw = ax = , ay = , az DIMENSION dt dt dt du ∂u  ∂x  ∂u  ∂y  ∂u ∂z ∂u For steady state condition ax = =   +   + . + dt ∂x  ∂t  ∂y  ∂t  ∂z ∂t ∂t (massin ) = (massout )

ρ1 A1 V1 = ρ2 A2 V2

ax =

For incompressible fluid

ρ1 = ρ2

A1 V1 = A 2 V2

Continuity equation is incompressible steady state.

valid

for

4.6 MOTION OF FLUID ELEMENT a) b) c) d)

linear motion Rotation motion linear deformation angular deformation

4.6.1 LINEAR MOTION

In pure translation motion the fluid particle retains its shape. It does not deform.

4.6.2 VELOCITY & ACCELERATION The velocity in flow field is given by

V = uˆi + vˆj + wkˆ

Where,

( ) v = f ( x,y,z ) w = f ( x,y,z )

u = f1 x,y,z 2

3

ay =

u.∂u v.∂u w.∂u ∂u + + + ∂x ∂y ∂z ∂t

u.∂v v.∂v w.∂v ∂v + + + ∂x ∂y ∂z ∂t

u.∂w v.∂w w.∂w ∂w + + + ∂x ∂y ∂z ∂t For steady state, dv =0 dt ∂u ∂v ∂w = 0,= 0,= 0 ∂t ∂t ∂t Hence acceleration u.∂u v.∂u w.∂u ax = + + ∂x ∂y ∂z az =

4.6.3 LOCAL ACCELERATION CONVECTIVE ACCELERATION

&

Local acceleration is defined as the rate of change of velocity with respect to time at a given point in a flow field.

∂u ∂v ∂w , , ∂t ∂t ∂t

4.6.4 CONVECTIVE ACCELERATION Convective acceleration is defined as change of velocity with change in position of fluid particle in a fluid flow.  u.∂u u.∂u u.∂v  + +   ∂y ∂z   ∂x

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4.6.5 FLUID ROTATION

ε xx=

∂u ∂v ∂w , ε yy= , εzz= ∂x ∂y ∂z

4.6.7 ANGULAR DEFORMATION/SHEAR STRAIN Rotation of fluid particle is a vector quantity given by

ω = ωx ˆi + ωy ˆj + ωz kˆ

Where,

Rate of shear strain is given by 1  ∂u ∂v  ε= +   xy 2  ∂y ∂x  4.7 FLOW PATTERNS

In vector notation

Fluid Mechanics is a subject with visualisations. Patterns of flow can be visualized in several ways. Basic types of line patterns used to visualize flow are streamline, path line, streak line and time line.

ωx is rotation about x axis,

ωyis rotation about y axis & ωz is rotation about z axis

 1 ω= (∇ × V) 2

Rotational Components are given by 1  ∂v ∂u  = ωz  −  2  ∂x ∂y  = ωy

= ωx

1  ∂u ∂w   −  2  ∂z ∂x 

1  ∂w ∂v  −   2  ∂y ∂z 

For irrotational fluid flow,

ωx =ωy =ωz =0

4.6.6 LINEAR STRAIN RATE It is defined as the rate of increase in length per unit length. Mathematically, the linear strain rate of a fluid element depends on the initial orientation or direction of the line segment upon which we measure the linear strain

Linear strain is given by

4.7.1 STREAM LINE

Stream lines are lines drawn in the flow field so that the tangent at any point gives the direction of velocity of particle at that point. Since the streamlines are tangent to the velocity vector at every point in the flow field, there can be no flow across a streamline. Differential equation of streamline ds =V dt dx dy dz dt = = = u v w dx dy dz = = u v w 4.7.11 STREAM TUBE

A bundle of neighboring streamlines may be imagined to form a passage through which the fluid flows. This passage is known as a stream-tube.

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4.8 STREAM FUNCTION ( Ψ )

Properties of Stream tube: 1) The stream-tube is bounded on all sides by streamlines. 2) Fluid velocity does not exist across a streamline; no fluid may enter or leave a stream-tube except through its ends. 3) The entire flow in a flow field may be imagined to be composed of flows through stream-tubes arranged in some arbitrary positions 4.7.2 PATH LINE

A path line means the path or line actually described by a single fluid particle as it moves during a period of time. The path line indicates direction of the velocity of the same fluid particle at successive instant of time. 4.7.3 STREAK LINE

Let at any instant these particles arrive at points Q, R and S.Q, R and S represent the end points of the trajectories of these three particles at the instant. The curve joining the points S, R, Q and the fixed point P will define the streak line at that instant. The fixed point P will also lie on the line, since at any instant; there will be always a particle of some identity at that point. In the steady flow, the velocity at each point in the flow field remains constant with time and, consequently, the streamlines do not vary from one instant to the next. This implies that a particle located on a given streamline will remain on the same streamline. Furthermore, consecutive particles passing through a fixed point in space will be in the same streamline and subsequently, will remain on this streamline. Thus in a steady flow, path line, streak lines, and streamlines are identical lines in the flow field

It is defined as scalar function of space & time, such that its partial derivative w.r.t. to any direction gives the velocity perpendicular to that direction. It is defined for two dimensional flows. ∂ψ ∂ψ , V= − ∂y ∂x

U=

Properties of stream function (Ψ ) : 1) If stream function exists, it is possible case of continuous, incompressible steady flow. 2) If stream function (Ψ ) satisfies Laplace Equation i.e., ∂ 2ψ ∂ 2ψ + = 0 ∂y 2 ∂x 2

then, it is possible case of irrotational flow. 4.8.1 CONSTANT STREAM FUNCTION

Ψ = const dΨ = 0

= dΨ

∂Ψ ∂Ψ dx + dy ∂x ∂y

= 0 vdx − udy dy v = dx u

4.9 VELOCITY POTENTIAL FUNCTION It is defined as a scalar function of space & time, such that its negative derivative with respect to any direction gives the fluid velocity in that direction. φ = f ( x, y, z ) for steady flow. u=-

dϕ dϕ dϕ , v= , w= dx dy dz

Properties of Potential Function: 1. If velocity potential ( φ ) exists, the flow should be irrotational.

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2. If ( φ ) satisfy the Laplace equation, the flow is continuous, incompressible, and steady. ∂ 2φ ∂ 2φ ∂ 2φ + + = 0 ∂x 2 ∂y 2 ∂z2

4.10 EQUIPOTENTIAL LINE

It is defined as line along which the velocity potential is constant. φ = const

d φ=0

( )

φ = f x,y

= dφ

for steady, 2D flow

dφ dφ dx + dy dx dy

= 0 udx + vdy

dy u =− dx v

4.10.1 RELATION B/W STREAM FUNCTION & VELOCITY POTENTIAL FUNCTION ∂ψ ∂ϕ u= =− ∂x ∂y ∂ψ ∂ϕ = v = ∂y ∂x Hence,

∂φ ∂Ψ = − ∂x ∂y

∂ϕ ∂ψ = ∂y ∂x

Stream line and equipotential line are orthogonal to each other  dy   dy  = −1 .     dx equipotential  dx stream line

Example: The following cases represents the two velocity components, determine the third component of velocity such that they satisfy the continuity equation: 2 2 2 2 2 i) u = x + y + z , v = xy − yz + xy

ii) v = 2 y 2 , w = 2 xyz Solution: The continuity equation for incompressible fluid is given by equation as δu δv δw + + =0 δx δy δz Case 1: u = x 2 + y 2 + z 2    δu ∴ = 2x δx v = xy 2 − yz 2 + xy δv ∴ = 2 xy – z 2 + x δy δv δu Substituting the values of and in δy δx continuity equation 2x + 2xy – z 2 + x +

δw = 0 δz

Or δw = −3x – 2xy + z 2 or δz

δw = ( −3x – 2xy + z 2 ) δz Integration of both sides gives,

∫δw = ∫ ( −3x – 2xy + z ) δz 2

z3 ) +C 3 Where constant of integration cannot be function of z, but can be a function of x and y that is f(x,y). 3  z  w =  −3 xz – 2 xyz+  + f ( x, y ) 3  w = (−3xz – 2xyz +

Case 2:

δv = 4y δy δw w = 2xyz ∴ = 2 xy δz

v = 2y 2 ∴

Substituting the values of

continuity equation, we get δu + 4y + 2xy = 0 δx or

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δv δw and in δy δz

δu = −4y − 2xy δx

Substituting the values x =2, y = 1 and z = 3 in velocity field, we get 2  2  2  V = x yi + y zj − (2xyz + yz )k

or

δu = ( −4y − 2xy ) δx

= (22 × 1)i+ (12 × 3)j− (2 × 2 × 1 × 3 + 1 × 32 )k

Integrating we get, x2 u= −4 xy − 2 y + f ( y, z ) 2 2 = −4 xy − x y+ f(y, z)

− 21k  = 4i + 3j

And resultant velocity = =

Example: A fluid flow field is given by

− (2xyz + yz2 )k  V = x yi + y 2zj 2

Prove that it is a case of possible steady incompressible fluid flow. Calculate the velocity and acceleration at the point (2,1,3). Solution: For the given fluid flow field ∂u u = x2y ∴ = 2xy ∂x ∂v ∴ = 2yz v = y 2z ∂y

42 + 32 + ( −21)2

16 + 9 + 441 =

466 = 21.587 units

Acceleration (2, 1, 3) The acceleration

a x ,a y ,and a z

for steady flow are ∂u ∂u ∂u ax = u + v + w ∂x ∂y ∂z ay = u

az = u

components

∂v ∂v ∂v +v +w ∂x ∂y ∂z

∂w ∂w ∂w +v +w ∂x ∂y ∂z

u = x2y ∂ w ∂u ∂u 2 ∂u = −2xy − 2yz w = −2xyz − yz2 ∴ = 2xy, = x= , 0 ∂z ∂x ∂y ∂z For a case of possible steady v = y 2z incompressible fluid flow, the continuity equation should be satisfied ∂v ∂v ∂v = 0,= 2yz,= y 2 ∂u ∂v ∂w ∂x ∂y ∂z + + = 0 ∂x ∂y ∂z w = −2xyz − yz2 , Substituting the value of we get, δu δv δw + + δx δy δz

∂u ∂v ∂w , and , ∂x ∂y ∂z

= 2 xy + 2 yz − 2 xy − 2 yz = 0 Hence,

the velocity V = x yiˆ + y zjˆ − 2xyz + yz2 kˆ 2

2

(

)

is a possible case of fluid flow Velocity at (2, 1, 3)

field

∂w ∂w = −2xyz, = −2xz − z2 , ∂x ∂y

∂w = −2xy − 2yz ∂z Substituting these values in acceleration components, we get acceleration at (2, 1, 3)

a= x2y(2xy) + y 2z(x2 ) − (2xyz + yz2 )(0) x

= 2x3y 2 + x2y 2z

= 2(23 )(12 ) + (2)2 (1)2 x 3 = 2 × 8 + 12 = 16 + 12 = 28 units

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a y = x2y(0) + y 2z(2yz) − (2xyz + yz2 )(y 2 )

Example: The stream function for a two-dimensional flow is given by ψ =2xy . Calculate the velocity at the point P ( 2, 3) . Also, find the

= 2y 3z2 − 2xy 3z − y 3z2

= 18 − 12 − 9 = −3units

velocity potential function φ Solution: Given:

a z = x 2 y(−2yz) + y 2 z(−2xz − z 2 )

− (2xyz + yz 2 )(−2xy − 2yz) = −2x 2 y 2 z − 2xy 2 z 2 − y 2 z3 +[4x 2 y 2 z + 2xy 2 z 2 + 4xy 2 z 2 + 2y 2 z3 ]

ψ =2xy

The velocity components u and v in terms of ψ are

=− ( 2 × 22 × 12 × 3) − (2 × 2 × 12 × 32 ) − 12 × 33 + {(4 × 22 × 12 × 3) + (2 × 2 × 12 × 32 ) + (4 × 2 × 12 × 32 ) + (2 × 12 × 33 )}

= −24 − 36 − 27 + (48 + 36 + 72 + 54) = 123







 



∴ Acceleration= a x i + a y j + az k = 28i − 3j + 123k Or magnitude of resultant acceleration 2 282 + ( −3)2 + 123 =

= 126.18units

784 + 9 + 15129 =

15922

Example: The velocity potential function is given by

= φ 5(x2 − y 2 ) . Calculate the velocity component at the point ( 4, 5) . Solution:

= φ 5(x 2 − y 2 )

δφ = 10x δx δφ = −10y δy But the velocity components u and v are given by equation as δφ u = − = −10x δx δφ v =− =−( −10y) =10y δy The velocity components at the point (4, 5), i.e., at= x 4,y = 5 are u =−10 × 4 =−40units v = 10 × 5 = 50units

δψ δ u= − = − (2xy) = −2x δy δy δψ δ = v = (2xy) = 2y δx δx At the point P ( 2, 3) . , we get u =−2 × 2 =−4units

v = 2× 3 = 6 ∴ Resultant velocity @P =

u2 + v 2 =

42 + 62 =

52 = 7.21units / sec

Velocity potential function φ We know δφ =−u =−( −2x) =2x ..... ( i ) δx

δφ =−v =−2y ..... ( ii ) δy Integrating equation ( i ) , we get

∫ dφ =∫ 2xdx

2x 2 + C= x2 + C ..... ( iii ) 2 Where C is a constant which is independent of x but can be a function of y Differentiating equation (iii) w.r.t. ‘y’, we φ=

get

δφ δC = δy δy

But from ( ii ) ,

δφ = −2y δy

δC = −2y δy Integrating this equation, we get



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2y 2 C= − 2y dy = − = −y 2 ∫ 2 Substituting this value of C in equation ( iii ) , 2 2 we get φ= x − y

Example: In a two dimensional incompressible flow, the fluid velocity components are given by u = x − 4y and v = − y − 4x . Show that velocity potential exists and determine its form. Find also the stream function. Solution: Given: u =x − 4yandv =−y − 4x



∂u ∂v = 1 and = −1 ∂x ∂y

∂u ∂v + =1−1 = 0 ∂x ∂y Hence flow is continuous and velocity potential exists. Let φ =Velocity potential. Let velocity components in terms of velocity potential is given by δφ =−u =−(x − 4y) =−x + 4y....(i) δx δφ =−v =−(y − 4x) =y + 4x .......(ii) δy Integrating equation ( i ) , we get



φ=−

x2 + 4xy + C .....(iii) 2

Where C is constant of integration, and is independent of x. This constant can be a function of y. Differentiating the above equation, i.e., equation (iii) w.r.t. ‘y’, we get δφ δC =0 + 4x + δy δy But from equation ( iii ) , we have

δφ = y + 4x δy

Equating the two values of

δφ , we get δy

δC =y δy

Integrating the above equation, we get y2 = C + C1 2 Where C1 is a constant of integration, which is independent of x and y. y2 Taking it equal to zero, we get c = 2 Substituting the value of C in equation ( iii ) , we get x2 y2 φ = − + 4xy + 2 2

Value of Stream functions Let ψ =stream function The velocity components in terms of stream function are δψ =v =−y − 4x ..... ( iv ) δx and

δψ =−u =−(x − 4y) =−x + 4y ...... ( v ) δy Integrating equation ( iv ) w.r.t. x, we get

4x2 +k ....... ( vi ) 2 Where k is a constant of integration which is independent of x but can be function of y Differentiating equation ( vi ) w.r.t. y, we get ψ = −yx −

δψ δk =−x − 0 + δy δy But from equation

δψ =−x + 4y δy

(v) ,

Equating the two values of δk = 4y δy

we

δψ , we get δy

Integrating the above equation, we get

k =

4y 2 = 2y 2 2

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have

Substituting the values of k in equation ( vi ) , we get

ψ = − yx − 2x 2 + 2y 2

Example: A fluid flow is given by= V 8x3 $ i − 10x 2 yj$ Find the shear strain rate and state whether the flow is rotational or irrotational.

Solution: Given:  = V 8x3i − 10x 2yj

δu 3 2 δu = u 8x = , 24x = , 0 δx δy δv δv v= −10x2y, = −20xy, = −10x2 δx δy ( i ) Shear strain rate is given by equation as

1  δv δu  1 = +  =( −20xy + 0) = −10xy 2  δx δy  2 ( ii ) Rotation in x-y plane is given by

Also

v=xy 2 − 2y −

δv ∴= 2xy − 2 δy

x3 3

i) For a two-dimensional flow, continuity δu δv + = 0 equation is δx δy Substituting

the

ii) Rotation, ωz is given by

= ωz

As rotation ωz ≠ 0 . Hence flow is rotational.

u=

in

y3 x3 + 2x − x 2y and v=xy 2 − 2y − 3 3

a

twoare

Show that these components represent a possible case of an irrotational flow. Solution: Given u =

δu δv and , δx δy

1  δv δu  1 2 2 2 [(y − x ) − (y 2 − x= )] 0  − = 2  δx δy  2

1  δv δu  1 ωz =  −  = ( −20xy − 0) =−10xy 2  δx δy  2

components flow

of

we get δu δv + = 2 − 2xy + 2xy − 2 = 0 δx δy ∴ It is a possible case of a fluid flow.

equation or

Example: The velocity dimensional

value

y3 + 2x − x 2y 3

δu = 2 − 2xy δx δu 3y 2 = − x2 = y 2 − x2 δy 3

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GATE QUESTIONS Q.1

A fluid flow is represented by the �= axı̂+ ayȷ̂, where a is velocity field V constant. The equation of stream line passing through a point (1, 2) is a) x − 2y = 0 b) 2x + y = 0 c) 2x − y = 0 d) x + 2y = 0 [GATE–2004]

Q.2

The velocity components in the x and y directions of a two dimensional potential flow are u and ∂u v , respectively. Then is equal to ∂x ∂v ∂v a) b) − ∂x ∂x ∂v ∂v c) d) − ∂y ∂y [GATE–2005]

Q.3

A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of 120 m from the centre of the whirlpool. The whirlpool can be described by the following velocity  60 ×103  distribution: Vr = −  m / s  2πr 

300 ×103 m / s where r (in 2πr metres) is the distance from the centre of the whirlpool. What will be the distance of the leaf from the centre when it has moved through half a revolution ? a) 48 m b) 64 m c) 120 m d) 142 m [GATE–2005]

Q.5

A two-dimensional flow field has velocities along the x and y directions given by u = x2t and v =

In a two-dimensional velocity field with velocities u and v along the x and y directions respectively, the convective acceleration along the x direction is given by ∂u ∂v ∂v ∂u a) u + v b) u + v ∂x ∂y ∂x ∂y ∂u ∂u ∂u ∂u c) u + v d) v + u ∂x ∂y ∂x ∂y [GATE–2006]

Q.6

In a steady flow through a nozzle, the flow velocity on the nozzle axis is given by v = u0(1+ 3x/L)𝚤𝚤⃗, where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is L L a) b) ln4 3u o uo L L c) d) 4u o 2.5u o [GATE–2007]

Q.7

Which combination of the following statements about steady incompressible forced vortex flow is correct ? P: Shear stress is zero at all points in the flow. Q: Vorticity is zero at all points in the flow.

and Vθ =

Q.4

−2xyt respectively, where tis time. The equation of stream line is a) x2y= constant b) xy2= constant c) xy= constant d) not possible to determine [GATE–2006]

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R: Velocity is directly proportional to the radius from the center of the vortex. S: Total mechanical energy per unit mass is constant in the entire flow field. a) P and Q b) R and S c) P and R d) P and S [GATE–2007] Q.8

For the continuity equation given �⃗ is by ∇. �V⃗ = 0 to be valid, whereV the velocity vector, which one of the following is a necessary condition? a) steady flow b) irrotational flow c) inviscid flow d) incompressible flow [GATE–2008]

Common Data For Q.9and Q.10 The gap between a moving circular plate and a stationary surface is being continuously reduced, as the circular plate comes down at a uniform speed V towards the stationary bottom surface, as shown in the figure. In the process, the fluid contained between the two plates flows out racially. The fluid is assumed to be incompressible and inviscid.

Q.9

The radial velocity Vr at any radius r, when the gap width is h, is Vr Vr a) Vr = b) Vr = 2h h 2Vh Vh c) Vr = d) Vr = r r [GATE–2008]

Q.10 The radial component of the fluid acceleration at r = R is

3V 2 R 4h 2 V2R c) 2h 2

V2R 4h 2 V2h d) 2R 2 [GATE–2008] Q.11 You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application. The following three flow choices, expressed in terms of the two dimensional velocity fields in the xy -plane, are made available P: u= 2y,v=-3x Q: u= 3xy, v=0 R: u=−2x, v=2y Which flow(s) should be recommended when the application requires the flow to be incompressible and irrotational ? a) P and R b) Q c) Q and R d) R [GATE–2009] a)

b)

Q.12 Velocity vector of a flow field is �⃗ = 2xy𝐢𝐢⃗ − x2z�⃗. given as V 𝐣𝐣 The vorticity vector at (1, 1, 1) is a) 4ı⃗ − ⃗ȷ b) 4ı⃗ − �⃗ k �⃗ c) ⃗ı − 4ȷ�⃗ d) ⃗ı − 4k [GATE–2010] Q.13 A streamline and an equipotential line in a flow field a) are parallel to each other b) are perpendicular to each other c) intersect at an acute angle d) are identical [GATE–2011] Q.14 For an incompressible flow field, V, which one of the following conditions must be satisfied? r a) ∇.V = 0 r b) ∇ × V = 0 r r c) (V.∇)V = 0 r r ∂V r d) + (V.∇)V = 0 ∂t [GATE–2014 (2)]

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Q.15 A flow field which has only convective acceleration is a) a steady uniform flow b) an unsteady uniform flow c) a steady non-uniform flow d) an unsteady non-uniform flow [GATE–2014 (4)] Q.16 Consider the following statements regarding streamline(s): i) It is a continuous line such that the tangent at any point on it shows the velocity vector at that point ii) There is no flow across streamlines dx dy dz iii) = = is the differential u v w equation of a streamline, where u, v and w are velocities in directions x, y and z, respectively iv) In an unsteady flow, the path of a particle is a streamline Which one of the following combinations of the statements is true? a) (i), (ii), (iv) b) (ii), (iii), (iv) c) (i), (iii), (iv) d) (i), (ii), (iii) [GATE–2014 (4)]

�⃗ = K(yı̂ + Q.17 Consider a velocity field V � xk). where K is a constant. The vorticity, Ωz ,is a) − K b) K c) − K/2 d) K/2 [GATE–2014 (4)] Q.18 If the fluid velocity for a potential flow is given by V(x, y) = u(x, y)ı�+ v(x, y)ȷ̂ with usual notation then the slope of the potential line at (x, y) is a) v/u b) —u/v c) v2/u2 d) u/v [GATE–2015 (2)] Q.19 Match the following pairs:

a) P – IV, Q – I, R – II, S – III b) P – IV, Q – III, R – I, S – II c) P – III, Q – I, R – IV, S – II d) P – III, Q – I, R – II,S-IV [GATE–2015 (1)] Q.20 The velocity field of an incompressible flow is given by V = (a1x + a2y + a3z)ı̂ + (b1x + b2y + b3z)ȷ̂ + (c1x + c2y + c 3z)k� , where a1= 2 and c3 = − 4. The value of b2 is ______. [GATE–2015 (1)] Q.21 The volumetric flow rate (per unit depth) between two streamlines having stream functions Ψ1 and Ψ2 is a) |Ψ1 + Ψ2 | b)Ψ1 Ψ2 c) Ψ1 /Ψ2 d) |Ψ1 − Ψ2 | [GATE–2016 (2)] Q.22 For a certain two-dimensional incompressible flow, velocity field is given by 2xy �ı − y 2 ȷ̂ . The streamlines for this flow are given by the family of curves a) 𝑥𝑥2𝑦𝑦2 = constant b) 𝑥𝑥𝑥𝑥2 = constant c) 2𝑥𝑥𝑥𝑥 − 𝑦𝑦2 = constant d) 𝑥𝑥𝑥𝑥 = constant [GATE–2016 (3)] Q.23 For a two-dimensional flow, the velocity field is x $ y $ r = u i+ 2 j , where 𝚤𝚤̂ and x 2 + y2 x + y2 𝚥𝚥̂ are the basis vectors in the x-y Cartesian coordinate system. Identify the CORRECT statements from below.

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1) The flow is incompressible. 2) The flow is unsteady. 3) y-component of acceleration, −y ay = 2 ( x 2 + y2 )

drainage of the fluid between the plate and the wall as shown in the figure. Assume two-dimensional incompressible flow and that the plate remains parallel to the wall. The average velocity, uavg of the fluid (in m/s) draining out at the instant shown in the figure is _____________(correct to three decimal places).

4) x-component of acceleration, −(x + y) ax = 2 ( x 2 + y2 ) a) (2) and (3) c) (1) and (2)

b) (1) and (3) d) (3) and (4) [GATE–2016, Set (3)]

Q.24 Consider the two dimensional velocity field given by  V = ( 5+a1 x+b1 y ) ˆi+ ( 4+a 2 x+b 2 y ) ˆj, where a1 , b1 , a 2 , b 2 are constants. Which one of the following conditions needs to be satisfied for the flow to be incompressible? a) a1 +b1 = 0 b) a1 +b 2 = 0 c) a 2 +b 2 = 0

d) a 2 +b1 = 0

[GATE–2017, Set (1)] Q.25 For a steady flow, the velocity field is  V = ( − x 2 +3y ) ˆi+ ( 2xy ) ˆj The magnitude of the acceleration of a particle at (1, -1) is a) 2 b) 1 c) 2 5

d) 0

[GATE–2017, Set (1)] Q.26For a two-dimensional incompressible r flow field given by = u A x$ i − x$j

(

)

, where A > 0 which one of the following statements is FALSE? a) It satisfies continuity equation. b) It is unidirectional when x → 0 and y → ∞ c) Its streamlines are given by x = y . d) It is irrotational [GATE–2018, Set (1)] Q.27 A flat plate of width L = 1 m is pushed down with a velocity U = 0.01 m/s towards a wall resulting in the

[GATE–2018, Set (1)]

Q.28 In a Lagrangian system, the position of a fluid particle in a flow is described as x = x0 e − kt and y = y0 e kt where t is the time while 𝑥𝑥𝑜𝑜, 𝑦𝑦𝑜𝑜, and k are constants. The flow is a) unsteady and one-dimensional b) steady and two-dimensional c) steady and one-dimensional d) unsteady and two-dimensional [GATE–2018, Set (1)] Q.29A sprinkler shown in the figure rotates about its hinge point in a horizontal plane due to water flow discharged through its two exit nozzles.

The total flow rate Q through the sprinkler is 1 litre/sec and the crosssectional area of each exit nozzle is 1 cm2. Assuming equal flow rate through both arms and a frictionless hinge, the steady state angular speed of rotation (in rad/s) of the sprinkler is ______ (correct to two decimal places). [GATE–2018, Set (1)]

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ANSWER KEY: 1 (c) 15 (c) 29 (c)

2 (d) 16 (d)

3 (b) 17 (a)

4 (a) 18 (b)

5 (c) 19 (c)

6 (b) 20 2

7 (b) 21 (d)

8 (d) 22 (b)

9 (a) 23 (b)

10 (a) 24 (b)

11 (d) 25 (c)

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12 (d) 26 (b)

13 (b) 27 0.05

14 (a) 28 b

EXPLANATIONS Q.1

(c) �= axı̂+ ayȷ̂ ....(i) Given V The equation of stream line is. dx dy dy = = ....(ii) ux uy uz

Q.4

Vr 1 = − Vθ 5 V ⇒ Vr = − θ 5 dr ⇒ Vr = dt dθ  dθ  ⇒ V= r Q V= rω = r  θ θ dt  dt  dr r dθ ⇒ = − dt 5 dt dr dθ ⇒ = − r 5 Integrating both sides, we get r π dr 1 = − ∫ r 5 ∫0dθ 120 π r ⇒ ln r 120 = − 5 r π ⇒ ln = − 120 5 r ⇒ = e − π/5 120 ∴ r=64m

Q.5

(c)



From equation (i) Ux = ax ,Uy = ay and Uz=0 Substitute there values in equation (ii), we get dx dy dy dy = ⇒ = ax ay x y Integrating both sides, we get dx dy ∫ x =∫ x log x = log y + log c = log yc ⇒x = yc.... (iii) 1 At point (1, 2) 1 = 2c ⇒ C = 2 From equation (iii) y x = ⇒ 2x − y = 0 2 Q.2

Q.3

(d) For two dimensional flow, continuity equation should be satisfied. ∂u ∂v + = 0 i.e. ∂x ∂y ∂u ∂v = − ∴ ∂x ∂y (b)  300 ×103  Vr 2πr = − × 3 Vθ  2πr  (60 ×10 )

(a) Equation of streamline, dx dy = u v dx dy = xy −2xy dx dy = x −2y dx dy 2 + = 0 x y 2 ln x + ln y = ln c 2 ln yx = ln c x2 y = c

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Q.6

Q.7

Convective Acceleration is defined as the rate of change of velocity due to the change of position of fluid particles in a fluid flow. In Cartesian coordinates, the components of the acceleration vector along the x direction is given by. ∂u ∂u ∂u ∂u ax = u + v + w + ∂x ∂y ∂z ∂t In above equation term ∂u/∂t is known as local acceleration and terms other than this, called convective acceleration. Hence for given flow. Convective acceleration along x-direction. ∂u ∂u ∂u = a x u + v (neglecting w ∂x ∂y ∂z because it is a two dimensional flow) (b) Given dx  3x  u = u 0 1 +  = x ( L + 3x ) dt L L  L 1 dt = × dx u 0 ( L + 3X )

ω2 2 2 ( r2 − r1 ) −ρg ( Z2 − Z1 ) = 0 2 Δ K.E − ΔP.E = 0 Δ K.E = ΔP.E So, total mechanical energy per unit mass is constant. ρ

Q.8

(d) The continuity equation is given by r ∂ρ ∇. ρV + =0 ∂t Now if ρ remains constant then only we can write the above equation as �⃗ = 0 ∇. V So, the above equation represents the incompressible flow

( )

Q.9

(a) Let us consider for small time dt, then for any cylindrical control volume between 2 plates of radius ‘r’ volume reduced by the movement of upper plate in time ‘dt’ is equal to the volume of fluid moving out ∴ ( V × dt ) × πr 2 = (Vr × dt) × 2πrh ∴ Vr =

V.r 2h

On integration both the sides within Q.10 (a) limits ⇒ 0 to t and x ⇒0 to L we get Radial acceleration L L L 1 dt = dx ∂V ∂V ∫0 u 0 ∫0 ( L + 3X ) = a r Vr . r + r ∂r ∂t L L V.r ∂ V.r   ∂  V.r  t= [ln(1 + 3x / L)]0 = ar .  +   3u o 2h ∂r  2h  ∂t  2h  L For convective part r is variable and t= ln4 3u o for local acceleration h is variable V.r V V V.r  −1 ∂h  ∴ ar = . . + .  (b) 2h 2h 2h 2  h 2 ∂t  For forced Vortex flow the relation ∂h Also, − = V is given by, V = rω ...(i) ∂t From equation (i) it is shown easily V 2 .r V.r  V  that velocity is directly proportional ar + .  ∴= 4h 2 2  h2  to the radius from the centre of the vortex. 3V 2 .r a = r (Radius of fluid particle from the 4h 2 axis of rotation) At r=R And also for forced vortex flow,

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ar =

3V 2 .R 4h 2

Q.11 (d) Incompressible flow satisfy continuity equation ∂u ∂v + = 0 ∂x ∂y P. u=2y, v= −3x ∂ (2y) ∂ (−3x) + ∂x ∂y Incompressible flow ⇒ 0+0=0 Q. u=3xy, v=0 ∂ (3xy) ∂ (0) + ∂x ∂y Compressible flow ⇒ 3y+0 ≠0 R. u=−2x,v= 2y ∂ (−2x) ∂ (2y) + ∂x ∂y Incompressible flow −2+2=0 For irrotational flow condition is 1  ∂v ∂u  wz =  − = 0 2  ∂x ∂y  For P : u=2y, v=−3x 1  ∂ (−3x) ∂ (2y)  = wz − 2  ∂x ∂y  1 = [−3 − 2] ≠ 0 2 (Rotational flow) For Q: u=3xy,v=0 1  ∂ (0) ∂ (3xy)  = wz − 2  ∂x ∂y  1 = [0 − 3x] ≠ 0 2 (Rotational flow) For R: u = −2x, v=2y 1  ∂ (2y) ∂ (−2x)  = wz − 2  ∂x ∂y  1 = [0 − 0]= 0 2 (irrotational flow)

Q.12 (d) GivenV = 2xyı⃗ - x2z�⃗ȷ  i j  ∂ ∂ Vorticity =   dx dy  v u

k  ∂ dz   w

Substitute u = 2xy, v = − x 2 z, w = 0

 i j  ∂ ∂ So Vorticity=   ∂x ∂y  2  2xy − x z =x 2⃗ı − 0 + �⃗ k(−2xz−2x) Vorticity vector at P(1,1,1) = ⃗ı + �⃗ k [- 2 – 2] = ⃗ı ̶ 4 �⃗ k

k  ∂ ∂z   0

Q.13 (b)

Q.14 (a) For an incompressible flow field, r divergence must be zero i.e. ∇.V = 0 Q.15 (c) Total r acceleration, r r DV ∂V r = + (V.∇)V Dt ∂t r r (V.∇)V → Convective acceleration r ∂V → Local acceleration ∂t In steady non-uniform flow, conditions change from point to point in the stream but do not change with time. Q.16 (d)

Q.17 (a) Vorticity about z-axis: ∂v ∂u Ω= − z ∂x ∂y ∂v ∂u = 0,= K ∂x ∂y ∴ Ω z =− 0 K= −K Q.18 (b)

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Here V(x, y) = u(x, y)ı̂+ v(x, y)ȷ̂ As we know that u =

− ∂ϕ ∂x

∂∅ ∂∅ dx + = dy 0 ∂x ∂y ∴−udx − vdy = 0 dy u = − dx v

= d∅

& v=

−∂φ ∂y

Q.19 (c) Q.20 (1.9 to 2.1) ∂u ∂v ∂w + + = 0 ∂x ∂y ∂z a1 + b 2 + c3 = 0 2 − 4 + b2 = 0 b2 = 2 Q.21 (d) Q.22 (b) Equation of the streamline is given by dx dy = u v dx dy = 2 2xy − y dx 2dy = x −y ⇒ lnx = −2lny + lnc ln x + lny 2 = ln c 2 ∴ xy = c

Q.23 (b) For the flow to be incompressible ∂u ∂v + = 0 [Continuity equation] ∂x ∂y ∂u ∂v ∂  x  ∂  y  ∴= + + ∂x ∂y ∂x  x 2 + y 2  ∂y  x 2 + y 2 

(x =

2

+ y 2 ) − x(2x)

(x

2

+ y2 )

2

(x +

2

+ y 2 ) − y(2y)

(x

2

+ y2 )

2

=

y2 − x 2 x 2 − y2 + x 2 + y 2 ( x 2 + y 2 )2

∂u ∂v + = 0 ∂x ∂y ∴ (1) statement is correct r ∂u Also, =0 ∂t ∴ Flow is steady So the correct answer is (b) Q.24 (b) ∴

For flow to be incompressible ∂u ∂v + = 0...........(1) ∂x ∂y Here u=5 + a1x +b1 y & v=4 + a2 x +b2 y ∂u ∂v = ∴ a1 = & b2 ∂x ∂y put value in equation (1) a1 + b 2 = 0

Q.25 (c) Given u = − x 2 + 3y ∂u = −2x ∂x ∂u =3 ∂y v = 2xy ∂v = 2y ∂x ∂v = 2x ∂y ∂u ∂u = ax u +v ∂x ∂y 2 a x =− ( x + 3y)(−2x) + 2xy(3) a x = (−1 + 3)(−2) + 2(3) ax = 8 − 6 = 2 ∂v ∂v = ay u + v ∂x ∂y

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=− ( x 2 + 3y)(2y) + 2xy(2x) = (−1 − 3)(−2) − 2(2) =+8 − 4 =4

at =

a 2x + a 2y =

22 + 42

at = 2 5

1. Here flow is two dimensional because dx dy ≠ ≠0 dt dt dx = u= x0  ( −k ) e − kt dt

= − x0 ke − kt = −kx

Q.26(c) C is the false statement

dy = v= = ky0 e kt ky dt As, both u and v are not depends on time so, 2D incompressible flow continuity equation. it is steady flow ∂u ∂v + = 0 Q.29(10) ∂x ∂y

∂ ( Ax ) ∂ ( − Ay ) + = 0 ∂x ∂y A − A = 0 it satisfies continuity equation. ur ⇒ As V = Axi$− Ayi$

A →∞ dx dy = u v dx dy = Ax − Ay ln x = − ln y + ln c

litre/sec, A = 1 cm 2 ⇒ First find the velocity of flow Q = A1 V1 2 110−3 1    )  V1 ⇒ = ( 2 5m/s ⇒ V1 = If the angular velocity of the sprinkler is ω then the absolute velocities of flow through the nozzle are

ln xy = ln c xy= c → streamline equation Q.27 Volume swept by the plate per unit time = Discharge going out or drainage from two sides ∴ U  (L W ) = 2  uavg  d  W (Assuming width of the plate (W) to be very long, the discharge perpendicular to the plane can be neglected.) UL 0.01  1 = = 0.050 m/s u= avg 2d 2  0.1

Q.28(b) x = x0 e − kt

V1ab =5 + 0.1 ω V2ab =5 − 0.2 ω

Since the moment of momentum of flow entering is zero and there is no friction the momentum leaving the sprinkler must also be zero.

( ρQ) 

0 ( 5 + 0.1ω) 0.1 − ( 5 − 0.2ω) 0.2 = 2  (1) 0.5 + 0.01ω − 1 + 0.04ω = 0 ⇒ [ ] 2 ⇒ − 0.5 + 0.05ω = 0 ω =10 rad / sec ⇒

y = y0 e kt

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5

BERNOULLI’S EQUATION & ITS APPLICATIONS

5.1 INTRODUCTION Consider a small element of fluid in flow field. The energy in the element as it moves in the flow field is conserved. This principle of conservation of energy is used in the determination of flow parameters like pressure, velocity and potential energy at various locations in a flow. The concept is used in the analysis of flow of ideal as well as real fluids. Energy can neither be created nor destroyed. It is possible that one form of energy is converted to another form.



P V2 const + + zg = ρ 2 Multiplying the Bernoulli’s equation by the density (ρ). Each term in this equation has units of pressure. ρV 2 P+ + ρzg =const 2



5.2 EULER’S EQUATION

Euler’s equation is obtained from the conservation of momentum for a fluid particle moving along a streamline. The forces due to gravity & pressure are taken into consideration. dP + VdV + gdz = 0 ρ This equation is known as Euler’s equation of motion. The assumptions involved are: 1) Steady flow 2) Motion along a stream line 3) Ideal fluid (frictionless) In the case of incompressible flow, this equation can be integrated to obtain Bernoulli Equation. 5.3 BERNOULLI’S EQUATION

The Bernoulli’s equation states that the sum of the flow, kinetic, and potential energies of a fluid particle along a streamline is constant. Therefore, the kinetic and potential energies of the fluid can be converted to flow energy (and vice versa) during flow. By integrating Euler’s equation for incompressible flow

dP + VdV + ∫ gdz = ∫0 ρ ∫





P is the static pressure (it does not incorporate any dynamic effects); it represents the actual thermodynamic pressure of the fluid. This is the same as the pressure used in thermodynamics and property tables. P is known as pressure head ρg

ρV 2 2

is the dynamic pressure; it

represents the pressure rise when the fluid in motion is brought to stop isentropically. ρV 2 is known as dynamic head 2

ρgz

is the hydrostatic pressure, which is not pressure in a real sense since its value depends on the reference level selected. It accounts for the elevation effects, i.e., of fluid weight on pressure. Z is gradient or datum head. The sum of the static, dynamic, and hydrostatic pressures is called the total pressure. Therefore, the Bernoulli’s equation states that the total pressure along a streamline is constant. The sum

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of the static and dynamic pressures is called the stagnation pressure, and it is expressed as Stagnation pressure = static pressure + dynamic pressure. ρv 2 Stagnation pressure= P + 2 5.4 APLLICATION EQUATION

OF

BERNOULLI’S

Venturimeter and Orifice meters are the obstruction type meters commonly used for the measurement of flow through pipes. In each case the meter acts as an obstacle placed in the path of the flowing fluid causing local changes in pressure and velocity i) Venturimeter ii) Orifice plate iii) Pitot tube 5.4.1 VENTURIMETER

A venturimeter consist of a short converging part, throat & diverging part. The liquid undergoes gradual contraction & expansion, therefore it has lesser losses. By applying Bernoulli’s equations at 1 & 2

P1 V12 P2 V22 + + z1 = + + z2 ρg 2g ρg 2g

Points 1 & 2 are at same horizontal level,

∴ z1 = z 2

P P  V 2 − V12 h= 1 − 2 + 2 2g  ρg ρg  = Q a= v a2v 2 1 1

v2 =

2gh

a  1− 2   a1 

2

2gh

Q= a= v a2 theo 2 2

2

a  1− 2   a1  In case of ideal fluid, the above equation is valid. To modify the equation for real fluid, coefficient of discharge (Cd) is multiplied to theoretical flow. It accounts for viscous loss, expansion loss & boundary roughness. It is defined as actual discharge cd = theoretical discharge

Q act c= a v cd a2 = d 2 2

2gh

a  1− 2   a1 

2

 PA − PB  ρm =  − 1 x ρg  ρ  Where, x = the difference in mercury level ρm = Density of heavy liquid ρ = Density of flowing fluid Case1: Manometric fluid is lighter then liquid flowing in pipe (in case of inverted manometer)  ρ  = h x1 − m  ρ   Case2: Inclined/vertical venturimeter with differential U-tube Manometer  P1  P   ρm  − 1 x h =+ z1  −  2 + z 2  =    ρg   ρg   ρ 

= h

5.4.2 ORIFICE METER OR ORIFICE PLATE

It is a device used for measuring the rate of flow of a fluid through a pipe. It is a cheaper device as compared to venturimeter. The orifice diameter is kept generally 0.35 times the diameter of pipe. Due to sudden expansion & contraction, loss is high in orificemeter.

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Due to the viscous effects, the actual flow velocity through the orifice will always be less than the theoretical possible velocity. The velocity coefficient Cv is defined as A differential Manometer is connected to measure the pressure difference at section (1), which is at distance of about 1.5-2.0 times the pipe diameter and section (2), which is at a distance of about half diameter of the orifice on the downstream side. By applying Bernoulli’s equations at 1 & 2

P1 V12 P2 V22 + + z1 = + + z2 ρg 2g ρg 2g

Points 1 & 2 are at same horizontal level, ∴ z1 = z2

 P1 P2  V22 − V12 − h= + 2g  ρg ρg 

= Q a= v a2v 2 1 1

As the liquid comes out of orifice it contracts further and the area just outside the orifice is lower compared to the area of the orifice. This section is called as venacontracta. Area of jet at the vena-contracta is less than the area of the orifice itself due to convergence of stream lines. The coefficient of contraction Cc is defined as follows. area of jet at vena contracta(a2 ) cc = area of orifice(a o ) The value of coefficient of contraction varies from 0.61 to 0.69 depending on the shape and size of the orifice

a2 = cc .a0

follows.

Cv =

Actual of jet at venna contacta Therotcial veocity of jet at orifice

The value of coefficient of contraction varies from 0.61 to 0.69 depending on the shape and size of the orifice.

2gh

Vact = c v 2

a  1− 0   a1 

Q act a= v c c a 0c v = 2 act 2

Q act = cda0 Where, Cd = Cc C v

2gh

a  1− 0   a1 

2

2gh

a  1− 0   a1 

2

2

Average value of Cd for orifices is 0.62. 5.4.3 PITOT TUBE

It is a device used for measuring flow velocity at any point in a pipe. Principle: If velocity of flow at any point becomes zero, the pressure is increased due to conversion of kinetic energy to pressure energy.

Q a= v a= v cc .a0 v 2 = 1 1 2 2 v2 =

2gh

2

a  1 −  2  cc2  a1 

By applying Bernoulli’s equations at 1 & 2 Stagnation pressure = static pressure + Dynamic pressure

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= R Ucos θgt 2Ucos θgUsin θ = g

P1 V12 P2 + = ρg 2g ρg

V12 P2 P1 = − =h 2g ρg ρg

=

Vact = c v 2gh

 P2 − P1  ρm =  − 1 x ρg  ρ  Where, Cv is velocity coefficient

= h

5.5 BERNOULIS EQUATION FOR REAL FLUIDS 2 1

2 2

P1 V P V + + z1 = 2 + + z2 + h f ρg 2g ρg 2g

Where, hf is head loss due to viscous force & minor losses. 5.6 FREE LIQUID JETS

U2 sin2θ g

4. Value of 𝜃𝜃 for max range U2 sin2θ R= g sin2θ = 1

2θ =90o Or θ =45o

Example: Water is flowing through a pipe of 5cm diameter under a pressure of 2 29.43N / cm (gauge) and with mean

velocity of 2m / s . Find the total head or total energy per unit weight of the water at a cross–section, which is 5m above the datum line.

Solution: Free liquid jet is defined as the jet of water Given: coming out from the nozzle in atmosphere. Diameter of pipe= 5 cm = 0.05m The path travelled by the free jet is Pressure, parabolic, given by equation of parabola = P 29.43N /= cm2 29.43 × 104 N / m2 1 y x tan θ − gx 2 sec2 θ = Velocity, V = 2.0m / s 2 1. Max height Datum head, Z = 5m 2 2 Total head 2gy max ( Usin θ) − (0) = =Pressure head+kinetic head+datum head U2 sin2 θ P 29.43 × 104 y max = Pressure head = = = 30m 2g ρg 1000 × 9.81 2.

Time of flight

1 y Usin θgt − gt 2 = 2 For full flight, y = 0 1 Usin θgt − gt 2 = 0 2 2Usin θ t= g 3. Max Distance (Range)

Kinetic head =

∴ Total head =

V2 2× 2 = = 0.204m 2g 2 × 9.81

P V2 + + z = 30 + 0.204 + 5 ρg 2g =35.204m

Example: An oil of S.G. 0.8 is flowing through a venturimeter having inlet diameter 20 cm and throat diameter 10cm. The oil– mercury differential manometer shows a

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reading of 25 cm. Calculate the discharge of oil through the horizontal venturimeter. Take Cd = 0.98 Solution: Given: S.G. Of oil, S0 = 0.8

S.G. Of mercury, Sh =13.6 Reading of differential manometer, x =25cm ∴ Difference of pressure head,

S   13.6  − 1  cm of oil h= x  h − 1=  25  0.8   So 

=25(17-1) =400 cm of oil. Dia at inlet, d1 = 20cm

π π ∴ a1 = × d12 = × 202 =314.16cm2 4 4 d2 =10cm

π ∴ a2 = × 102 = 78.54cm2 4 Cd = 0.98 ∴ The discharge Q is given by equation a 1a 2 = × 2gh Or Q cd 2 2 a1 −a2

0.98

314.16 × 78.54

(314.16)2 − (78.54)2

× 2 × 981 × 400

21421375.7 = 70422.4cm3 / s 98696 − 6168

= 70.422litres / sec

Dia at throat, d2 =15cm

∴ = a2

2 π = 15 176.7cm2 ( ) 4

S   13.6  − 1 h= x  h − 1=  20  S  1   o 

= 20 × 12.6 = 252cm of water Cd = 0.98 a 1a 2

Discharge, Q cd = = 0.98× =

a12 −a22

706.85×176.7

× 2gh

(706.85) − (176.7) 2

86067593.36

=

2

× 2×981×252

86067593.36 684.4

499636.3-31222.9 = 125756cm3 / s = 125.756lit/ s

Example: In a vertical pipe conveying oil of specific gravity 0.8, two pressure gauges have been installed at A and B where the diameters are 16cm and 8cm respectively. A is 2 meters above B. The pressure gauge readings have shown that the pressure at B is greater than at A by 0.981N/ cm2 . Neglecting all losses, calculate the flow rate. If the gauges at A and B are replaced by tubes filled with the same liquid and connected to a U-tube containing mercury, calculate the difference of level of mercury in the two limbs of the U-tube. Solution: Given: S.G. of oil, S0 = 0.8

Example: A 30cm ×15cm venturimeter is inserted in kg vertical pipe carrying water, flowing in the ∴ Density, ρ = 0.8×1000 = 800 3 upward direction. A differential mercury m manometer connected to the inlet and Dia. at = A, da 16cm = 0.16m throat gives a reading of 20cm. Find the ∴ Area at A, discharge. Take Cd = 0.98 2 π = a1 = 0.16 0.0201m2 Solution: 4 Given: Dia at B,= db 8cm = 0.08m Dia at inlet, d1 = 30cm ∴ Area at B, 2 π 2 2 ∴ = = 706.85cm a1 30 π = a2 = 0.08 0.005026m2 4 4

( )

(

)

(

)

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∴ Rate of flow, Q= V × A

0.99 × 0.0201 = 0.01989m3 / s =

ii) Difference of level of mercury in the Utube. Let h=Difference of mercury level. Where S  Then h x  h − 1  =  So  Where

i) Difference of pressure, pB − pA = 0.981N / cm2

h=

0.981 × 104 N / m2 = 9810N / m2 =

Difference of pressure head pB − pA 9810 ∴= = 1.25 ρg

800 × 9.81

Applying Bernoulli’s equation at A and B and taking the reference as line passing through section B, we get, PA VA 2 PB VB2 + + zA = + + zB ρg 2g ρg 2g Or

2

2

PA PB V V − + z A − zB = B − A 2g 2g ρg ρg

 P −P  V2 V2 Or  A B  + 2.0 − 0.0 = B − A 2g 2g  ρg  Or  V2 V2   P −P −1.25 + 2.0 = B − A Q A B = 1.25 ρg 2g 2g    = 0.75

()

VB2 VA 2 ----- i − 2g 2g

Now applying continuity equation at A and B, we get VA × A1 = VB × A 2

V ×A Or VB = A 1 A2

Substituting the value of VB in equation (i), we get 0.7= 5

∴ VA =

16VA 2 VA 2 15VA 2 − = 2g 2g 2g

0.75 × 2 × 9.81 = 0.99m / s 15

PA PB − +z −z ρg ρg A B

 P −P  = −1.25+2.0 − 0 Q B A =1.25  ρg   = 0.75 



 0.8



∴ 0.75 =x  13.6 − 1  =x × 16 ∴= x

0.75 = 0.04687m 16

= 4.687cm

Example: A pitot- static tube placed in the centre of a 300mm pipe line has one orifice pointing upstream and other perpendicular to it. The mean velocity in the pipe is 0.80 of the central velocity. Find the discharge through the pipe if pressure difference between the two orifices is 60 mm of water .Take the coefficient of pitot tube as Cv = 0.98 Solution: Given: Dia of pipe, d=300mm=0.30m Diff of pressure head,

h = 60mm of water = 0.06m of water

Cv = 0.98

= V 0.80 × centre line velocity Mean velocity, Centre line velocity is given by equation

= Cv 2gh = 0.98 × 2 × 9.81 × 0.06 = 1.063m / s ∴ V= 0.80 × 1.063= 0.8504m / s

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= Q area of pipe × V Discharge, Q=

2 π 0.30) × 0.8504= 0.06m3 / s ( 4

Example: A pitot–tube is inserted in a pipe of 300mm diameter. The static pressure in pipe is 100mm of mercury (vacuum). The stagnation pressure at the centre of the pipe, recorded by the pitot–tube is 0.981N/ cm2 . Calculate the rate of flow of water through pipe, if the mean velocity of flow is 0.85 times the central velocity. Take Cv = 0.98. Solution: Given: Dia of pipe

d=300mm = 0.30m

22.263 N/cm2, while the datum head at A and B are 28m and 30m. Find the loss of head b/w A & B. Solution: D = 400 m = 0.4 m VA= V= 25m / s Total energy at A PA VA 2 + + zA ρg 2g Total energy at B

=

PB VB2 + + zB ρg 2g

H= EA − EB 2

 PA VA 2   PB VB2  + + zA  −  + + zB     ρg 2g   ρg 2g    2

PA 29.43 × 10 VA 2 π 31.85 = = 30, = 2g ρg 1000 × 9.81 4 Static pressure head = 100mm of mercury PB 22.563 × 104 VB2 (vacuum) = = 23, = 31.85 g 1000 9.81 2g ρ × 100 =− × 13.6 = −1.36m of water EA − EA = (30 + 28) − (23 + 30) = 5m 1000 Stagnation pressure 2 Example: = .981N / cm = 981 × 104 N / m2 A jet of water from a 25 mm diameter ∴ Stagnation pressure head nozzle is directed vertically upwards. .981 × 103 0.981 × 103 Assuming that the jet remains circular and = = = 1m ρg 1000 × 9.81 neglecting any loss of energy, what will be h = Stagnation pressure head –static the diameter of the jet at a point 4.5 m pressure head above the nozzle, if the velocity with which = 1.0 − ( −1.36)= 1.0 + 1.36= 2.36m of water the jet leaves the nozzle is 12 m/s. Solution: ∴ Velocity at centre V = Cv 2gh Given: Diameter of nozzle, = d1 25mm = 0.025m = 0.98 × 2 × 9.81 × 2.36 = 6.668m / s

∴ Area, = A = (0.3) 0.07068m2

Mean velocity,

V =0.85 × 6.668 =5.6678m / s

∴ Flow rate ofwater= V × area of pipe = 5.6678 × 0.07068 =0.4006m3 / s

4

V1 = 12m / s Velocity of jet at nozzle Height of point A h = 4.5m Let the velocity of the jet at a height of 4.5 m = V2

Example: A pipe of dia 400 mm carries water at velocity 25 m/s. The pressure at the points A & B are given as 29.43 N/cm2 and © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

Considering the vertical motion of the jet from the outlet of the nozzle to the point A (neglecting any loss of energy.) Initial velocity,

u V= 12m / s = 1

Final velocity,

V = V2

Value of g = 9.81m / s2 and h=4.5m

2gh, we get Using, V 2 − U2 = V22 − 122 = 2 × ( −9.81) × 4.5

∴ V= 2

122 − (2 × 9.81 × 4.5)

= 144 − 88.29 =7.46m / s

Now, applying continuity equation to the outlet of nozzle and at point A, we get

a1 v 1 = a 2 v 2

Or

π 2 A1 V1 4 D1 × V1 A2 = = V2 V2

π× 0.0252 × 12 = 0.0007896 4 × 7.46

Let D2 = Diameter of jet at point A Then, π 2 = A2 = D 0.0007896 4 2

= D2

0.0007896 × 4 = 0.0317m = 31.7mm π

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6

DYNAMICS OF FLUID FLOW

6.1 INTRODUCTION In case the surfaces cause a change in the magnitude and direction of the velocity of the fluid particles, the fluid particles exert a force on the surface. In turn the surfaces exert an equal and opposite force on the fluid particles. The force exerted by moving fluid particles on the surface is called dynamic force. Dynamic force always involves a change in the magnitude and direction of the velocity of the fluid. 6.2 IMPULSE MOMENTUM PRINCIPLE

When applied to a single body Newton’s second law can be started as “The sum of forces on the body equals the rate of change of momentum of the body in the direction of the force. ∑ F = d(mV) / dt This can also be written as ∑ Fdt = d(mV)

Where, M is the mass of the body V is the velocity of the body dt is the time. This also means the impulse Fdt equals the change in momentum of the body during the time dt. For flowing fluid, the principle can be stated as “The sum of forces on the fluid equals the difference between the momentum flowing in and momentum flowing out and the change in momentum of the fluid inside the control volume. Under steady flow conditions, the last term vanishes. So the forces in the fluid is given by Dynamics of Fluid Flow ∑ F=d ( mV ) / dt out – d ( mV ) / dt in In other words, the net force on the fluid mass is equal to the net rate of out flow of

momentum minus net rate of in-flow across the control surface. This can also be written as ρ2 Q 2 V2 – ρ2 Q 2 V2 ∑F = Where, ⍴ is the density ( Kg / m3 )

Q is the volume flow rate ( m3 / s )

If the fluid is incompressible, then ρQ ( ∆V ) ∑F = 6.3 FORCE EXERTED BY A FLUID ON A PIPE BEND

Free body of fluid element Force exerted on fluid element along x direction P1A1 − A 2 A 2 cos θ − Fx By impulse momentum principle for incompressible fluid ∑ Fx =ρ ( ∆V )x

P1A1 − P2 A 2 cos θ − Fx = rate of change of momentum = ρQ[V2 cos θ − V1 ] Fx =ρQ[V1 − V2 cos θ] + P1A1 − P2 A 2 cos θ ΣFy =ρQ ( ∆V ) y

P2 A2 sin   Fy  Q V2 sin   Fy = −(ρQV2 sin θ + P2 A 2 sin θ)

-ve sign indicates that direction of force is opposite i.e upwards 6.4 FORCE EXERTED BY A FLUID ON VERTICAL STATIONARY PLATE

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Force exerted on fluid in x direction is equal to change in momentum  1 = ρa V12 Fx = mV Fy = 0

6.5 FORCE EXERTED BY A FLUID ON STATIONARY BLADE The direction of the velocity is changed. There is negligible change in the magnitude. In the case considered pressure forces are equal both at inlet and outlet.

Force exerted by fluid in x direction with the assumed direction.

−Fx = ρQ(V2 cosβ − V1cosα) Q = AV1 Fx = ρQ(V1cosα − V2 cosβ)

Force exerted by fluid in y direction with the assumed direction.

Fy = ρQ(V2sinβ − V1sinα)

equals ρAV1 kg/s. But when the vane moves away from the direction of the jet with a velocity u, then the mass of water striking the vane equals ρA(V1–u). (V-u) is the relative velocity between the jet and the vane. r2= r For tangential balde r= 1  ( Vw1  ± Vw 2 ) Force on blade = F m   [ Vw1  ± Vw 2 ] U Power = P m  

 = m ρA(V1 − U) U = πDN / 60

Where D is mean diameter & N is speed of rotation in rpm

Example: A 45° reducing bend is connected in a pipe line, the diameters at the inlet and outlet of the bend being 600mm and 300mm respectively. Find the force exerted by water on the bend if the intensity of pressure at inlet to bend is 8.829N/ cm2 and rate of flow of water is 600 liters/s. Solution:

6.5.1 FORCE EXERTED BY A FLUID ON MOVING BLADE In the case of the moving vane it is necessary to consider both the absolute and relative velocities. The other difference is that the amount of fluid that strikes a moving vane at any time interval is different from that which strikes the stationary vane. If a jet of area A with a velocity V1 strikes a stationary vane, the mass impinging per unit time on the vane

Given: θ = 45° Angle of bend, Dia at inlet, D1 = 600mm = 0.6m

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2 π π A1 = D12 = ( .6 ) = 0.2827 m2 4 4

Dia at outlet, D2 = 300mm = 0.30m

∴ A2 = π ( .3) = 0.07068m2 2

4

∴ Resultant force,

FR = Fx2 +Fy2

 1911.42  6322.2

2

 20890.9N

Pressure at inlet

p1 = 8.829 N/ cm2 = 8.829×104 N/ m2

Q = 600lit/ s = 0.6m3 / s V1 =

V2 =

Q 0.6 = = 2.122m/ s A1 .2827

Q 0.6 = = 8.488m/ s A2 .07068

Applying Bernoulli’s equation at section (1) and (2), we get p1 V12 p V2 + +z1 = 2 + 2 +z2 ρ g 2g ρ g 2g But z1 = z2 p p1 V12 p2 V22 8.4882 8.829×104 2.1222 = 2+ + = + or + ρ g 2g ρ g 2g 1000×9.81 2×9.81 ρ g 2×9.81

9+ .2295= p2 / ρ g +3.672



p2 = 9.2295 − 3.672= ρg

=5.5575m of water

∴ p2 =5.5575×1000×9.81N/ m2 4

=5.45×10 N/ m

2

Forces on the bend in x- and y- directions are given by equations as

Fx = ρ Q  V1 − V2cos θ  + p1 A1 − p2 A2cos θ

= 1000×0.6 2.122 − 8.488cos45°

The angle made by resultant force with xaxis is given by equation F ∴ 6322.2 tan θ = y = = 0.3175 Fx 19911.4

θ = tan-1 (.3175)=17°36'

Example: A nozzle of diameter 20 mm is fitted to a pipe of diameter 40mm. Find the force exerted by the nozzle on the water which is flowing through the pipe at the rate of 1.2m3 / minute . Solution: Given: Dia of pipe, D1 = 40mm = 40×10−3 m = .04m

2 π 2 π 2 ∴ Area, A1 = D1 = .04 = 0.001256m 4 4 Dia of nozzle, D2 = 20mm = 0.02m

( )

∴ Area,

Discharge,

A2 =

2 π .02) = .000314m2 ( 4

1.2 3 3 Q =1.2m / minute = m / s = 0.02m3 / s +8.829×10 ×.2827 − 5.45×10 ×.07068×cos45° 60 4

4

= −2327.9+24959.6 − 2720.3= 2459.6 − 5048.2 =19911.4N And Fy  Q V2 sin    P2 A2 sin  = 1000×0.6  −8.488sin 45°  − 5.45×104

= −3601.1 − 2721.1= −6322.2N

- ve sign means

downward direction

Fy

×.07068×sin45°

is acting in the

Applying continuity equation at section (1) and (2) A1 V1 = A 2 V2 = Q

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∴ And

V1 =

V2 =

Q 0.2 = =15.92m/ s A1 .001256

Q 0.2 = = 63.69m/ s A 2 .000314

Applying Bernoulli’s equation at section (1) and (2) ,we get

P1 V12 P2 V22 + + z1 = + + z2 ρg 2g ρg 2g

Now z1 = z2 ,

p2 = atmospheric pressure=0 ρg

p1 V12 V22 + = ∴ ρ g 2g 2g

(

) (

)

63.692 15.922 p1 V22 V12 ∴ = − = − = ρ g 2g 2g 2×9.81 2×9.81 = 206.749 − 12.917 = 193.83m of water

∴ p1 =193.83×1000×9.81 N2 m N = 1901472 2 m

Let the force exerted by the nozzle on water = Fx Net force in the direction of x=rate of change of momentum in the direction of x ∴ p1 A1 − p2 A 2 + Fx = ρ Q ( V2 − V1 ) Where p2 =atmospheric pressure =0 and

ρ =1000

∴ 1901472×.001256 − 0+Fx =1000× 0.02( 63.69 − 15.92)

Or 2388.24+Fx = 916.15 ∴ Fx = −2388.24+ 916.15= −1472.09 -ve sign indicates that the force exerted by the nozzles on water is acting from right to left.

Example: A lawn sprinkler shown in Fig has 0.8cm diameter nozzle at the end of a rotating arm and discharges water at the rate of 10m/s velocity. Determine the torque required to hold the rotating arm stationary. Also determine the constant

speed of rotation of the arm, if free to rotate.

Solution: Given: Dia of each nozzle =0.8cm =0.008 m ∴ Area of each nozzle =

2 π .008 ) = 0.00005026m2 ( 4

Velocity of flow at each nozzle = 10m / s ∴ Discharge through each nozzle Q = Area× Velocity = .00005026×10 = .0005026m3 / s

Torque exerted by water coming through nozzle A on the sprinkler =moment of momentum of water through A = rA × ρ ×Q× VA = 0.25×1000×.0005026×10 clockwise

Torque exerted by water coming through nozzle B on the sprinkler = rB × ρ ×Q× VB = 0.20×1000×.0005026×10 clockwise

∴ Total torque exerted by water on sprinkler = .25×1000×.0005026×10 + .20×1000×.0005026×10 =1.2565+1.0052= 2.26Nm

∴ Torque required to hold the rotating arm stationary =Torque exerted by water on sprinkler = 2.26Nm Speed of rotation of arm, if free to rotate Let ω = speed of rotation of the sprinkler The absolute velocity of flow of water at the nozzles A and B are V1 = 10.0 − 0.25× ω and V2 = 10.0 − 0.20× ω

Torque exerted by water coming out at A, on sprinkler = rA × ρ ×Q× V1 = 0.25×1000×.0005026× (10 − 0.25ω) = 0.12565(10 − 0.25ω)

Torque exerted by water coming out at B, on sprinkler

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= rB × ρ ×Q× V2 = 0.20×1000×.0005026× (10.0 − 0.2ω)

Example: A jet of water of 30mm diameter strikes a = 0.10052(10.0 − 0.2ω) hinged square plate at its centre with a ∴ Total torque exerted by water velocity of 20m/s. The plate is deflected  0.12565(10.0  0.25 )  0.10052(10.0  0.2 ) through an angle of 20° . Find the weight of Since moment of momentum of the flow the plate. entering is zero and no external torque is If the plate is not allowed, to swing, what applied on sprinkler, so the resultant will be the force required at the lower torque on the sprinkler must be zero. edges of the plate to keep the plate in  0.12565(10.0  0.25 )  0.10052(10.0  0.2 )  0 vertical position. 1.2565 − 0.0314ω +1.0052 − 0.0201ω = 0 1.2565 + 1.0052 = ω ( 0.0314 + 0.0201)

2.2617 = 0.0515ω

2.2617 = 43.9rad/ s 0.0515 60× ω 60× 43.9 and N = = = 419.2 r.p.m. 2π 2π

∴ ω=

Example: Water is flowing through a pipe at the end of which a nozzle is fitted. The diameter of the nozzle is 100 mm and the head of water at the centre of nozzle is 100mm. Find the force exerted by the jet of water on a fixed vertical plate. The co-efficient of velocity is given as 0.95. Solution: Given: Diameter of nozzle d = 100mm = 0.1m Head of water H = 100 Co-efficient of velocity Cv = 0.95 Area of nozzle, a = π ( .1) = 0.007854m2 4

2

Theoretical velocity of jet water is given as

Vth = 2gH = 2×9.81×100

But

Cv =

Actual velocity Theoretical velocity

∴ Actual velocity of jet of water, V = Cv × Vth = 0.95× 44.294 = 42.08 m/ s

Force on a fixed vertical plate is given by equation F= ρaV 2 = 1000 × .00785 × 442.082 1000kg/ m ) (In S.I. units ρ for water =

= 13907.2N = 13.9kN

3

Solution: Given: Diameter of the jet, d=30mm=3cm=0.03m ∴ Area, π 2 π 2 2 a=

4

d =

4

(.03)

= .0007068 m

V = 20 m/ s Velocity of jet, θ = 20° Angle of swing Using equation for angle of swing.  AV 2 sin   W Or sin 20° = 1000×

.0007068× 202 282.72 = W W

282.72 = 826.6 N sin 20° If the plate is not allowed to swing, a force p will be applied at the lower edge of the plate as shown in fig. The weight of the plate is acting vertically downward through the C.G. of the plate. Now, let F=Force exerted by jet of water H=Height of plate =Distance of P from the hinge The jet strikes at the centre of the plate and hence distance of the center of the jet from h hinge = 2 Taking moments about the hinge O,

∴ W=

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Ph  F 

h 2

F× h F ρ aV 2 = = 2× h 2 2 .0007068× 202 = 1000× = 141.36 N 2

∴ P=

Example: A jet of water of diameter 10 cm strikes a flat plate normally with a velocity of 15 m/s. The plate is moving with a velocity of 6 m/s in the direction of the jet and away from the jet. Find the power and efficiency of the jet. Solution: The given data from problem is a = .007854m2 , V = 15m/ s, u = 6m/ s

Also work done per second by the jet 2 W   a V  u  u =3817.02Nm/s

(i)Power of the jet in kW Work done per second 3817.02 = = = 3.817 kW 1000 1000 (η) (ii)Efficiency of the jet Output of the jet per second = Input of the per second Where output of jet/ sec = Work done by jet per second = 3817.02 Nm/ s And input per second

 = Kinetic energy of the jet / sec

1  mass  2 1 1 2   V  (  AV )  V   AV 3  2  sec  2 2 1  1000 0.007854153  13253.6 Nm / s 2 3817.02 = 0.288 = 28.8% ∴ η of the jet = 13253.6

Example: A jet of water from a nozzle is deflected through 60° from its original direction by a curved plate which it enters tangentially without shocks with a velocity of 30m/s and leaves with a mean velocity of 25m/s. If the discharge from the nozzle is 0.8kg/s calculate the magnitude and direction of the resultant force on the vane, if the vane is stationary.

Solution: Given: Velocity at inlet V1 = 30 m/ s

Velocity at outlet, V2 = 25 m/ s Mass per second = 0.8kg/ s

Force in the direction of jet, Fx = Mass per second× ( V1x - V2 x )

Where V1x = Initial velocity in the direction of x=30m/s V2 x = Final velocity in the direction of x 1 = 25cos 60° = 25× = 12.5 m/ s 2 ∴ Fx = 0.8 [30 − 12.5] = 0.8×17.5 = 14.0 N Similarly, force normal to the jet, Fy = Mass per second× ( V1y − V2 y )

= 0.8 [ 0 − 25sin 60°] = −17.32 N

− ve sign means the force , Fy is acting in

the vertically downward direction . ∴ resultant force on the vane

= Fx2 + Fy2 = 142 + ( −17.32 ) = 22.27 N 2

The angle made by the resultant with x-axis Fy −17.32 tan θ = = = −1.237 Fx 14.0 − ve sign means the angle θ is in the clockwise direction with x-axis as shown in fig . −1 ∴ θ = tan 1.237 = 51°2.86 ' Example: A jet of water of diameter 50mm, having a velocity of 20m/s strikes a curved vane which is moving with a velocity of 10m/s in the direction of the jet . The jet leaves the

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vane at an angle of 60° to the direction of motion of vane at outlet .Determine: i) The force exerted by the jet on the vane in the direction of motion ii) Work done per second by the jet Solution: Given: Diameter of the jet D=50mm=0.5m π 2 ∴ Area, a = ( 0.5 ) = .001963m 2 4 Velocity of jet, V1 = 20 m/ s Velocity of vane, u1 = 10 m/ s

As jet and vane are moving in the same direction ∴ α =0 Angle made by the leaving jet, with the direction of motion= 60° ∴ β = 180° - 60° = 120° For this problem ,we have u1 = u 2 = u = 10 m/ s Vr1 = Vr2

From fig ,we have

Vr1 = AB− AC = V1 − u1

= 20 − 10 = 10 m/ s

Vw1 = V1 = 20 m/ s

2

2

is an obtuse angle) = 1000×.001963×10 [ 20 − 5] N = 294.45 N

(ii) Work done per second by the jet Fx × u = 294.45×10 = 2944.5 Nm/ s = 2944.5 W

1

Now in ∆EFG, EG = Vr = 10 m/ s, GF = u 2 = 10 m/ s , ∠ GEF = 180° − ( 60° + φ ) = (120° − φ ) From sine rule , we have

 Nm/ s = W ( watt ) 

Example: A jet of water having a velocity of 15m/s strikes a curved vane which is moving with a velocity of 5m/s in the same direction as that of the jet at inlet. The vane is so shaped that the jet is deflected through 135° .The diameter of jet is 100mm. Assuming the vane to be smooth, find i) Force exerted by the jet on the vane in the direction of motion, ii) Power exerted on the vane, and iii) Efficiency of the vane Solution: Given: V1 = 15 m/ s Velocity of jet, u = u1 = u 2 = 5 m/ s Velocity of vane At inlet jet and vane are in the same direction , hence α = 0 Diameter of jet, d=100mm=0.1m π 2 ∴ Area, a = ( 0.1) = .007854 m 2 4 Angle of deflection of the jet = 135° = 180° − φ . ∴

∴ Vr = Vr = 10 m/ s 2

Now Vw = HF = GF− GH = u 2 − Vr cos φ = 10 − 10×cos 60° = 10 − 5 = 5 m/ s (i) The force exerted by the jet on the vane in the direction of motion is given by equation Fx = ρ aVr1  Vw1 − Vw 2  ((-ve sign is taken as β

φ = 180° − 135° = 45°

2

EG GF 10 10 = or = sin 60° sin (120° − φ ) sin 60° sin (120° − φ )

sin 60° = sin (120° − φ ) Or ∴ 60° = 120° − φ or120° − 60° = 60°

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As vane is given smooth hence

Vr1 = Vr2

From the inlet velocity triangle, which is a straight line in this case, we have

From Problem,

V1 = 15 m/ s

u = u1 = u 2 = 5 m/ s

Vr1 = V1 − u1 = 15 − 5 = 10 m/ s

α = 0 a = .007854 m 2 φ = 45°,

Vw1 = V1 = 15 m/ s

Vw1 = 15 m/ s and Vw 2 = 2.07 m/ s

From the outlet velocity triangle DEG, we have Vr2 = Vr1 = 10 m/ s u 2 = u1 = u = 5 m/ s Vr2 cos φ = u 2 + Vw 2 or 10 cos 45° = 5 + Vw 2

∴ Vw 2 = 10 cos 45° − 5 = 7.07 − 5 = 2.07 m/ s

i) Force exerted by the jet on the vane in the direction of motion is given by equation as Fx = ρ aVr1  Vw1 − Vw 2  ((-ve sign is taken as β is an obtuse angle)  1000 0.0078541015  (2.07) 

=1340.6N

ii) Power of the vane is given as

Fx × uNm/ s = 1340.6×5 = 6703 W = 6.703kW

iii) Efficiency of the vane work done on vane per second  Kinetic energy supplied by jet per second

=

=

For the series of vanes, mass of water striking per second =Mass of water coming out from nozzle

= ρ aV1 = 1000×.007854×15 = 117.72

i) Force exerted by the jet on the vane in the direction of motion Fx = ρ aV1  Vw + Vw  1

2

= 117.72 [15 + 2.07 ] = 2009.5 N

ii) Power of the vane in kW

Work done per second Fx × u 2009.5×5 = kW = 1000 1000 1000 = 10.05 kW

=

iii) Efficiency

Work done per second 1 × ( mass per second ) × V12 2 2009.5× 5.0 = = 0.7586 or 75.86% 1 ×117.72×152 2

η=

Fx × u Fx × u = 1 1 × ( mass per second ) × V 2 × ( ρ aV1 ) × V12 2 2 1340.6×5.0 = 0.505 = 50.5% 1 × (1000×.007854×15 ) ×152 2

Example: If in above problem, the jet of water instead of striking single plate, strikes a series of curved vanes, find for the data given in the above problem. i) Force exerted by the jet on the vane in the direction of motion, ii) Power exerted on the vane and iii) Efficiency of the vane Solution: Given:

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GATE QUESTIONS Q.1

Q.2

Q.3

A water container is kept on a weighing balance. Water from a tap is falling vertically into the container with a volume flow rate of Q; the velocity of the water when it hits the water surface is U . At a particular instant of time the total mass of the container and water is m. The force registered by the weighing balance at this instant of time is a) mg + ρQU b) mg + 2ρQU c) mg + ρQU2/2 d) ρQU2/2 [GATE–2003] The following data about the flow of liquid was observed in a continuous chemical process plant:

a) ( ρU 2 / 2k ) πDS2

 D2  b) ( ρU 2 / 8k )  2 − 1 πDS2  Dt  2 D  c) ( ρU 2 / 2k )  2 − 1 πDS2  Dt  4 D  d) ( ρU 2 / 8k )  4 − 1 πDS2  Dt 

[GATE–2003]

Q.4

Mean flow rate of the liquid is a) 8.00 litres/sec b) 8.06 litres/sec c) 8.16 litres/sec d) 8.26 litres/sec [GATE–2004] Air flows through a venturi and into atmosphere. Air density is ρ; atmospheric pressure is 𝑝𝑝𝑎𝑎 ; throat diameter is Dt ; exit diameter is D and exit velocity is U . The throat is connected to a cylinder containing a frictionless piston attached to a spring. The spring constant is k . The bottom surface of the piston is exposed to atmosphere. Due to the flow, the piston moves by distance x. Assuming incompressible frictionless flow, x is

Q.5

A venturimeter of 20 mm throat diameter is used to measure the velocity of water in a horizontal pipe of 40 mm diameter. If the pressure difference between the pipe and throat sections is found to be 30 kPa then, neglecting frictional losses, the flow velocity is a) 0.2 m/s b) 1.0 m/s c) 1.4 m/s d) 2.0 m/s [GATE–2005] A U-tube manometer with a small quantity of mercury is used to measure the static pressure difference between two locations A and B in a conical section through which an incompressible fluid flows. At a particular flow rate, the mercury column appears as shown in the figure. The density of mercury is 13600 kg/m3 and g = 9.81 m/s2. Which of the following is correct ?

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a) Flow direction is A to B and PA− PB= 20 kPa b) Flow direction is B to A and PA − PB = 1.4 kPa c) Flow direction is A to B and PB − PA= 20 kPa d) Flow direction is B to A and PB − PA= 1.4 kPa [GATE–2005] Q.6

Q.7

Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. The pressure in the 20 cm pipe just upstream of the reducer is 150kPa. The fluid has a vapour pressure of 50 kPa and a specific weight of 5 kN/m3. Neglecting frictional effects, the maximum discharge (in m3/s) that can pass through the reducer without causing cavitation is a) 0.05 b) 0.16 c) 0.27 d) 0.38 [GATE–2009] Figure shows the schematic for the measurement of velocity of air (density =1.2 kg/m3) through a constant area duct using a pitot tube and a water tube manometer. The differential head of water (density = 1000 kg/m3) in the two columns of the manometer is 10 mm. Take acceleration due to gravity as 9.8 m/s2. The velocity of air in m/s is

a) 6.4 c) 12.8

b) 9.0 d) 25.6 [GATE–2011]

Q.8

Water is coming out from a tap and falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/s2 . Assuming steady, inviscid flow, constant atmospheric pressure everywhere & neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately a) 10 b) 15 c) 20 d) 25 [GATE–2013]

Q.9

Within a boundary layer for a steady incompressible flow, the Bernoulli equation a) holds because the flow is steady b) holds because the flow is incompressible c) holds because the flow is transitional d) does not hold because the flow is frictional [GATE–2015 (2)]

Q.10 A Prandtl tube (Pitot-static tube with C = 1) is used to measure the velocity of water. The differential manometer reading is 10 mm of liquid column with a relative density of 10. Assuming g = 9.8 m/s2, the velocity of water (in m/s) is_____. [GATE–2015 (3)]

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Q.11 Water (ρ = 1000 kg/m3) flows through a venturimeter with inlet diameter 80 mm and throat diameter 40 mm. The inlet and throat gauge pressures are measured to be 400 kPa and 130 kPa respectively. Assuming the venturimeter to be horizontal and neglecting friction, the inlet velocity (in m/s) is _____. [GATE–2015 (1)] Q.12 The water jet exiting from a stationary tank through a circular opening of diameter 300 mm impinges on a rigid wall as shown in the figure. Neglect all minor losses and assume the water level in the tank to remain constant. The net horizontal force experienced by the wall is ___________ kN. Density of water is 1000 kg/m3. Acceleration due to gravity g = 10 m/s2.

[GATE–2016 (3)] Q.13 The arrangement shown in the figure measures the velocity V of a gas of density 1kg / m3 flowing through a pipe. The

acceleration due to gravity is 9.81m / s2 . If the manometric fluid is water (density 1000kg / m3 ) and the velocity V is 20 m/s, the differential head h (in mm) between the two arms of the manometer is ____________

[GATE–2017 (2)] Q.14 A tank open at the top with a water level of 1 m, as shown in the figure, has a hole at a height of 0.5 m. A free jet leaves horizontally from the smooth hole. The distance X (in m) where the jet strikes the floor is

a) 0.5 c) 2.0

b) 1.0 d) 4.0 [GATE–2018 (1)]

Q.15 Air flows at the rate of 1.5 m3 / s through a horizontal pipe with a gradually reducing cross section as shown in the figure. The two crosssections of the pipe have diameters of 400 mm and 200 mm. Take the air density as 1.2 kg / m3 and assume inviscid incompressible flow. The change in pressure (P2 − P1 ) (in kPa) between sections 1 and 2 is

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a) - 1.28 c) - 2.13

b) 2.56 d) 1.28 [GATE–2018 (2)]

ANSWER KEY: 1 (a)

2 (c)

3 (d)

4 (d)

5 (a)

6 (b)

7 (c)

8 (b)

9 (d)

10 1.32

11 6

12 13 14 15 8.76 20.4 (b) (a)

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EXPLANATIONS Q.1

Q.2

(a)

Q.3

Given : Flow rate = Q Velocity of water when it strikes the water surface = U Total Mass (container + water) = m Force on weighing balance due to water striking the surface = Change in momentum = Initial Momentum−Final momentum = ρQU– 0 ( Since final velocity is zero) Weighing balance also experiences the weight of the container and water. So, Weight of container and water = mg. Therefore, total force on weighing balance = ρQU + mg (c)

Flow Rate

7.5-7.7 7.7-7.9 7.9-8.1 8.1-8.3 8.3-8.5 8.5-8.7

Mean value flow rate (x) 7.6 7.8 8.0 8.2 8.4 8.6

Σf = 80 Σfx = 652.8

Frequency (f)

fx

1 5 35 17 12 10

7.6 39 280 139.4 100.8 86

Σfx 652.8 ∴ Mean flow rate == = 8.16 Σx 80

(d).

First of all we have to take two sections (1) and (2). Applying Bernoulli’s equation at sections (1) and (2). p1 V12 p V2 + + Z1 = 2 + 2 + Z2 ρg 2g ρg 2g 2 p 2 V2 p 2 V22 (Since Z1 = Z2) + = + ρ 2 ρ 2 ρ 2  V2 − V12  ...(i) p1 − p= 2 2 Apply continuity equation, we get. A1V1 = A2V2 π 2 π D t V1 = D 2 U 4 4 2

D ∴ V1 =   × U ...(ii)  Dt  Substitute the value of V1 from equation (ii) into the equation (i) 4 ρ  2  D  2 p1 – p 2 =  U −   U  2   Dt   4 ρ  D   2 1 −    U ....(iii) = 2   Dt     From the figure, we have Spring force = Pressure force due to air π 2 −= kx A s ( p1 – = p2 ) Ds × ( p1 – p 2 ) 4 From equation (iii)

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Q.4

4  π 2 ρ  D  = kx Ds ×   − 1 U 2 4 2  D t    4  ρU 2  D    − 1 πDs2 x= 8k  D t    (d).

Q.6 For continuity equation, we get A1V1 = A 2 V2 π 2 d1 A1 V2 V1 4 × V1 = = π 2 A2 d2 4 2

 d1   40  = =  V1   V1  20   d2  Applying Bernoulli’s equation, we get p1 V12 p V2 + + z1 = 2 + 2 + z 2 ρg 2g ρg 2g For horizontal pipe; z1 = z 2 2

p1 − p 2 V22 − V12 = ρg 2g ∆p 1 2 = V2 − V12 ρ 2

(

Q.5

)

30 ×103 1  2 ⇒ = ( 4V1 ) − V12   1000 2 2 15V1 = 30 2 ∴ V1 = 2m/s

(a) It is a U -tube differential Manometer. In this manometer A and B at different level and the liquid contain in manometer has the same specific gravity (only mercury is fill in the manometer)

Given: ρmercury = 13600 kg/m3, Δh = 150 mm = 0.150 m Static pressure difference for U-tube differential manometer is given by, PA−PB= ρg(hA − hB ) = 13600×9.81×0.15 = 20.01 × 103 Pa ≈ 20 kPa PA – PB comes out to be positive. Therefore, flow is from A to B. (b).

Given: PV= 50 kPa, w = 5 kN/m3= ρg Consider steady, incompressible and irrotational flow and neglecting frictional effects. Applying continuity equation at section (1) and (2). A1 V1= A2 V2 x x 2 2 V1 ( d 2 ) × V2 ( d1 ) ×= 4 4 Substitute the values of d1 and d2, we get x x 2 2 V1 ( 20 ) ×= (10 ) × V2 4 4 400 V1 = 100 V2 ⇒ V2 = 4V1 …(i) Cavitation is the phenomenon of formation of vapor bubbles in a flowing liquid in a region where the pressure of liquid falls below the vapor pressure[PL
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Q.7

Here Z1 = Z2 for horizontal pipe And w = ρg = 5 kN/m2 V2 150 V12 50 (4V1 ) 2 + = + 20 = 15 1 5 2g 5 2g 2g V1 = 5.114 m/s And V2=4V1 =4×5.114=20.46m/ s Maximum discharge, Qmax = A1 V1 π = D12 V1 4 π = × 0.22 × 5.114 4 =0.16 m3/s

(c) Given: ρs = 1.2 kg/m3, ρM = 1000 kg/m3 x = 10 × 10-3m, g = 9.8 m/s2 If the difference of pressure head ‘h’ is measured by knowing the difference of the level of the manometer liquid say x .Then  SG   pw  = h x  w − 1=  x  − 1  SG 2   p2 

1000  = 10 ×10 − 3  − 1  1.2  = 8.32 m Velocity of air V = 2gh Q.8

= 2 × 9.8 × 8.32 = 12.8 m/s

(b)

Q.9

π π 2 = d12 × u= d2 × v 4 4 d2 × u d 22 = 1 2 v (0.02) 2 × 2 2 d2 = (3.716) d2 = 14.67 ≈ 15mm

(d) Bernoulli equation does not hold because the flow is frictional.

Q.10 (1.3 to 1.34) Velocity of water = C v 2gh Cv = 1 (Given)  sg  = h x  − 1  s0  = 0.01(10-1) = 0.09m ∴ Velocity of flow =√2 × 9.8 × 0.09 = 1.328 m/s

Q.11 (6) p1 400 ×103 Pa  p1 v12 p 2 v 22 = + = +   ρg 2g ρg 2g  = p 2 130 ×103 Pa  v1 × 802 =v 2 × 402

v 2 = 4v1 Substituting v2 and solving for v1 we get v1 = 6m/s

Q.12 (8.76 to 8.78) Force, F = [ρAV ] v

= ρAV 2 Where velocity, V = 2gh

v 2 = u2 + 2gh v 2 = 22 + 2 × 9.81 × 0.5 v = 3.716 m/s By continuity equation Q1 = Q 2

= 2 ×10 × 6.2 = 11.136 m/s π ∴ F = 103 × × 0.32 ×11.1362 4 = 8765.78 N = 8.765 KN Q.13 (20.4)

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ρ  2gX  m − 1 ρ  ρ  = V 2 2gX  m − 1 ρ  1000  400 = 2 × 9.81× h  − 1  1  h = 20.4mm

= v

Q.14 (b)

Applying Bernoulli’s equation for points (1) and (2) : V12 V2 + Z1 = P2 + ρair 2 + Z 2 .......(1) 2 2 But Z1 = Z2 P1 + ρair

and = V1

4Q 4 1.5 = = 11.937 m/s 2 πd1 π  0.42

& V2 4= =  V1 47.746 m/s From equation (1) ,

(P

2

− P= 1 )

ρair 2

(V

2 1

− V22 )

1.2 11.937 2 − 47.7462 ) ( 2 = −1.28 kPa

=

The velocity of jet at the exit of hole is given by u = 2 gh The jet performs projectile motion with x component of velocity u = 2 gh and y component v = 0 at the exit of hole. The time‘t’ taken by jet to reach the ground can be obtained by kinematic equation of jet motion as: 1 1 = y  0  t + gt 2 2 2 2y i.e., t = g The horizontal distance traveled by jet in the same time is t = x u=

2 gh 

2y g

h  y 2 0.5  0.5 1m = 2= =

Q.15 (b)

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7

FLOW THROUGH CONDUITS/PIPES

7.1 INTERNAL FLOW Fluids are conveyed (transported) through closed conduits in numerous industrial processes. It is found necessary to design the pipe system to carry a specified quantity of fluid between specified locations with minimum head loss. The flow may be laminar with fluid flowing in an orderly way, with layers not mixing macroscopically. The momentum transfer and consequent shear induced is at the molecular level by pure diffusion. Such flow is encountered with every viscous fluids. The flow turns turbulent under certain conditions with macroscopic mixing of fluid layers in the flow. At any location, the velocity varies about a mean value. The flow is controlled by (i) pressure gradient (ii) the pipe diameter or hydraulic mean diameter (iii) the fluid properties like viscosity and density and (iv) the pipe roughness.

7.1.2 ENTRY LENGTH In internal flow, the boundary layer develops all over the circumference. At some distance from the entrance, the boundary layers merge. The velocity profile beyond this point remains unchanged. The distance upto this point is known as entry length.

7.1.1 BOUNDARY LAYER

When fluid flows over surface, the molecules near the surface are brought to rest due to the viscosity of the fluid. The adjacent layers also slow down, but to a lower and lower extent. This slowing down is found limited to a thin layer near the surface. The fluid beyond this layer is not affected by the presence of the surface. The fluid layer near the surface in which velocity of fluid is less than free stream velocity is known as boundary layer.

7.2 LAMINAR FLOW/VISCOUS FLOW 7.2.1 ANALYSIS OF FULLY DEVELOPED LAMINAR FLOW IN CIRCULAR DUCT i) Shear stress

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C= −

1  dP  2  R 4µ  dx 

1  dP  −   R 2 − r2  U= 4µ  dx 

2) for, r = 0, U =Umax 1  dP  Umax = −   R 2 4µ  dx  At low velocity the fluid moves in layers. The shear stress in laminar flow is given by du τw =µ dy force balance on fluid element gives Pgπr2 − τ× 2πrdx − (P + dP)πr2 = 0

dP 2 πr dx dP r τw =− × dx 2

τw 2πr =−

ii) Velocity distribution du τw =µ dy y= R − r (y is measured from the pipe wall) dy = −dr

∴τw = µ

du dr

du dP r =− × dr dx 2 1  dP  r ∴ ∫ du =   dr µ ∫  dx  2 ∴−µ

1  dP  1 ∴ U =   r2 + C µ  dx  4

On applying boundary conditions: 1) For r = R, U = 0 1  dP  2 0=  R + C 4µ  dx 

  r 2  U Umax 1 −    =   R  

iii) Average velocity dQ = UgdA dA = 2πrdr

  r 2  dQ = Umax 1 −    × dA   R     r 2  ∫ dQ = ∫ Umax 1 −  R   × 2πrdr     π −dP 4 Q=  R 8µ  dx 

U=

Q = A

π  −dP  4  R 8µ  dx 

Umax =2 U

πR 2

=−

1  dP  2  R 8µ  dx 

iv) Pressure drop for length (l) Average velocity is given by U=

1  dP  2  R 8µ  dx 

By rearranging above equation P

L

1 8µU − ∫ dP = ∫ 2 P2 0 R

Pressure drop across length ‘L’ is 32µUL P1 − P2 = 2 D

= hf

P1 − P2 32µUL = ρg ρgD2

The above equation Poiseuille equation

is

Hagen

7.2.2 FLOW OF VISCOUS FLUID BETWEEN TWO PARALLEL PLATES

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1  dP  i) U = −    ty − y 2  2µ  dx 

Where, t is the distance between parallel plates Velocity is max at y = t/2 Umax = −

1  dP  2  t 8µ  dx 

 y y2  U= 4Umax ×  − 2  t t 

ii) Shear stress distribution 1  dP  τ = −    t − 2y  2  dx 

at centre, shear stress will be zero.

iii) U = Avg. velocity= −1  dP  t 2 12µ  dx  Umax 3 = U 2 iv) Pressure head for a given length 12µUL P1 − P2 = 2 t

In turbulent flow, instantaneous values of the velocity fluctuate about an average value, which suggests that the velocity can be expressed as the sum of an average value u and a fluctuating component u’,

u= u + u'

When a liquid is flowing through a pipe, the velocity of the liquid layer adjacent to wall is zero. The friction resistance for turbulent flow is. 1) ∝ V n where n varies from 1.5-2.0 2) ∝ ρ 3) ∝ A Area of contact 4) ∝ P Pressure 5) ∝ Nature of surface . 7.3.1 EXPRESSION FOR COEFFICIENT OF FRICTION IN TERMS OF SHEAR STRESS Consider a fluid element of length ‘L’ & diameter ‘d’

7.3 TURBULENT FLOW IN PIPES

Most flows encountered in engineering practice are turbulent, and thus it is important to understand how turbulence affects the wall shear stress. Turbulent flow is characterized by random and rapid fluctuations of swirling regions of fluid, called eddy, throughout the flow. These fluctuations provide an additional mechanism for momentum and energy transfer. In laminar flow, fluid particles flow in an orderly manner along pathlines, and momentum and energy are transferred across streamlines by molecular diffusion. In turbulent flow, the swirling eddies transport mass, momentum, and energy to other regions of flow much more rapidly than molecular diffusion, thus greatly enhancing mass, momentum, and heat transfer. As a result, turbulent flow is associated with much higher values of friction, heat transfer, and mass transfer coefficients

P1 A − P2 A − F = 0

(P1 − P2 )A = F F = force due to shear stress. F= τO πdL

(

)

(P1 − P2 )A = τO ( πdL )

(

P1 − P2 4fLV 2 τO πdL = = ρg ρg 2gd

A=

f=

)

π 2 d 4 2τo

ρV 2

7.3.2 SHEAR STRESS IN TURBULENT FLOW Shear stress in turbulent flow is sum of shear stress due to viscous flow & turbulent flow

τ = τv + τt

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Viscous/laminar shear stress is given by du τV =µ dy J. Boussinesq expressed turbulent shear in terms of average velocity gradient du τt =η dy Where, η is Eddy viscosity, du is average velocity. 7.3.3 REYNOLD’S EXPRESSION FOR TURBULENT SHEAR STRESS

Reynold expressed turbulent shear stress between two layers of a fluid at a small distance in terms of velocity fluctuations τt =ρu'v'

Where, u’ and v’ are fluctuating velocity components in the direction of x & y due to turbulence. τt will also be varying, Hence to find the shear stress time avg on both side is considered τ = ρu'v' 7.3.4 PRANDTL MIXING LENGTH THEORY FOR TURBULENT SHEAR STRESS

 du  du τ =µ + ρl2   dy  dy 

2

7.3.5 VELOCITY DISTRIBUTION TURBULENT FLOW IN PIPES  du  du τ =µ + ρl2   dy  dy 

IN

2

Viscous shear stress is negligible except near the boundary. Prandtl assumed that mixing length is proportional to ‘y’ l∝y l = ky Where, K is Karman factor = 0.4 2

2

 du   du  τt =ρ ( ky )   =ρk 2y 2    dy   dy  by rearranging the above equation 2

τ du = dy ρk 2y 2

du 1 τ = dy ky ρ

U = ( ln y )

τ 1 g +C ρ k

τ (U* is known as Shear velocity) U∗ = The turbulent shear stress can only be ρ calculated if u’ and v’ is known. According Since y is distance from the surface of pipe to Prandtl, the mixing length ‘l’, is that U = Umax at y=R distance between two layers in the transverse direction such that the lumps of U* = Umax lnR + C fluid particles from one layer could reach k the other layer and the lumps are mixed in U* = C Umax − lnR k the other layer in such a way that momentum of particle in the direction of U*  y  = U ln   + Umax motion is same. k R Prandtll expressed velocity fluctuations in U*  y  = U ln   + Umax terms of mixing length 0.4  R  du du y U' = l & V' = l = U Umax + 2.5U* ln   dy dy R 2

 du  τt =ρl2    dy  Total shear stress is given by





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This is Prandtl’s universal velocity distribution equation for turbulent flow in pipes. (Umax − U) is known as velocity defect. 7.3.6 HYDRO DYNAMICALLY SMOOTH & ROUGH BOUNDARIES

For turbulent flow analysis along a boundary, the flow is divided in two portions. Viscous portion near the surface is known as laminar sub layer. The boundaries are considered as smooth or rough on the basis of ratio of average irregularities ‘k’ and laminar sub layer thickness ‘δ’ if k < δ Smooth if k > δ Rough k < 0.25 boundary is smooth if δ k >6 if boundary is rough δ 7.3.7 VELOCITY DISTRIBUTION FOR TURBULENT FLOW IN SMOOTH PIPES U =

U* ln y + C k

at y 0, U=-∞ =

It means at some finite distance from wall, the velocity will be equal to zero. Let y’ be the distance from wall where velocity is zero y = y ', U=0 U= 0=

U* ln y '+ C k

−U * ln y ' k U*  y  U = ln   k  y' 

C=

For smooth boundary y ' ∝ δ' Where, δ' is thickness of laminar sub layer From experiments, y’& δ' is given by

δ' 107 11.6ν δ' = * U * U  y.100  U = ln   k  δ  y' =

7.3.8 REYNOLDS’ NO IN TERMS OF ROUGHNESS

U* k Re = ν

Where, k is average height of irregularities ν is kinematic viscosity τ U* = Shear velocity = ρ

U* k < 4, boundary is smooth ν U* k ii) 4 < < 100, boundary is in transition ν U* k iii) > 100, the boundary is rough. ν i)

7.3.9 VELOCITY IN ROUGH PIPES

(Nikurddse’s Experiment) show that k y' = 30 u* u = In(y / y ') k  y  u = 2.5u *In    k / 30   30y  u = 2.5u *In    k  Example: A Pipe carrying water has avg irregularities of 0.15 mm. What type of boundary it is? (Shear stress developed is 2 4.9N / m , ν =0.01strokes ) Solution: = k 0.15 ×10−3 m τ0 =4.9N / m 2

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ν = 0.01strokes = 0.01' cm 2 / s = ∆10−6 m 2 / s . ρ =1000Kg / m3 11.6ν δ= u* 11.6 ×10−6 δ= u* τ0 4.9 = u* = = 0.07m / s ρ 1000 Roughness Reynold No.: u *k 0.07 ×15 ×103 = v 10−6 7.4 LOSS OF ENERGY IN FLUID FLOW 7.4.1 MAJOR LOSS Darcy-Weisbach Formula: 4f .L.V 2 hf = d.2g h f = Loss of head due to viscosity. 16 f= for Re<2000 (Viscous flow). Re 0.079 = for Re varying from 4000 to 106. 1 Re 4 (for turbulent flow in smooth pipes). Where, L=Length, V = mean velocity of flow, d = diameter of pipe. 7.4.2 MINOR ENERGY LOSS

1) Loss of head due to sudden enlargement: Due to sudden change of diameter the liquid flowing from the smallest pipe is not able to follow the abrupt change of boundary. Thus the flow separates from the boundary and turbulent eddies are formed.

Head loss is ‘m’ is given by

he

(v − v ) = 1 2

2

2g

2) Loss due to sudden contraction: As the liquid comes out of orifice it contract further and the area just outside the orifice is lower compared to the area of the orifice. This section is called as vena-contracta. The liquid expands from vena contracta due to which the loss of energy takes place.

Energy loss is given by v 2 .k he = 2g Where, 2

1  = k  − 1  Cc  Cc is coefficient of contraction

if= Cc 0.62, = k 0.375

Cc = 0.62, then k=0.375 v2 v 2 .k = 0.375 2 2g 2g Cc is not given

∴ h e= if

h e = 0.5

v 22 2g

3) Loss of head at entrance of a pipe:

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h i = 0.5

Q=A2 v2= 0.03275m3/s = 32.75lit/s

v2 2g

7.5 FLOW THROUGH PIPES IN SERIES OR FLOW THROUGH COMPOUND PIPES

4) Loss of head due to exit: v2 ho = 2g

7.5.1 EQUIVALENT PIPE

5) Loss of Head due to an obstruction in a pipe: Due to sudden enlargement of the area of flow beyond the obstruction, head loss takes place. After vena contracta, sudden enlargement takes place. a =max area of obstruction A = area of pipe V = velocity of liquid. Head loss =

 v  A − 1  2g  cc (A − a)  2

2

6) Loss of head due to bend in pipe: k.v 2 hb = 2g The value of ‘k’ depends on • Angle of bend • Radius of curvature of bend • Diameter of pipe Example: At a sudden enlargement of a water main from 240 mm to 480 mm dia the hydraulic gradient rises by 10 mm. Estimate the rate of flow. Solution: Hydraulic Gradient  P1   P2   + z1  −  + z 2   ρg   ρg 

P1 v2 P v2 ( v − v ) + z1 + 1 = 2 + z 2 + 2 + 1 2 ρg 2g ρg 2g 2g d v = d v2 2 1 1

2 2

d1 = 240mm d2 = 480mm v1=4v2 v2=0.1808=0.181m/s

2

This is defined as the pipe of uniform diameter having loss of head and discharge equal to the loss of head and discharge of a compound pipe consisting of several pipes of different lengths and diameters.

If no minor loss is assumed H = ZA − ZB = hf total

4f L V 4f L V 2 4f L V 2 + 2 2 2 + 3 3 3 2gd1 2gd2 2gd3 2 1 1 1

H=

πd32 πd22 πd2 = = .V1 V2 V 4 4 3 4

= Q

For friction coefficient, 4 × 16fQ2  L1 L2 L3  = H + +   π2 .2g  d15 d25 d35  = H

Leq

deq

4fLV12 4 × 16Q2f  Leq  =   2g d π2 2g  deq5 

5

L L  L =  15 + 25 + 35   d1 d2 d3 

7.5.2 POWER TRANSMISSION THROUGH PIPES Total head available at outlet of pipe (Minor loss neglected)

= H − hf

= H−

4f lv 2 d.2g

Weight of water flowing Q=

πd2 .v 4

= W ρgQ = ρg

πd2 .v 4

Power transmitted = W × head available at outlet

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 4f lv 2  ρg πd2 = P H − v  2gd  4 

7.6 FLOWS THROUGH NOZZLES Total head at inlet of pipe = total head energy + losses

v2 2g By Bernoulli’s equation for real fluids Assuming minor loss to be negligible v2 H − hf = 2g But, total head at outlet of nozzle =

4f lv 2 2gd By continuity equation AV = av hf =

v 2 4f L  a  = + H  v 2g 2gD  A 

2

Viscosity, µ = 0.1Ns/ m2 Relative density = 0.9 density of oil = 0.9×1000= 900kg/ m3 D =50mm = .05m L =300m 3.5 Q =3.5liters/ S = = .0035m3 / s 1000 (i) Pressure drop p1 − p2 =

32µuL Q .0035 .0035 , where u = = = =1.782m/ s 2 Area π 2 π D2 D (.05) 4 4

The Reynolds number (R e ) is given by ρ VD Re = µ Where ρ= 900kg / m3 , averagevelocity= u= 1.782m / s



R e = 900×

1.782×.05 = 801.9 0.1

As Reynolds number is less than 2000, the flow is viscous/laminar 2gh v= 32 × 0.1×1.782 × 300  4f L a2  684288N / mm 2 = ∴ p1 − p 2 = × 2 1 + 2 0.05 D A   2 = 68.43N / m ii) Shear Stress at the pipe wall ( τ0 ) 7.6.1 POWER TRANSMITTED THROUGH NOZZLE The shear stress at any radius r is given by the equation 1 2 −δ p r K.E of the jet at the outlet of Nozzle = mv τ= i.e., 2 δx 2 Mass flow rate = ρav ∴ Shear stress at pipe wall, where r=R is 1 given by K.E.= ρav3 2 −δ p R τ= power of outlet of Nozzle 1 / 2ρav3 δx 2 η = Power at the inlet of pipe ρg Q.H δp −(p 2 − p1 ) p1 − p 2 p1 − p 2 = = Now= δx x 2 − x1 x 2 − x1 L Example: 2 684288 N/ m = = 2280.96N/ m3 An oil of viscosity 0.1Ns/ m2 and relative 300 m density 0.9 is flowing through a circular And R = D = .05 = .025m pipe of diameter 50mm and of length 2 2 .025 N 300m. The rate of flow of fluid through the τ0 = 2280.96× = 28.512N/ m2 2 pipe is 3.5 lit/s. Find the pressure drop in a 2 m length of 300m and also the shear stress at the pipe wall. Example: Solution: A laminar flow is taking place in a pipe of Given: diameter 200mm. The maximum velocity is © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

1.5m/s. Find the mean velocity and the radius at which this occurs. Also calculate the velocity at 4cm from the wall of the pipe. Solution: Given: Dia of pipe, D=200mm=0.20m Umax =1.5m/ s (i)Mean velocity , u Ratio of

Umax 1.5 1.5 = 2.0 or = 2.0 ∴ u = = 0.75m/ s u u 2.0

(ii)Radius at which 𝐮𝐮 occurs The velocity, u, at any radius ‘r’ is given by u=−

1 δp 2 2 1 δp 2  r2  R − r  = − − R 1    4µ δ x  4µ δµ  R 2 

But from equation Umax is given by 1 δp 2 Umax = − R 4µ δ x

  r 2  ∴ u = Umax 1 −      R  

Now the radius r at which

.....(i)

  r 2    .06 2  = Umax 1 −    =1.5 1 −      R     .1   =1.5 1.0 − .36 =1.5×.64 = 0.96m/ s

Example: Crude oil of µ =1.5 poise and relative density 0.9 flows through a 20mm diameter vertical pipe. The pressure gauges fixed 20m apart read 58.86N/cm2 and 19.62N/cm2 as shown in Fig. Find the direction and rate of flow through the pipe.

u = u = 0.75m/ s

2     r   ∴ 0.75 = 1.5 1 −   D     2  

  r 2  ∴ 0.75 = 1.5 1 −      0.1   0.75  r  ∴ = 1−   1.5  0.1 

2

2

0.75 1 1  r  ∴ =1 − =  =1 − 1.50 2 2  0.1 



r 1 = = 0.5 0.1 2

∴ r = 0.1× .5 = 0.1×707 = .0707m =70.7mm

(iii) Velocity at 4cm from the wall ∴ r = R − 4.0=10 − 4.0=6.0cm =0.06m The velocity at a radius =0.06m Or 4cm from pipe wall is given by equation(i)

Solution: Given:

1.5 = 0.15Ns/ m2 10 Relative density = 0.9 ∴ Density of oil ∴ = 0.9×1000= 900kg/ m3 Dia. Of pipe, D= 20mm = 0.02m L = 20m PA =58.86N/ cm2 =58.86×104 N/ m2 µ =1.5poise =

PB =19.62N/ cm2 =19.62×104 N/ m2 Solution: (i)Direction of flow. To find the direction of flow,  p  v2 the total energy  + +Z  at the lower  ρ g 2g 

end A and at the upper end B is to be calculated. The direction of flow will be given from the higher energy to the lower energy. As here the diameter of the pipe is same and hence kinetic energy at A and B

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will be same. Hence to find the direction of  p  flow, calculating  +Z  at A and B.  ρg 

Taking the level at A as datum. p  The value of  A +Z  at A  ρg  4 6 ×10 × 9.81 = = + 0 66.67m 900 × 9.81

2×104 ×9.81 = +20 = 22.22+20 = 42.22m 900×9.81  p  As the value of  +Z  is higher at A.  ρg  Hence, flow takes place from A to B.

(ii) Rate of flow. The loss of pressure head for viscous flow through circular pipe is given by

32µuL ρ gD2

For a vertical pipe hf = Loss of peizometric head p  p  =  A +ZA  −  B +ZB  = 66.67 − 42.22= 24.45m  ρg   ρg 



Or

24.45=

2 π π = u× D2 = 0.9× × ( .02) m3 / s = 2.827×10−4 m3 / s 4 4

= 0.2827lit res/ s

 p  +Z  at B The value of   ρg 

hf =

As Reynolds number is less than 2000, the flow is laminar. ∴ Rate of flow = avg. velocity * area

32×0.15×u×20.0 900×9.81× ( .02)

2

24.45×900×9.81×.0004 u= = 0.889 ; 0.9m/ s 32×0.15×20.0 The Reynolds number should be calculated. If Reynolds number is less than 2000, the flow will be laminar and the above expression for loss of pressure head for laminar flow can be used. ρ VD Now Reynolds number = µ 0.9×.02 = 900× =108 0.15

Example: A shaft having a diameter of 50mm rotates centrally in a journal bearing having a diameter of 50.15mm and length 100mm. The annular space between the shaft and the bearing is filled with oil having viscosity of 0.9 poise. Determine the power absorbed in the bearing when the speed of rotation is 60 rpm. Solution: Given: Dia. of shaft, D=50mm or .05m Dia of bearing D1 = 50.15mm or 0.05015m L =100mm or 0.1m Length , 0.9 Ns µ of oil = 0.9 poise = 10 m2

N = 600r.p.m.

Power = ?

∴ Thickness of oil film, t = =

D1 − D 50.15 − 50 = 2 2

0.15 = 0.075mm = 0.075×10−3 m 2

Tangential speed of shaft, π DN π ×0.05×600 V= = = 0.5× π m/ s 60 60

∴ Shear stress τ=µ

du V 0.9 0.5× π =µ = × =1883.52N/ m2 −3 dy t 10 0.075×10

∴ Shear force (F)

= τ × Area =1883.52× π D×L

=1883.52× π ×.05×0.1 = 29.586N

Resistance torque D .05 T = F× = 29.586× = 0.7387Nm 2 2 2π N 2π NT Power, P = T× ω = T× = watts 60 60

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Power =

2π NT 2π ×600×0.7387 = = 46.41W. 60 60

Example: An oil of S.G. 0.7 is flowing through a pipe of diameter 300 mm at the rate of 500 liters/s. Find the head lost due to friction and power required to maintain the flow for a length of 1000m. Take ν = .29 stokes. Solution: Given: S.G. of oil S = 0.7 Dia. of pipe d =300mm = 0.3m Discharge Q = 500 litres/ s = 0.5m3 / s Length of pipe L =1000m Velocity,

V =

Q 0.5 0.5 × 4 = = = 7.073m / s Area π 2 π× 0.32 d 4

Reynolds number, vd 7.073 × 0.3 Re = = = 7.316 ×104 ν 0.29 ×10−4

∴ Co-efficient of friction, 0.79 0.79 f= 1 = = 0.0048 1 4 4 4 Re 7.316×10

(

= v 0.01stoke = .01 × 10−4 m2 / s

Reynolds number, vd 3 × 0.2 Re= = = 6 ×105 −4 ν 0.01×10 Value of f, 0.09 0.09 f =0.02 + 0.3 =0.02 + 0.3 Re ( 6 ×105 )

=0.02 +

0.09 =0.02166 54.13

Head lost due to friction:

4 × f × L × V 2 4.0 × 0.02166 × 5.0 × 32 hf = = = 0.993 m of water d × 2g 0.20 × 2.0 × 9.81

)

∴ Head lost due to friction,

4 × f × L × V 2 4 × 0.0048 × 1000 × 7.0732 hf = = = 163.18m d × 2g 0.3 × 2 × 9.81 Power required ρg . Q . hf = kW 1000 Where = ρ density of oil = 0.7 × 1000 = 700kg / m3 ∴ Power required 700 × 9.81 × 0.5 × 163.18 = 560.28kW 1000

number. The kinematic viscosity of water = 0.1 Stokes. Solution: Given: Dia. of pipe, d = 200mm = 0.20m     Velocity, V = 3m/ s Length, L =5 m Kinematics viscosity,

Example: Water is flowing through a pipe of diameter 200mm with a velocity of 3m/s. Find the head lost due to friction for a length of 5m if the co-efficient of friction is given by .09 = f 0.02 + 0.3 where R e is Reynolds Re

Example: Determine the rate of flow of water through a pipe of diameter 20 cm and length 50 m when one end of the pipe is connected to a tank and other end of the pipe is open to the atmosphere. The pipe is horizontal and height of water in the tank is 4 m above the centre of the pipe. Consider all minor losses and take f = .099 in the formula hf =

4 × f × L × V2 . d × 2g

Solution: Given Dia. of pipe Length of pipe, Height of water

= d 20cm = 0.20m L = 50m H = 4m

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Co-efficient of friction f = .009

Let the velocity of water in pipe = V m / s Applying Bernoulli’s equation at the top of the water surface in the tank and at the outlet of pipe, we have[Taking point 1 on the top and 2 at the outlet of pipe].

P1 V12 P2 V22 + + z1 = + + z2 + all losses ρg 2g ρg 2g

Consider datum line passing through the centre of pipe V2 0 + 0 + 4.0 = 0 + 2 + 0 + (hi + hf ) 2g Or

4.0 =

V22 + hi + hf 2g

But the velocity in pipe= V , ∴V = V2

V2 + hi + hf 2g From equation V2 4 × f × L × V2 = hi 0.5 = and hf 2g d × 2g Substituting these values, we have 4.0 =

elevation of 15m. The total length of the syphon is 600m and the summit is 4 m above the water level in the upper reservoir. If the separation takes place at 2.8m of water absolute, find the maximum length of syphon from upper reservoir to the summit. Take f=0.004 and atmospheric pressure =10.3m of water. Solution: Given = d 200m = 0.2m Dia. of syphon Difference of level in two reservoirs = 15m Total length of pipe = 600m Height of summit from upper reservoirs

= 4m

Pressure head at summit pc = 2.8m of water absolute ρg Atmospheric pressure head pc = 10.3m of water absolute ρg Co-efficient of friction, f = .004

V2 V2 4 × f × L × V2 4.0 = + 0.5 + 2g 2g d × 2g

v2  4 × 0.009 × 50  = ∴4 1.0 + 0.5 +  2g  0.2 

Applying Bernoulli’s equation to points A and C and taking the datum line passing 2 v through, A and C 1.0 + 0.5 + 9.0 = ∴4 2g pC VC2 pA VA 2 + + z = + + z + Losses of head ρg 2g A ρg 2g C v2 ∴4 = 10.5 due to friction between A and C 2g Substituting the values of pressures in terms of absolute, we have 4 × 2 × 9.81 ∴ V = 2.734 m / sec = V2 10.5 10.3 + 0 + 0= 2.8 + + 4.0 + hf 1 2g ∴ Rate of flow, = [Q VC velocity = in pipe V] π Q = A × V = × (0.2)2 × 2.734 = 0.08589 m3 / s 4 V2 V2 ∴ h = 10.3 − 2.8 − 4.0 − = 3.5 − ..... ( i ) = 85.89 litres / s. f1 2g 2g Example: A syphon of diameter 200 mm connects two reservoirs having a difference in

Applying Bernoulli’s equation to points A and B and taking the datum line passing through B,

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pA VA 2 p V2 + + z A = B + B + zB ++ Losses of head ρg 2g ρg 2g due to friction between A and B pB A But , p= = atmospheric pressure ρg

ρg

VA 0,= VB 0,= z A 15,= zB 0 And,= ∴ 0 + 0 + 15 = 0 + 0 + 0 + hf hf

4 × f × L × V2 = 15 or 15 d × 2g

4 × .004 × 600 × V 2 Or = 15 0.2 × 2 × 9.81 15 × 0.2 × 2 × 9.81 Or V = 2.47m / s = 4 × .004 × 600

Substituting this value of V in equation ( i ) we get

hf 1 = 3.5 −

But

hf 2 =

( )

2.472 = 3.5 − 0.311 = 3.189m ..... ii 2 × 9.81

4 × f × L1 × V 2 d × 2g

.... ( iii )

Where L1 =inlet leg of syphon or length of syphon from upper reservoir to the summit.

= hf 1

4 × .004 × L1 × (2.47)2 = 0.0248 × L1 0.2 × 2 × 9.81

Substituting this value in equation ( ii ) ,

0.0248L1 =  3.189 

Difference of water levels, H = 16m Length & dia. of pipe1, L1 = 400m and d1 = 400 mm = 0.4 m Length & dia of pipe 2 L2 = 200m and d2 = 200m = 0.2m Length & dia of pipe 3 L3 = 200m and d3 = 300m = 0.3m Also, f1= f2= f3= 0.005

(i) Discharge through the compound pipe first neglecting minor losses. Let V1 , V2 , V3 are the velocities in the

1st 2nd and 3rd pipe respectively. From continuity, we A= V A= V A3V3 1 1 2 2

have

π 2 2 A1 V1 4 d1 d12  0.4  ∴ V= = × V= V=   V= 4V1 2 π 2 1 d22 1  0.2  1 A2 d 4 2 And π 2 2 A1 V1 4 d1 d12  0.4  V= = × V= V=   V= 1.77V1 3 π 2 1 d32 1  0.2  1 A3 d 4 3 Now using equation of 4 × f1 × L1 × V12 4 × f2 × L2 × V22 4 × f3 × L3 × V32 H= + + d1 × 2g d2 × 2g d3 × 2g

= 16

4 × 0.005 × 400 × V12 4 × 0.005 × 200 × ( 4V1 ) + 0.4 × 2 × 9.81 0.2 × 2 × 9.81

3.189 = 128.58m 0.0248

2

4 × 0.005 × 00 × (1.77V1 ) + 0.3 × 2 × 9.81 2 V1 V12 Example: 16 = (20 + 320 + 63.14) = × 403.14 Three pipes of 400 mm, 200 mm and 300 2 × 9.81 2 × 9.81 mm diameters have lengths of 400m, 200m 16 × 2 × 9.81 V1 = 0.882m / s = 300m respectively. They are connected in 403.14 series to make a compound pipe. The ends ∴ Discharge, of this compound pipe are connected with π two tanks whose difference of water levels Q = A1 × V1 = (0.4)2 × 0.882 = 0.1108 m3 / s 4 is 16m. If co-efficient of friction for these pipes is same and equal to 0.005, (ii) Discharge through the compound determine the discharge through the pipe considering minor losses also compound pipe neglecting first the minor Minor losses are: losses and then considering them. 0.5V12 Solution: hi = 2g Given: (a) At inlet,

= ∴ L1

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2

(b) Between 1st and 2nd pipe, due to contraction,

= hC

=

0.5V22 0.5(4V1 )2 = 2g 2g

= (Q V2 4V1 )

0.5 × 16 × V12 V2 = 8× 1 2g 2g

(c) Between 2nd and 3rd pipe, due to sudden enlargement, 0.5V12 hi = 2g = he

(V2 − V3 )2 (4V1 − 1.77V1 )2 = 2g 2g

V2 V2 = (2.23)2 × 1= 4.973 1 2g 2g

(Q V3 = 1.77V1 )

(d)At the outlet of 3rd pipe,

V 2 (1.77V1 )2 V2 V2 ho = 3 = = 1.772 × 1 = 3.1329 1 2g 2g 2g 2g

The major losses are 4 × f1 × L1 × V12 4 × f2 × L2 × V22 4 × f3 × L3 × V32 = + + d1 × 2g d2 × 2g d3 × 2g

4 × 0.005 × 400 × V12 4 × 0.005 × 200 × (4V1 )2 + 0.4 × 2 × 9.81 0.2 × 2 × 9.81 4 × 0.005 × 300 × (1.77V1 )2 + 0.3 × 2 × 9.81 V12 = 403.14 × 2 × 9.81 ∴ Sum of minor losses and major losses

 0.5V12 V2 V2 V2 V2 =  + 8 × 1 + 4.973 1 + 3.1329 1  + 403.14 1 2g 2g 2g  2g  2g

= 419.746

V12 2g

But total loss must equal to H (or 16m) V2 ∴ 419.746 × 1 = 16 2g

= ∴ V1

16 × 2 × 9.81 = 0.864m / s 419.746

Two pipes have a length L each. One of them has diameter D, and the other has diameter d. If the pipes are arranged in parallel, the loss of head, when a total quantity of water Q flows through them is h, but, if the pipes are arranged in series and the same quantity Q flows through D 2

them, the loss of head is H. If d = , find the ratio of H to h, neglecting secondary losses and assuming the pipe co-efficient has a constant value. Solution: Given: Length & dia. of pipe1 L1 = L and d1 = D Length & dia. of pipe2

L2 = L and d2 = d

Total Discharge =Q Head loss when pipes are arranged in parallel = h Head loss when pipes are arranged in series = H D d= and f is constant 2 1st case: When pipes are connected parallel Q = Q1 +Q2 …(i) Loss od head in each pipe =h 4fL1 V12 For pipe AB, = h, d1 ×2g

where V1 =

d1 = D

∴ Discharge

p Q = A1 V1 = ×(0.4)2 ×0.864 = 0.1085m3 / s 4

Example:



Q1 Q 4Q = 1 = 12 A1 π 2 π D D 4

 4Q  4fL  12  2 2  π D  = h or 32fLQ1 = h D× 2g π2 D5×g

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.....(ii)

32fLQ 22

For pipe AC,



or



π2d5 × g

H = Q2 ×

=h

32fLQ12 32fLQ22 Q15 Q25 or = = π2D5 × g π2d5 × g D5 d5

 Q1  D5 (2d ) 5 ∴D = 2d    = 5 = 5 = 2 = 32 d  Q2  d Q1 = 32 = 5.657 or Q1 = 5.657Q2 Q2 5

2

Substituting the value of Q1 in equation (i), we get Q =5.657Q2 +Q2 =6.657Q2

Q =0.15Q 6.657 From (i) ∴ Q1 = Q − Q2 = Q-0.15Q = 0.85Q

∴Q 2 =

 Q2 Q2  h h Q2 Q2 + Q2 × 2 = 2 h + 2 h = h  2 + 2  2 Q1 Q 2 Q1 Q2  Q1 Q2 

H Q2 Q2 = + h Q12 Q22

But from equations (iv) and (v)

, Q1 = .85Q and Q 2 = 0.15Q H Q2 Q2 1 1 = 2 2+ 2 2= 2+ 2 h .85 Q .15 Q .85 .15 =1.384 + 44.444 =45.828

.....(iv

.....(v

2nd case: When pipes are connected in series Total loss = sum of head losses on two pipes

= H

4f .L.V12 4f .L.V22 + d1 × 2g d2 × 2g

where= V1

Q 4Q Q 4Q ,= V2 = = π 2 πD2 π 2 πd2 D d 4 4

2

 4Q   4Q  4f.L.  2  4f.L.  2  D π   +  πd  ∴= H D × 2g d × 2g

= or H

32fLQ 2 32fLQ 2 + 5 2 5 2 D π ×g d π ×g

From equation(ii),

2

....(vi)

32fl h = 2 5 π D ×g Q1 2

and frome quation (iii),

32fl h = 2 π d ×g Q 2 2 5

Substituting these values in equation (vi), where

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GATE QUESTIONS Q.1

For a fluid flow through a divergent pipe of length L having inlet and outlet radii of R1 and R2 respectively and a constant flow rate of Q, assuming the velocity to be axial and uniform at any cross-section, the acceleration at the exit is 2Q(R1 − R 2 ) 2Q 2 (R1 − R 2 ) a) b) πLR 32 πLR 32 c)

Q.2

Q.4

2Q 2 (R 2 − R1 ) π 2 LR 52 [GATE–2004]

d)

A siphon draws water from a reservoir and discharges it out at atmospheric pressure. Assuming ideal fluid & the reservoir is large, the velocity at point P in the siphon tube is

a)

Q.3

2Q 2 (R1 − R 2 ) π 2 LR 52

c)

2gh1 2g ( h 2 − h1 )

b)

2gh 2

Q.5

Q.6

d 2g ( h 2 + h1 )

[GATE–2006]

Water at 25℃ is flowing through a 1.0 km long C.I. pipe of 200 mm diameter at the rate of 0.07 m3/s. If value of Darcy friction factor for this pipe is 0.02 and density of water is 1000 kg/m3 , the pumping power (in kW) required to maintain the flow is a) 1.8 b) 17.4 c) 20.5 d) 41.0 [GATE–2009] A smooth pipe of diameter 200 mm carries water. The pressure in the

Q.7

pipe at sectionS1 (elevation: 10 m) is 50 kPa. At sectionS2 (elevation: 12m) the pressure is 20 kPa and velocity is 2 m/s. Density of water is 1000 kg/m3 and acceleration due to gravity is 9.8 m/s2. Which of the following is TRUE a) flow is fromS1 to S2 and head loss is 0.53 m b) flow is fromS2 to S1 and head loss is 0.53 m c) flow is fromS1 to S2 and head loss is1.06 m d) flow is fromS2 to S1 and head loss is 1.06 m [GATE–2010] Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225) of length 500 m. The volumetric flow rate is 0.2 m3/s. The head loss (in m) due to friction is (assume g = 9.81 m/𝑠𝑠 2 ) a) 116.18 b) 0.116 c) 18.22 d) 232.36 [GATE–2013] For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, the pressure drop ∆p over a length L and the wall shear stress τW are related by ∆pD 2 ∆pD a)τw= b)τw= 4L2 4L ∆pD 4∆pL c)τw= d)τw= 2L D [GATE–2013] Water flows through a 10 mm diameter and 250 m long smooth pipe at an average velocity of 0.1 m/s. The density and the viscosity of water are 997 kg/m3 and 855 ×

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Q.8

Q.9

10−6 Ns/m2, respectively. Assuming fully-developed flow, the pressure drop (in Pa) in the pipe is ____. [GATE–2014(2)] A fluid of dynamic viscosity 2 × 10−5 kg/ms and density 1 kg/m3 flows with an average velocity of 1 m/s through a long duct of rectangular (25mmx 15 mm) crosssection. Assuming laminar flow, the pressure drop (in Pa) in the fully developed region per meter length of the duct is _____. [GATE–2014(3)] A siphon is used to drain water from a large tank as shown in figure below. Assume that the level of water is maintained constant. Ignore frictional effect due to viscosity and losses at entry and exit. At the exit of the siphon, the velocity of water is

a) c)

2g(ZQ − ZR )

2g(ZO − ZR )

b)

d)

2g(ZP − ZR ) 2gZQ

[GATE–2014(3)]

Q.10 Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, density and friction factor to be constant, if the length of the pipe is doubled and the diameter is halved, the head loss due to friction will increase by a factor of a) 4 b) 16 c) 32 d) 64 [GATE–2015(1)]

Q.11 Three parallel pipes connected at the two ends have flow-rates Q1, Q2 and Q3 respectively, and the corresponding frictional head losses are hL1, hL2 and hL3 respectively. The correct expressions for total flow rate (Q) and frictional head loss across the two ends (hL) are a) Q= Q1+ Q2 + Q3; hL = hL1+ hL2+ hL3 b) Q = Q1+ Q2 + Q3; hL= hL1= hL2= hL3 c) Q= Q1= Q2 = Q3; hL = hL1+ hL2+ hL3 d) Q = Q1 = Q3; hL = hL1= hL2= hL3 [GATE–2015(3)] Q.12 The head loss for a laminar incompressible flow through a horizontal circular pipe is h1. Pipe length and fluid remaining the same, if the average flow velocity doubles and the pipe diameter reduces to half its previous value, the head loss is h2. The ratio h2/h1 is a) 1 b) 4 c) 8 d) 16 [GATE–2015(2)] Q.13 A channel of width 450 mm branches into two sub-channels having width 300 mm and 200 mm as shown in figure. If the volumetric flow rate (taking unit depth) of an incompressible flow through the main channel is 0.9 m3/s and the velocity in the sub-channel of width 200 mm is 3 m/s, the velocity in the sub-channel of width 300 mm is _____________ m/s. Assume both inlet and outlet to be at the same elevation.

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[GATE–2016(3)] Q.14

For steady flow of a viscous incompressible fluid through a circular pipe of constant diameter, the average velocity in the fully developed region is constant. Which one of the following statements about the average velocity in the developing region is TRUE? a) It increases until the flow is fully developed. b) It is constant and is equal to the average velocity in the fully developed region. c) It decreases until the flow is fully developed. d) It is constant but always lower than the average velocity in the fully developed region.

Q.16 Water (density =1000 kg/m3) at ambient temperature flows through a horizontal pipe of uniform cross section at the rate of 1 kg/s. If the pressure drop across the pipe is 100 kPa, the minimum power required to pump the water across the pipe, in watts, is _______

[GATE–2017(1)] Q.15

Consider steady flow of an incompressible fluid through two long and straight pipes of diameters d1 and d2 arranged in series. Both pipes are of equal length and the flow is turbulent in both pipes. The friction factor for turbulent flow though pipes is of the form, f =K ( Re )

−n

where K

and n are known positive constants and Re is the Reynolds number. Neglecting minor losses, the ratio of the frictional pressure drop in pipe 1 to that in pipe 2, is given by

d  (A)  2   d1 

d  (C)  2   d1 

( 5− n )

( 3− n )

d  (B)  2   d1 

5

d  (D)  2   d1 

( 5+ n )

[GATE–2017(1)]

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[GATE–2017(1)]

ANSWER KEY: 1 (c) 14 (b)

2 (c) 15 (a)

3 (a) 16 100

4 (c)

5 (a)

6 (a)

7 6840

8 1.82

9 (b)

10 (d)

11 (b)

12 (c)

13 1

EXPLANATIONS Q.1

(c)

∂u ∂u ∂u ∂u +v +w + ∂x ∂y ∂z ∂t ∂u ∴ ax = u ∂x A×u =Q Q u= A Q u= πR 2

∂u ∂x ∂u a x =L = u ∂x x = L and also replacing a and b ax = u

ax = u

R 2 − R1 .x L R − R1 Let. R1 a= = and 2 b L ∴R = a+bx Q u= π(a + bx) 2 ∂u −2bQ = ∂x π(a + bx)3 R = R1 +

a x =L =

Q.2

−2Q 2 (R 2 − R1 ) π 2 LR 52

(c)

In a steady and ideal flow of incompressible fluid, the total energy at any point of the fluid is constant. So applying the Bernoulli’s Equation at section (1) and (2) P1 V12 P2 V22 + + Z1 = + + Z2 ρg 2g ρg 2g V1= 0 = Initial velocity at point (1) Z2 = 0 = At the bottom surface P1= P2= Patm

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And Z1= h2−h1 V12 = 2g ( h1 − h 2 ) 2g V2 So, h 2 − h1 =1 2g

P1 P2 = + Z1 + Z2 + h L ρg ρg P −P  = h 2  1 2  + ( z1 − z 2 )  pg  ( 50 − 20 ) ×102 + 10 − 12 = ( ) = V2 2g ( h 2 − h1 ) (100 × 9.8) = 3.058 − 2 =1.06m = +ve Since velocity of fluid is same inside Thus flow is from S1 to S2 and head the tube loss is 1.06 m. V= V= 2g ( h 2 − h1 ) p 2

Q.3

(a) Given: L = 1 km = 1000 m, D =200 mm = 0.2m, Q = 0.07 m3/ s F = 0.02, ρ= 1000 kg/m3 Head loss is given by, FLV 2 8FLQ 2 = hf = D × 2g π 2 D5g

Q.5

Q.6

(a) From Darcy Weisbach equation head loss fLV 2 hf = D × 2g 200 Given that L= 500 m, D = = 1000 0.2m. f=0.0225 Since volumetric flow rate is given by Q = Area × velocity 0.2 V= = 6.366 π / 4 × 0.22 0.0225 × 500 × 6.3662 Hence, h f = 0.2 × 2 × 9.81 hf = 116.18 m

Q.7

(6840)

8 × 0.02 ×1000 × 0.07 2 = 25.30 m π 2 × 0.25 × 9.81 Pumping power required. P = ρgQhf = 1000×9.81×0.07×25.3 =17373.51 = 17.4 kW hf =

Q.4

(c) Given: p1= 50 kPa, Z1= 10 m, p2= 20 kPa, Z2= 12 m,V2 =2 m/ sec, ρ= 1000 kg/m3, g = 9.81 m/ sec2

Applying continuity equation at section S1 and S2, A1 V1= A2V2 V1= V2, D1= D2 so A1= A2... (i) Applying Bernoulli’s equation at section S1 and S2 with head loss hL, P1 V12 P2 V22 = + + Z1 + + Z2 + h L ρg 2g ρg 2g As area is constant, velocity head will be same

Q.8

(a) For steady, fully developed flow inside a straight pipe, the pressure drop and wall shear stress are ∆pD related by τw= 4L

32μuL P1 − P2 = 2 D 32 × 855 ×10−6 × 0.1× 250 = (0.01) 2 ∴ P1 − P2 = 6840Pa (1.82) 32μuL ∆P = 2 D

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∆P 32μu = L D2 4A 4(25 ×15) = D= P 2[25 + 15] 4 × 25 ×15 = = 18.75mm 80 ∆P 32 × 2 ×10−5 ×1 = ∴ L (18.75 ×10−3 ) 2 = 1.82 Pa/m

h2 =8 h1



Q.9

(b) Applying Bernoulli’s equation between P and R: V2 V2 Patm + P + ZP = R + ZR + Patm 2g 2g

VR2 = (ZP — ZR ) 2g (∵VP = 0for still water) ∴ VR = 2g(ZP — ZR ) Or

Q.15

2

( )

Q.11

Q.12

(b) Total flow rateQ = Q1+ Q2 + Q3 Head loss = h h= h= h L3 L1 L2 (c)

Head loss, h ∝

u avg D2

(a)

k k = ⇒ f ∝ Dn n n (Re)    ρ.Q.D  π   D 2 .µ  4  2 ∆P flQ 8 Q= × 2 5 ρg gD π f ∆P ∝ 5 D ∆P ∝ D n −5

= f

Q.10 (d)

Q FL   2 FLV A = Head loss = 2gd 2gd L ⇒ head loss ∝ 5 d L h1* ∝ 5 d 2.L L =× h *2 ∝ 64 5 d5 d 2 h1 1 h ⇒ 2 = 64 = h 2 64 h1

Q.13 (1) Let the velocity in the sub-channel of width 300mm be V. ∴ 0.9 = (0.3V) + (0.2×3) ∴ 0.3V= 0.3 V= 1 m/s Q.14 (b) Q Vavg = A As given steady flow means Q → constant Answer also given constant diameter pipe, therefore area also constant. ∴ Vavg = Cons tan t

∆P  D1  ∴ 1 =   ∆P2  D 2  or

n −5

 ∆P1   D 2  = =     ∆P2   D1 

5− n

Q.16 (100) Q power= Q × ∆P m = × ∆P ρ

=

1×100 ×103 = 100 watt 1000

2

h 2  D1  u avg 2 ∴ = = 22 × 2   h1  D 2  u avg1

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8

EXTERNAL FLOW

8.1 BOUNDARY LAYER FORMATION The variation of velocity from zero to free stream velocity in the direction normal to the boundary takes place in a narrow region in the vicinity of solid boundary. Boundary layer is a very thin layer of the fluid, in the immediate neighbourhood of the solid boundaries where the variation of velocity is from zero to free stream velocity.

du dy Fluid exerts a shear stress on the wall in the direction of motion. The value of shear µdu stress is given by τ = dy The velocity gradient is set up in the fluid near the surface of the plate. This velocity gradient develops shear resistance which retards the fluid. Downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded. The velocity gradient =

8.2 REGIONS OF BOUNDARY LAYER

Boundary layer can be divided in three major regions: 1. Laminar 2. Transition 3. Turbulent 4. Laminar sub-layer

8.2.1 LAMINAR BOUNDARY LAYER In laminar region the fluid flows in streamline. The viscous force is higher than inertia force. The Reynold’s No. for flat plate is given by U x Re x    Where, U ∞ = free stream velocity

X = distance from leading edge ν  = kinematic viscosity of fluid

For laminar flow R ex < 5 × 105

8.2.2 TURBULENT BOUNDARY LAYER When the length of plate is more than the critical length ‘x cr ’ , calculated from  Re 5105   equation    U then, transition from laminar to turbulent takes place. 8.2.3 LAMINAR SUBLAYER

It is region in the turbulent zone adjacent to the solid surface of the plate the velocity variation is influenced only by viscous effects. Though velocity distribution is

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parabolic for small thickness, we can reasonably assume that velocity variation is linear here  du   ∆u  τo = µ   ⇒ µ    dy   ∆y 

τo is constant in laminar sub-layer.

8.3 BOUNDARY LAYER THICKNESS It is defined as the distance from the boundary of the solid measured in the ydirection to the point, where the velocity of fluid is approximately equal to 0.99 times free stream velocity ( U ∞ )

Slam ,Stur ,S' laminar sub layer

8.3.1 DISPLACEMENT THICKNESS (δ) * It is defined as displacement of surface in the direction normal to the surface to compensate for the reduction in the flow rate, due to boundary layer formation.

8.3.3 ENERGY THICKNESS It is defined as the distance measured normal to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in K.E. of the flowing fluid. 2   u  1 u dy  **   0 U  U 2  



8.4 DRAG FORCE ON FLAT PLATE DUE TO BOUNDARY LAYER 8.4.1 VON KARMAN INTEGRAL EQUATION τ0 dθ = 2 ρU∞ dx

MOMENTUM

 dθ  = τ0   ρU ∞ 2  dx  τ0 is shear stress. θ is momentum thickness

The above equation is valid for 1. Zero pressure gradient. 2. Laminar & turbulent flow 3. Incompressible steady flow.

The plate is displaced by distance (δ) * given by δ  u  * δ= ∫0 1 − U∞ dy 8.3.2 MOMENTUM THICKNESS It is defined as the distance, measured normal to the boundary of the solid body, by which the boundary should be displaced to compensate for the reduction in momentum of the flowing fluid on account of boundary layer formation. δ u  u  = θ ∫ 1 − dy U∞  U∞  0

8.4.2 LOCAL COEFFICIENT OF DRAG τ0 cd* = 1 2   ρU∞  2  Local shear stress in laminar flow is obtained by du τ0 =µ dy Average coefficient of drag Favg cd = 1 2   ρU∞  2  L

Favg =

∫ τ .b.dx o

o

b is the width of plate

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L is the length up to which average force is to be evaluated 8.5 BOUNDARY CONDITION FOR THE VELOCITY PROFILE du y 0,= U 0, 1) at= has some finite value. dy U∞ , δ, u = 2) at y = du δ, = 0, 3) at y = dy Assume velocity profile to be 2

2U d  2δ  2 µ=   ρU δ dx  15  15µdx ρdδ = δU

δ2 15µ = x +C 2 ρU

at= x 0,δ = 0,C = 0 at 

δ=

30µx2 ρUx

y y u = a + b  + c   U∞ δ δ ρux Using boundary conditions, value of = Re coefficients obtained are μ a = 0 , b = 2 , c = −1 , x2 δ = 30 . The velocity profile for laminar boundary Re x layer flows is given by Shear stress in terms of Reynolds No u = 2(y / δ) − (y / δ) 2 2µU U∞ τo = δ Equation involved τ0 do µU ρUx 2µU = = τo = 0.365 2 ρU dx x µ x 5.48 U is free stream velocity Rex µdu τo = Coefficient of drag dy F0 2 cD = du d   y   y   1 2  −    = 2  ρAU  dy dy   δ   δ     2  L du  2 2Y  = U  − 2  at y=0 FD = ∫τo .b.dx dy δ δ  o du 2U 2u Where b is thickness of plate = , τ=µ .... 1 dy δ δ L µU ρUx δ FD = ∫ 0.365 gbdx u  u 0 x µ = θ ∫ 1 −  dy

()

o

U

U

2   Y   Y 2   Y Y     = θ ∫ 2  −   1 − 2  +    dy δ   δ    δ   δ   o    δ

2δ θ= 15

dθ d  2δ  =   … (2) dx dx  15 

= FD 0.365µU

= FD 0.73bµU

L 1 ρU dx gb∫ 0 µ x

ρUL µ

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Cd =

ρUL µ

0.73bµU

1 ρAU2 2 1.46µ ρUL Cd = ρU µ

Cd = 1.46 Cd =

1.46

µ ρUL

If R e > 5 × 105 & less then107 δ=

0.37x

(R )

CD =

ex

1/5

0.072

,

(R ) eL

1/5

Where, x = distance from leading edge R e = Reynold’s No. for length x x

R e = Reynold’s No at the length ‘L’ of plate L

For R e > 107 ,R e > 109

ReL

8.6 ANALYSIS OF TURBULENT BOUNDARY LAYER

8.6.1 TURBULENT BOUNDARY LAYER ON FLAT PLATE Blasius, on basis of his experiments gave expression for velocity profile in turbulent flow over falt plate n

u y =  U∞  δ  Where = n 1 / 7,R e < 107

CD =

0.455

( log10R ) eL

2.58

Example: Determine the thickness of the boundary layer at the trailing edge of smooth plate of length 4m & width 1.5 m, when the plate is moving with a velocity of 4 m/s. In stationary air, kinematic viscosity ν 1.5 × 10−5 m2 / s . =

Solution: UL 4× 4 Re = = = 10.66 × 105 L ν 1.5 × 10−5 Turbulent 1/7 u y 0.37x =  = δ = 92.19mm U∞  δ  1 5 Rex Since shear stress in turbulent flow is given by 1/4 8.7 BOUNDARY LAYER SEPARATION  μ 

(

τ o = 0.0225ρu2 

  ρδu 

8.6.2 BOUNDARY LAYER THICKNESS IN TURBULENT FLOW 1

 µ  5 45 δ =0.37   gx U ρ  ∞ 0.37x δ= 1 ( Rex ) 5

8.6.3 COEFFICIENT OF DRAG

)

Separation of flow is said to occur when the direction of the flow velocity near the surface is opposite to the direction of the free stream velocity, which means (du/dy) ≤ 0. If (dp/dx) increases to the extent that it can overcome the shear near the surface, then separation will occur. Such a pressure gradient is called adverse pressure gradient. In the case of incompressible flow in a nozzle a favourable pressure gradient exists & Separation will not occur in such flows. In the case of diverging section of a diffuser,

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separation can occur if the rate of area increase is large.

2      y   y    1 2 − −  ∫0   δ   δ    dy      δ

* δ=

δ 2   2y 2 y 3   y   y   =∫ 1 − 2  +   dy = y − +  2δ 3δ2  0 δ δ   0    δ

δ2 δ3 δ δ + 2 = δ−δ+ = δ 3δ 3 3 ( ii ) Momentum thickness θ is given by =δ−

δ

= θ

u u  dy ∫0 U  1 − U = 

 2y y 2    2y y 2   ∫0  δ − δ2  1 −  δ − δ2  dy   δ

 2y y 2   2y y 2  ∫0  δ − δ2  1 − δ + δ2  dy δ

=

 2y 4y 2 2y 3 y 2 2y 3 y 4  ∫0  δ − δ2 + δ3 − δ2 + δ3 − δ4 dy δ

8.7.1 LOCATION OF SEPARATION POINT

=

 ∂u    < 0 Separated flow  ∂y y =0

 δ2 5δ3 δ4 δ5  δ 5δ =  − 2 + 3 − 4  =δ− +δ− δ 3 5 δ δ δ 3 5   15δ − 25δ + 15δ − 3δ 30δ − 28δ 2δ = = = 15 15 15

 ∂u    = 0 Verge of separation  ∂y  y =0  ∂u    > 0 No separation  ∂y  y =0

Example: Find the displacement thickness, the momentum thickness and energy for the velocity distribution in the boundary layer

y y u = 2  −   given by U δ δ Solution: Given:

2

by

δ= *

δ



u

∫  1 − U  dy 0

( iii ) Energy thickness u  u2  ∫0 U 1 − U2  dy= δ

* δ *=

δ * * is given by

2  2y y 2    2y y 2    1 −  −   dy −   ∫0  δ δ2    δ δ2     δ

 2y y 2    4y 2 y 4 4y 3   1 − −  ∫0  δ δ2    δ2 + δ4 − δ3   dy   δ

=

 2y y 2  4y 2 y 4 4y 3  1 − − − +   dy ∫0  δ δ2  δ2 δ4 δ3  δ

=

 2y 8y 3 2y 5 8y 4 y 2 4y 4 y 6 4y 5  ∫0  δ − δ3 − δ5 + δ4 − δ2 + δ4 + δ6 − δ5 dy δ

=

y y u Velocity distribution, = 2  −   U δ δ

( i ) Displacement thickness

δ

 2y 5y 2 4y 3 y 4   2y 2 5y 3 4y 4 y 5  ∫0  δ − δ2 + δ3 − δ4 dy =  2δ − 3δ2 + 4δ3 − 5δ4  0 δ

=

2

 2y y 2 8y 3 12y 4 6y 5 y 6  ∫0  δ − δ2 − δ3 + δ4 − δ5 + δ6 dy δ

=

δ * is given

δ

 2y y 3 8y 4 12y 5 6y 6 y 7  = − 2− 3+ − +  5δ4 6δ5 7δ6  0  2δ 3δ 4δ

= 2

y y u Substituting the value of,= 2  −   we U δ δ have

=

δ2 δ3 2δ4 12δ5 δ6 δ7 δ δ 12 − 2 − 3 + 4 − 5 + 6 = δ − − 2δ + δ − δ + δ 3δ 3 5 7 δ δ 7δ 5δ

δ 12 δ −210δ − 35δ + 252δ + 15δ = −2δ − + δ + = 3 5 7 105 −245δ + 267δ 22δ = 105 105

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Example: Air is flowing over a smooth plate with a velocity of 10 m/s. The length of the plate is 1.2m and width is 0.8m. If laminar boundary layer exists up to a value of R e = 2 × 105 , find the maximum distance from the leading edge upto which laminar boundary layer exists. Find the maximum thickness of laminar boundary layer if the velocity profile is given by 2

y y u = 2  −   U δ δ Take kinematics viscosity for air = 0.15 stokes. Solution: Given: U =10m/ s Velocity of air, Length of plate, L = 1.2m Width of plate, b= 0.8m Reynolds number upto which laminar

plate is moving with a velocity of 4m/s in stationary air. Take kinematics viscosity of air as 1.5×10−5 m2 / s Solution: Given: Length of plate, Width of plate, Velocity of plate,

b=1.5m U = 4m/ s Kinematics viscosity,   1.5105 m 2 / s

Reynolds number, 4 4 U L Re L    10.66105 5 1.510  As the Reynolds number is more than

5 × 105 and hence the boundary layer at the

trailing edge is turbulent. The boundary layer thickness for turbulent boundary layer is given by δ=

= boundary exists = 2 × 10 ν for = air 0.15 stokes = 0.15 × 10−4 m2 / s 5

= Reynolds number, Re L

ρUx Ux = µ ν

If R e = 2 × 10 , then x denotes the distance x

5

from leading edge upto which laminar boundary layer exists 10 × x ∴ 2 × 105 = 0.15 × 10−4

∴x =

2 × 105 × 0.15 × 10−4 = 0.30m = 300mm 10

Maximum thickness of the laminar boundary for the velocity profile, 2

y y u = 2  −   Is given by U δ δ

= δ

L = 4m

(R ) ex

(

(

Here x 0.37x= 1

5

L= and R e R e

L

)

0.37 × 4.0 = 0.0921m = 92.1mm 1 10.66 × 105 5

)

Example: Water is flowing over a thin smooth plate of length 4m and width 2m at a velocity of 1.0m/s. If the boundary layer flow changes from laminar to turbulent at a Reynolds number 5 × 10 , find ( i ) the distance from 5

leading edge upto which boundary layer is laminar, ( ii ) the thickness of boundary layer at the transition point, and ( iii ) the

drag force on one side of the plate. Take viscosity of water µ = 9.81×10−4 Ns/ m2 .

5.48 × x 5.48 × 0.30 = = 0.00367m = 3.67mm 5 Re 2 × 10 x

Example: Determine the thickness of the boundary layer at the trailing edge of smooth plate of length 4m and of width 1.5m, when the

x

Solution: Given:

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L = 4m Length of plate, b= 2m Width of plate, U =1.0m/ s Velocity of flow, Reynolds number for laminar boundary

layer = 5 × 10 Viscosity of water, µ = 9.81×10−4 Ns/ m2 i) Let the distance from leading edge upto which laminar boundary layer exists = x Ux 10001 x  5105    9.81104 5

−4 5 ∴ x = 5 × 10 × 9.81 × 10 1000 = 0.4900m = 490mm

b) Drag force due to turbulent boundary layer from F to G =Drag force due to turbulent boundary layer from E to G − Drag force due to turbulent flow from E to F = FEG − FEF

( )

Now,

( )

turb

turb

1 AU2 × CD ( FFG )turb =ρ 2

Where CD is CD =

But

0.072

(R ) eL

1

5

ρUL

1.0 × 4.0

=× R e ==× 1000 40.77 105 ii) Thickness of boundary layer at the −4 µ 9.81 × 10 point where the boundary layer 0.072 changes from laminar to turbulent i.e.,= = 0.00343 ∴ C L

at Reynolds number = 5 × 10 , is given by Blasius’s solution as 5x 5g0.49 = δ = = 3.46mm Rex 5 × 105

D

5

(Here x=

49cm= 0.49m, R e = 5 × 10

5

x

)

iii) Drag force on the plate on one side = Drag due to laminar boundary layer + Drag due to turbulent boundary a) Drag due to laminar boundary layer (i.e., from E to F) 1 FEF = ρ AU2×CD ...... ( i ) 2 Where CD is given by Blasius solution for laminar boundary layer as CD =

1.328 Re

x

=

1.328

5

5×10

( for EF,R

ex

5

=5×10

)

= 0.001878 A=Area of plate upto laminar boundary layer = 0.49×b= 0.49×2= 0.98 m2

Substituting the value of CD and A in equation ( i ) , we

get

1 FEF = ×1000×0.98×1.02 ×.001878 = 0.92N 2

.... ( ii )



(F )

( 40.77 × 105)

EG turb

1

1 =ρAU2 × CD 2

5

1 = × 1000 × (4 × 2) × 12 × .00343 13.72N = 2 1 FEF = ρA EF × U2 × CD Also turb 2 Where A EF =Area of plate upto

( )

EF = EF× b = 0.49×2= 0.98 m2 And 0.072 0.072 CD = = = .00522 1 1 5 5 5 5×10 ( R EF )

(

(F )

EF turb

)

1 = ×1000×0.98×12 ×.00522= 2.557N 2

∴ Drag force due to turbulent boundary

layer from F to G = ( FEG ) - ( FEF ) =13.72-2.557 =11.163N turb

turb

∴ Drag force on the plate on one side =Drag force due to laminar boundary layer upto F + Drag force due to turbulent boundary layer from F to G = 0.92+11.163 = 12.083N

Example: For the following velocity profiles, determine whether the flow has separated

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or on the verge of separation or will attach with the surface. 3

u 3 y  1 y  = i)  −   U 2 δ  2 δ  2

y y u −2  +   iii) = U δ δ Solution: Given 1st velocity profile

separation.

3rd velocity profile

3

y y u = 2  −   ii) U δ δ

 δu   = 0 , the flow is on verge of δ y   y =0

As 

y y u = −2  +   U δ δ

2

2

y y ∴u = −2U   + U   δ δ

3

3

u 3 y  1 y  3U  y  U  y  =−     or u =  −   U 2 δ  2 δ  2 δ 2δ

Differentiating w.r.t., y, the above equation becomes, 2

δ u 3U 1 U  y  1 = × − ×3  × δy 2 δ 2 δ δ

At y =0 ,

2

 δu  3U 3U  0  1 3U −   × =   =  δ y  y =0 2δ 2  δ  δ 2δ  u  As   is positive. Hence flow will not   y 

2

1 y 1 δu = −2U   + 2U   × δy δ δ δ

At y=0,  δu  0 1 2U 2U − + 2U   × = −   = δ δ  δy  y =0 δ δ

 δu    is negative the flow  δy  y =0 separated.

As

y 0

separate or flow will remain attached with the surface. 2nd velocity profile 2

3

y y u = 2  −   U δ δ 2

3

y y = ∴ u 2U   − U   δ δ

2

y 1 y 1 δu = 2U × 2  × − U × 3  × ∴ δy δ δ δ δ

 δu    δ y  y =0

At y = 0, 

2

0 1 0 1 = 2U × 2  × − U × 3  × = 0 δ δ δ δ

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has

GATE QUESTIONS Q.1

Flow separation in flow past a solid object is caused by a) a reduction of pressure to vapour pressure b) a negative pressure gradient c) a positive pressure gradient d) the boundary layer thickness reducing to zero [GATE–2002]

Q.2

If x is the distance measured from the leading edge of a flat plate, the laminar boundary layer thickness varies as 1 a) b) x 4/5 x c) x 2 d) x1/2 [GATE–2002]

Common Data Questions Q.3 to Q.4 A syringe with a frictionless plunger contains water and has at its end a 100 mm long needle of 1 mm diameter. The internal diameter of the syringe is 10 mm. Water density is 1000 kg/m3. The plunger is pushed in at 10 mm/s and the water comes out as a jet.

Q.3

Assuming ideal flow, the force F in newtons required on the plunger to push out the water is a) 0 b) 0.04 c) 0.13 d) 1.15 [GATE–2003]

Q.4

Neglect losses in the cylinder and assume fully developed laminar viscous flow throughout the needle;

the Darcy friction factor is 64/Re where Re is the Reynolds number. Given that the viscosity of water is 1.0 × 10−3 kg/s-m, the force F in newtons required on the plunger is a) 0.13 b) 0.16 c) 0.3 d) 4.4 [GATE–2003] Q.5

Q.6

An incompressible fluid (kinematic viscosity, 7.4 × 10−7 m2/s , specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surfaces of top plate is a) 0.651×103 b) 0.651 c) 6.51 d) 0.651×103 [GATE–2004] For air flow over a flat plate, velocity (U) and boundary layer thickness (δ) can be expressed respectively, as 3

U 3 y 1 y 4.64x =−   ;δ = U∞ 2 δ 2  δ  Re x If the free stream velocity is 2 m/s, and air has kinematic viscosity of 1.5 × 10−5 m2/s and density of 1.23 kg/m3, the wall shear stress at x = 1 m, is a) 2.36 × 102 N/m2 b) 43.6 × 10-3 N/m2 c) 4.36 × 10-3 N/m2 d) 2.18 ×10-3 N/m2 [GATE–2004]

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Q.7

The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u= u0(1 -4r /D2), where r is the radial distance from the center. If the viscosity of the fluid is μ, the pressure drop across a length L of the pipe is 4µu 0 L µu 0 L a) b) 2 D2 D 16µu 0 L 8µu 0 L c) d) 2 D2 D [GATE–2006]

Common Data For Q.8 and Q.9 A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U=10 m/s. The thickness of the boundary layer at section r-s is 10 mm, the breadth of the plate is 1m (into the paper) and the density of the gas ρ = 1.0 kg/m3. Assume that the boundary layer is thin, twodimensional, and follows a linear velocity distribution, u = U(y/δ), at the section r-s, where y is the height from plate.

plate to the drag force on the rear half, then a) F < 1/2 b) F = 1/2 c) F= 1 d) F> 1 [GATE–2007] Q.11 Consider steady laminar incompressible anti-symmetric fully developed viscous flow through a straight circular pipe of constant cross-sectional area at a Reynolds number of 5. The ratio of inertia force to viscous force on a fluid particle is a) 5 b) 1/5 c) 0 d) [GATE–2007] Common Data For Q.12 and Q.13 : Consider a steady incompressible flow through a channel as shown below.

The velocity profile is uniform with a value of U0at the inlet section A. The velocity Q.8 The mass flow rate (in kg/s) across profile at section B downstream is the section q - r is y  0≤y≤δ a) zero b) 0.05  Vm δ ,  c) 0.10 d) 0.15 = u  Vm , δ ≤ y ≤ H−δ [GATE–2006]  H−y Vm H−δ ≤ y ≤ H δ Q.9 The integrated drag force (in N) on  the plate, between p-s, is a) 0.67 b) 0.33 Q.12 The ratio Vm/U0is c) 0.17 d) zero 1 a) b) 1 [GATE–2006] 1 − 2(δ − H) 1 1 Q.10 Consider an incompressible laminar c) d) 1 − (δ / H) 1 + (δ / H) boundary layer flow over a flat plate [GATE–2007] of length L, aligned with the direction of an incoming uniform free stream. If F is the ratio of the drag force on the front half of the © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

pA − pB (where pAand 1 2 ρU 0 2 pBare the pressures at section A and Brespectively, and ρis the density of the fluid) is 1 1 a) b) −1 2 2 [1 + (δ / H)] [1 − (δ / H)]

Q.13 The ratio

c)

1 1 − 1 d) 2 1 + (δ / H) (1 − (2δ / H) [GATE–2007]

Q.14 The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression R 2  dp   r2  u(r) = −    1 − 4μ  dx   R 2 

dp is a constant. The dx average velocity of fluid in the pipe is where

R 2  dp  a) −   8μ  dx  c) −

R 2  dp    2μ  dx 

R 2  dp  b) −   4μ  dx 

R 2  dp    μ  dx  [GATE–2009]

d) −

Q.15 A phenomenon is modeled using n dimensional variables with k primary dimensions. The number of non-dimensional variables is a) k b) n c) n− k d) n+k [GATE–2010] Q.16 The maximum velocity of a onedimensional incompressible fully developed viscous flow, between

two fixed parallel plates, is 6 m/s. The mean velocity (in m/s) of the flow is a) 2 b) 3 c) 4 d) 5 [GATE–2010] Q.17 Match the following P.Compressible flow Q. Free surface flow R.Boundary layer flow S.Pipe flow T. Heat convection

U.Reynolds number V.Nusselt number W. Weber number X.Froude number Y. Mach number

a) P-U; Q-X; R-V; S-Z; T-W b) P-W; Q-X; R-Z; S-U; T-V c) P-Y; Q-W; R-Z; S-U; T-X d) P-Y; Q-W; R-Z; S-U; T-V [GATE–2010(ME)] Q.18 An incompressible fluid flows over a flat plate with zero pressure gradient. The boundary layer thickness is 1 mm at a location where the Reynolds number is 1000. If the velocity of the fluid alone is increased by a factor of 4, then the boundary layer thickness at the same location, in mm will be a) 4 b) 2 c) 0.5 d) 0.25 [GATE–2012] Q.19 Consider the turbulent flow of a fluid through a circular pipe of diameter, D. Identify the correct pair of statements. I. The fluid is well-mixed II. The fluid is unmixed III. ReD< 2300 IV. ReD> 2300 a) I and III b) II and IV c) II and III d) I and IV [GATE–2014(3)] Q.20 Water flows through a pipe having an inner radius of 10 mm at the rate of 36 kg/hr at 25°C. The viscosity of water at 25°C is 0.001 kg/m.s. The Reynolds number of the flow is_____. [GATE–2014 (1)]

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Q.21 For a fully developed flow of water in a pipe having diameter 10 cm, velocity 0.1 m/s and kinematic viscosity 10−5 m2/s, the value of Darcy friction factor is _____. [GATE–2014 (1)] Q.22 In a simple concentric shaft-bearing arrangement, the lubricant flows in the 2 mm gap between the shaft and the bearing. The flow may be assumed to be a plane Couette flow with zero pressure gradient. The diameter of the shaft is 100 mm and its tangential speed is 10 m/s. The dynamic viscosity of the lubricant is 0.1 kg/ms. The frictional resisting force (in newton) per 100 mm length of the bearing is ____. [GATE–2014 (1)] Q.23 Consider laminar flow of water over a flat plate of length 1 m. If the boundary layer thickness at a distance of 0.25 m from the leading edge of the plate is 8 mm, the boundary layer thickness (in mm), at a distance of 0.75 m, is _____ [GATE–2014 (2)] Q.24 Couette flow is characterized by a) steady, incompressible, laminar flow through a straight circular pipe b) fully developed turbulent flow through a straight circular pipe c) steady, incompressible, laminar flow between two fixed parallel plates d) steady, incompressible, laminar flow between one fixed plate and the other moving with a constant velocity [GATE–2015 (3)] Q.25 Air (ρ= 1.2 kg/m3 and kinematic viscosity, υ = 2 × 10−5 m2/s) with a velocity of 2 m/s flows over the top surface of a flat plate of length 2.5 m.

If the average value of friction 1.328 coefficient is Cf = the total Re x drag force (in N) per unit width of the plate is _____. [GATE–2015 (1)] Q.26 For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm, the axial pressure gradient is −10Pa/m. The magnitude of axial velocity (in m/s) at the radial location of 0.2 cm is __________. [GATE–2015 (2)] Q.27 The instantaneous stream-wise velocity of a turbulent flow is given as follows: u(x, y, z, t) = u(x, y, z) + u′(x, y, z, t) The time-average of the fluctuating velocity u’(x, y, z, t) is a) 𝑢𝑢′/2 b) − u /2 c) zero d) 𝑢𝑢 /2 [GATE–2016 (1)] Q.28 Oil (kinematic viscosity, ν𝑜𝑜𝑜𝑜𝑜𝑜 = 1.0 × 10−5 m2/s) flows through a pipe of 0.5 m diameter with a velocity of 10 m/s. Water (kinematic viscosity, ν𝑤𝑤 = 0.89 × 10−6 m2/s) is flowing through a model pipe of diameter 20 mm. For satisfying the dynamic similarity, the velocity of water (in m/s) is __________ [GATE–2016 (1)] Q.29 A steady laminar boundary layer is formed over a flat plate as shown in the figure. The free stream velocity of the fluid is Uo. The velocity profile at the inlet a-b is uniform, while that at a downstream location c-d is   y   y 2  given by 𝑢𝑢 =  2   −    .   δ   δ  

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a)

2

b) 1 −

π

c) 1 +

2

2

π

d) 0

π

[GATE–2017 (1)] The ratio of the mass flow rate,ṁ bd , leaving through the horizontal section b-d to that entering through the vertical section a-b is ___________ [GATE–2016 (1)]

Q.30 Consider fluid flow between two infinite horizontal plates which are parallel (the gap between them being 50 mm). The top plate is sliding parallel to the stationary bottom plate at a speed of 3 m/s. The flow between the plates is solely due to the motion of the top plate. The force per unit area (magnitude) required to maintain the bottom plate stationary is _________ N/m2. Viscosity of the fluid μ = 0.44 kg/m-s and density ρ = 888 kg/m3. [GATE–2016 (2)] Q.31 Consider a fully developed steady laminar flow of an incompressible fluid with viscosity μ through a circular pipe of radius R. Given that the velocity at a radial location of R/2 from the centerline of the pipe is U1, the shear stress at the wall is KμU1/R, where K is __________ [GATE–2016 (3)] Q.32 The velocity profile inside the boundary layer for flow over a flat u π y plate is given as = sin   U∞ 2 δ where U ∞ is the free stream velocity and δ is the local boundary layer thickness. If δ * is the local displacement thickness, the value of δ* is δ

Q.33 Consider a laminar flow at zero incidence over a flat plate. The shear stress at the wall is denoted by τw The axial positions

x1 and x 2 on the plate are measured

from the leading edge in the direction of flow. If x 2 > x1 , then

a) τw x1 = τw x 2 = 0

b) τw x1 = τw x 2 ≠ 0

c) τw x1 > τw x 2

d) τw x1 < τw x 2

[GATE–2017 (2)]

Q.34 For the laminar flow of water over a sphere, the drag coefficient CF is defined

(

)

as = CF F / ρU2D2 where F is the drag

force, ρ is the fluid density, U is the fluid velocity and D is the diameter of the sphere. The density of water is

1000kg / m3 . When the diameter of the

sphere is 100mm and the fluid velocity is 2m/s, the drag coefficient is 0.5. If water now flows over another sphere of diameter 200mm under dynamically similar conditions, the drag force (in N) on this sphere is _____________

[GATE–2017 (2)] Q.35 The viscous laminar flow of air over a flat plate results in the formation of a boundary layer. The boundary layer thickness at the end of the plate of length L is δL . When the plate length

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is increased to twice its original length, the percentage change in laminar boundary layer thickness at the end of the plate (with respect to δL ) is ________ (correct to two decimal places). [GATE–2018 (2)]

ANSWER KEY: 1

2

3

(c) (d) (b) 15 16 17 (c) (c) (d) 29 30 31 0.33 26.4 2.67

4

5

(c) 18 (c) 32 (b)

(b) 19 (d) 33 (c)

6

7

8

9

(c) (d) (b) (c) 20 21 22 23 636.93 0.06 15.7 13.8 34 35 20 41.42

10 (d) 24 (d)

11

12

(a) (c) 25 26 0.015 0.64

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13

14

(b) 27 (c)

(a) 28 22.25

EXPLANATIONS Q.1

Q.2

(c) The pressure is minimum at point C. Along the region CSD of the curved surface, the area of flow increases and hence velocity of flow along the direction of Fluid decreases.

Due to decrease of velocity, the pressure increases in the direction dp of flow and pressure gradient is dx dp positive or >0 dx (d) For laminar flow δ 1 ∝ x R ex δ∝

x Vx υ

δ∝ x ⇒δ∝x Q.3

Apply continuity section (1) and (2), A1V1 = A2V2

Q.4

2

(b) Given: L = 100 mm, d = 1mm, D =10 mm, V1 = 10 / s We have to take the two sections of the system (1) and (2).

on

π × 0.012 A1 4 V2 = = V1 ×= 0.01 1 m / s π A2 2 × 0.001 4 Again applying the Bernoulli’s equation at section (1) and (2) p1 V12 p 2 V22 + + Z1 = + + Z2 ρg 2g ρg 2g Since z1 = z2. Taking gauge pressure p2 =0 (Atmospheric pressure) p1 V22 V12 = − ρg 2g 2g Putting the values of V1 and V2 , We get p1 = 499.95 N/m2 Force required on plunger = p1 × A π = 499.95× × 0.012 4 =0.04 N (c)

ρV2 d 2 μ 1000 ×1× 0.001 = 1×10−3 Re=1000 Now Darcy’s friction factor, 64 64 f= = Re 1000 f = 0.064 Head loss in needle, fLV 2 hf = 2gd Re =

1

equation

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= 0.6512 N/m2 0.064 × 0.1× (1) 2 = 2 × 9.81× 0.001 Q.6 (c) = 0.3262m of water Given relation is, Applying Bernoulli’s equation at 3 U 3 y 1 y 4.64x points 1 to 2, we get = −   ;δ = 2 2 U∞ 2 δ 2  δ  p1 V1 p V Re x + + z1 = 2 + 2 + z 2 + h f ρg 2g ρg 2g U∞ = 2 m/s, υ = 1.5 × 10−5 m2/s Since z1 = z2 ρ = 1.23 kg/m3 And p2 = 0 (atmospheric) Kinematic viscosity 2 2 µ p1  V2 − V1  = ⇒ µ = v × p = 1.5 ×10−5 ×1.23 v = + h   f p ρg  2g  = 1.845×10-5 kg/m.sec ρ 2 2 p1= ( V2 − V1 ) + ρgh f Reynolds Number is given as, 2 pUx 1.23 × 2 ×1 1000 Re x = 1.33 ×105 = = −5 (1) 2 − (0.01) 2  + 1000 × 9.81× 0.3262 = μ 1.845 ×10 2 4.64 × 1 p1 = 3699.97 N/m2 = δ = 0.0127 Force required on plunger 1.33 ×102 2 p1 × A1 u 3y 1  y  And = −   π = 3699.97 × × (0.01) 2 u ∞ 2δ 2  δ  4 3 dU d  3y 1  y   = 0.29 N = U∞  −    dy dy  2δ 2  δ   Q.5 (b)  3 1 3 y2  du = − U ∞  × 3  Shear stress τ = μ 2 δ 2 δ  dy Where U ∞ = Free stream velocity

For linear velocity profile,

τ = μ.

u y

μ ρ ⇒μ= υ×ρ =7.4 ×10−7 × (0.88 ×1000) = 6.512 ×10−4 Pa − s 0.5 ∴ τ = 6.512 ×10−4 × 0.5 ×10−3 υ=

 dU   3   3U ∞  Now,   = U∞   =   2δ   2δ   dy  y =0 Now, shears stress  dU  3U ∞ −5 τ0 =µ   =1.8 ×10 × 2δ  dy  y =0

Q.7

and we Substitute the values of get 3× 2 = 1.845× 10−5 × 2 × 0.0127 = 435.82 × 10−5 N/m2 = 4.36 × 10−3 N/m2 (d) Given

 4r 2   r2  u = u 0 1 − 2  = u 0 1 − 2   D   R  Drop of pressure for a given length (L)of a pipe is given by,.

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32µuL D2 Where u is the average velocity. We know that for a laminar flow through a pipe. uo = 2u or u=uo/2 Where uois the maximum velocity. Substituting the value of u, we get 32µL u 0 32µu 0 L ∆= ×= P D2 2 D2 ∆P = P1 − P2 =

Q.8

Q.9

1 Drag force F= CD × ρAV 2 D 2 1.328 1 = × ρAV 2 Re L 2

Mass flow rate entering (PQ) = ρ × volume m1 = ρ × A × U = 1 × (1 × δ) × 10 = 1 × 10 × 10-3 × 10 = 0.1 kg/ sec Mass flow rate entering (SR) 1 m 2 = ρ×1× ( δ × U ) 2 1 = 1×1× ×10 ×10−3 ×10 2 = 0.05 kg/sec ∴ Mass leaving(QR) = m1– m2 = 0.1 – 0.05 = 0.05 kg/sec

(c) Drag force on plate = rate of change of momentum of the flowing fluid. Momentum thickness δ u  u  = θ ( ) ∫ 1 −  dy u u∞  0 ∞  δ

y

y

∫ δ 1 − δ  dy 0

= θ

=

FD = 0.1666N Q.10 (d)

(b)

= θ

δ δ δ − = 2 3 6 ∴force on plate per unit width = ρu ∞2 × (θ ×1) Here, u∞ = U = 10m/s  10 ×10−3  × 1 FD = 1× (10) 2 ×  6   =

δ

 y y2  ∫0  δ − δ2  dy 2

3 δ

y y − 2 2δ 3δ

0

=

1.328 1 × ρ × A × L × V2 ρLV 2 μ

i.e. FD ∝ L Now drag force on front half L FD/2 ∝ 2 F ∴ FD/2 = D 2 Drag force on rear half F'D/2= FD − FD/2 1   = 1 −  FD 2  F Now F = D/2 F'D/2 FD 2= = 1   1 −  FD 2  ∴ F >1

1 >1 2 −1

Q.11 (a)

Reynolds Number =

Inertia force =5 Viscous force

Q.12 (c) Assume width of channel is b. Applying continuity equation at A and B

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ρU 0 ( H ×= b ) ρVm (H − 2δ) × ba  δ  +2 ρ × ∫u.dy.b   0 

Vm y 2 U 0 H= Vm ( H − 2δ ) + 2 . δ 2

δ

0

∴ U 0 H= Vm ( H − 2δ ) + Vm .δ Vm H 1 ∴= = U0 H − δ 1 − δ H

Q.13 (b) Applying Bernoulli’s equation at A and B p A VA2 p B VB2 + = + ρg 2g ρg 2g p − p B VB2 − VA2 = ⇒ A ρg 2g

p A − p B Vm2 − U o2 = ⇒ ρ 2 2

p − p B  Vm  ⇒ A= −1 1 2  U 0  ρU 0 2 V Using the value of m from first U0 part of question, we get, PA − PB 1 = −1 2 1 2  δ  ρU 0 1 −  2  H Q.14 (a)

The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in figure, is given by the expression R 2  dp   r 2  − u (r) =   1− 4μ  dx   R 2 

 r2  = U max 1 − 2   R  Where U max = −

R 2  dP    4μ  dx 

dP is a constant dx The average velocity of fluid in the pipe is U R 2  dp  Vavg = max = −   2 8μ  dx 

Where

Q.15 (c) From Buckingham’s p-theorem It states “If there are n variable (Independent & dependent variables) in a physical phenomenon and if these variables contain k fundamental dimensions (M,L,T), then variables are arranged into (n −k) dimensionless terms. Here n = dimensional variables k = Primary dimensions (M, L, T) So, non-dimensional variables is n-k Q.16 (c) In case of two parallel plates, when flow is fully developed, the ratio of Vmax and Vavg is a constant. Vmax 3 = , Vmax = 6m / sec Vavg 2 Vav =

2 2 × Vmax = × 6 = 4m / sec 3 3

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Q.17 (d) Here type of flow is related to the dimensionless numbers (Nondimensional numbers). P. Compressible flow Q. Free surface flow R. Boundary layer S. Pipe flow T. Heat convection

Y. Mach number W.Webernumber Z.Skin friction coefficient U.Reynolds number V.Nusselt number Q.22

So, correct pairs are P-Y, Q-W, R-Z, S-U, T-V Q.18 (c) For flat plate with zero pressure gradient and Re = 1000 (laminar flow). Boundary layer thickness varies as δ ∝ V −1/2 where V = velocity of fluid V1 δ2 = = δ1 V2

δ2 =

δ2 =

V1

4V1

× δ1 = ×

1 ×1 = 0.5 mm 2

Q.19 (d) For turbulent flow, Re > 2300 Turbulent flow is considered as well mixed flow. Q.20 (636.94) d =0.02 m ṁ = 36 kg/hr = 0.01 kg/s μ = 0.001kg/ms & ρVd md & = = (Q ρAV m) Re= μ μA & & 4md 4m = Re = 2 μπd μπd 4 × 0.01 = 3.14 × 0.001× 0.02 = 636.94 Q.21 (0.064) Given, D = 10 cm V=0.1 m/s υ= 10−5 m2/s

VD 0.1× 0.1 = υ 10−5 =1000 64 64 f= = Re 1000 = 0.064

Re =

(15.7) Given, d = 100 mm L =100 mm U = 10 m/s μ = 0.1 kg/ms y= 2 ×10−3 m U τ=μ y

(Linearization of Newton’s law of viscosity) 10 = 0.1 × 2 ×10−3 = 500 N/m2 Frictional area = πdL = 3.14×0.1×0.1 Frictional resistance force = τ × πdL = 500×0.0314 = 15.7 N

Q.23 (13.85) δ ∝ √x δ2 x2 = = δ1 x1

0.75 = 0.25

3

∴ δ2 = δ1 √3 = 1.732×8 =13.8564 mm

Q.24 (d) Couette flow is steady incompressible, laminar flow between one fixed plate and other moving with constant velocity.

Q.25 (0.0159) 1.328 Cf = Re x

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ρVx Vx = μ υ V = 2m/s L =2.5m υ =2×10−5 m2/s 1 F = Cf ρAV 2 2 A=2.5×1 On substituting we get F = 0.0159 N = R ex

Q.26 (6.24) −1 ∂P 2 2 = u (R − r ) 4μ ∂x −1 = × ( −10 ) (0.05) 2 − (0.002) 2  4 × 0.001 u = 6.24m/s Q.27 (c)

Q.28 (22.25) For dynamic similarity – Reynold’s Number should be equal V D VD ∴ oil oil = ω ω Voil ω

10 × 0.5 0.89 ×10−6 × 10−5 0.02 Vω = 22.25m / s

= ∴ Vω

Q.29 (0.32 to 0.34)

 δ 2 = ρU 0 δ −  = ρδU 0  3 3 &= &ab − m &cd m ∴ m bd 2 = ρδU 0 − ρδU 0 3 1 = ρδU 0 3 1 &bd 3 ρδU 0 1 m ∴ = 0.33 = = &ab m ρδU 0 3

Q.30 (26.4) The force required to hold the bottom plate stationary will be equal to drag force exerted by the upper plate. F U ∴ =μ A h 3 = 0.44 × = 26.4 N/m2 50 ×10−3

Q.31 (2.67) We know that −1 ∂P 2 2 = u [R − r ] 4μ ∂x R at = r = , u U1 (given) 2 −1 ∂P  2 R 2  = U1 R − 4μ ∂x  4 

U1 =

Considering unit width of the plate &ab =ρ ( δ ×1) U 0 =ρδU 0 m

&cd ρ ( dy ×1) u dm =

  y   y 2  &cd ρ ∫U 0  2   −    dy = ∴m   δ   δ   0 δ

δ

 y 2 y3  & = m cd ρU 0 x  − 2  2δ 3δ  0 2

−1 ∂P  3R 2    4μ ∂x  4 

∂P −16μU1 ----- (1) = ∂x 3R 2 −∂P r Also, τ = ∂x 2 At r = R ∂P R τ= − ∂x 2  −16μ  R τ= − U1  × [From equation 2  3R  2 (1)] 8 μU1 ∴ τ= 3 R

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μU1 R ∴ K = 2.67 τ = 2.67

Q.32 (b)

δ

0

δ

δ*=

u

∫ (1 − u

Q δ*=



)dy



 π y 

∫ 1 − sin  2 . δ dy 0

δ

  π y  2δ  δ =  y + cos  .    2 δ  π 0  2δ δ* = δ + 0 − 0 − π  2 δ* =δ 1 −   π * δ 2 = 1− δ π *

Q.33 (c) 1 x1/2 ∴ as the distance from the leading edge increases shear stress decreases. τ∝

Q.34 (20) For Dynamic similarity Re1 = Re 2 ρV1D1 ρV2 D 2 = µ µ ----- (1) V1D1 = V2 D 2 And we know 24 (For flow over sphere ) CD = Re ∴ CD1 = CD2 Now = F2 CD2 ρU 2 2 D 22 = CD1 ρ ( U1D1 )

2

F2 = 0.5 ×1000(2 × .1) 2

F2 = 20N

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Q.35(41.42) In laminar flow over a flat plate, δ ∝ x1/2

Given, δ = δ L at x = L = Let, δ δ= 2L 2 L at x

δ δL

2L Then, =

2L = L

= 2 1.414

Therefor % change in δ =

δ2L − δ L  100 δL

= (1.4142 − 1)  100 = 41.42 %

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9

HYDRAULIC TURBINES

9.1 INTRODUCTION Turbines have been used for the centuries to convert freely available mechanical energy from rivers and wind into useful mechanical work, usually through a rotating shaft. The rotating part of a pump is called the impeller, whereas the rotating part of a hydro turbine is called the runner. When the working fluid is water, the turbomachines are called hydraulic turbines or hydro-turbines. When the working fluid is air, and energy is extracted from the wind, the machine is properly called a wind turbine. The word windmill should technically be applied only when the mechanical energy output is used to grind grains, as done in ancient times However, most people use the word windmill to describe any wind turbine, whether used to grind grain, pump water, or generate electricity. In coal or nuclear power plants, the working fluid is usually steam; hence, the turbo machines that convert energy from the steam into mechanical energy of a rotating shaft are called steam turbines. A more generic name for turbines that employ a compressible gas as the working fluid is gas turbine.

This type of turbine is found suitable when the available potential energy is high and the flow available is comparatively low. In this turbine, the fluid flow is along the tangential direction. In reaction turbines the available potential energy is progressively converted in the turbines rotor and the reaction of the accelerating water causes the rotation of the wheel. These are again divided into radial flow, mixed flow and axial flow machines. Radial flow machines are found suitable for moderate levels of potential energy and medium quantities of flow. The axial machines are suitable for low levels of potential energy and large flow rates. The potential energy available is generally denoted as “head available”. With this terminology plants are designated as “high head”, “medium head” and “low head” plants. 9.3 TURBINE EFFICIENCIES

9.3.1 HYDRAULIC EFFICIENCY

9.2 CLASSIFICATION OF TURBINES

It is defined as the ratio of the power produced by the turbine runner and the power supplied by the water at the turbine inlet.

The main classification depends upon the type of action of the fluid on the turbine. These are 1) Impulse turbine 2) Reaction Turbine.

9.3.2 VOLUMETRIC EFFICIENCY

In impulse turbine all the potential energy is converted to kinetic energy in the nozzles. The impulse provided by the jets is used to turn the turbine wheel. The pressure inside the turbine is atmospheric.

ηH = RP / WP

Water power produced is given by ρQ g H Where, Q is the volume flow rate H is the net or effective head. The power produce by runner is evaluated by Euler’s equation. This reflects the runner design effectiveness. Volumetric efficiency is defined as the ratio between the volume of water flowing through the runner and the total volume of water supplied to the turbine. The

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reduction in water flowing through runner is because of leakage through the clearance between the runner and casing without passing through the runner 9.3.3 MECHANICAL EFFICIENCY

The power produced by the runner is always greater than the power available at the turbine shaft. This is due to mechanical losses at the bearings, friction. ηm =Shaft power / Runner power

9.3.4 OVERALL EFFICIENCY

The moment of momentum at inlet  w1r1 = mV

The

moment

 w2r2 = mV

of

momentum

at

exit

The torque on the rotor equals the rate of change of moment of momentum of the fluid as it passes through the runner. Torque, T  m Vw1 r1  Vw2 r2  P  m Vw1 r1  Vw2 r2  



P  m Vw1 r1    Vw2 r2  

 w1U1 ± Vw2U2 ] = P m[V

Power

per

unit

weight

is

given

This is the ratio of power output at the = [Vw1U1 + Vw2U2 ]/ g shaft and power input by the water at the turbine inlet. 9.5 PELTON TURBINE η0 = Shaft Power / Water Power The hydraulic turbine of the impulse type Also the overall efficiency is the product of in common use, is named after an American the other three efficiencies defined engineer Laster A Pelton. Therefore this ηo =ηH ηm ηv machine is known as Pelton turbine or Pelton wheel. It is an efficient machine 9.4 POWER DEVELOPED BY TURBINE particularly suited for high heads. The (EULER’S EQUATION) rotor consists of a large circular disc or wheel on which a number (usually less The fluid velocity at the turbine entry and than 15) of buckets are spaced uniformly exit can have three components in the round its periphery. The high speed jet tangential, axial and radial directions of the then impinges on bucket-shaped vanes that rotor. This means that the fluid momentum transfers energy to the turbine shaft. The can have three components at the entry buckets of a Pelton wheel are designed so and exit. This also means that the force as to split the flow in half, and turn the flow exerted on the runner can have three nearly 165° around. components. Out of these the tangential force only can cause the rotation of the runner and produce work. The axial component produces a thrust in the axial direction, which is taken by suitable thrust bearings. The radial component produces a bending of the shaft which is taken by the journal bearings.

U1    is the tangental velocity at entry, U 2 is the tangential velocity at exit, Vw1 is the tangential component of the

absolute velocity of the fluid at inlet,  Vw 2 is the tangential component of the absolute velocity of the fluid at exit

Pelton turbine 9.5.1 POWER DEVELOPED

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{

}

{

}

The bucket splits the jet into equal parts =   V1 ± k ( V1 − u ) cos β − u  u P m   and changes the direction of the jet by Whether case1 or case 2 the power can be about 165° . The velocity diagram for Pelton written as turbine is shown in figure. The force acting on blade is equal to rate of change of =   V1 ± k ( V1 − u ) cos β − u  u P m   momentum of fluid passing through runner   V1 − u 1 + k cosβ  u P= m  w1 ± Vw2 ) = F m(V   In case of blades mounted around  is given by ρAV1 at periphery of wheel, m 9.5.2 HYDRAULLIC EFFICIENCY entry. As already discussed in previous chapter the mass flow rate for single blade  ( V1 – u )(1 + k cos β )  u m  ηH =  is given by ( V1 − U1 )  mgh

(

)(

)

 = 1 / 2mV  12 mgh For maximum ηH , dηH =0 du

V1 = Cv 2gH

T  m Vw1  Vw2  r

 w1 ± Vw2 )rω P = m(V

U = V1 / 2

Substituting the value U = V1 / 2 , maximum efficiency is 1 + k cos β ηH = 2 Speed Ratio: u ku = 2gH Vr2 cos β2 > u

Vr2 cos β2 < u

9.6 REACTION TURBINE

Velocity Triangle of Pelton wheel When,

Vr2cos β > u,Vw1 + Vw2 Vr2cos β > u,Vw1 − Vw2

From velocity triangle, Vw1  V1 Vr1= V1 − u

The relative velocity Vr 2 becomes slightly less than Vr1 mainly because of the friction in the bucket. Vr2 = kVr1

In the ideal case = K 1,V = Vr1 r2

= Vw2 Vr2 cos β − u

V = k(V1 − u)cos β − u w2

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The other main type of energy-producing hydro-turbine is the reaction turbine, which consists of fixed guide vanes called stay vanes, adjustable guide vanes called wicket gates, and rotating blades called runner blades. Flow enters tangentially at high pressure, is turned toward the runner by the stay vanes as it moves along the spiral casing or volute, and then passes through the wicket gates with a large tangential velocity component. Momentum is exchanged between the fluid and the runner as the runner rotates, and there is a large pressure drop. Unlike the impulse turbine, the water completely fills the casing of a reaction turbine. For this reason, a reaction turbine generally produces more power than an impulse turbine of the same diameter, net head, and volume flow rate. 9.6.1 COMPONENTS TURBINE

OF

REACTION

9.6.11 SPIRAL CASING The spiral casing surrounds the runner completely. Its area of cross section decreases gradually around the circumference. This leads to uniform distribution of water all along the circumference of the runner. Water from the penstock pipes enters the spiral casing and is distributed uniformly to the guide blades placed on the periphery of a circle. The casing should be strong enough to withstand the high pressure. 9.6.12 GUIDE BLADES

Water enters the guide blades circumferentially. the guide blades guide the water in the proper direction. They also act as nozzle. The velocity if water is increased by the pressure drop in the passage. 9.6.13 RUNNER

The runner is circular disc and has the blades fixed on one side. In high speed runners in which the blades are longer a circular band may be used around the blades to keep them in position. The shape of the runner depends on the specific speed of the unit. These are classified as i) slow runner ii) medium speed runner iii) high speed runner iv) very high speed runner

9.6.14 DRAFT TUBE After passing through the turbine runner, the exiting fluid still has kinetic energy, and perhaps swirl. To recover some of this kinetic energy (which would otherwise be wasted), the flow enters an expanding area diffuser called a draft tube, which turns the flow horizontally and slows down the flow speed, while increasing the pressure prior to discharge into the downstream water

The head recovered by the draft tube will equal the sum of the height of the turbine exit above the tail water level and the difference between the kinetic head at the inlet and outlet of the tube less frictional loss in head V12 V22  Hd  H   hf 2g Different types of draft tubes are used as the location demands. These are i) Straight diverging tube ii) Bell mouthed tube iii) Elbow shaped tubes of circular exit or rectangular exit. 9.6.2 VELOCITY TRIANGLE OF REACTION

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TURBINE

accelerating component and this will be present only in the reaction turbines. Reaction of turbine is given by R=

(

(

) (

)

1 2 U1 – U 22 + Vr22 – Vr12  2

) (

) (

)

1 2 V1 – V22 + U12 – U 22 + Vr22 – Vr12  2

For b 90, = = Vw 2 0 and V2 = Vf 2 cot α R= 1− 2 ( cot α − cot θ ) Vf 1 & Vf 2 is flow velocity

V1 & V2 is absolute velocity Vr1 & Vr2 is relative velocity

9.6.3 POWER DEVELOPED REACTION TURBINE

BY

1) Torque developed by reaction turbine is given by 2) T  m Vw1 r1  Vw2 r2   = ρQ m

3) Power is given by  ( Vw1r1  ± Vw2r2 ) ω = P m

Or  ( Vw1U1  ± Vw2U2 )  = P m

4) Discharge is given by Q  V f1  D1b1  V f2  D2b2 Q is volume flow rate b is the thickness D is the diameter

9.7 FRANCIS TURBINE The inward flow reaction turbine having radial discharge is known as francis turbine. Since the discharge is radial at outlet, the whirl velocity at outlet will be zero. Hence the work done by water on the runner per second will be  ( Vw1U1 ) P=m 9.8 KAPLAN TURBINE The popular axial flow turbines are the Kaplan turbine and propeller turbine. In propeller turbine the blades are fixed. In the Kaplan turbines the blades are mounted in the boss in bearings and the blades are rotated according to the flow conditions by a servo mechanism maintaining constant speed. In this way a constant efficiency is achieved in these turbines. The system is costly and where constant load conditions prevail, the simpler propeller turbines are installed.

5) Speed of rotation

U1 = πD1N / 60     U2 = πD2N / 60

N is rotation per minute By geometry

Vw1U1=   ± Vw2U2 1 / 2 ( V12 – V22 ) + (U12 – U22 ) + ( Vr22 – Vr12 )

Where,

½ (V12 – V2 2 ) is the dynamic component

of work done ½ ( U12 – U 2 2 ) is the centrifugal component of work and this will be present only in the radial flow machines ½ (Vr2 2 – Vr12 ) is the

9.8.1 SPECIFIC SPEED The specific speed is given by

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N P H 5/4 Where, P is power in KW N is speed of rotation in rpm H is head in ‘m’ Ns =

9.9 SPECIFIC SPEED SIGNIFICANCE Specific speed does not indicate the speed of the machine. It can be considered to indicate the flow area and shape of the runner. When the head is large, the velocity when potential energy is converted to kinetic energy will be high. The flow area required will be just the nozzle diameter. This cannot be arranged in a fully flowing type of turbine. Hence the best suited will be the impulse turbine. When the flow increases, still the area required will be unsuitable for a reaction turbine. So multi jet unit is chosen in such a case. As the head reduces and flow increases purely radial flow reaction turbines of smaller diameter can be chosen. As the head decreases still further and the flow increases, wider rotors with mixed flow are found suitable. The diameter can be reduced further and the speed increased up to the limit set by mechanical design. As the head drops further for the same power, the flow rate has to be higher. Hence axial flow units are found suitable in this situation.

Dimensionles Dimensionles Types of turbine having specific speed specific speed the best efficient at range in SI system these values 0.015-0.053 8-29 Single jet pelton turbine 0.047-0.072 26-40 Twin jet pelton turbine 0.72-0.122 40-67 Multiple jet pelton turbine 0.122-1.819 67-450 Radial flow turbine francistype(H<350m) 0.663-1.66 364-910 Axial flow haplan turbine(H<60m)

9.10 MODEL TESTING (DIMENSIONLESS TURBINE PARAMETERS) Hydraulic turbines are mainly used for power generation and because of this, these are large and heavy. The operating conditions in terms of available head and

load fluctuation vary considerably. In spite of design methodology, it is found the designs have to be validated by actual testing. In addition to the operation at the design conditions, the characteristics of operation under varying input & output conditions should be established. It is found almost impossible to test a full size unit under laboratory conditions. In case of variation of the operation from design conditions, large units cannot be modified or scrapped easily. We define dimensionless groups (Pi groups) for turbines. Dimensional variables: • gravity times net head ( gH ) • • • • •

volume flow rate (V) diameter of the runner blades(D) runner rotational speed (N) output power (P) fluid density. (ρ)

The dimension less parameters obtained are 1. The head coefficient, gH / N 2 D 2 2. The flow coefficient, Q / ND3 3. The power coefficient, P / ρN 3 D5

9.10.1 UNIT QUANTITIES

The dimensionless constants can also be used to predict the performance of a given machine under different operating conditions. As the linear dimension will be the same, the same will not be taken into account in the calculation. Thus Head coefficient is given by gH1 / N12 D 2 = gH 2 / N 2 2 D 2

Or

H1 / N12 = H 2 / N 2 2

Flow coefficient is given by Q1 / N1D3 = Q 2 / N 2 D3

Or Q1 / N1 = Q 2 / N 2 Using previous two relations Q1 Q  12 1 H1 2 H 2 2 Where,

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Q / H1/2 = constant known as unit discharge N / H1/2 = constant is known as unit speed P / H 3/2 = constant known as unit power

Example: The penstock supplies water from a reservoir to the Pelton wheel with a gross head of 500m. One third of the gross head is lost in friction in the penstock. The rate of flow of water through the nozzle fitted at the end of the penstock is 2.0m3 /s. The angle of deflection of the jet is 165° Determine the power given by the water to the runner and also hydraulic efficiency of the Pelton wheel .Take speed ratio=0.45 and Cv =1.0 Solution: Given: Gross head,

H g = 500 m

∴ Vr = V1 − u1 = 80.86 − 36.387 1

= 44.473m/ s Also

VW1 = V1 = 80.86 m/ s

From outlet velocity triangle , we have Vr2 = Vr1 = 44.473

Vr2 cos φ = u 2 + VW2

Or 44.473cos15° = 36.387 + VW2

Or VW2 = 44.473cos15° − 36.387 = 6.57 m/ s

Work done by the jet on the runner per second is given by equation as Fx = ρ aV1  Vw1 − Vw 2  × u = ρ Q  Vw1 + Vw 2  × u

= 1000× 2.0× [80.86 + 6.57 ] ×36.387 =

= 6362630 Nm/ s

∴ Power given by the water to the runner in kW =

Work done per second 6362630 = = 6362.63kW 1000 1000

Hydraulic efficiency of the turbine is given by equation as 2  Vw + Vw  × u 2 [80.86 + 6.57 ] ×36.387 = ηh =  2 1

2

V1

80.86×80.86

= 0.9731 or 97.31%

Head lost in friction H g 500 hf = = = 166.7 m 3 3 ∴ Net head ,

H = H g − h f = 500 − 166.7 = 333.30 m

Q = 2.0 m3 / s Discharge, Angle of deflection = 165° ∴ Angle, φ = 185° − 165° = 15° Speed ration = 0.45 Co-efficient of velocity, C v = 1.0 Velocity of jet, V1 = C v 2 gH = 1.0× 2×9.81×333.3 =

= 80.86 m/ s

Velocity of wheel, u = speed ratio× 2 gH Or u = u1 = u 2 = 0.45× 2×9.81×333.3 = 36.387 m/ s

Example: A Pelton wheel is working under a gross head of 400m. The water is supplied through penstock of diameter 1 m and length 4km from reservoir to the Pelton wheel. The co-efficient of friction for the penstock is given as .008.The jet of water of diameter 150mm strikes the buckets of the wheel and gets deflected through an angle of 165° . The relative velocity of water at outlet is reduced by 15% due to friction between inside surface of the bucket and water. If the velocity of the buckets is 0.45 times the jet velocity at inlet and mechanical efficiency as 85% determine: i) Power given to the runner ii) Shaft power, iii) Hydraulic efficiency & overall efficiency Solution: Given:

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Gross head,

Diameter of penstock, Length of penstock

H g = 400 m

 0.0033V12  0.51V12 or 400=0.0543V12

D = 1.0 m

400 = 85.83m/ s .0543 Now velocity of bucket, u1 = 0.45 V1 = 0.45×85.83 = 38.62 m/ s From inlet velocity triangle, ∴ V1 =

L = 4 km = 4×1000 = 4000 m

f = .008 Co-efficient of friction , d = 150 m = 0.15 m Diameter of jet, Angle of deflection = 165° F = 180° − 165° = 15° ∴ Angle, Relative velocity at outlet, Vr = 0.85 Vr 2

Vr1 = V1 − u1 = 85.83 − 38.62 = 47.21m/ s

Vw1 = V1 = 85.83 m/ s

1

Velocity of bucket, u = 0.45× Jet velocity Mechanical efficiency, ηm = 85% = 0.85 Let V* = Velocity of water in penstock, and V1 = Velocity of jet of water.

From outlet velocity triangle

Vr2 = 0.85× Vr1 = 0.85× 47.21 = 40.13 m/ s

Vw 2 = Vr2 cos φ − u 2 = 40.13cos15° − 38.62

= 0.143m/ s

( u = u1 = u 2 = 38.62 )

Discharge through nozzle is given as Q = Area of jet× Velocity ofjet = a× V1 π 2 π 2 d × V1 = (.15 ) ×85.83 = 1.516 m3 / s 4 4 Work done on the wheel per second is given by equation as = ρ aV1  Vw + Vw  × u = ρ Q  Vw + Vw  × u 1

2

1

2

= 1000×1.516 [85.83 +.143] ×38.62 = 5033540 Nm/ s

i) Power given to the runner in kW =

Using continuity equation, we have Area of penstock × velocity of water in penstock = Area of jet ×V1 π 2 π Or D × V* = d 2 × V1 4 4 2 d 0.152 ∴V * = × V1 = × V1 = .0225V1 D2 1.02 Applying Bernoulli’s equation to the free surface of water in the reservoir and outlet of the nozzle, we get V2 H g = Head lost due to friction+ 1 2g Or *2

400 =

2 1

*2

2 1

V V 4 fLV 4×.008× 4000× V + = + D× 2 g 2 g 1.0× 2×9.81 2g

Substituting the value of V * from equation . we get V12 4×.008× 4000 2 400 = × ( 0.0225 V1 ) + 2×9.81 2g

Work done per second 5033540 = = 1000 1000

= 5033.54 kW

ii) Using equation for mechanical efficiency, Power at the shaft S.P. = ηm = Power given to the runner 5033.54 ∴ S.P = ηm ×5033.54 = 0.85×5033.54 = 4278.5 kW

iii) Hydraulic efficiency equation as 2  Vw + Vw  × u ηh =  1 2 2  V1 =

2 [85.83 +1.43] ×38.62 85.83×85.83

is

given

by

= 0.9014 = 90.14%

Overall efficiency is given by equation as η0 = ηm ×ηh = 0.85×.9014 = = 0.7662 or 76.62%

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Example: A 137 mm diameter jet of water issuing from a nozzle impinges on the buckets of a Pelton wheel and the jet is deflected through an angle of 165° by the buckets The head available at the nozzle is 400m. Assuming co-efficient of velocity as 0.97, speed ratio as 0.46, and reduction in relative velocity while passing through buckets as 15% find: i) The force exerted by the jet on buckets in tangential direction, ii) The power developed Solution: Given: Given of jet,

d = 137 mm = 0.137 m

∴ Area of jet, a = 0.01474m2 Angle of deflection = 165° ∴ Angle, φ = 180° − 165 = 15° H = 400 m Head of water, Co-efficient of velocity, C v = 0.97

Speed ratio = 0.46 Relative velocity at outlet = 0.85× relative velocity at inlet Or Vr = 0.85 Vr 2

1

Now velocity of jet,

V1 = C v 2 gH = 0.97 2×9.81× 400 = 85.93m/ s

Speed ratio

=

u1 u1 or 0.46 = 2 gH 2×9.81× 400

∴ u1 = 0.46× 2×9.81× 400 = 40.75 m/ s Hence Vr = V1 − u1 = 85.93 − 40.75 = 45.18 m/ s And Vr = 0.85 Vr = 0.85× 45.18 = 38.40 m/ s 1

1

1

For Pelton turbine, u1 = u 2 = u = 40.75 m/ s Vr cos φ = 38.40× cos15° = 37.092 2

Here Vr2 cos φ

is less than u2 . Hence

velocity triangle at outlet will be as shown in Fig ∴ Vw = u 2 − Vr cos φ = 40.75 − 37.092 = 3.658 m/ s 2

2

i) Force exerted by jet on buckets in tangential direction is given by , Fx = ρ aV1  Vw − Vw  1

2

(Here –ve sign is taken as Vw and Vw 1

2

are in the same direction ) ∴ Fx = 1000× 0.01474×85.93 ( 85.93 − 3.658 ) N = 104206 N ii) Power developed is given by, Power =

Fx × u 104206× 40.75 kW = = 4246.4 kW 1000 1000

Example: An inward flow reaction turbine has external and internal diameters as 0.9m and 0.45m respectively .The turbine in running at 200r.p.m. and width of turbine at inlet is 200mm. The velocity of flow through the runner is constant and is equal to 1.8m/s. The guide blades make an angle of 10° the tangent of the wheel and the discharge at the outlet of the turbine is radial. Draw the inlet and outlet velocity triangles and determine i) The absolute velocity of water at inlet of runner ii) The velocity of whirl at inlet, iv) The relative velocity at inlet. vi) Mass of water flowing through the runner per second. v) Width of the runner at outlet, vii) Head at the inlet of the turbine viii)Power developed and hydraulic efficiency of the turbine Solution: Given: External Dia: D1 = 0.9 m Internal Dia., D 2 = 0.45 m Speed N = 200 r .p.m Width at inlet , B1 = 200 mm = 0.2 m Velocity of flow, Vf = Vf = 1.8 m/ s 1

Guide of flow, Discharge at outlet

2

α = 10° = Radial

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∴ β = 90° andVW = 0

∴ B2 = D1B1

2

=

D2

0.90× 0.20 = 0.40 m = 400 mm 0.45

vi) Mass of water flowing through the runner per second. The discharge,

Q = π D1B1Vf1 = π × 0.9× 0.20×1.8 = 1.0178 m3 / s

∴ Mass = ρ × Q = 1000×1.0178 kg/ s = 1017.8 kg/ s vii) Head at the inlet of turbine, i.e., H Using equation, we have H−

Tangential velocity of wheel at inlet and outlet are: π D1 N π ×.9× 200 u1 = = = 9.424 m/ s 60 60 π D 2 N π ×.45× 200 u2 =

=

60

= 4.712 m/ s

60

i) Absolute velocity of water at inlet of the runner i.e. V1 . From inlet velocity triangle, V1 sin α =Vf 1 V1 =

Vf1

=

sin α

18 = 10.365 m/ s sin10°

ii) Velocity of whirl at inlet , ie., Vw

1

Vw1 = V1cos α = 10.365× cos10° = 10.207 m/ s

iii) Relative velocity at inlet, i.e. Vr

(

Vr1 = Vr31 + Vw1 − u1

)

2

1

= 1.82 + (10.207 − 9.424 )

2

= 3.24 + 613 = 1.963m/ s iv) The runner blade angles means the angle θ and Φ Now tan θ =

(V

Vf1

w1

− u1

)

=

1.8 = 2.298 (10.207 − 9.424 )

∴ θ = tan -1 2.298 = 66.48° or 66°29.' From outlet velocity triangle, we have Vf 1.8 tan φ = = = tan 20.9° u 2 4.712 ∴ φ = 20.9° or 54.4 ' v) Width of runner at outlet, i.e, B2 From equation, we have π D1B1Vf = π D 2 B2 Vf or D1B1 = D 2 B2 2

 V

1

f1

2

 V f2 as V f1  V f2 

(

( hereV

w2

H =

)

(

V22 1 1 = Vw1 u1 ± Vw 2 u 2 = Vw1 u1 2g g g

Vw1 u1 g



=0

)

)

V22 2g

10.927 9.424 1.82   V2  V f2  9.81 29.81

= 9.805 + 0.165 = 9.97m

viii) Power developed, i.e.,

work done per second on runner 1000 ρ Q  Vw1 u1 

P=

=

1000

= 1000×

1.0178×10.207×9.424 = 97.9 kW 1000

Hydraulic efficiency equation as ηh =

Vw1 u1 gH

=

is

given

by

10.207×9.424 = 0.9834 = 98.34% 1000

Example: An outward flow reaction turbine has internal and external diameters of the runner as 0.6m and 1.2m respectively. The guide blade angle is 15° and velocity of flow through the runner is constant and equal to 4m/s. If the speed of the turbine is 2000 rpm, head on the turbine is 10m and discharge at outlet is radial, determine: i) The runner vane angles at inlet & outlet ii) Work done by the water on the runner per second per unit weight of water striking per second, iii) Hydraulic efficiency, and iv) The degree of reaction Solution: Given:

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Internal diameter, External diameter Guide blade angle, Velocity of flow,

D1 = 0.6 m D 2 = 1.2 m

α = 15°

Vf1 = Vf2 = 4 m/ s

Speed N = 200 r .p.m H = 10 Head, Discharge at outlet =Radial ∴ VW = 0.Vf = V2 2

2

iii) Hydraulic efficiency is given by equation Vw u1 14.928×6.283 ηh = 1 = = 0.9561or 95.61% gH 9.81×10 iv) Given : In this question, the velocity of flow is constant through the runner (ie., Vf = Vf ) and the discharge is radial at 1

2

(ie., β = 90° orVw = 0),

outlet

2

the

degree of reaction (R ) is given by equation R = 1−

cot α 2 ( cot α - cot θ )

Here α = 13.928° and θ = 41.09° ( calculated ) Substituting the value of α and θ, we get

R = 1−

Tangential velocity of runner at inlet and outlet are: π D1 N

π × 0.6× 200

= 6.283m/ s 60 60 π D 2 N π ×1.2× 200 u2 = = = 12.566 m/ s 60 60 u1 =

=

From the inlet velocity triangle, Vf tan α = 1 Vw

∴ Vw = 1

1

Vf1

tan α

=

4.0 = 14.928 m/ s tan15°

i) Runner vane Angles at inlet and outlet are θ and φ tan θ =

Vf1

Vw1 − u1

=

4.0 = 0.4627 (14.928 − 6.283)

θ = tan −1 .4627 = 24.83 or 24°49.8' From outlet velocity triangle, Vf 4.0 tan φ = 2 = = 0.3183 u2 12.566

∴ φ = tan −1 .3183 = 17.65° or17°39.4 ' ii) Work done by water per second per unit weight of water striking per second 1 = Vw1 u1 ( Vw = 0 ) g 2

1 = × 14.928 × 6.283 =9.561Nm / N 9.81

cot13.928° 4.032 = 1− 2 ( cot13.928° − cot 41.09° ) 2 ( 4.032 − 1.146 )

= 1 − 0.698 = 0.302  0.3

For Francis turbine , the degree of reaction varies from 0 to 1 i.e., 0 ≤ R ≤ 1

Example: A pelton turbine develops 3000kW under a head of 300m. The overall efficiency of turbine is 83%. If speed ratio =0.46, Cv = 0.98 and specific speed is 16.5, then find: i) Diameter of the turbine, and ii) Diameter of the jet Solution: Given: P = 3000 kW Power H = 300 m Net head Overall efficiency, ηo = 83% or 0.83 Speed ratio Value of Cv'

= 0.46 = 0.98

Specific speed *, Using equation, Ns =

N s = 16.5

Ns H5/4 16.5×3005/4 N P or N = = = 375r.p.m H 5/4 P 3000

The velocity (V) at the outlet of nozzle is given by V = C v 2× g× H = 0.98 2×9.81×300 = 75.1m/ s

Now speed ratio,

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=

u or u = Speed ratio× 2 gH 2 gH

= 0.46× 2×9.81×300 = 34.95 m/ s i) Diameter of the turbine (D) using, π DN

u=

60

60× u 60×34.95 or D = = = 1.78 m π ×N π ×37.5

ii) Diameter of the jet (d) Let Q = Discharge through turbine in

m3 / s Using the relation, ηo =

P1 ,  π × g× Q× H    1000  

where π × g = 1000×9.81N/ m3 for water

∴ 0.83 =

∴ Q=

3000  1000×9.81× Q×300    1000  

3000 = 1.23m3 / s 9.81×300× 0.83

But discharge through a Pelton turbine is given by, Q = Area of jet× Velocity π Or 1.23 = d 2 × 75.1 4 4×1.23 ∴d= = 0.142 m = 142 mm π × 75.1 Example: A turbine develops 9000 kW when running at 10r.p.m. The head on the turbine is 30m.If the head on the turbine is reduced to 18m; determine the speed and power developed by the turbine. Solution: Given: Power developed P1 = 9000 kW Speed N1 = 100 r .p.m. Head, H1 = 30 m Let for a head H 2 = 18 m Speed = N2 Power = P2 Using equation, N2 =

N1 H 2 H1

=

N1 N = 2 H1 H2

100 18 100× 4.2426 = = 77.46 r.p.m. 5.4772 30

Also we have

P1 P2 = 3/2 3/2 H1 H2

3/ 2

3/ 2

∴ P2 = P1H3/22 = 9000×18 3/ 2

H1 = 4182.84 kW

30

=

687307.78 164.316

Example: A pelton wheel is revolving at a speed of 190r.p.m. and develops 5150.25 kW when working under a head of 220m with an overall efficiency of 80% Determine unit speed, unit discharge and unit power. The speed ratio for the turbine is given as 0.47. Find the speed. Discharge and power when this turbine is working under a head of 140m Solution: Given: N1 = 190 r .p.m. Speed, P1 = 5150.25 kW Power, H1 = 220 m Head, Overall efficiency, ηo = 80% = 0.80 = 0.47 Speed ratio New head of water, H 2 = 140 m Overall efficiency is given by ηo =

P1

ρ × g× Q1× H1

=

1000× P1 ρ × g× Q1× H1

1000

∴ Q1 = 1000× P1 = 1000×5150.25 = h o × ρ ×g× H1 0.80×1000×9.81× 220 = 2.983m3 / s

Unit speed is given by equation Nu =

N1 19 = = 12.81 r .p.m. H1 220

Unit discharge is given by equation Qu =

Q1 2.983 = = 0.201m3 / s H1 220

Unit power is given by equation . P1 5150.25 Pu = 3/2 = = 1.578 kW H1 2203/2 When the turbine is working under a new head of 140m, the speed, discharge and power are given by equation as N N For speed, 1 = 2 H1 H2

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H2

 N 2  N1

H1

For discharge ,  Q2  Q1

For power,  P2  P1

H2 H1

 190

140  151.6rpm 220

Q1 Q = 2 H1 H2

 2.983

140  2.379m3 / s 220

P1 P2 = 3/2 3/2 H1 H2 3

H2 2 3

H1 2

3

 140  2  5150.25   2614.48kW  220 

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GATE QUESTIONS Q.1

Q.2

Q.3

If there are m physical quantities and n fundamental dimensions in a particular process, the number of non-dimensional parameters is a) m + n b) m × n c) m− n d) m / n [GATE–2002] A centrifugal pump running at 500 rpm and at its maximum efficiency is delivering a head of 30 m at a flow rate of 60 litres per minute. If the rpm is changed to 1000, then the head H in metres and flow rate Q in litres per minute at maximum efficiency are estimated to be a) H=60, Q=120 b) H=120, Q=120 c) H = 60, Q=480 d) H = 120, Q = 30 [GATE–2003] Match List-I with List-II and select the correct answer using the codes given below the lists : List-I P.Curtis Q.Rateau R.Kaplan S.Francis

Codes : P a) 2 b) 3 c) 1 d) 3 Q.4

List-II 1.Reaction steam turbine 2. Gas turbine 3. Velocity compounding 4.Pressure compounding 5.Impulse water turbine 6. Axial turbine 7. Mixed flow turbine 8. Centrifugal pump

Q 1 4 3 4

R S 1 6 6 7 1 5 7 6 [GATE–2003]

A centrifugal pump is required to pump water to an open water tank situated 4 km away from the location of the pump through a pipe

Q.5

Q.6

of diameter 0.2 m having Darcy’s friction factor of 0.01. The average speed of water in the pipe is 2 m/s. If it is to maintain a constant head of 5 m in the tank, neglecting other minor losses, then absolute discharge pressure at the pump exit is a) 0.449 bar b) 5.503 bar c) 44.911 bar d) 55.203 bar [GATE–2004] At a hydroelectric power plant site, available head and flow rate are 24.5 m and 10.1 m3/s respectively. If the turbine to be installed is required to run at 4.0 revolutions per second (rps) with an overall efficiency of 90%, the suitable type of turbine for this site is a) Francis b) Kaplan c) Pelton d) Propeller [GATE–2004] Match List-I with List-II and select the correct answer using the codes given below the lists: List-I P. Reciprocating pump Q. Axial flow pump R. Microhydel plant S. Backward curved vanes List-II 1. Plant with power output below 100 kW 2. Plant with power output between 100 kW to 1 MW 3. Positive displacement 4. Draft tube 5. High flow rate, low pressure ratio 6. Centrifugal pump impeller Codes : P Q R S a) 3 5 6 2 b) 3 5 2 6

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c) d) Q.7

Q.8

Q.9

3 4

5 5

1 6 1 6 [GATE–2004]

In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0.1 m3/s. If the jet deflection angle is 120° and the flow is ideal, the power developed is a) 7.5 kW b) 15.0 kW c) 22.5 kW d) 37.5 kW [GATE–2006] A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m. For initial testing, a 1:4 scale model of the turbine operates under a head of 10 m. The power generated by the model (in kW) will be a) 2.34 b) 4.68 c) 9.38 d) 18.75 [GATE–2006] A horizontal-shaft centrifugal pump lifts water at 65℃. The suction nozzle is one meter below pump center line. The pressure at this point equals 200 kPa gauge and velocity is 3 m/s. Steam tables show saturation pressure at 65℃ is 25 kPa, and specific volume of the saturated liquid is 0.001020 m3 /kg. The pump Net Positive Suction Head (NPSH) in meters is

a) 24 c) 28

b) 28 d) 30 [GATE–2006] Q.10 The inlet angle of runner blades of a

Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant throughout the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is a) 25% b) 50% c) 80% d) 89% [GATE–2007] Q.11 A model of a hydraulic turbine is tested at a head of 1/4th of that under which the full scale turbine works. The diameter of the model is half of that of the full scale turbine. If N is the RPM of the full scale turbine, the RPM of the model will be a) N/4 b) N/2 c) N d) 2N [GATE–2007] Q.12 Match List-I with List-II and select the correct answer using the codes given below the lists : List-I P. Centrifugal compressor Q. Centrifugal pump R. Pelton wheel S. Kaplan turbine List-II 1. Axial flow 2. Surging 3. Priming 4. Pure impulse Codes : P Q R S a) 2 3 4 1 b) 2 3 1 4 c) 3 4 1 2 d) 1 2 3 4 [GATE-2007] Q.13 Water, having a density of 1000 kg/m3, issues from a nozzle with a velocity of 10 m/s and the jet strikes a bucket mounted on a Pelton wheel. The wheel rotates at

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10 rad/s. The mean diameter of the wheel is 1 m. The jet is split into two equal streams by the bucket, such that each stream is deflected by 120° as shown in the figure. Friction in the bucket may be neglected. Magnitude of the torque exerted by the water on the wheel, per unit mass flow rate of the incoming jet, is

fluid and U denotes the blade velocity. Subscripts 1 and 2 refer to inlet and outlet respectively. If V2 = W1 and V1 = W2 , then the degree of reaction is

a) 0 c) 0.5

a) 0 (N-m)/(kg/s) b)1.25(N-m)/(kg/s) c) 2.5 (N-m)/(kg/s) d)3.75(N-m/(kg/s)

[GATE–2008]

Q.14 A hydraulic turbine develops 1000 kW power for a head of 40 m. If the head is reduced to 20 m, the power developed (in kW) is a) 177 b) 354 c) 500 d) 707 [GATE–2010] Q.15 A pump handing a liquid raises its pressure from 1 bar to 30 bars. Take the density of the liquid as 990 kg/m3. The isentropic specific work done by the pump in kJ/ kg is a) 0.10 b) 0.30 c) 2.50 d) 2.93 [GATE–2011] Q.16 The velocity triangles at the inlet and exit of the rotor of a turbo machine are shown. V denotes the absolute velocity of the fluid, W denotes the relative velocity of the

b) 1 d) 0.25 [GATE–2012]

Q.17 In order to have maximum power from a Pelton turbine, the bucket speed must be a) equal to the jet speed b) equal to half of the jet speed. c) equal to twice the jet speed d) independent of the jet speed. [GATE–2013] Q.18 Kaplan water turbine is commonly used when the flow through its runner is a) axial and the head available is more than 100 m b) axial and the head available is less than 10m c) radial and the head available is more than 100 m d) mixed and the head available is about 50 m [GATE–2014 (4)] Q.19 An ideal water jet with volume flow rate of 0.05 m3/s strikes a flat plate placed normal to its path and exerts a force of 1000 N. Considering the density of water as 1000 kg/m3, the diameter (in mm) of the water jet is___ [GATE–2014(1)] Q.20 At the inlet of an axial impulse turbine rotor, the blade linear speed

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required to _________

is 25 m/s, the magnitude of absolute velocity is 100 m/s and the angle between them is 25°. The relative velocity and the axial component of velocity remain the same between the inlet and outlet of the blades. The blade inlet and outlet velocity triangles are shown in figure. Assuming no losses, the specific work (in J/kg) is ___.

hold

the

plate

is

[GATE–2017(2)]

[GATE–2014(3)] Q.21

Q.22

Which of the following statements are TRUE, when the cavitation parameter σ = 0? i. the local pressure is reduced to vapor pressure ii. cavitation starts iii. boiling of liquid starts iv. cavitation stops a) i, ii and iv b) only ii and iii c) onlyi and iii d) i, ii and iii [GATE–2015(3)]

Consider two hydraulic turbines having identical specific speed and effective head at the inlet. If the speed ratio (N1/N2) of the two turbines is 2, then the respective power ratio (P1/P2) is _____________. [GATE–2016(1)] Q.23 A 60 mm-diameter water jet strikes a plate containing a hole of 40mm diameter as shown in the figure. Part of the jet passes through the hole horizontally, and the remaining is deflected vertically. The density of water is 1000kg / m3 . If velocities are as indicated in the figure, the magnitude of horizontal force (in N)

Q.24

Which one of the following statement is TRUE? a) Both Pelton and Francis turbines are impulse turbines. b) Francis turbine is a reaction turbine but Kaplan turbine is an impulse turbine. c) Francis turbine is an axial – flow reaction turbine. d) Kaplan turbine is an axial – flow reaction turbine. [GATE–2017(2)]

Q.25 For a Pelton wheel with a given water jet velocity, the maximum output power from the Pelton wheel is obtained when the ratio of the bucket speed to the water jet speed is _______ (correct to two decimal places). [GATE–2018(1)] Q.26 Select the correct statement for 50% reaction stage in a steam turbine. a) The rotor blade is symmetric. b) The stator blade is symmetric. c) The absolute inlet flow angle is equal to absolute exit flow angle. d) The absolute exit flow angle is equal to inlet angle of rotor blade.

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[GATE–2018(2)]

Q.27 A test is conducted on a one-fifth scale model of a Francis turbine under a head of 2 m and volumetric flow rate of 1 m3 / s at 450 rpm. Take the water density and the acceleration due to gravity as 103 kg / m3 and 10 m / s2 , respectively. Assume no losses both in model and prototype turbines. The power (in MW) of a full sized turbine while working under a head of 30 m is _______ (correct to two decimal places). [GATE–2018(2)]

ANSWER KEY: 1 (c) 15 (d)

2 (b) 16 (c)

3 (b) 17 (b)

4 (b) 18 (b)

5 (a) 19

6 7 (b) (c) 20 21 56.43 3280.5 (d)

8 (a) 22 0.25

9 (a) 23 628.32

10 11 (c) (c) 24 25 (d) 0.5

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12 13 14 (a) (d) (b) 26 27 (d) 29.05

EXPLANATIONS Q.1

Q.2

(c) From Buckingham’sπ -theorem, “If there are m variables (Independent and dependent variables)in a physical phenomenon and if these variables contain n fundamental dimensions (M, L, T) then variables are arranged into (mn) dimensionless terms. (b)

= U Speed,

πDN = 60

Q.4

∴ N∝

Q.5

2

H 2  1000  =  30  500  ∴ H2 = 30 × 4 = 120m We also know that specific speed will be constant. Q1 Q ∴ = 32 3 D1 N1 D 2 N 2



Q.3

N  ∴ = Q 2  2  × Q1  N1  constant]  1000  = Q2   × 60  500  Q 2 = 120 lit/min

[Diameter

(b) Curtis→Velocity compounding Rateau→ Pressure compounding Kaplan → Axial flow turbine Francis →Mixed flow turbine

is

fLV 2 0.01× 4000 × ( 2 ) = hf = = 40.77m 2gd 2 × 9.81× 0.2 of water. Now total pressure (absolute discharge pressure) to be supplied by the pump at exit = Pressure loss by pipe + pressure head of tank + Atmospheric pressure head Total pressure, p= ρghf + ρgH +ρghatm= 1000 × 9.81 [40.77 + 5 + 10.3] = 5.503 × 105 N/m2 = 5.503 bar 2

2gH

H D For constant diameter, N ∝ √H 2 H2  N2  ∴ =  H1  N1 

(b) Given: L = 4 km = 4×1000 = 4000 m. d = 0.2 m, f = 0.01, V = 2m/s. Head loss due to friction in the pipe,

(a) Given H = 24.7 m, Q = 10.1 m3/s, ηo = 90 %. N = 4 rps = 4 × 60 = 240 rpm. Shaft power in kW η0 = Water Power in kW P =  p×g×Q× H    1000   p×p×g×Q×H P= 1000 0.90 ×1000 × 9.81×10.1× 24.5 P= 1000 = 2184.74 kW For turbine Specific speed N P Ns = 5/4 H 240 2184.75 = = 216.54 24.55/4 Hence 51
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Q.6 Q.7

(b) So, correct pairs are P-3, Q-5, R-2, S-6 (c) The velocity triangle for the pelton wheel is given below.

Q.8

Jet deflection angle = 1200 ϕ = 1800 – 1200 = 600 Power developed by pelton wheel = ρQ(V ̶ u)( 1+ cos∅ )u =1000×0.1×( 25 ̶10)(1+cos60 ͦ)×10 = 22500 W = 22.5 kW

Q.9

⇒Nm = 2000 Now specific speed will be same N p Pp N P ∴ m 5/4 m = Hm H 5/4 p ⇒ Pm = 2.34 kW

(a) For similar turbines specific power will be same N m Dm N p Dp = Hm Hp

(a) Net positive suction head, NPSH = Pressure head + static head Net Pressure difference,=200 -(-25) = 225 kPa Δp =2.25 bar = 22.95 m of water Static head = 1m (Given) Now, NPSH = 22.95 + 1 = 23.95 m =24 m of water

Q.10 (c) Given figure shows the velocity triangle for the pelton wheel.

Velocity Triangle For Francis Turbine Given : Flow Velocity at Inlet Vf1 = flow velocity at outlet Vf2

V= V= f1 f2

V2 = Vf2

u1 (blade velocity) 2

θ =900 From inlet velocity triangle, 2 2 2 5 2 u  V12 = ( Vf1 ) + ( Vf1 ) =  1  + ( u1 ) = u12 4  2 2 2 v −v Blade efficiency = 1 2 2 ×100 v1

=

5 / 4u12 − u12 / 4 × 100 = 80% 5 / 4u12

Q.11 (c)

πDN = 2gH 60 From this equation √H ∝ DN H = Constant DN So using this relation for the given model an prototype.  H  H   =    DN p  DN m

= u

N p Dm Hp = × N m Dp Hm

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Np

1 × 4 Nm 2 Nm =Np=N Q.12 (a) So, correct pairs are P-2,Q-3,R-4, S-1 =

Q.13 (d) Velocity of pelton wheel = w×r ∴u = 10× 0.5 = 5m/s Torque =𝛒𝛒Q(V-u)(1+cos 𝛉𝛉)×R Torque/mass = (V-u)(1+cos 𝛉𝛉) ×R = (10-5)(1+cos 60°)×0.5 = 3.75(N-m)/kg/s

Q.14 (b) Given: P1 = 103 kW, H1= 40 m, H2= 20 m. If a turbine is working under different heads, the behavior of turbine can be easily known from the values of unit quantities i.e. from the unit power. P1 P2 = 3/2 3/2 H1 H2 P2 =1000 × (20/40)3/2 P2 =353.5 kW

Q.15 (d) Given: P1 = 1bar, P2 = 30 bar, ρ = 990 kg/m3 Isentropic work down by the pump is given by, m m = W vdp = dp, = v p p w 1 1 = dp= × ( 30 − 1) ×105 m p 990 =2929.29 J/kg = 2.93 kJ/kg Q.16 (c) Degree of reaction Enthalpy drop in moving blade = Total enthalpy drop W22 − W12 2 R= 2 2 V1 − V2 W22 − W12 + 14 22 43 14 22 43

(sinceU = U= U) 1 2 Fixed Moving ∵ V1 = W2 andV2 = W1 W22 − W12 R= 2 W2 − W12 + W22 − W12 ∴R=0.5

Q.17 (b) V u= 2 Condititon for maximum power in pelton wheel. Q.18 (b)

Q.19 (56.43)

Q = 0.05 m3/s F= 1000 N F = ρAV2 The above expression can also be written as 4ρQ 2 F= πd 2 4ρQ 2 4 ×1000 × 0.052 = d2 = πd 2 3.14 ×1000 2 ∴ d = 3.1847 × 10−3 Or d=0.05643 Or = 56.43 mm

Q.20 (3280.5)

Specific work = u Cw Given, u=25m/s Cw= EB+BF

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Q.21 Q.22

Let, ∠CBF = θ As axial velocity is same, DE =CF … (i) DE = 100 Sin 25°… (ii) Also, CF= 58.6 Sin θ… (iii) From equation (i), (ii) and (iii), we get 100 Sin25° Sin θ= 58.6 ⇒ θ =46.153° EB= Cw1=100 Cos25° =90.63m/s BF= Cw2=58.6 Cos θ = 58.6 × Cos46.153° =40.59m/s ∴ Cw= Cw1+ Cw2 =90.63 + 40.59 =131.22m/s ∴ Specific work = 25×131.22 = 3280.5 J/kg

&1V1 − m &2 V2 = F m F = ρA1V1V1 − ρA 2 V2 V2 ∴F= ρV12 [A1 − A 2 ] QV = V= 2m / sec 1 2

π π  = 1000 × 202  (0.06) 2 − (0.04) 2  4 4  F = 628.31N

Q.24 In Pelton Wheel turbine for maximum efficiency, u 1 ηmax= = = 0.50 v1 2

Q.25 (d)

(d) σ = 0 implies (i), (ii) and (iii) (0.25) N P N s = 5/4 H

2

P  N  ∴  1 = 2   P2   N1  Q N s1 N= = H 2  s 2 andH1 P1 1 = = 0.25 P2 4 Q.23 (628.31)

50 % reaction stage θ =

α =φ Q.26

It is know that,

(

)

Power, P ∝ Q H ∝ D 2 H H

Holding force in horizontal direction (F) = change in momentum in horizontal direction

⇒ P ∝ D 2 H 3/2 Thus, we can write for similarity between model and prototype Francis turbine.

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D  = p Pm  D m  Dm where, = DP Pp

2

 Hp     Hm 

3/2

1= :5 H P 30 m, = H m 2= m, Q m 1m3 /s and

Pm = ρgQ m H m  Dp  Thus Pp = ρgQ m H m   D   m

2

 Hp     Hm 

2  30  = 104 10     ( 5 )     2  = 29.047 MW = 29.05 MW

3/2

3/2

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ASSIGNMENT QUESTIONS Q.1

Q.2

Q.3

What is the dimension of kinematic viscosity of a fluid a) LT-2 b) L2T-1 -1 -1 c) ML T d) ML-2T-2

The general form of expression for the continuity equation in a Cartesian co-ordinate system for incompressible or compressible flow is given by ∂u ∂v ∂w a) + + = 0 ∂x ∂y ∂z ∂ ∂ ∂ b) (ρu) + (ρv) + (ρw) = 0 ∂x ∂y ∂z ∂ρ ∂ ∂ ∂ c) + (ρu) + (ρv) + (ρw) = 0 ∂t ∂x ∂y ∂z ∂ρ ∂ ∂ ∂ d) + (ρu) + (ρv) + (ρw) = 1 ∂t ∂x ∂y ∂z Point of application of a horizontal force on a curved surface submerged in liquid is

IG −h Ah Ah +h c) IG a)

Q.4

Q.5

2

IG + Ah Ah I d) G + Ah Ah b)

The resultant pressure on a body immersed in a fluid is called a) Fluid pressure b) Up thrust c) Buoyancy d) Gravity The center of pressure of a surface subject to fluid pressure is the point a) On the surface at which resultant pressure acts b) On the surface at which gravitational force acts c) At which all hydraulic forces meet d) Similar to metacenter

Q.6

Q.7

Q.8

Q.9

An odd shaped body weighing 7.5 kg and occupying 0.01m3 volume will be completely submerged in a fluid having specific gravity of a) 1 b) 1.2 c) 0.8 d) 0.75 The flow profile of a fluid depends upon a) Velocity of the fluid only b) the diameter of the tube only c) the Reynolds number d) the surface roughness

Stream line, path line and streak line are identical when the a) Flow is steady b) Flow is uniform c) Flow velocities do not change steadily with time d) Flow is neither steady nor uniform The convective acceleration of fluid in the x-direction is given by ∂u ∂v ∂w a) u +v +w ∂x ∂y ∂z ∂u ∂v ∂w b) + + ∂t ∂t ∂t ∂u ∂v ∂w c) u + u + u ∂x ∂y ∂z ∂u ∂u ∂u d) u + v + w ∂x ∂y ∂z

Q.10 In a two-dimensional flow of a viscous fluid Couette flow is defined for a) pressure gradient driven laminar flow between fixed parallel plates b) pressure gradient driven laminar flow through non-circular duct c) pressure gradient driven laminar flow through pipe

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d) Laminar flow between a fixed and a moving plate

Q.11 Most efficient channel section is a) half hexagon in the form of trapezoid b) triangular c) rectangular d) semicircular Q.12 Which one of the following is a typical example of non-Newtonian fluid of Pseudo plastic variety a) Milk b) Air c) Water d) Printing ink

Q.13 Match List-I (Physical properties of fluid) with List-II (Dimensions/definitions) and select the correct answer using the codes given below the lists: List-I A. Absolute viscosity B. Kinematic viscosity C. Newtonian fluid D. Surface tension List-II 1. Viscosity is constant 2. Newton per meter 3. Poise 4. Stress/strain is constant 5. Stroke Codes: A B C D a) 5 3 1 2 b) 3 5 2 4 c) 5 3 4 2 d) 3 5 1 2 Q.14 Capillarity is due to a) Adhesion of liquid particle to a surface b) Cohesion of liquid particles c) Cohesion and Adhesion both d) Surface tension Q.15 For a Glass tube of diameter d, height of capillary is given by 4wd 4σ cos θ a) b) σ cos θ wd

4σ σ cos θ d) wd cos θ 4wd Where, w= specific weight of liquid σ= surface tension θ= angle of contact between liquid and surface c)

Q.16 The approximate value of θ (angle of contact) for Mercury is a) 132 b) 182 c) 152 d) None of these

Q.17 What should be the property of a good manometer fluid. a) low vapour pressure, low density b) low vapour pressure, high density c) high vapour pressure, high densit d) high vapour pressure, low density Q.18 Fluid is a substance which offers no resistance to change of a) Volume b) Pressure c) Shape d) Flow

Q.19 When a piezometer cannot be used for pressure measurement in pipes a) the pressure difference is low b) the velocity is high c) the fluid in the pipe is a gas d) the fluid in the pipe is highly viscous Q.20 The instrument preferred in the measurement of highly fluctuating velocities in air flow is a) Pitot-static tube b) Propeller type anemometer c) Three cup anemometer d) Hot wire anemometer Q.21 The property by virtue of which a liquid opposes relative motion between its different layers is called a) Surface tension b) Cohesion c) Viscosity d) Caillarity Q.22 The Bulk modulus of elasticity of a fluid is defined as dV / V dP a) − b) − dP dV / V

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c)

dP dV

d)

dP / 8dρ

Q.23 In the phenomenon of cavitation, the characteristic fluid property involved is a) Surface tension b) Viscosity c) Bulk modulus of elasticity d) Vapour pressure Q.24 The difference of pressure between the inside and outside the soap bubble is 4σ 2σ a) b) d d 6σ 8σ c) d) d d Q.25 The difference of pressure between the inside and outside the liquid jet 4σ 2σ a) b) d d 6σ 8σ c) d) d d

Q.26 The difference of pressure between the inside and outside the liquid drop is 4σ 2σ a) b) d d 8σ 6σ c) d) d d Q.27 The buoyant force is a) equal to volume of liquid displaced b) force necessary to maintain equilibrium of a submerged body c) the resultant force acting on a floating body d) the resultant force on a body due to the fluid surrounding it

Q.28 The metacenter is a) centroid of the displaced fluid volume

b) midpoint between centre of gravity and centre of Buoyancy c) the point of intersection of the line of action of Buoyant force and the centre line of the body d) the point of intersection of the line of action of Buoyant force and that of gravitational force

Q.29 The total pressure on a horizontally immersed surface is a) wA b) wx� wA c) wAx� d) x� Where, w = specific weight of the liquid A = area of the immersed surface x� = depth of the center of gravity of the immersed surface from the liquid surface Q.30 When a body, floating in a liquid is given a small angular displacement, it starts oscillating about a point known as a) Center of pressure b) Center of gravity c) Center of buoyancy d) Metacenter Q.31 The time period of oscillation of a floating body is given by

K2 a) 2π hg c)

1 K2 2π hg

b) 2π d)

hg K2

1 hg 2π K 2

Q.32 The time period of oscillation of a floating body with increase in metacentric height will a) Increase b) decrease c) unaltered d) lower/higher depending on weight of body Q.33 A floating body equilibrium when

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is

in

stable

a) Its center of gravity is below the center of buoyancy b) its metacentric height is zero c) its metacentric height is +ve d) Its metacentric height is -ve

Q.34 Which one of the following is the condition for stable equilibrium of a floating body a) The metacenter coincides with the center of gravity b) The metacenter is above the center of gravity c) The metacenter is below the center of gravity d) None of these

Q.35 The pressure center is a) The centroid of the pressure prism b) a point of the line of action of resultant force c) at the centroid of the submerged area d) Always above the centroid of the area Q.36 A submerged body will be in stable equilibrium if a) The centre of buoyancy B is below the centre of gravity G b) The centre of buoyancy B coincides with G c) The centre of buoyancy B is above the metacenter M d) The centre of buoyancy B is above G

Q.37 How is the metacentric height (GM) expressed?  I a) GM = BG −   V V b) GM =   − BG  I  I c) GM =   − BG V V d) GM = BG −    I

Where I = Moment of inertia of the plan of the floating body at the water surface. V = Volume of the body submerged in water. BG = Distance between the center of gravity (G) and the center of Buoyancy (B).

Q.38 The continuity equation represents the conservation of a) Mass b) Momentum c) Energy d) Vorticity

Q.39 The Bernoulli’s equation refers to conservation of a) Mass b)Momentum c) Force d) Energy Q.40 How are the velocity coefficient Cv, the discharge coefficient Cd and the contraction coefficient Cc of an orifice related? a) Cv= CvCd b) CC = CvCd c) Cd= CCCv d) CCCvCd = 1 Q.41 The coefficient of velocity (Cv) for an orifice is a) C v =

4x 2 yH

x2 c) C v = 4yH

b) C v =

2x 4yH

d) None of above

Q.42 The equation of motion for a viscous fluid is known as a) Euler's equation b) Reynolds equation c) Navier strokes equation d) Hagen Poiseuille equation

Q.43 The rate of flow through venturimeter varies as a) H b) H 3/2 c) H d) H 5/2

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Q.44 How is the difference of pressure head h measured by mercury oil differential manometer expressed  Sg  a)= b)h= x[Sg - So] h x 1 −   S0 

 Sg  d) h x  − 1 =  S0  Where x = manometer reading Sg and So are the specific gravities of mercury and oil respectively c) h = x[So - Sg]

Q.45 Differential manometer is used to measure a) very low pressure b) pressure in pipes, channels etc c) atmospheric pressure d) difference of pressure between two points

Q.46 The diameter of the nozzle (d) for maximum transmission of power is given by

D  a) d =    8fl  5

1 2

1

D  b) d =    8fl  5

1 3

1

 D5  4  D5  5 c) d =  d) d =     8fl   8fl  Where D = Diameter of pipe, f =Darcy's friction factor l= Length of pipe

Q.47 The critical value of Reynolds number for transition from laminar to turbulent boundary layers in external flow is taken a) 2300 b) 4000 5 c) 5 × 10 d)3 × 106

Q.48 For pipes, laminar flow occurs when Reynolds number is a) less than 2000 b) between 2000 and 4000 c) more than 4000 d) less than 4000 Q.49 In a two-dimensional flow, the velocity components in x and y

directions in terms of stream function ( ) are ∂ψ ∂ψ a) u = and v = ∂x ∂y ∂ψ ∂ψ b) u = and v = ∂x ∂y ∂ψ ∂ψ c) u = and v = ∂x ∂y ∂ψ ∂ψ d) u = and v =∂x ∂y Q.50 It is recommended that the diffuser angle of a venturimeter should be kept less than 6° because a) Pressure decreases in the flow direction and flow separation may occur. b) Pressure decreases in the flow direction and flow may become turbulent. c) Pressure increases in the flow direction and flow separation may occur. d) Pressure increases in flow direction and flow may become turbulent.

Q.51 If is the velocity potential function in 2-D flow field, then the velocity components u and v are defined as ∂φ ∂φ a) u = and v = ∂x ∂y ∂φ ∂φ b) u =- and v = ∂x ∂y ∂φ ∂φ c) u =and v =∂x ∂y ∂φ ∂φ d) u = and v = ∂x ∂y Q.52 Boundary layer separation is caused by a) Adverse pressure gradient b) laminar flow changing to turbulent flow c) reduction in pressure to vapour pressure d) none of these

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Q.53 Laminar sub-layer acts as a) An insulating medium b) good conductor of heat c) refractory substance d) heat absorber

Q.54 In order to increase .sensitivity of Utube manometer, one leg is usually inclined by angle 𝜃𝜃, sensitivity of inclined tube to sensitivity of U-tube equal to 1 a) sinθ b) sinθ 1 c) cos θ d) cosθ Q.55 At the point of boundary layer separation a) Shear stress is maximum b) shear stress is zero c) Velocity is positive d) density variation is maximum Q.56 The energy thickness δe is given by δ

u  u  1 −  dy u∞  0 ∞  δ u  u2  b) = δe ∫ 1 − 2  dy u u∞  0 ∞  a) = δe

∫u δ

u  u  1−  dy 2  u∞  0 ∞  δ  u  d) δ= − 1  e ∫0  u ∞  dy c) = δe

∫u

Q.57 The displacement thickness δ* is δ u  u  * a) = δ ∫ 1 −  dy u u∞  0 ∞  δ

b) = δ*

u  u2  ∫0 u ∞ 1 − u ∞2  dy

δ

u  u  1−  dy 2  u∞  0 ∞  δ  u  * d) δ= ∫ 1 −  dy u∞  0 c) = δ *

∫u

Q.58 A stream line is a line

a) Which is along path of the particle b) Which is always parallel to the main direction of flow c) Along which there is no flow d) On which tangent drawn at any point gives the direction of velocity Q.59 For a fully developed laminar flow through a pipe the volumetric flow is given by (symbols have the usual meaning) π π 4  −dp   −dp  a) R 4  b) R    8µ  dz  4µ  dz  π 4  −dp  π 4  −dp  c) d) R  R    32µ  dz  16µ  dz  Q.60 Maximum efficiency of transmission of power through a pipe is a) 25% b) 66.66% c) 33.3% d) 50%

Q.61 Power transmitted through a pipe is maximum when H H a) H L = b) H L = 2 3 H c) H L = d) HL = H 4 Where H = total head supplied HL = head loss due to friction

Q.62 Power transmitted through a pipe is given by a) W × Q × H b) W ×Q × HL c) W × Q × (H- HL) d)W×Q×(H+HL) Where W = specific weight of the fluid flowing through pipe Q = discharge Q.63 Boundary layer flow separates from the surface if dp du a) =0 = 0 and dy dx dp du b) >0 = 0 and dy dx dp du c) >0 = 0 and dy dx d) Boundary layer thickness is zero

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Q.64 The hydraulic mean depth for a circular pipe of diameter d when running full is d a) d b) 2 d d c) d) 4 3 Q.65 Which one of the dimensionless numbers identifies the compressibility effect of a fluid a) Euler number b) Froude number c) Mach number d) Weber number Q.66 Capillary tube viscometers used for measurement of viscosity are based on a) Stroke's law b) Hagen-Poiseuille c) Darcy Weisbach equation d) none of these

Q.67 The hydrodynamic boundary layer thickness is defined as the distance from the surface where the a) Velocity equals to the local external velocity b) Velocity equals the approach velocity c) Momentum equals 99% of the momentum of the tree stream d) Velocity equals 99% of the local external velocity Q.68 Laminar boundary layer thickness (δ) at any point x for flow over a flat plate is given by 1.328x 0.664x a) b) Re x Re x c)

1.75x Re x

d)

5.0x Re x

Q.69 The velocity profile for turbulent layer over a flat plate is u x y a)= sin  −  u∞ 2 δ

1

u  y 7 b) =  u∞  δ 

1

u  y   y 7 c) = 2  −   u∞ δ δ

3

u 3 y 1 y d)=  −   u∞ 2  δ  2  δ  Q.70 Given that 𝛅𝛅 = boundary layer thickness 𝛅𝛅* = displacement thickness 𝛅𝛅e = energy thickness θ = momentum thickness The shape factor H is given by

δ a) H = e δ δ c) H = θ

δ* θ δ d) H = * δ b) H =

Q.71 According to Blasius, the local skin friction coefficient in the boundary layer over a flat plate is given by 0.332 0.664 a) b) Re x Re x c)

0.647 Re x

d)

1.328 Re x

Q.72 The thickness of laminar boundary layer at a distance 'x' from the leading edge over a flat plate varies as a) x b) x1/2 c) x1/5 d) x4/5 Q.73 The von-karman momentum integral equation expressed as (where θ is momentum thickness) τ0 ∂θ τ0 ∂θ a) b) = = 2 1 2 ∂x 2ρU ∞ ∂x ρU ∞ 2 τ0 ∂θ τ ∂θ b) 0 2 = d) = 1 2 ∂x ρU ∞ ∂x ρU ∞ 3 Q.74 Geometric similarity between model and prototype means the similarity in a) Motion b) Discharge c) Linear dimensions d) Forces

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Q.75 Kinematic similarity between model and prototype is the similarity of a) Shape b) Discharge/motion c) Stream d) Forces

Q.76 Euler number is defined as the square root of ratio of inertia forces to a) Viscous force b) Elastic force c) Pressure force d) Gravity force Q.77 If 'n' variables in a physical phenomenon contain 'm' fundamental dimensions, then the variables can be arranged into a) n dimensionless terms b) m dimensionless terms c) (n − m) dimensionless terms d) (n + m) dimensionless terms

Q.78 Froude number is the square root of ratio of inertia force to a) Viscous force b) Surface tension force c) Elastic force d) Gravity force Q.79 For laminar flow in pipes the momentum correction factor is a) >1 b) 1.03 c) 1.33 d) 2.00

Q.80 For turbulent flow in pipes the momentum correction factor is a) > 1 b) 1.2 c) 1.33 d) 2.00 Q.81 The logarithmic velocity distribution is observed in a) open channel flow b) laminar boundary layer c) turbulent boundary layer d) laminar boundary layer over a flat plate

Q.82 The specific speed of a turbine is given by N Q N P a) b) 3  32   2 H H       

c)

N P  54  H   

d)

N Q  54  H   

Q.83 Surge tank is used in pipe line a) to reduce frictional loss in pipe b) to ensure uniform flow in pipe c) to relieve the pressure due to water hammer d) to reduce cavitations

Q.84 Which one of the following is not a positive displacement pump a) Reciprocating pump b) Centrifugal pump c) Vane pump d) Lobe pump Q.85 Which one of the following turbine exhibits a nearly constant efficiency over a 60% to 140% of design speed a) Pelton wheel b) Francis turbine c) Deriaz turbine d) Kaplan turbine

Q.86 Which phenomenon will occur when the valve at the discharge end of a pipe connected to a reservoir is suddenly closed a) Cavitation b) Erosion c) Hammering d) Surging

Q.87 The function of guide vanes in a reaction turbine is to a) Allow the water to enter the runner without shock b) Allow the water to flow over theory, without forming eddies c) Allow the required quantity of water to enter the turbine d) all of the above Q.88 The degree of reaction of turbine is defined as the ratio of a) Static pressure drop to total energy transfer b) Total energy transfer to static pressure drop

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c) Change of energy across the turbine to the total energy transfer d) Velocity energy to pressure energy

Q.89 The maximum number of Jets generally employed in impulse turbine without Jet interference is a) 4 b) 8 c) 6 d) Not defined

Q.90 A draft tube is used in reaction turbine a) To guide water downstream without splashing b) To convert residual pressure energy into kinetic energy c) To convert residual kinetic energy into pressure energy d) To stream line the flow in tailrace Q.91 A centrifugal pump gives maximum efficiency when its blades are a) Bent forward b) Bent backward c) Straight d) Wave shaped Q.92 The number of buckets on the periphery of a pelton wheel is given by m m a) 5+ b) 10+ 2 2 m m c) 15+ d) 20+ 2 2

Q.93 If α is the blade angle at the outlet, then the maximum hydraulic efficiency of an ideal impulse turbine is 1 + cos α 1 − cos α a) b) 2 2 1 + sin α 1 − sin α c) d) 2 2 Q.94 Which one of the following is an example of the pure (100%) reaction machine a) Pelton wheel b) Francis turbine c) Modern gas turbine d) Lawn sprinkler

Q.95 Hydraulic efficiency of a turbine is defined as a) Power available at the inlet of turbine to power given by water to runner b) Power at the shaft of the turbine to power given by water to the runner c) Power at the shaft of the turbine to the power at the inlet of turbine d) Power given by water to the runner of a turbine to the power supplied by the water at the inlet of the turbine Q.96 The flow of water leaving the impeller in a centrifugal pump casing is a) Free vortex flow b) Forced vortex flow c) Centrifugal flow d) Coutte flow

Q.97 Operating characteristic curves of a turbine means a) curve drawn at constant speed b) curve drawn at constant efficiency c) curve drawn at constant head d) curve drawn at constant load

Q.98 Air vessel in a reciprocating pump is used a) To obtain a continuous supply of water at uniform rate b) To reduce suction head c) To increase the delivery head d) all of above Q. 99 The stream function is given by Ψ=3xy, then the velocity at the point (2, 3) is a) 9 b) -6 c) 117 d) 10.8 Q.100 Why is multi-staging in centrifugal pumps used a) For high flow rate b) For high head c) For high speed d) For high efficiency

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Q.101 Match List-I(Properties of fluids) with List-II(Definition) and select the correct answer using the codes given below the lists: List-I A. Ideal fluid B. Newtonian fluid C. μ/⍴ D. Mercury in glass List-II 1. Viscosity does not vary with rate of deformation 2. Fluid of zero viscosity 3. Dynamic viscosity 4. Capillary depression 5. Kinematic viscosity 6. Capillary rise Codes: A B C D a) 1 2 4 6 b) 1 2 3 4 c) 2 1 3 6 d) 2 1 5 4

Q.102 The capillary rise at 20°C in clean glass tube of 1 mm diameter containing water is (Take σw−air = 0.0736 N/m) a) 15 mm b) 50 mm c) 20 mm d) 30 mm

Q.103 Consider a soap film bubble of diameter D. If the external pressure is P0 and the surface tension of the soap film is , the expression for the pressure inside the bubble is 2σ a) P0 b) P0 + D 8σ 4σ c) P0 + d) P0 + D D Q.104 The relation between the stress and the strain rate (dux/dy) for the rapid flow of a granular material is 2

 du  given by τ =B  x  where B is a  dy  constant. If M, L and T are the mass,

length and time dimension respectively, what is the dimension of the constant B? a) ML-1T-1 b) ML-1T-2 c) MT-1 d) ML-1

Q.105 An increase in pressure of 2 bars decreases the volume of a liquid by 0.01 percent. The bulk modules of elasticity of the liquid is a) 2×105 N/m2 b)2×107 N/m2 9 2 c) 2×10 N/m d) 2×1011 N/m2

Q.106 The specific weight of ocean water changes according to the equation W=w0+c h where w0 =specific weight of ocean water at the free surface, then pressure as a function of depth is 5 2 a) w0h b) w 0 h + ch 5 2 3 3 2 2 c) w 0 h + ch d) w 0 h + ch 2 3

Q.107 The pressure inside a soap bubble of 50 mm diameter is 25 N/m2 above the atmospheric pressure. What is the surface tension in soap film a) 0.156N/m b) 0.312N/m c) 0.624 N/m d) 0.948 N/m Q.108 The work done in blowing a soap bubble of diameter 12 cm. Assume the surface tension of soap solution =0.040 N/n a) 36.2×10-4 N-m b) 51.8×10-4 N-m c) 36.2×10-3N-m d) 58.8×10-4 N-m Q.109 Match List-I (fluid parameters) with List-II (basic dimensions) and select the correct answer using the codes given below the lists: List-I A. Dynamic viscosity B. Chezy's roughness coefficient C. Bulk modulus of elasticity D. Surface tension ( ) List-II

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1. M/T2 2. M/LT2 3. M/LT 4.�

L

T

Codes: A B C D a) 3 1 4 2 b) 1 4 2 3 c) 3 4 2 1 d) 1 2 4 3 Q.110 Match List-I (phenomenon) with List-II (causes) and select the correct answer using the codes given below the lists: List-I A. Shock wave B. Flow separation. C. Capillary rise D. Cavitation List-II 1 . Surface tension 2. Vapour pressure 3. Compressibility 4. Adverse pressure gradient Codes: A B C D a) 3 1 2 4 b) 4 2 1 3 c) 3 4 1 2 d) 4 1 2 3

Q.111 In an experiment, the following shear stress-time rate of shear strain values are obtained for a fluid Time rate of shear (1/s) Shear stress (kPa)

V2 D μVL c) 2 D

a) Lρ

0 0

2 1.4

3 2.6

How can the fluid be classified a) Newtonian fluid b) Bingham plastic c) Pseudo plastic d) Dilatant

4 4

Q.112 A Newtonian liquid (ρ = density, μ=viscosity) is flowing with velocity V in a tube of diameter D. Let P be the pressure drop across the length L. For a laminar flow, P is proportional to

b) Dρ

V2 L

d) μv/L

Q.113 Consider the following statements 1. Gases are considered incompressible when Mach number is less than 0.2. 2. A Newtonian fluid is incompressible and non-viscous. 3. An ideal fluid has negligible surface tension. Which of these statement(s) is/are correct? a) 2 and 3 b) 2 alone c) 1 alone d) 1 and 3

Q. 114 The standard atmospheric pressure is 762 mm of Hg. At a specific location, the barometer reads 700 mm of Hg. At this place, what does the absolute pressure of 380 mm of Hg corresponds to? a) 320 mm of Hg vacuum b) 382 mm of Hg vacuum c) 62 mm of Hg vacuum d) 62 mm of Hg gauge Q.115 The depth of centre of pressure for a rectangular lamina immersed vertically in water up to height h is given by h h a) b) 2 4 2h 3h c) d) 3 2

Q.116 What is the vertical component of pressure force on submerged curved surface equal to a) its horizontal component b) the product of the pressure at centroid and surface area c) the force on a vertical portion of the curved surface d) the gravity force of liquid vertically above the curved surface up to the free surface

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Q.117 The flow in a pipe whose valve is being opened or closed gradually is an example of a) steady flow b) nonsteady flow c) steady uniform flow d) steady nonuniform flow

Q.118 While measuring the velocity of air ρ =1.2 kg/m3, the difference in the stagnation and static pressure of a Pitot static tube was found to be 380Pa. The velocity at that location in m/sec is a) 24.03 b) 4.02 c) 17.8 d) 25.17 Q.119 A pipe flow system with flow direction is shown in above figure. The following table gives the velocities & the corresponding areas

The value of V2 is a) 2.5 cm/sec b) 5 cm/sec c) 7.5 cm/sec d) 10 cm/sec Q.120 The correct sequence of the centrifugal pump components though which the fluid flows is a) Impeller, suction pipe, foot valve and strainer delivery pipe b) Foot valve and strainer, suction pipe, impeller & delivery pipe c) Impeller, suction pipe, delivery pipe, foot valve and strainer d) suction pipe delivery pipe, Impeller , foot valve and strainer Q.121

Consider flow of oil and water through a channel; the boundary conditions at the interface are a) Velocity & shear stress are continuous b) Shear stress is continuous and velocity is discontinuous c) Shear stress is zero and velocity is continuous d) Shear stress is zero Q.122 A float valve of the ‘ball-cock’ type required to close an opening of a supply pipe feeding a cistern is as shown in the given figure

The buoyant force FB required to be exerted by the float to keep the value closed against a pressure of 0.28 N/mm2 a) 4.4 N b) 5.6 N c) 7.5 N d) 9.2 N

3 Q.123 Given ϕ= 3xy and Ψ= (y2 –x2), the 2 discharge passing between the stream lines through the point (1, 3) and (3, 3) is a) 2 unit b) 4 unit c) 8 unit d) 12 unit

Q.124 A centrifugal pump having an impeller of 10cm diameter discharges 40 litre/second when running at 1000 rpm. The corresponding speed of a geometrically similar pump having an impeller of 40 cm diameter and 0.8 m3/s discharges will be a) 276.4 rpm b) 298.3 rpm c) 312.5 rpm d) 358.2 rpm

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Q.125 A U-tube manometer is used to measure the pressure in an oil pipe A as shown in the figure. The specific gravity of the oil is 0.8 and that of mercury is 13.6. The equivalent gauge pressure nearly.

a) 8.53 kN/M2 c) 13.34 kN/M2

b) 11.77 kN/M2 d) 15.00 kN/M2

Q.126 A multi-tube manometer filled with water up to level A, B and C as shown in the figure is rotated about the vertical axis at A. The water levels at A, B and C will all lie on

a) a circle c) a hyperbola

b) an ellipse d) a parabola

Q.127 What is the absolute pressure at A measured by an open tube manometer as in the figure above? Assume atmospheric pressure as 103 kN/m2 (Where SA and SB are the specific gravities of the two fluids)

a) 78.5 kN/M2 c) 1030 kN/M2

b) 180 kN/M2 d) 103 kN/M2

Q.128 The head loss in a sudden expansion from 8 cm diameter pipe to 16cm diameter pipe in terms of velocity V1 in the smaller pipe is 1  V2  3  V12  a)  1  b)   4  2g  16  2g  c)

1  V12    64  2g 

d)

9  V12    16  2g 

Q.129 If a fluid jet discharging from a 50 mm diameter orifice has a 40 mm diameter at its vena contracts, then its coefficient of contraction will be a) 0.32 b) 0.64 c) 0.96 d) 1.64

Q.130 A U-tube manometer shown attached to an air pipe reads a height of 20 cm of mercury as shown. What is the pressure in kPa in the air pipe?

a) 26.7 b) 32.4 c) 38.6 d) 42.5 Q.131 Pressure drop in a 100 mm diameter horizontal pipe is 50 kPa over a length of 10m. The shear stress at the pipe wall is a) 0.25 kPa b) 0.125 kPa c) 0.50 kPa d) 25.0 kPa

Q.132 In flow through a pipe, the transition from laminar to turbulent flow does not depend on a) density of fluid b) length of pipe c) diameter of pipe d) velocity of the fluid Q.133 Two pipe lines of equal length are connected in series. The diameter of second pipe is two times that of first pipe. The ratio of head loss between first and second pipe

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a) 1 : 32 c) 1: 8

b) 32: 1 d) 1 : 4

Q.134 Darcy-Weisbach equation for the head loss in a flow through a pipe is 4fLV 2 given by h f = (The symbols 2gd have the usual meaning) for the laminar flow through a circular pipe, how does the friction factor f vary with a Reynolds number (Re)? 8 16 a) f= b) f= Re Re 32 64 c) f= d) f= Re Re Q.135 In a fully turbulent flow through a rough pipe, the friction factor 'f' is (Re is Reynolds number and ξSD is relative roughness) a) a function of Re b) a function of Re and ξS D c) a function of ξS D d) independent of Re and ξS D Q.136 The Chezy's constant c is related to Darcy-weishbach friction factor 'f' as

g  a) C =    8f 

 8g  c) C =    f 

  b) C =  g  1  4 f 

 f  d) C =    8g 

Q.137 The stream function for a twodimensional flow is given by ψ = 2xy. The velocity at (2,2) is a) 4 2 b) 4 c) 2 2 d) 2 Q.138 The velocity distribution in a turbulent boundary layer is given by 1

u  y 7 =   What is the displacement v δ thickness δ* δ a) δ b) 7 7 δ c) δ d) 8 8

Q.139 In a laminar boundary layer, the velocity distribution can be assumed to be given in usual notation as u y = . Which one of the following is v δ the correct expression for the displacement thickness (δ*) for this boundary layer δ a) δ*= δ b) δ*= 2 δ δ c) δ* = d) δ* = 6 4

Q.140 Using the Prandtl's mixing length concept, how is the turbulent shear stress expressed? du du a) ρl b) ρl2 dy dy 2 2  du   du  c) ρlv   d) ρl2    dy  dy   Q.141 If the stream function is given by ψ = 3xy then the velocity at a point (2 , 3) will be a) 7.21 unit b) 18 unit c) 10.82 unit d) 54 unit

Q.142 Both free vortex and forced vortex can be expressed mathematically in terms of tangential velocity v at the corresponding radius r. Choose the correct combination Free vortex Forced vortex a) v = r (constant) vr = (constant) b) v2r = (constant) v = r (constant) c) vr = (constant) v2=r (constant) d) vr = (constant) v = r (constant)

Q.143 Air flows past a golf ball of 20 mm radius. It is observed that the boundary layer becomes turbulent at Reynolds number of 2 × 105. If the kinematic viscosity of air is 1.5 × 10-5 m2/sec then the speed of air when the flow becomes turbulent is a) 150 m/sec b) 90 m/sec c) 75 m/sec d) 62.5 m/sec

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Q.144 An open circular cylinder of 1.2m height is filled with a liquid to its top. The liquid is given a rigid body rotation about the axis of the cylinder & the pressure at the centre line at the bottom surface is found to be 0.6 m of liquid. What is the ratio of volume of liquid spilled out of the cylinder to the original volume a) 1/4 b) 3/8 c) 1/2 d) ¾

Q.145 By which one of the following, a small quantity of water may be lifted to a great height a) Hydraulic ram b) Hydraulic crane c) Hydraulic lift d) Hydraulic coupling Q.146 As water flows through the runner of a reaction turbine, pressure acting on it would vary from a) more than atmospheric pressure to vacuum b) less than atmospheric pressure to zero gauge pressure c) atmospheric pressure to more than atmospheric pressure d) atmospheric pressure to vacuum

Q.147 Which of the following statements is NOT correct in the context of laminar flow through with pipeline? a) Shear stress is zero at the centre & varies linearly with pipe radius b) Head loss is proportional to square of the average flow velocity c) The friction factor varies inversely with flow Reynolds number d) No dispersion of die injected into the flow stream Q.148 A penstock pipe of 10 m diameter carries water under a pressure head of 100 m. If the wall thickness is 9 mm, what is the tensile stress in the pipe wall in MPa? a) 27.25 b) 544.4 c) 272.5 d) 1090 Q.149

The characteristics of a pump are as shown in the given figure. Based on this figure, match List-I with List-II choose the correct answer using the codes given below the list: List-I A. Curve P B. Curve Q C. Curve R List-II 1. Discharge versus head 2. Head versus discharge 3. Power versus discharge. 4. Efficiency versus discharge Codes: A B C a) 2 4 3 b) 1 3 2 c) 1 4 3 d) 4 3 1

Q.150 A pump running at 1000 rpm consumes 1 kW and generates head of 10m of water. When it is operated at 2000 rpm. Its power consumption and head generated would be a) 4 kW, 50 m of water b) 6 kW, 20m of water c) 3 kW, 30 m of water d) 8 kW, 40 m of water Q.151 Match List-I (Type of pump) with List-II (Liquid handled) and select the correct answer using the codes given below the lists: List-I A. Closed impeller pump B. Semi-open impeller pump C. Open impeller List-II 1. Sandy water 2. Acids 3. Sewage water Codes: A B C

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a) b) c) d)

1 3 2 1

3 1 1 2

2 2 3 3

Q.152 The transition Reynolds number for flow over a flat plate is 5 × 105. What is the distance from the leading edge at which transition will occur for flow of water with a uniform velocity of 1m/sec (for water the kinematic viscosity, υ=0.86×10-6 m2/sec) a) 0.43 m b) 1 m c) 43 m d) 103 m

Q.153 A hydraulic press has a ram of 15 cm diameter and plunger of 1.5 cm. It is required to lift a mass of 100 kg. The force required on plunger is nearly equal to a) 100 N b) 1000 N c) 10,000 N d) 10 N

Q.155 Priming is necessary in a) Centrifugal pumps to lift water from a greater depth. b) Centrifugal pumps to remove air in the suction pipe and casing c) Hydraulic turbine to remove air in the turbine casing d) Hydraulic turbine to increase the speed of turbine and to generate more power.

Q.156 A pressure drop for a relatively low Reynolds number flow in a 600 mm, 30 m long pipe line is 70 kPa. What is the wall shear stress a) 0 Pa b) 1400 Pa c) 700 Pa d) 350 Pa

Q.154 Match List-I (Hydraulic turbine) with List-II (Application area) and select the correct answer using the codes given below the lists. List-I A. Pelton turbine B. Francis turbine C. Kaplan turbine List-II 1. Low head, high discharge 2. Medium head, medium discharge 3. High head , low discharge Codes: A B C a) 2 3 1 b) 2 1 3 c) 3 1 2 d) 3 2 1

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ANSWER KEY: 1 (b) 15 (b) 29 (b) 43 (b) 57 (d) 71 (b) 85 (d) 99 (d) 113 (d) 127 (b) 141 (c) 155 (b)

2 (c) 16 (a) 30 (d) 44 (d) 58 (d) 72 (b) 86 (c) 100 (b) 114 (a) 128 (d) 142 (d) 156 (d)

3 (b) 17 (b) 31 (a) 45 (d) 59 (a) 73 (c) 87 (d) 101 (c) 115 (c) 129 (b) 143 (c)

4 (c) 18 (c) 32 (b) 46 (c) 60 (b) 74 (c) 88 (a) 102 (d) 116 (d) 130 (a) 144 (a)

5 (a) 19 (c) 33 (c) 47 (c) 61 (b) 75 (b) 89 (c) 103 (d) 117 (b) 131 (b) 145 (a)

6 (d) 20 (d) 34 (b) 48 (a) 62 (c) 76 (c) 90 (c) 104 (d) 118 (d) 132 (b) 146 (a)

7 (c) 21 (c) 35 (b) 49 (c) 63 (b) 77 (c) 91 (b) 105 (c) 119 (b) 133 (b) 147 (b)

8 (a) 22 (b) 36 (d) 50 (c) 64 (c) 78 (d) 92 (c) 106 (c) 120 (b) 134 (d) 148 (b)

9 (d) 23 (d) 37 (c) 51 (c) 65 (c) 79 (c) 93 (a) 107 (a) 121 (b) 135 (b) 149 (a)

10 (d) 24 (c) 38 (a) 52 (a) 66 (b) 80 (b) 94 (b) 108 (a) 122 (a) 136 (c) 150 (d)

11 (a) 25 (b) 39 (d) 53 (a) 67 (d) 81 (c) 95 (d) 109 (c) 123 (d) 137 (a) 151 (c)

12 (a) 26 (a) 40 (c) 54 (b) 68 (d) 82 (c) 96 (b) 110 (c) 124 (c) 138 (d) 152 (a)

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13 (d) 27 (d) 41 (c) 55 (b) 69 (b) 83 (c) 97 (a) 111 (d) 125 (b) 139 (b) 153 (d)

14 (c) 28 (c) 42 (c) 56 (b) 70 (b) 84 (b) 98 (a) 112 (c) 126 (d) 140 (d) 154 (d)

EXPLANATIONS Q.1

Q.2

(b) Unit of kinematic viscosity is m2/s. ∴Dimension of kinematic viscosity is L2T-1

(c) The general form of expression for the continuity equation in a Cartesian co-ordinate system ∂ρ ∂ ∂ ∂ + (ρu) + (ρv) + (ρw) = 0 ∂t ∂x ∂y ∂z

Q.3

(b)

Q.4

I C.P = G + h Ah Where h is distance of centre of gravity from free surface and Ic area moment of inertia.

Q.5

Q.6 Q.7

(c)

(a) Centre of pressure is defined as the point of application of the total pressure on the surface. (d)

Q.8

(c) Reynolds number is the factor by which type of flow is decided i.e. laminar or turbulent.

Q.9

(d)

(a)

ax = u

∂u ∂u ∂u ∂u +v +w + ∂x ∂y ∂z ∂t

Where

u

∂u ∂u ∂u +v +w ∂x ∂y ∂z

convective acceleration and the temporal acceleration.

is the

∂u is ∂t

Q.10 (d) Coutte flow is characterized as flow of very low value of Reynolds number between two parallel plate, one is fixed and other is movable. Q.11 (a) Q.12 (a) Q.13 (d) Q.14 (c) Q.15 (b) Weight of liquid raised or lowered in the capillary to be = (area of tube × rise or fall) × specific weight

π 2  d hw 4 

=

Vertical components of surface tension force = σ cos θ ×circumference = σ cos θ × πd = π d σ cos θ In equilibrium, the downward weight of the liquid column h is balanced by the vertical component

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of the force of surface tension. Hence π 2 d hw =πdσ cos θ 4 Height of capillarity 4σ = h cos θ wd Q.16 (a)

Q.17 (b) Manometric fluid should have low vapor pressure and high density to have less height of piezometric tube. Since, Mercury has both the properties, this is the reason, Mercury is used as a manometric fluid. Q.18 (c) A fluid does not offer any resistance to change of shape and gets deformed under the action of very small shear force.

Q.19 (c) Limitations of Piezometer:• Piezometer cannot be used to measure pressure which is considerably excess of atmospheric pressure. Use of very long glass tube would be unsafe. It being both fragile and unmanageable. • Gas pressure cannot be measured as gas does not form any free surface with atmosphere. • Measurement of negative pressure is not possible due to flow of atmospheric air into the container through the tube. Q.20 (d) Hot wire anemometer is used in the measurement of highly fluctuating velocities in air flow.

Q.21 (c) From the definition of viscosity, viscosity is the resistance offered by one layer of fluid to another layer & opposes relative motion between them. Q.22 (b)

dp dv v It represents the change in volume with the change in pressure.

Bulk modules of elasticity, K= −

Q.23 (d) In any flow system, if the pressure at any point in the liquid approaches the vapor pressure, vaporization of liquid starts, resulting in the pockets of dissolved gasses and vapors. The bubbles of vapor thus formed are carried by the flowing liquid into a region of high pressure where they collapse, giving rise to high impact pressure. This phenomenon is known as cavitation.

Q.24 (c) A spherical soap bubbles has surfaces in contact with air, one inside and other outside, each one of which contributes the same amount of tensile force due to surface tension. As such, on a hemispherical section of a soap bubble of radius r, the tensile force due to surface tension is equal to 2σ(2πr). However the pressure force acting on the hemispherical section of the soap bubble is same as in the case of a droplet and it is equal to ΔP(πr2). Thus equating these two forces for equilibrium, we have ΔP(πr2) = 2σ (2πr) 4σ 8σ ∆P= = r d Q.25 (b)

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Consider a jet of liquid of radius r, length l and having internal pressure intensity p in excess of the outside pressure intensity. P(2rl)=σ(2l) σ 2σ or P= = r d Q.26 (a) Consider a spherical droplet of radius r having internal pressure intensity P in excess of the outside pressure intensity. If the droplet is cut into two halves, then the forces acting on one half will be those due to pressure intensity P in the projected area (πr2) and the tensile force due to surface tension σ acting around the circumference (2πr). These two forces will be equal and opposite for equilibrium and hence we have P(πr2) = σ(2πr) 2σ 4σ Or= P = r d Q.27 (d) A body submerged in a fluid experiences an upward thrust due to fluid pressure. This force is called buoyant force and a body immersed in a fluid is buoyed or lifted up by a force equal to the weight of the fluid displaced by the body. The body apparently loses as much of its weight as the weight of the fluid displaced by it. Q.28 (c) Metacentre is the point of intersection of normal axis of the body with the new line of action of buoyant force when body is tilted. Q.29 (b) F = wAx P = wx Where w is the weight density. Q.30 (d)

Q.31 (a) Time of oscillation of a floating body

k2 gh Where, k = radius of gyration h = metracentric height Q.32 (b) Time of oscillation is T = 2π

k2 T = 2π gh

1 h as h increases, time of oscillation decreases. T∝

Q.33 (c) For stable equilibrium

Q.34 (b) For stable equilibrium

For stable equilibrium metacenter is above the center of gravity.

Q.35 (b) Q.36 (d) Q.37 (c)

Metacentric height GM = BM – BG

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I V Where I = Area moment of inertia of top view about longitudinal axis V = Volume of the fluid displaced Hence, I GM =  -BG V Q.38 (a) The continuity equation represents the conservation of mass. BM =

Q.39 (d) Bernoulli’s equation refers conservation of energy.

to

Q.40 (c) The relation between Cd , Cc and Cv is Cd = Cc Cv Q.41 (c)

V C = = v Vth

gx 2 1 × 2y 2gh

 Sg  = h x  − 1  S0  Where, Sg=Specific gravity Mercury S0 = Specific gravity of Oil

of

Q.45 (d)

Q.46 (c) For maximum power transmission through nozzle: 1/4

 D5  Nozzle diameter, d =    8fl  Q.47 (c)

Q.48 (a) For internal flow or pipe flow (Re)cr = 2000 i.e. Re<2000 – Flow is laminar Re>2000 – Flow is turbulent Q.49 (c)

Q.50 (c) If diffuser angle is more, eddies formation may take place and also flow separation may occur. Q.51 (c) 2

Cv

x x ⇒ 4yH 4yH

Q.42 (c) In Naiver-stroke equation pressure force, gravity force and viscous force are considered.

Q.43 (b) Rate of flow through a venturimeter A1A 2 is given as: Q = 2gh A12 −A 22 Hence Q ∝ H

Q.44 (d)

Q.52 (a) Boundary layer separation is caused due to adverse pressure gradient or positive pressure gradient.

Q.53 (a) In turbulent flow analysis along the boundary layer, a thin layer of fluid in the immediate neighborhood of the boundary develops where viscous shear stress is predominant while the shear stress due to turbulence is negligible. This portion is known as laminar sub layer. In turbulent flow hear transfer is more as compared to laminar flow and also in this laminar sub layer heat transfer is very less while, it is

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present in turbulent flow hence it acts as insulating medium in turbulent flow.

Q.54 (b) Q.50 (b)

At the verge of separation

∂u =0 ∂y y =0

 ∂u  Shear stress τ =  . Hence τ = 0  ∂y  y =0 Q.56 (b) Q.57 (d) Q.58 (a) Stream line is an imaginary line drawn in space such that tangent drawn to it at any point gives velocity at that point. Q.59 (a)

Vmax   Q= A × Vavg  Vavg =  2   1  1  dp   Q= πR 2   −  R 2  2  4µ  dz   π 4  −dp  Q= R   8µ  dz 

Q.60 (b) Efficiency of power transmission is given by H − HL η= H H For maximum efficiency, H L = 3 H−H/3 We get, ηmax = H ηmax = 66.66% Q.61 (b) For maximum power transmission H HL = 3 Q.62 (c)

Q.63 (b) Boundary layer separation occurs at adverse pressure gradient i.e. du dp =0 > 0 and dy dx Q.64 (c)

Q.65 (c) At higher Mach number the compressibility effect is predominant. Q.66 (b)

Q.67 (c) Hydrodynamic boundary layer thickness is defined as the distance measured in the y direction from boundary to the point where the velocity is 99% of the free stream velocity. Q.68 (d)

From the blasius solution δ =

5x Re x

Q.69 (b) Turbulent velocity profile follows U y 1/7 power law. Hence =  U∞  δ  will be velocity distribution.

Q.70 (b) Shape factor is given by Displacement thickness H= Momentum thickness Q.71 (b) Q.72 (b)

= δ δ∝x

1 1− 5x ⇒δ∝x 2 Re x 1 2

Q.73 (c)

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1

7

Von-karman

momentum

integral τ0 ∂θ equation is expressed as = 2 ρU ∞ ∂x

Q.74 (c) Geometric similarity ⇒ Similarly in linear dimensions. Q.75 (b) If the ratio of velocity and acceleration at the corresponding points in the model and prototype are same then there exists kinematic similarity. Q.76 (c)

Euler Number =

Q.84 (b)

Ineria force Pr essure force

Q.77 (c) If ‘n’ is the number of variables and ‘m’ is the fundamental dimensions, then variable can be arranged into (n - m) dimensionless group known as π terms. Q.78 (d)

Froude number =

during speed regulation of a turbine. Due to rapid velocity fluctuations, large magnitude pressure transients are set-up and these excessive pressures may lead to bursting of pipe. This phenomenon is known as water hammer. Some arrangement needs to be made to convert the rapid velocity fluctuations into slow velocity fluctuations and for this surge tank is provided just upstream of the power unit in a hydroelectric plant.

Ineria force Gravity force

Q.79 (c)

Q.80 (b) Q.81 (c) Logarithmic profile turbulent flow.

exist

is

in

Q.82 (c)

Q.83 (c) Rapid velocity fluctuations occur in pipe line during 1. Sudden opening and closing of valves 2. Start and shut down of a turbine or closure of wicket GATE’s

Note that centrifugal pump is not positive displacement pump. It is a roto dynamic or dynamic pressure pump.

Q.85 (d) A special type of propeller turbine is called Kaplan turbine in which the individual runner blades are pivoted on the hub so that their inclination may be adjusted automatically rotating about pivots with the help of a governor servo-mechanism. In Kaplan turbine, because of the arrangement for automatic variation of inlet blade angle with variation in load, the turbine can be run at max efficiency at all loads.

Q.86 (c) When valve at the discharge end is closed suddenly, this causes a sudden rise in pressure. This phenomenon of sudden rise in pressure is called

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water hammer. To reduce the effect of water hammer surge tank is provided.

Q.87 (d) Guide vanes are the vanes which are attached to the casing and guides the water to enter into the runner.

Q.88 (a) Degree of reaction is defined as the ratio of pressure head drop within the runner to the total energy drop in the runner.

Q.89 (c) Q.90 (c) It is a diverging tube which is attached at outlet of the runner to carry water from the exit of runner to tail race. The diverging tube helps to increase the pressure on the account of conversion of kinetic head into pressure head. Q.91 (b) Backward blades are used for maximum efficiency. It gives better performance over wide range of application where forward blades are used for higher pressure ratio.

Q.92 (c) Number of buckets on the periphery of a pelton wheel is given by an empirical relation ship D m Z = 15 + , where m = d 2 Q.93 (a)

Q.94 (b) Lawn sprinkler is an example of 70% reaction machine. While pelton when is an example of 100% impulse machine. Q.95 (d)

Hydraulic Efficiency: It is defined as the ratio of power given by the water to runner of a turbine (runner is a rotating part of a turbine and on the runner vanes are fixed) to the power supplied by the water at the inlet of the turbine.

Q.96 (b) Q.97 (a)

Q.98 (a) Functional or air reciprocating pump

vessel

in

a

1. To obtain a continuous supply of liquid at a uniform rate. 2. To save a considerable amount of work in overcoming the frictional resistance in suction and delivery pipes. 3. To run the pump at a high speed without separation.

Q.99 (d) Ψ = 3xy −∂Ψ U= = −3x ∂y =−3 ×2=−6 m/s ∂Ψ ∴v = = 3y = 9m / s ∂x

= v

u 2 + v2

= 62 + 92 = 36 + 81 = = = 117 10.817m / s

Q.100 (b)

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Multi-staging (series combination) of pump is used to develop high head.

Q.101 (c)

Q.109 (c)

Q.102 (d) Heights of Capillary (h) =

=

4σ cos θ wd

4 × 0.0736 = 0.03 m = 30 mm 9810 ×10−3

Q.103 (d)

P = P0 +

Q.104 (d)

2

 12  = 0.040 × 4π  ×10−2  × 2  2  -4 = 36.2 × 10 N-m

8σ D

 du  τ = B   dy 

2

 du  F = B  A  dy 

2

⇒B= ML−1

Q.111 (d) Q.112 (c)

Q.105 (c) Bulk modulus of elasticity K =

−dp dv v

−(2 ×105 ) K= = 2 X 109 N / m 2 −0.01 100

Q.106 (c) dp dp =w ⇒ =w 0 + c h dh dh = ∫ dp ∫ (w 0 + c h )dh

2 = P w 0 h + ch 3

N −s M = m2 LT Chezy’s roughness coefficient L → g= T Bulk modules of elasticity N M → 2 → LT 2 m N M Surface tension → → 2 m T Q.110 (c) Dynamic viscosity →

(given)

Q.113 (d) Q.114 (a) Reference is always taken from atmospheric pressure which is indicated by Barometer. Barometer reads the local atmospheric pressure. So pressure = 700 – 380 = 320 mm of Hg vacuum. Q.115 (c)

3 2

Q.107 (a)

8σ 8× σ ⇒ 25= d 0.05 25 × 0.05 = σ = 0.156N / m 8 ∆p=

Q.108 (a) W.D= surface tension × total surface area

IG Ax b × h3

C.P = C.P= x +

h + 2 12 × b × h × h 2 h h 2h + = 2 6 3

C.P =

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Q.116 (d) Vertical component of pressure force on the submerged curved surface is equal to the gravity force of liquid vertically above the curved surface upto the free surface. Q.117 (b)

Q.118 (d) P = ρgh P h= ρg 380 h= 1.2 × 9.812 V = 2gh V=

2 × 9.812 × 380 1.2 × 9.812

V = 633.33 V = 25.17 m/sec

Q.119 (b) Qin = Qout Q1 + Q2 = Q3 + Q4 (A1V1)+(A2V2) = (A3V3)+(A4V4) 500 + 50 V2 = 400 + 350 50 V2= 250 V2= 5 cm/sec Q.120 (b)

Q.123 (d) Note the difference between two stream line is discharge per unit width dQ = ψ 2 − ψ 1 3 ψ (at1,3)= (9 − 1)= 12 unit 2 Ψ (at 3,3) = 0 ⇒ dQ = 12 unit

Q.124 (c) = D1 10 = cm, Q1 40 lit / s= 40 ×10−3 m3 / s N1 = 1000 rpm D2 = 40 cm,Q2 = 0.8 m3/s From the model relationship Q1 Q2 = 3 N1D1 N 2 D32 N 2 = 312.5rpm Q.125 (b) Pressure at A= ρm gh1 - ρ0gh2 = [13.6 × 9.81× 0.1 − 0.8 × 9.81× 0.2] ×103

= 11.77 ×103 N / m 2 = 11.77 kN / m 2

Q.126 (d) The free surface joining the tube will make a parabola and the water level in tubes will be highest in C and lowest in A. It means that water level in A will decrease from original level and increase in B and C from their original level. Q.127 (b) PA – ( p A gh A ) – ( p m gh m ) – Pa = 0 PA − (0.8 ×103 × 9.812 ×1 .25) −

(13.6 10 × 3 × 9.812 × 0.5) − 103 ×103 = 0 3 PA= [9.812 + 66.7216 + 103] ×10             Q.121 (b) Q.122 (a) Taking moment about hinge. We get π Fg × 500= 0.28 × ×102 ×100 4 Fg = 4.4N

PA = 179.53 KPa Q.128 (d)

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h L1 32D5 32 = = h L2 D5 1 Q.134 (d)

 f =

∴ A1V1 = A 2 V2 V ∴ V2 = 1 4

Q.135 (b) 2

V  V1 − 1   2 2 ( V1 − V2 = )  4  V1 3 h= = × h 2g 2g 2g 42 2

hL =

64 Re

9  V12    16  2g 

Q.129 (b) Area of orifice Area of jet π 2 ( 40 ) = CC 4= 0.64 π 2 ( 50 ) 4 CC =

Q.130 (a)

Q.136 (c) Q.137 (a)

dψ u= − ⇒ dy u = -2x u = -4 at (2,2) ∂ψ v= ⇒ v= 2y ∂x v = 4 at (2,2) Resultant =

16 + 16 = 4 2

Q.138 (d) Displacement thickness δ

 u δ*= ∫ 1 − dy= v 0

 y 7 ∫0 1 −  δ  dy

δ

Q.131 (b)  dp  R τ=-    ∂x  2 ∵ dp = −ve 50 ×1000 × 50 ×10−3 ∴τ = 10 × 2

= 125Pa = 0.125KPa

Q.132 (b) Transitional flow depends upon Reynolds number ∴ Re depends on density of fluid Velocity of fluid Viscosity of fluid Diameter of pipe Q.133 (b)

hL ∝

1

δ

1 D5

= y−

1 +1 7

1( y ) 1 1  (δ ) 7  + 1 7  0 8

8 1 − 7 (δ ) 7 7 7 7 = = − δ− δ δ ( ) 1 8 8 (δ ) 7 7 δ = δ − ( δ) = 8 8

Q.139 (b) = δ*

δ

 u ∫0 1 − v  dy=

δ



0

δ

= y−

y2  δ δ = δ− =  2δ  0 2 2

Q.140 (d) Turbulent shear stress

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y

∫ 1 − δ  dy

  τ = ρu ' v ' ⇒ u ' = v ' = l  du  ⇒τ= ρl    dy 

du dy

2

2

Q.141 (c)

dψ dψ ;v = dy dx δ ⇒u= − ( 3xy ) ⇒ u = −3x  δy ⇒ at(2,3) ⇒ u =−3 × 2 =−6 unit ∂ = v (3xy)= ⇒ v 3y ∂x ⇒ at(2,3)v = 3 × 3 = 9 unit u=−

Resultant = u 2 + v2 = 36 + 81 = 10.82 unit Q.142 (d) Q.143 (c)

ρVD vD Re =× 2 105 = = μ υ

V × 40 ×10−3 1.5 × 10−5 3 ×103 = = 75m / sec 4 ⇒=

Q.144 (a) Q.145 (a) Q.146 (a) At inlet of reaction turbine water is admitted with high pressure head (above atmospheric pressure) with some kinetic energy. The pressure head of water gradually decreases from inlet to exit. Finally water leaves the runner with low pressure and small kinetic energy to tail race. Quit often, the pressure head of the water at turbine exit fall below the ambient pressure. A draft tube attached to the exit of the turbine, working like a diffuser converts the remaining

kinetic energy of water discharged by the runner into the pressure energy so that water is discharged to the tail race with a positive pressure head to avoid cavitation.

Q.147 (b) Laws of fluid friction for laminar flow are as follows: Head loss is i) proportional to the velocity of flow ii) independent of the pressure iii) proportional to the area of surface in contact iv) independent of the nature of the surface in contact. v) greatly influenced by the variation of the temperature of the flowing fluid. So, option (b) is wrong and valid for turbulent flow. Q.148 (b) Tensile stress in the pipe wall = Circumferential stress in pipe wall Pd Where, P=ρgH=980000 N/m2 = 2t 980000 ×10 ∴ Tensile stress = 2 × 9 ×10−3 = 544.4 × 106 N/m2 = 544.4 MN/m2 = 544.4 MPa Q.149 (a) Operating characteristic curve of a centrifugal pump

Q.150 (d) For head H ∝ N2

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H 2 N 22 = H1 N12 ∴ H2 = 40 m For power P ∝ N3 P N3 ∴ 2 =23 P N1 ∴P2 = 8 kW ∴

Q.151 (c) Closed impeller provides better guidance for the liquid and most suited when the liquid to be pumped is pure and comparatively free from debris. Hence acid is the suitable candidate. Semi-open type impeller is suitable even if the liquids are charged with some debris. Hence sandy water is the suitable this for semi-open impellers. Open impellers are useful in the pumping liquids containing suspended solid matter such as paper pulp, sewage and water containing sand or grit. These impellers are less liable to clog when handling liquids charged with a large quantity of debris.

Kaplan Turbine: It is low head (less than 60m) and high discharge turbine.

Q.155 (b) Filling the pump casing and the suction pipe with the liquid before it is started is known as priming. Unless the centrifugal pump is primed, it will not start as there is not centrifugal head impressed on the liquid. Positive displacement type reciprocating or rotary pump do not needed priming because the air if any in the cylinder or casing is driven out by the piston or vanes. Q.156 (d)

−dp  R    dx  2  ∵ Pressure drop is there ∴ dp = -70 KPa −[−70 ×103 ] 0.3 = ∴τ = × 350Pa 30 2 τ=

Q.152 (a)

Ux 5 ×105 = υ 5= ×105

x = 0.43m 0.86 ×10−6

Q.153 (d) F F =     A  Punger  A  ram F 100 ×10 = 2 π / 4(1.5) π / 4(15) 2 F = 10 N

Q.154 (d) Pelton wheel: it is high head and low discharge turbine. Francis turbine: it is medium head (60-250m) and medium discharge turbine. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

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