FLUID FLUID MECHANICSMECHANICS

ENGR. RICHARD T. BARNACHEACE Instructor

REFERENCES:OFluid Mechanics 8th Edition

by: Fox, McDonald, PritchardOEssentials of Fluid Mechanics by:

John M. Cimbala Yunus A. Cengel

OMechanics of Fluids 4th Editionby: Potter, Wiggert, Ramadan

OMechanics of Fluids 6th Editionby: Frank M. White

What is Fluid Mechanics?What is Fluid Mechanics?OFluid MechanicsFluid Mechanics is defined as the science

that deals with the behavior of fluids at rest (fluid statics) or in motion (fluid dynamics).

O It deals with liquids and gases in motion or at rest.

OMechanicsMechanics is the oldest physical science that deals with both stationary and moving bodies under the influence of forces.

OFluid Mechanics itself is also divided into several categories.

The study of the motion of fluids that are practically incompressible (such as liquids, especially water, and gases at low speeds) referred to as hydrodynamicshydrodynamics.

A subcategory of hydrodynamics is hydraulicshydraulics, which deals with liquid flows in pipes and open channels.

A subcategory of hydrodynamics is hydraulicshydraulics, which deals with liquid flows in pipes and open channels.

Gas dynamicsGas dynamics deals with the flow of gases that undergo significant density changes, such as flow of gases through nozzles at high speeds.

The category of aerodynamicsaerodynamics deals with the flow of gases (especially air) over bodies such as aircraft, rockets, and automobiles at high or low speeds.

What is a What is a FluidFluid??a substance in the liquid or gas

phase. a substance that deforms

continuously under the application of the shear stress no matter how small the shear stress may be.

Difference in behavior of a solid and a fluid due to a shear force.

F

Fa. Solid or Fluid b. Solid or Fluid

c. Fluid only

Classification of Fluid FlowsClassification of Fluid FlowsThere is a wide variety of fluid flow problems encountered in practice, and is usually convenient to classify them on the basis of some common characteristics to make it feasible to study them in groups.Internal versus External FlowCompressible versus Incompressible FlowLaminar versus Turbulent FlowSteady versus Unsteady Flow

Internal versus External Flow A fluid flow is classified as being internal

or external, depending on whether the fluid flows is confined space or over a surface.

The flow of an unbounded fluid over a surface such as a plate or a wire is external flowexternal flow. Example airflow over a ball.

The flow in a pipe or duct is internal internal flowflow if fluid is completely bounded by solid surfaces. Example water flow in a pipe.

Compressible versus Incompressible

A flow is classified as being compressible or incompressible, depending on the level of variation of density during flow. Incompressibility is an approximation in which a flow is said to be incompressibleincompressible if the density remains nearly constant throughout.

Laminar versus Turbulent FlowSome flows are smooth and orderly while

others are rather chaotic.The highly ordered fluid motion characterized by smooth layers of fluid is called laminarlaminar.The flow of high-viscosity fluids such as oil at low velocities is typically laminar.The highly disordered fluid motion that typically occurs at high velocities and is characterized by velocity fluctuations is called turbulentturbulent.The flow of low viscosity fluids such as air at high velocities is typically turbulent.

Steady versus Unsteady Flow

The terms steady and uniform are used frequently in engineering, and it is important to have a clear understanding of their meanings.

The term steadysteady implies no change at a point with time. The opposite of steady is unsteady.

The term uniformuniform implies no change with location over a specified region.

Methods of AnalysisMethods of AnalysisThe first step in solving a problem is to define the system that you are attempting to analyze.In basic mechanics, we made extensive use of the free-body diagram.We will use a system or a control volume, depending on the problem being studied.

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

O Systems may be considered to be closed or open, depending on whether a fixed mass or a volume in space is chosen for study.

SYSTEMSYSTEM

Surroundings

Boundary

OThe mass or region outside the system is called the SURROUNDINGSSURROUNDINGS.

OThe real or imaginary surface that separates the system from its surroundings is called the BOUNDARYBOUNDARY.

OThe boundary of a system can be fixed or movable.

The gas in the cylinder is the system. If the gas is heated, the piston will lift the weight; the boundary of the system thus moves. Heat and work may cross the boundaries of the system, but the quantity of matter within the system boundaries remains fixed. No mass crosses the system boundaries.

Gas2 kg1m³

Gas2 kg1m³

Moving Boundary

Fixed Boundary

Weight

Weight

OControl volumeControl volume is an arbitrary volume in space through which fluid flows.

OThe geometric boundary of the control volume is called the control surfacecontrol surface.

OThe control surface may be real or imaginary; it may be at rest or in motion.

Control Volume

Control Surface(real boundary)

Imaginary boundary

Importance of Dimensions and Importance of Dimensions and UnitsUnits

OAny physical quantities can be characterized by dimensions.

OThe magnitudes assigned to the dimensions are called unitsunits.

OSome basic dimensions such as mass m, length L, time t, and temperature T are selected as primary or primary or fundamental dimensionsfundamental dimensions.

Six Fundamental Six Fundamental Quantities:Quantities:

OMeter (m) for lengthOKilogram(kg) for massOSecond (s) for timeOAmpere (A) for electric currentODegree Kelvin (°K) for

temperatureOCandela (cd) for luminous

intensity (amount of light)

Chapter Two

PROPERTIES OF FLUIDS

O In this chapter, we discuss properties that are encountered in the analysis of fluid flow.

O First we discuss intensive and extensive properties and define density and specific gravity.

O This is followed by a discussion of the properties of vapor pressure, energy and its various forms, and the specific heat of ideal gases and incompressible substances.

O Then we discuss the property viscosity, which plays the dominant role in most aspects of fluid flow.

O Finally, we present the property surface tension and determine the capillary rise from static equilibrium conditions.

O Any characteristic of a system is called propertyproperty.Some familiar properties are pressure, temperature, volume, and mass.

O Properties are considered to be either intensive or extensive.

O Intensive propertiesIntensive properties are those that are independent of the mass of the system, such as temperature and pressure.

O Extensive propertiesExtensive properties are those whose values depend on the size or extent of the system. Total mass, total volume, and total momentum are some examples of extensive properties.

Density Density Ois defined as mass per unit volume.density, ρ = m/V (kg/m³)The reciprocal of density is the specific volume, υ. υ = V/m = 1/ρ

Ois defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ρ = 1000 kg/m³).

OSpecific Gravity: SG = ρfluid/ρwater

OThe specific gravity of a substance is a dimensionless quantity.

Specific Gravity or Relative Specific Gravity or Relative DensityDensity

OThe weight of a unit volume of a substance is called specific weight, or weight density, and expressed as

γs = ρg (N/m³)where gg is the gravitational acceleration

Table 2-1: The specific gravity of some substance at 20°C and 1 atm unless stated otherwiseSubstance Spec

. Grav

., Water 1.0Blood(at 37°C)

1.06

Seawater 1.025Gasoline 0.68Ethyl Alcohol 0.790Mercury 13.6Balsa Wood 0.17

Substance Spec. Grav.,

Dense Oak Wood

0.93

Gold 19.3Bones 1.7 – 2.0Ice (at 37°C) 0.916Air 0.00120

4

Density of Ideal GasDensity of Ideal GasO Any equation that relates the pressure,

temperature, and density (or specific volume) of a substance is called an equation of stateequation of state.

O The simplest and best-known equation of state for substance in the gas phase is the ideal-gas equation of state, expressed asPυ = RT or P = ρRT

where P is the absolute pressure, υ is the spec. volume, T is the thermodynamic (absolute) temperature, ρ is the density, and R is the gas constant

The gas constant R is different for each gas and is determined fromR = RR = Ruu/M/Mwhere Ru is the universal gas constant, and M is the molar mass ( also called molecular weight) of the gas.Ru = 8.314 kJ/kmol*K (SI System)Ru = 1.986 Btu/lbmol*R (English System)

The thermodynamic temperature scale in the SI is the Kelvin scale, and the temperature unit on this scale is the kelvin, designated by K. In the English system, it is Rankine scale, and the temperature unit on this scale is rankine, R. Various temperature scales are related to each other by T(K) = T(°C) + 273.15 = T(R)/1.8T(K) = T(°C) + 273.15 = T(R)/1.8T(R) = T(°F) + 459.67 = 1.8 T(K)T(R) = T(°F) + 459.67 = 1.8 T(K)

Example:Example:Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4m x 5m x 6m at 100 kPa and 25°C.

AIRP = 100kPa

T = 25°C

Solution:Solution:O Gas constant of air is

R = 0.287 kPa*m³/kg*KO Density :ρ = P/RTρ = 100kPa/(0.287 kPa*m³/kg*K)

(25+273.15)Kρ = 1.17 kg/m³

OSpecific Gravity:SG = ρ / ρH2O

SG = 1.17 kg/m³ / 1000 kg/m³SG = 0.00117Mass of airV = (4m x 5m x 6m) = 120m³m = ρVm = (1.17 kg/m³)(120m³)m = 140 kg

Vapor PressureVapor PressureAt a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature, Tsaturation temperature, Tsatsat. Likewise, at a given temperature, the pressure at which a pure substance changes phase is called the saturation saturation pressure, Ppressure, Psatsat.For example:At an absolute pressure of 1 standard atmosphere (1 atm or 101.325 kPa), the saturation of water is 100°C.

The vapor pressure (Pv)vapor pressure (Pv) of a pure substance is defined as the pressure exerted by its vapor in phase equilibrium with its liquid at a given temperature. Vapor pressure is a property of the pure substance, and turns out to be identical to the saturation pressure of the liquid (Pv = Pv = PsatPsat). We must be careful not to confuse vapor pressure with partial pressure. Partial pressurePartial pressure is defined as the pressure of a gas or vapor in a mixture with other gases.

Energy and Specific HeatEnergy and Specific HeatO Energy can exist in numerous forms such

as thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear, and their sum constitute the total energy total energy E E of a system.

O The forms of energy related to the molecular structure of a system and the degree of the molecular activity are referred to as the microscopic energymicroscopic energy.

O The sum of all microscopic forms of energy is called the internal energyinternal energy of a system and is denoted by UU.

The macroscopic energy of a system is related to motion and the influence of some external effects such as gravity, magnetism, electricity, and surface tension. The energy that a system possesses as a result of its motion relative to some reference frame is called kinetic kinetic energyenergy.The energy that a system possesses as a result of its elevation in a gravitational field is called potential energypotential energy.

OThe international unit of energy is the joule or kilojoulejoule or kilojoule.1 kJ = 1000 J1 J = 1 N*m1 Btu (British thermal unit) = 1.0551 kJ1 cal (calorie) = 4.1868 J

Viscosity Viscosity OThere is property that represents the

internal resistance of a fluid to motion or the fluidity, and the property is the viscosityviscosity.

OThe force a flowing fluid exerts on a body in the flow direction is called the drag forcedrag force, and the magnitude of this force depends on viscosity.

O To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates ( or equivalently, two parallel plates immersed in a large body of fluid) separated by a distance ℓ.

O Now a constant parallel force F is applied to the upper plate while the lower plate is held fixed.

Fℓ

V

u=0

u=V

dβy

x

Area A

After the initial transients, it is observed that the upper plate moves continuously under the influence of this force at a constant velocity VV. The fluid in contact with the upper plate sticks to the plate surface and moves with it at the same velocity, and the shear stress Ʈ(tau)Ʈ(tau) acting on this layer is Ʈ = F/AƮ = F/Awhere A is the contact area between plate and the fluid.

OThe rate of deformation of a fluid element is equivalent to the velocity gradient du/dydu/dy.

O It can be verified experimentally that for most fluids the rate of deformation is directly proportional to the shear stress Ʈ Ʈ,

Ʈ α dβ/dy or Ʈ α du/dy Fluids for which the rate of deformation is

proportional to the shear stress are called Newtonian fluidsNewtonian fluids after Sir Isaac Newton, who expressed it first in 1687.

OMost common fluids such as water, air, gasoline, and oils are Newtonian fluids. Blood is an example of non-Newtonian fluid.

O In one-dimensional shear flow of Newton fluids, shear stress can be expressed by the linear relationship Ʈ = μdu/dy (N/m²)where the constant of proportionality μ is called coefficient of viscosity or dynamic (or absolute) viscosity of the fluid whose unit is kg/m*s or N*s/m².

OThe shear force acting on a Newtonian fluid layer ( or by Newton’s third law, the force acting on the plate) isF = ƮA = μAdu/dy (N)

Then the force F required to move the upper plate at a constant velocity V while the lower plate remains stationary is F = μAV/ℓ

O In fluid mechanics the ratio of absolute viscosity to density often arises.

OThis ratio is given the name kinematic viscosity kinematic viscosity and is represented by the symbol υ.

υυ = = μμ / / ρρ (m²/s)

Surface Tension and Capillary Surface Tension and Capillary EffectEffect

OYou can tell when you car needs waxing: Water droplets tend to appear somewhat flattened out. After waxing, you get a nice beading (edging) effect.

OWe define a liquid as wetting a surface when the contact angle θ < 90°. By definition, the car’s surface was wetted before waxing, and not wetted after.

OThis is an example of effects due to surface tension.

OWhenever a liquid is in contact with other liquids or gases, or in this case a gas/solid surface, an interface develops that acts like a stretched elastic membrane (skin or covering), creating surface tension.

θ < 90°θ > 90°Water droplet

a. A wetted surface b. A nowetted surface

OThe pulling force that causes this tension acts parallel to the surface and is due to the attractive forces between the molecules of the liquid.

OThe magnitude of this force per unit length is called surface surface tension ( or coefficient of tension ( or coefficient of surface tension) surface tension) σσss and is expressed in the unit N/m (lbf/ft).

ODroplet or air bubble:

where Pi and Po are the pressures inside and outside the droplet, ΔP excess pressure.

(2πR)σσss

(πR²) ΔPdroplet

(2πR)σσss = (πR²) ΔPdroplet

ΔPdroplet = Pi – Po = 2σσss / R / R

Capillary EffectCapillary EffectOAnother interesting consequence of

surface tension is the capillary capillary effecteffect, which is the rise and fall of a liquid in a small-diameter tube inserted into the liquid.

OSuch narrow tubes or confined flow channels are called capillariescapillaries.

OThe curved free surface of a liquid in a capillary tube is called the meniscusmeniscus.

OCapillary rise:(2πR)σσss

W

h

2Rliquid

Φ

h = 2σσs s coscosΦρgR

Example #1:OA 0.6mm diameter glass tube is

inserted into water at 20°C in a cup. Determine the capillary rise of water in the tube.

Example #2:OA 1.9-mm-diameter tube is

inserted into an unknown liquid whose density is 960 kg/m³, and it is observed that the liquid rises 5 mm in the tube, making a contact angle of 15°. Determine the surface tension of the liquid.

O In a diesel engine, the piston compress air at 305 K to a volume that is one-sixteenth of the original volume and a pressure that is 48.5 times the original pressure. What is the temperature of the air after the compression?

Example #3:

OThe pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m³, determine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.

Example #4:

Example #5:OThe velocity distribution for laminar

flow between parallel plates is given byu/umax = 1 – (2y/h)²

where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at 20°C with maximum speed of 0.05 m/s and h=0.1mm. Calculate the force on a 1m² section of the section of the lower plate and give its direction.

Example #6:O A 50-cm X 30-cm X 20-cm block weighing

150 N is to be moved at a constant velocity of 0.8 mls on an inclined surface with a friction coefficient of 0.27. (a) Determine the force F that needs to be applied in the horizontal direction. (b) If a O.4-mm-thick oil film with a dynamic viscosity of 0.012 Pa . s is applied between the block and inclined surface, determine the percent reduction in the required force.

PRESSURE AND PRESSURE AND FLUID STATICSFLUID STATICS

Objectives:At the end of this chapter, the

students must be able to:1.Determine precisely the variation of pressure in a fluid at rest;2.Calculate the forces exerted by a fluid at rest on plane or curved submerged surfaces; and3.Analyze the stability of floating and submerged bodies.

OThis chapter deals with forces applied by fluids at rest.

OThe fluid property responsible for those forces is pressure, which is a normal force exerted by a fluid per unit area.

OWe start this chapter with a detailed discussion of pressure, including absolute and gauge pressures, the pressure at a point, the variation of pressure with depth in a gravitational field, the manometer, the barometer, and other pressure measurement devices.

OThis is followed by a discussion of the hydrostatic forces applied on submerged bodies with plane or curved surfaces.

OWe then consider the buoyant force applied by fluids on submerged or floating bodies, and discuss the stability of such bodies.

PRESSUREPRESSUREOPressurePressure is defined as a normal force

exerted by a fluid per unit area. We speak of pressure only when we deal with gas or a liquid.

OThe counterpart of pressure in solids is normal stress. Since pressure is defined as force per unit area, it has the unit of newton per square meter (N/m²), which is called a Pascal (Pa). That is1 Pa = 1 N/m²

OThree other pressure units commonly used in practice, especially in Europe, standard atmosphere, and kilogram-force per square centimeter:1 bar = 10 = 0.1 MPa = 100 kPa1 atm = 101.325 kPa = 1.01325 bars

1 kgf/cm² = 9.807 N/cm² = 9.807x10 Pa1 kgf/cm² = 0.9807 bar1 kgf/cm² = 0.9679 atm

5

4

OThe actual pressure at a given position is called absolute pressureabsolute pressure, and it is measured relative to absolute vacuum.

OMost pressure measuring devices are calibrated to read zero in the atmosphere, and so they indicate the difference between the absolute pressure and the local atmospheric pressure.

OThis difference is called the gage gage pressurepressure.

OPPgagegage can be positive or negative, but pressures below atmospheric pressure are sometimes called vacuum vacuum pressurespressures.

OAbsolute, gage, and vacuum pressures are related to each other byPPgagegage = P = Pabsabs – P – Patmatm

PPvacvac = P = Patmatm – P – Pabsabs

Pressure at a PointPressure at a PointO Pressure is the compressive force per

unit area. O Pressure at any point in a fluid is the

same in all directions. PP PP

PPPP

PP

That is, it has magnitude but does not have a specific direction and thus it is a scalar quantity.

In other words, the pressure at any point in a fluid has the same magnitude in all directions.

Variation of Pressure with Variation of Pressure with DepthDepth

O To obtain a relation for the variation of pressure with depth, consider a rectangular fluid element of height Δz, length Δx, and unit depth (Δy = 1 unit) in equilibrium.zz

xxPP22

PP11

ΔzΔx

W

OAssuming the density of the fluid ρρ to constant, a force balance in the vertical z -direction givesΣFz = maz =0

P2ΔxΔy – P1ΔxΔy – W = 0 where W = mg = ρVg = ρg

ΔxΔyΔz P2ΔxΔy – P1ΔxΔy – ρg ΔxΔyΔz = 0 P2 – P1 = ρgΔz; ΔP = P2 – P1; γ = ρg

ΔP = P2 – P1 = γΔz

O If we take point 1 to be at the free surface of a liquid open to the atmosphere, then the pressure at a depth hh from the free surface becomes

P = PP = Patmatm + + ρρgh or Pgh or Pgagegage = = ρρgh gh 11

22

hh

Pressure Measurement Devices

OThe ManometerThe ManometerO Manometers use the relationship

between pressure and head to measure pressure.

OThe simplest manometer is an open tube.

OThis is attached to the top of a container with liquid at pressure.

OThe tube is open to the atmosphere,OThe pressure measured is relative to

atmospheric so it measures gauge pressure

B

A

h1h2

Piezometer

OPressure at A = pressure due to column of liquid h1

PPAA = = ρρghgh11

OPressure at B = pressure due to column of liquid h2

PPBB = = ρρghgh22O Problems with the Piezometer:

1. Can only be used for liquids2. Pressure must above atmospheric3. Liquid height must be convenient

i.e. not be too small or too large

The “U”-Tube ManometerO“U”-Tube enables the pressure of

both liquids and gases to be measured.

A

B C

h2

D

h1

Fluid density ρ ρ Manometric fluid ρ ρmanman

OPressure in a continuous static fluid is the same at any horizontal level.pressure at B = pressure at C

PB = PC

OFor the left hand arm:pressure at B = pressure at A + pressure of height of liquid being measured

PB = PA + ρgh1

OFor the right hand arm:pressure at C = pressure at D + pressure of height of manometric liquid PC = ρmangh2 + PD(atmospheric)

OMeasuring gauge pressure we can subtract Patmospheric giving

PB = PC PPAA = = ρρmanmanghgh22 - - ρρghgh11

Example #1: (3-38) The 500-kg load on the hydraulic lift shown in Fig.

P3–43 is to be raised by pouring oil (ρ = 780 kg/m³) into a thin tube. Determine how high h should be in order to begin to raise the weight.

Example #2: (3-127) The pressure of water flowing through a pipe is

measured by the arrangement shown in Fig. P3–127. For the values given, calculate the pressure in the pipe.

Example:OA differential manometer is

connected across the vertical pipe as shown in the figure. If the manometric fluid is mercury and the readings and dimensions are as shown in the figure, what is the difference in pressure between points A and B?

FLUID FLUID STATICSSTATICS

O Fluid staticsFluid statics deals with problems associated with fluids at rest.

O Fluid statics is generally referred to as hydrostaticshydrostatics when the fluid is a liquid and aerostaticsaerostatics when the fluid is a gas.

O In fluid statics, there is no relative motion between adjacent fluid layers, and there is no shear stresses (tangential) in the fluid trying to deform it.

O The only force we deal with in fluid statics is the normal stress, which is the pressure, and the variation of pressure due to the weight of the fluid.

Forces on Submerged Forces on Submerged Surfaces in Static FluidsSurfaces in Static Fluids

Fluid pressure on a surfaceFluid pressure on a surfaceOPressure is defined as force per unit

area. If a pressure pp acts on a small area δδAA then the force exerted on that area will be F = pF = pδδAA

OSince the fluid is at rest the force will act at right-angles to the surface.

General Submerged General Submerged Plane Plane

O Consider the plane surface shown in the figure. O The total area is made up of many elemental

areas. O The force on each elemental area is always normal

to the surface but, in general, each force is of different magnitude as the pressure usually varies.

F1

F2

Fn

OThe total or resultant force, FR, on the plane is the sum of the forces on the small elements i.e.

FFRR = pp11δδAA11 + p + p22δδAA22 + … + p + … + pnnδδAAnn = = ΣΣ p pδδAA and this resultant force will act through the center of pressure.

OFor a plane surface all forces acting can be represented by one single resultant force, acting at right-angles to the plane through the center of pressure.

Horizontal submerged Horizontal submerged planeplane

OThe pressure, pp, will be equal at all points of the surface.

OThe resultant force will be given by FRFR = pressure x area of planeFFRR = pA = pA

Inclined Surfaces Inclined Surfaces

x

y

FR

Take pressure as zero at the surface.

P

Q

O

zz x S

Sc

OMeasuring down from the surface, the pressure on an element δA, depth z,p = ρgz

OSo force on elementF = ρgz δA

OResultant force on planeFR = ρg ΣzδA (assuming ρ and g as constant).

O ΣzδA is known as the 1st Moment of Area of the plane PQ about the free surface.

OAnd it is known that ΣzδA = Az A is the area of the plane z is the distance to the center of

gravity (centroid)O In terms of distance from point O

ΣzδA = Ax sinθ= 1st moment of area x sinθ

about a line through O (as z = x sinθ)

OThe resultant force on a planeFR = ρgAzFR = ρgAx sinθ

OThe position of the center of pressure (Sc) along the plane measure from the point O is:

Sc = 2nd Moment of area about a line through O 1st Moment of area about a line through O

Sc = Sc = IIGGGG + x + x

AxAx

andODepth to the center of pressure is

D = Sc sinD = Sc sinθθ D = D = IIGGGG + x sin + x sinθθ

AxAx

O2nd moment of area about O: Io = Σ s²δA Io = IGG + Ax ²

O1st moment of area about O = Ax ²

Example #1: (3-65)The water side of the wall of a 100-m-long dam is a quarter circle with a radius of 10 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim.

Example #: (3-66)OA 4-m-high, 5-m-wide rectangular plate

blocks the end of a 4-m-deep freshwater channel, as shown in Fig. P3–66. The plate is hinged about a horizontal axis along its upper edge through a point A and is restrained from opening by a fixed ridge at point B. Determine the force exerted on the plate by the ridge.

Example #: (3-68)OThe flow of water from a reservoir

is controlled by a 5-ft-wide L-shaped gate hinged at point A, as shown in Fig. P3–68E. If it is desired that the gate open when the water height is 12 ft, determine the mass of the required weight W.

Example: (3-71)OThe two sides of a V-shaped water

trough are hinged to each other at the bottom where they meet, as shown in Fig. P3–71, making an angle of 45° with the ground from both sides. Each side is 0.75 m wide, and the two parts are held together by a cable and turnbuckle placed every 6 m along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.

Buoyancy and StabilityBuoyancy and StabilityO It is a common experience that an object feels

lighter and weighs less in a liquid than it does in air.

O This can be demonstrated easily by weighing a heavy object in water by a waterproof spring scale.

O These and other observations suggest that a fluid exerts an upward force on a body immersed in it.

O This force that tends to lift the body is called the buoyant force buoyant force and is denoted by FB.

O The buoyant force is caused by the increase of pressure with deep in a fluid.

Archimedes’ Principle Archimedes’ Principle O The buoyant force acting on a body

immersed in a liquid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.

OFB = Fv2 – Fv1= fluid weight above Surface 2 (ABC)– fluid weight above Surface 1 (ADC)= fluid weight equivalent to body vol. V

OFFBB = = ρρgV gV , V = submerged volumeOLine of action is through centroid of V

= center of buoyancyONet Horizontal forces are zero since

FBAD = FBCD

O For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body. That isFFBB = W = WFFBB = = ρρgVgVsub

W =W = ρρavg,body gVgVtotal

VVsub/VVtotal = ρρavg,body / ρρffO Therefore, the submerged volume fraction of

a floating body is equal to the ratio of the average density to the density of fluid.

ρρ<< ρρff

ρρ==ρρff

ρρ>> ρρff

Sinking Body

Floating body

Suspended body

Fluid, ρρff

Example:

OA crane is used to lower weights into the sea (density =1025 kg/cu.m) for an underwater construction project. Determine the tension in the rope of the crane due to a rectangular 0.4m x 0.4m x 3m concrete block (density = 2300 kg/cu.m) when it is a) suspended in the air and b) completely immersed in water.

MASS, MASS, BERNOULLI, BERNOULLI,

CONTINUITY, AND CONTINUITY, AND ENERGY ENERGY

EQUATIONSEQUATIONS

Objectives:

At the end of this chapter, the student must be able to:1.familiarize precisely the Bernoulli’s equation by memorizing the formula and through problem solving; and2.solve correctly any problems that involves Bernoulli’s equation through board work.

Bernoulli’s Equation: Bernoulli’s Equation: IntroductionIntroduction

Daniel Bernoulli(1700-1782)

Swiss mathematician, son of Johann Bernoulli, who showed that as the velocity of a fluid increases, the pressure decreases, a statement known as the Bernoulli principle. He won the annual prize of the French Academy ten times for work on vibrating strings, ocean tides, and the kinetic theory of gases. For one of these victories, he was ejected from his jealous father's house, as his father had also submitted an entry for the prize. His kinetic theory proposed that the properties of a gas could be explained by the motions of its particles.

Energy Conservation (Bernoulli’s Equation)

If one integrates Euler’s eqn. along a streamline, between two points , &

Which gives us the Bernoulli’s Equation

Constant 22 2

222

1

211 gzVpgzVp

Flow work + kinetic energy + potential energy = constant

02

1

2

1

2

1 gdzVdVdp

0 gdzVdVdp

Recall Euler’s equation:

Also recall that viscous forces were neglected, i.e. flow is inviscid (having zero or negligible viscosity)

We get :

Bernoulli’s Equation (Continued)

p

A

x

tW

AVp

pAVtxAp

txpA

tW

1

,

Flow Work (p/) : It is the work required to move fluid across the control volume boundaries.

Consider a fluid element of cross-sectional area A with pressure p acting on the control surface as shown.

Due to the fluid pressure, the fluid element moves a distance x within time t. Hence, the work done per unit time W/t (flow power) is:

Flow work per unit mass

Flow work or Power

1/mass flow ratepvp

Flow work is often also referred to as flow energy

t)unit weighper (energy g where,22 2

222

1

211

z

gVpz

gVp

Very Important: Bernoulli’s equation is only valid for :incompressible fluids, steady flow along a streamline, no energy loss due to friction, no heat transfer.

Application of Bernoulli’s equation - Example 1:

Determine the velocity and mass flow rate of efflux from the circular hole (0.1 m dia.) at the bottom of the water tank (at this instant). The tank is open to the atmosphere and H=4 m

H

1

2

p1 = p2, V1=0

)/(5.69

)85.8()1.0(4

*1000

)/(85.84*8.9*2

2)(2

2

212

skg

AVm

sm

gHzzgV

Bernoulli’s Equation (Cont)

Bernoulli’s Eqn/Energy Conservation (cont.)Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time required to drain the tank to level 2.

h(t)

1

2

First, choose the control volume as enclosedby the dotted line. Specify h=h(t) as the waterlevel as a function of time.

0 20 40 60 80 1000

1

2

3

4

time (sec.)

wat

er h

eigh

t (m

)

4

2.5e-007

h( )t

1000 tsec 90.3 t

0.0443t- 20

4

h

Energy exchange (conservation) in a thermal system

1

211

2z

gVp

2

222

2z

gVp

Energy added, hA

(ex. pump, compressor)

Energy extracted, hE

(ex. turbine, windmill)

Energy lost, hL

(ex. friction, valve, expansion)

pump turbine

heat exchanger

condenser

hEhA

hL, friction lossthrough pipes hL

loss throughelbows

hL

loss throughvalves

Energy conservation(cont.)

2

222

1

211

22 z

gVphhhz

gVp

LEA

Example: Determine the efficiency of the pump if the power input of the motoris measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.

Mercury (m=844.9 lb/ft3)water (w=62.4 lb/ft3)

1 hp=550 lb-ft/s

No turbine work and frictional losses, hence: hE=hL=0. Also z1=z2

6-in dia. pipe 4-in dia.pipe Given: Q=300 gal/min=0.667 ft3/s=AVV1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/s

kinetic energy head gain

V Vg

ft22

12 2 2

27 54 3 33

2 32 20 71

( . ) ( . )

* .. ,

p z z p z zp p z

lb ft

w o m w o w

m w

1 2

2 1

29 62 1 25 97813

( )

(844. .4)* . . /

pump

Z=15 in

1 2

zo

If energy is added, removed or lost via pumps turbines, friction, etc.then we use

Extended Bernoulli’s Equation

Looking at the pressure term:

Energy conservation (cont.)Example (cont.)

Pressure head gain:

pump work

p p ft

h p p V Vg

ft

w

Aw

2 1

2 1 22

12

9781362

15 67

216 38

.

.4. ( )

. ( )

Flow power delivered by pumpP =

Efficiency = PP

w

input

Qhft lb s

hp ft lb sP hp

A

( .4)( . )( . ). ( / )

/.

..

. .

62 0 667 16 38681 7

1 5501 24

1 241 5

0 827 82 7%

Static, Stagnation, Dynamic, and Total Pressure: Bernoulli Equation

Static Pressure

Dynamic Pressure

Hydrostatic Pressure

Static Pressure: moves along the fluid “static” to the motion.

Hydrostatic Pressure: potential energy due to elevation changes.

Dynamic Pressure: due to the mean flow going to forced stagnation.

Follow a Streamline from point 1 to 2

hp 1

12

1122

22 21

21 zVpzVp

Following a streamline:

0 0, no elevation 0, no elevation2

112 21 Vpp Hp 2

H > hNote:

hHV 1In this way we obtain a measurement of the centerline flow with piezometer tube.

“Total Pressure = Dynamic Pressure + Static Pressure”

Stagnation Point: Bernoulli Equation

Stagnation point: the point on a stationary body in every flow where V= 0

Stagnation Streamline: The streamline that terminates at the stagnation point.

Symmetric:

Axisymmetric:If there are no elevation effects, the stagnation pressure is largest pressure obtainable along a streamline: all kinetic energy goes into a pressure rise:

2

2Vp

streamlineaontconspzVp T tan21 2

Total Pressure with Elevation:

Stagnation Flow I:

Stagnation Flow II:

Pitot-Static Tube: Speed of Flow

H. De Pitot(1675-1771)

p2

p2

p2p2

p1

p1

p1

p1

p1 = p4

p2 = p3

Stagnation Pressure occurs at tip of the Pitot-static tube:

32

2 21 pVpp

Static Pressure occurs along the static ports on the side of the tube:41 ppp (if the elevation differences are negligible, i.e. air)

Now, substitute static pressure in the stagnation pressure equation:2

43 21 Vpp 2

43 21 Vpp

Now solve for V:

432 ppV

Air Speed:

Uses of Bernoulli Equation: Free Jets

New form for along a streamline between any two points:

If we know 5 of the 6 variable we can solve for the last one.

Free Jets: Case 1

Following the streamline between (1) and (2):

0 gage 0 gage0 h V 0

Torricelli’s Equation (1643):

Note: p2 = p4 by normal to the streamline since the streamlines are straight.

As the jet falls:

42

4432

33 21

21 zVpzVp

Uses of Bernoulli Equation: Free Jets

Free Jets: Case 2

=(h-l) 0 gage0 l V 0

Then,

Physical Interpretation:

All the particles potential energy is converted to kinetic energy assuming no viscous dissipation.

The potential head is converted to the velocity head.

Uses of Bernoulli Equation: Free Jets

Free Jets: Case 3 “Horizontal Nozzle: Smooth Corners”

Slight Variation in Velocity due to Pressure Across Outlet

However, we calculate the average velocity at h, if h >> d:

Free Jets: Case 4 “Horizontal Nozzle: Sharp-Edge Corners”vena contracta: The diameter of the jet dj is less than that of the hole dh due to the inability of the fluid to turn the 90° corner.

The pressure at (1) and (3) is zero, and the pressure varies across the hole since the streamlines are curved.

The pressure at the center of the outlet is the greatest.

However, in the jet the pressure at a-a is uniform, we can us Torrecelli’s equation if dj << h.

Torricelli Flow:

Uses of Bernoulli Equation: Free JetsFree Jets: Case 4 “Horizontal Nozzle: Sharp-Edge Corners”

Vena-Contracta Effect and Coefficients for Geometries

Uses of the Bernoulli Equation: Confined FlowsThere are some flow where we cannot know the pressure because the system is confined, i.e. inside pipes and nozzles with changing diameters.

In order to address these flows, we consider both conservation of mass (continuity equation) and Bernoulli’s equation.

The mass flow rate in must equal the mass flow rate out for a steady state flow:

Consider flow in and out of a Tank:

and

With constant density,

ContinuityContinuity

Station 1Density 1

Velocity V1

Area A1

Station 2Density 2

Velocity V2

Area A2

Mass Flow Rate In = Mass Flow Rate Out1 V1 A1 = 2 V2 A2

The Venturi MeterThe Venturi Meter

It is used to measure Flow rates. Gas companies, Water works, and aircraft fuel monitors all use this device.

How does the Venturi Meter work?How does the Venturi Meter work?

222

211

1

2

2

1

21

222111

21

21:

,:_

VpVpBernoulli

AA

VVThus

FlowibleIncompressAVAV

11

22

21

121

1

1222

212

1

122

1

222

1

122

22

1

_:rate flow Compute

1

2

:Vfor Solve

21

21

21

21

AVrateFlow

AAppV

ppAAV

ppVVV

ppVV

That’s all folks!!!That’s all folks!!!