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UNIVERSITY OF MANCHESTERSCHOOL OF MECHANICAL, AEROSPACE AND CIVIL ENGINEERING

Civil Engineering, Hydraulics 1

Semester 2, Course UnitTwo Lectures and one Examples’ class per week, over 12 WeeksOne three-hour laboratory class

Course Leader: Dr G-L Serff (Weeks 4-12)Contributor: Prof H. Iacovides (Weeks 1-3)

AssessmentAssignmentsLab ReportEnd-of-Semester Exam

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CONTENTS

Properties of fluids – density, viscosity (H.I.)

Hydrostatics – static pressure, forces (H.I.)

Kinematics – motion, velocity, acceleration (G-L.S.)

Energy and momentum – basic principles (G-L.S.)

Pipe flow – application of above (G-L.S.)

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• Definition of a Fluid- Substances can be broadly classified into :

Solids and Fluids

- A solid can resist a force by Static Deformation

- A fluid would continuously move and deform under the action of a force.

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- Fluids are further sub-divided into :Liquids and Gases

Liquids retain a Gases expand to occupyconstant volume all available space

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• A MICROSCOPIC VIEW.

- A fluid consists of a collection of ‘freely’ and randomly moving atoms/molecules which require containment to stay together.

- For the purpose of engineering analysis, a fluid is treated as a continuous substance.

- Exception: Nano-Applications

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Fluid Properties

Continuum – large number of molecules representedthough bulk properties below :

Density - kg/m3 (mass/volume)

Pressure – N/m2 or Pa (normal stress)

Viscosity – kg/(m.s) (causes shear stress)

Compressibility (volume changes due to pressure)

Surface tension (small force on interface between two fluids)

Gas (special law)

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Density, ρ

ρ mass/volume – kg/m3

γ ≡ ρ.g specific weight (weight/volume) – N/m3

Relative density or specific gravity – dimensionless ratio of density of fluid to density of water.

Specific Gravity ≡ density/density of water

In Gases density changes with temperature and pressure

In Liquids density changes only with temperature

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Hydrostatic pressure

Normal force on an area due to weight of water (fluid) above the areaPressure = force/area (N/m2 or Pa)

BUT pressure acts in all directions – it is a scalar.

For example a horizontal force is exerted on a dam due to vertical depth of water

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Static water (or water with velocity but ZERO acceleration)

px Δz = pn Δs sin(θ), Since Δz = Δs sin(θ) px = pn

Same for z direction.

PROOF

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• Vapour Pressure

It is the pressure at which a liquid boils, and is in equilibrium with its own vapour.

Its value increases with temperature

When, at a given temperature, the liquid pressure falls below the vapour pressure, bubbles begin to appear in the liquid.

This phenomenon called ‘cavitation’.

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Viscosity- Property which slows down fluid motion.- It opposes the relative motion within the fluid.- Water is slightly viscous. Treacle is very viscous.- When different fluid layers move at different speeds, SHEAR force is produced.- Shear stress is shear force / tangential area

Gravity force downhill

Retarding shear dueto viscosity

flow

Viscometer

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Newtonian fluid like waterμ is viscosityKinematic viscosity ν = μ/ρ

Shear stress proportional to rate of strain(distortion)

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Because of viscosity, the fluid velocity u is zero at a solid boundary.

This is the so-called ‘no-slip’ boundary condition, true for any viscous fluid.

Boundary Layer

The layer across which the velocity increases from zero at the wall to the constant free stream value is the Boundary Layer.

The velocity gradient, du/dz, usually greatest at the wall, generates a shear stress, which opposes the fluid motion.

Boundary Layer behaviour, determined by the value of the flow Reynolds number.

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• Reynolds number

U and L are, respectively, the characteristic velocity and length of a flow.

A professor of engineering at Manchester University,

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Compressibility

Water is slightly compressibleAir is much more compressible

When pressure is exerted on a volume of fluid, the volume decreases.

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Surface tension

A fluid surface or interface experiences a very small tension.

Across any line drawn on the interface there is exerted a forceof magnitude per unit length in a direction normal to the line and tangential to the interface.

σ is surface tension force per unit length. F = σ . ΔL

The existence of surface tension means that there is a pressure jumpacross a curved surface of fluids.

The pressure being higher on the concave side.

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Apply Force Equilibrium on Each Half.Pressure Force on Inner Side (P+ΔP)πR2

Pressure Force on Outer Side - PπR2

Surface Tension Force - 2πRσForce Balance

If fluid volume is small enough it forms into drops.

- Consider a circular liquid droplet

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For the more general case of an arbitrarily curved interface whose principal radii of curvature are R1 and R2, a force balance normal to the surface will show that

P

P+ΔP

When R1=R2=R, then the equation for the sphere is recovered (Example 2 in notes).

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Contact angleIn the case of a liquid interface intersecting with a solid surface, another important surface effect, in addition to surface tension, is the contact angle θ.

If < 90, the liquid ’wets’the solid.

If > 90, the liquid is non–wetting.

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- From a simple force balance:

Weigh of Water Column = Vertical Component of S.T. Force

Weigh of Water Column: ρgπD2 h/4

Surface Tension Force: πDσ

Vertical Component of S.T. Force: πDσ cosθ

Equilibrium: πDσ cosθ = ρgπD2h/4 Thus h = 4σ cosθ / (ρgD)

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• Equation of state for perfect gas p = RρT

- T is absolute temperature (degree Kelvin, ToK = ToC + 273.16)- R is a constant (independent of temperature).- If M is the relative molecular mass of a gas, i.e. the ratio of

the mass of the molecule to the mass of a normal hydrogen atom, then:

MR = 8314 J/(kg K) – the universal gas constant

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HYDROSTATICSForces due hydrostatic pressure – in a still/static fluid.Consider a small tube in a fluid.

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Pressure with uniform densityHydrostatic equation

Integrates to

With p+ρgz called the piezometric pressure

When z=0, p=pair

p + ρgz = pair

p = pair + ρgh

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Absolute pressure is that relative to zero pressure, i.e. a vacuum

e.g. pabsolute = pair + ρgh in the previous case

Gauge pressure is that relative to atmospheric ~105 Pa = 1 bar

pgauge = ρgh in the previous case

Vacuum pressure is that below atmospheric – negative gauge pressure

Pressure head or head is given by h = pgauge/ ρg

Convenient as in metres of water and easy to visualise.

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Pressure measurement 1

piezometer

p = pair + ρgh

ρgh = p - pair

ρgh = pgauge

h = pgauge/ ρg

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Pressure measurement 2

U tube manometer – two fluids ρA < ρB

Static so pressures at X-X same

p + ρA gh1 = pair + ρB gh2

Relative to atmospheric

pgague = ρB gh2 - ρA gh1

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Differential manometer

Pressures at X-X same

p1+ρAg (x+y) = p2+ ρA g y+ ρBg x

p1 – p2 = (ρB – ρA) g x

Head loss: (p1-p2)/ (ρAg) = (ρB/ ρA – 1) x

This head loss is due to wall friction. A bigger downstream pressure is required to drive water upstream .

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Inclined tube manometer for measuring small pressures

Measuring l thus gives pressure

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Hydrostatic force on a plane surface

Force per unit width = ∫p.1.dh = ∫ ρ g h dh = ρ g (h22-h1

2)/2

= area under pressure diagram= (ρ g (h2+h1)/2) × (h2-h1)= pressure at centroid × surface length

In general force = average pressure × area = ρ g hCG A

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Inclined rectangle has same formula where h1 and h2are still the vertical distances below the surface at thetop and bottom of the rectangle

F = b.d.ρ.g.hCG

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(a)

(b)

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NORMAL Force on an element = p × dA = ρ g h dAForce on surface F = ∫ dF = ∫ p dA = ∫ ρ g h dA = ρ g ∫ ℓ sin α dA

= ρ g sin α ∫ ℓ dA = ρ g sin α ℓ CG A = ρ g hCG A pressure at centroid = ρ g hCGForce normal to surface = area × pressure at centroid

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Where does the pressure force act?NOT through centroid but through CENTRE OF PRESSURE

Take moments about OM = F. ℓCP - definition of ℓ CP = hCP/sin(α)

Moment of dF = p. dA× ℓ = ρ g h ℓ dA= ρ g ℓ sin(α). ℓ.dA

Total moment = ∫ ρ g ℓ sin(α). ℓ.dA = ρ g sin(α) ∫ ℓ2.dA F = ρ g sin(α) ∫ ℓ.dA = ρ g sin(α) ℓCG AM = ρ g sin(α) ℓCG A ℓ CP = ρ g sin(α) ∫ ℓ2.dA

ℓCP = ∫ ℓ2.dA / (A lCG)

ℓCP = [second moment of area about O,IO] / (A ℓCG)

I =∫ ℓ2dA = 2nd Moment of Area

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How to calculate second moment of area

Use parallel axes theorem to relate 2nd Moment of area about axis through centroid, ICG, to that about another parallel axis

I≡∫ ℓ2.dA = ICG + A ℓ CG2 = ℓ CPA ℓCG (from before)

∫ ℓ 2.dA /(A ℓCG) = ℓCP = ℓCG + ICG/(A ℓCG)

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Hydrostatic force on a curved surface

FH = ∫ p. ds. cos(α) = ∫ ρ g h dh = ρ g hCG AV Acts through centre of pressure AV – projected areaAs for plane surface

FV = ∫ p ds sin(α) = ∫ ρ g h dx = ρ g V ……dx = ds sin(α) FV equals weight of water above surface and acts throughthe centre of gravity of volume V.

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BuoyancyConsider a solid of rectangular shape submerged in a liquid

The vertical pressure forces on the upper and lower horizontal surfaces, FU and FL are:

FU = b.c ρgy1 FL = b.c.ρ.g (y1+a)Net vertical force, FB:

FB = FL - FU = b.c [ ρg(y1+a)] - b.c ρgy1

FB = a . b . c . ρ.g = ρ.g .Volume

Archimedes principle: A solid body experiences an upward force equal to the weight of the fluid it displaces.

NOTE: The above also applies to a body of any shape

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Solution

-Vertical Force FV, weight of water in virtual volume above gate.

-Horizontal force FH, force on vertical surface of height h and width 3.5 m at 2 m below the free surface.

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h = R sin(α) = 5 sin(30) m = 2.5 m

FH = ρ g (2m+h/2) (h 3.5m )

FH = 1000 x 9.81 x (2+2.5/2) x2.5 x 3.5 N

FH = 279 kN

Vol1 =3.5m x 2mx {R-[R2-h2]1/2}

Vol1 = 4.69 m3

Vol2 = 3.5m {πR2(α/360)-h[R2- h2]1/2/2}

Vol2 = 3.97 m3

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FV = ρg(Vol1+Vol2) = 1000 x 9.81 x (4.69+3.97) NFV = 85 kNResultant Force FR = {FH

2+FV2)1/2 = 291.6 kN

Angle with horizontal θ = atan (FV/FH)θ = atan (85/279) = atan (.295) = 17o