mek 307 fluid mechanics - eskisehir.edu.trinsaat.eskisehir.edu.tr/aocelik/mek 307/icerik/1_4th...

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MEK 307 Fluid Mechanics

A. Ozan Celik www.hku.hk

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Basics of fluid flow

Basics of fluid flow, kinematics

Mechanics

Statics

Dynamics

Kinematics

Kinetics

Kinematics: deals with motion apart from considerations

of mass, force or energy

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Dimensions and units (Chapter 1.6)

SI BG

F=ma

Basic dimensions: length [L], mass [M], force [F], time [T]

[F] = [M][LT-2]

lb = Slug.ft/s2

Slug = lb.s2/ft

N = kg.m/s2

base units

derived units

(derivative dimensions)

SI: International System of Units

BG: British Gravitational System

We use sets of units to properly define, express certain physical properties/

phenomenon

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Dimensions and units, example:

SI BG

Energy =Force x Distance

= [M][LT-2][L]

=(kg.m/s2)(m)

=Nm

=Joule

= [F][L]

=lb.ft

(1.3558 J )

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Unit homogeneity:

2+2=?4

Manning’s Equation:

Q=(1/n)A.R2/3.S1/2

L3/T=?[L2][L]2/3

??

Where:

Q = Flow Rate [L3/T]

A = Flow Area, [L2]

n = Manning’s Roughness

Coefficient

R = Hydraulic Radius, [L]

S = Channel Slope, [L/L]

Has units:

[L1/3/T]

(1 meter = 3.2808399 feet)

For those who are interested: http://en.wikipedia.org/wiki/Robert_Manning_(engineer)

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Properties of Fluids (Chapter 2.2)

Density, ρ = mass per unit volume, [M/L3]

Specific volume, v = 1/ ρ, volume occupied by unit mass [L3/M]

Specific weight, γ = weight per unit volume, [F/L3] (depends on latitude)

Specific gravity, SG (s) = ρ/ ρwater

http://www.agen.ufl.edu/~chyn

density of fluids vary with

temperature

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Properties of Fluids (Chapter 2)

Compressibility of Liquids

v

vE

pp

v

vvdp

dv

v

dv

dpvE 12

1

12

Bulk Modulus (volume modulus of elasticity) Units?

We consider liquids to be incompressible but

why?

pressure volume,specific pv

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Properties of Fluids

Fluids Solids

Deforms continuously or

flows when subject to

shear force

Can resist a shear force at

rest (with some

displacement but tend to

retain shape)

Shear force: Force component tangent to a surface

Ft

F Fn Shear Stress: τ = Ft /A

Pressure: p= Fn /A

A

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The strain in a solid is independent of the time over

which the force is applied and (if the elastic limit

is not reached) the deformation disappears when the

force is removed.

A fluid continues to flow for as

long as the force is applied and will not recover its

original form when the force is removed.


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Shear force, Ft: Force component tangent to a surface

Normal Force, Fn: Force component normal to a surface

Shear Stress: τ= Ft/A Pressure: p=Fn/A

Ft

F Fn

A

Deformation Compressibility

(dynamic)

Viscosity, μ

Some liquids are

more viscous than

others

3120 psi needed to

compress a unit

volume of water

1%

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Properties of Fluids (Chapter 2.6)

Viscosity is the means by which the fluid resists friction

(shear stress)

Constant of

proportionality, viscosity

(dynamic, absolute)

For fluids, the rate of strain

(angular deformation) is

proportional to the applied

stress.

Newtonian Fluids: μ is constant

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Viscosity (Chapter 2)

For gases μ as T

For liquids μ as T

μ => units?

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Laminar shear flow velocity profile Uniform flow velocity profile

No Shear

In practice we are concerned with flow past solid (sometimes rough)

boundaries; pipe walls, river bed etc. and there will always be shear stress

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Shear Stress and Viscosity (Chapter 2)

Viscosity, μ is a measure of resistance to relative motion of

adjacent fluid layers (angular deformation)

τ

τ

τ = μdu/dy

FLOW

y

u du

u

u+du dy

Velocity profile

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Shear Stress and Viscosity (Chapter 2)

Viscosity, μ Fluids resist to shear lightly

τ

Ideal fluids Viscosity, μ=0 Only normal stresses

present

Newtonian fluids Viscosity, μ=const. (for const. temperature)

τ=0

du/dy=0

u=0, fluid is at rest Only normal stresses

present

u=u(y)=const.

No slip condition: Fluid in the immediate contact

with a solid boundary has the same velocity as the

solid boundary

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Surface Tension (Chapter 2)

Exists whenever there is a density discontinuity

(Interfaces between a gas and a liquid; two different liquids…etc)

Cohesion: Attraction between like molecules, enables a fluid to

resist tensile stresses

Adhesion: Attraction between unlike molecules, enables a fluid to

stick to another body

http://www.bcscience.com

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Vapor Pressure (Chapter 2)

Pressure of a vapor in equilibrium with

its non-vapor phases

Vapor pressure depends on

temperature

To avoid boiling, pressure on top of

liquid should be higher than vapor

pressure

http://www.unit5.org/

http://www.chem.purdue.edu

Towards the

perfect boiled egg

on top of Mount

Everest

On top of Mount Everest the pressure is about

260 mbar (26.39 kPa, 0.25 atm)

69°C

You need water temperature higher than 70°C to

cook an egg (for a hard yolk)

Conclusion: you can not boil an egg

on top of Mount Everest. http://blog.khymos.org/

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Vapor Pressure, Cavitation (Chapter 2)

Cavitation damage on an impeller of a BW5000 pump

www.lightmypump.com

http://www.lightmypump.com/pump_glossary.htm

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Fluid Statics (Chapter 3)

Fluid at rest (no shear force)

Normal Force, Fn: Force component normal to a surface

(Fn :surface force component)

Pressure: P = Fn / A

Fn

A: area

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Fluid Statics

Absolute and gage pressures

Absolute pressure = pressure measured relative to absolute zero (perfect

vacuum)

Gage pressure = pressure measured relative to local atmospheric

pressure

Pressure usually doesn’t affect liquid properties and atmospheric pressure

usually appears on both sides of an equation. So we commonly use gage

pressure in problems dealing with liquids.

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Pressure at a point!

A

Variation of pressure in a static Fluid:

Pressure increases linearly with depth

is same in all directions

A dz

dp

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Variation of Pressure in a static Fluid

Pressure increases linearly with depth (for an

incompressible fluid)

p = -γz + patm

z h

patm

Pascal’s law: a surface of equal pressure for a

liquid at rest is a horizontal plane. It is a surface

everywhere normal to the direction of gravity.

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Pressure expressed in height of fluid

[L] = [F]/[L]2 x [F]/[L]3 = [L]

h is the height of fluid at the bottom of which the pressure will be p

Also called pressure head

1 Atm = 76 cm Hg

= 10.33 m H2O

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Fluid Statics

Pressure head, an important property

h2 h1

1) Identify a reference point (datum)

2) Identify elevation of points 1 and 2

1

2

z1

z2

3) Sum of elevation of a point and pressure head at that point

is constant

h1 + z1 = h2 + z2 = const.

that is

p1/ γ + z1 = p2/ γ + z2 = const.

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Fluid Statics

Remarks

p = γh or h = p/γ when

• Specific weight, γ = const.

• fluid is at rest

• fluid is incompressible

• the pressure of air or vapor acting on the free

surface is neglected (for now)

• distances (h) measured vertically downward

from the free surface

z h

p

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Fluid Statics

Pressure head, examples

pA

pB

pC

pD

pA = pB = pC = pB = γh

γ γ γ γ h

Which of these pressures is the highest?

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Fluid Statics


water

gasoline

A B C

Which of these points

(A, B, C) will

experience the highest

pressure?

pA = pB = pC = γgasolineh

h

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Fluid Statics


Pressure at the same depth is the same, moving sideways

does not change the pressure

Pressure distribution

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Fluid Statics


γ2

γ1 A

B

pB = ?

pB =

h1

h2

γ1h1

γ2h2

+

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Fluid Statics


γ1

γ2

p1

p2

p2 - p1 = ?

h1

h2

h3

h4

p1 + γ1(h1-h2) = p2+ γ1(h4-h3) + γ2(h3-h2)

p2 - p1 = γ1(h1-h2) - γ2 (h3-h2) - γ1(h4-h3)

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Fluid Statics

Absolute and gage pressures

Absolute pressure = pressure measured relative to absolute zero (perfect

vacuum)

Gage pressure = pressure measured relative to local atmospheric

pressure

Throughout the course, unless otherwise specified, pressure will mean

gage pressure.

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Fluid Statics

Absolute and gage pressures, example

γ2

γ1 A

B

pB (gage) =

h1

h2

γ1h1

γ2h2

+

γ2

γ1 A

B

h1

h2

pB (abs.) =

γ1h1

γ2h2

+

+patm

patm

pB (gage)

pB (abs.) =

pB (gage) + patm

pB (abs.)

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Fluid Statics

Pascal's Principle

Any external pressure applied to a fluid

is transmitted undiminished throughout

the liquid and onto the walls of the

containing vessel

http://www.ac.wwu.edu/~vawter

This is the basic idea of hydraulic machines!

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Fluid Statics

Summary of last lecture

Pressure increases linearly with depth

dp/dz=-γ =>

liquid with free surface pb=γh

h

Pressure at a point is the same in

all directions

A

Pressure is constant on a horizontal plane

pb

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Fluid Statics

Pressure head, Example

An open tank contains water 1.4 m deep covered by a 2-m-thick layer of

oil (s = 0.855). What is the pressure head at the bottom of the tank, in

terms of a water column?

ooi hp

wwoowob hhppp

ooi hp

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Fluid Statics

Measurement of pressure, barometer

p0 = γy + pvapor

pa = p0

pa = patm

po = pa = patm= γy + pvapor

patm= γy

/37

Fluid Statics

Measurement of pressure, other methods

Bourdon gage

Pressure transducers

Converts energy

from the pressure

system to

displacement of a

mechanical system

http://www.pchemlabs.com

http://www.pcb.com http://www.honeywell.com http://www.fiso.com

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Fluid Statics

Measurement of pressure, manometer

Piezometer column

pA = γh + RM(sM/sF)γ +patm

Simple manometer

Liquids only

Liquids and gases

We can omit portions of the

same fluid with the same end

elevations (pressure doesn’t

change on a horizontal

plane).

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Fluid Statics


Vacuum

pA = γh - RM(sM/sF)γ +patm

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Fluid Statics


Vacuum

pA /γF = -((sM/sF)Rm+h)

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Fluid Statics

Measurement of pressure, differential manometer

pA /γ - hA - RM(sM/sF) +hB = pB/ γ

pA /γ - pB/ γ = hA - hB + RM(sM/sF)

Liquid A and B have the same

density!

pA /γ - pB/ γ = zB – zA+RM - RM(sM/sF)

m

F

M Rs

sz

p

1

m

F

M Rs

sz

p

1

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Fluid Statics

Manometer problem

The air pressure in a tank is measured

by an oil manometer. For a given oil-

level difference between the two

columns, determine the absolute

pressure in the tank.

(density of oil is given to be = 850

kg/m3)

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Fluid Statics

Differential manometer problem

Fresh and seawater flowing in

parallel horizontal pipelines are

connected to each other by a

double U-tube manometer.

Determine the pressure

difference between the two

pipelines.

hw=0.6 m, hHg=0.1 m, hsea=0.4 m

The densities of seawater and mercury

are given to be sea = 1035 kg/m3 and Hg

= 13,600 kg/m3. The density of water w

=1000 kg/m3.

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Fluid Statics

Piston problem

F1=200 N, F2=?

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Fluid Statics

Measurement of pressure, inclined manometer

Example

http://www.gnw.co.uk/

The actual vertical

rise is still the same as it is in a vertical

tube, but here it is much easier to read

accurately when the

scale has been expanded!

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Fluid Statics

Example

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Fluid Statics

Forces on plane areas, applications: Panama Canal locks

http://sems1.cs.und.edu/~sems/

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Fluid Statics

Forces on plane areas, applications: Foundation failure

http://www.helitechonline.com/

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Fluid Statics

Forces on plane areas, applications: Gates

http://sems1.cs.und.edu/~sems/

Feeder Gates for Canal

Gate Valves for

Spillway Control

http://users.owt.com/chubbard/gcdam/html/photos

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Fluid Statics

Forces on plane areas, applications: Dams

Hoover Dam

A beaver dam

http://commons.wikimedia.org/wiki/User:Leoboudv http://snailstales.blogspot.com

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Fluid Statics

Forces on plane areas, horizontal surface

If fluid is at rest:

No tangential forces exist

All forces are normal to the surfaces

If the pressure is uniformly distributed over the area A (e.g.

submerged horizontal area for liquids and gases) , then:

pAdAp pdA F

A

p = γh h F

F: static fluid force, acts at the

centroid of area A

If fluid is gas: pressure

variation with depth over a

surface can be neglected!

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Fluid Statics

Forces on plane areas, vertical plane

The deeper the plane is submerged, the

closer the resultant (F) moves to the

centroid of the surface

When the fluid at rest is liquid, then the

pressure distribution is not uniform

F: static fluid force, always acts

below the centroid of area A

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Fluid Statics

Forces on plane areas, magnitude of F

If the width (into the page), x is constant, then F = pA = 0.5(pMJ+pNK)(MNx)

If x varies, dA=xdy, p = γh, h=ysinθ, then the force dF on the

horizontal strip:

dAyhdApdAdF sin

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Fluid Statics

Forces on plane areas, magnitude of F

dAyhdApdAdF sin

Integrating

AyydAsin Fd F c sin

Ah F c

A

Total force on any plane area (submerged in liquid) can

be found by multiplying the specific weight by the

product of the area and the depth of its centroid.

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Fluid Statics

Forces on plane areas, center of pressure

The point where the line of action of the resultant force goes through

Pressure prism

is the centroid of the pressure prism (volume).

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Fluid Statics


hp = 2/3 MN

Pressure prism

concept is useful

to determine the

center of pressure

for simple

geometries.

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Fluid Statics


If the shape of the area is not regular we have to

take moments and integrate: Axis of

moment

dAyydF sin2

dAydAyyypdAFyA

p 2sin)sin(

58

Fluid Statics


dAyydF sin2

dAydAyyypdAFyA

p 2sin)sin(

AyF c sin

Ay

I

Ay

Iy

c

O

c

Op

sin

sin

AyII

dAyI

ccO

A

O

2

2

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Fluid Statics

Forces on plane areas, location of center of pressure

Ay

Iy

Ay

IAy

Ay

Iy

c

cc

c

cc

c

Op

2

Ic : moment of inertia of an area about its centroidal axis

Parallel axis theorem

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Fluid Statics

Forces on plane areas, center of pressure, two methods

1) The point where the line of action of the resultant force goes through is

the centroid of the pressure prism (volume).

F

Ay

Iyy

c

ccp

2) Consider the moments

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Fluid Statics

Forces on plane areas, Steps

Ahc FAy

Iyy

c

ccp

1) Determine the area (A) of the submerged surface

2) Find the centroid of the area, yc, and then hc

3) Calculate the magnitude of the resultant hydrostatic force, F

4) Calculate the line of action of the hydrostatic force, yp

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Fluid Statics

Forces on plane areas, applications: Gates

mek 307 fluid mechanics - eskisehir.edu.trinsaat.eskisehir.edu.tr/aocelik/mek 307/icerik/1_4th...

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