fluid properties

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FLUID PROPERTIES. Independent variables. SCALARS. VECTORS. TENSORS. , w. REFERENCE FRAME. , v. , u. SCALARS. Need a single number to represent them: P , T , ρ. besttofind.com. Temperature. May vary in any dimension x , y , z , t. www.physicalgeography.net/fundamentals/7d.html. - PowerPoint PPT Presentation

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FLUID PROPERTIES

Independent variables

SCALARS

VECTORS

TENSORS

x

z

y

dy

dz

dx, u

, w

, v

REFERENCE FRAME

SCALARS

Need a single number to represent them: P, T, ρ

besttofind.com

Temperature

May vary in any dimension x, y, z, t

www.physicalgeography.net/fundamentals/7d.html

VECTORS

Have length and directionNeed three numbers to represent them: zyxx ,,

wvuV ,,

http://www.xcrysden.org/doc/vectorField.html

Unit vector = vector whose length equals 1

100

ˆ,010

ˆ,001

ˆ kji

]ˆ,ˆ,ˆ[ kji

x

y

z

i

j

k

VECTORS

In terms of the unit vector: zkyjxix ˆˆˆ

wkvjuiV ˆˆˆ

txVV ,

CONCEPTS RELATED TO VECTORS

Nabla operator:

zyx

,,

Denotes spatial variability

kz

jy

ix

ˆˆˆ

Dot Product: 321321 ,,,, vvvuuuvu

zw

yv

xuV

332211 vuvuvu

CONCEPTS RELATED TO VECTORS

CrossProduct: 321321 ,,,, vvvuuuvu

321

321

ˆˆˆ

vvvuuukji

122131132332ˆˆˆ vuvukvuvujvuvui

wvuzyxv ,,,,

wvu

zyxkji ˆˆˆ

],,[

yuxvxwzuzvyw

INDICIAL or TENSOR NOTATION

3,2,1 iAA i

Vector or First Order Tensor

iiBABA

Vector Dot Product

332211 BABABA

3,2,1;3,2,1333231

232221

131211

jiC

CCCCCCCCC

C ij

Matrix

or Second Order Tensor

txVV iii ,

INDICIAL or TENSOR NOTATION

3,2,1

ixCC

iGradient of Scalar

j

i

xBB

Gradient of Vector Second Order Tensor

jiji

ij 01

Special operator – Kronecker Delta

jkijik CC

333333323332313331

232223222222212221

131113121112111111

CCCCCCCCCCCCCCCCCC

TENSORS

x

z

y

Need nine numbers to represent them: txiijij ,

3,2,1, ji

333231

232221

131211

ij

zyxji ,,,

zzzyzx

yzyyyx

xzxyxx

ij

For a fluid at rest:

33

22

11

000000

ij

Normal (perpendicular) forces caused by pressure

MATERIAL (or SUBSTANTIAL or PARTICLE) DERIVATIVE zyxtSS ,,,

tz

zS

ty

yS

tx

xS

tS

DtDS

zSw

ySv

xSu

tS

DtDS

Fluids

Deform more easily than solids

Have no preferred shape

Deformation, or motion, is produced by a shear stress

ndeformatioofrate

zu

z

x

u

μ = molecular dynamic viscosity [Pa·s = kg/(m·s)]

Continuum Approximation

Even though matter is made of discrete particles, we can assume that matter is distributed continuously.

This is because distance between molecules << scales of variation

tx,

ψ (any property) varies continuously as a function of space and time

space and time are the independent variables

In the Continuum description, need to allow for relevant molecular processes – Diffusive Fluxes

Diffusive Fluxes

e.g. Fourier Heat Conduction law:

TkQ

z

x

t = 0

t = 1

t = 2

Continuum representation of molecular interactions

This is for a scalar (heat flux – a vector itself)

but it also applies to a vector (momentum flux)

x

z

y

dy

dz

dx

Shear stress has units of kg m-1 s-1 m s-1 m-1 = kg m-1 s-2

Shear stress is proportional to the rate of shear normal to which the stress is exerted zu

zu

at molecular scales

µ is the molecular dynamic viscosity = 10-3 kg m-1 s-1 for water is a property of the fluid

or force per unit area or pressure: kg m s-2 m-2 = kg m-1 s-2

xu

dxxu

xxu

yu

dyyu

yyu

zu

dzzu

zzu

Diffusive Fluxes (of momentum)

x

z

y

dy

dz

dx

xu

dx

xu

xxu

yu

dyyu

yyu

zu

dzzu

zzu

Net momentum flux by u

zu

zyu

yxu

x

Diffusive Fluxes (of momentum)

For a vector (momentum), the diffusion law can be written as (for an incompressible fluid):

i

j

j

iij x

uxu

2

Shear stress linearly proportional to strain rate – Newtonian Fluid (viscosity is constant)

Boundary Conditions

Zero Flux

No-Slip [u (z = 0) = 0]

z

x

u

Hydrostatics - The Hydrostatic Equation

z

g

A

z = z0

z = z0 + dzdz

p

p + (∂p/∂z ) dz

dzzpzpdzzp

z000

AdzgforceGravity

00

00

AdzgAdzzpzpAzp

z

Adzbydividing

gzp

g

dzdp

Czgp

Integrating in z:

Example – Application of the Hydrostatic Equation - 1

z

H

Find hDownward Force?

Weight of the cylinder = W

Upward Force?Pressure on the cylinder = F

aircAhgF ;

c

c

AgWh

AhgW

Same result as with Archimedes’ principle (volume displaced = h Ac) so thebuoyant force is the same as F

Pressure on the cylinder = F = W

AC

h

Example – Application of the Hydrostatic Equation - 2

z

W

D

Find force on bottom and sides of tank

On bottom?

On vertical sides?

Same force on the other side

TD ADgF x

L

AT = L W

dFx gWzdzdzWpdFx

Integrating over depth (bottom to surface)

22

2020 DgWzgWgWzdzF

DDx

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