CH.10. FLUID MECHANICSmmc.rmee.upc.edu/documents/Slides/Ch10_v13.pdf · v 0 1 eqn. Linear Momentum Balance. First Cauchy’s Motion Equation. ... tn n n 1 0 VV dS p dS

CH.10. FLUID MECHANICS Continuum Mechanics Course (MMC) - ETSECCPB - UPC

Overview

Governing Equations Newtonian Fluids Barotropic fluids

Hydrostatics. Fluids at rest Hydrostatic problem Archimedes´ Principle Equilibrium of Floating Solids

Barotropic Perfect Fluids Fluid Mechanics Equations Bernoulli’s Trinomial Steady State Solution Transient State Solution

2

Overview (cont’d)

Newtonian Viscous Fluids Navier-Stokes Equations Energy Equation Reduced System of Equations Physical Interpretations Reduced System of Equations for Particular Cases

Boundary Conditions BC in velocities BC in pressures Mixed BC BC on free surfaces

Laminar and Turbulent Flows

3

4

Ch.10. Fluid Mechanics

10.1. Governing Equations

Balance equations of the thermo-mechanical problem:Conservation of Mass. Continuity Equation. 1 eqn.

Reminder – Governing Eqns.

0 v

Linear Momentum Balance. Cauchy’s Motion Equation. 3 eqns. b v

Angular Momentum Balance. Symmetry of Cauchy’s Stress Tensor. 3 eqns.

T

Energy Balance. First Law of Thermodynamics. 1 eqn.

:u r d q

Second Law of Thermodynamics.

2 restrictions 0u s :d

2

1 0

q

8 PDE + 2 restrictions

Clausius-Planck Inequality.Heat flux Inequality.

5

Constitutive equations of the thermo-mechanical problem in a Newtonian fluid:

Grand total of 20 PDE with 20 unknowns:

Thermo-Mechanical Constitutive Equations. 6 eqns.

Newtonian Fluids

Thermal Constitutive Equation. Fourier’s Law of Conduction. 3 eqns.

State Equations. 2 eqns.

2p Tr d d 1 1

, ,s s d 1 eqn.

K q

, , 0F p ,u f

Kinetic

Caloric

Entropy Constitutive Equation.

12 PDE

1, 3, 9, 1, 3, 1, 1, 1u s p v q

6

A barotropic fluid is characterized by the kinetic state equation:

The uncoupled mechanical problem becomes:

11 scalar unknowns: , , , . (Considering the symmetry of Cauchy stress tensor, , will have 6 unknowns).

Barotropic Fluids

v p

, , 0F p p

Thermo-Mechanical Constitutive Equations. 6 eqns. 2p Tr d d 1 1

1 eqn.Kinetic State Equation p

Conservation of Mass. Continuity Equation. 1 eqn.0 v

Linear Momentum Balance. First Cauchy’s Motion Equation. 3 eqns. b v

7

8


10.2. Hydrostatics. Fluids at Rest

Uniform velocity, :

Thus,

Uniform and stationary velocity, :

Thus,

Fluid at rest, . A particular hydrostatic case (where the name comes from)

,t t v x v v v v 0

2p Tr d d 1 1

0vvvd 21S

0 0p p p 1

,t cntv xddt t

v va v v 0

0p 1 03Tr p HYDROSTATIC CASE0p p p

,t cnt v x 0

Hydrostatic stress state vs. Hydrostatic problem

HYDROSTATIC STRESS SATE

9

A hydrostatic problem ( ) is characterized by:

Substituting the constitutive and the continuity eqn. into the Cauchy eqn.:

Hydrostatic Problem

,t cntv x

0 0

0 0 0 00

( )0 1,2,3i

i

pp p p p b i

x

b 0 1 1

Thermo-Mechanical Constitutive Equations. 6 eqns.0p 1


Linear Momentum Balance. First Cauchy’s Motion Equation.3 eqns. b v

0,t cnt X X

2p Tr d d 1 1

b 0

FUNDAMENTAL EQUATION OFHYDROSTATICS

10

For a fluid subjected to gravity forces,

the momentum eq. can be written as:

If the surface pressure is considered zero, then:

Gravity forces. Triangular pressure distribution

0( , ) 0t

g

b x

0 0, , ,p x y z p y z

0 0p gz C

0

0

00

, ,0

,0

0

p x y zx

p y zy

p zg

z

dx

dy 0 0,p y z p zdz

0 0 00 0 z hp g h C C g h 0 0 p g h z

Triangularpressure

distribution

0 0

00 0 1,2,3i

i

pp b ix

b 0

11

Any fluid applies a buoyant force to an object that is partially or completely immersed in it.

The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

The resultant of the buoyant force on a floating object acts at the center of mass of the displaced fluid (center of buoyancy).

Archimedes´ Principle

12

Consider a solid with a volume V and density ρ within a fluid in a hydrostatic state. Then, The traction vector on the solid boundary :

The resultant force exerted by the fluid on the solid :

Archimedes´ Principle - Proof

0 0

0 0( ) p p

p z g h z

t n n n 1

0 V V

dS p dS

R t n

13

only depends on the hydrostatic pressure distribution on the boundary of the solidR

Consider the same fluid without the solid in it and replaced by fluid . Then, Pressures on the boundary of the “replacing” fluid are the same than in the

immersed solid case (and, therefore, the resulting force, )

The divergence theorem can be applied: (The pressure distribution is continuous in space)

Finally,

Archimedes´ Principle (first part proof)

0 0 0

0ˆ

ˆ zV V V

zW

p dS p dV dV W

b

e

R n b e

ˆ ˆz zE W E W R e e

Up-thrust on the body = weight of the fluid displaced by the body

Volume of the displaced fluid

14

R

0 V V

dS p dS

R t n

0 0p b 0

( )E( )W

Consider the moment of the up-thrust forces at the center of gravity of the volume of displaced fluid:

Substituting the fundamental eq. of hydrostatics ,

Archimedes´ Principle (second part proof)

0

0 0

0

(GEV V

V

pp dS p dV

p dV

M x n x

x

Divergence Theorem

0 0p b

0 G GE WV

dV M x b M 0GWM Moment of the weight of the displaced fluid with respect to its center of gravity (by definition it must

be zero)The up-thrust force, E, passes through the CG

of the volume of the displaced fluid (center of buoyancy).15

0 0p b 0

The equilibrium can be:

Stable: the solid’s CGis below the CG of the displaced fluid.

Unstable: the solid’s CGis above the CG of the displaced fluid.

Equilibrium of Floating Solids

16

17


10.3. Fluid Dynamics.Barotropic Perfect Fluids

A perfect fluid is a Newtonian fluid with null viscosity, :

Therefore,

In a barotropic fluid temperature does not intervene in the kinetic state equation:

Barotropic Perfect Fluids

0

, , 0F p p

2p Tr d d 1 1 p 1

REMARK Do not confuse a hydrostatic stress state (spherical stress tensor) witha hydrostatic movement regime (null or uniform velocity).

hydrostatic stress state

p

p pTr : d : d d

1

18

The mechanical problem for a barotropic perfect fluid:

5 scalar unknowns: , , .

Barotropic Perfect Fluids: Field Equations

v p


Linear Momentum Balance. Euler’s Equation. 3 eqns. b v p b v

1 eqn.Kinetic State Equation p , , 0F p

19

The thermal problem for a barotropic perfect fluid:

Once the mechanical problem is solved, the thermal problem can be calculated as there are 5 scalar unknowns:

Energy Balance. First Law of Thermodynamics.

1 eqn.:u r d q

1, 3, 1u q

Thermal Constitutive Equation. Fourier’s Law of Conduction.

3 eqns. , K q q v

2u p r K v

1 eqn.Caloric State Equation ,u u , , ,u f p v

Barotropic Perfect Fluids: Field Equations

20

Consider a barotropic fluid with potential body forces:

And, consider the following lemmas: Lemma 1. For a barotropic fluid there exists a function

which satisfies:

Lemma 2. The convective term of the acceleration can be written as:

Bernoulli’s Trinomial

Body forces potential

0, , , 0

T

t gz t tx y z

g

x b x x

ˆ( , ) ( ( , ))t p tx xP P

p P

212 v2

v v v

Where is the vorticity vector.2 v

0

1 1( )( ) ( )

pp dp p

p p P PProof :

21

Taking the Euler equation and substituting,

Rearranging,

Bernoulli’s Trinomial

212

v 2t

v v P

Bernoulli’s TrinomialEQ. OF MOTION for a barotropic perfect fluid under potential body forces

212 2 vt

v v P

212

1

1 ;

2 v )

t

dp pdt

p

v v v

vb v b

b

v v v

(

P

22

Steady flow: the spatial description spatial of its properties is not dependent on time:

Pathlines and streamlines coincide.

Generally, any mechanical problem will have an initial transient state with unsteady flow:

Steady Flow

0v t

23

t

0

t

σ 0

The equation of motion for a steady flow becomes:

Considering a stream line parameterized in terms of its arc length :

Barotropic perfect fluid with potential forces:Steady state solution

212

v 2t

v v P 2

1 v 22

vP

: s x xs

24

Then, projecting the eq. of motion in the direction of the streamline’s tangent , t

212 v 2 v P 2M v w v v ( )M s x

0

d dsds dt

x

0 ( )M

d dMM s sds ds

x

xx x ( )M cnt x x

212 v cnt x xPBeurnoulli’s trinomial remains constant on the same streamline.

Barotropic perfect fluid with potential forces:Steady state solution

25

Incompressible fluid:

Potential gravitational forces:

0 0 00 0

0

1 1( )( )

p p pp cnt p dp dpp

P

00 ( ) gzg

b x x x

2 21 12 20

v vp gz cnt

x xP

2

0

1 v2

defpz H cntg g

x BERNOULLI’S THEOREM

26

Barotropic perfect fluid with potential(gravitational)forces: steady state solution

Bernoulli’s Theorem can be interpreted as:

It is a statement of the conservation of mechanical energy:

2

0

1 v2

p gz cnt

x

2

0

1 v2

defpz H cntg g

x

Piezometric or hydraulic head, h

elevation

pressure head

velocity head

total or energetic

headpressure

head

velocity head

elevation

pressure energy potential

energykinetic energy

27

Barotropic perfect fluid with potential(gravitational)forces: steady state solution

Determine the velocity and mass flow rate of water from the circular hole (0.1m dia.) at the bottom of the water tank (at this instant). The tank is open to the atmosphere and h = 4m. Consider a steady state regime.

Example

h

28

Example - Solution

h

2 21 1 2 2

1 20 0

v v2 2

p pz zg g g g

1 2

21 1 2 1 2

1

0

v 0 ( v v 0)

atmp p pSS SS

2 1 2v 2g z z 2

21 2

v2

z zg

Velocity at the bottom hole of the tank: 2v 2gh

h

29

1

cross section area at 1

S

2

cross section area at 2

S

The equation of motion for an unsteady flow is:

This expression may be simplified for: Potential (irrotational) flow Potential and incompressible flow

212

v 2t

v v P

REMARK A movement is said to be irrotational(or potential) if the rotational of the velocity field is null at any point:

, ,t t v x 0 x

Barotropic perfect fluid with potential forces:Transient solution

30

In an irrotational flow:

There will exist a scalar function (named velocity potential ) which satisfies:

Then, the equation of motion becomes:

Rearranging,

, ,t t v x 0 x 1, , ,2

t t t x v x 0 x

, t tv x x t x

212

v 2t

v v P

tt

x

212 v )t

P (

,M t x21

2 v t

0 P , ,M t t x 0 x

31

Barotropic perfect fluid with potential forces and irrotational flow: Transient solution

The momentum equation can be trivially integrated:

Defining a modified velocity potential, :

Finally,

, ,M t t x 0 x 212, vM t tt

x P

t x

0

, ,tdef

t t d x x

, t

tt t

v x

Differential equation of hydraulic transients

21 0 ,2

tt

xP

32

Barotropic perfect fluid with potential forces and irrotational flow: Transient solution

The mechanical problem for a potential (irrotational) flow:

3 scalar unknowns: , , . Once the potential is obtained, the velocity field can be easily calculated:

p

,t tv x x

Conservation of Mass. Continuity Equation. 1 eqn. 0t x

Linear Momentum Balance. Hydraulic Transients Equation. 1 eqn.

1 eqn.Kinetic State Equation p , , 0F p

2 0

21, 0 ,2p tt

xP

barotropic fluid

, t tv x x

Barotropic perfect fluid with potential forces:Transient solution in irrotational flows

33

In an incompressible flow:

Then, the term in the equation of motion becomes:

And, the equation of motion is:

20

1 ]2

pt

0[

0ddt 0

0 00 01 1,

p p pt dp dpp

xP

P

34

Incompressible perfect fluid with potential forces: transient solution in irrotational flows

The mechanical problem for a potential (irrotational) and incompressible flow:

2 scalar unknowns: , . Once the potential is obtained, the velocity field can be easily calculated:

Fluid Mechanics Equations

p

Conservation of Mass. Continuity Equation. 1 eqn. 0t x

Linear Momentum Balance. Hydraulic Transients Equation. 1 eqn.

2

0

1 0 ,2

p tt

x

, t tv x x

2 0not

35

36


10.4. Newtonian Viscous Fluids

Governing equations of the general fluid mechanics problem:

17 scalar unknowns: , , , , , , , .

Governing Equations


Linear Momentum Balance. Equation of Motion. 3 eqns. b v

Energy Balance. First Law of Thermodynamics. 1 eqn.:u r d q

Mechanical Constitutive Equations. 6 eqns. 2p Tr d d 1 1

Thermal Constitutive Equation. Fourier’s Law of Conduction. 3 eqns.K q

Caloric and Kinetic State Equations. 2 eqns. , , 0F p ,u u

, ,s s d 1 eqn.Entropy Constitutive Equation.

v u q s p Too large to solve analytically !!

37

Consider the following lemmas: Lemma 1.

Lemma 2.

Introducing the constitutive equation into the divergence of the stress tensor, ,and taking into account these lemmas:

Navier-Stokes Equations

1 12 2

d v v

1

Where is the deformation rate tensor

, td x

Where is a scalar function. , t x

2p Tr

v

d d 1 1

( ))

( ( ) 2 )

( ( )) (

p Tr

p Tr

v

v v

d d

d

1 1

38

Then, the linear momentum balance equation is rearranged:

The Navier-Stokes equations are essentially the equation of motion (Cauchy’s equation) expressed solely in terms of velocity and pressure.

Navier-Stokes Equations

ddt

vb dp dt v vvb

NAVIER-STOKES EQUATIONS

2 2v v v {1,2,3}j i ii

i i j j j

dpdt

p db ix x x x x dt

vvv b

39

Consider the definition of stress power:

And the Fourier’s Law:

Introducing these into the energy balance equation:

Energy Equation

2: 2 :pTr Tr d d d d dRECOVERABLE POWER, .WR

DISSIPATIVE POWER, . 2WD

K q K q

:u r d q 2( ( ) ( ) 2 : ) ( )du pTr Tr r K

dt d d d d

40

Then, the energy balance equation is reduced to:

The energy equation is essentially the energy balance equation expressed solely in terms of velocity and pressure.

Energy Equation

ENERGY EQUATION

22

i i

i i i i

:

v v {1,2,3}ij ij

du p r K Trdt

du p r K d d idt x x x x

v d d d

2( ( ) 2 : )du pTr Tr r Kdt

vd d d d

DISSIPATIVE POWER, . 2WD

41

Governing equations of the general fluid mechanics problem are reduced to:

8 scalar unknowns: , , , , , .

Reduced System of Equations


Energy Balance. 1 eqn. 2 :u p r K Tr v d d d

Caloric State Equation. 1 eqn. ,u u

Momentum Balance. Navier-Stokes Equations. 3 eqns. p v v b v

Kinetic State Equation. 1 eqn. , , 0F p

, ,s s v 1 eqn.Entropy Constitutive Equation.

v u p s

42

REMARK For a barotropic fluid, the mechanic and thermal problems are uncoupled, reducing the mechanical problem to 5 unknowns.

The Navier-Stokes equations can be physically interpreted as:

Navier-Stokes Equations: Physical Interpretation

dp dt >0

vvv b 0

Forces due to the pressure

gradientViscous forces due to the contact with neighbour particles

Body forces

Inertial forces

NOTE: All forces are per unit of volume.

43

The energy equation can be physically interpreted as:

2( ) ( ) :du p r K Trdt

v d d d

Variation of internal energy

Mechanical work of the thermodynamic

pressure:

( ) )1 ( )

1 ( )

( dVd dV

d dt dVdV

d dVdt

d dt

p

V

p

v

vv

Variation of volume per unit of volume

and per unit of time

Heat generated by the internal sources and conduction per unit

volume per unit time

Dissipative power, . 2WD

NOTE: All terms are per unit of volume and unit of time

Navier-Stokes Equations: Physical Interpretation

44

CONTINUITY EQUATION Cartesian Coordinates

Cylindrical Coordinates

Spherical Coordinates

Fluid Mechanics Equationsin Curvilinear Coordinates

( ) 0t

v

v v v 0x y zt x y z

1 1v v v 0r zrt r r r z

221 1 1v v sin v 0sin sinrrt r r r r

45

NAVIER-STOKES EQUATIONS for an incompressible fluid with and constants Cartesian Coordinates

component

component

component

x

y

z

2 2 2

2 2 2

v v v v v v vv vx x x x x x xx y z xpv b

t x y z x x y z

2 2 2

2 2 2

v v v v v v vv v vy y y y y y yx y z y

p bt x y z y x y z

2 2 2

2 2 2

v v v v v v vv v vz z z z z z zx y z zp b

t x y z z x y z

46

p va ;

t

v 0

v bva v v


NOTE: for “slow” motions left-hand-side term is zero ( )a 0

NAVIER-STOKES EQUATIONS for an incompressible fluid with and constants

Cylindrical Coordinates component

component

component

r

z

2 2 2

2 2 2 2

v v vv v v v v v1 1 2v v vr r r r r rr z r rp r b

t r r r z r r r r r r z

2 2

2 2 2 2

v v v v v v v v vv1 1 1 2v v vr rr zp r b

t r r r z r r r r r r z

2 2

2 2 2

vv v v v v v v1 1v vz z z z z z zr z zp r b

t r r z z r r r r z

47

p va ;

t

v 0

v bva v v


NOTE: for “slow” motions left-hand-side term is zero ( )a 0

2

22 2 2 2 2 2 2

v v v v v v v v vv 1vsin sin

v v vv1 1 1 1 2 2 θv sinsin sin sin sin

rr

r

pcotgt r r r r r r

cotgr br r r r r r r

NAVIER-STOKES EQUATIONS for an incompressible fluid with and constants Spherical Coordinates

component

component

component

r

2 2

22

2 2 2 2 2 2 2

v v vvv v v vvsin

vv v1 1 1 2 2v sin v sinsin sin sin sin

r r r rr

r rr r

pt r r r r r

r br r r r r r r

2

22

2 2 2 2 2 2 2

v vv v v v v v v 1vsin

vv v v1 1 1 1 2 2v sinsin sin sin

rr

r

cotg pt r r r r r r

cotgr br r r r r r r

48

p va ;

t

v 0

v bva v v


STRESS TENSOR for Newtonian fluids Cartesian Coordinates

2( )3

K

v 22 · ·3

xx px

v v K

vv ··32v2 K

p

yy

y

vv ··32v2 K

pzz

z

xyyx

yxxy

vv

yzzy

zyyzvv

zxxz

xzzxvv

49

22- ; ( )3

p Tr

Tr

d d

d v

1 1

KFluid Mechanics Equationsin Curvilinear Coordinates

NOTE: for incompressible fluids ( )v 0

STRESS TENSOR for Newtonian fluids Cylindrical Coordinates

v 22 · ·3

rr pr

v v K

1 v v 22 · ·3

r pr r

v v K

v 22 · ·3

zz pz

v v K

r

rr rrrr

v1v

z

zz rzv1v

zrrz

rzzrvv

1 1 v v· v zrrr r r z

v

50

2( )3

K

22- ; ( )3

p Tr

Tr

d d

d v

1 1



STRESS TENSOR for Newtonian fluids Spherical Coordinates

v 22 · ·3

rr pr

v v K

1 v v 22 · ·3

r pr r

v v K

v1 v v 22 · ·

sin 3r cotg p

r r r

v v K

r

rr rrrr

v1v

v1vsinrsinr

sin

rrr

sinrr

rr

vv1

22v1 1 1· v v sin

sin sinrr

r r r r

v

51

2( )3

K

22- ; ( )3

p Tr

Tr

d d

d v

1 1



Reduced governing equations of the general fluid mechanics problem for an incompressible viscous fluid are:

Reduced System of Equations for the incompressible case




Kinetic State Equation. 1 eqn. p

0ddt


0 Tr d vReminder:


52

Mechanical Constitutive Equations. 2p Tr d d 1 1 6 eqns.

Reduced governing equations of the general fluid mechanics problem for an incompressible viscous fluid are:

6 scalar unknowns: , , , .

0, ,s s v 1 eqn.Entropy

Constitutive Equation.

Energy Balance. 1 eqn. 0 0 :u r K d d

Caloric State Equation. 1 eqn. 0,u u

v u p

Momentum Balance. Navier-Stokes Equations. 3 eqns.0 0p v b v

Conservation of Mass. Continuity Equation. 1 eqn.0 vMechanical

problem

Thermal problem

0ddt 0 Tr d v

Reminder:

Reduced System of Equations for the incompressible case

53

Mechanical Constitutive Equations.

2p d 1 6 eqns.

Reduced governing equations of the general fluid mechanics problem for a fluid with null volumetric viscosity are:







23 0 2

3

13

Tr d vReminder:

Reduced System of Equations for the null bulk viscosity case

54


Reduced governing equations of the general fluid mechanics problem for a fluid with null volumetric viscosity are:

Energy Balance. 1 eqn. :u p r K v d d


Momentum Balance. Navier-Stokes Equations. 3 eqns.

13

p v v b v




2 03

23

13

Tr d vReminder:

Thermo-mechanical problem

Reduced System of Equations for the null bulk viscosity case

55


2 23

d dp Tr 1 1 6 eqns.

Reduced governing equations of the general fluid mechanics problem for a perfect fluid are:







0 Tr d vReminder:

Reduced System of Equations for the perfect fluid case

56


Reduced governing equations of the general fluid mechanics problem for a perfect fluid are:

Energy Balance. 1 eqn. u p r K v


Momentum Balance. Navier-Stokes Equations. 3 eqns.p b v




0 Tr d vReminder:

Thermo-mechanical problem

Reduced System of Equations for the perfect fluid case

57


p 1 6 eqns.

, , , ,

Reduced governing equations of the general fluid mechanics problem for a fluid at rest (HYDROSTATIC CASE) are:







Tr d vReminder:


0va dtd

0v 0 0pp 0p 1

58


Reduced governing equations of the general fluid mechanics problem for a fluid at rest (HYDROSTATIC CASE) are:


Energy Balance. 1 eqn. 0 0u r K


Momentum Balance. Navier-Stokes Equations. 3 eqns.0 0p b 0

Mechanical problem



Thermal problem

Kinetic State Equation. 1 eqn. 0, op

0va dtd

, , , , 0v 0 0pp 0p 1 Tr d vReminder:

59

Mechanical Constitutive Equations.0

p 1 6 eqns.

, , , ,

Reduced governing equations of the general fluid mechanics problem for a barotropic fluid at rest (HYDROSTATIC CASE) are:






Tr d vReminder:


0va dtd

0v 0 0pp 0p 1

Kinetic State Equation. 1 eqn. p

60


Reduced governing equations of the general fluid mechanics problem for a barotropic fluid at rest (HYDROSTATIC CASE) are:





Mechanical problem



Thermal problem

0va dtd

, , , , 0v 0 0pp 0p 1 Tr d vReminder:

61


p 1 6 eqns.

62


10.5. Boundary Conditions

To solve the governing equations in a fluid mechanics problem, adequate boundary conditions must be imposed.

Fluid mechanics problems typically use: Spatial (or Eulerian) description of the equations. Control volume (fixed in space).

Boundary conditions are applied on the contour of this control volume. The most frequent are: In velocities In traction vectors (or pressures) Mixed On free surfaces

Boundary Conditions

63

Prescribed velocity Velocities are known in certain parts of the control volume’s contour :

Impervious wall condition Part of the control volume’s contour is formed by impervious walls which

cannot be penetrated by the fluid. The relative velocity of the fluid w.r.t. the wall (which may be mobile

and have a velocity ) in the normal direction to the contour must be null:

Boundary Conditions in Velocities

v, ,t t v x v x xv

vn

rv*v

n

*n vv , t x v n v n x

n

*v0r v n v v n x

64

Adherence condition In a viscous fluid in contact with a wall the fluid is considered to adhere to

the wall. The relative velocity of the fluid w.r.t the wall is null:

Boundary Conditions in Velocities

rv

v, r t *v x v v 0 x

v *v v x

65

Prescribed traction vector The traction vector’s value is prescribed in certain parts of the control

volume’s contour :

Sometimes, only part of the traction vector is prescribed, such as the thermodynamic pressure.

For a Newtonian fluid:

Boundary Conditions in Pressures

*, ,t t t x n t x x

( ) 2p Tr d d 1 1 ( ) 2p Tr t n n d n d n

*p( , ) ( , ) p t p t x x x

66

Prescribed traction vector and velocities Pressure (part of the normal component of the traction vector) and the

tangential components of velocity are prescribed :

This boundary condition is typically used in problems involving entrance and exit sections of pipes.

Mixed Boundary Conditions

67

The contact surface between air and fluid (generally water) is a free surface.

Boundary Conditions on Free Surfaces

68

HYPOTHESIS: The free surface is a material surface. This implicitly establishes certain boundary conditions on the velocity field

of the material surface . Consider the free surface:

Impose the condition for a material surface (null material derivative):


sl

: { | , , , , , 0}sl x y z t z x y t x

x y zv v v 0ddt t t x y z

v

1

z x yv ( , ) v v slt t x y

x x

69

Another boundary condition typically used on free surfaces is:

This allows identifying the position of the free surface once the pressure field is known:


, atm slp t P x x

: { | , 0 }sl atmp t P x x

70

71


10.6. Laminar and Turbulent Flows

Flow persists as unidirectional movement. Particles flow in parallel layers which do not mix. A flow’s laminar character is identified by the Reynolds number: The governing equations of the fluid mechanics problem are valid for this

type of flow.

Laminar Flow

def

eV LR

, Flow’s characteristic velocity, Domain’s characteristic length, Kinematic viscosity:

VL

1000eR

72

High values of the Reynolds number. Highly distorted and unstable flow: Stress and velocity at a given spatial point fluctuate randomly and very

fast along time about a mean value. Specific models (turbulence models) are used to characterize this regime.

Turbulent Flow

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NOTE: Turbulent flow is out of the scope of this course

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Summary

Governing equations of the general fluid mechanics problem:

17 scalar unknowns: , , , , , , , .

Summary (cont’d)


Linear Momentum Balance. Equation of Motion. 3 eqns. b v

Energy Balance. First Law of Thermodynamics. 1 eqn.:u r d q

Mechanical Constitutive Equations of a Newtonian Fluid. 6 eqns. 2p Tr d d 1 1

Thermal Constitutive Equation. Fourier’s Law of Conduction. 3 eqns.K q

Caloric and Kinetic State Equations. 2 eqns. , , 0F p ,u u

, ,s s d 1 eqn.Entropy Constitutive Equation.

v u q s p

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Simplifications in the governing equations: BAROTROPIC FLUID:

UNIFORM VELOCITY:

UNIFORM AND STATIONARY VELOCITY:

Summary (cont’d)

, , 0F p

Thermo-Mechanical Constitutive Equations. 6 eqns. 2p Tr d d 1 1



Linear Momentum Balance. First Cauchy’s Motion Equation. 3 eqns. b v

Uncoupled mechanical problem

p 1 p p

0p 1 0p p p HYDROSTATIC

CASE

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Simplifications in the governing equations: HYDROSTATIC PROBLEM

Summary (cont’d)

HYDROSTATIC FUNDAMENTAL EQUATION

Thermo-Mechanical Constitutive Equations. 6 eqns.0p 1

Conservation of Mass. Continuity Equation. 1 eqn.

Linear Momentum Balance. First Cauchy’s Motion Equation.3 eqns.

0,t cnt X X

b 0

0 0p b 0

0 0 p g h z

For a fluid exposed to gravity forces and considering surface pressure zero, the solution is:

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Archimedes’ PrincipleAny fluid applies a buoyant force to an object that is partially or completely immersed in it.

The magnitude of the buoyant force is equal to the weight of the fluid that the object displaces.

The resultant of the buoyant force on a floating object acts at the center of mass of the displaced fluid (center of buoyancy).

Summary (cont’d)

ˆ ˆz zW E F e e

78

Barotropic perfect fluids. (hydrostatic stress state)

Summary (cont’d)

p 1 p



Mechanical problem


Thermal Constitutive Equation. Fourier’s Law of Conduction.

, K q q v 3 eqns.

Energy Balance. First Law of Thermodynamics. 1 eqn.

2u p r K v

1 eqn.Caloric State Equation ,u u

Thermal problem

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Barotropic perfect fluids.

Steady state solution:

If the fluid is incompressible: and the potential is defined as :

Summary (cont’d)

21 v 22 t

v vP

Bernoulli’s TrinomialEQ. OF MOTION for a barotropic perfect fluid under potential body forces:

0

, , 0t tg

b x x

21 v2

cnt

x xPBeurnoulli’s Trinomial remains constant on

the same streamline.

001p pdpp

xP

2

0

1 v2

defpz H cntg g

x BERNOULLI’S THEOREM

gz

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Barotropic perfect fluids. Transient state solution:

For potential (irrotational) flow:

For an incompressible flow:

Summary (cont’d)

21 v 22 t

v vP

, ,M t t x 0 x

,M t x

21 0 ,2

tt

xP

, t

tt t

v x Differential eqn. of hydraulic transients

20

1 0 ,2

p tt

x

0

, pt

xP

2

,1 v2

t M t

t

x

P

81

Newtonian (viscous) fluids.

8 scalar unknowns: , , , , , .

Summary (cont’d)







v u p s

82

Simplifications in the governing equations of Newtonian fluids: INCOMPRESSIBLE BAROTROPIC FLUID: ,

Summary (cont’d)

, , 0F p 0



Energy Balance. 1 eqn. 0 0 :u r K d d


Momentum Balance. Navier-Stokes Equations. 3 eqns.0 0p v b v

Conservation of Mass. Continuity Equation. 1 eqn.0 vMechanical

problem

Thermal problem

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2p d 1 6 eqns.

Simplifications in the governing equations of Newtonian fluids: FLUID WITH NULL VOLUMETRIC VISCOSITY:

Summary (cont’d)

2 03

Energy Balance. 1 eqn. :u p r K v d d


Momentum Balance. Navier-Stokes Equations. 3 eqns.

13

p v v b v



, ,s s v 1 eqn.Entropy Constitutive Equation.Thermo-

mechanical problem

84


2 23

p Tr d d 1 1 6 eqns.

Simplifications in the governing equations of Newtonian fluids: PERFECT FLUID:

Summary (cont’d)

0

Energy Balance. 1 eqn. u p r K v





, ,s s v 1 eqn.Entropy Constitutive Equation.Thermo-

mechanical problem

85


p 1 6 eqns.

Simplifications in the governing equations of Newtonian fluids: FLUID AT REST (HYDROSTATIC CASE):

Summary (cont’d)

0va dtd

0v 0 0pp 0p 1




Mechanical problem



Thermal problem

Kinetic State Equation. 1 eqn. 0, op

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p 1 6 eqns.

Simplifications in the governing equations of Newtonian fluids: BAROTROPIC FLUID AT REST (HYDROSTATIC CASE):

Summary (cont’d)

0va dtd

0v 0 0pp 0p 1




Mechanical problem



Thermal problem

, , 0F p

87


p 1 6 eqns.

Boundary Conditions Prescribed velocity

Impenetrability condition

Adherence condition

Prescribed traction vector

Sometimes, just the thermodynamic pressure is prescribed. For a Newtonian fluid:

Summary (cont’d)

v, ,t t v x v x x

n

*v0r v n v v n x

v

v

v *v v x

*, ,t t t x n t x x

* p, , p t p t x x x

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Boundary Conditions Mixed: Pressure and the tangential components of velocity are prescribed.

On free surfaces HYPOTHESIS: The free surface is a material surface.

Summary (cont’d)

z x yv , v v slt t x y

x x

If ,then:

, atm slp t P x x

: { | , 0 }sl atmp t P x x

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Laminar Flow: Molecules flow in parallel layers which do not mix. Reynolds number:

Turbulent Flow: Stress and velocity at a given spatial point fluctuate randomly and very fast

along time about a mean value. Turbulence models are used to characterize this regime.

Summary (cont’d)

def

eV LR

, Flow’s characteristic velocity, Domain’s characteristic length, Kinematic viscosity:

VL p

90

Documents

CH.10. FLUID MECHANICSmmc.rmee.upc.edu/documents/Slides/Ch10_v13.pdf · v 0 1 eqn. Linear Momentum Balance. First Cauchy’s Motion Equation. ... tn n n 1 0 VV dS p dS