Phy 352: Fluid Dynamics, Spring 2013

Phy 352: Fluid Dynamics, Spring 2013

Prasad [email protected]

Indian Institute of Science Education and Research (IISER), Pune

Subramanian Fluid Dynamics

Introduction

Course webpage:http://www.iiserpune.ac.in/∼p.subramanian/FluidsSpring2013.html


Introduction


3 credit course offered to students of the 6th and 8thsemesters at IISER Pune. Also offered over the NationalKnowledge Network to students at IISER Kolkata


Introduction



11:30 am - 12:25 pm, Tuesdays and Fridays


Introduction




Useful for students wishing to gain an overview of the vastfield of fluid dynamics. Applications: astrophysics,aerodynamics, biofluid dynamics, computational fluiddynamics, etc.


Introduction




Useful for students wishing to gain an overview of the vastfield of fluid dynamics. Applications: astrophysics,aerodynamics, biofluid dynamics, computational fluiddynamics, etc.

Pre-requisites: Classical mechanics, a sound knowledge of(basic) vector calculus (e.g., gradient, divergence, curl, etc.)and some familiarity with tensors


Course organization - I

The continuum hypothesis, kinematics, conservation laws:continuity equation, Euler and Navier-Stokes equations(chapters 1, 3 and 4, Kundu & Cohen, 3 weeks)




Dimensionless numbers, dynamic similarity, aerodynamics(chapters 8 and 15, Kundu and Cohen, 2 weeks)





Quiz 1 (15 %)





Quiz 1 (15 %)

Compressible flows, speed of sound, shocks (2 weeks, chapter16, Kundu and Cohen, also relevant parts from Physics ofFluids and Plasmas, Arnab Rai Choudhuri)





Quiz 1 (15 %)

Compressible flows, speed of sound, shocks (2 weeks, chapter16, Kundu and Cohen, also relevant parts from Physics ofFluids and Plasmas, Arnab Rai Choudhuri)

Mid-term exam (30 %)


Course organization - II

Fluid instabilities and turbulence (chapters 12 and 13, Kunduand Cohen, also relevant parts from Physics of Fluids andPlasmas, Arnab Rai Choudhuri, 2 weeks)




A quick introduction to computational fluid dynamics(chapter 11, Kundu and Cohen, 1 week)





Quiz 2 (15 %)





Quiz 2 (15 %)

Applications of fluid dynamics in Astrophysics: e.g.,Astrophysical jets, the de Laval nozzle, spherical accretiononto compact objects, the solar wind (2 weeks)





Quiz 2 (15 %)


Applications of fluid dynamics in geophysics (chapter 14,Kundu and Cohen, 1 week)





Quiz 2 (15 %)


Applications of fluid dynamics in geophysics (chapter 14,Kundu and Cohen, 1 week)

Final exam (40 %)


Reference Material

Fluid Mechanics by Kundu and Cohen, 4th edition, 2008,Elsevier (Indian edition): primary text


Reference Material


The Physics of Fluids and Plasmas, Arnab Rai Choudhuri,Cambridge University Press (Indian edition)


Reference Material


The Physics of Fluids and Plasmas, Arnab Rai Choudhuri,Cambridge University Press (Indian edition)

Fluid Mechanics, Landau & Lifshitz, 2nd edition, PergamonPress (Indian edition)


The continuum hypothesis

Sub-branch of continuum mechanics (as opposed to themechanics of point bodies)




We don’t concern ourselves with microscopics (like collisionsbetween atoms/molecules); materials are only characterizedby bulk properties like conductivity, elasticity (for solids),viscosity (for fluids)





These bulk properties are characterized by the relevanttransport coefficients, which can be derived frommicroscopic properties





These bulk properties are characterized by the relevanttransport coefficients, which can be derived frommicroscopic properties

To operate in the continuum limit, we need (roughly speaking)N (number of particles in some macroscopic volume) ≫ 1.


Bulk properties from kinetics: examples

Consider a gas whose molecules are in thermodynamicequilibrium. It has a definite distribution f (say,Maxwell-Boltzmann distribution)




Number density n (cm−3) =∫

f d3xd3v





f d3xd3v

The average velocity (if any) would be 〈v〉 =∫

v f d3xd3v





f d3xd3v


v f d3xd3v

The average kinetic energy would be〈(1/2)mv2〉 =

∫

(1/2)mv2 f d3xd3v . For a thermaldistribution, (i.e., a Maxwell-Boltzmann f ) we know that thisis equal to (3/2)kT





f d3xd3v


v f d3xd3v

The average kinetic energy would be〈(1/2)mv2〉 =

∫

(1/2)mv2 f d3xd3v . For a thermaldistribution, (i.e., a Maxwell-Boltzmann f ) we know that thisis equal to (3/2)kT

The pressure of the gas is thus a macroscopic (fluid) concept,representing a statistical average of the force per unit areadue to molecules striking the walls of a container.


Solids vs fluids

Fluids are characterized mainly by their response to shear forces;they cannot resist shear; they deform/flow continuously forarbitrarily small shear. Solids, however, can be elastic.Complications: viscoelesticity, etc.


How do fluids respond to shear?

They flow. Viscosity tends to reduce the velocity gradient (thesolid line velocity profile tends towards the dashed line)


Stress

Given an area element dA, one can define a

normal stress τn ≡ dFn/dA (the scalar pressure we are familiarwith) and a


Stress

Given an area element dA, one can define a

normal stress τn ≡ dFn/dA (the scalar pressure we are familiarwith) and a

shear stress τs ≡ dFs/dA


Shear stress, viscosity

For a wide class of fluids called Newtonian fluids, the shear stressis experimentally observed to be linearly proportional to thevelocity gradient:

τs = µdu

dy

where µ (g cm−1 s−1) is called the coefficient of dynamic viscosity.In fact, the components of stress form a tensor


Fluids: kinematics

A study of the appearance of motion (e.g., displacement, velocity,acceleration, rotation) of fluid elements without explicitlyconsidering the forces acting on them. There are two alternativedescriptions possible:


Fluids: kinematics


The Eulerian description: fluid flow as seen by an observer inthe lab frame (the “field” picture): A fluid variable F (likedensity, velocity) is expressed in terms of (lab) position x andtime t: F (x, t)


Fluids: kinematics



The Lagrangian description: fluid flow as seen by an observersitting on a fluid parcel (the “particle” picture): F is afunction of the position of the fluid parcel x0 at a referencetime t = t0 and time t: F (x0, t)


Fluids: kinematics



The Lagrangian description: fluid flow as seen by an observersitting on a fluid parcel (the “particle” picture): F is afunction of the position of the fluid parcel x0 at a referencetime t = t0 and time t: F (x0, t)

the two pictures are related via a (typically Galilean) frametransformation involving the bulk fluid velocity


The “material/substantive/particle” derivative

What is the rate of change of F experienced by a Lagrangian

observer (i.e.; one sitting on a fluid parcel) expressed in regular lab(Eulerian) variables?





dF =∂F

∂tdt +

∂F

∂xidxi

: ∂/∂t gives the local (time) rate of change at a point x





dF =∂F

∂tdt +

∂F

∂xidxi


dF

dt=∂F

∂t+ ui

∂F

∂xi





dF =∂F

∂tdt +

∂F

∂xidxi


dF

dt=∂F

∂t+ ui

∂F

∂xi=∂F

∂t+ u .∇F





dF =∂F

∂tdt +

∂F

∂xidxi


dF

dt=∂F

∂t+ ui

∂F

∂xi=∂F

∂t+ u .∇F

dF/dt (often written as DF/Dt is the material derivative; thetotal rate of change in quantity F ) felt by a Lagrangianobserver


Streamlines

If |u| = q, and we define a “streamline coordinate” s that pointsalong the local direction of u, the material derivative dF/dt can bewritten as


Streamlines


dF

dt=∂F

∂t+ u .∇F


Streamlines


dF

dt=∂F

∂t+ u .∇F =

∂F

∂t+ q

∂F

∂s

Alternatively, if ds = (dx , dy , dz) and u = (ux , uy , uz), astreamline can be defined by


Streamlines


dF

dt=∂F

∂t+ u .∇F =

∂F

∂t+ q

∂F

∂s


dx

ux=

dy

uy=

dz

uz


Streamlines


dF

dt=∂F

∂t+ u .∇F =

∂F

∂t+ q

∂F

∂s


dx

ux=

dy

uy=

dz

uz

Equivalently, ds× u = 0.


Laminar flow around a sphere: streamlines


Turbulent flow around a sphere: streamlines


Streamlines, pathlines and streaklines

Pathline: trajectory of a fluid parcel of fixed identity




Streakline: Current location of all fluid parcels that havepassed through a fixed spatial point (e.g., injectdye/smoke/some kind of tracer)




Streakline: Current location of all fluid parcels that havepassed through a fixed spatial point (e.g., injectdye/smoke/some kind of tracer)

Streamline, pathline, streakline all equivalent for steady flow(what precisely does “steady” mean?)


The streamfunction

For 2-dimensional flows (i.e., no z-dependence), its often useful todefine a streamfunction ψ:


The streamfunction


u = −∇× [zψ(x , y)]


The streamfunction


u = −∇× [zψ(x , y)]

Using this definition, and the definition of streamlines, one canshow that (work it out!)


The streamfunction


u = −∇× [zψ(x , y)]


δψ =∂ψ

∂xdx +

∂ψ

∂ydy


The streamfunction


u = −∇× [zψ(x , y)]


δψ =∂ψ

∂xdx +

∂ψ

∂ydy = 0 ,

where dx and dy are along a streamline.


The streamfunction


u = −∇× [zψ(x , y)]


δψ =∂ψ

∂xdx +

∂ψ

∂ydy = 0 ,

where dx and dy are along a streamline. In other words thestreamfunction remains constant along a streamline.


Vorticity

The vorticity ω of a flow is defined as

ω = ∇× u


Vorticity


ω = ∇× u

If the vorticity of a flow equals zero, the flow is irrotational


Vorticity


ω = ∇× u


For irrotational flows, one can define a potential φ, such thatu = −∇φ (why?)


Vorticity


ω = ∇× u



and then ∇φ .∇ψ = 0


Vorticity


ω = ∇× u



and then ∇φ .∇ψ = 0

Why, and can you think of parallels in electrostatics? What

about the Cauchy-Riemann conditions in complex algebra?


Mass conservation: the equation of continuity

First, define the mass flux across a bounding surface: theamount of mass per unit area per unit time flowing out of (orinto) the surface:




Mass flux = ρu, where ρ is the matter density and u is theflow velocity





The mass contained in a volume∫

ρ dV can change only dueto mass flux through the bounding surface (mass can’t becreated/destroyed within the volume);






ρ dV can change only dueto mass flux through the bounding surface (mass can’t becreated/destroyed within the volume); in other words,






ρ dV can change only dueto mass flux through the bounding surface (mass can’t becreated/destroyed within the volume); in other words,

∂

∂t

∫

ρ dV = −∫

ρu . dS


Equation of continuity (Mass conservation)

∂

∂t

∫

ρ dV = −∫

ρu . dS

Use Gauss’s law on RHS to get



∂

∂t

∫

ρ dV = −∫

ρu . dS

Use Gauss’s law on RHS to get∫[

∂ρ

∂t+∇ . (ρu)

]

dV = 0



∂

∂t

∫

ρ dV = −∫

ρu . dS


∂ρ

∂t+∇ . (ρu)

]

dV = 0

Since this is true for an arbitrary volume,

∂ρ

∂t+∇ . (ρu) = 0



∂

∂t

∫

ρ dV = −∫

ρu . dS


∂ρ

∂t+∇ . (ρu)

]

dV = 0


∂ρ

∂t+∇ . (ρu) = 0

This is the conservative form of the mass continuity equation



∂

∂t

∫

ρ dV = −∫

ρu . dS


∂ρ

∂t+∇ . (ρu)

]

dV = 0


∂ρ

∂t+∇ . (ρu) = 0

This is the conservative form of the mass continuity equationIn general, the conservative form of any quantity goes as



∂

∂t

∫

ρ dV = −∫

ρu . dS


∂ρ

∂t+∇ . (ρu)

]

dV = 0


∂ρ

∂t+∇ . (ρu) = 0

This is the conservative form of the mass continuity equationIn general, the conservative form of any quantity goes as(Partial) time derivative of quantity + divergence of flux ofthat quantity = 0



∂

∂t

∫

ρ dV = −∫

ρu . dS


∂ρ

∂t+∇ . (ρu)

]

dV = 0


∂ρ

∂t+∇ . (ρu) = 0

This is the conservative form of the mass continuity equationIn general, the conservative form of any quantity goes as(Partial) time derivative of quantity + divergence of flux ofthat quantity = 0Alternatively, using the Lagrangian derivative d/dt, (show!)

dρ

dt+ ρ∇ .u = 0


Incompressibility

The mass continuity equation can be rewritten as

1

ρ

dρ

dt+∇ .u = 0


Incompressibility


1

ρ

dρ

dt+∇ .u = 0

One frequently encounters situations where (to a fairapproximation) the first term is negligible;


Incompressibility


1

ρ

dρ

dt+∇ .u = 0

One frequently encounters situations where (to a fairapproximation) the first term is negligible; i.e.,

1

ρ

dρ

dt≈ 0


Incompressibility


1

ρ

dρ

dt+∇ .u = 0


1

ρ

dρ

dt≈ 0

From the continuity equation, the condition forincompressibility then becomes

∇ .u = 0


Incompressibility


1

ρ

dρ

dt+∇ .u = 0


1

ρ

dρ

dt≈ 0

From the continuity equation, the condition forincompressibility then becomes

∇ .u = 0

Applicable when flow speeds ≪ sound speed


Incompressibility...some more

In the Boussinesq approximation, density variations in thefluid can be neglected everywhere except where the density ismultiplied by g




Consider a (gravitationally) vertically stratified atmosphere. Ithas a vertical scale height L ≈ c2s /g ; this is the e-folding scaleheight for the density (show that this is so, at least for

isothermal flows)





isothermal flows)

For the Boussinesq approximation to be valid, the scale of theflow should be ≪ L





isothermal flows)

For the Boussinesq approximation to be valid, the scale of theflow should be ≪ L

The Boussinseq approximation often just reduces to ∇ .u = 0,but not always


Momentum conservation

Newton’s second law of motion: F = m a




The RHS (m a) is simply the rate of change of momentum





The momentum can change because of “intrinsic” (∂/∂t)change inside the volume:

∫

∂(ρu)

∂tdV





The momentum can change because of “intrinsic” (∂/∂t)change inside the volume:

∫

∂(ρu)

∂tdV

..and also due to the flux of momentum through the boundingsurface

∫

(ρu)u . dA


Momentum conservation - momentum flux

Using Gauss’s divergence theorem (for tensors, since uu is asecond rank tensor), the surface integral

∫

(ρu)u . dS

becomes




∫

(ρu)u . dS

becomes

∫

∇ .(ρuu) dV




∫

(ρu)u . dS

becomes

∫

∇ .(ρuu) dV

So that the rate of change of momentum is




∫

(ρu)u . dS

becomes

∫

∇ .(ρuu) dV

So that the rate of change of momentum is

∫[

∂(ρu)

∂t+∇ .(ρuu)

]

dV


Momentum conservation - forces

The rate of change of momentum is equal to the force, ofcourse



The rate of change of momentum is equal to the force, ofcourseThe force on the fluid element too can have a volume

component (due to body forces),∫

ρg dV





ρg dV..note, g represents the acceleration due to any kind of bodyforce (like gravity, for instance)





ρg dV..note, g represents the acceleration due to any kind of bodyforce (like gravity, for instance)Surface forces are a little trickier. Recall, one can have forceson a surface area element that are normal to it, as well astangential to it:


Momentum conservation - the pressure tensor

The ijth component of the pressure tensor represents the forcein the ith direction felt by an area element whose outwardnormal points in the jth direction:

(d Fsurface)i = −Pij d Aj

So that





So that

(Fsurface)i = −∫

Pij d Aj





So that


Pij d Aj = −∫

∂Pij

∂xjdV

So the total force on a fluid element is∫

ρg dV −∫

p dA =

∫[

ρg −∇ .P

]

dV





So that


Pij d Aj = −∫

∂Pij

∂xjdV

So the total force on a fluid element is∫

ρg dV −∫

p dA =

∫[

ρg −∇ .P

]

dV

where

(∇ .P)i =∂Pij

∂xj


Momentum conservation - putting it together

Taken together, the momentum conservation equation is




∂

∂t(ρu) +∇ . (ρuu) = −∇ .P+ ρ g




∂

∂t(ρu) +∇ . (ρuu) = −∇ .P+ ρ g

Note, uu is a second order tensor, in the sense of

a b =

a1 b1 a1 b2 a1 b3a2 b1 a2 b2 a2 b3a3 b1 a3 b2 a3 b3

.


Momentum conservation: “conservative” form

Neglect body forces (g)




The “conservative” form of the momentum conservationequation is

∂

∂t(ρu) +∇ . (ρuu+ P) = 0




The “conservative” form of the momentum conservationequation is

∂

∂t(ρu) +∇ . (ρuu+ P) = 0

which is of the form: partial time derivative of quantity +

divergence of flux of that quantity = 0


Momentum conservation for inviscid fluids: The Eulerequation

For an inviscid fluid, its enough to consider the scalar pressure

Pij = p δij




Pij = p δij

; i.e., only the diagonal elements of the pressure tensor arenon-zero




Pij = p δij


Furthermore, we expand

∂

∂t(ρu) +∇ . (ρuu) = 0




Pij = p δij



∂

∂t(ρu) +∇ . (ρuu) = 0

and use the mass continuity equation




Pij = p δij



∂

∂t(ρu) +∇ . (ρuu) = 0


∂ρ

∂t+∇ . (ρu) = 0




Pij = p δij



∂

∂t(ρu) +∇ . (ρuu) = 0


∂ρ

∂t+∇ . (ρu) = 0

to get the Euler equation (work it out!)

ρ∂u

∂t+ ρ (u .∇)u = −∇ p + ρ g


Some alternatives

If this treatment doesn’t appeal to you, consider following thethe Lagrangian treatment (sit on top of a fluid parcel andwork out momentum conservation); e.g., Arnab RaiChoudhuri’s book


Some alternatives


Keep in mind, the Euler equation

ρ∂u

∂t+ ρ (u .∇)u = −∇ p + ρ g

is valid only for inviscid fluids (ones for which Pij = p δij).


Some alternatives


Keep in mind, the Euler equation

ρ∂u

∂t+ ρ (u .∇)u = −∇ p + ρ g

is valid only for inviscid fluids (ones for which Pij = p δij).

Else, there are additional terms in the pressure tensor(involving viscosity), and we get the Navier-Stokes equation


Before we move onto Navier-Stokes..I

Lets consider flows that are incompressible (∇ .u = 0) andirrotational (ω = ∇ × u = 0)




By virtue of irrotationality, we can define a (scalar) velocitypotential φ defined by





u = ∇φ





u = ∇φ

This also has to with inviscid flows, as we’ll see later.





u = ∇φ


By virtue of incompressibility, the velocity potential satsifiesLaplace’s equation

∇2 φ = 0





u = ∇φ



∇2 φ = 0

and we can use the solutions familiar to us from electrostatics





u = ∇φ



∇2 φ = 0

and we can use the solutions familiar to us from electrostatics

Recall, the scalar potential has to be featureless


Before we move onto Navier-Stokes..II

By virtue of incompressibility, we can define a (scalar)streamfunction ψ



By virtue of incompressibility, we can define a (scalar)streamfunction ψ (for 2D velocity fields, at any rate)



By virtue of incompressibility, we can define a (scalar)streamfunction ψ (for 2D velocity fields, at any rate) ..andrecall, streamlines are lines along which the streamfunction isconserved




u(x , y) = −∇× (zψ)




u(x , y) = −∇× (zψ)

Taken together, ∇φ .∇ψ = 0; i.e., equipotential lines andstreamlines are orthogonal.


Application of complex variables - I

For 2D flows, from incompressibility (∇ .u = 0) we get




∂ux∂x

+∂uy∂y

= 0




∂ux∂x

+∂uy∂y

= 0

..and this can be related to the stream function as




∂ux∂x

+∂uy∂y

= 0


ux ≡ ∂ψ

∂yuy ≡ ∂ψ

∂x




∂ux∂x

+∂uy∂y

= 0


ux ≡ ∂ψ

∂yuy ≡ ∂ψ

∂x

From irrotationality we get




∂ux∂x

+∂uy∂y

= 0


ux ≡ ∂ψ

∂yuy ≡ ∂ψ

∂x


ux ≡ −∂φ∂x

uy ≡ −∂φ∂y




∂ux∂x

+∂uy∂y

= 0


ux ≡ ∂ψ

∂yuy ≡ ∂ψ

∂x


ux ≡ −∂φ∂x

uy ≡ −∂φ∂y

Which of course gives the Cauchy-Riemann (like) condition




∂ux∂x

+∂uy∂y

= 0


ux ≡ ∂ψ

∂yuy ≡ ∂ψ

∂x


ux ≡ −∂φ∂x

uy ≡ −∂φ∂y

Which of course gives the Cauchy-Riemann (like) condition

∂φ

∂x=∂ψ

∂yand

∂φ

∂y= −∂ψ

∂x


Application of complex variables - II

Its as though ux and uy are components of a single complexvariable;



Its as though ux and uy are components of a single complexvariable; one can therefore apply powerful techniques fromcomplex analysis to solve for the flow field




Note, you can also easily verify ∇ψ .∇φ = 0 from theCauchy-Riemann condition, and





Also show that ∇2 ψ (in addition to ∇2 φ) = 0






So both the streamfunction ψ and the velocity potential φ aresolutions to Laplace’s equation






So both the streamfunction ψ and the velocity potential φ aresolutions to Laplace’s equation

In practice, though, one usually identifies an already knownsolution to the Laplace’s equation and then hunts for aphysical situation where its applicable!


Inviscid, incompressible, irrotational flow around a smoothsphere

The general solution for ∇2 φ = 0 in 2D polar coordinates is




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)

(Cn cos n θ+Dn sin n θ)




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)


For a smooth sphere of radius a, the boundary conditions are

∂φ

∂r= 0 at r = a




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a

i.e., normal component of velocity vanishes on the surface of thesphere (tangential slip allowed);




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a

i.e., normal component of velocity vanishes on the surface of thesphere (tangential slip allowed); how do you think the boundary

condition at the sphere’s surface would look like for a viscous fluid?




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a



and




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a



andu∞ = −U x




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a



andu∞ = −U x or φ = U r cos θ




φ = (A0+B0 lnr) (C0+D0 θ)+Σ

(

An rn+

Bn

rn

)



∂φ

∂r= 0 at r = a



andu∞ = −U x or φ = U r cos θ

i.e., the velocity is “unchanged” at large distances


..and the particular solution is..

φ = U cos θ

(

r +a2

r

)



φ = U cos θ

(

r +a2

r

)

This incorporates both the boundary conditions mentioned earlier.



φ = U cos θ

(

r +a2

r

)

This incorporates both the boundary conditions mentioned earlier.The velocity field is (from u = −∇φ)

u = −U x+ Ua2

r2(cos θ r + sin θ θ)


Laminar flow around a sphere: streamlines


Laplace’s eq: numerical solution

Finite-difference representation of ∇2φ = 0 (assuming equal stepsin x and y)


Laplace’s eq: numerical solution

Finite-difference representation of ∇2φ = 0 (assuming equal stepsin x and y)

φi , j =1

4[φi−1 , j + φi+1 , j + φi , j−1 + φi , j+1]


The Bernoulli “constant” - I

Start with the Euler equation

∂u

∂t+ (u .∇)u = −1

ρ∇ p + g ,




∂u

∂t+ (u .∇)u = −1

ρ∇ p + g ,

characterize the body acceleration g using a potential g = −∇Φ,and




∂u

∂t+ (u .∇)u = −1

ρ∇ p + g ,

characterize the body acceleration g using a potential g = −∇Φ,and recognize that (u .∇)u = 1/2∇(u .u)− u× (∇× u)




∂u

∂t+ (u .∇)u = −1

ρ∇ p + g ,

characterize the body acceleration g using a potential g = −∇Φ,and recognize that (u .∇)u = 1/2∇(u .u)− u× (∇× u) to write(for steady flows)

∇(1

2u2)− u× (∇× u) = −1

ρ∇ p −∇Φ




∂u

∂t+ (u .∇)u = −1

ρ∇ p + g ,

characterize the body acceleration g using a potential g = −∇Φ,and recognize that (u .∇)u = 1/2∇(u .u)− u× (∇× u) to write(for steady flows)

∇(1

2u2)− u× (∇× u) = −1

ρ∇ p −∇Φ

Lets us now (line) integrate this whole equation along astreamline element d l


The Bernoulli “constant” - II

Since d l is along a streamline, (i.e., along u) d l is always ⊥u× (∇× u)




∫

d l .

[

∇(1

2u2)− u× (∇× u) +

1

ρ∇ p +∇Φ

]

yields




∫

d l .

[

∇(1

2u2)− u× (∇× u) +

1

ρ∇ p +∇Φ

]

yields

1

2u2 +

∫

d p

ρ+Φ = Constant




∫

d l .

[

∇(1

2u2)− u× (∇× u) +

1

ρ∇ p +∇Φ

]

yields

1

2u2 +

∫

d p

ρ+Φ = Constant

This is essentially a statement of energy conservation along astreamline




∫

d l .

[

∇(1

2u2)− u× (∇× u) +

1

ρ∇ p +∇Φ

]

yields

1

2u2 +

∫

d p

ρ+Φ = Constant

This is essentially a statement of energy conservation along astreamline (no surprise, since we integrated force along a lineelement)




∫

d l .

[

∇(1

2u2)− u× (∇× u) +

1

ρ∇ p +∇Φ

]

yields

1

2u2 +

∫

d p

ρ+Φ = Constant

This is essentially a statement of energy conservation along astreamline (no surprise, since we integrated force along a lineelement)

The Bernoulli constant is yet another label for a streamline


Application of Bernoulli constant: flow from an orifice

Speed of water out of a hole in a water tank at depth h:uout =

√2 g h


Application of Bernoulli constant: flow over airfoil

Pbelow > Pabove, hence airfoil experiences a lift


Vorticity: some applications

Recap: Euler equation

∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,




∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,

and use(u .∇)u = 1/2∇(u .u)− u× (∇× u)

to write it as




∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,

and use(u .∇)u = 1/2∇(u .u)− u× (∇× u)

to write it as

∂u

∂t+ 1/2∇(u .u)− u× (∇× u) = −1

ρ∇ p −∇Φ




∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,

and use(u .∇)u = 1/2∇(u .u)− u× (∇× u)

to write it as

∂u

∂t+ 1/2∇(u .u)− u× (∇× u) = −1

ρ∇ p −∇Φ

Take the curl of the whole equation to get the evolution ofvorticity ω = ∇× u:




∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,

and use(u .∇)u = 1/2∇(u .u)− u× (∇× u)

to write it as

∂u

∂t+ 1/2∇(u .u)− u× (∇× u) = −1

ρ∇ p −∇Φ


∂ ω

∂t= ∇× (u× ω) +

1

ρ2∇ ρ×∇ p




∂u

∂t+ (u .∇)u = −1

ρ∇ p −∇Φ ,

and use(u .∇)u = 1/2∇(u .u)− u× (∇× u)

to write it as

∂u

∂t+ 1/2∇(u .u)− u× (∇× u) = −1

ρ∇ p −∇Φ


∂ ω

∂t= ∇× (u× ω) +

1

ρ2∇ ρ×∇ p

show!Subramanian Fluid Dynamics

Vorticity equation for incompressible, barotropic fluids

If p = f (ρ), (barotropic fluid) then the vorticity equation becomessimpler (because ∇ ρ and ∇ p are parallel):




∂ ω

∂t= ∇× (u× ω)




∂ ω

∂t= ∇× (u× ω)

In principle, this is a dynamical equation for vorticity, sinceω = ∇× u




∂ ω

∂t= ∇× (u× ω)

In principle, this is a dynamical equation for vorticity, sinceω = ∇× u ..and specifying both the curl and divergence(∇ .u = 0) of a vector field specifies it uniquely, (caveats?) so wehave the entire velocity dynamics specified.


Kelvin’s vorticity theorem

Define the circulation K =∮

u . d l =∫

(∇× u) . d A




u . d l =∫

(∇× u) . d A

Write the Euler equation using the material derivative:




u . d l =∫

(∇× u) . d A


d u

dt=

1

ρ∇p +∇Φ




u . d l =∫

(∇× u) . d A


d u

dt=

1

ρ∇p +∇Φ

From the definition of K , we can write its material derivativeas

d K

dt=

∮

d u

dt. d l+

∮

u .d

dtd l


When is the circulation conserved?

Using the Euler equation,

d K

dt=

∮

∇Φ . d l+

∮

dp

ρ+

∮

u .d l

dt




d K

dt=

∮

∇Φ . d l+

∮

dp

ρ+

∮

u .d l

dt

Each term is a perfect differential (show!), so the integral overa closed path is zero.




d K

dt=

∮

∇Φ . d l+

∮

dp

ρ+

∮

u .d l

dt

Each term is a perfect differential (show!), so the integral overa closed path is zero.

So circulation is conserved in an inviscid, barotropic fluid. Ifcirculation = 0 to begin with, it will always remain that way.


Irrotational flow around a cylinder


Irrotational flow + circulation around a cylinder: themagnus effect



Pbelow > Pabove:



Pbelow > Pabove: so spinning ball experiences a “lift” (why?).



Pbelow > Pabove: so spinning ball experiences a “lift” (why?). Thedirection of the force is ⊥ the rotation axis as well as the (local)flow. Remind you of Lorentz forces?


Viscosity


Viscosity

Consider viscous flow between two (unbounded) parallel plates.


Viscosity

Consider viscous flow between two (unbounded) parallel plates.Due to viscosity, the flow doesn’t slip at the boundaries;


Viscosity

Consider viscous flow between two (unbounded) parallel plates.Due to viscosity, the flow doesn’t slip at the boundaries; it sticks.


Momentum equation with viscosity



Fluid flows along the x-direction due to the pressure gradientdp/dx




But there is also a “frictional” opposing force (in steadystate), else the fluid would accelerate





Recall, for Newtonian fluids, the viscous stress is proportionalto the velocity gradient (strain): τ = µ du/dy






So, per unit length in the z-direction (i.e., per dz), the forcebalance reads







[

µ

(

du

dy

)

y+dy

− µ

(

du

dy

)

y

]

=







[

µ

(

du

dy

)

y+dy

− µ

(

du

dy

)

y

]

= µd2u

dy2dy =







[

µ

(

du

dy

)

y+dy

− µ

(

du

dy

)

y

]

= µd2u

dy2dy =

dp

dxdy


The Navier-Stokes equation

So

µd2u

dy2=

dp

dx



So

µd2u

dy2=

dp

dx

More generally,



So

µd2u

dy2=

dp

dx

More generally,

µ∇2u = ∇ p



So

µd2u

dy2=

dp

dx

More generally,

µ∇2u = ∇ p

With this additional term due to viscous stresses, the fullmomentum equation reads



So

µd2u

dy2=

dp

dx

More generally,

µ∇2u = ∇ p

With this additional term due to viscous stresses, the fullmomentum equation reads (Euler equation plus viscous stressterm)



So

µd2u

dy2=

dp

dx

More generally,

µ∇2u = ∇ p


ρ∂u

∂t+ ρ (u .∇)u = −∇ p + µ∇2u+ ρ g



So

µd2u

dy2=

dp

dx

More generally,

µ∇2u = ∇ p


ρ∂u

∂t+ ρ (u .∇)u = −∇ p + µ∇2u+ ρ g

This is the Navier-Stokes equation.


Navier-Stokes ...a little more formally

Remember, we can have normal, as well as tangential stresses oneach face in a fluid volume element:




...so instead of Pij = p δij , which is okay for inviscid fluids, we’dhave Pij = p δij + σij for viscous fluids.




...so instead of Pij = p δij , which is okay for inviscid fluids, we’dhave Pij = p δij + σij for viscous fluids. As always, Pij is the forcein the ith direction on the face whose outward normal is in the jthdirection.




...so instead of Pij = p δij , which is okay for inviscid fluids, we’dhave Pij = p δij + σij for viscous fluids. As always, Pij is the forcein the ith direction on the face whose outward normal is in the jthdirection. p is still the thermodynamic pressure, but there are somecaveats


So what would σij look like?



Recall, for Newtonian fluids,

τ = µdu

dy



Recall, for Newtonian fluids,

τ = µdu

dy

..so σij will involve velocity derivatives like dui/dxj


..σij ..some more..

∂ui∂xj

=1

2

(

∂ui∂xj

+∂uj∂xi

)

+1

2

(

∂ui∂xj

− ∂uj∂xi

)



∂ui∂xj

=1

2

(

∂ui∂xj

+∂uj∂xi

)

+1

2

(

∂ui∂xj

− ∂uj∂xi

)

The second term represents rigid body rotation (note, u = Ω× ris rigid body rotation) and therefore doesn’t involve shear stresses(show)



∂ui∂xj

=1

2

(

∂ui∂xj

+∂uj∂xi

)

+1

2

(

∂ui∂xj

− ∂uj∂xi

)

The second term represents rigid body rotation (note, u = Ω× ris rigid body rotation) and therefore doesn’t involve shear stresses(show) If we insist on Newtonian fluids, the most general secondrank tensor involving velocity gradients is of the form

σij = a

(

∂ui∂xj

+∂uj∂xi

)

+ b δij ∇ .u


...furthermore..

Recall Pij = pδij + σij .


...furthermore..

Recall Pij = pδij + σij . If we take p to be the (scalar)thermodynamic pressure, it contains the trace of Pij ; i.e.,


...furthermore..


p =1

3Pii


...furthermore..


p =1

3Pii

...in which case σij had better be traceless.


...furthermore..


p =1

3Pii

...in which case σij had better be traceless. The only way this canhappen is if b = −(2/3)a, by which

σij = −µ(

∂ui∂xj

+∂uj∂xi

− 2

3δij∇ .u

)


..putting it all together

ρdui

dt= ρgi −

∂p

∂xi+

∂

∂xj

[

µ

(

∂ui∂xj

+∂uj∂xi

− 2

3δij ∇ .u

)]



ρdui

dt= ρgi −

∂p

∂xi+

∂

∂xj

[

µ

(

∂ui∂xj

+∂uj∂xi

− 2

3δij ∇ .u

)]

..if µ is isotropic, it can be considered to be a scalar, and we get



ρdui

dt= ρgi −

∂p

∂xi+

∂

∂xj

[

µ

(

∂ui∂xj

+∂uj∂xi

− 2

3δij ∇ .u

)]


ρdu

dt= ρg −∇p + µ

[

∇2u+1

3∇(∇ .u)

]



ρdui

dt= ρgi −

∂p

∂xi+

∂

∂xj

[

µ

(

∂ui∂xj

+∂uj∂xi

− 2

3δij ∇ .u

)]


ρdu


[

∇2u+1

3∇(∇ .u)

]

..and for incompressible flows we recover



ρdui

dt= ρgi −

∂p

∂xi+

∂

∂xj

[

µ

(

∂ui∂xj

+∂uj∂xi

− 2

3δij ∇ .u

)]


ρdu


[

∇2u+1

3∇(∇ .u)

]

..and for incompressible flows we recover

du

dt= g − 1

ρ∇p +

µ

ρ∇2u


Finally, the energy equation

We have already seen energy conservation in the guise of theBernoulli constant, but now lets include viscosity (we followLandau & Lifshitz here)




Recall, Pij = pδij + σij , and for incompressible fluids, theviscous stress tensor is





σij = −µ(

∂ui∂xj

+∂uj∂xi

)





σij = −µ(

∂ui∂xj

+∂uj∂xi

)

The Navier-Stokes equation can be equivalently be written as





σij = −µ(

∂ui∂xj

+∂uj∂xi

)

The Navier-Stokes equation can be equivalently be written as

∂ui∂t

= −uk∂ui∂xk

− 1

ρ

∂p

∂xi+

1

ρ

∂σik∂xk


Energy equation ..cont’d

We’re interested in knowing how the kinetic energy 1/2ρu2 of aparcel of fluid evolves with time.



We’re interested in knowing how the kinetic energy 1/2ρu2 of aparcel of fluid evolves with time. In other words, evaluate




∂(1/2ρu2)

∂t= ρui

∂ui∂t




∂(1/2ρu2)

∂t= ρui

∂ui∂t

Using the form of Navier-Stokes




∂(1/2ρu2)

∂t= ρui

∂ui∂t


∂ui∂t

= −uk∂ui∂xk

− 1

ρ

∂p

∂xi+

1

ρ

∂σik∂xk




∂(1/2ρu2)

∂t= ρui

∂ui∂t


∂ui∂t

= −uk∂ui∂xk

− 1

ρ

∂p

∂xi+

1

ρ

∂σik∂xk

we get

∂(1/2ρu2)

∂t= −ρu . (u .∇)u− u .∇p + ui

∂σik∂xk




∂(1/2ρu2)

∂t= ρui

∂ui∂t


∂ui∂t

= −uk∂ui∂xk

− 1

ρ

∂p

∂xi+

1

ρ

∂σik∂xk

we get

∂(1/2ρu2)

∂t= −ρu . (u .∇)u− u .∇p + ui

∂σik∂xk

=

−ρ(u .∇)

(

1

2u2 +

p

ρ

)

+ div(u . σ)− σik∂ui∂xk


Energy equation..cont’d

Since we’re considering incompressible fluids (∇ .u = 0),




∂(1/2ρu2)

∂t= −∇ .

[

ρu

(

1

2u2 +

p

ρ

)

− u . σ

]

− σik∂ui∂xk




∂(1/2ρu2)

∂t= −∇ .

[

ρu

(

1

2u2 +

p

ρ

)

− u . σ

]

− σik∂ui∂xk

Integrating over a macroscopic volume (and bounding area),




∂(1/2ρu2)

∂t= −∇ .

[

ρu

(

1

2u2 +

p

ρ

)

− u . σ

]

− σik∂ui∂xk


∫

∂

∂t(1/2ρu2) dV = −

∮[

ρu

(

1

2u2+

p

ρ

)

−u . σ

]

. dA−∫

σik∂ui∂xk

dV




∂(1/2ρu2)

∂t= −∇ .

[

ρu

(

1

2u2 +

p

ρ

)

− u . σ

]

− σik∂ui∂xk


∫

∂

∂t(1/2ρu2) dV = −

∮[

ρu

(

1

2u2+

p

ρ

)

−u . σ

]

. dA−∫

σik∂ui∂xk

dV

The first term on the RHS vanishes (why?)


..leaving, finally

Ekin = −∫

σik∂ui∂xk

dV =


..leaving, finally

Ekin = −∫

σik∂ui∂xk

dV = − 1

2

∫

σik

(

∂ui∂xk

+∂uk∂xi

)

dV


..leaving, finally

Ekin = −∫

σik∂ui∂xk

dV = − 1

2

∫

σik

(

∂ui∂xk

+∂uk∂xi

)

dV

and substituting for σik ,

σik = −µ(

∂ui∂xk

+∂uk∂xi

)


..leaving, finally

Ekin = −∫

σik∂ui∂xk

dV = − 1

2

∫

σik

(

∂ui∂xk

+∂uk∂xi

)

dV

and substituting for σik ,

σik = −µ(

∂ui∂xk

+∂uk∂xi

)

we get

Ekin = −1

2µ

∫(

∂ui∂xk

+∂uk∂xi

)2

dV


Phy 352: Fluid Dynamics, Spring 2013

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Transcript of Phy 352: Fluid Dynamics, Spring 2013