22.581 Module 7: Stress, Viscosity, and The Navier-Stokes ...faculty.uml.edu/dwillis/MECH5810/Lecture/2016_22_581_Module7.pdf22.581 Module 7: Stress, Viscosity, and The Navier-Stokes

22.581 Module 7: Stress, Viscosity, and TheNavier-Stokes Equations

D.J. Willis

Department of Mechanical EngineeringUniversity of Massachusetts, Lowell

22.581 Advanced Fluid DynamicsFall 2016

Tuesday/Thursday 8:00am - 9:15am

1 / 58

Outline

1 Administrative & Schedule Issues2 Review of (inviscid) fluids covered thus far

Differential conservation laws3 Review and Formalization of Flow Kinematics

Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor

Vorticity4 Stresses in a continuum

Introduction to Stress

Equation of motion with full stress tensor

Constitutive Equations

Putting it all together: General Viscous Equation of Motion5 Non-Dimensional Navier Stokes

Introduction2 / 58

Administrative & Schedule Issues

Midterm Exam: Not graded yet....probably Thursday/next week?

Schedule

Today : Introduce Navier-Stokes Equations

3 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction4 / 58

Conservation of mass and momentum

Conservation of mass statement:∂

∂tρ+∇ · (ρ~u) = 0

Incompressible conservation of mass:

∇ · (~u) = 0

We will see that if we introduce a scalar potential Φ where

∇Φ = ~u, that the incompressible conservation of mass reduces to:

∇2 (Φ) = 0

Which leads to irrotational/potential flow theory (a few weeks

time). 5 / 58


Conservation of momentum in a non-accelerating reference frame:

ρ~a = ρD~uDt

= −∇p + ρ~g +~Fvisc

Vol︸︷︷︸=? Current Module

If we neglect viscous effects, this leads to different manifestations

of the Bernoulli equation in non-accelerating reference frames:

s− direction : p +12ρu2 + ρgh = Const.

n− direction :∂

∂n(p + ρgz) =

ρu2

R

unsteady Bernoulli :

∫ 2

1ρ∂~u∂t· d~s + p1 +

12ρu2

1 + ρgh1

= p2 +12ρu2

2 + ρgh2

6 / 58

ASIDE: Conservation of momentum in anaccelerating reference frame

Let’s assume we wish to apply conservation of momentum in a

general (accelerating) reference frame, then:

ρ~a = ρD~uXYZ

Dt= ρ~aref + ρ

D~uxyz

Dt= −∇p + ρ~g +

~Fvisc

Vol

Here ~uXYZ corresponds to the velocity as seen in the

non-accelerating reference frame, and ~uxyz corresponds to the

velocity measured in the accelerating reference frame.

7 / 58


In general, the accelerating reference frame may be accelerating

due to translation (linear acceleration) or rotation (centripetal,

angular and Coriolis accelerations):

ρ(~atrans + ~ω ×

(~ω × ~R

)+ 2~ω ×~urel + ~α× ~R

)+ ρ

D~uxyz

Dt=

−∇p + ρ~g +~Fvisc

Vol.

2-examples from previous year’s midterms (straw pump and the

cart problem). Try the straw pump bonus question, and also

determining an explicit relationship for the velocity at the exit of the

accelerating cart using this approach. Aditionally, try the

accelerating cart problem using unsteady Bernoulli.8 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction9 / 58

Kinematics: Linear Strain Rate – ε

Similar idea to solid mechanics - normalized change in length

The difference is strain rate – continues to deform

Linear strain rate: represents the normalized rate of change in

length of a fluid element with respect to time:

Strain Rate =1lo

D(l)Dt

=∂u∂x

So, how do we get ∂u∂x ?

10 / 58

Kinematics : Linear Strain Rate

Strain Rate =

(1lo

D(l)Dt

)'(

1lo

∆l∆t

)=

(1lo

(u + ∂u

∂x lo)

∆t − (u) ∆t∆t

)

Strain Rate = εxx =

(1

��lo

(�u + ∂u

∂x��lo)��∆t − (�u)��∆t

��∆t

)=

(∂u∂x

)

εxx, εyy, εzz: Linear strains in x, y, z

11 / 58

Kinematics : Volumetric/Bulk Strain Rate

Volumetric strain rate = Rate of change of the volume per unit

volume

The infinitesimal volume δV = δxδyδz.

Volumetric Strain Rate =1δV

DDt

(δV) =1

δxδyδzDDt

(δxδyδz)

=1δx

DDt

(δx) +1δy

DDt

(δy) +1δz

DDt

(δz)

Simplifying, we get:

Volumetric Strain Rate =∂u∂x

+∂v∂y

+∂w∂z

=∂ui

∂xi︸︷︷︸indicial notation

Which can be represented using the divergence operator:

Volumetric Strain Rate = ∇ ·~u =∂u∂x

+∂v∂y

+∂w∂z

εxx, εyy, εzz: Linear strains in x, y, z

12 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction13 / 58

Kinematics : Shear Strain Rate

Shear strain rate: Rate of decrease of the angle formed by two

mutually perpendicular lines on the element

The shear strain rate value depends on the orientation of the line

pair

εij =dα+ dβ

dt

14 / 58

Kinematics : Shear Strain RateLet’s look at dα first:

dα = atan(OA

) ' OA

' ∆llo

'∂u∂y · lo · dt

lo

' ∂u∂y

dt

15 / 58

Kinematics : Shear Strain Rate

So:dαdt

=∂u∂y

Similarly, it can be shown that:

dβdt

=∂v∂x

εxy =dα+ dβ

dt=∂u∂y

+∂v∂x

16 / 58

Vortex Example

17 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction18 / 58

Kinematics : Strain Rate Tensor

εij =12

(∂ui

∂xj+∂uj

∂xi

)

∂u∂x

12

(∂u∂y + ∂v

∂x

)12

(∂u∂z + ∂w

∂x

)12

(∂v∂x + ∂u

∂y

)∂v∂y

12

(∂v∂z + ∂w

∂y

)12

(∂w∂x + ∂u

∂z

) 12

(∂w∂y + ∂v

∂z

)∂w∂z

The diagonal is the normal strain rate terms (εii)

The off-diagonals are half the shear strain rate terms

εij = εji

19 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction20 / 58

Kinematics : Vorticity and Curl - curl(~u) = ∇× ~u

Rotation or angular velocity is an important component of

kinematics as we shall see

We define vorticity to be twice the angular velocity of the fluid

element

21 / 58

Kinematics : Vorticity and Curl - curl(~u) = ∇× ~u

The curl of a vector indicates whether there is rotationality in the vector

field.

curl(~u) = ∇×~u =

∂∂x∂∂y∂∂z

×

u

v

w

=

∂w∂y −

∂v∂z

∂u∂z −

∂w∂x

∂v∂x −

∂u∂y

=

ωx

ωy

ωz

= ~ω

Pure rotation of an element:

22 / 58

Kinematics : Summary of Possible Fluid Motions

Vorticity, Rotation Rate

Shear Strain Rate

Linear Strain Rate

Volumetric Strain Rate

23 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction24 / 58

Stresses in a continuum

Definition: Stress is force per unit area

Consider an arbitrarily oriented differential surface in a continuum:

25 / 58


In an inviscid fluid (what we have covered up until now), the stress

acting on the surface is the normal stress or pressure. This can be

integrated over the area to get forces.

ρD~uDt

= −∇p︸︷︷︸normal stress

+ρ~g

In general, the force due to integration of stress over area has

components in both the normal direction (due to pressure or

normal stress) and the tangential direction (due to shear

stress(es))

26 / 58


Consider a cartiesian fluid element (dimensions ∆x, ∆y, ∆z):

27 / 58


Convention for a positive stress:1 For normal stresses, the element is in tension when a positive

stress is applied.2 For shear stresses, both the normal and shear stress direction are

positive in a positive stress.

Each face will have a normal stress σii and a pair of shear

stresses τij.

For this derivation, the normal stress, σii, will include pressure and

any relevant viscous effects.

The shear stress direction on a given surface is determined as

follows: τij is the i−component of stress on a surface element with

a normal pointing in the j-direction.

28 / 58

The Stress Tensor

Therefore, for a cartesian differential element, the stress tensor

will have 9-entries (some of which will be identical).

Π =

σxx τxy τxz

τxy σyy τyz

τxz τyz σzz

Note: For all fluids we will consider, τij = τji

The above stress tensor is symmetric (hence only 6 independent

quantities at a given location in the fluid).

29 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction30 / 58

Viscous equation of motion

Starting point: The general form of the conservation of

momentum:

D~uDt︸︷︷︸

Lagrangian Acceleration

= ~F︸︷︷︸Volume/Body Forces

+ ~G︸︷︷︸Surface Forces

1 Volume/Body Forces: Such as gravity, magnetohydrodynamic(MHD)

2 Surface Forces: Stresses (shear and pressure).

31 / 58


Consider the stresses acting on a 3-D differential element (with

stress gradients in the flow field):

32 / 58


As an example, let us consider all forces acting on a fluid element

in the x-direction due to the shear and normal stresses:

~Gx ·∆x ·∆y ·∆z =

(σxx +

∂σxx

∂x·∆x− σxx

)·∆y∆z

+

(τxy +

∂τxy

∂y∆y− τxy

)·∆x∆z

+

(τxz +

∂τxz

∂z∆z− τxz

)·∆x∆y

Note: We multiply ~Gx ·∆x ·∆y ·∆z because, we are considering

the forces on the differential volume and ~G is the force per unit

volume.

33 / 58


Simplifying:

~Gx ·∆x ·∆y ·∆z =

(∂σxx

∂x·∆x

)·∆y∆z

+

(∂τxy

∂y∆y)·∆x∆z

+

(∂τxz

∂z∆z)·∆x∆y(2)

Dividing through by the volume of the differential element (we

want force per unit volume):

~Gx =∂σxx

∂x+∂τxy

∂y+∂τxz

∂z

34 / 58


The surface forces per unit volume due to fluid stresses acting on

a differential element in the x, y, and z-directions respectively, are:

Gx =∂σxx

∂x+∂τxy

∂y+∂τxz

∂z

Gy =∂τxy

∂x+∂σyy

∂y+∂τyz

∂z

Gz =∂τxz

∂x︸︷︷︸yz−plane

+∂τyz

∂y︸︷︷︸zx−plane

+∂σzz

∂z︸︷︷︸xy−plane

(3)

The simpler, vector notation:

~G = ∇ ·Π

35 / 58


The equation of motion:

D~uDt︸︷︷︸



+ ~G︸︷︷︸Surface Forces

Becomes:

D~uDt︸︷︷︸



+ ∇ ·Π︸︷︷︸Surface Forces

Problem: We have introduced new unknowns, namely the shear

stresses τij, but have not included any new information/equations.

36 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction37 / 58

What are constitutive equations?

Constitutive equations are stress-strain relationships

ASIDE: Hooke’s Law: Linear Elasticity

In its simplest (1-D) form, Hooke’s Law states σ = Eε

The complete (general) relationship between stress-and-strain in anelastic solid is:

σij = Cijklekl

Where Cijkl is a 4-order tensor, so there are 81 elements.

38 / 58

Aside: Constitutive relations in elasticity/solidmechanics

For isotropic materials, this can be reduced to a much simpler

constitutive relationship:

σxx = λ(εxx + εyy + εzz) + 2Gεxx

σyy = λ(εxx + εyy + εzz) + 2Gεyy

σzz = λ(εxx + εyy + εzz) + 2Gεzz

σxy = 2Gεxy

σyz = 2Gεyz

σxz = 2Gεxz

Where λ = 2Gν1−2ν is one of the Lame constants and G = λ(1−2ν)

2ν is

the shear modulus.

Hence, if we were interested in the force-acceleration relationship

in an elastic solid, we could simply use this constitutive relation

appropriately in the giverning equations (solid dynamics).

39 / 58

Stress-strain relationship in fluids

In this course we will exclusively deal with Newtonian Fluids.1 The relationship between the stress and the strain rate is linear2 Stress has a value of zero when strain rate is zero3 The constant of linear proportionality is the fluid viscosity µ

Ie. The viscosity in a Newtonian fluid does not depend on the

foces that are acting on it. Viscosity is only a function of pressure

and temperature.

Newtonian fluids include: Air, water, oil etc.

Non-newtonian fluids include: Blood (biological fluids and

mucouses in general), long chain polymers, etc.

40 / 58

Stress-strain relationship in Newtonian fluids

Let’s consider a simple example of a 2-D Poiseille flow:

From experiments we know that:

τxy = µ∂u∂y

The shear stress is linearly proportional to the deformation of the

fluid.

We can also say that the shear stress is proportional to the rate of

change of what was originally a right angle:

τ = µ∂γ

∂t

41 / 58

Stress-strain relationship in Newtonian fluids

Let’s look at a general 2-D element and examine the shear stress:

τxy = µ∂γ

∂t= µ

(∂γ1

∂t+∂γ2

∂t

)= µ

(∂u∂y

+∂v∂x

)

Applying similar analyses to the remaining shear stresses:

τxy = µ

(∂u∂y

+∂v∂x

); τyz = µ

(∂v∂z

+∂w∂y

); τxz = µ

(∂w∂x

+∂u∂z

)

42 / 58

Normal Stress to Strain relationship :Compressible Fluid

Unfortunately, in general, the same trick does not apply for the

normal stresses in a compressible fluid. The constitutive

relationship is a little more complicated.

σxx = −p +

(µB −

23µ

)(∂u∂x

+∂v∂y

+∂w∂z

)+ 2µ

∂u∂x

σyy = −p +

(µB −

23µ

)(∂u∂x

+∂v∂y

+∂w∂z

)+ 2µ

∂v∂y

σzz = −p +

(µB −

23µ

)(∂u∂x

+∂v∂y

+∂w∂z

)+ 2µ

∂w∂z

The bulk viscosity µB is typically hard to determine. Luckily, for a

monatomic gas it is zero.

Note: The divergence term above is zero for incompressible flow.43 / 58

Stress-Strain relationship in fluids

By combining the shear and normal stress - strain rate

relationships, a general constitutive equation giving the

relationship between the fluid stress and the fluid deformation is:σxx τxy τxz

τxy σyy τyz

τxz τyz σzz

= µ

∂u∂x

∂v∂x

∂w∂x

∂u∂y

∂v∂y

∂w∂y

∂u∂z

∂v∂z

∂w∂z

+ µ

∂u∂x

∂u∂y

∂u∂z

∂v∂x

∂v∂y

∂v∂z

∂w∂x

∂w∂y

∂w∂z

−

p 0 0

0 p 0

0 0 p

+ (µB −23µ)

∇ ·~u 0 0

0 ∇ ·~u 0

0 0 ∇ ·~u

Discussion: Write down the expression relating σxx to the fluid

kinematics. Perform a similar exercise for τxy.44 / 58

Stress-Strain relationship in fluids

45 / 58

ASIDE: Recall the strain rate tensor from Module 2

ASIDE: The strain rate tensor:∂u∂x

12

(∂u∂y + ∂v

∂x

)12

(∂u∂z + ∂w

∂x

)12

(∂v∂x + ∂u

∂y

)∂v∂y

12

(∂v∂z + ∂w

∂y

)12

(∂w∂x + ∂u

∂z

) 12

(∂w∂y + ∂v

∂z

)∂w∂z

Compared with the two tensors in the constitutive relations:

µ

∂u∂x

∂v∂x

∂w∂x

∂u∂y

∂v∂y

∂w∂y

∂u∂z

∂v∂z

∂w∂z

+ µ

∂u∂x

∂u∂y

∂u∂z

∂v∂x

∂v∂y

∂v∂z

∂w∂x

∂w∂y

∂w∂z

We can see that the strain rate tensor from before is used directly

in the contritutive relations for a fluid (as expected).46 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction47 / 58

Putting it all together

At this point we can take our general equation of motion from

earlier:

ρD~uDt

= ~F + ~G = ρg +∇ ·Π

And insert the constitutive relation from the previous slide:

Π = µ(∇~u +∇T~uT)− p− 2/3µ∇ ·~u

The manipulation and simplification gives the Navier-Stokes

Equations.

48 / 58

Navier Stokes Equation

Discovered by Navier in 1827 and Poisson in 1831, re-discovered

(more consistent) by St. Vennant (1843) and Stokes (1845)

x−momentum is:

ρ

(∂u∂t

+ u∂u∂x

+ v∂u∂y

+ w∂u∂z

)= −∂p

∂x+

13µ∂

∂x

(∂u∂x

+∂v∂y

+∂w∂z

)+ µ

(∂2u∂x2 +

∂2u∂y2 +

∂2u∂z2

)+ ρgx

y−momentum is:

ρ

(∂v∂t

+ u∂v∂x

+ v∂v∂y

+ w∂v∂z

)= −∂p

∂y+

13µ∂

∂y

(∂u∂x

+∂v∂y

+∂w∂z

)+ µ

(∂2v∂x2 +

∂2v∂y2 +

∂2v∂z2

)+ ρgy

The z−momentum expression is similar

49 / 58

Navier Stokes Equation

In vector form the expression is more compact:

ρD~uDt

= −∇p +13µ∇ (∇ ·~u) + µ∇2~u + ρ~g

For comparison, the y−momentum is:

ρ

(∂v∂t

+ u∂v∂x

+ v∂v∂y

+ w∂v∂z

)= −∂p

∂y+

13µ∂

∂y

(∂u∂x

+∂v∂y

+∂w∂z

)+ µ

(∂2v∂x2 +

∂2v∂y2 +

∂2v∂z2

)+ ρgy

50 / 58

Incompressible Flow: Navier-Stokes Equation

If the flow is incompressible, the Navier-Stokes equation becomes:

ρD~uDt

= −∇p + µ∇2~u + ρ~g

Fully expanded in the x−direction:

ρ

(∂u∂t

+ u∂u∂x

+ v∂u∂y

+ w∂u∂z

)= −∂p

∂x+µ

(∂2u∂x2 +

∂2u∂y2 +

∂2u∂z2

)+ρgx

y−momentum is:

ρ

(∂v∂t

+ u∂v∂x

+ v∂v∂y

+ w∂v∂z

)= −∂p

∂y+ µ

(∂2v∂x2 +

∂2v∂y2 +

∂2v∂z2

)+ ρgy

51 / 58

Discussion of the incompressible, viscousequations of motion

The Navier-Stokes equations represent an unsteady

convection-diffusion system of equations. Convection-diffusion

equations are a common occurence in physical systems/laws.

ρD~uDt

= −∇p + µ∇2~u + ρ~g

The convection component relates to the carrying of momentum

by the net flow of the fluid. The LHS of the equation governs this

behavior (the fluid acceleration).

The µ∇2~u ont he RHS indicates that the momentum is diffused

into the flow by viscous stresses via the viscosity.

When fluid elements are in relative motion to each other, the

momentum of the fluid is redistributed/diffused.52 / 58

Discussion of the incompressible, viscousequations of motion

Zones of influence

Elliptic equations

Hyperbolic equations

53 / 58

Outline



Linear Strain Rate

Shear Strain Rate

Strain Rate Tensor






Introduction54 / 58

Non-dimensional Navier Stokes

Let’s consider the Navier-Stokes Equations:

ρD~uDt

= −∇p + µ∇2~u + ρ~g

It would be convenient to compare the relative importance of one

part of the equation vs. some other part of the equation for a

particular flow/region of the flow.

To do this, we consider a non-dimensional form of the N-S

equations.

55 / 58


To non-dimensionalize the N-S equations, we use the following

relations:

~u∗ = ~u/U → ~u = ~u∗U

~x∗ = ~x/L→ ~x = ~x∗L

t∗ = t · UL→ t = t∗

LU

p∗ =pρU2 → p = p∗ρU2

Try non-dimensionalizing the N-S equations:

ρD~uDt

= −∇p + µ∇2~u + ρ~g

56 / 58


in the x-direction, for a 2-D flow:

ρ

(∂u∂t

+ u∂u∂x

+ v∂u∂y

)= −∂p

∂x+

(∂2u∂x2 +

∂2u∂y2

)

Substitute in the expressions from previous slide:

ρ

(∂Uu∗

∂ LU t∗

+ Uu∗ ∂Uu∗

∂Lx∗+ Uv∗

∂Uu∗

∂Ly∗

)= −∂ρU2p∗

∂Lx∗+µ

(∂2Uu∗

∂(Lx∗)2 +∂2Uu∗

∂(Ly∗)2

)

Simplifying:

ρU2

L

(∂u∗

∂t∗+ u∗

∂u∗

∂x∗+ v∗

∂u∗

∂y∗

)= −ρU2

L∂p∗

∂x∗+

UL2µ

(∂2u∗

∂(x∗)2 +∂2u∗

∂(y∗)2

)

57 / 58


Final simplifications:(∂u∗

∂t∗+ u∗

∂u∗

∂x∗+ v∗

∂u∗

∂y∗

)= − L

ρU2

ρU2

L∂p∗

∂x∗+

UL2

LρU2µ

(∂2u∗

∂(x∗)2 +∂2u∗

∂(y∗)2

)

The end result is the non-dimensional N-S equations:(∂u∗

∂t∗+ u∗

∂u∗

∂x∗+ v∗

∂u∗

∂y∗

)= −∂p∗

∂x∗+

µ

ρUL︸︷︷︸=Re−1

(∂2u∗

∂(x∗)2 +∂2u∗

∂(y∗)2

)

Discussion.

58 / 58

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22.581 Module 7: Stress, Viscosity, and The Navier-Stokes ...faculty.uml.edu/dwillis/MECH5810/Lecture/2016_22_581_Module7.pdf22.581 Module 7: Stress, Viscosity, and The Navier-Stokes