22.581 Module 7: Stress, Viscosity, and The Navier-Stokes...
Transcript of 22.581 Module 7: Stress, Viscosity, and The Navier-Stokes...
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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
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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
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Administrative & Schedule Issues
Midterm Exam: Not graded yet....probably Thursday/next week?
Schedule
Today : Introduce Navier-Stokes Equations
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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
Introduction4 / 58
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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
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Conservation of mass and momentum
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
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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.
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Conservation of mass and momentum
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
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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
Introduction9 / 58
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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 ?
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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
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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
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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
Introduction13 / 58
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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
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Kinematics : Shear Strain RateLet’s look at dα first:
dα = atan(OA
) ' OA
' ∆llo
'∂u∂y · lo · dt
lo
' ∂u∂y
dt
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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
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Vortex Example
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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
Introduction18 / 58
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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
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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
Introduction20 / 58
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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
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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:
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Kinematics : Summary of Possible Fluid Motions
Vorticity, Rotation Rate
Shear Strain Rate
Linear Strain Rate
Volumetric Strain Rate
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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
Introduction24 / 58
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Stresses in a continuum
Definition: Stress is force per unit area
Consider an arbitrarily oriented differential surface in a continuum:
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Stresses in a continuum
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))
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Stresses in a continuum
Consider a cartiesian fluid element (dimensions ∆x, ∆y, ∆z):
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Stresses in a continuum
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.
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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).
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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
Introduction30 / 58
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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).
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Viscous equation of motion
Consider the stresses acting on a 3-D differential element (with
stress gradients in the flow field):
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Viscous equation of motion
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.
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Viscous equation of motion
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
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Viscous equation of motion
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
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Viscous equation of motion
The equation of motion:
D~uDt︸︷︷︸
Lagrangian Acceleration
= ~F︸︷︷︸Volume/Body Forces
+ ~G︸︷︷︸Surface Forces
Becomes:
D~uDt︸︷︷︸
Lagrangian Acceleration
= ~F︸︷︷︸Volume/Body Forces
+ ∇ ·Π︸ ︷︷ ︸Surface Forces
Problem: We have introduced new unknowns, namely the shear
stresses τij, but have not included any new information/equations.
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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
Introduction37 / 58
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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.
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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).
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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.
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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
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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
)
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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
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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
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Stress-Strain relationship in fluids
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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
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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
Introduction47 / 58
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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.
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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
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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
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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
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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
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Discussion of the incompressible, viscousequations of motion
Zones of influence
Elliptic equations
Hyperbolic equations
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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
Introduction54 / 58
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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.
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Non-dimensional Navier Stokes
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
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Non-dimensional Navier Stokes
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
)
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Non-dimensional Navier Stokes
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