School of Mechanical Aerospace and Civil Engineering
3rd Year Fluid Mechanics
The Navier Stokes Equations
Derivation of The Navier Stokes Equations
◮
Here, we outline an approach for obtaining the Navier Stokes equations
that builds on the methods used in earlier years of applying mass
conservation and force-momentum principles to a control volume.
◮
The approach involves:
◮
T. J. Craft
◮
George Begg Building, C41
◮
Contents:
◮ Navier-Stokes equations
◮ Inviscid flows
◮ Boundary layers
◮ Transition, Reynolds averaging
◮ Mixing-length models of turbulence
◮ Turbulent kinetic energy equation
◮ One- and Two-equation models
◮ Flow management
Reading:
F.M. White, Fluid Mechanics
J. Mathieu, J. Scott, An Introduction to Turbulent Flow
P.A. Libby, Introduction to Turbulence
P. Bernard, J. Wallace, Turbulent Flow: Analysis Measurement & Prediction
S.B. Pope, Turbulent Flows
D. Wilcox, Turbulence Modelling for CFD
Notes: http://cfd.mace.manchester.ac.uk/tmcfd
- People - T. Craft - Online Teaching Material
Defining a small control volume within the flow.
Applying the mass conservation and force-momentum principle to
the control volume.
Considering what happens in the limit as the control volume
becomes infinitesimally small.
◮
Although the derivation can be done using any arbitrarily shaped control
volume, for simplicity we consider here a rectangular control volume.
◮
We will first derive the equations for two-dimensional, unsteady, flow
conditions, and it should then be apparent how these extend to
three-dimensional flows.
The Navier Stokes Equations
◮
Mass Conservation (Continuity)
◮
The mass conservation principle is
i
i
h
i
h
h
Rate of mass accu= Rate of mass − Rate of mass
◮
For a two-dimensional control volume
of dimensions ∆x and ∆y as shown:
mulation within CV
Mass accumulation rate = ∂ (ρ ∆x ∆y )/∂ t
Mass inflow = (ρ U)x ∆y + (ρ V )y ∆x
Mass outflow = (ρ U)x +∆x ∆y + (ρ V )y +∆y ∆x
◮
◮
◮
( ρU)x
( ρU)x+∆x
∆y
(ρ V )y +∆y → (ρ V )y + ∆y
∂ (ρ V )
∂y
Substituting these into equation (2) gives
or
∂ ρ ∂ (ρ U) ∂ (ρ V )
+
+
=0
∂t
∂x
∂y
◮
In three-dimensional flows this is easily extended to
◮
In steady-state flows, ∂ ρ /∂ t = 0, so
∂ ρ ∂ (ρ U) ∂ (ρ V ) ∂ (ρ W )
+
+
+
=0
∂t
∂x
∂y
∂z
∆x
( ρV)y
The mass conservation equation thus gives
∂ (ρ ∆x ∆y )
= (ρ U)x ∆y + (ρ V )y ∆x − (ρ U)x +∆x ∆y − (ρ V )y +∆y ∆x
∂t
Division by ∆x ∆y and rearrangement leads to
∂ρ
(ρ U)x − (ρ U)x +∆x (ρ V )y − (ρ V )y +∆y
=
+
∂t
∆x
∆y
The Navier Stokes Equations
∂ (ρ U)
∂x
∂ρ
∂ (ρ U) ∂ (ρ V )
=−
−
∂t
∂x
∂y
( ρV) y+∆y
3 / 22
(5)
In incompressible flows the density is constant, so we obtain
∂U ∂V ∂W
+
+
=0
∂x
∂y
∂z
(2)
2008/9
(3)
(4)
∂ (ρ U) ∂ (ρ V ) ∂ (ρ W )
+
+
=0
∂x
∂y
∂z
(1)
◮
2 / 22
In the limit as ∆x , ∆y → 0, the control volume becomes infinitesimally
small, and using Taylor series expansions we have
(ρ U)x +∆x → (ρ U)x + ∆x
flow out of CV
flow into CV
2008/9
The Navier Stokes Equations
(6)
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4 / 22
Force-Momentum Principle
◮
◮
◮
Accumulation rate = ∆x ∆y [∂ (ρ U)/∂ t]
( ρUU)x
Px
(τxx )x
Mom. flux out = ∆y (ρ UU)x +∆x + ∆x (ρ VU)y +∆y
∆y
ρFx
(ρUU)x+
∆x
Px+ ∆x
∆x
(τxx )x+∆x
◮
As before, as ∆x and ∆y → 0, for any quantity φ we have:
(τ xy) y
φx +∆x → φx + ∆x
( ρVU)y
◮
Surface forces arise from the pressure and viscous stresses:
Net surface force = [(P+τxx )x − (P+τxx )x +∆x ] ∆y + (τxy )y − (τxy )y +∆y ∆x
◮
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5 / 22
∂φ
∂x
and φy +∆y → φy + ∆y
∂φ
∂y
Applying these to equation (8) the U-momentum balance becomes
∂ (ρ U)
∂ (ρ UU) ∂ (ρ VU) ∂ P ∂ τxx ∂ τxy
=−
−
−
−
−
+ ρ Fx
∂t
∂x
∂y
∂x
∂x
∂y
Body force = ρ Fx ∆x ∆y
The Navier Stokes Equations
(7)
Dividing by ∆x ∆y gives:
∂ (ρ U) (ρ UU)x − (ρ UU)x +∆x (ρ VU)y − (ρ VU)y +∆y Px − Px +∆x
=
+
+
∂t
∆x
∆y
∆x
(τxx )x − (τxx )x +∆x (τxy )y − (τxy )y +∆y
+
+
+ ρ Fx (8)
∆x
∆y
( ρVU) y+∆y
(τxy )y+∆y
Consider the U momentum equation, on a
control volume of dimensions ∆x and ∆y :
Mom. flux in = ∆y (ρ UU)x + ∆x (ρ VU)y
+ ρ Fx ∆x ∆y
◮
within CV
on CV faces
The U-momentum balance then gives
∆x ∆y [∂ (ρ U)/∂ t] = [(ρ UU)x − (ρ UU)x +∆x ] ∆y + (ρ VU)y − (ρ VU)y +∆y ∆x
+ [(P + τxx )x − (P + τxx )x +∆x ] ∆y + (τxy )y − (τxy )y +∆y ∆x
The force-momentum principle is
h
i
h
i
h
i
Accumulation of mo= Rate of momen- − Rate of momenmentum within CV
tum flow into CV
tum flow out of CV
i
i
h
h
Body
forces
Forces
acting
+
+
The Navier Stokes Equations
(9)
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6 / 22
The Viscous Stresses
◮
Rearranging gives the usual form of the U-momentum equation:
∂ (ρ U) ∂ (ρ U 2 ) ∂ (ρ VU)
∂ P ∂ τxx ∂ τxy
+
+
=−
−
−
+ ρ Fx
∂t
∂x
∂y
∂x
∂x
∂y
◮
(10)
◮
As with the continuity equation, the U momentum equation is also a
differential equation.
◮
The corresponding V -momentum equation is
τxy = −µ
◮
∂ (ρ V ) ∂ (ρ UV )
+
+
∂t
∂x
∂ (ρ V 2 )
∂y
=−
∂ P ∂ τxy ∂ τyy
−
−
+ ρ Fy
∂y
∂x
∂y
In a simple shear flow, Stoke’s law states that
the viscous shear stress, τxy , is obtained from
(∂ U/∂ y )∆y ∆t
∆y
For small θx , tan(θx ) ≈ θx , so
◮
In their above forms, however, the U and V -momentum equations still
contain additional unknown variables, namely the viscous stresses, τxx ,
τyy and τxy .
∂ θx
∂U
≈
∂t
∂y
◮
The Navier Stokes Equations
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U
τxy
τxy
y
This equation can be obtained by considering how, in a simple case, the
rate at which a fluid element is deformed is opposed by the fluid viscosity.
tan(θx ) =
(11)
∂U
∂y
(dU/dy) ∆y ∆t
∆x
θx
∆y
t
t+ ∆ t
For many common fluids we have τ ∝ ∂ θx /∂ t.
The Navier Stokes Equations
2008/9
8 / 22
◮
In the more general case, expressions for the viscous stresses can again
be derived by considering the deformation caused by the flow field to an
initially rectangular fluid element.
◮
For Newtonian fluids these general stress-strain relations can be
expressed as the viscous stresses being linearly related to the strain
rates, with the constant of proportionality being the viscosity µ .
◮
Hence, in 2-D, we obtain the viscous stresses as
τxx = −2µ
∂U
∂x
τyy = −2µ
∂V
∂y
τxy = −µ
∂U ∂V
+
∂y
∂x
A similar equation can be derived for the V momentum component.
◮
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In three-dimensional flows the equations are expanded to:
∂ ρ ∂ (ρ U) ∂ (ρ V ) ∂ (ρ W )
Continuity:
+
+
+
=0
∂t
∂x
∂y
∂z
9 / 22
(13)
The U-momentum equation:
∂U
∂ (ρ U) ∂ (ρ U 2 ) ∂ (ρ VU)
∂P
∂
∂
∂U ∂V
2µ
+
+ ρ Fx
µ
+
+
=−
+
+
∂t
∂x
∂y
∂x ∂x
∂x
∂y
∂y ∂x
(14)
The V -momentum equation:
∂V
∂ (ρ V ) ∂ (ρ UV ) ∂ (ρ V 2 )
∂P
∂
∂V ∂U
∂
µ
+
2µ
+ ρ Fy
+
+
=−
+
+
∂t
∂x
∂y
∂y ∂x
∂x ∂y
∂y
∂y
(15)
The Navier Stokes Equations
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Convection and Diffusion Terms
(16)
◮
U-Momentum:
∂ (ρ U) ∂ (ρ UU) ∂ (ρ VU) ∂ (ρ WU)
∂P
+
+
+
=−
+ ρ Fx
∂t
∂x ∂y ∂ z ∂ x ∂
∂
∂U ∂V
∂
∂U ∂W
∂U
+
+
+
+
+2
µ
µ
µ
∂x
∂x
∂y
∂y
∂x
∂z
∂z
∂x
V -Momentum:
∂ (ρ V ) ∂ (ρ UV ) ∂ (ρ VV ) ∂ (ρ WV )
∂P
+
+
+
=−
+ ρ Fy
∂t
∂x
∂
y
∂z ∂ y ∂V
∂
∂V ∂U
∂
∂
∂V ∂W
+2
+
+
+
µ
µ
µ
+
∂x
∂x
∂y
∂y
∂y
∂z
∂z
∂y
W -Momentum:
∂ (ρ W ) ∂ (ρ UW ) ∂ (ρ VW ) ∂ (ρ WW )
∂P
+
+
+
=−
+ ρ Fz
∂t
∂x
∂y
∂z
∂z
∂
∂W ∂U
∂
∂W ∂V
∂
∂W
+
+
+
µ
+
µ
+2
µ
∂x
∂x
∂z
∂y
∂y
∂z
∂z
∂z
The Navier Stokes Equations
The above set of equations that describe a real fluid motion are
collectively known as the Navier Stokes equations. In 2-D they can be
written as:
∂ ρ ∂ (ρ U) ∂ (ρ V )
+
+
=0
∂t
∂x
∂y
Substituting the expressions for τxx and τxy into the U momentum
equation gives:
∂U
∂ (ρ U) ∂ (ρ U 2 ) ∂ (ρ VU)
∂P
∂
∂
∂U ∂V
+
+
=−
+
+
2µ
+
µ
+ ρ Fx
∂t
∂x
∂y
∂x ∂x
∂x
∂y
∂y ∂x
(12)
◮
◮
The continuity equation:
◮
The Navier Stokes Equations
The Navier Stokes Equations
2008/9
The term
(17)
◮
The terms on the right hand sides of the equations involving the viscosity
represent viscous diffusion.
◮
The general form of the momentum transport equations is thus seen to be
◮
The combination of time derivative and convection terms represents the
total rate of change of a quantity following a fluid path line. It is often
written in shorthand notation as D φ /Dt:
Time derivative + Convection terms = Forcing terms + Diffusion terms
(18)
Dφ
∂ φ ∂ (U φ ) ∂ (V φ ) ∂ (W φ )
≡
+
+
+
Dt
∂t
∂x
∂y
∂z
(19)
11 / 22
∂ (ρ U φ ) ∂ (ρ V φ ) ∂ (ρ W φ )
+
+
∂x
∂y
∂z
where φ stands for any of the velocity components (U, V or W )
represents convection of φ by the fluid.
The Navier Stokes Equations
(20)
2008/9
12 / 22
◮
The time derivative and convection terms are sometimes written as above
(with ρ , U, V , W inside the derivatives), and sometimes in the form
ρ
◮
◮
∂φ
∂φ
∂φ
∂φ
+ ρU
+ ρV
+ ρW
∂t
∂x
∂y
∂z
These are, in fact, entirely equivalent, since differentiating by parts gives
∂ ρφ ∂ ρ U φ ∂ ρ V φ ∂ ρ W φ
+
+
+
∂t
∂x
∂y
∂z
∂φ
∂φ
∂φ
∂φ
∂ρ
∂ ρU
∂ ρV
∂ ρW
=ρ
+φ
+ ρU
+φ
+ ρV
+φ
+ ρW
+φ
∂t
∂t
∂x
∂x
∂y
∂y
∂z
∂z
∂φ
∂φ
∂φ
∂ ρ ∂ ρU ∂ ρV ∂ ρW
∂φ
+ ρU
+ ρV
+ ρW
+φ
+
+
+
=ρ
∂t
∂x
∂y
∂z
∂t
∂x
∂y
∂z
If the viscosity is constant the diffusion terms can be simplified by taking
µ outside the derivatives. In 2-D, for example:
∂
∂
∂U ∂V
∂U
2
µ
µ
+
+
∂x
∂x
∂y
∂y
∂x
∂ ∂U
∂ ∂U
∂ ∂U
∂ ∂V
=µ
+µ
+µ
+µ
∂x ∂x
∂y ∂y
∂x ∂x
∂y ∂x
∂ 2U
∂ 2U
∂ ∂U ∂V
+µ
+µ
=µ
+
∂x ∂x
∂y
∂x2
∂y2
=µ
∂ 2U
∂ 2U
+µ
∂x2
∂y2
and the term in brackets multiplied by φ is zero from the continuity
equation.
The Navier Stokes Equations
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Further Simplifications
The Navier Stokes Equations
14 / 22
Other Transport Equations
◮
In many flows that we will consider certain additional simplifications can
be introduced.
◮
In steady flows the time derivatives become zero:
◮
The governing equations for other quantities transported by a flow often
take the same general form of transport equation to the above
momentum equations.
◮
For example, the transport equation for the evolution of temperature in a
fluid flow can often be written (in 2-D for simplicity) as
∂T
∂T
∂ T ∂ (UT ) ∂ (VT )
∂
∂
α
α
+
+
+
=
∂t
∂x
∂y
∂x
∂x
∂y
∂y
∂ (ρ U)/∂ t = ∂ (ρ V )/∂ t = ∂ (ρ W )/∂ t = 0
◮
The body force terms, Fx , Fy , Fz , are, in many cases, negligible.
◮
These simplifications lead to the momentum equations for a 2-D steady,
incompressible, constant viscosity, flow without body forces being given
by
ρ
2008/9
where α is the molecular thermal diffusivity.
◮
∂ (U 2 )
∂ (VU)
∂ 2U
∂ 2U
∂P
+ρ
=−
+µ 2 +µ
∂x
∂y
∂x
∂x
∂y2
Notice the general form of
Time derivative + Convection terms = Diffusion terms + Source terms
(in this case, the source, or forcing, terms are zero).
∂ (UV )
∂ (V 2 )
∂ 2V
∂ 2V
∂P
+ρ
=−
+µ
ρ
+µ
∂x
∂y
∂y
∂x2
∂y2
The Navier Stokes Equations
◮
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15 / 22
We will meet transport equations for other, turbulence-related, quantities
later in the course.
The Navier Stokes Equations
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Solving the Navier Stokes Equations
◮
◮
◮
◮
◮
◮
Tensor/Summation Notation
The Navier Stokes equations form a system of differential equations:
◮
Although the equations have been presented for a Cartesian coordinate
system (x , y , z), they can also be transformed mathematically to other
coordinate systems, (eg. cylindrical, or spherical, polars).
In principle, therefore, the Navier Stokes
equations can be integrated over a flow
domain of interest, with appropriate
boundary conditions, to produce detailed
velocity and pressure fields.
◮
◮
◮
Tensor notation and, in particular, the Einstein summation convention is
often used to write the equations in a more compact, shorthand, form.
◮
One can use subscripts to denote the elements of a vector or tensor.
◮
The summation convention means that if a subscript letter is repeated in
an expression, there is an implied summation over it.
◮
So, for example,
dP/dy=0
U=V=0
U=U in
dU/dx=0
V=0
dV/dx=0
P=Pin
dP/dx=Const
U=V=0
Although analytical solutions can be obtained for a few cases, in practice
the equations must usually be solved using numerical methods.
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17 / 22
Using the same notation, the three momentum equations from the Navier
Stokes system can be written compactly as
∂ ρ Ui ∂ ρ Ui Uj
∂P
∂
∂ Ui
µ
(22)
+ ρ Fi
=−
+
+
∂t
∂ xj
∂ xi ∂ xj
∂ xj
Here, the 2nd and 4th terms contain a repeated “j”, so one sums from j
equals 1 to 3 in them. For example, the convection term expands to
The continuity equation for incompressible flow can then simply be
written as
∂ Ui / ∂ x i = 0
◮
Note that the “i” in the above expression could have been replaced by “j”
or “k ” (or anything else). It is purely a dummy index indicating summation.
The Navier Stokes Equations
18 / 22
There are a few, very simple, laminar flows for which the Navier Stokes
equations can be solved analytically.
◮
For example, for steady, incompressible, fully
developed flow in a plane channel as shown, we
have V = 0 and U does not depend on x .
◮
Equations (21) and (22) are far more compact and convenient than using
the expansions of equations (16) to (19) for the Navier Stokes system.
◮
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◮
◮
The Navier Stokes Equations
(21)
Analytical Solutions of The Navier Stokes Equations
The subscript i is not repeated, and is being used to denote a component
of the velocity vector (one gets the U1 momentum equations by setting
i = 1, the U2 one by setting i = 2, etc).
For much of this course we can relatively easily write equations out in
terms of x , y components etc., and will not have to use summation
notation. It is widely used in textbooks and papers on fluid mechanics
and turbulence, and we will use it in a few places, for convenience.
∂ Ui
∂ Ui
∂ U1 ∂ U2 ∂ U3
≡ ∑
=
+
+
∂ xi
∂ xi
∂ x1
∂ x2
∂ x3
i=1,3
◮
dP/dy=0
∂ ρ Ui Uj
∂ ρ Ui U1 ∂ ρ Ui U2 ∂ ρ Ui U3
≡
+
+
∂ xj
∂ x1
∂ x2
∂ x3
◮
Although it is sometimes appropriate to write the Navier Stokes equations
out in their expanded Cartesian form as above, in general it becomes
rather cumbersome.
In three-dimensional flows there are four variables and four
differential equations.
The Navier Stokes Equations
◮
◮
In two-dimensional flows there are three variables (U, V , P) and
three differential equations (Continuity, U and V -momentum).
Continuity (∂ U/∂ x + ∂ V /∂ y = 0) is satisfied.
y=h
y
x
U(y)
y=−h
The V momentum equation reduces to ∂ P/∂ y = 0, so P is constant
across the channel.
The U momentum equation becomes
0=−
∂P
∂
+
∂x ∂y
µ
∂U
∂y
(23)
with boundary conditions U = 0 at y = ±h.
19 / 22
The Navier Stokes Equations
2008/9
20 / 22
◮
Since P is not a function of y , we can easily integrate this:
µ
∂U
∂P
=y
+A
∂y
∂x
(24)
◮
The pressure gradient can be related to the bulk (average) velocity, since
Ub =
for some constant of integration A. Integrating a second time gives
y2 ∂P
µU =
+ Ay + B
2 ∂x
◮
h2 ∂ P
+ Ah + B
2 ∂x
and
0=
(25)
◮
h2 ∂ P
− Ah + B
2 ∂x
(26)
U(y )dy = −
U = (3/2)Ub (1 − y 2 /h2 )
1 ∂P
4h µ ∂ x
Z h
−h
(h2 − y 2 )dy
y=h
Ub
U(y)
(29)
y=−h
h2 ∂ P
A=0
and
B=−
2 ∂x
The velocity profile is therefore given by the parabola
(27)
◮
1 ∂P 2
(h − y 2 )
U =−
2µ ∂ x
The Navier Stokes Equations
−h
Hence ∂ P/∂ x = −3µ Ub /h2 , and the velocity
profile can be written as
This gives
◮
Z h
ih
1 ∂P h 2
h2 ∂ P
=−
h y − y 3 /3
=−
4h µ ∂ x
−h
3µ ∂ x
To determine the constants A and B, we apply the boundary conditions
that U = 0 at y = ±h:
0=
1
2h
A similar analysis can be applied to some other simple 1-D flows, such as
fully-developed pipe flow, flow between moving infinite flat plates, etc.
(28)
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The Navier Stokes Equations
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