Navier strokes equation

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Navier stokes eqn

Derivation of the Navier-Stokes equation
Euler’s equation
The fluid velocity u of an inviscid (ideal) fluid of density ρ under the action
of a body force ρf is determined by the equation:
ρ
Du
Dt
= − p + ρf, (1)
known as Euler’s equation. The scalar p is the pressure. This equation is
supplemented by an equation describing the conservation of mass. For an
incompressible fluid this is simply
· u = 0. (2)
Real fluids however are never truly inviscid. We must therefore see how
equation (1) is changed by the inclusion of viscous forces.
The stress tensor
We first introduce the stress tensor σij as follows:
σij is the i-component of stress on a surface element δS that has a normal n
pointing in the j-direction.
Then if t is the stress on a small surface element δS with unit normal n it is
straightforward to demonstrate that
ti = σijnj. (3)
Proof of (3)
Take δS to be the large face of the tetrahedron shown in the figure on the
following page. We shall apply Newton’s 2nd law to the fluid in the tetra-
hedron. For definiteness consider the i-component of the force. The force of
surrounding fluid on the large face is tiδS. The i-component of the force on
the ‘back’ face is −σi1 (since the normal is in the −e1 direction). Similarly
1
for the two faces with normals −e2 and −e3. Thus the i-component of the
total force exerted by surrounding fluid on the tetrahedron is
(ti − σijnj) δS (4)
(summation over j).
This force (plus any body force) will be equal, from Newton’s 2nd law, to the
mass of the element (ρδV ) multiplied by its acceleration. If is the linear
length scale of the tetrahedron then δV ∼ 3
and δS ∼ 2
. Thus if we let
→ 0 then the term (4) clearly dominates and hence ti = σijnj.
The equation of motion
The ith component of force exerted by the surrounding fluid on a fluid blob
with surface S and volume V is given by
S
tidS =
S
σijnjdS =
V
∂σij
∂xj
dV,
where we have used relation (3) together with the divergence theorem. Ap-
plying Newton’s 2nd law to an arbitrary fluid blob then leads to the equation
of motion:
ρ
Dui
Dt
=
∂σij
∂xj
+ ρfi. (5)
2
Applying an argument involving angular momentum to a tetrahedron, which
is similar in spirit to the argument above using linear momentum, leads to
the result that the tensor σij is symmetric.
Newtonian fluids
In order actually to solve equation (5) it is necessary to relate the stress
tensor σij to the fluid velocity u. We shall restrict our attention to fluids
that are incompressible and for which the stress tensor is assumed to take
the form:
σij = −pδij + µ
∂ui
∂xj
+
∂uj
∂xi
, (6)
where µ is the viscosity and p is the pressure. Note that σij is symmetric,
as required. Fluids for which the linear relation (6) holds are said to be
Newtonian, in recognition of the fact that Newton proposed such a relation for
a simple shearing motion. For water and for most gases under non-extreme
conditions the linear relation (6) holds. There are though many important
fluids that are non-Newtonian, i.e. the simple relation (6) does not hold.
These include paint, tomato ketchup, blood, toothpaste and quicksand.
Note that, since · u = 0, then
p = −
1
3
(σ11 + σ22 + σ33) ,
i.e. −p is the mean of the three normal stresses.
The Navier-Stokes equation
The final step in deriving the Navier-Stokes equation is to substitute ex-
pression (6) for σij into equation (5). This leads to the equation (assuming
constant viscosity µ),
ρ
Du
Dt
= − p + ρf + µ 2
u. (7)
Equation (7) is known as the Navier-Stokes equation for an incompressible
fluid; it is to be solved in conjunction with the incompressibility condition
(2). For compressible fluids — with which we shall not be concerned in this
course — the viscous term is a little more complicated.
Assuming that the density ρ is constant, equation (7) can be written as
Du
Dt
= −
1
ρ
p + f + ν 2
u, (8)
3
where ν = µ/ρ is the kinematic viscosity.
It is often helpful to express the Navier-Stokes equation in dimensionless
form. Suppose a flow has characteristic length scale L and characteristic
velocity scale U. Then we may introduce dimensionless variables (denoted
by tildes) as follows:
u = U ˜u, t =
L
U
˜t,
∂
∂x
=
1
L
∂
∂˜x
, p = ρU2
˜p, f =
U2
L
˜f.
On substituting into (8), and dropping the tildes, we obtain the dimensionless
Navier-Stokes equation
Du
Dt
= − p + f +
1
Re
2
u,
where Re = UL/ν is the Reynolds number.
Further reading
The most comprehensive derivation of the Navier-Stokes equation, covering
both incompressible and compressible fluids, is in An Introduction to Fluid
Dynamics by G.K. Batchelor (Cambridge University Press), §3.3.
4

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Navier strokes equation

  • 1. Derivation of the Navier-Stokes equation Euler’s equation The fluid velocity u of an inviscid (ideal) fluid of density ρ under the action of a body force ρf is determined by the equation: ρ Du Dt = − p + ρf, (1) known as Euler’s equation. The scalar p is the pressure. This equation is supplemented by an equation describing the conservation of mass. For an incompressible fluid this is simply · u = 0. (2) Real fluids however are never truly inviscid. We must therefore see how equation (1) is changed by the inclusion of viscous forces. The stress tensor We first introduce the stress tensor σij as follows: σij is the i-component of stress on a surface element δS that has a normal n pointing in the j-direction. Then if t is the stress on a small surface element δS with unit normal n it is straightforward to demonstrate that ti = σijnj. (3) Proof of (3) Take δS to be the large face of the tetrahedron shown in the figure on the following page. We shall apply Newton’s 2nd law to the fluid in the tetra- hedron. For definiteness consider the i-component of the force. The force of surrounding fluid on the large face is tiδS. The i-component of the force on the ‘back’ face is −σi1 (since the normal is in the −e1 direction). Similarly 1
  • 2. for the two faces with normals −e2 and −e3. Thus the i-component of the total force exerted by surrounding fluid on the tetrahedron is (ti − σijnj) δS (4) (summation over j). This force (plus any body force) will be equal, from Newton’s 2nd law, to the mass of the element (ρδV ) multiplied by its acceleration. If is the linear length scale of the tetrahedron then δV ∼ 3 and δS ∼ 2 . Thus if we let → 0 then the term (4) clearly dominates and hence ti = σijnj. The equation of motion The ith component of force exerted by the surrounding fluid on a fluid blob with surface S and volume V is given by S tidS = S σijnjdS = V ∂σij ∂xj dV, where we have used relation (3) together with the divergence theorem. Ap- plying Newton’s 2nd law to an arbitrary fluid blob then leads to the equation of motion: ρ Dui Dt = ∂σij ∂xj + ρfi. (5) 2
  • 3. Applying an argument involving angular momentum to a tetrahedron, which is similar in spirit to the argument above using linear momentum, leads to the result that the tensor σij is symmetric. Newtonian fluids In order actually to solve equation (5) it is necessary to relate the stress tensor σij to the fluid velocity u. We shall restrict our attention to fluids that are incompressible and for which the stress tensor is assumed to take the form: σij = −pδij + µ ∂ui ∂xj + ∂uj ∂xi , (6) where µ is the viscosity and p is the pressure. Note that σij is symmetric, as required. Fluids for which the linear relation (6) holds are said to be Newtonian, in recognition of the fact that Newton proposed such a relation for a simple shearing motion. For water and for most gases under non-extreme conditions the linear relation (6) holds. There are though many important fluids that are non-Newtonian, i.e. the simple relation (6) does not hold. These include paint, tomato ketchup, blood, toothpaste and quicksand. Note that, since · u = 0, then p = − 1 3 (σ11 + σ22 + σ33) , i.e. −p is the mean of the three normal stresses. The Navier-Stokes equation The final step in deriving the Navier-Stokes equation is to substitute ex- pression (6) for σij into equation (5). This leads to the equation (assuming constant viscosity µ), ρ Du Dt = − p + ρf + µ 2 u. (7) Equation (7) is known as the Navier-Stokes equation for an incompressible fluid; it is to be solved in conjunction with the incompressibility condition (2). For compressible fluids — with which we shall not be concerned in this course — the viscous term is a little more complicated. Assuming that the density ρ is constant, equation (7) can be written as Du Dt = − 1 ρ p + f + ν 2 u, (8) 3
  • 4. where ν = µ/ρ is the kinematic viscosity. It is often helpful to express the Navier-Stokes equation in dimensionless form. Suppose a flow has characteristic length scale L and characteristic velocity scale U. Then we may introduce dimensionless variables (denoted by tildes) as follows: u = U ˜u, t = L U ˜t, ∂ ∂x = 1 L ∂ ∂˜x , p = ρU2 ˜p, f = U2 L ˜f. On substituting into (8), and dropping the tildes, we obtain the dimensionless Navier-Stokes equation Du Dt = − p + f + 1 Re 2 u, where Re = UL/ν is the Reynolds number. Further reading The most comprehensive derivation of the Navier-Stokes equation, covering both incompressible and compressible fluids, is in An Introduction to Fluid Dynamics by G.K. Batchelor (Cambridge University Press), §3.3. 4