Continuum Mechanics
Lecture 5
Ideal fluids
Prof. Romain Teyssier
http://www.itp.uzh.ch/~teyssier
Continuum Mechanics 15/05/2013
Romain Teyssier
Outline
- Helmholtz decomposition
- Divergence and curl theorem
- Kelvin’s circulation theorem
- The vorticity equation
- Vortex dynamics and vortex flow
- Bernoulli theorem and applications
Continuum Mechanics 15/05/2013
Romain Teyssier
Helmholtz decomposition of the velocity field
For a continuous and differentiable velocity field, we have the following unique
decomposition:
→
− →
−
→
−
→
− →
−
�v = ∇φ + ∇ × A
with Gauge condition ∇ · A = 0
→
−
→
−
→
−
The scalar and vector potential are solutions of ∆φ = ∇ · �v and ∆ A = − ∇ × �v
with appropriate boundary conditions.
The source terms for these 2 Poisson equations are respectively
→
−
∇ · �v : the divergence of the velocity field
→
−
∇ × �v : the curl of the velocity field
Limiting cases:
→
−
1- ∇ · �v = 0 for an incompressible flow. The velocity field is solenoidal or
divergence free.
→
−
2- ∇ × �v = 0 for a potential flow, because in this case the velocity field
derives from a scalar potential. The velocity is said to be curl free.
Continuum Mechanics 15/05/2013
Romain Teyssier
Physical interpretation of the divergence
We have seen in the previous lecture that the variation of a Lagrangian
�
volume is given by
→
−
dVt
=
dx3 ∇ · �v
dt
Vt
→
−
1 DV
The rate of change of the specific volume V = 1/ρ is
= ∇ · �v
V Dt
Using the divergence theorem, we can express the total volume variation
as the net flux of volume across the outer surface as:
�
dVt
=
�v · �ndS
dt
St
Let’s consider the case of a point source (or sink) of divergence at r=0.
→
−
∇ · �v = Q δ(�x = 0)
We have a spherically symmetric velocity field around the source. Using
the divergence theorem, we have: Q = 4πr2 vr
Q
A source (or sink) velocity field is thus �v =
e�r
4πr2
Continuum Mechanics 15/05/2013
Romain Teyssier
Physical interpretation of the curl
→
−
−
We define the vorticity as the following vector field →
ω = ∇ × �v
�
�
� ∂y vz − ∂z vy �
�
�
→
−
�
�
∂
v
−
∂
v
ω
=
In components, we have
x z �
� z x
� ∂x vy − ∂y vx �
→
− →
−
→
−
→
−
→
−
ω
=
∇
×
Ω
×
�
r
=
2
Ω
�
v
=
�
v
+
Ω
×
�
r
For a rigid body motion
, we have
0
The vorticity is thus twice the local rotation rate in the fluid.
−
−
ω /|→
ω|
A vortex is a vorticity line along the axis �t = →
We introduce Stoke’s theorem or curl’s theorem.
�
We define the circulation Γ as the integral of the
Γ=
�v · d�l =
parallel velocity along a closed contour.
L
�
�
→
−
�v · d�l = ( ∇ × �v ) · �ndS
We have the following identity Γ =
L
Let us consider a vortex line at r=0.
Using Stoke’s theorem, we have
A vortex velocity field is thus
Continuum Mechanics 15/05/2013
�v =
�
L
v� dl
S
→
−
∇ × �v = Ωz �ez δ(�r = 0)
Γ = 2πrvθ = Ωz
Ωz
�eθ
2πr
Romain Teyssier
Velocity field induced by a vortex distribution
This is the Biot-Savart law for vortices.
→
−
We consider an incompressible fluid for which ∇ · �v = 0
→
− →
− →
−
We know from the Helmholtz decomposition that �v = ∇ × A + ∇φ
→
− →
−
together with the Gauge condition ∇ · A = 0
→
−
ω
The potential vector satisfies the Poisson equation ∆ A = −�
�
�
�
�x − x��
1
3
�
v
(�
x
)
=
−
×
ω
�
(�
x
)dx
The solution is
� 3
4π V |�x − �x |
We need to add the scalar potential contribution, solving ∆φ = 0
with the appropriate boundary conditions (see lecture on potential flows).
�
� (�x ) = Ωδ(r = 0) using the curvilinear
We consider a filament of vorticity ω
�
�
coordinate s.
�
�x − �x
Ω
�t(�x )ds
�v (�x) = −
×
4π L |�x − �x� |3
�
�
For a vertical filament, we have �x − �x = r�er + (z − z )�ez
�
Ω +∞
rdz
Ω
�
e
=
�v (�x) =
�eθ
θ
4π −∞ (r2 + z 2 )3/2
2πr
Continuum Mechanics 15/05/2013
with �t = �ez
and �er × �ez = −�eθ
Romain Teyssier
Kelvin’s circulation theorem
Lt = φ(t,0) (L0 )
L0
We consider a closed contour evolving with the flow. We use the inverse
Lagrangian mapping to compute the time derivative of the circulation.
�
�
d
d
d
∂�x
Γ(t) =
�v · d�x =
�v (�x(�y , t), t) · ( d�y )
dt
dt Lt
dt L0
∂�y
�
�
�
� 2�
�
d
D�v ∂�x
D�v
∂�v
v
Γ(t) =
· ( d�y ) +
· d�x
�v · ( d�y ) =
�v · d�v =
=0
dt
Dt
∂�
y
∂�
y
Dt
2
L0
L0
Lt
L0
−
−
D�v →
1→
We now inject the Euler equation for an ideal fluid
= F − ∇P
Dt
ρ
�
�
→
−
−
→
−
d
1→
Γ(t) =
F · d�x +
∇ρ × ∇P · �ndS
2
dt
ρ
Lt
St
→
−
→
−
F
=
−
∇Φ
If the external force derives from a potential
−
→
−
1→
d
and if the fluid is barotropic
∇P = ∇Π then
Γ=0
ρ
dt
Lagrange theorem: if initially the vorticity is zero, then it remains zero everywhere.
Continuum Mechanics 15/05/2013
Romain Teyssier
The vorticity equation
−
−
D�v →
1→
= F − ∇P
Dt
ρ
−
→
−
→
− →
− →
−
−
∂�
ω →
1→
Taking the curl leads to
+ ∇ × (�v · ∇�v ) = ∇ × F − ∇ × ( ∇P )
∂t
ρ
→
−
→
− v2
� × �v we have
Using the identity �v · ∇�v = ∇( ) + ω
2
−
→
− →
−
−
→
−
∂�
ω →
1→
+ ∇ × (�
ω × �v ) = ∇ × F + 2 ∇ρ × ∇P
∂t
ρ
→
−
→
−
→
−
→
−
∇
×
(�
ω
×
�
v
)
=
(
∇
·
�
v
)�
ω
+
(�
v
·
∇)�
ω
−
(�
ω
·
∇)�v
Using the identity
We start with the Euler equation for ideal fluids
we find the vorticity equation:
→
−
→
−
→
− →
−
−
→
−
D�
ω
1→
= (�
ω · ∇)�v − ( ∇ · �v )�
ω + ∇ × F + 2 ∇ρ × ∇P
Dt
� �ρ �
�
→
−
D
ω
�
ω
�
For a barotropic fluid under gravity, we have
=
· ∇ �v
Dt ρ
ρ
Helmholtz theorem: vortex lines move with the fluid.
→
−
D � ��
Proof: a line element that moves with the fluid satisfies
δ � = (δ �� · ∇)�v
Dt
Continuum Mechanics 15/05/2013
Romain Teyssier
Vortex dynamics
For a barotropic fluid, the vorticity equation writes in component form:
→
−
Dωi
= ωx ∂x vi + ωy ∂y vi + ωz ∂z vi − ( ∇ · �v )ωi
Dt
Let’s consider a vertical vortex line ω
� = ωz �ez
Dωy
= ωz ∂ z v y
Dt
vortex tilting due to shear































Dωz
= −ωz (∂x vx + ∂y vy )
Dt































Dωx
= ωz ∂ z v x
Dt
vortex stretching due to 2D divergence
The 2D divergence is the rate of change of the section of the vortex tube
1 DS
= (∂x vx + ∂y vy )
S Dt
For a 2D velocity field, the total vorticity in the vortex tube is conserved.
ωz S = constant
Continuum Mechanics 15/05/2013
Romain Teyssier
First Bernoulli Theorem
→
−
−
D�v
1→
∇P
=
−
∇Φ
−
We start with the Euler equations in Lagrangian form
Dt
ρ
with equations for the thermodynamical variables
→
−
Dρ
−
D�
P→
= −ρ ∇ · �v
= − ∇ · �v
Dt
Dt
ρ
P
Multiplying by velocity and defining the specific enthalpy as h = � +
, we have
ρ
� �
→
−
−
D v2
�v →
= −�v · ∇Φ − · ∇P
Dt 2
ρ
−
D
1 ∂P
�v →
→
−
D
∂Φ
(h)
=
+
·
∇P
and
(Φ) =
+ �v · ∇Φ
Dt
ρ
∂t
ρ
Dt
∂t
Collecting everything, we have the following relation:
� 2
�
D v
1 ∂P
∂Φ
+Φ+h =
+
Dt 2
ρ ∂t
∂t
v2
Theorem follows trivially: in a stationary flow, the total enthalpy H =
+Φ+h
2
is conserved along streamlines.
Validity: no viscosity, no dissipation (reversible isentropic flow)
Continuum Mechanics 15/05/2013
Romain Teyssier
Second Bernoulli Theorem
−
→
−
1→
→
−
∇P
=
∇Π
We consider a curl free flow �v = ∇φ in a barotropic fluid
ρ
→
−
→
− v2
� × �v
Using the now well known vector relation �v · ∇�v = ∇( ) + ω
2
�
�
→
− ∂φ v 2
+
+Φ+Π =0
the Euler equation becomes ∇
∂t
2
The theorem follows:
For a potential flow, we have everywhere in the flow (not only along streamlines):
∂φ v 2
+
+ Φ + Π = C(t)
∂t
2
The constant depends only on time. The flow doesn’t have to be stationary.
v2
+ Φ + Π is uniform everywhere.
For a stationary flow, the quantity H =
2
v2
P
For a curl free incompressible fluid, we have H =
+Φ+
2
ρ
Continuum Mechanics 15/05/2013
Romain Teyssier
Application of the Bernoulli Theorem: Pitot tube
Ram pressure
We would like to measure the velocity of the fluid at infinity.
We consider a probe with section AC equal to section ED.
v2
The flow is stationary and incompressible: ρ + P = constant
2
Mass conservation implies vA SA = vD SD so that vA = vD = v∞
Point B, however, is a stagnation point with vB = 0 .
2
v∞
+ P∞ . Using the probe, we measure ∆P = PB − P∞
We conclude that PB = ρ
2
�
2∆P
The velocity is just v∞ =
and ∆P is called the ram pressure.
ρ
These probes (also called Pitot tube) are used in planes to measure the velocity.
Continuum Mechanics 15/05/2013
Romain Teyssier
Application of the Bernoulli Theorem: Venturi tube
We would like to measure the incoming velocity in a pipe.
We modify slightly the section of the pipe around point B.
Mass conservation implies vA SA = vB SB .
2
2
vB
vA
+ PB = ρ
+ PA
Bernoulli theorem implies ρ
2
2
Assuming that SB = SA (1 − �), if we measure ∆P , we have:
�
∆P
vA =
ρ�
This probe is called a Venturi tube.
Continuum Mechanics 15/05/2013
Romain Teyssier
Hugoniot theorem
For an stationary incompressible fluid, mass conservation implies vS = constant .
dv
dS
If the section decreases, the velocity increases
=− .
v
S
For a compressible fluid, we now have ρvS = constant .
dv dρ
dS
+
=−
v
ρ
S 1
The stationary Euler equation gives us vdv = − dP .
ρ
dP
2
Introducing the sound speed c =
,
�
�
dρ
dv
v2
dS
1
−
=
−
combining the 2 equations results in
v
c2
S
v
The dimensionless number M = is called the Mach number of the flow.
c
If M < 1 , the fluid behaves qualitatively like an incompressible fluid.
If M > 1 , it is reversed: the velocity will increase if the section increases.
Continuum Mechanics 15/05/2013
Romain Teyssier