Viscous Fluids
Amanda Meier
December 14th, 2011
Abstract
Fluids are represented by continuous media described by mass density, velocity and pressure. An Eulerian description of uids focuses on
the transport of a small xed volume. Real uids have the ability to resist
stress and are known as viscous uids. Motion of these uids is governed
by the Euler equation, where the Navier-Stokes equation is a simplication of the Euler equation due to imcompressibilty. These equations are
derived from conservation laws of mass, momentum and energy.
1
1 Introduction
Viscosity is a measure of the resistance of a uid to a stress and diers among
materials that may have similar properties such as density and weight.
For
example, water ows very dierently than oil. In order to analyze this, we can
think of placing a material between two plates, where the bottom plate is xed
and the top plate is allowed to move.
If we have a uid between the plates
and we apply a pressure to the upper plate, the plate will move continuously
with a certain velocity. The uid in contact with the upper plate moves with a
specic velocity, but the uid in contact with the lower plate has a zero velocity
due to the solid boundary condition on the lower plate. This tells us that the
uid between the plates developes a velocity gradient. The viscosity is the uid
property that relates the shearing stress, or pressure per area, to the velocity
gradient. If the shearing stress is linearly related to the velocity gradient, then
the uid is known as Newtonian, whereas if the shearing stress is not linearly
related to the velocity gradient, the uid is Non-Newtonian. In order to understand how the velocity varies, we need to be able to examine an innitesimal
volume within the ow, which we do through dierential analysis. The equations of motion for uids are governed by conservation of mass, momentum and
energy. For incompressible, Newtonian uids, the Navier-Stokes equation is the
dierential equation of motion. What follows is the derivation of these equations
and their use in a simple example example describing uid ow in a pipe.
2 Derivations
2.1
Conservation Laws
In order to look at what happens to a particular part of a uid as it moves
through a system we need to look at the changes of an extensive property of
the uid over a control volume, which is done using the Reynolds Transport
theorem. The theorem states that these changes are equal to what is lost (or
gained) through the boundary of the volume and what is created (or consumed)
by the sources or sinks in the control volume. This is shown in the following
integral where L is an extensive property of the uid,
Ω
is the control volume,
v is the velocity and Q is the sink or source in the uid.
d
dt
ˆ
ˆ
ˆ
L dV = −
Lv · n dA −
Ω
∂Ω
Q dV
(1)
Ω
The divergence theorem can then be applied to change the surface integral,∂Ω,
to a volume integral,Ω.
d
dt
ˆ
ˆ
L dV = −
Ω
ˆ
∇ · (Lv) dV −
Ω
Q dV
(2)
Ω
Applying Leibniz's rule for dierentiation under an integral we get the following.
2
ˆ
Ω
∂L
dV = −
∂t
ˆ
ˆ
∇ · (Lv) dV −
Ω
Q dV
(3)
Ω
And then combining the integrals we get,
ˆ Ω
∂L
+ ∇ · (Lv) + Q
∂t
dV = 0
(4)
where the integrand is equally zero.
∂L
+ ∇ · (Lv) + Q = 0.
∂t
(5)
This relationship gives us a general continuity equation in which we can then
examine the conservation of mass, momentum and energy.
2.1.1 Conservation of Mass
The continuity equation that describes liquids arises from looking at the conservation of mass. Assuming
at the mass density,
ρ
Q=0
or no sources or sinks of mass and looking
is the extensive property, the general continuity equation
becomes
∂ρ
+ ∇ · (ρv) = 0
∂t
For an incompressible uid,
ρ
(6)
is constant and this equation simplies to
∇·v =0
(7)
which is a statement of conservation of volume.
2.1.2 Conservation of Momentum
Conservation of momentum leads to discussion of a stress tensor and euler's
equations of motion for the uid. Applying the continuity equation (Eq. 5) to
momentum,
ρv ,
Expanding the
we get the following equation.
∂
(ρv) + ∇ · (ρvv) + Q = 0
∂t
derivatives and taking b to be a body
(8)
force as a sink or source
of momentum,
∂ρ
∂v
v+ρ
+ ∇(ρv) · v + ρv∇ · v = b
∂t
∂t
(9)
we have expanded the divergence of a second rank tensor which gives a rst
rank tensor
∂v
∂ρ
v+ρ
+ ∇(ρ)v · v + ρ∇(v) · v + vρ∇ · v = b
∂t
∂t
3
(10)
if we rearrange this equation to be of the form
∂ρ
∂v
+ρ
+ vv · ∇ρ + ρv · ∇v + ρv∇ · v = b
∂t
∂t
out v and ρ,
∂ρ
∂v
v
+ v · ∇ρ + ρ∇ · v + ρ
+ v · ∇v = b
∂t
∂t
v
and pulling
v · ∇ρ + ρ∇ · v = ∇ · (ρv).
∂ρ
∂v
+ ∇ · (ρv) + ρ
+ v · ∇v = b
∂t
∂t
(11)
(12)
we can then recognize
v
(13)
The left most term in Eq. 13 is the conservation of mass we saw earlier that is
equal to zero (Eq. 6,7). Therefore, we are left with
ρ
where
D
Dt
∂v
+ v · ∇v
∂t
⇒
=b
ρ
Dv
=b
Dt
(14)
is the convective derivative. This equation is just an expression for
Newton's second law where we have body forces instead of point forces. And
Euler's equations of motion follow directly from this, where
b
can be broken up
into two terms, one describing forces resulting from stresses (σ is stress tensor)
and the other describing forces such as gravity (f ).
ρ
Dv
=∇·σ+f
Dt
(15)
2.1.3 Conservation of Energy
Conservation of energy will lead to information pertaining to the thermodynamics of the system, which I am choosing not to focus on in this paper. For
those interested, the extensive property energy density which has two parts, the
1
2
2 ρv , associated with the center-of-mass motion, and the
internal energy density, ρ, where is the internal energy per unit mass will be
kinetic energy density,
used as
2.2
L
in Eq. 5.
Viscous Stress Tensor
If we have a uid in a steady shear ow, where
vx
increases linearly with
v = vx (y)x̂
we have incompressible ow because the divergence is zero.
acting on uid lying below area
dA
y,
(16)
The drag force
is then
Fx = η
∂vx
dA
∂y
4
(17)
where
η
is the viscosity. Refer to Fig. 1 for a schematic on viscous drag force.
Figure 1: Viscous drag force exerted by uid above surface y=y0 on uid below
for linear shear ow.
Since viscosity contributes to the surface stress, it is added in to the stress
tensor. Newton's second law for a xed volume, subject to an external force
f
is as follows
ˆ
V
3
X
∂ (ρvk ) 3
d x=−
∂t
ˆ
ˆ
A
l=1
Tkl dA +
ρfk d3 x
(18)
V
And in dierential form this is exactly what we got out of applying the continuity
equation to momentum (Eq. 15).
3
X ∂Tkl
∂ (ρvk )
=−
+ ρfk
∂t
∂xl
(19)
l=1
The symmetric stress tensor for nonviscous or ideal uids has the form
ideal
Tkl
= ρvk vl + pδkl
Where
δkl
(20)
is the Kronecker delta and we just need to add the part that describes
the viscous forces. Since the uid is stationary, i.e. isotropic with no preferred
spatial direction, the only terms that remain in the stress tensor are
5
∂vk
∂vl
+
∂xl
∂xk
(21)
3
X
∂vj
= δkl (∇.v)
∂xj
j=1
δkl
(22)
We assume the viscous stress tensor is a linear combination of these two.
v
Tkl = −η
η
Where
∂vk
∂vl
2δkl ∇.v
+
−
∂xl
∂xk
3
is the viscosity and
ζ
− ζδkl (∇.v)
(23)
is the bulk viscosity. This simplies for incom-
pressible uids where the divergence of the velocity is zero meaning the bulk
viscosity no longer contributes.
v
Tkl
= −η
∂vl
∂vk
+
∂xl
∂xk
(24)
So for our example shown in Figure 1, the viscous stress tensor becomes the
following.
v
v
Txy = Tyx = −η
∂vx
∂y
(25)
The total stress tensor is then the sum of the ideal and viscous stress tensors.
Tij = Tijideal + Tijv
Tij = ρvi vj + pδij − η
∂vi
∂vj
2δij ∇.v
+
−
∂xj
∂xi
3
(26)
− ζδij (∇.v)
(27)
Using the expression for Newton's second law (Eq. 18) for a particular stress
tensor.
3
X
∂ (ρvi vj )
j=1
∂xj
+
3
3
X
∂ (ρvi ) η X ∂ 2 vj
∂ 2 vi
∂p
= ζ+
+η
+ ρfi −
2
∂t
3 j=1 ∂xi ∂xj
∂x
∂x
j
i
j=1
(28)
The left side simplies with the continuity equation (Eq. 5)
3
X
∂ (ρvj )
j=1
∂xj
+
∂ρ
=0
∂t
(29)
to give the following result.
3
X
j=1
vj
3
3
∂vi
1
η X ∂ 2 vj
η X ∂ 2 vi
1 ∂p
∂vi
+
=
ζ+
+
+ fi −
∂xj
∂t
ρ
3 j=1 ∂xi ∂xj
ρ j=1 ∂xj 2
ρ ∂xi
6
(30)
And in vector form we get the generlized Euler equation for a vicous uid.
(v.∇)v +
η
η
∂v
1
1
ζ+
∇(∇.v) + ∇2 v + f − ∇p
=
∂t
ρ
3
ρ
ρ
(31)
With the simplication for incompressible ow, we get the Navier-Stokes equation.
(v.∇)v +
∂v
η
1
= ∇2 v + f − ∇p
∂t
ρ
ρ
υ,
Where we dene the kinematic viscosity,
(32)
as the ratio of viscosity to mass
density.
ν=
η
ρ
(33)
3 Important Quantities in Characterizing Flow
3.1
Reynolds number
The Reynolds number is important in characterizing the relative eect of inertial
forces to viscous forces, so the dimensionless parameter is dened as the ow
velocity,
viscosity,
v , times
ν = ηρ .
the characteristic length scale,
Re =
L,
divided by the kinematic
vL
ν
(34)
A Reynolds number much less than 2100 describes completely laminar or smooth
ow. A Reynolds number around 2100 is where vortices begin to appear. And
a Reynolds number much greater than 2100 describes a regime of turbulence.
3.2
Mach number
Another dimensionless parameter that is important in characterizing the uid
ow is the mach number which is ratio of the speed of an object moving through
a uid to the speed of sound in that uid for its particular physical conditions.
Specically it is dened as the ow velocity,
cs =
√
c
kRT ,where k = cvp
v,
divided by the speed of sound
is the ratio of specic heat at constant pressure to
specic heat at constant volume,
R
is the universal gas constant, and
temperature.
Ma =
v
cs
T
is the
(35)
The uid is considered compressible for a Mach number greater than 0.3 and
incompressible for a Mach number of less than 0.3. Since we are already assuming incompressible ow in deriving the equations we do not need to look further
at this parameter.
7
3.3
Knudsen number
The Knudsen number is helpful in classifying how rareed a ow is or how far
apart the molecules of the ow are relative to the scale of the ow. It is dened
√ 1
, where d is the diameter of the
2πd2 n
is the density of the gas, divided by the characteristic length
λ=
as the mean free path of the atom,
gas molecule and
scale of the ow,
n
L.
Kn =
A Knudsen number of less than
10−3
the Navier-Stokes Equations, between
conditions, between
than
101
10
−1
and
1
10
λ
L
(36)
indicates no-slip boundary conditions in
10−3
and
10−1
indicates slip boundary
indicates slip transitional ow, and greater
indicates free molecular ow. The Knudsen number can be related to
the Reynolds and Mach number by the following, where we once again have the
ratio of specic heats included .
Ma
Kn =
Re
r
kπ
.
2
(37)
4 Boundary Conditions
Very important in understanding a physical situation are the boundary conditions of the system. Some examples are as follows:
1. For slow motion, a viscous uid in contact with a moving surface moves
with the velocity of the surface, whereas the velocity is zero for a viscous
uid in contact with a xed surface.
2. On a surface separating two uids the velocities have to be the same in
order to maintain continuity of the velocity and its derivatives.
3. Continuity of force requires Eq. 38 to be continuous across any surface,
including the one separating two uids.
3
X
(pδij + Tijv )n̂
(38)
j=1
4. At a free surface adjoining a vacuum, the tangential component of velocity
is not restricted and the nonviscous or ideal condition of zero pressure in
the comoving frame is given by the following.
3
X
(pδij + Tijv )n̂ = 0
j=1
8
(39)
5 Example: Fluid ow in channel or pipe
5.1
Analytic Solution From Navier-Stokes Equation
Figure 2: Steady viscous ow in a channel of width d.
Following the steady ow in a channel or pipe example in Fetter and Walecka,
an incompressible uid owing in the z-direction has the following ow velocity.
v(r, t) = vz (r)ẑ
(40)
The assumption that ow is invariant in the z-direction is in order to have steady
ow.
∂vz
=0
∂t
(41)
∂vz
=0
∂z
(42)
Where the Navier-Stokes equations for incompressible ow reduce down to two
terms.
∇2 vz =
1 ∂p
η ∂z
(43)
Assuming a uniform pressure gradient, we get the total change in pressure over
length, where the minus sign comes from a decreasd pressure along the direction
of ow.
9
∂p
∆p
=−
∂z
L
(44)
Then the velocity satises the 2-D Poisson equation.
∇2 vz = −
With the boundary condition that
vz
1 ∆p
η L
(45)
must be zero at the walls of the channel,
the solution to the dierential equation is given by,
vz (x) = −
∆px2
+c
2ηL
(46)
and by solving forc at the boundaries, refer to Fig. 2,
vz
d
±
2
=0
(47)
The result is a parabolic prole that is a maximum at the center and decreases
to zero at the walls,
∆p
2ηL
vz (x) =
d2
− x2
4
(48)
which the drag force follows from.
Fid = −
3 ˆ
X
j=1
ˆ
A
Tij dA = −
dydz
Tix |x=d/2 − Tix |x=−d/2
(49)
The pressure contributions cancel and leave only the viscous stress.
Tzx = −η
∂vz
∆px
=
∂x
L
(50)
So the total drag force is simplied to the following.
Fzd
=−
d
2
A
− − d2 ∆p
= −∆pY d
L
(51)
And the total mass ow through the channel per unit length in the y-direction
is as so.
ˆ
0
d
2
M =ρ
vz (x) dx =
−d
2
d3 ∆p
12Lv
(52)
In order to avoid edge corrections is it common to examine a long pipe with
a circular cross section of radius
R
where the velocity if axisymmetric.
The
equations for velocity prole, drag force and mass owrate are then as follows.
vz (r) =
∆p
R2 − r 2
4ηL
10
(53)
Fzd = −∆pπR2
M0 =
(54)
πR4 ∆p
8Lv
(55)
Eq. 55 is the Poiseuille formula, that gives the pressure drop in the uid owing through a long cylindrical pipe and is only valid for viscous, laminar, and
incompressible ow. One simplication of the above velocity prole is to look
at the centerline velocity, dened as the change in pressure times the diameter
of the pipe squared, divided by the dynamic viscosity times the characteristic
length of ow.
Vc (∆P, µ, l) =
∆p D 2
16µL
(56)
And the average velocity is the owrate divided by the cross-sectional area.
V =
5.2
πR4 ∆p
8Lv
πR2
=
∆p D 2
Vc
=
32µL
2
(57)
Numerical Solution
In most physical situations, there are not as many simplications and we are
not able to solve the Navier-Stokes equation analytically.
Numerical analysis
in software like COMSOL (nite element analysis software) is able to solve the
Navier-Stokes equation with an input set of boundary conditions dened by the
given parameters above (the Mach, Knudsen, and Reynold numbers).
5.2.1 Specic Boundary Conditions
In COMSOL, the ow is dened to be laminar and incompressible, with noslip boundary conditions (zero velocity at xed walls of channel) to model get
the correct form of the Navier Stokes equation the program will solve based
on the ow parameters of our system (usually dened by the Reynolds, Mach
and Knudsen numbers). We then have to draw the channel geometry we want
and give input parameters, such as the specic uid we are owing down the
channel and inow and outow velocities or pressures it has. A mesh or grid is
dened over the entire geometry that examines the velocity in dierent volumes
specied by the mesh size (ne, coarse, etc.). If the mesh is Physics Controlled,
there is usually a smaller grid size closer to the boundaries of the system, so the
walls of the channel in our system. The solver is then run, and the results show
a velocity prole in the specied geometry. Referring to Fig. 3, we can see the
parabolic velocity prole in the channel that matches the analytic description
we found above.
11
Figure 3: Velocity magnitude and eld prole of ow through channel.
More complicated systems can then be input to COMSOL and convergence
studies run to make sure the equations and boundary conditions make physical
sense. Having an analytic check with the pipe example helps in understanding
what the nite element analysis is doing with the specied input and boundary
conditions.
6 Conclusion
Derivation and examinination of the Euler and Navier-Stokes equations leads
to understanding of the continuous motion of viscous uids.
Once equipped
with the equations of motion, it is important to know the ow regime and
physical boundaries of the system in order to simplify the equations of motion
as much as possible. When we simplied our ow to be viscous, laminar, and
incompressible, we were able to get an analytic description of the velocity, drag
force and mass ow of the uid. Comparison with numerical analysis veried
our analytic solution.
Now that the most simple case of viscous uid ow is
understood, we can analyze more complicated uid systems which are more
prevalent in the real world.
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References
Theoretical Mechanics
of Particles and Continua. Dover Publications, 2003.
[1]
Alexander L. Fetter and John Dirk Walecka.
[2]
Nam-Trung Nguyen and Steven T. Wereley.
[3]
Bruce R. Munson, Donald F. Young and Theodore H. Okiishi.
Fundamentals and Applications fo Microuidics, 2nd Edition. Artech House, 2006.
Fundamentals of Fluid Mechanics, 4th Edition. John Wiley & Sons,
2002.
[4]
Ain A. Sonin
Equations of Motions for Viscous Fluids, 8th Edition.
Massachusetts Institute of Technology, 2001.
[5]
COMSOL Multiphysics, registered trademark of COMSOL AB,
Stockholm, Sweden
[6]
Derivation
of
the
Navier-Stokes
Equations
http://en.wikipedia.org/wiki/Derivation_of_the_Navier%E2%80%93Stokes_equations
Date accessed: December 2011
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