"XXIII.
FLUID DYNAMICS
By definition, a fluid is a material continuum that is unable to withstand a static shear stress. Unlike an elastic solid which responds to a shear stress with a recoverable deformation, a fluid responds with an irrecoverable flow. Examples of fluids include gases and liquids. Typically, liquids are considered to be incompressible, whereas gases are considered to be compressible. However, there are exceptions in everyday engineering applications. Fluid flow can be either laminar or turbulent. The factor that determines which type of flow is present is the ratio of inertia forces to viscous forces within the fluid, expressed by the nondimensional Reynolds Number, R= ρV D η
where V and D are a fluid characteristic velocity and distance. For example, for fluid flowing in a pipe, V could be the average fluid velocity, and D would be the pipe diameter. Typically, viscous stresses within a fluid tend to stabilize and organize the flow, whereas excessive fluid inertia tends to disrupt organized flow leading to chaotic turbulent behavior. Fluid flows are laminar for Reynolds Numbers up to 2000. Beyond a Reynolds Number of 4000, the flow is completely turbulent. Between 2000 and 4000, the flow is in transition between laminar and turbulent, and it is possible to find subregions of both flow types within a given flow field. Laminar fluid: An organized flow field that can be described with streamlines. In order for laminar flow to be permissible, the viscous stresses must dominating. Tubulent fluid: A flow field that cannot be described with streamlines in the absolute sense. However, time-averaged streamlines can be defined to describe the average behavior of the flow. In turbulent flow, the inertia stresses dominate over the viscous stresses, leading to small-scale chaotic behavior in the fluid motion. Newtonian fluid: A Newtonian fluid is a viscous fluid whose shear stresses are a linear function of the fluid strain rate.
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A.
Transport equation
The Leibnitz’s formula d dt
(b(t) (b(t)
f (x, t)dx =
a(t) a(t)
∂b ∂a ∂f (x, t) dx + f (b, t) − f (a, t), ∂t ∂t ∂t
can be generalized to d dt F (x, t)dV =
V (t) V (t)
∂F (x, t) dV + ∂t
F (x, t)v · ndA,
A(t)
where v is the velocity field of the time dependent movement of the surface A(t). This equation is called the general transport theorem. This can also be written in an equivalent form: d dt
B. Hydrostatics
F (x, t)dV =
V (t) V (t)
(
dF (x, t) + F (x, t) t
· v)dV
In this subsection we assume that we have a liquid of gas in equilibrium and there are no flows or currents. The stress tensor takes the form 1 σ (r) = −p(r)ˆ ˆ that is
this means that there are to shear forces. Newton’s equation in this case v = 0: ρf + that is ρf − p(r) = 0 · σ (r) = 0 ˆ
p00 ˆ σ (r) = − 0 p 0 00p
this is the fundamental equation of hydrostatics. Example: ideal gas ρ= M p RT
2
v dA
C.
Ideal Fluid
Newton’s equation for an ideal fluid ρ dv = ρf + dt · σ (r) ˆ
The total time derivative of the velocity is dvi ∂vi = + dt ∂t or in vector form ∂v dv = + (v · dt ∂t )v
3
j=1
∂vi ∂vi ∂xj = + ∂xj ∂t ∂t
3
j=1
∂vi vj ∂xj
this is sometimes called material derivative. Newton’s second law now for continuous medium ∂v + (v · ∂t )v = f − 1 p ρ
(this is valid for ideal fluid, no shear, no viscosity). Using the identity (v · we can also write it as ∂v + ∂t 12 v 2 −v×( × v) = f − 1 p ρ )v = 12 v −v×( 2 × v)
Note that the equation is nonlinear in the velocity field.
D.
Conservation of matter: The continuity equation
The mass in a volume V (t) m=
V (t)
ρdV
3
where both ρ and V (t) can change in time but they must do so in such a way that m is unchanged, it is contant in time d dm = dt dt By applying the general transport theorem: ∂ρ dV + ∂t ρv · dA = 0
A
ρdV = 0.
V (t)
V
(in other words, the change of matter in unit time in a given volume is equal to the net in/out flow through the surface of the volume). Using the divergence theorem ρv · dA =
A V
· (ρv)dV
and then we have
V
∂ρ + ∂t ∂ρ + ∂t
· (ρv) dV = 0
or · (ρv) = 0.
This is the continuity equation: ∂ρ + ∂t Equivalently, ∂ρ + ∂t ρ·v+ρ ·v =0 ·j = 0 j = ρv
where the first two terms are the total derivative of ρ, so dρ +ρ dt For incompressible fluids dρ =0 dt therefore · v = 0. ∂ρ =0 ∂t · v = 0.
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E.
Momentum conservation
Newton’s law reads as d dt ρvdV =
V (t) A(t)
σ ndA + ˆ
V (t)
ρf dV.
using the general transport equation, the left hand side can be rewritten as d dt so locally we can write dρv + ρv dt which can be simplified to ρ ∂v dv = ρ( + (v · dt ∂t )v) = ρf + σ ˆ · v = ρf + σ ˆ ρvdV =
V (t) V (t)
dρv + ρv dt
· v dV
this is Newton’s equation, also called as Euler’s equation of the non-viscous hydrodynamics.
F.
Viscous fluids
The drag force acting on the fluid lying below the element of area dA Fx = η where η is the viscosity coefficient (dimension ∂vx dA ∂y
m ) lt
The generalization of this simple relation
to the stress tensor states that the viscous contribution should be linear in the velocity gradients and should be symmetric. There are two appropriate combinations of the velocity gradient that satisfies this requirement ∂vl ∂vk + ∂xl ∂xk and δkl
j=1 3
∂vj = δkl ( ∂xj
· v).
Taking the linear combinations of these the viscous stress tensor is
v σkl = −η
∂vl 2 ∂vk + − δkl ( ∂xl ∂xk 3 5
· v) − ξδkl (
· v).
where η is the viscosity coefficient and ξ is the bulk (second) viscosity coefficient. For incompressible fluids · v = 0 so the second part is zero. The first part is constructed to
be traceless. For an incompressible shear flow v σkl = −η
0
∂vx ∂y
0
∂vx ∂y
0
0 0 00
gives back the starting formula.
G.
Navier-Stokes equation
The stress tensor of a viscous fluid is ˆˆ σ = σi + σv ˆ substituting this into Newton’s equation ρ and using the continuity equation ∂ρ + ∂t one can show that ∂v + (v · ∂t )v = f − 1 η p+ ρ ρ
2
dv = ρf + dt
σ ˆ
· (ρv) = 0
1 η v + (ξ + ) ( ρ 3
· v).
(1)
This is Euler’s equation for viscous fluids. For nonviscous fluids, this equation becomes ∂v + (v · ∂t )v = f − 1 η p+ ρ ρ
2
v.
(2)
which is the celebrated Navier-Stokes equation.
H.
Archimedes’ Principle
The (bouyant) force acting on a submerged object due to the pressure p in the fluid is Fb = 6 pdA,
where the surface integral is over the surface of the object. This can be rewritten (using the divergence theorem) as Fb =
V
pdV =
V
ρgdV = mg
where the volume integral is over the volume of the object, m is the mass of the fluid in the volume of the object, and we assume that the volume force is due to the gravitation. In a similar way, we can calculate the torque acting on the object: Γb = pr × dA = R × Fb
where R is the center-of-mass of the fluid in the volume of the object.
I.
Energy density
In this section we will derive an equation for the change of kinetic energy density in a given point. The kinetic energy density is defined as 1 ek = ρv2 . 2 Our starting equation is Newton’s equation ρ dv = ρf + dt · σ (r). ˆ
Multiplying this equation by v and using the identities vρ 1 dv2 dv = , dt 2 dt 7
1 d 2 ρv2 v2 dρ dek 1 dv2 ρ = − = + ek · v, 2 dt dt 2 dt dt ∂ek dek + ek · v = + · (ek v) dt ∂t
and v where D[v]ij = one can show that ∂ek + (ek v − σ v) = ρf · v − Trace(ˆ D[v]) ˆ σ ∂t (energy balance equation). The right hand side is the source. For an ideal fluid, σ v = −pv ˆ therefore ∂ek + ((ek + p)v) = ρf · v + p ∂t If the volume forces are conservative f =− u then ρfv = −ρv u = − (ρuv) + u (ρv) where the last term can be substituted using the continuity theorem ρfv = − (ρuv) + u Defining the energy density em = ek + ρu and the energy current density jm = (em + p)v we can have ∂em + ((em + p)v) = p · v. ∂t For incompressible fluids the energy source term is zero: ∂em + ∂t ((em + p)v) = 0. 8 ∂uρ ∂ρ = − (ρuv) + . ∂t ∂t · v. Trace(ˆ D[v]) = −p σ ·v ∂vi , ∂xj ·σ = ˆ · (ˆ v) − Trace(ˆ D[v]) σ σ
XXIV.
BERNOULLI’S EQUATION
Using Newton’s equation ∂v + ∂t and assuming that (1) rotationless fluid ×v=0 that is the velocity field can be derived from a scalar velocity potential: v=− Φ (2) conservative forces f =− u (3) incompressible fluid ρ = constant we have p 1 ∂φ + u + v2 − ρ 2 ∂t This equation can be readily integrated and we obtain 1 ∂φ p + u + v2 − =0 ρ 2 ∂t Bernoulli’s equation. =0 12 v 2 −v×( × v) = f − 1 p ρ
A.
Example: Steady flow in a channel
Assume that there is a steady flow in the z direction in a viscous fluid in a channel of width h v = v z e3 vz = vz (x, y) The Navier-Stokes equation in this case
2
∂vz =0 ∂t ∂vz =0 ∂z
vz = 9
1 ∂p η ∂z
where
∂p ∂z
should be independent of z (otherwise
∂vz ∂z
would not be zero). We assume
p = αz and then we have
2
vz =
α . η
This is a Poisson equation for vz . Specifying the boundary conditions vz |boundary = 0 this equation can be solved. Assuming a channel of width h in the x direction vz (x = −d/2) = vz (x = d/2) = 0 the solution is vz (x) = This is a parabolic velocity profile. α12 ( d − x2 ). 2η 2
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