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Governing Equations and Const i tut ive Relat ionships for Biolog i ca l ly Relevant Fluids
While the balance laws are no different from the equations derived in chapter 3, it is
sometimes useful to consider alternative forms. In fluid mechanics, the balances of mass,
linear momentum, and angular momentum are usually written,
,
or, in indicial form,
.
For the first law of thermodynamics and the Clausius-Duhem form of the second law of
thermodynamics, we have,
!ρ + ρ∇iv =
∂ρ∂t
+∇ρiv + ρ∇iv =∂ρ∂t
+∇iρv = 0
ρ !v = ρ ∂v
∂t+∂v∂x
v⎛⎝⎜
⎞⎠⎟= divT + ρb
T = TT
!ρ + ρ
∂vi
∂xi
=∂ρ∂t
+∂ρ∂xi
vi + ρ∂vi
∂xi
=∂ρ∂t
+∂ρvi
∂xi
= 0
ρ !vi = ρ
∂vi
∂t+∂vi
∂x j
v j
⎛
⎝⎜
⎞
⎠⎟ =
∂Tij
∂x j
+ ρbi
Tij = Tji
ρ !ε = ρ ∂ε
∂t+∇εiv⎛
⎝⎜⎞⎠⎟= TiD + ρr − divq
2
,
which may also be represented in indicial form,
,
ρ0 !η = ρ0
∂η∂t
+∇ηiv⎛⎝⎜
⎞⎠⎟=ρ0rθ
−Divq0θ
ρ !ε = ρ ∂ε
∂t+∂ε∂xi
vi⎛⎝⎜
⎞⎠⎟= TijDij + ρr − ∂qi
∂xi
ρ0 !η = ρ0
∂η∂t
+∂η∂xi
vi⎛⎝⎜
⎞⎠⎟=ρ0rθ
−1θ∂q0i∂Xi
3
Newtonian Flows and the Navier-Stokes Equations In fluid mechanics, many texts derive the balance laws with the inherent assumption
that the flow is Newtonian, making it difficult for students to derive the governing equations
for more complicated rheological behaviors. And, while Newtonian flows are important,
there are numerous examples in biomedical engineering where the flows are distinctly non-
Newtonian. These include, but are not limited to turbulence in the upper airways, mucus and
bile flows, and flow through the gut (Table 8.1). We will start by considering incompressible
and compressible Newtonian flows and illustrate the process for obtaining the Navier-Stokes
equations. Subsequently, we will consider nonlinear fluids and utilize the same process to
derive the governing equations.
Table 8.1 – Examples of biological fluid flows and their characteristics
Flow Flow Regime Linear vs.
Nonlinear
Compressibility
Blood in ?? mm arteries Laminar Nonlinear Incompressible
Blood in ?? mm arteries Laminar Linear Incompressibility
Plasma skimming layer Laminar Linear Incompressible
Upper airways Turbulent Nonlinear Compressible
Lower airways Laminar Linear Slightly compressible
Sweat/tears Laminar Linear Incompressible
Mucus Laminar Nonlinear Incompressible
Bile Laminar Nonlinear Incompressible
Synovial fluid Laminar Linear Incompressible
Bioreactors Laminar or turbulent Linear Incompressible
Flow through the gut Laminar Nonlinear Incompressible
Aqueous and vitreous
humors of the eye
Laminar Linear Incompressible
Urine Laminar Linear Incompressible
Canaliculi Laminar Linear Incompressible
Interstitial flows Porous media Darcy flows Incompressible at
microscale, compressible
at macroscale
4
For incompressible, Newtonian fluids, the constitutive law has the form,
T = − pI + 2µD
where p is the fluid pressure and µ is the viscosity. Clearly this equation satisfies the required
invariance under superposed rigid body motions. Inserting it into our governing differential
equations, we note that the Balance of Mass equation is unchanged. But, introducing this
equation into the Balance of Linear Momentum yields,
ρ ∂v
∂t+∂v∂x
v⎛⎝⎜
⎞⎠⎟= −∇p + 2µdiv D( ) + ρb ,
or, in component form,
ρ
∂vi
∂t+∂vi
∂x j
v j
⎛
⎝⎜
⎞
⎠⎟ = −
∂p∂xi
+ µ∂2vi
∂x j2 +
∂2v j
∂xi∂x j
⎛
⎝⎜
⎞
⎠⎟ + ρbi .
The Balance of Angular Momentum is identically satisfied. THERMO
If we then decide to consider a constitutive law that allows for compressibility of the fluid,
we might choose something like,
,
where p is the fluid pressure, λ is the second viscosity coefficient (sometimes called the
dilatational viscosity or the volume viscosity), and µ is the dynamic viscosity. To
demonstrate that this satisfies invariance requirements, we first recall the definition of a
superposed rigid body motion,
T = − pI + λtr D( )I + 2µD
5
While Newtonian fluids can be used to model a wide range of biomedically relevant
fluids including, but not limited to, plasma, tears, cell culture medium, air in the lower
airways, and lymph. But, to describe flows in the upper airways, and the flows of mucus and
bile, and flow through the gut, it is necessary to introduce more complicated constitutive
laws. First, we will consider the potential variables that can be included in the constitutive
laws and then examine the possible ways that they can be combined.
Bingham Fluids
Need to check the tau term. Is it invariant? (Actually a discontinuous
change at tau_0). How do you characterize the objectivity of that? Should be okay
Power Law Fluids
others??
T = − pI + τ0 + 2µD
T = − pI + kDn
6
Planar Couette Flow Couette type flows are generated by one surface moving past another with a layer of
fluid in between (Fig. ??). One of the best biological applications is the lubrication layer of
fluid that separates two diarthroidal joints in the absence of actual contact between the two
joints. While the actual process of cartilage lubrication is considerably more complicated,
Couette flow provides a good place to start.
Figure ?? (A) When the knee is extending, the lower leg moves relative to the femur and the two cartilage surfaces move past each other, creating a thin layer of fluid lubrication (exaggerated in the exploded view). While the actual mechanisms of joint lubrication are considerably more complicated, and the joint exhibits some level of curvature, planar Couette flow provides an approximation for flows between two cartilage surfaces (B). Adapted from an image initially created by Patrick J. Lynch, medical illustrator.
7
For planar flows of this type, we begin by describing the velocity field using
Cartesian coordinates,
,
where forms the usual fixed orthonormal basis. The components of the acceleration
vector can then be written by employing the chain rule for differentiation,
,
and the rate of deformation tensor has the general form,
.
For planar Couette flow it is usually assumed that the flow has no velocity components in
the E2 and E3 directions, there is no variation in the E3 direction (i.e. partial derivatives with
respect to x3 are negligible) and that the flow is steady. The latter assumption implies that the
partial derivatives with respect to time are identically zero. One may certainly argue that,
over the time span that most joint loading problems occur, the assumption of steady state
flow simply cannot apply. But it is useful starting point and will at least provide us with
v = v1E1 + v2E2 + v3E3
Ei{ }
a1 =∂v1
∂t+∂v1
∂x1
dx1
dt+∂v1
∂x2
dx2
dt+∂v1
∂x3
dx3
dt
a2 =∂v2
∂t+∂v2
∂x1
dx1
dt+∂v2
∂x2
dx2
dt+∂v2
∂x3
dx3
dt
a3 =∂v3
∂t+∂v3
∂x1
dx1
dt+∂v3
∂x2
dx2
dt+∂v3
∂x3
dx3
dt
Dij =12
2∂v1
∂x1
∂v1
∂x2
+∂v2
∂x1
∂v1
∂x3
+∂v3
∂x1
∂v2
∂x1
+∂v1
∂x2
2∂v2
∂x2
∂v2
∂x3
+∂v3
∂x2
∂v3
∂x1
+∂v1
∂x3
∂v3
∂x2
+∂v2
∂x3
2∂v3
∂x3
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
8
approximations of the actual flow. Given these assumptions, the acceleration field simplifies
to,
and the rate of deformation tensor can be written as,
9
Once we have simplified the kinematics as far as we dare, the next step is to apply the balance of mass equation which, in Cartesian coordinates, takes the form,
,
and when we employ the previously discussed kinematic assumptions it becomes,
!ρ + ρdivv = !ρ + ρ
∂v1
∂x1
+∂v2
∂x2
+∂v3
∂x3
⎛
⎝⎜⎞
⎠⎟= 0
10
The next step is to introduce our constitutive law. For all the archetypal flows
discussed in this section, we will utilize the relationship for an incompressible Newtonian
fluid, , where µ is the fluid viscosity. The components of the stress tensor
can then be written as,
where p is the hydrostatic pressure. It is now possible to incorporate the results so far into
the balance of linear momentum, ,
,
T = − pI + 2µD
ρ !v = divT + ρb
0 =∂T11
∂x1
+∂T12
∂x2
+∂T13
∂x3
= −∂p∂x1
+ µ∂2v1
∂x22
0 =∂T21
∂x1
+∂T22
∂x2
+∂T23
∂x3
= µ∂2v1
∂x1∂x2
−∂p∂x2
0 =∂T31
∂x1
+∂T32
∂x2
+∂T33
∂x3
= −∂p∂x3
11
It is important to note that, at this stage, it is usually assumed that the velocity of the moving surface propagates into the body of the liquid because of its viscosity, generating the flow
field without any external pressure gradient, (i.e. ). Provided that this is a valid
assumption, the pressure gradient in every direction is zero and the total fluid pressure must remain constant. The remaining equation can then be integrated directly with respect to x2,
,
where A1 and A2 are constants of integration that can be solved by applying the appropriate
boundary conditions,
,
which ultimately leads to,
.
This result is interesting for a few reasons. First, the velocity in the E1 direction is a linear
function of the x2 coordinate. Second, even though the flow field initially propagates due to
viscous interactions, the fluid viscosity is not part of the final solution. It should be noted,
however, that viscosity will play a role in a number of different aspects of the flow as it
develops including the amount of force required to accelerate the moving surface and how
quickly the flow field reaches steady state.
∂p∂x1
= 0
∂2v1
∂x22 = 0
∂v1
∂x2
= A1
v1 = A1x2 + A2
v1 x2 = 0( ) = 0
v1 x2 = h( ) =U0
v1 =
x2
hU0
12
Planar Poiseuille Flow Pressure-driven flow between two plates has two important biomedical applications.
The first is flow through the alveolar capillaries where the blood vessels appear to be
stretched around the alveolus and form two nearly parallel plates through which advected
transport of oxygen and carbon dioxide occurs almost continuously. The second application
is the parallel plate flow chamber, introduced in chapter ??. These devices are commonly
used to measure cell adhesion and to evaluate the effects of fluid shear on the short-term
response of adherent and suspended cells.
A
Figure ?? Parallel
plate flow chambers
were introduced in
chapter 1 (A) and
they are used to
measure cell adhesion
and characterize the
cellular response to
fluid shear stress.
Planar Poiseuille flow
(B) approximates flow
through parallel plate
flow chambers and
the capillaries
surrounding alveoli.
B
,
v = v1E1 + v2E2 + v3E3
13
where forms the usual fixed orthonormal basis. The general components of the
acceleration vector are,
,
and the rate of deformation tensor has the general form,
.
Ei{ }
a1 =∂v1
∂t+∂v1
∂x1
dx1
dt+∂v1
∂x2
dx2
dt+∂v1
∂x3
dx3
dt
a2 =∂v2
∂t+∂v2
∂x1
dx1
dt+∂v2
∂x2
dx2
dt+∂v2
∂x3
dx3
dt
a3 =∂v3
∂t+∂v3
∂x1
dx1
dt+∂v3
∂x2
dx2
dt+∂v3
∂x3
dx3
dt
Dij =12
2∂v1
∂x1
∂v1
∂x2
+∂v2
∂x1
∂v1
∂x3
+∂v3
∂x1
∂v2
∂x1
+∂v1
∂x2
2∂v2
∂x2
∂v2
∂x3
+∂v3
∂x2
∂v3
∂x1
+∂v1
∂x3
∂v3
∂x2
+∂v2
∂x3
2∂v3
∂x3
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
14
For planar Poiseuille flow, we typically assume that there is no flow in the E2 or E3
directions and that the flow is steady. This implies that the acceleration field can be
simplified considerably,
and the rate of deformation tensor takes the form,
In addition, we can make the obvious assumption that the velocity gradients in the E3
direction should also be zero, which yields,
15
Once again, we have simplified the kinematics as far as possible, the next step is to apply the balance of mass equation which, in Cartesian coordinates, takes the form,
,
and when we employ the previously discussed kinematic assumptions it becomes,
.
.
For an incompressible, Newtonian fluid, the constitutive law has the form, ,
where µ is the fluid viscosity. Inserting the rate of deformation tensor into this expression
yields,
!ρ + ρdivv = !ρ + ρ
∂v1
∂x1
+∂v2
∂x2
+∂v3
∂x3
⎛
⎝⎜⎞
⎠⎟= 0
!ρ + ρ
∂v1
∂x1
⎛
⎝⎜⎞
⎠⎟= 0
Dij =
012∂v1
∂x2
0
12∂v1
∂x2
0 0
0 0 0
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥
T = − pI + 2µD
16
This is precisely the same form that we obtained for planar Couette flow. This implies that
any difference we observe in the final solution is due to the differences in the
boundary conditions. When we incorporate the results so far into the balance of linear
momentum, , we obtain,
,
provided we assume that the viscosity is constant. Recall that the balance of mass provided
us with the restriction, , which requires that the term must also be
identically zero,
ρ !v = divT + ρb
0 =∂T11
∂x1
+∂T12
∂x2
+∂T13
∂x3
= −∂p∂x1
+ µ∂2v1
∂x22
0 =∂T21
∂x1
+∂T22
∂x2
+∂T23
∂x3
= µ∂2v1
∂x1∂x2
−∂p∂x2
0 =∂T31
∂x1
+∂T32
∂x2
+∂T33
∂x3
= −∂p∂x3
∂v1
∂x1
= 0 µ
∂2v1
∂x1∂x2
17
.
Adding them together provides B2,
,
and subtracting them gives us B1 = 0 so that the final solution has the form,
.
Unlike the solution for planar Couette flow, this one is parabolic and depends expressly on
the viscosity of the fluid.
v1 x2 = −h2
⎛⎝⎜
⎞⎠⎟=
18µ
∂p∂x1
h2 −h2
B1 + B2 = 0
v1 x2 =h2
⎛⎝⎜
⎞⎠⎟=
18µ
∂p∂x1
h2 +h2
B1 + B2 = 0
14µ
∂p∂x1
h2 + 2B2 = 0
B2 = −1
8µ∂p∂x1
h2
v1 =
12µ
∂p∂x1
x22 −
18µ
∂p∂x1
h2 =1
8µ∂p∂x1
4x22 − h2( )
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