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Often a power-law is assumed:
n− 1
F
η = K
D
,
(8.2)
where n = 1 corresponds to a Newtonian fluid, n> 1 a shear-thickening
fluid where apparent viscosity increases as you try to deform the fluid
faster (e.g., cornstarch suspended in water), and 0 <n< 1ashear-
thinning fluid where apparent viscosity decreases (e.g., paint). Granular
materials such as sand can even be modeled as the limit n =0ofshear-
thinning where the magnitude of “viscous” stress depends instead on pres-
sure, not the magnitude of the strain rate, making it more akin to dry
Coulomb friction; see Zhu and Bridson [Zhu and Bridson 05] for more on
this subject.
Getting back to simple Newtonian fluids, the relationship for incom-
pressible flow is
τ =2 ηD + λ tr( D ) δ,
(8.3)
where η is the coecient of dynamic viscosity . The second term, involv-
ing tr( D )=
u , is of course zero for incompressible flow for any λ
(which is termed the second coecient of viscosity ). For compressible flow
λ is often taken to be
∇·
2
3 η , though theoretically this is only an ideal-
ization of monatomic gases. However, for an incompressible fluid we are
free to choose λ as we please, 6
and thus for simplicity's sake we'll set
λ =0.
Plugging this into the momentum equation, we get
ρ ∇· η ( ∇u + ∇u T ) .
Du
Dt
+ 1
1
ρ ∇p =
(8.4)
You may notice that this isn't quite the same as our first statement of the
momentum equation, Equation (1.1). It turns out that for the common
case where η is constant, the correct equation (8.4) does in fact simplify to
Equation (1.1): but be aware, for simulations with variable viscosity, only
Equation (8.4) is correct. For example, Equation (1.1) doesn't conserve
angular momentum in the variable viscosity case.
2
3 η , basic thermodynamics are violated, with viscosity
actually accelerating expansion or contraction, increasing the energy of the system. In a
numerical method, where divergence probably isn't exactly zero even for incompressible
flow, problems are bound to arise.
6 This isn't quite true: for λ<−
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