12.005 Lecture Notes 24
Fluids (continued)
For a general stress-strain, for a Newtonian fluid
σ ij = − p ' δ ij + Dijkl ε kl
where Dijkl is viscosity tensor and ε kl is strain rate tensor.
For an isotropic fluid
σ ij = − p ' δ ij + λε kk δ ij + 2 µε ij
For volumetric strain rate
σ kk = −3 p '+ (3λ + 2µ )ε kk
σ kk
2
= − p '+ (λ + µ )ε kk
3
3
Mean normal stress:
2
where λ + µ is bulk viscosity.
3
2
For many applications, λ + µ = 0 (Stokes fluid)
3
2
3
σ ij = − p ' δ ij + 2 µε ij − µε kk δ ij
For many applications, ε kk
0
σ ij = − p ' δ ij + 2 µε ij
Often η is used for viscosity
σ ij = − p ' δ ij + 2ηε ij
Sometimes η → 0
(“perfect fluid”)
σ ij = − p 'δij
1
Material Derivative
Laws of physics – conservation of mass, conservation of energy, etc.
Express in reference frame of material, e.g. rod
T=0
T = T0
x
L
Figure 24.1
Figure by MIT OCW.
Steady state: T = T0 x / L;
Lagrangian frame: ρ c p
∂T
=0
∂t
∂T
= −k∇ 2T + A
∂t
Eulerian frame – material is moving. There would be a
∂T
for the above rod moving
∂t
through.
Marching band example.
Need to account for “non physical” change due to motion.
Above example:
∂T
∂T
∂T
where −v
is advection term.
= −v
∂x
∂x
∂t
2
Material derivative:
D ∂
= + v ⋅∇
Dt ∂t
Heat conduction
DT ∂T
=
+ v∇T = − k ∇ 2T + H
∂t
Dt
Conservation of Mass – Continuity Equation
Consider motion in x2 direction:
X3
dx3
ρv2
] dx
ρv2 + ∂ (ρv2) dx2
∂x2
1
dx2
X2
X1
Figure 24.2
Figure by MIT OCW.
Sides: mass in - mass out = -
∂
∂
(ρ v2 )dx2 dx1dx3 = (ρ v2 )dV
∂x 2
∂x 2
Front, back: mass in - mass out = -
∂
(ρ v1 )dV
∂x1
Top, bottom: mass in - mass out = -
∂
(ρ v3 )dV
∂x 3
3
For all 3 directions: -
∂
∂
∂ρ
∂
(ρ v1 )(ρ v2 )( ρ v3 ) =
∂x 2
∂x 3
∂t
∂x1
∂ρ ∂
+
(ρ vi ) = 0
∂t ∂x i
∂ρ
∂ρ
∂v
+ vi
+ρ i =0
∂t
∂x i
∂x i
Dρ
∂v
+ ρ i = 0 (Law of conservation of mass)
Dt
∂x i
For an incompressible fluid with constant properties
−∇p + µ∇ 2 v + ρ x = ρ
Dv
Dt
or, with ν ≡ µ / ρ (dynamic viscosity)
1 ∂p
Dv
∂2 vi
−
+ν
+ xi = i
ρ ∂xi
∂x j ∂x j
Dt
(Navier-Stokes equation)
“Plane strain”
V0
BOAT
X2
ν (water)
X1
Figure 24.3
Figure by MIT OCW.
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t = 0, v = 0, v( x1 = 0) = (0, v0 , 0)
only have v2 ≠ 0
∞ in x2 direction ⇒
∂
=0
∂x2
Subtract out hydrostatic
∂v2
∂2 v
= ν 22
∂t
∂x1
The solution becomes v = v0 (1 − erf
where erf (y) =
2
π
y
x1
2 νt
)
∫ e ξ dξ .
−
2
0
Velocity propagates downward a characteristic depth, x1 = 2 ν t .
Example: canoe 5 meters long,
v0 = 5 m/sec ⇒ t : 1 sec
water ν : 10 −2 cm2/sec ⇒ x1 : 2 10 −2 = 2 mm
A canoe will drag along about 2 mm water.
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