Poiseuille Flow Through a Duct in 2D
Assumptions:
•
Velocity is independent of
0
,
=
∂
∂
=
∂
∂
x
v
x
u
x
•
Incompressible flow
•
Constant viscosity,
µ
•
Steady
•
Pressure gradient along length of pipe is nonzero, i.e.
0
≠
∂
∂
x
p
Boundary conditions:
•
No slip:
()
0
walls are not moving
0
uy
h
vy
h
=±
=
⇐
=
To be clear, we now will take the compressible, unsteady form of the NS
equations and carefully derive the solution:
Conservation of mass:
0
)
(
=
⋅
∇
+
∂
∂
V
t
K
ρ
But
0
=
∂
∂
t
because flow is steady and incompressible.
Also, since
=
constant, then
V
V
K
K
⋅
∇
=
⋅
∇
)
(
0
=
∂
∂
+
∂
∂
=
⋅
∇
⇒
y
v
x
u
V
K
Finally,
0
=
∂
∂
x
u
because of assumption #1
→
long pipe.
0
=
∂
∂
⇒
y
v
h
y
+
=
h
y
−
=
y
x
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View Full DocumentPoiseuille Flow Through a Duct in 2D
16.100
2002
2
Now, integrate this:
C
v
=
=
constant
Apply boundary conditions:
()0
(
)0
vh
v
y
±
=⇒
=
We expect this but it is good to see the math confirm it.
Now, let’s look at
momentum
−
y
.
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 Fall '03
 willcox
 Dynamics, Fluid Dynamics, Aerodynamics, µ, dx

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