Poiseuille Flow Through a Duct in 2-D
y
y = +h
x
y = −h
Assumptions:
∂u ∂v
=
=0
∂x ∂x
•
Velocity is independent of x,
•
•
•
Incompressible flow
Constant viscosity, µ
Steady
•
Pressure gradient along length of pipe is non-zero, i.e.
∂p
≠0
∂x
Boundary conditions:
•
u ( y = ± h ) = 0
No slip: 
⇐ walls are not moving
 v( y = ±h) = 0
To be clear, we now will take the compressible, unsteady form of the N-S
equations and carefully derive the solution:
Conservation of mass:
∂ρ
+ ∇ ⋅ ( ρV ) = 0
∂t
∂ρ
= 0 because flow is steady and incompressible. Also, since ρ =
∂t
constant, then ∇ ⋅ ( ρV ) = ρ∇ ⋅ V
But
⇒ ∇ ⋅V =
Finally,
∂u ∂v
+
=0
∂x ∂y
∂u
= 0 because of assumption #1 → long pipe.
∂x
⇒
∂v
=0
∂y
Poiseuille Flow Through a Duct in 2-D
Now, integrate this:
v = constant = C
Apply boundary conditions: v ( ± h ) = 0 ⇒ v ( y ) = 0
We expect this but it is good to see the math confirm it.
Now, let’s look at y − momentum .
Conservation of y − momentum :
∂τ
∂p ∂τ
Dv
= − + xy + yy
∂y
∂x
∂y
Dt
∂τ
∂v
∂p ∂τ
ρ + V ⋅ ∇v = − + xy + yy
∂t
∂y
∂x
∂y
ρ
∂τ
∂v
∂v
∂p ∂τ
∂v
= − + xy + yy
+u
+v
∂t
∂y
∂x
∂y
∂x =0 ∂y
ρ
=0
steady
=0
Now, what about τ xy &τ yy
τ xy


∂u
 ∂u ∂v 
→ τ xy = µ
= µ +

∂y
 ∂y ∂x 
=0 

τ yy = 2 µ
∂v
+λ
∂y
v =0
 ∂u ∂v 
 ∂x + ∂y  → τ yy = 0


∇ iV =0
So y − momentum becomes:
0=−
But
∂p ∂  ∂u 
µ
+
∂y ∂x  ∂y 
∂u
∂  ∂u 
µ  = 0 because
= 0 & µ = constant

∂x  ∂y 
∂x
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Poiseuille Flow Through a Duct in 2-D
⇒
0=−
∂p
⇒ p ( x, y , t ) = p ( x )
∂y ∂p
∂t
=0 (steady)
Conservation of x − momentum :
ρ
∂τ
Du
dp
∂τ
=−
+ xx + yx
Dt
dx
∂x
∂y
p= p( x )
only


∂u
∂u
∂u
∂u  ∂  ∂u 
dp ∂ 
+u
=−
+
ρ
+v
2µ
µ
+
∂t
∂x =0 ∂y
∂x  ∂y  ∂y 
dx ∂x 


=0
=0
=0 

steady
0=−
dp ∂  ∂u 
µ
+
dx ∂y  ∂y 
Now, we just need to solve this…
∂  ∂u  dp
µ
=
∂y  ∂y  dx
µ = const & u = u ( y ) so,
d 2u
1 dp
=
⇒ must be constant
2
µ dx
dy
only f ( y )
⇒
only f ( x )
dp
= const. ⇒ pressure can only be a linear function of x !
dx
Now, integrating twice in y gives:
u( y ) =
1 dp 2
y + C1 y + C0
2 µ dx
Finally, apply BC’s:
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Poiseuille Flow Through a Duct in 2-D
u( ± h ) = 0
1
u( + h ) =
2µ
1
u( −h ) =
2µ
dp 2
h + C1h + C0 = 0
dx
dp 2
h − C1h + C0 = 0
dx
Solve for C0 & C1 gives:
C0 = −
1 dp 2
h
2 µ dx
C1 = 0
⇒
2
−1 dp   y  
u( y ) =
1 −   
2 µ h 2 dx   h  
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