•
•
•
Assume steady
∂
⇒ =0
∂t
L
Assume >> 1
Kh
∂V
⇒
=0
∂x
Assume 2-D
∂
⇒ w = 0, = 0
∂z
y= h
y
pL
y=-h
Incompressible N-S equations:
1.
∂u ∂v
+
=0
∂x ∂y
 ∂ 2u ∂ 2u 
∂u
∂u
∂u
1 ∂p
+u +v
=−
+ν  2 + 2 
∂t
∂x
∂y
∂y 
ρ ∂x
 ∂x
 ∂ 2v ∂ 2v 
∂v
∂v
∂v
1 ∂p
3.
+u +v
=−
+ν  2 + 2 
∂t
∂x
∂y
∂y 
ρ ∂y
 ∂x
2.
BC’s
v ( x,± h ) = 0
u ( x, − h ) = 0
u ( x, + h ) = u w
Turning the crank:
∂u ∂v
∂v
+
=0⇒
= 0 ⇒ v = v( x )
∂x ∂y
∂y
N
∂
=0
∂x
∂v
= 0 ⇒ v = const
∂x
Apply bc's ⇒ v = 0
but
Now, y − momentum : Since v = 0 , we have:
∂p
= 0 ⇒ p( x, y ) = p( x )
∂y
uw
x
2h
pR
Lecture 28
Note: since the pressure does change from pL to pR over the length L, p = p( x ) .
Finally x − momentum :


 2

1 dp
∂u
∂u
∂u
∂ u ∂ 2u 

+ u + vN
=−
+ν
+
N
ρ dx
∂t
∂x =0 ∂y
∂x 2 ∂y 2 
N
N
 ∂ =0

steady
=0
∂
 ∂x

=0
∂t
⇒
1 dp
µ
∂ 2u
, where ν ≡
=−
2
µ dx
ρ
∂y
Observe that LHS = f ( y ) and RHS = g ( x)
⇒ f ( y ) = g ( x) = const.
p − pL
dp
⇒
= const = R
dx
L
For this problem, I’ll just use the gradient
dp
but realize this is specified by the
dx
end pressures.
Next, integrate in y :
 d 2u
1 dp 
∫  dy 2 = − µ dx dy
 du

1 dp
⇒ ∫ = −
y + C1  dy
µ dx
 dy

1 dp 2
⇒u=−
y + C1 y + Co
2 µ dx
Now, apply bc’s:
1
2µ
1
u ( y = + h) = −
2µ
u ( y = − h) = −
dp 2
h − C1 h + C o = 0
dx
dp 2
h + C1 h + C o = u w
dx
Solving for C o & C1 :
16.100 2002
2
Lecture 28
Co =
1
1 dp 2 
uw +
h

2
µ dx 
C1 =
uw
2h
2
 uw  y 
h 2 dp  y 
⇒ u( y ) = −
  − 1 +  + 1
2 µ dx  h 

 2  h
Suddenly started flat plate (Stokes 1st Problem)
IC:
u = 0
t = 0, 
v = 0
BC:
u ( x, 0) = uw
t > 0, 
 v ( x,0) = 0
y
Assume infinite length,
p∞
uw
∂
=0
∂x
Continuity:
∂u ∂v
+
= 0 ⇒ v = v( x )
∂x ∂y
N
=0
but
∂v
= 0 so v = 0
∂x
y − momentum :
 ∂ 2v ∂ 2v 
1 ∂p
∂v
∂v
∂v
+u +v
=−
+ν  2 + 2 
ρ ∂y
∂t N
∂x N
∂y
∂x
∂y 
N

=0
=0
=0
=0
∂p
= 0 ⇒ p = p( x ) = p∞
∂y
16.100 2002
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Lecture 28
x − momentum :


 2
2 
∂u
∂u
∂u
∂u ∂u
1 ∂p
+u
+ vN
=−
+ν  2 + 2 
N
∂t
∂x =0 ∂y
∂y
∂x
∂y 
ρ N
N


∂
∂
=
p
p
∞
=0
∂x
 ∂x =0

⇒
∂u
∂ 2u
=ν 2
∂t
∂y
This is the diffusion equation (also known as heat equation).
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•
There are many ways to solve this equation
We’ll use a similarity solution approach used in boundary layer theory.
Similarity Solution
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•
Assume that u (t , y ) = u (η ) where η = η (t , y ). Reduce PDE to ODE.
Usually, the assumption is made that:
⇒
⇒
η = Ct a y b
∂η
a −1
= aCt
aη
t
bη
=
y
yb =
∂t
∂η
= bCt a y b −1
∂y
∂u du ∂η aη du
=
=
∂t dη ∂t
t dη
16.100 2002
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Lecture 28
∂ 2u ∂  ∂u  ∂  du dη 
=
=
∂y 2 ∂y  ∂y  ∂y  dη dy 
=
∂  du bη 
∂y  dη y 
=
bη ∂  du  du ∂  bη 
+
y ∂y  dη  dη ∂y  y 
2
 bη  ∂ 2u du ∂
(bCt a y b−1 )
= 
+
2
dη ∂y
 y  ∂η
2
 bη  ∂ 2u du
b(b − 1)Ct a y b−2
= 
+
2
dη
 y  ∂η
∂ 2u
bη 2 d 2 u
η du
=
(
)
+ b(b − 1) 2
2
2
y dη
∂y
y dη
Thus:
∂u
∂ 2u
= v 2 becomes
∂t
∂y
2
 bη  d 2u
aη du
η du
=ν  
+ ν b(b − 1) 2
2
t dη
y dη
 y  dη
Re-arranging:


d 2u  a
=
dη 2 ν b 2


y2
tη
N
needs to be
f (η ) only
 y 
⇒η = C  
 t


b − 1  du
−
bη  dη


b
For simplicity, b = 1 and C =
⇒a=−
⇒η =
16.100 2002
1
2 ν
1
2
y
νt
5
Lecture 28
⇒
d 2u
du
= −2η
2
dη
dη
Note: bc is
u (0) = u w
u (η → ∞) = 0 ← Also is correct initial condition
2
du
= Ce −η
dη
Integrate again
η
u (η ) = C ∫ e − β d β + Co
2
0
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