Problem #1
Assume:
• Incompressible
•
•
•
ω1
∂
2-D flow ⇒ Vz = 0, = 0
∂z
∂
Steady ⇒
=0
∂t
Parallel ⇒ Vr = 0
r1
r0
a) Conservation of mass for a 2-D flow is:
1 ∂
1 ∂
( r Vr ) +
(Vθ ) = 0
r ∂r N r ∂θ
=0
∂
(Vθ ) = 0 ⇒ Vθ does not depend on θ
∂θ
⇒ Vθ = Vθ ( r )
⇒
b) θ-mometum equation is:
K
∂Vθ
VV
1 ∂p
2 ∂Vr Vθ
+ (V ⋅ ∇)Vθ + r θ = −
− 2)
+ ν (∇ 2Vθ + 2
∂t
∂θ
r
ρ r ∂θ
r N
r
N
N
Vr =0
steady
Vr =0
In cylindrical coordinates:
K
∂ 1
∂
+ Vθ
(V ⋅ ∇) = Vr
N ∂r r ∂θ
=0
Thus,
K
1
∂Vθ
=0
(V ⋅ ∇)Vθ + Vθ
r N
∂θ
=0 from
continuity
Also,
1 ∂  ∂Vθ  1 ∂ 2Vθ
∇ Vθ =
r
+ 2
2
r ∂r  ∂r  r
∂θ
2
=0
ω0
Problem #1
Combining all of these results gives:
1 ∂p
 1 ∂  ∂Vθ  Vθ 
=ν 
r
−
ρ r ∂θ
r ∂r  ∂r  r 2 

this side is independent of θ
Since the right-hand-side (RHS) is independent of θ , this requires that
∂p
= constant for fixed r . But as θ varies from 0 → 2π , it must be equal at
∂θ
0 & 2π , that is p(θ = 0) = p (θ = 2π ) . If not, the solution would be discontinuous.
∂p
= 0 ⇐ constant must be zero!
Thus,
∂θ
The differential equation for Vθ is:
V
1 ∂ ∂Vθ
(r
) − θ2 = 0
r ∂r
∂r
r
A little rearranging gives:
d 1 d

(rVθ ) = 0

dr  r dr

Integrating once gives:
1 d
(rVθ ) = C1
r dr
Integrating again gives:
rVθ =
1
C1 r 2 + C 2
2
1
C
⇒ Vθ = C1r + 2
r
2
16.100 2002
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Problem #1
Next, we must apply the no-slip boundary conditions to find Vθ . Specifically,
at r = ro , Vθ = ω o ro
at r = r1 , Vθ = ω1 r1
because flow velocity equals wall velocity in a viscous flow.
So, apply r = ro & r = r1 bc’s:
r
r

ω1 1 − ωo o
1
ro
r1
C 
1
C =
ωo ro = C1ro + 2   2 1
r1 r0
−
ro  
2
r
⇒
 
⇒ 0 r1
C2  
1
ω1r1 = C1r1 +
r r (ω − ω1 )
r1   C2 = o 1 o
2
r1 r0

−
r0 r1

Or, rearranged a little gives:
r − ro
r
r
+ r1ω1 o
Vθ = roωo
ro
ro
r1
r1
−
−
ro
r1
ro
r1
r1
r
− r
r1
c) The radial momentum equation is:
K
∂Vr
1
1 ∂p
V
2 ∂V 

+ (V ⋅ ∇)Vr − Vθ2 = −
+ ν  ∇ 2Vr − 2r − 2 θ 
∂t
r
r
r ∂θ 
ρ ∂r

But Vr = 0 &
∂Vθ
= 0 so this reduces to:
∂θ
∂p ρVθ2
=
∂r
r
Since
ρVθ2
r
≥ 0 always, then clearly
∂p
≥ 0.
∂r
Thus, pressure increases with r .
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Problem #1
d) On the inner cylinder, the moment is a result of the skin friction due to the fluid
shear stress. For this flow in which only Vθ ≠ 0 and is only a function of r , the
only non-zero shear stress is τ rθ and has the following form:
 ∂ V
 ∂Vθ Vθ 
−  = µ r  θ
∂r
r 

 ∂r  r
τ rθ = µ 



=ε rθ , the only
non-zero strain
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Problem #1
Rotating Cylinders
ω1
ωo
For the problem you studied in the homework:
1. What direction is the fluid element acceleration?
2. What direction are the net pressure forces on a fluid element?
3. What direction are the net viscous forces on a fluid element?
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