MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 6.04a
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
Consider a steady, fully developed laminar flow in an annulus with inside radius R2 and outside radius R1 .
• (a) Find a relation between the pressure gradient
2
and R
R1 .
dp
dx ,
the volume flow rate Q, the fluid viscosity µ, R1 ,
• (b) Fin the limiting form of the relation for a very thin annulus by expressing it in terms of R1 and
h
h
R1 , where h = R1 − R2 , and taking the limit R1 → 0. Compare with the formula for fully developed
laminar flow between parallel flat plates separated by a distance h.
2
• (c) In the opposite limit R
R1 → 0, does the relation of (a) reduce to the formula for Hagen-Poiseuille
flow in a circular pipe of radius R1 ? Discuss your answer.
2.25 Advanced Fluid Mechanics
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Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04a
Solution:
• From the N-S in cylindrical coordinates, the equation can be reduced to
)
(
1 ∂p 1 ∂
∂vx
0=−
+
r
,
∂r
µ ∂x r ∂r
(6.04aa)
where the first term is approximately a constant across the space between the cylinders (long cylinder
approximation), then
(
)
1 ∂
∂vx
0 = −K +
r
,
(6.04ab)
r ∂r
∂r
then, integrating,
Krdr =
(
)
∂
∂vx
r
dr,
∂r
∂r
⇒ K
r2
∂vx
+ C1 = r
,
2
∂r
⇒ K
r
C1
∂vx
+
=
.
∂r
2
r
(6.04ac)
Now, integrating again
(K
r
C1
+
)dr =
2
r
(
∂vx
)dr,
∂r
⇒ K
r2
+ C1 ln(r) + C2 = vx ,
4
(6.04ad)
Then, applying the boundary conditions,
vx (R1 ) = 0, vx (R2 ) = 0,
(6.04ae)
the constants can be obtained. Then,
K
R2
R12
+ C1 ln(R1 ) + C2 = 0, OR K 2 + C1 ln(R2 ) + C2 = 0.
4
4
(6.04af)
Now, substracting the solutions to obtain C1 ,
K 2
R1
= 0,
(R − R22 ) + C1 ln
R2
4 1
(6.04ag)
then,
C1 = −
K
2
4 (R1
ln
− R22 )
R1
R2
(6.04ah)
.
Now, re-expressing in terms of the requested variables,
C1 = −R12
K
4 (1
− Φ2 )
,
− ln Φ
⇒ C1 = −R12
K
2
4 (Φ
− 1)
,
ln Φ
(6.04ai)
− C1 ln(R2 ).
(6.04aj)
where, Φ= R2 /R1 .
Now, for C2 , we can use any of the two equations,
C2 = −K
2.25 Advanced Fluid Mechanics
R12
− C1 ln(R1 ),
4
C2 = −K
2
R22
4
c 2008, MIT
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Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04a
Upon substitution of C1 ,
C2 = −K
R12
+
4
K
2
4 (R1
ln
− R22 )
R1
R2
ln(R1 ),
C2 = −K
R22
4
+
K
4
(R12 −R22 )
ln
R1
R2
ln(R2 ),
(6.04ak)
simplifying,
)
(R12 − R22 )
ln(R
)
,
1
1
ln R
R2
)
(
K
(R2 − R2 )
C2 =
− R22 + 1 R1 2 ln(R2 ) .
4
ln R2
C2 =
K
4
(
− R12 +
(6.04al)
(6.04am)
Then the velocity is
vx =
[
( )]
1 dp 2
R12 − R22
r
r − R22 −
ln
R2
4µ dx
ln(R1 /R2 )
(6.04an)
Now, to obtain the flux, let’s integrate this expression,
2π
0
2π
R1
(
K
0
R2
R1
2π
vx rdrdθ =
R2
0
R1
R2
(
)
r2
+ C1 ln(r) + C2 rdrdθ = 2π
4
)
r2
K + C1 ln(r) + C2 rdrdθ,
4
R1
R2
(
K
(6.04ao)
)
r3
+ C1 r ln(r) + C2 r dr
4
(6.04ap)
After integration,
2π
R1
R2
(
K
)
(
[
])
r3
Kr4
r2
r2
1
+ C2 + C1
ln(r) −
+ C1 r ln(r) + C2 r dr = 2π
4
16
2
2
2
R1
,
(6.04aq)
R2
then, finally,
Q = 2π
(
[
]
[
])
R2
(R2 − R22 )
1
R2
1
K(R14 − R24 )
+ C2 1
+ C1 1 ln(R1 ) −
− C1 2 ln(R2 ) −
.
16
2
2
2
2
2
(6.04ar)
Now, substituting C1 and C2 ,
πK
Q=
2
(
R4
R2 R2
R2
(R22 − R12 )2
R14
+ 2 − 1 2 − (R12 − R22 ) 1 +
4
4
2
2
4 ln(R1 /R2 )
)
(6.04as)
Now, re-expressing in terms of the requested variables,
C2 =
KR12
4
(
−1+
)
1 − Φ2
ln(R1 ) ,
− ln Φ
C2 =
KR12
4
(
−1+
Φ2 −1
ln Φ
)
ln(R1 ) .
(6.04at)
And simplifying again,
Q=
2.25 Advanced Fluid Mechanics
πKR14
2
(
(Φ2 −1)2
4
3
(
1−
1
ln Φ
)
+
(Φ2 −1)
2
)
(6.04au)
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Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04a
D
Problem Solution by MC, Fall 2008
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Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04b
Problem 6.04b
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
Solution:
• Now, factorizing taking into account that R1 = R2 + H, then R2 = R1 − H,
Q=
Kπ
8
(
1
1
−2(R1 − H)2 R12 + (R1 − H)4 ln( R1R−H
) + (R1 − H)4 − R14 ln( R1R−H
) + R14
1
ln( R1R−H
)
)
(6.04ba)
Now, let’s substitute H using F = H/R1 ,
Q=
Kπ
8
(
R1
R1
4
4
4
−2(R1 − F R1 )2 R12 + (R1 − F R1 )4 ln( R1 −F
R1 ) + (R1 − F R1 ) − R1 ln( R1 −F R1 ) + R1
R1
ln( R1 −F
R1 )
)
(6.04bb)
Now, taking the limit as F → 0, but keeping the higher order terms,
Q=−
2 4
R KπF 3 ,
12 1
(6.04bc)
and substituting the value of K, and the original variables,
Q=−
Q=
1 dP
R1 πH 3 ,
6µ dx
H3
− 12µ
(
dP
dx
)
(2πR1 ),
(6.04bd)
(6.04be)
which corresponds to the solution of a pressure driven flow between two plates separated by a distance
H, over a length equal to the average circumference of the annulus.
NOTE: SEE PLOTS OF THE SOLUTIONS USING THE ATTACHED MATLAB FILES, PLAY
WITH THE SOLUTIONS TILL THE LIMITS MAKE SENSE TO YOU.
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Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04b
D
Problem Solution by MC, Fall 2008
2.25 Advanced Fluid Mechanics
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Copyright ©
Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04c
Problem 6.04c
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
Solution:
• Now, taking the limit asΦ → 0 of part a) solution,
(
(
)
)
1
(Φ2 − 1)
πKR14 (Φ2 − 1)2
lim Q = lim
1−
+
,
Φ→0
Φ→0
2
4
lnΦ
2
(
)
−1
πKR14 1
lim Q = lim
(1) +
,
Φ→0
Φ→0
2
4
2
πKR14
lim Q = −
,
Φ→0
8
(
)
R13 dP
lim Q = −
(2πR1 ),
Φ→0
16µ dx
(6.04ca)
(6.04cb)
(6.04cc)
(6.04cd)
which is the solution for Poiseuille flow for a simple tube. You may have guessed that the solution did
not converge to this value, i.e. the velocity profile had a hole in the center, but this is wrong. The
solution converges to the simple tube flow because as the inner cylinder becomes smaller, the area that
it uses to transmit vorticity decreases, and as the area decreases, its influence decreases too (Think of
a small string (hot wire) inside the tube for measuring flow, and think how small are the disturbances
that it creates in the flow).
To further verify that the solution makes physical sense, let’s look at the product r ∗ τviscous to show
that the viscous force per unit length decreases as r → 0. Using the velocity profile, the viscous stress
can be obtained,
µ
)
(
dvx
R12 − R22 1
= µK 2r −
,
dr
ln(R1 /R2 ) r
(6.04ce)
now, let’s evaluate at r = R2 , and multiply by R2 ,
µR2
now, taking the limit as R2 → 0,
&
(
)
dvx &&
R12 − R22
2
=
µK
2R
−
,
2
ln(R1 /R2 )
dr &R2
lim µR2
R2 →0
&
dvx &&
= lim µK(2R22 ) = 0,
dr &R2 R2 →0
(6.04cf)
(6.04cg)
then, the net viscous force goes to 0 as the radius approaches 0.
NOTE: SEE PLOTS OF THE SOLUTIONS USING THE ATTACHED MATLAB FILES, PLAY
WITH THE SOLUTIONS TILL THE LIMITS MAKE SENSE TO YOU.
2.25 Advanced Fluid Mechanics
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c 2008, MIT
Copyright ©
Viscous Flows
A.H. Shapiro and A.A. Sonin 6.04c
D
Problem Solution by MC, Fall 2008
2.25 Advanced Fluid Mechanics
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c 2008, MIT
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2.25 Advanced Fluid Mechanics
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