# 2.25 MIT Problem ```MIT Department of Mechanical Engineering
Problem 10.05
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
An inviscid, incompressible ﬂuid ﬂows steadily through a circular pipe with a contraction. At the entrance
section, the velocity is purely in the axial direction and is given by :
�
u1 (r) = Vo
1−
�
r
R1
�2 �
(a) What does the vorticity ﬁeld look like at the entrance section?
(b) What is the velocity proﬁle at the exit?
1
c 2010, MIT
Vorticity Theorems
A.H. Shapiro and A.A. Sonin 10.05
Solution:
(a) In the cylindrical coordinate, the vorticity is deﬁned as
= ∇&times;V
∂
1 ∂
∂
eˆθ +
eˆz
∇ =
eˆr +
∂r
r ∂θ
∂z
ω
(10.05a)
(10.05b)
Since the ﬂow is axially symmetric, vθ = ∂/∂θ = 0. And the radial velocity in this problem is zero, i.e.,
vr = 0. Therefore,
ω=−
∂vz
eˆθ ,
∂r
where vz = u
(10.05c)
2r
eˆθ
R12
(10.05d)
Substituting the given velocity proﬁle gives
ω = Vo
The vortex distribution looks like as following. The vortex tube looks like a ring.
(b) Kelvin’s Circulation Theorem
The Kelvin’s theorem represents that the circulation remains at a constant in an inviscid, barotropic ﬂow
with conservative body forces.
dΓ
=0
dt
⇒ ω1 A1 = ω2 A2 = const
f ollow same tube
(10.05e)
The mass conservation of the tube gives
A1 2πr1 = A2 2πr2 ⇒
2
A2
r1
ω1
=
=
A1
r2
ω2
(10.05f)
c 2010, MIT
Vorticity Theorems
A.H. Shapiro and A.A. Sonin 10.05
Hence, the Kelvin’s theorem is
ω1
ω2
=
r2
r1
(10.05g)
Using the fact above and the ω1 = ωθ at section (1), the velocity at section (2) becomes
−
2Vo r
Vo r 2
∂u2
=
⇒ u2 (r) = − 2 + c
2
R1
∂r
R1
(10.05h)
Let’s obtain the constant c by mass conservation between section (1) and (2).
�
0
�
R1
Vo
1−
�
r
R1
�
�2 !
�
�
Vo r 2
=
− 2 + c 2πr dr
R1
0
�
�
Vo R24
1 2
= 2π −
+
cR
2 2
4R12
�
2πr dr
πVo 2
R
2 1
R2
(10.05i)
(10.05j)
Then the constant c and the velocity proﬁle at section (2) are
1 R2
1 R12
Vo 2 + Vo 22
2 R1
2 R2
2 �
�
2 �
�
� �
R2
V o 2 V o R1
= − r +
1+
2 R2
R1
R1
c =
⇒ u2 (r)
(10.05k)
(10.05l)
D
Problem Solution by J.Kim, Fall 2009