MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 10.11
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
The steady sink flow in the sketch is set up by injecting water tangentially through a narrow channel near
the periphery and letting it drain through a hole at the center. The vessel has a radius R. At the point of
injection, the water has a velocity V and depth h0 ; the width of the injection channel, b, is small compared
with R. In what follows, we consider the region of the flow not too close to the drain, and assume that
everywhere in that region (i) the flow is essentially incompressible and inviscid, (ii) the radial velocity
component |vr | i small compared with the circumferential velocity component vtheta , and (iii) the water
depth does not differ much from its value h0 at the periphery.
• (a) Starting with Kelvin’s theorem on circulation, show that
vθ =
VR
.
r
(10.11a)
This equation states that the angular momentum of a fluid particle remains constant in this flow. Is
the angular momentum of a particle always constant? Why is it constant in this case.
• (b) Obtain result (a) from Helmoltz’s vortex laws.
• (c) Obtain the result of (a) directly from Euler’s equation of motion.
• (d) Show that the assumption that |vr | « vθ is satisfied if b « R.
• (e) Derive an expression for the actual distribution of water depth, given the velocity distribution of
part (a), and show that the water depth is essentially constant, as we assume, provided that
r
R
2.25 Advanced Fluid Mechanics
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»
1
V2
.
2gh0
(10.11b)
c 2008, MIT
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Vorticity Theorems
A.H. Shapiro and A.A. Sonin 10.11
Solution:
• (a)
Using Kelvin’s Theorem with ρ = Const and μ = 0, and using a circular material line as shown in the
figure,
ΓC(t2 ) = vθ (rm (t2 ))rm (t2 )2π = vθ (rm (t1 ))rm (t1 )2π = ΓC(t1 ) ,
(10.11c)
since DΓ
Dt . Now, let’s choose rm (t1 ) = R and rm (t2 ) = r. At these positions the velocities are U and
vθ (r) respectively . Then,
U R = vθ (r)r,
(10.11d)
then,
vθ =
UR
.
r
(10.11e)
• (b)
Since the fluid starts with null vorticity ω = 0, and Dω
Dt = 0 (Helmholtz’s Vorticity Equation), then it
has to remain null as the particle travels through the container, then,
ω=
∂vθ
vθ
+
= 0,
∂r
r
(10.11f)
C
,
r
(10.11g)
then, after integrating,
vθ =
2.25 Advanced Fluid Mechanics
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c 2008, MIT
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Vorticity Theorems
A.H. Shapiro and A.A. Sonin 10.11
but vθ (r = R) = U , then C = U R, finally
vθ (r) =
UR
.
r
(10.11h)
• (c) Since |vr | «|vθ |, we know, from Euler in cylindrical cordinates, that
∂p
v2
∼ρ θ
∂r
r
(10.11i)
and, since (from Helmholtz) ω = 0, then from Bernoulli,
1 2
p + ρvθ
= Const,
2
(10.11j)
then, differentiating this equation, we can get the value of the pressure derivative,
∂p
∂vθ
+ ρvθ
= 0,
∂r
∂r
(10.11k)
then equating the value of the derivatives,
ρ
vθ2
∂vθ
= −ρvθ
,
r
∂r
(10.11l)
∂vθ
vθ
+
= 0,
∂r
r
(10.11m)
then, we obtain,
as before. Hence,
vθ (r) =
UR
.
r
(10.11n)
• (d) By continuity,
U bh0 = −vr 2πrh0 ,
(10.11o)
or
|vr | =
where the second term
b
2πR
Ub
UR b
=
,
2πr
r 2πR
(10.11p)
|vr | « 1.
(10.11q)
∂p
= −ρg,
∂z
(10.11r)
« 1 and then,
• (e) Assuming a 2D flow,
where p(z = h(r)) = patm , then
2.25 Advanced Fluid Mechanics
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Vorticity Theorems
A.H. Shapiro and A.A. Sonin 10.11
p = patm + ρg(h(r) − z).
(10.11s)
∂p
v2
U 2 R2
dh
dh
U 2 R2
= ρ θ = ρ 3 = ρg , ⇒
=
,
dr
dr
gr3
∂r
r
r
(10.11t)
Also, from Euler-n,
then after integrating, and dividing by h0 ,
(h0 − h(r))
U 2 R2
=
h0
2gh0
!
"
1
1
−
,
r2
R2
(10.11u)
but since r « R,
U 2 R2 1
(h0 − h(r))
=
,
2gh0 r2
h0
(10.11v)
then h ∼ h0 if
U 2 R2
« 1, ⇒
2gh0 r2
!
r
R
"2
»
U 2
.
2gh0
(10.11w)
Note: We could also obtain the exact (Potential Flow) solution without assuming |vr | «|vθ | by combining a
sink and an irrotational vortex,
Φ = U Rθ −
∂
where ∇ = îr ∂r
+ îθ
1 ∂
r ∂θ ,
Ub
UR
Ub
îr ,
ln r, ⇒ V = ∇Φ =
îθ −
2π
r
2πr
(10.11x)
and then the Bernoulli constant is
p+
!
!
"2 "
1 U 2 R2
b
1
+
.
2πR
2 r2
(10.11y)
D
Problem Solution by ATP/MC, Fall 2008
2.25 Advanced Fluid Mechanics
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2.25 Advanced Fluid Mechanics
Fall 2013
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