Fluid Dynamics questions
Fluid Dynamics I
Course Work
(Due before 4pm, 11th November)
1. Consider an incompressible viscous fluid motion through a tube of elliptic cross section, given by the equation
y2
a2 +
z2
b2 = 1.
The pressure difference between the tube ends is ∆p. Assuming that the tube length L is large, find the velocity distribution in the tube.
Suggestion: Use the Cartesian coordinates with x-axis directed along the centre- line of the tube. You may assume without proof that the y- and z-components of the velocity are zeros. Seek the solution for the x-component in the form
u = C1 + C2y 2 + C3z
2,
where C1, C2 and C3 are constants to be found.
2. A layer of viscous fluid of thickness h is sliding down a flat slope in the gravitational field g. The angle between the slope and horizon is α; see Figure E1.1. Find the velocity distribution across the layer.
x
y
h
α
g
Figure E1.1: Fluid layer on the downslope.
Hint: Use Cartesian coordinates with the x-axis measured down the slope, and notice that the tangential stress
τyx = µ
(
∂v
∂x +
∂u
∂y
)
is zero at the upper edge of the fluid layer.
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3. Consider a single circular cylinder of radius R surrounded by viscous fluid of density ρ and dynamic viscosity µ. The cylinder rotates around its axis with angular velocity Ω. Assuming that the fluid remains at rest far from the cylinder (r → ∞), prove that the velocity field and pressure are given by the “potential vortex” solution
Vr = 0, Vφ = Γ
2πr , Vz = 0, p = p∞ −
ρΓ2
8π2r2 . (e3.1)
How does the circulation Γ depend on the cylinder radius R and angular velocity Ω?
Hint: This may be done through direct substitution of (e3.1) into the Navier-Stokes equations (e3.1)–(e3.4).
4. It is known that the Earth is not a perfect sphere. It is also known that the pressure in the ocean increases with depth much faster than it decreases in the atmosphere. Keeping this in mind, find the shape of the Earth by assuming that it may be thought of as a rotating volume of fluid surrounded by vacuum. The fluid is kept together through the action of the gravitational force. Assume that this force has only a radial component, which is proportional to the distance from the Earth’s centre, namely
fr = −αr.
Given that the angular velocity of the Earth’s rotation is Ω, and the Earth’s radius at the North Pole is R0, show that at any other meridional angle ϑ (measured from the North Pole), the distance R from the Earth’s surface to the centre is given by
R = R0
√
1 − Ω2
α sin2 ϑ
. (e3.2)
Suggestion: Thanks to the fact that the fluid motion is symmetric with re- spect to the Earth’s axis, it is convenient to use spherical polar coordinates (see Figure E1.2),
Mφ r
ϑ
x
y
z
O
Vr
Vφ
Vϑ
C
S
Figure E1.2: Spherical polar coordinates.
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where the Navier-Stokes equations are written as
∂Vr
∂t + Vr
∂Vr
∂r +
Vϑ
r
∂Vr
∂ϑ +
Vφ
r sin ϑ
∂Vr
∂φ −
V 2ϑ + V 2
φ
r = fr −
1
ρ
∂p
∂r +
+ ν
(
∂2Vr
∂r2 +
1
r2 ∂2Vr
∂ϑ2 +
1
r2 sin2 ϑ
∂2Vr
∂φ2 +
2
r
∂Vr
∂r +
+ 1
r2 tan ϑ
∂Vr
∂ϑ −
2
r2 ∂Vϑ
∂ϑ −
2
r2 sin ϑ
∂Vφ
∂φ −
2Vr r2
−
2Vϑ r2 tan ϑ
)
,
(e3.3a)
∂Vϑ
∂t + Vr
∂Vϑ
∂r +
Vϑ
r
∂Vϑ
∂ϑ +
Vφ
r sin ϑ
∂Vϑ
∂φ +
VrVϑ
r −
V 2φ
r tan ϑ = fϑ −
1
ρr
∂p
∂ϑ +
+ ν
(
∂2Vϑ
∂r2 +
1
r2 ∂2Vϑ
∂ϑ2 +
1
r2 sin2 ϑ
∂2Vϑ
∂φ2 +
2
r
∂Vϑ
∂r +
+ 1
r2 tan ϑ
∂Vϑ
∂ϑ −
2 cos ϑ
r2 sin2 ϑ
∂Vφ
∂φ +
2
r2 ∂Vr
∂ϑ −
Vϑ
r2 sin2 ϑ
)
,
(e3.3b)
∂Vφ
∂t + Vr
∂Vφ
∂r +
Vϑ
r
∂Vφ
∂ϑ +
Vφ
r sin ϑ
∂Vφ
∂φ +
VrVφ
r +
VϑVφ
r tan ϑ = fφ −
1
ρr sin ϑ
∂p
∂φ +
+ ν
(
∂2Vφ
∂r2 +
1
r2 ∂2Vφ
∂ϑ2 +
1
r2 sin2 ϑ
∂2Vφ
∂φ2 +
2
r
∂Vφ
∂r + (e3.3c)
+ 1
r2 tan ϑ
∂Vφ
∂ϑ +
2
r2 sin ϑ
∂Vr
∂φ +
2 cos ϑ
r2 sin2 ϑ
∂Vϑ
∂φ −
Vφ
r2 sin2 ϑ
)
,
∂Vr
∂r +
1
r
∂Vϑ
∂ϑ +
1
r sin ϑ
∂Vφ
∂φ +
2Vr r
+ Vϑ
r tan ϑ = 0. (e3.3d)
Take into account that
Vφ = Ωr sin ϑ, Vr = 0, Vϑ = 0,
and, using (e3.3), show that the pressure is zero on the surface given by (e3.2).
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