These LaTeX notes contain full solutions to the problems from the past papers for the Part B MPhys examination in B1: Flows, Fluctuations and Complexity. Features colour images of diagrams, as well as full explanations of the steps taken.
B1: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins
June 22, 2017
Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at
toby.adkins@merton.ox.ac.uk
2011
Question 1
Consider the Navier Stokes equation
∂u
ρ + (u · ∇)u = −∇p + η∇2 u − ∇χ
∂t
for some conservative potential χ. Using the vector identity
u2 u2
(u · ∇)u = ∇ − u × (∇ × u) = ∇ −u×ω
2 2
Assume that the
ow is incompressible, so ∇p = 0. Then,
u2 p χ
∂u
+∇ + + = u × ω + ν∇2 u
∂t 2 ρ ρ
Suppose that χ is due to a gravitational force, such that χ = ρgz . Then,
∂u
+ ∇H = u × ω + ν∇2 u
∂t
where
u2 p
H= + + gz
2 ρ
If inviscid and steady, ν = 0, ∂t u = 0. If H is constant along a streamline, when we require
that
(u · ∇)H = u · ∇(u × ω) = 0
1
,Toby Adkins B1 Past Papers
Thus,
u2 p
H= + + gz
2 ρ |{z}
|{z} Gravitational Potential Energy
Kinetic energy
|{z}
Work done against pressure per unit mass
per unit volume per unit volume
Irrotational: ∇ × u = ω = 0
Incompressible: ∇ · u = 0.
De
ne u = ∇ × (ψey ), for streamfunction ψ , which clearly satis
es ∇ · u = 0. Then, we
have that
ωi = ijk k`m ∂j ∂` ψe`m
= ∂i ∂m ψem − ∂`2 ψei
= ∂i (∂y ψ) − ∂`2 ψei
However, ∂y ψ = 0 by de
nition, such that ωi = −∇2 ψei . However, we have already stated
that ω = 0 for all components, meaning that
∇2 ψ = 0
Taking derivatives:
∂2ψ π 2 πx πz
= U0 a cos e− b
∂x2 b a
∂2ψ π 2 πx πz
= −U0 a cos e− b
∂z 2 b a
Thus,
∂2ψ ∂2ψ
+ =0
∂x2 ∂z 2
As z → ∞, e− b → 0, so ψ = ψ0 z . This is uniform
ow along the ex direction. The
ow
πz
must satisfy the no slip and no perpendicular
ow conditions on the surface ie. we must
have ψ = constant at zs (x, y).
πx πx πa
ψ(zs ) = U0 a cos − U0 a cos e− b ≈0
b b
where we have used the approximation that a
b. The
ow at the surface is given by
∂ψ aπ πx
u= = U0 1 + cos
∂z b b
Consider an aerofoil of length b in the uniform
ow U0 . Let pb be the pressure on the
bottom surface of the aerofoil, and pt be the pressure on the top surface of the aerofoil.
Ignore the contribution of gravity to H . Then:
ρ
pb = p0 − U02
2
ρ 2 ρ aπ πx
pt = p0 − u = p0 − U02 1 + 2 cos
2 2 b a
where p0 is some reference pressure, and we have ignored terms of O( aπ/b)2 . The
dierence between these pressures is
ρ 2 aπ πx ρ aπ πx
pb − pt = U0 1 + 2 cos − U02 = ρU02 cos
2 b a 2 b b
2
,Toby Adkins B1 Past Papers
The force per unit length is thus given by
Z Z b/2
aπ πx
F = dA ∆p = dx ρU02 cos = 2aρU02
−b/2 b b
By the Kutta-Joukowski Theorem:
F = −ρU0 Γ −→ Γ = −2aU0
This means that the circulation is negative (resulting in lift), and that there must be a
trailing vector on the edge of the wing.
F = 2(0.1)(2)2 (103 ) = 800 Nm−1
Evaluating the Reynolds number for this
ow:
U0 `0 U0 a
Re = v v 105
103
ν ν
This means that the Laminar
ow assumption is not valid, as the
ow is turbulent.
Question 2
Consider the Navier Stokes equation
∂u
ρ + (u · ∇)u = −∇p + η∇2 u − ∇χ
∂t
Assume that the
ow is steady (∂t u = 0), viscous (ν
1), and ignore the eects of gravity
(χ = 0). As ν
1, Re
1, meaning that we can neglect the inertial terms in the Navier
Stokes. Then:
1
∇p = ν∇2 u
ρ
Assume that the
ow is only along x (we can neglect
ow along y and z due to symmetry
constraints). Then, as ∇p = −f , we have that
1 ∂u
(−f ) = ν∇2 u = ν 2
ρ ∂z
assuming that the
ow varies along z (cannot vary along y due to symmetry, nor along x
due to the fact that the
ow is steady, and well-developed). Integrating:
1 ∂p 2
u= z + c1 z + c2
2ρν ∂x
Boundary conditions: u = 0 at z = 0 and z = b.
c2 = 0
1 ∂p
c1 = − b
2ρν ∂z
1 ∂p f
→u= z(z − b) = − z(z − b)
2ρν ∂z 2ρν
This is a valid solution for the
ow if Re
1, meaning that we require that
f b2 f b3
ν
u0 `0 = b −→ ν2
8ρν 8ρ
The remainder of this question appears to be no longer on syllabus, so I will not provide
any solution here
3
, Toby Adkins B1 Past Papers
Question 3
Consider a dynamical system described by the system of equations
ṙ = f (r)
for some non-linear function f (r).
• Fixed Point - A point in phase space that satis
es ṙ = 0. Trajectories on this point
remain on this point
• Unstable Spiral - Trajectories diverge from a point, spiralling outwards. The eigen-
values of the Jacobean J for such a point are imaginary, with positive real part
• Attractor - In 2D, an attractor is either a stable
xed point, or a stable limit cycle,
to which trajectories converge. In higher dimensions, this can be a strange attractor;
a region of space to which all trajectories are con
ned, but do not settle down into
periodic orbits.
Setting ṙ = 0 for
xed points:
y = −z
x = −ay
0 − a − y(x − c)
= a − y(−ay − c)
= a + ay 2 + cy
r
c c 2
→y=− ± −1
2a 2a
Thus, we have that the
xed points are located at
r
c c 2
z= ± −1
2a 2a
y = −z
x = az
Consider the Jacobean:
0 −1 −1
∂ ẋi
J = = 1 a 0
∂xj
z 0 x−c
If c
1, a
1, then we can approximate that ż ≈ −cz . For c
1, this means that
trajectories will exponentially converge to the x-y plane. As such, we can now consider
the 2D system in the x-y plane:
0 −1
J =
1 a
Finding eigenvalues:
−λ −1 !
=0
1 a−λ
4
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