Two-Dimensional Motion

Question 1

Helmholtz’s Equation

Answer 1

Dω/Dt = (ω.∇)u
u and ω are coupled together.

Question 2

The streamfunction

Answer 2

(u, v, 0) = (∂ψ/∂y , − ∂ψ/∂x , 0)
the incompressibility condition is automatically satisfied.

Question 3

Steamlines and the stream function

Answer 3

ψ= const

Question 4

The streamfunction and curl

Answer 4

u = (u, v, 0) = (∂ψ/∂y , − ∂ψ/∂x , 0) = ∇ × (0, 0, ψ)

Question 5

Cylindrical polar coordinates independent of z

Answer 5

u has a radial component ur and an azimuthal component uθ

Question 6

The streamfunction in cylindrical polars independent of z.

Answer 6

u = 1/r ∂ψ/∂θ ˆr − ∂ψ/∂r θˆ = ∇ × (0, 0, ψ).

Question 7

The Cauchy-Riemann Equations

Answer 7

u = ∂ϕ/∂x = ∂ψ/∂y
v = ∂ϕ/∂y = −∂ψ/∂x

Question 8

What do the Cauchy-Riemann Equations mean?

Answer 8

If both ϕ and ψ are smooth
functions, the combination ϕ+iψ is a differentiable, complex function of z = x + iy, w(z).

Question 9

What is the complex potential and complex position?

Answer 9

The function w(z) is referred to as the ‘complex potential’, and we sometimes refer to z = x + iy as the ‘complex position’.

Question 10

What is ∇ϕ.∇ψ

Answer 10

0, Hence ∇ϕ is perpendicular to ∇ψ, streamlines (lines of constant ψ) are orthogonal to lines of constant ϕ (equipotentials); the two sets of lines form an system of orthogonal coordinates for the plane.

Question 11

What is the complex velocity?

Answer 11

dw/dz = u − iv

Question 12

How can we work out u and v?

Answer 12

a) Re[w(z)] = velocity potential ϕ, ∇ϕ = (u, v)
b) Im[w(z)] = streamfunction ψ, ∇ψ = (−v, u)
c) Re[dw/dz] = u, Im[dw/dz] = −v

Question 13

ur − iuθ and the complex potential

Answer 13

ur − iuθ = exp[iθ] dw/dz

Question 14

What are the 5 special 2-D flows?

Answer 14

a) the Uniform Stream
b) the Source/Sink
c) the Vortex
d) the Dipole
e) the Corner Flow

Question 15

The uniform stream

Answer 15

Flow is of speed U at an angle α to the positive x-axis (measured anti-clockwise)
(u, v) = (U cos α, U sin α)
dw/dz = u − iv = U cos α − iU sin α = U exp[−iα]
w(z) = Uz exp[−iα] + a constant.

Question 16

What is a source? What is a sink?

Answer 16

A source is a point where fluid is inserted into a flow.
A sink is a negative source, and is thus a point where fluid is extracted from the flow.

Question 17

Complex potential for a source at the origin

Answer 17

w(z) = m/2π ln(z)

Question 18

What can a source or sink be referred to in 3-D?

Answer 18

“Line Source” or “Line Sink”

Question 19

Fluid velocity of a Source

Answer 19

u = f(r)ˆr

Question 20

Fluid velocity of a Vortex

Answer 20

u = f(r)θˆ

Question 21

Complex potential for a vortex at the origin

Answer 21

w(z) = −iK/2π ln(z)

Question 22

Complex potential of a dipole at the origin

Answer 22

w(z) = −µ exp[iα] / 2πz
µ is dipole strength

Question 23

What is a dipole?

Answer 23

Combination of a source m and sink -m, imagine the sink at the origin and the source a small distance δexp[iα] away.

Question 24

What happens if we replace any streamline by a fixed boundary?

Answer 24

The flow is unchanged.

Question 25

Complex potential for flow in a corner

Answer 25

w(z) = A z^α α > 1/2 θ = nπ/α

Question 26

What is the complex velocity in cartesian and polars?

Answer 26

dw/dz = u − iv = exp[−iθ(ur − iuθ)]

Question 27

What is the method of images?

Answer 27

We can guarantee that u = 0 on x = 0 if we have a flow which is symmetric about x = 0.

Question 28

Method of images (vortex by a wayy)

Answer 28

If there is a vortex of strength K at (a, 0), in the presence of the rigid wall at x = 0; we put an image vortex at (−a, 0), we need to reverse the flow) the image must have strength −K.

Question 29

Method of images (dipole by a wall)

Answer 29

In the presence of the rigid wall at x = 0, there is a Dipole of strength µ, placed at (a, b) and with direction given by exp[iα]. Symmetry requires the image to be another Dipole, placed at (−a, b), strength µ, direction −e −iα. (The imaginary part of this, corresponding to the part in the y direction and so parallel to the wall, is the same as in the real Dipole; however, the real part, corresponding to the part in the x direction and so perpendicular to the wall, is the opposite of what is in the real Dipole.)

Question 30

Method of images (source near a right angle corner)

Answer 30

a) an image Source of strength m at (a, −b), b) an image Source of strength m at (−a, b), and c) an image Source of strength m at (−a, −b)

Question 31

Method of images (vortex near a right angle corner)

Answer 31

For a Vortex of strength K at (a, b) in the right-angled corner, we require Vortices −K at (a, −b) and (−a, b) and a Vortex of strength +K at (−a, −b).

Question 32

Complex potential of a source in a uniform stream

Answer 32

w(z) = Uz + m/2π ln(z).

Question 33

Complex potential of a dipole in a uniform stream

Answer 33

w(z) = Uzexp[−iα] − µexp[i(α+π)] / 2πz

Question 34

What can the flow of a uniform stream past a dipole be regarded as?

Answer 34

The flow of a Uniform Stream past a circle of radius a.

Question 35

What is the circle theorem?

Answer 35

w(z) = f(z) + (f(a^2 / z*))*