Question:

The marks for this “case study” will constitute 40% of your final mark in this unit. • Marks may be allocated for technical merit, clarity and appropriateness of argument and clarity of setting out. You should show enough working in each question in order for the marker to easily follow your solution.

• You should submit your answers in your own handwriting where possible. Any graphs you generate should be printed off and attached where appropriate with proper captions.

• You may use any references that you deem appropriate. Where references have been used they should be fully cited within the text of your solution.

• The written part of this case study should be handed by 12pm on 22 May 2018 (Lecture, Week 12). Any MATLAB (or equivalent) files you use should be submitted via blackboard by this time. If you are unable to make the lecture at this time, submitting a scanned copy of your written part along with the code is possible but not preferred.

• You should submit your answers in your own handwriting where possible. Any graphs you generate should be printed off and attached where appropriate with proper captions.

• You may use any references that you deem appropriate. Where references have been used they should be fully cited within the text of your solution.

• The written part of this case study should be handed by 12pm on 22 May 2018 (Lecture, Week 12). Any MATLAB (or equivalent) files you use should be submitted via blackboard by this time. If you are unable to make the lecture at this time, submitting a scanned copy of your written part along with the code is possible but not preferred.

This case study is concerned with calculating the water wave pattern that forms at the stern of a steadily moving ship. We start in Part A with deriving the fundamental relationship between the wavelength of a water wave with its speed (have you noticed that the speed of waves that you surf on have different speeds, depending on the wavelength?). In Part B we use a simple toy model for generating waves in two dimensions while in Part C we analyse the full threedimensional version. Finally in Part D we look at the wake angle property of three-dimensional ship waves.

Throughout the case study we shall assume that viscosity can be ignored so we are concerned with potential flow. That is, we assume that the velocity field is related to the velocity potential via

q = ∇φ.

Assuming the density is constant, we have the continuity equation

∇ · q = 0.

Combining these two equations gives Laplace’s equation

∇2φ = 0

throughout the flow domain.

Throughout the case study we shall assume that viscosity can be ignored so we are concerned with potential flow. That is, we assume that the velocity field is related to the velocity potential via

q = ∇φ.

Assuming the density is constant, we have the continuity equation

∇ · q = 0.

Combining these two equations gives Laplace’s equation

∇2φ = 0

throughout the flow domain.

For this part let’s consider a train of periodic 2D waves that is travelling along the surface of water with speed U. We are interesting in relating the speed of these waves to the wavelength. One way to proceed is to move in a reference frame of the wave so that the flow is steady. Weset up a coordinate system so that the y-axis points upwards and the x-axis coincides with the horizontal bottom. The as-yet unknown upper surface (the free surface) is denoted by y = η(x)

(note t does not appear as the flow is steady in this frame of reference). The average height of the wave-train is at y = H. With all this mind, we need to study the following system:

∇2φ = 0, 0 < y < η(x), (1)

∂φ

∂y =

∂φ

∂x

dη

dx

on y = η(x), (2)

1

2

|∇φ|

2 + gy =

1

2

U

2 + gH on y = η(x), (3)

∂φ

∂y = 0 on y = 0. (4)

1. Give an explanation for how each of the three boundary conditions (2)-(4) are derived. Note that one is a kinematic condition (similar to that used for our example of spreading of a thin droplet), one comes from Bernoulli’s equation and one is a condition on the bottom.

2. The system (1)-(4) is highly nonlinear and too challenging for this case study. To make progress, we linearise the system by assuming the amplitude of the waves A is small compared to the depth H. That is, we set = A/H and write the solution out as a perturbation series

φ(x, y) = Ux + φ1(x, y) + O(

2

η(x) = H + η1(x) + O(

2

By substituting these expressions into the governing equations, derive the linear system

∇2φ1 = 0, 0 < y < H, (5)

∂φ1

∂y = U

dη1

dx

on y = H, (6)

U

∂φ1

∂x + gη1 = 0 on y = H, (7)

∂φ1

∂y = 0 on y = 0. (8)

Explain each step clearly.

3. Use a separation of variables argument to write φ1 = F(y)G(x) to derive general solutions for F and G. Assuming the solution has wavelength λ, apply the boundary conditions (7)-(8) to show that

Φ = Ux +

gλA

2πU

cosh(2πy/λ)

cosh(2πH/λ)

cos

2πx

λ

+ O(

2

)

η(x) = H + A sin

2πx

λ

+ O(

2

).

3

4. We are nearly there. (a) Now apply the kinematic condition (6) to derive the dispersion relation for periodic waves on a finite-depth fluid:

U

2 =

gλ

2π

tanh

2πH

λ

. (9)

Note this is our relationship between the speed of waves U and their wavelength λ. (b) Are longer waves faster or slower?

(c) How does (9) behave in the infinite-depth limit H → ∞? How about in the other limit, (the shallow-water limit) H → 0?

(d) Now divide both sides of (9) by gH. By using the Froude number F = U/√ gH and the dimensionless wave number κ = 2πH/λ, write out a dimensionless version of (9). (e) Draw a graph of F against κ for κ > 0. For what value of F are periodic waves possible and not possible?

(note t does not appear as the flow is steady in this frame of reference). The average height of the wave-train is at y = H. With all this mind, we need to study the following system:

∇2φ = 0, 0 < y < η(x), (1)

∂φ

∂y =

∂φ

∂x

dη

dx

on y = η(x), (2)

1

2

|∇φ|

2 + gy =

1

2

U

2 + gH on y = η(x), (3)

∂φ

∂y = 0 on y = 0. (4)

1. Give an explanation for how each of the three boundary conditions (2)-(4) are derived. Note that one is a kinematic condition (similar to that used for our example of spreading of a thin droplet), one comes from Bernoulli’s equation and one is a condition on the bottom.

2. The system (1)-(4) is highly nonlinear and too challenging for this case study. To make progress, we linearise the system by assuming the amplitude of the waves A is small compared to the depth H. That is, we set = A/H and write the solution out as a perturbation series

φ(x, y) = Ux + φ1(x, y) + O(

2

η(x) = H + η1(x) + O(

2

By substituting these expressions into the governing equations, derive the linear system

∇2φ1 = 0, 0 < y < H, (5)

∂φ1

∂y = U

dη1

dx

on y = H, (6)

U

∂φ1

∂x + gη1 = 0 on y = H, (7)

∂φ1

∂y = 0 on y = 0. (8)

Explain each step clearly.

3. Use a separation of variables argument to write φ1 = F(y)G(x) to derive general solutions for F and G. Assuming the solution has wavelength λ, apply the boundary conditions (7)-(8) to show that

Φ = Ux +

gλA

2πU

cosh(2πy/λ)

cosh(2πH/λ)

cos

2πx

λ

+ O(

2

)

η(x) = H + A sin

2πx

λ

+ O(

2

).

3

4. We are nearly there. (a) Now apply the kinematic condition (6) to derive the dispersion relation for periodic waves on a finite-depth fluid:

U

2 =

gλ

2π

tanh

2πH

λ

. (9)

Note this is our relationship between the speed of waves U and their wavelength λ. (b) Are longer waves faster or slower?

(c) How does (9) behave in the infinite-depth limit H → ∞? How about in the other limit, (the shallow-water limit) H → 0?

(d) Now divide both sides of (9) by gH. By using the Froude number F = U/√ gH and the dimensionless wave number κ = 2πH/λ, write out a dimensionless version of (9). (e) Draw a graph of F against κ for κ > 0. For what value of F are periodic waves possible and not possible?

PART B

Now we are interested in a toy model for a 2D “ship”. Modelling an actual ship-like hull in MXB321 is much too complicated, so instead we suppose that the effects of a ship can be approximated by an applied pressure on the surface of the water. Think of a hovercraft, for example.

Suppose an applied pressure is moving at speed U and generating a wave pattern behind it. Then, moving in the reference frame of the applied pressure, the flow is steady. The governing equations are

∇2φ = 0, 0 < y < η(x), (10)

∂φ

∂y =

∂φ

∂x

dη

dx

on y = η(x), (11)

1

2

|∇φ|

2 + gy +

P0

ρ

P(x) = 1

2

U

2 + gH on y = η(x), (12)

∂φ

∂y = 0 on y = 0, (13)

where ρ is the water density and the pressure distribution, P(x), is some arbitrary function. In the literature, the most common function used is the Gaussian

P(x) = e−x

2/L2

. (14)

Note here L is a length scale. We can linearise these equations if the strength of the pressure P0 is small compared to ρU2

That is, by setting = P0/(ρU2

) 1 and performing the same perturbation as in part A, we find

∇2φ1 = 0, 0 < y < H, (15)

∂φ1

∂y = U

dη1

dx

on y = H, (16)

U

∂φ1

∂x + gη1 + U

2P(x) = 0 on y = H, (17)

∂φ1

∂y = 0 on y = 0. (18)

4

5. Show that by scaling the lengths by H and the velocities by U, the equations (14)–(18) can be nondimensionalised to

∇2φ1 = 0, 0 < y < 1, (19)

∂φ1

∂y =

dη1

dx

on y = 1, (20)

∂φ1

∂x +

η1

F2

+ P(x) = 0 on y = 1, (21)

∂φ1

∂y = 0 on y = 0, (22)

P(x) = e−x

2/`2

, (23)

where F = U/√

gH is the Froude number and ` = L/H is a nondimensional length. 6. Solve the problem (19)–(23) with Fourier ransforms to give η = 1 − F2P +

F2

2π

Z ∞

−∞

kP˜(k)

k − tanh(k)/F2

e

ikx dk, (24)

where P˜(k) = `

√

π exp(−`

2k

2/4) is the Fourier transform of the pressure distribution.

Now we are interested in a toy model for a 2D “ship”. Modelling an actual ship-like hull in MXB321 is much too complicated, so instead we suppose that the effects of a ship can be approximated by an applied pressure on the surface of the water. Think of a hovercraft, for example.

Suppose an applied pressure is moving at speed U and generating a wave pattern behind it. Then, moving in the reference frame of the applied pressure, the flow is steady. The governing equations are

∇2φ = 0, 0 < y < η(x), (10)

∂φ

∂y =

∂φ

∂x

dη

dx

on y = η(x), (11)

1

2

|∇φ|

2 + gy +

P0

ρ

P(x) = 1

2

U

2 + gH on y = η(x), (12)

∂φ

∂y = 0 on y = 0, (13)

where ρ is the water density and the pressure distribution, P(x), is some arbitrary function. In the literature, the most common function used is the Gaussian

P(x) = e−x

2/L2

. (14)

Note here L is a length scale. We can linearise these equations if the strength of the pressure P0 is small compared to ρU2

That is, by setting = P0/(ρU2

) 1 and performing the same perturbation as in part A, we find

∇2φ1 = 0, 0 < y < H, (15)

∂φ1

∂y = U

dη1

dx

on y = H, (16)

U

∂φ1

∂x + gη1 + U

2P(x) = 0 on y = H, (17)

∂φ1

∂y = 0 on y = 0. (18)

4

5. Show that by scaling the lengths by H and the velocities by U, the equations (14)–(18) can be nondimensionalised to

∇2φ1 = 0, 0 < y < 1, (19)

∂φ1

∂y =

dη1

dx

on y = 1, (20)

∂φ1

∂x +

η1

F2

+ P(x) = 0 on y = 1, (21)

∂φ1

∂y = 0 on y = 0, (22)

P(x) = e−x

2/`2

, (23)

where F = U/√

gH is the Froude number and ` = L/H is a nondimensional length. 6. Solve the problem (19)–(23) with Fourier ransforms to give η = 1 − F2P +

F2

2π

Z ∞

−∞

kP˜(k)

k − tanh(k)/F2

e

ikx dk, (24)

where P˜(k) = `

√

π exp(−`

2k

2/4) is the Fourier transform of the pressure distribution.

Note: the transformed variable is k (i.e. F{f(x)} = ˜f(k)). 7. Alter the code provided on Blackboard to plot the surface height for both a subcritical (F < 1) and a supercritical (F > 1) solution. For the subcritical solution, ensure that multiple wavelengths are visible (roughly four or more peaks and troughs). Otherwise, you are free to choose the domain and the exact values of , F and `. Comment on any differences between the two solutions.

PART C

Part C is concerned with genuinely 3D ship waves. This is a difficult problem but we are able to extend the tools from Part B to make real progress in computing ship waves. Without including all the details, if we follow the same method to that presented in Part B, we arrive at the following dimensionless linear problem for 3D ship waves:

∇2φ1 = 0, 0 < z < 1, (25)

∂φ1

∂z =

∂ζ1

∂x on z = 1, (26)

∂φ1

∂x +

ζ1

F2

+ P(x, y) = 0 on z = 1, (27)

∂φ1

∂z = 0 on z = 0, (28)

P(x, y) = e−(x

2+y

2

)/`2

, (29)

where z = ζ(x, y) is the wave height, φ(x, y, z) is the velocity potential and, as before, F is the Froude number and ` is a dimensionless length. Note: in this case the perturbation used was

Part C is concerned with genuinely 3D ship waves. This is a difficult problem but we are able to extend the tools from Part B to make real progress in computing ship waves. Without including all the details, if we follow the same method to that presented in Part B, we arrive at the following dimensionless linear problem for 3D ship waves:

∇2φ1 = 0, 0 < z < 1, (25)

∂φ1

∂z =

∂ζ1

∂x on z = 1, (26)

∂φ1

∂x +

ζ1

F2

+ P(x, y) = 0 on z = 1, (27)

∂φ1

∂z = 0 on z = 0, (28)

P(x, y) = e−(x

2+y

2

)/`2

, (29)

where z = ζ(x, y) is the wave height, φ(x, y, z) is the velocity potential and, as before, F is the Froude number and ` is a dimensionless length. Note: in this case the perturbation used was

φ(x, y, z) = x + φ1(x, y, z) + O(

2

ζ(x, y) = 1 + ζ1(x, y) + O(

2

5

8. Solve the problem (25)–(29) with Fourier transforms to give

ζ = 1 − F2P(x, y) + F2

4π

2

Z π

−π

Z ∞

0

k

2P˜(k, ψ)

k − sec2 ψ tanh(k)/F2

e

ik(x cos ψ+y sin ψ) dk dψ, (30)

where P˜(k, ψ) = `

2π exp(−`

2k

2/4) is the Fourier transform of the pressure distribution in polar coordinates. Note: the transformed variables are ξ and η (i.e. F{f(x, y)} = ˜f(ξ, η)) which are then converted into polar coordinates ξ = k cos ψ, η = k sin ψ.

9. It can be shown that if P˜(k0, ψ) = P˜(k0, ψ + π) the wave train downstream is given by

ζ = 1 −

F2H(x)

π

Z ∞

k0

kP˜(k, ψ0)

tan ψ0

sin(kx cos ψ0) cos(ky sin ψ0) dk (31)

where H(x) is the Heaviside function, k0 is the real positive root of k − tanh(k)/F2 = 0 for F < 1 and k0 = 0 otherwise, and

ψ0 = arccos r

tanh k

F2k

(if you are interested in the details, just ask). For this question, alter the 3D wave train code from Blackboard and plot a subcritical and supercritical solution. Again, you are free to choose the domain and the exact values of , F and `. Comment on any differences between the two solutions.

2

ζ(x, y) = 1 + ζ1(x, y) + O(

2

5

8. Solve the problem (25)–(29) with Fourier transforms to give

ζ = 1 − F2P(x, y) + F2

4π

2

Z π

−π

Z ∞

0

k

2P˜(k, ψ)

k − sec2 ψ tanh(k)/F2

e

ik(x cos ψ+y sin ψ) dk dψ, (30)

where P˜(k, ψ) = `

2π exp(−`

2k

2/4) is the Fourier transform of the pressure distribution in polar coordinates. Note: the transformed variables are ξ and η (i.e. F{f(x, y)} = ˜f(ξ, η)) which are then converted into polar coordinates ξ = k cos ψ, η = k sin ψ.

9. It can be shown that if P˜(k0, ψ) = P˜(k0, ψ + π) the wave train downstream is given by

ζ = 1 −

F2H(x)

π

Z ∞

k0

kP˜(k, ψ0)

tan ψ0

sin(kx cos ψ0) cos(ky sin ψ0) dk (31)

where H(x) is the Heaviside function, k0 is the real positive root of k − tanh(k)/F2 = 0 for F < 1 and k0 = 0 otherwise, and

ψ0 = arccos r

tanh k

F2k

(if you are interested in the details, just ask). For this question, alter the 3D wave train code from Blackboard and plot a subcritical and supercritical solution. Again, you are free to choose the domain and the exact values of , F and `. Comment on any differences between the two solutions.

PART D

Notice that the ship’s wake appears to be contained in a V-shaped wedge and that the angle of this wedge depends on the Froude number. We can determine this “wake angle” (specifically the half-angle of the V-shape) through geometric arguments. The basis of this geometric method are the following physical properties:

• When waves travel through water (of finite depth) they travel with phase velocity

cp =

?(k)

k

, (32)

and group velocity

cg =

d?

dk

, (33)

where ?(k) = p

k tanh(k)/F is the dispersion function and k is the wavenumber. • When a ship travels with dimensionless speed of unity (that is, speed is 1) the phase velocity of the waves generated depends on the direction they travel in ψ relative to the

direction the ship moves. This relation is cp = cos ψ. (34)

• The waves that form the stationary ship waves travel at the group velocity. The above properties are illustrated in the following schematic

6

t

cgt

cpt

k

θwake ψ

O M

Here we have a ship moving from the point M to the point O with a dimensionless speed 1 over the time period t. The waves generated by the ship at the point M propagate at angle ψ from the sailing line, and wavenumbers k with phase velocity, cp, and group velocity, cg. The group velocity curve (dark blue) originating from the point M is traced out using the group velocity for angle ψ. The wake angle, θwake, is defined as the smallest angle that encloses the entire group velocity curve.

10. Combine equations (32) and (34) to derive a relation between k and ψ. Where has this relation previous appeared in this case study?

11. If we suppose the point O in the schematic is fixed to be the origin with coordinates (0,0), give a parametric expression for the blue curve in terms of cg, t and ψ. Hence, give an expression for the wake angle θwake.

12. Plot the wake angle against the Froude number for 0 < F < 3. Plot the vertical axis in degrees.

13. For supercritical flow (F > 1) the wake angle occurs at ψ = arccos(1/F). Use this to give the wake angle θwake explicitly in terms of F for F > 1.

14. Plot the solutions from question 9 again in plan (bird’s eye) view and overlay the appropriate wake angle wedge. Are the wake angles a good indicator of the edge of the wake?

Notice that the ship’s wake appears to be contained in a V-shaped wedge and that the angle of this wedge depends on the Froude number. We can determine this “wake angle” (specifically the half-angle of the V-shape) through geometric arguments. The basis of this geometric method are the following physical properties:

• When waves travel through water (of finite depth) they travel with phase velocity

cp =

?(k)

k

, (32)

and group velocity

cg =

d?

dk

, (33)

where ?(k) = p

k tanh(k)/F is the dispersion function and k is the wavenumber. • When a ship travels with dimensionless speed of unity (that is, speed is 1) the phase velocity of the waves generated depends on the direction they travel in ψ relative to the

direction the ship moves. This relation is cp = cos ψ. (34)

• The waves that form the stationary ship waves travel at the group velocity. The above properties are illustrated in the following schematic

6

t

cgt

cpt

k

θwake ψ

O M

Here we have a ship moving from the point M to the point O with a dimensionless speed 1 over the time period t. The waves generated by the ship at the point M propagate at angle ψ from the sailing line, and wavenumbers k with phase velocity, cp, and group velocity, cg. The group velocity curve (dark blue) originating from the point M is traced out using the group velocity for angle ψ. The wake angle, θwake, is defined as the smallest angle that encloses the entire group velocity curve.

10. Combine equations (32) and (34) to derive a relation between k and ψ. Where has this relation previous appeared in this case study?

11. If we suppose the point O in the schematic is fixed to be the origin with coordinates (0,0), give a parametric expression for the blue curve in terms of cg, t and ψ. Hence, give an expression for the wake angle θwake.

12. Plot the wake angle against the Froude number for 0 < F < 3. Plot the vertical axis in degrees.

13. For supercritical flow (F > 1) the wake angle occurs at ψ = arccos(1/F). Use this to give the wake angle θwake explicitly in terms of F for F > 1.

14. Plot the solutions from question 9 again in plan (bird’s eye) view and overlay the appropriate wake angle wedge. Are the wake angles a good indicator of the edge of the wake?

Answer:

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