1. Laminar flow in a water channel

Solutions to problem sheet IV Fluid Dynamics

1. Laminar flow in a water channel

Balance between gravitational force and vicous stress:

u (z) = ΓHz − 1

2 Γz ² where Γ = g · sin α ν Discharge of water:

Q = Z H

0

ρ · u (z) dz

(a) What is the percentage change in depth, if the discharge is increased by 20%:

First derive the discharge:

Q = Z H

0

ρ

ΓHz − 1 2 Γz ²

dz

= ρ 1

2 ΓH ³ − 1 6 ΓH ³

= 1

3 ρΓH ³

Get now the change in discharche by 20%:

Q ⁰ = 1.2 · Q 1

3 ρΓH ^{0 3} = 1.2 · 1 3 ρΓH ³ H ^{0 3} = 1.2 · H ³

H ⁰ = √

1.2 H ⁰ ∼ 1.063 H The height of the water channel increases by 6.3%.

(b) The change of depth is neither dependent upon the viscosity nor the temperature. Therefore

the 6.3% change is valid whatever the fluid properties are.

2. Water flow

One solution of the governing equations is:

u (z) = U 1 − e ⁻

Boundary layer: Layer over which the effects of the surface (e.g. frictional effects) can be felt.

= ν

W where

( ν dynamical viscosity W = 10 ^{− 2} ms ^{− 1} ν [m ² s ^{− 1} ] ε [m]

Water 1.1 · 10 ^{− 6} 1.1 · 10 ^{− 4} = 0.1 mm Air 1.5 · 10 ^{− 5} 1.5 · 10 ^{− 3} = 1.5 mm

So the boundary layer of the atmosphere is generally speaking much deeper than that of the ocean.

3. 2 D flow field

Streamfunction:

ψ = A sin (kx) e ^{− ly} (a) Streamfunction and streamlines

−10 −8 −6 −4 −2 0 2 4 6 8 10

0 0.5 1 1.5 2

−1

−0.8

−0.6

−0.4

−0.2 0 0.2 0.4 0.6 0.8 1

Figure 1: Streamfunction (left) and the some corresponding streamlines (right).

(b) Velocity components u and v:

1.2 1.4 1.6 1.8 2

(c) Vertical vorticity:

ζ = ∂v

∂x − ∂u

∂y

∂v

∂x = − Ak ² sin (kx) e ^{− ly}

∂u

∂y = − Al ² sin (kx) e ^−ly ζ = l ² − k ²

A sin (kx) e ^{− ly} = l ² − k ² ψ

(d) Values of k and l to be a solution of:

Dζ

Dt = ν · ∇ ² ζ

D

Dt (l ² − k ² )ψ = ν ∇ ² (l ² − k ² )ψ (l ² − k ² )

Dψ

Dt − ν ∇ ² ψ

= 0

Two possibilities:

(l ² − k ² ) = 0 l = ± k or

Dψ

Dt − ν ∇ ² ψ = 0

k = 0 (trivial solution)

4. Shallow water system

Set of equations (6.16) in lecture notes on page 65. 1-dimensional variation in y, z direction

∂

∂y , _∂z ^∂ = 0.

(a) Normalisation and parameter ∃

∂u

∂t + u ∂u

∂x − f v = − g ∂h

∂x (1)

∂v

∂t + u ∂v

∂x − f u = 0 (2)

∂h

∂t + u ∂h

∂x + h ∂u

∂x = 0 (3)

Normalize with: ˜ u = _U ^u , ˜ h = _H ^h , ˜ x = _L ^x , ˜ t = ^U _L t

• (1)

∂u

∂t = U ² L

∂˜ u

∂ ˜ t , u _t = U ² L u ˜ t ˜

∂u

∂x = U L

∂ u ˜

∂ x ˜ , u x = U L u ˜ ˜ x

∂h

∂x = H L

∂ ˜ h

∂ x ˜ , h _x = H L ˜ h x ˜

U ²

L (˜ u ^˜ _t + ˜ u x ^˜ ) − f U ˜ v = − g H L ˜ h ^˜ x

With R ⁰ = _{f L} ^U follows:

R ⁰ (U f (˜ u ^˜ _t + ˜ u x ^˜ )) − f U v ˜ = R ⁰

− gf H U

˜ h ^˜ x

R 0 (˜ u ˜ t + ˜ u x ˜ ) − v ˜ = R 0

− g H U ²

˜ h ˜ x

Therefore: ∃ = − g _U ^H

• (2)

U ²

L ˜ v ˜ t + U ²

L u˜ ˜ v x ˜ + f U u ˜ = 0 R ⁰ (f U v ˜ _t ^˜ + f U ˜ u˜ v x ^˜ ) + f U u ˜ = 0 R 0 (˜ v ^˜ _t + ˜ u˜ v x ˜ ) + ˜ u = 0

• (3)

U H

L ˜ h ^˜ _t + U H

L ˜ h x ^˜ u ˜ + U H

L u ˜ x ^˜ ˜ hu = 0

˜ h ^˜ _t + ˜ h x ˜ u ˜ + ˜ u x ˜ ˜ hu = 0 (b) Characteristic values

R 0 = U

f L = 10 ^{− 2} ms ^{− 1}

10 ^{− 4} s ^{− 1} 10 ⁴ m = 10 ^{− 2}

→ Corioliseffect is important

∃ = − gH

U ² = 10 ms ^{− 2} · 10 m

(10 ^{− 2} ms ^{− 1} ) ² = 10 ⁶

(c) Dynamical similarity

The combination of some key elements is more important than elements themselves sepa- rately. Different configuration but same combination ⇒ same flow response similar for same

∃ , R 0 as (b).

Assumptions: R 0 = 10 ^{− 2} , ∃ = 10 ⁶ Lake: U _L = 10 ^{− 2} ms ^{− 1} , L _L = 10 ⁴ m Model: U _M = 10 ^{− 3} ms ^{− 1} , L _M = 1 m

Depth H Model ,f Model =?

f Model = U _M R ⁰ L M

= 10 ^{− 3} ms ^{− 1}

10 ^{− 2} · 1 m = 10 ^{− 1} s ^{− 1} Since f = 2ω ⇒ ω = 5 · 10 ^{− 2} s ^{− 1}

⇒ T = ² _ω ^π ≈ 126s ∼ 2min

H ^Model = ∃ U _M ²

g = 10 ⁶ · 10 ^{− 3} ms ^{− 1 2}

10ms ^{− 2} = 10 ^{− 1} m