F G = − 1 ρ ∇ ~ p Due to:

Solutions to problem sheet III Fluid Dynamics

1. Pressure

Pressure gradient force:

F G = − 1 ρ ∇ ~ p Due to:

F = m · a F = dp · A m = A · dz · ρ

With A = area and ρ = density.

Acceleration:

Dv Dt = − 1

ρ ∇ ~ p = − 1 1kgm ^{− 3}

10 ³ Pa

2 · 10 ⁶ m ∼ = 5 · 10 ^{− 4} ms ^{− 2} Velocity after 1 hour:

v = a · t = 5 · 10 ^{− 4} ms ^{− 1} · 3600s = 1.8ms ^{− 1} ∼ = 6.5kmh ^{− 1}

Although pressure gradient not that strong (10Pa/2000km), acceleration taking place still signifi- cantly.

2. Snooker

Explanation coriolis force:

F ~ c = − 2

~ Ω ∧ ~v

with Ω = ~



 0 Ω cos ϕ Ω sin ϕ



 and ~v =



 u v w





Ω ~ ∧ ~v =





− Ω sin ϕv + Ω cos ϕw uΩ sin ϕ

− uΩ cos ϕ





2-dimensional acceleration ⇒ w = 0:

F ~ c = − 2Ω





− v sin ϕ u sin ϕ

− u cos ϕ





Coriolis parameter: f = 2Ω sin ϕ ∼ = 10 ^{− 4} at midlatitudes.

(a) t = 3s, v = 1ms ^{− 1}

1

Coriolis acceleration: ^Du _Dt = f v

Displacement: x = ¹ ₂ f vt ² ∼ = 5 · 10 ^{− 4} m ∼ = 0.5mm

or

Coriolis acceleration: ^Du _Dt = 2 (Ω~v) in horizontal plane: ^Du _Dt = 2 · Ω~v · sin α

Displacement: x = ¹ ₂ ^Du _Dt t ² ∼ = 5 · 10 ^{− 4} m ∼ = 0.5mm

(b) Northern hemisphere: F ~ c ⊥ ~v

Facing to the velocity vector, the coriolis force is aimed to the right.

Southern hemisphere: F ~ c ⊥ ~v but: ~ Ω 7→ − ~ Ω

Facing to the velocity vector, the coriolis force is aimed to the left.

Athletic track races are always anticlockwise (left turn). Therefore the race in Melbourne (southern hemisphere) is backed by F ~ c , in Zurich it is the opposite, respectively.

3. “Fun-fair” ride

The centrifugal force must be stronger than the gravitational force:

F cent > g ⇒ rω ² > g ω >

r g r

>

r 9.81 ms ^{− 2} 6 m

> 1.28 rad s ^{− 1} or

rω ² = v ² r > g

v = √ g · r = 7.7ms ^{− 1} The time of circulation is: τ = ² _ω ^π

τ < 2π

1.28 rad s ^{− 1} < 4.9 s

4. Adiabatic temperature change

2

Typical example of fall wind (e.g.“Fohn ⁰⁰ , Chinook) due to adiabatic compression. Compare script p. 33.

DH

Dt = c p dT − αdp = 0 α = 1

ρ = R ^∗ T p c p

dT

T = R ^∗ dp p c p ln T | ^T T

= R ^∗ ln p | ^p p

c p (ln T 1 − ln T 2 ) = R ^∗ (ln p 1 − ln p 2 ) c p

ln T 1

T 2

= R ^∗

ln p 1

p 2

ln T 1

T 2

= R ^∗ c p · ln p 1

p 2

T 1

T 2

= p 1

p 2

If the flow is considered as adiabatic: ^DΘ _Dt = 0. Therefore Θ Altdorf = Θ Gotthard .

T G

p 0

p G

κ

= T A

p 0

p A

κ

⇒ T A

T G

= p A

p G

κ

Assumptions: p 0 = 1000 hPA, p A = 960 hPa, p G = 780 hPa, T A = 273 K, κ = _c ^R

p

= 0.286 ???

T A

T G

= 1.06 ⇒ T A = 289 K

∆T ∼ = 16 K (only due to compression)

5. Laplace equation

∇ ² Φ = ∂ ² Φ

∂x ² + ∂ ² Φ

∂y ² + ∂ ² Φ

∂z ² 6. Cartesian coordinates:

Solve ∇ ² Φ with:

Φ = (x ² + y ² + z ² ) ^{− 1 / 2}

∂Φ

∂x = − x

(x ² + y ² + z ² ) ^3/2

∂ ² Φ

∂x ² = − 2x ² − y ² − z ² (x ² + y ² + z ² ) ^{5 / 2}

3

∂Φ

∂x = − x

(x ² + y ² + z ² ) ^{3 / 2}

∂ ² Φ

∂x ² = − 2x ² − y ² − z ² (x ² + y ² + z ² ) ^5/2

(a) Cylindrical polar coordinates:

∇ ² Φ = 1 r

∂

∂r

r ∂Φ

∂r

+ ∂

∂ϕ 1

r

∂Φ

∂ϕ

+ ∂

∂z

r ∂Φ

∂z

= ∂ ² Φ

∂r ² + 1 r

∂Φ

∂r + 1 r ²

∂ ² Φ

∂ϕ ² + ∂ ² Φ

∂z ² For Φ = ln r, only dependent on r:

∂ ² Φ

∂r ² + 1 r

∂Φ

∂r

− 1 r ² + 1

r ² = 0

4