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Hydrodynamics Continuity equation streamline streamtube A 1 r 1 v 1 A 2 r 2 v 2 Conservation of mass: ρ 1 v 1 A 1 ˙ m 1 = ρ 2 v 2 A 2 ˙ m 2 Mass flux Incompressible fluid: (ρ 1 = ρ 2 = const) Conservation of volume flux : v 1 A 1 ˙ Q 1 = v 2 A 2 ˙ Q 2 Volume flux

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Hydrodynamics

Continuity equation

streamline

streamtubeA

1r

1v

1

A2

r2

v2

Conservation of mass: ρ1v1A1︸ ︷︷ ︸

m1

= ρ2v2A2︸ ︷︷ ︸

m2

Mass flux

Incompressible fluid: (ρ1 = ρ2 = const)

Conservation of volume flux : v1A1︸ ︷︷ ︸

Q1

= v2A2︸ ︷︷ ︸

Q2

Volume flux

Hydrodynamics

Examples for stream tubes:pipe flow: A = const

closed stream tube

water jet

m

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3

m1

m2

closed control volumem1 = m2 + m3

Continuity

Important: the one-dimensional continuity equation conta insan average value of the velocity ~v. In reality ~v is not constantdue to friction, vortices, . . .

x

y

h

reality~v = ~v(y)

~v ist constantin the one-dimensional continuity equation

Mass flux must be constant

−→∫

ρv(y) dy = ρvh

Formulation of the Bernoulli equation

2nd Newtonian law:Mass × acceleration = sum of outer forces

m ·d~v

dt=

Fa

Equation of motion for an infinitesimal element along one streamline:

sz

g

ρd~v

dt= −

∂p

∂s− ρg

dz

ds− R‘

��

inertia

pressure

��

gravitation

friction

Formulation of the Bernoulli equation

Velocity along a streamline: v = v(s, t)

d~v =∂~v

∂tdt +

∂~v

∂sds

−→d~v

dt=

∂~v

∂t+

ds

dt

∂~v

∂s=

∂~v

∂t+ v

∂~v

∂s

Total

(substantial)

acceleration

of a particle

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local acceleration

convective acceleration

Examples

v(x)

A

v1(t) v2(t)

ρ

DiffusorRohrströmung

v0 = konst

A, ρ = const.

−→ v1(t) = v2(t)

only local acceleration only convective acceleration

Bernoulli equation

Assumptions for further simplifications:

• Incompressible flow (ρ = const)

• Frictionless flow (R‘ = 0)

• Steady flow ∂∂t = 0

• Constant gravitation (~g = const)

ρ

∂v∂t + v∂v

∂s

= −∂p∂s − ρgdz

ds −R‘

= 0 = 0

f (s) −→ ∂∂s = d

ds

1

dv2

ds= −

dp

ds− ρg

dz

ds−→

ρ

2v2 + p + ρgz = const

Pressure

Static pressure: p (Index: 1, 2, a, ∞)

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p

p1

Total pressure (pitot tube): p0, p01, p02, pt

p0 = p + 12ρv2 + ρgh

at constant height ∆h = 0

−→ p0 = p + 12ρv2

Pressures

Potential pressure: ppot = ρgh

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h

Dynamic pressure: pdyn = 12ρv2

kinetic energy is converted, when the flow is decelerated to ~v = 0

Example 7

Water flows from a large pressurized tank into the open air. Thepressure difference ∆p is measured between A1 and A2.

A1 = 0, 3 m2, A2 = 0, 1 m2,A3 = 0, 2 m2, h = 1 m,ρ = 103 kg/m3, pa = 105 N/m2,∆p = 0, 64 · 105 N/m2 g = 10 m/s2

Compute thea) velocities v1, v2, and v3,b) pressures p1, p2, and p3 and the pressure pB above the surface.

Example 7

Pressure tank with nozzle:

z

12 3

pB h = konst.

gut gerundeter Einlass

Venturi nozzle

Example 7

Qualitative sketch: conservation of total energy along a streamlinep

p

s

pp

B

1 2

p3 = pa

rho g h

1/2 rho v1**2

1/2 rho v2**2

1/2 rho v3**2

Bernoulli: p0 = pb + ρgh = pi +1

2ρv2

i

Example 7

Continuity (mass balance): → = m = ρQ = const.

ρ = const → v1A1 = v2A2 = v3A3 → A ↓ → v ↑ → p ↓

a) Measured ∆p = p1 − p2 Bernoulli: p1 + ρ2v

21 = p2 + ρ

2v22

→ ∆p = p1 − p2 =ρ

2(v2

2 − v21) > 0

v1 = v2A2

A1→ ∆p =

ρ

2

1 −A2

2

A21

v22 → v2 =

√√√√√√√√√√√√

2

ρ

∆p

1 −

A2A1

2

= 12m

s

v1 = v2A2

A1= 4

m

sv3 = v2

A2

A3= 6

m

s

Example 7

The Venturi nozzle is used to measure mass and volume flux!

Q = vA = v2A2

principle:

• measurement of ∆p

• computation of v2

• computation of volume and mass flux

Example 7

b) determination of pressures pB, p1, . . . , p3

p0 represents the energy that can be converted into kinetic energy

p0 = pB + ρgh = p1 +ρ

2v21 = p2 +

ρ

2v22 = p3 +

ρ

2v23

If we know one pressure, we can compute the other values by usingBernoulli’s equation

p3 in the exit cross section

Assumption: parallel streamlines at thesharp edged exit

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→ pexit = pambience = const.

Example 7

p3 = pa

Remark:

Bernoulli: 0 → 3

pB + ρgh = pa + 12ρv2

3

→ v3 =√√√√√2ρ (pB − pa + ρgh)

open tank pB = pa →

v3 =√

2gh 6= f (A3)

Theorem of Torricelli (∗15.10.1608 - † 25.10.1647)

Example 8

Two large basins located one upon the other are connected with aduct.

A = 1 m2 Ad = 0.1 m2 h = 5 m H = 80 m

pa = 105 N/m2 ρ = 103 kg/m3 g = 10 m/s2

a) Determine the volume rate.b) Outline the distribution of the static pressure in the duct.c) Which exit cross section is necessary to produce bubbles, if thevapour pressure equals pD = 0.025 · 105 N/m2?

Example 8

gh

HA

Ad

par

5

4

32

1

v

s

z

0

incompressible, frictionless, steady→ Bernoullia) volume flux: Q = vA = v5A5Bernoulli: 0 → 5

pa + ρgH = p5 + ρg(−s) + ρ2v

25

s

4

5

computation of p5:straight parallel streamlines→ pressure is constant

in the exit cross-section

→ p5 = pa + ρgs

Example 8

pa + ρgH = pa + ρgs − ρgs +1

2ρv2

5 → v5 =√

2gH 6= f (Ad, s)

→ Q = Adv5 = 4m3

s

b)

Example 8

c) minimum pressure between 2 and 3: p2 = p3 = pD

continuity: v∗5A∗d = v∗A

Bernoulli: pa = pD + ρgh +1

2ρv∗

2

→ Ad = A

√√√√√√√√

pa − pD

ρgH−

h

H= 0.244 m2

Extended Bernoulli

A Av1

2

rho = const.

Delta h ~ Delta p

2v

1

bottle neck

Theoretical volume flux: Qth for frictionless flow

1. Bernoulli: p1 +ρ

2v21 = p2 +

ρ

2v22

2. Continuity: v1A1 = v2A2

Extended Bernoulli

Ratio of areas: m =A2

A1: → conti v1 = v2m

→ Bernoulli:p1

ρ+

1

2v22m

2 =p2

ρ+

1

2v22

→ v22

1 − m2

= 2p1 − p2

ρ= 2

∆p

ρ

→ v2 =

√√√√√√√√√

2∆p

ρ(1 − m2)

→ Qth = A2

√√√√√√√√√

2∆p

ρ(1 − m2)

Extended Bernoulli

In reality losses from friction, vortices, bottle necks, . . . occur.→ the flow is no longer frictionless

The losses and the ratio of areas are put together in thedischarge coefficient α⋆

vortex, dissipation

Qreal = αA2

√√√√√√√√√

2∆p

ρ(1 − m2)

α⋆ = α

√√√√√√√√

1

1 − m2

α⋆ from experiments

Losses in pipe flows can be predicted similar.

Extended Bernoulli

Extended Bernoulli equation with pressure losses:

p1 +ρ

2v21 + ρgz1 = p2 +

ρ

2v22 + ρgz2 + ∆pe + ∆pv

Pressure losses...

• across a constructive element (ellbow, valve, . . .): ∆pe = ζ · 12ρv2

• due to friction in pipes (diameter D, length L): ∆pv = λLD · 1

2ρv2

loss coefficient: ζ =∆pe12ρv2

=pressure loss

dynamic pressure

friction factor: λ =∆pv12ρv2

·D

L=

pressure lossdynamic pressure

·D

L

(Experiments, standards → catalogue)

Hydrodynamics: Bernoulli for unsteady flow

Bernoulli from ”0” to ”1”, unsteady flow

pa +ρ

2v20(t) + ρgh(t) = pa +

ρ

2v21(t) +

1∫

0ρ∂v

∂tds

rho

h(t)v

A

A

v

0

1

0

1

1

0

Assumption quasi-stationary flowA1

A0≪ 1 → v0 ≪ v1

v0 is neglectablebut h = h(t) and v1 = v1(t)

Assumption v1(t) =√

2gh(t)

Continuity equation: v1(t)A1 = −dh

dtA0

→ differential equation for h(t)