Hydrodynamics
Continuity equation
A2, rho2, v2
A1, rho1, v1
Stromröhre
Conservation of mass:
ρ| 1v{z1A1} = ρ
v2A2}
| 2 {z
ṁ1
Mass flux
ṁ2
inkompressible fluid: (ρ1 = ρ2 = const)
Conservation of volume flux :
v| 1{zA1} = v| 2{zA2}
Q̇1
Q̇2
Volume flux
Hydrodynamics
Example
pipe flow: A = const
closed stream tube
water jet
m
3
m
1
m
2
closed control volume
ṁ1 = ṁ2 + ṁ3
Continuity
Important: In the 1-dimensional continuit equation ~v is an average value of the velocity. In reality ~v is not constant due to
friction, vortices, . . .!
h
y
x
~v ist constant
in the continuit equation
reality
~v = ~v (y)
mass flux must be constant
−→
Z
ρv(y) dy = ρv̄h
Bernoulli
2.Newtonian law:
mass × acceleration = sum of outer forces
d~v X
= Fa
m·
dt
Equation of motion for an infinitesimal element along one streamline
g
z
pressure
s
friction
∂p
dz
d~v
= −
− ρg
− R‘
ρ
dt
∂s
ds
inertia
gravitation
Bernoulli
along a streamline: v = v(s, t)
∂~v
∂~v
dt +
ds
d~v =
∂t
∂s
d~v
∂~v
ds ∂~v
∂~v
∂~v
−→
=
+
=
+ v
dt
∂t
dt ∂s ∂t
∂s
total
(substantial)
acceleration
of a particle
lokal acceleration
convective acceleration
example
Rohrströmung
Diffusor
A
v(x)
ρ
v1(t)
v2(t)
v0 = konst
A, ρ = konst
−→ v1(t) = v2(t)
only local acceleration
only convective acceleration
example
assumptions:
• incompressible (ρ = const)
• frictionless (R‘ = 0)
∂ =0
• steady ∂t
• constant gravitation (~g = const)


∂v  = − ∂p − ρg dz −R‘
+
v
ρ  ∂v
∂t
∂s
∂s
ds
=0
=0
∂ = d
f (s) −→ ∂s
ds
dp
dz
ρ 2
1 dv 2
ρ
= −
− ρg −→ v + p + ρgz = const
2 ds
ds
ds
2
pressure measurement
static pressure: p (Index: 1, 2, a, ∞)
1
p
at constant height ∆h = 0
p0 = p + 12 ρv 2 + ρgh
−→ p0 = p + 12 ρv 2
p
Total pressure (pitot tube): p0, p01, p02, pt
pressure measurement
potential pressure: ppot = ρgh
h
dynamical pressure: pdyn = 12 ρv 2
kinetic energy is converted, when the flow is decelerated to ~v = 0
6.4
Water flows from a large pressurized tank into the open air. The
pressure difference ∆p is measured between A1 and A2.
A1 = 0, 3 m2,
A2 = 0, 1 m2,
A3 = 0, 2 m2,
h = 1 m,
pa = 105 N/m2,
ρ = 103 kg/m3,
g = 10 m/s2
∆p = 0, 64 · 105 N/m2
Compute the
a) velocities v1, v2, v3,
b) pressures p1, p2, p3 and the pressure p above the water surface!
6.4
pressure tank with nozzle
pB
z
h = konst.
gut gerundeter Einlass
1
2
3
Venturidüse
6.4
conservation of total energy along a streamline → qualitatively
p
1/2 rho v3**2
1/2 rho v2**2
rho g h
p
B
1/2 rho v1**2
p3 = pa
p1
p2
s
1 2
Bernoulli: p0 = pb + ρgh = pi + ρvi
2
6.4
continuity (mass balance): =⇒ = ṁ = ρQ̇ = const.
ρ = const =⇒ v1A1 = v2A2 = v3A3 =⇒ A ↓ =⇒ v ↑ =⇒ p ↓
Bernoulli: p1 + ρ2 v12 = p2 + ρ2 v22
ρ 2
=⇒ ∆p = p1 − p2 = (v2 − v12) > 0
2
a) measured ∆p = p1 − p2


ρ 
A22  2
A2
→ ∆p = 1 − 2  v2 −→ v2 =
v1 = v 2
A1
2
A1
m
A2
v1 = v 2
=4
A1
s
v
u
u
u
u
u
u
u
u
u
u
t
∆p
m
2


= 12


2
ρ 1 −  A2  
s
m
A2
v3 = v 2
=6
A3
s
A1
6.4
The Venturi-nozzle is used to measure mass- and volume fluxes!
Q̇ = vA = v2A2
prinziple:
• measurement of ∆p
• computation of v2
• computation of volume- and massflux
6.4
b) determination of pressures pB , p1, . . . , p3
p0 represents the energy that can be converted into kinetic energy
ρ 2
ρ 2
ρ 2
p0 = pB + ρgh = p1 + v1 = p2 + v2 = p3 + v3
2
2
2
If we know one pressure, we can compute the other values by using
Bernoulli’s equation
Assumption: parallel streamlines at the
sharp edged exit
p3 in the exit cross section
equation of motion for an element
p(x+dx)dA
p(x)dA
z
x
g
equation of motion in x-direction for a moving control volume
dAdx (includes always the same particlesl)
equation of motion for an element
m du
dt = ẍρdAdx = p(x)dA − p(x + dx)dA


∂p 
dx dA
−→ ẍρdAdx = p(x)dA − p + ∂x
∂p
−→ ρẍ = − ∂x
Assumption: parallel stream lines
−→ ẋ = 0
velocity
−→
u = dx
dt = ẋ
∂p
=0
necessary condition: ẍ = 0 −→ ∂x
=⇒ the pressure in the exit cross-section is function of y
dp
flow into air:
= −ρg
dy
Neglect the potential energy −→ pexit = pambience = const.
6.4
p3 = p a
Remark:
Bernoulli: 0 −→ 3
pB + ρgh = pa + 12 ρv32
v
u
u
u
t
−→ v3 = ρ2 (pB − pa + ρgh)
open tank pB = pa
s
−→ v3 = 2gh 6= f (A3)
theorem of Torricelli (15.Okt. 1608 25.Okt. 1647)
6.5
Two large basins located one upon the other are connected with a
duct.
A = 1 m2 ,
Ad = 0, 1 m2,
pa = 105 N/m2,
h = 5 m,
ρ = 103 kg/m3,
H = 80 m,
g = 10 m/s2
a) Determine the volume rate!
b) Outline the distribution of static pressure in the duct!
c) At what exit cross section bubbles are produced, when the vapour
pressure is
pD = 0, 025 · 105 N/m2?
6.5
v
2
0
3
h
1
g
pa
r
A
H
4
z
4
s
5
Ad
s
5
incompressible, frictionless, steady
−→ Bernoulli
a) volume flux: Q̇ = vA = v5A5
Bernoulli: 0 −→ 5
computation of p5:
straight parallel streamlines
−→ pressure is constant
in the exit cross-section
pa + ρgH = p5 + ρg(−s) + ρ2 v52
−→ p5 = pa + ρgs
6.5
s
1 2
pa + ρgH = pa + ρgs − ρgs + ρv5 −→ v5 = 2gH 6= f (Ad, s)
2
m3
−→ Q̇ = Adv5 = 4
s
b)
p,p0
p2 = p3
pa
p4
p1
1/2 rho u^2
p5= pa+rho g s
2
rho g H
0
1
1/2 rho ud
rho g (H+h)
2
3
4
rho g 0
s
5 rho g (−s)
6.5
c) minimum pressure between 2 and 3: p2 = p3 = pD
continuity: v5∗A∗D = v ∗A
1 ∗2
Bernoulli: pa = pD + ρgh + ρv
2
v
u
u
u
u
u
u
t
h
pa − p D
− = 0.244 m2
−→ Ad = A
ρgH
H
extended Bernoulli
rho = const.
A1
v1
A2
v2
bottle neck
Delta h ~ Delta p
Theoretical volume flux: Q̇th for frictionless flow
ρ 2
ρ 2
1. Bernoulli: p1 + v1 = p2 + v2
2
2
2. continuity: v1A1 = v2A2
extended Bernoulli
A2
ratio of areas: m =
: −→ conti v1 = v2m
A1
p1 1 2 2 p2 1 2
+ v2 m = + v2
−→ Bernoulli:
ρ 2
ρ 2
−→
−→
∆p
p1 − p 2
2
2


=2
v2 1 − m = 2


ρ
v
u
u
u
u
u
u
u
t
2∆p
v2 =
ρ(1 − m2)
v
u
u
u
u
u
u
u
t
2∆p
−→ Q̇th = A2
ρ(1 − m2)
ρ
extended Bernoulli (Cont’d)
In reality losses from friction, vortices, bottle necks . . . occur.
−→ the flow is no longer frictionless
The losses and the ration of aread are put together in the
discharge coefficient α
v
u
u
u
u
u
u
u
t
2∆p
Q̇real = αA2
ρ(1 − m2)
v
u
u
u
1
u
?
u
u
α = αt
1 − m2
α? from experiments
vortex, dissipation
losses in pipe flows can be predicted similar.
extended Bernoulli (Cont’d)
pressure loss across a constructive element (ellbow, valve, . . .)
1 2
∆pv = ζ · ρv
2
∆pv
pressure loss
coeffizient: ζ = 1 2 =
dynamic pressure
2 ρv
v
u
u
u
u
u
u
u
t
v
u
u
u
u
u
u
u
t
1
2∆p
2∆p
1
=⇒ Q̇ = v · A = √ A
−→ v = √
2
ζ ρ(1 − m )
ζ
ρ(1 − m2)
(Experiments, standards −→ catalogue)
Hydrodynamics: unsteady Bernoulli
assumption
A1
1 −→ v0 v1
A0
v0 is neglectable
but h = h(t) and v1 = v1(t)
A0
0
v0
h(t)
rho
A1
1
v
1
unsteady Bernoulli from ”0” to ”1”
Z1
ρ 2
ρ 2
∂v
pa + v0 (t) + ρgh(t) = pa + v1 (t) + ρ ds
2
2
0 ∂t
s
Assumption v1(t) = 2gh(t)
dh
continuity equation: v1(t)A1 = − A0
dt
−→ differential equation for für h(t)
Hydrodynamics: unsteady Bernoulli
L
D
v1 (t)
“well rounded” inlet
s
assumption:
πs2
D
s < − √ : radial flow with Q̇ = v ·
2
8
D
√
s≥−
: v = v1
8
Hydrodynamics: unsteady Bernoulli
Potential flow without any losses −→ Bernoulligleichung
ZL
−∞
∂v
ds =
∂t
√
−D/
8 ∂v(s)
Z
−∞
dv1(t)
=
dt
∂t
ds +
ZL
√
−D/ 8
∂v1
ds =
∂t
√
−D/
8
Z
−∞
∂
∂t

√
−D/
8 D2
Z
−∞
dv1(t) ZL
ds +
ds
8s2
dt −D/√8
D  dv1(t)
D
D
dv1(t)



√
√
√




+L
=
+L+
·
=
dt
dt }
8
8
2
|
{z
well rounde inlet
ZL ∂v
dv1(t)
ds = L ·
if L D −→
dt
−∞ ∂t





D2 
v π
 1
4 



 ds

2πs2 
+
ZL
√
−D/ 8
∂v1
ds
∂t
Example: duct from a big tank
D
z
L
4
s
kg
m
ρ = 103 m
3 , g = 10 s2
g
1
L = 20m D, L1 = 5m
h
h = 5m
A
h1
L1
0
3
2
a) At what time after the immediately opening of the flap the flow
reaches 99 % of its final value?
b) At that time, what is the difference between the current pressure
and the final pressre at point “A”?
Example: duct from a big tank
sZ 4 ∂v
ρ 2
ds
a) Bernoulli: pa + ρg(h + s) = pa + ρgs + v4 + ρ
s0 ∂t
2
sZ 4 ∂v
dv4
ds = L
well rounded inlet:
s0 ∂t
dt
dv4
ρ 2
−→ ρgh = v4 + Lρ
2
dt
−→ 0T dt = L
Integration
T = 2L
√
0.99Z 2gh
0
dv4
2gh − v42
Z
dv4
2
gh − v2 4
√
√
L
2gh + v4 0.99 2gh
ln √
= √
= 10.6 s
2gh
2gh − v4 0
Example (cont’d)
the accelerated initial flow is dependent of L,
but v4(t → ∞) is independent of L.
b)
ρ 2
dv4
pa = pA + ρgh1 + v4 + L1
2
dt
ρ
t → ∞ : pa = pA,∞ + ρgh1 + 2gh
2


v42 

dv4 1 
= gh − 
from a)
dt
L
2
=⇒


L1 
N

2
 
pA − pA,∞ = ρgh 1 − 0.99 1 −  = 746 2
L
m



example: moving piston
P
L
s
D
1
pa
g
h
A piston is moving in a duct: s = s0 · sin ωt
pa = 1 bar L = 10m D h = 2m
kg
s0 = 0.1m ρ = 103 m
3
g = 10 sm2
N
pD = 2500 m
2
At what angular speed ω the pressure at the bottom of the piston
reaches the vapour pressure pD ?
example: moving piston
sZP ∂v
ρ 2
ds
pa = pP + ρgh + vP + ρ
s
2
1 ∂t
sZP ∂v
dvP
s0 L −→
ds = L
s1 ∂t
dt

s
pP = pa − ρgh + ρs0ω 2 L sin ωt − 0 cos 2ωt
2

pP,min = pD
dpP
=0
pD = pP,min at cos ωt = 0 −→
dt
v
u
u
u
u
u
u
t
pa − pD − ρgh
= 8.8 s−1
=⇒ ω =
ρs0L
example: balloons
2 balloons are connected with a pipe of length L and the crosssection A The pressure in the balloons depends linearly on the balloon volume. V0 is the volume at ambient pressure.
p = pa + C(V − V0)
At t = 0 one balloon is compressed by the volume ∆V .
A
u
1
2
L
example: balloons
a) Show, that the frictionless flow in the pipe is described by the
equation of oscillation
ü + K 2u = 0.
Determine the eigenfrequency of the system.
b) Determine the maximum of the velocity in the pipe.
c) Determine the maximum of the pressure difference between the
balloons.
Given: ∆V, L, A, ρ, C
Hint:
• General Ansatz for the eqation of oscillations
ẍ + a2x = 0:
x = C1 sin (at) + C2 cos (at)
example: balloons
a)
continuity for the balloons:
V̇1 = −uA
V̇2 = uA
Bernoulli (unsteady, frictionless)
ρLu̇ + p2 − p1 = 0
=⇒ ρLü + p˙2 − p˙1 = 0
∂
( )
∂t
p = pa + C(V − V0) ⇒ ṗ = C V̇
v
u
u
u
u
u
u
t
2CA
2CA
u=0⇒K=
=⇒ ρLü + 2CAu = 0 ⇒ ü +
ρL
ρL
example: balloons
b)
V2(t) = V + 0 − ∆V cos (Kt) ⇒ V̇2(t) = ∆V K sin (Kt) = uA
v
u
u
u
u
u
u
t
2C
∆V K
∆V K
sin (Kt) =⇒ umax =
= ∆V
⇒ u(t) =
A
A
ρAL
c)
|p2 − p1|max = |ρLu̇|max
∆V K 2
ρL∆V 2
cos (Kt) ⇒ |p2 − p1|max =
K = 2C∆V
u̇ =
A
A
6.6
The flap at the exit of the water pipe (constant width B) of a large container is opened abruptly. The appearing flow is without any
losses.
Given: H, h1, h2, g, L ;
L >> h1
6.6
Determine
a) the differential equation for the exit velocity v 3
b) - the local acceleration
- the convective acceleration
- the substantial acceleration
L
at x = when the exit velocity reaches half of its asymptotic final
2
value!
Hint: The computation of v(t) is not necessary for solving this problem.
6.6
Bernoulli from ”0” to ”3”
Z3
ρ 2
∂v
p a + ρ g H = p 3 + v3 +
ds
ρ
2
∂t
0
,
p 3 = pa
Splitting of the integral
∂v
ds ≈ 0(h1 << L)
ρ
∂t
0
Z2
∂v
h2 − h 1
h2
ds , v = v2
, h = h1 +
x
ρ
∂t
h
L
1
1
Z
dv2 h2 L
h2
dv2
h2
dv2 Z2
dx = ρ
ln
= ρ
L
⇒ρ
h
−
h
dt 1 h + 2
dt h2 − h1
h1
dt
1
x
1
L
6.6
3 ∂v
dv2
ρ
ds = ρ L
dt
2 ∂t
Z
introduce in Bernoulli
3 (L + L)
pa + ρ g H = pa + ρ2 v32 + ρ dv
dt
dv3
1
=
dt
L+L
t→∞:

gH − 2




v32 

√
1
2
g H − 2 v3e = 0 =⇒ v3e = 2 g H
local acceleration:
dv3 h2
3 2 h2
∂v
1
=
, bl (v = 1 v , x= L ) =
gH
bl =
3
3e
∂t
dt h
L+L
4 h1 + h 2
2
2
6.6
convective acceleration:
v = v2 hh2
∂v = −v h 1
2 2 h2
∂x
h22 dh
∂v
2
bk = v ∂x = − v3 h3 dx
,
dh = h2−h1
L
dx
g H h22 (h1−h2)
bk(v = 1 v , x= L ) = 4 L (h +h )3
3 2 3e
1
2
2
substantial acceleration:
h2
3 gH
g H h22 (h1 − h2)
+ 4
bs = b l + b k =
2 L + L h1 + h 2
L (h1 + h2)3