Turbulent flows
Reynolds averaging: splitting of the turbulent velocity ~v in an average
~
value ~v and a fluctuation v‘
~
~v
+
v‘
~v
=
6
6
total vector
6
time average
fluctuation
Example: Pipe
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v‘
u
r
fully turbulent
u‘
x
u(r)
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symmetric flow
Turbulent flows
u(r, φ, x, t) = u(r)+ u‘(r, φ, x, t)
v(r, φ, x, t) =
v‘(r, φ, x, t)
definition:
1 Z
u=
u(x, y, z, t)dt
TT
→ u = u(x, y, z) 6= f (t)
1.2
u(t)
u’
u_avg
1
0.8
u
0.6
0.4
0.2
0
-0.2
0
1
2
3
4
t
5
6
7
u‘ = u − u
Computational rules
f‘ = 0
f=
average of the fluctuation
f
average of the average
konst6=f (t)
|{z}
1 Z
1 Z
1 Z
(f + g)dt =
f dt +
g dt = f + g
f +g =
TT
TT
TT
f g=f g:
∂f ∂f
=
∂x ∂x
1 Z
1 Z
g=
6 g(t) →
f g dt = g f dt = f g
TT
T T
average of the derivative
Computational rules
1 Z
1 Z
fg =
f g dt =
(f + f ‘)(g + g‘) dt
TT
TT
1 Z
=
(f g + f ‘g + f g‘ + f ‘g‘)dt
TT
1 Z
1 Z
f ‘ dt +f
g‘ dt +f g‘
= fg + g
T T {z }
T T {z }
|
|
=0
= f g + f ‘ g‘
=0
usually 6= 0, z. B.. f = g → f ‘2 6= 0
level of turbulence
turbulent intensity











v
u
u
u
u
u
t
1 1 2
2 + w‘2)
Tu
=
+
v‘
(u‘





u∞ 3





3-D, incompressible, unsteady momentum equation
∂vk vj
∂uv ∂vw
konvective term:
z. B.:
;
∂xk
∂x ∂y
for turbulent flows: average of the complete equation
∂vk vj
∂
=
= (vk vj +
→
∂xk
∂xk
−ρv‘k v‘j
v‘k v‘j
)
|
{z
}
additional term
turbulent shear stress tensor
Bernoulli equation (Energy equation) for pipe flows with loss of the
total pressure
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1
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g
2
z
um
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∆ pv
p01 = p02 +
∆p
v
{z }
Total pressure loss
|
ρ
ρ
2
p1 + um1 + ρgz1 = p2 + um2 2 + ρgz2 + ∆pv
2
2
Li ρ
X
∆pv = (ξi + λi ) umi 2
Di 2
∧
ζi =
pressure loss coefficient for special places,
where losses occur
(inlet, unsteady enlargement of cross section, ellbow, . . . )
∧
λi =
loss coefficient in straight pipes
∧
umi =
average velocity
examples for pressure loss coefficients
uasually: determine ζ from experiments
ξ = ξ(Re, geometry)
Ro
Ro
Krümmer
Ri
Ri
Einlass
Ri
Ro
unsteady enlargement of cross section
A1
um,1
um,2
A2
Carnot eaution
∆p
A1 2
ζE = ρ 2 = (1 − )
A2
2 um1
∆p
laminar flow, inlet, circulare pipes
→ 1.12 ≤ ζe ≤ 1.45 experimental
pressure loss coefficients for pipes (smooth pipes)
uρD
Re =
η
C
• laminar: (Re ≤ 2.300) λ = Re
C = 64 for circular cross-sections (Hagen-Poisseuille)
• turbulent: Blasius (2.300 ≤ Re ≤ 105)
0.316
λ= √
4
Re
√
iterative solution: Prandtl: √1 = 2 log(Re λ) − 0.8
λ
Reference velocity
viscous effects in the pipe
∆p
L
um
Lρ 2
∆p = λ
um
D2
average pipe velocity
D
λ
um
um
inlet
ξ
e
ρ 2
∆pv = ξe um
2
average pipe velocity
Reference velocity
unsteady change of cross section
A1
um,1
ξ
um,2
E
A2
∆p
ρ 2
∆pv = ξE um1
2
incoming velocity
typical problem (losses)
typical problem (losses)
ξ
K
ξ
λ
ξ
λ
K
Valve
Bemerkung:
• mechanical losses are known (ζ, λ)
• the flow in the inlet and in the nozzle is lossfree
• the flow in the pipes is fully developed
typical problem (losses)
→ Bernoulli
p01
total available
energie
=
p02
+
still existing
total energy in ’2’
ρ 2
p0 = p + u + ρgz
2
Li ρ
X
∆pv = (ξi + λi ) umi 2
Di 2
∆pv
transformed
→ inner energy
total pressure loss
typical problem (losses)
Bernoulli from ’d’ → ’H’ (uH = 0)
ρ 2
pa + ud = pa + ρgH
2
u2d
→H=
→ unknown ud ?
2g
extended Bernoulli
L
ρ 2
ρ 2
p01 = pa + ρgh = pa + 2 ud + 2 umD (2ξK + ξv + λ )
D
nozzle velocity
continuity:
6
|
{z
K
}
pipe velocity
d 2
umD AD = udAd → umD = ud 
D






typical problem (losses)
lossfree
with losses
ξK = ξv = λ = 0


4
ρ 2
d
2


ρgh = 2 ud + ρ2ud D K
√
ud = 2gh
ud =
v
u
u
u
u
u
u
t
2gh!
d 4K
1+ D
Volume flux Q̇ = π4 d2ud
√
2
d
π
2
Q̇ = 4 2ghD D2 Q̇ =
2
d

Q̇ ∼ D



d 2
√
2
π 2ghD 2 d v D
u
!
4
4
D2 u
d
t
1+ D K
!
typical problem (losses)
Q
verlustfrei
verlustbehaftet
d2
D2
typical problem (losses)
ceiling of the fountain
u2d
H=
2g
no losses with losses
h!
H=h
H=
d 4
1+ D K
d
influence of D
d ↓→ H ↑
D
10.2
The velocity profile in a fully developed flow in a pipe with a smooth
surface can be approximated with the potential law:
1
r
= 1 −  n , mit n = n(Re).
v̄max
R
v̄
Re
1 · 105
6 · 105
1.2 · 106
2 · 106

n
7
8
9
10
10.2
a) Use the continuity equation to compute the relation between the
v̄m
=
average velocity v̄m and the maximum velocity v̄max, i. e.
v̄max
f (n).
r
b) At what position is v̄(r/R) = v̄m?
R
c) How can the results of a) and b) be used, if the volume flux shall
be measured?
10.2
The ratio between the average and the maximum velocity is
1
Z1
v̄m
r
2n2
n
= 2 ξ(1 − ξ) dξ =
mit ξ =
.
v̄max
(n + 1)(2n + 1)
R
0
The integral is solved using partial integration. The average velocity
is at a distance
rm
= 1 −
R
see table.





v̄m n


v̄max

10.2
Re
1 · 105
6 · 105
1.2 · 106
2 · 106
n
7
8
9
10
v̄m/v̄max
0.8166
0.8366
0.8526
0.8658
rm/R
0.7577
0.76
0.762
0.7633
Measuring v̄(r) at a distance R − rm from the wall, and with the
known v̄max the average velocity can be determined, and the volume
flux V̇ = vm π R2 can be computed.
10.6
The pressure decrease ∆p along L is measured in a fully developed
pipe flow with the volume flux V̇ .
V̇ = 0, 393 m3/s L = 100 m
900 kg/m3
η = 5 · 10−3 N s/m2
D = 0, 5 m
∆p = 12820 N/m2
ρ=
10.6
Determine
a) the skin-friction coefficient,
b) the equivalent roughness of the pipe,
c) the wall shear stress and the force of the support.
d) What is the pressure decrease, if the pipe is smooth?
10.6
10.6
11
00
00
11
00
11
F
D
τw
L
∆p
a)
=⇒
L ρ 2
∆p = λ
ūm
D 2
π D2
V̇ = ūm
4
π 2 ∆p D5
λ =
= 0, 0356
2
8 ρ L V̇
10.6
b)
ρ ūm D
Re =
= 1, 8 · 105
η
ks
= 0, 0083 (from Moody diagram)
D
=⇒ ks = 4, 2 mm
c)
momentum equation for the inner control surface:
π D2
− τW π D L = 0
∆p
4
=⇒
D
= 16 N/m2
τW = ∆p
4L
10.6
momentum equation for the outer control surface:
π D2
F = − ∆p
= − 2517 N
4
d)
λ = 0, 016
=⇒
(from diagram)
∆p = 5, 8 · 103 N/m2