laminar boundary layers
Why Boundary layer theory?
• Only pressure distributions can be computed with the potential
theory
→ Lift force
• The drag force cannot be described with the potentila theory. Viscous forces have to be considered
The pressure ditribution at slender bodies fits well with the theoretical distribution from the potential theory, if Re ≫ 1. The influence of
the viscous forces is limited to a thin layer at the wall → boundary
layer
laminar boundary layers
Beispiel
ua
ua
0.99 ua
Boundary layer edge
δ
No slip
laminar boundary layers
• The segmentation of the flow into two parts (the frictionless outer
part + the viscous boundary layer) allows a complete description
of the flow field.
• The boundary layer theory is not valid in the nose region!
1111111111111111
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laminar boundary layers
• Due to the deceleration in the boundary layer the streamlines are
pushed away from the wall. The streamlines are no longer parallel to the wall.
11111111111111111
00000000000000000
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• The line δ(x), describing the edge of the boundary layer (boundary layer thickness) is not a streamline. It denotes the line where
the velocity reaches the value of the outer flow up to a certain
amount. Usually 99 %
u(y)
= 0.99
ua
arbitrary
Boundary layer thickness
In the boundary layer:
O(Inertia) ≈ O(viscous forces)
∂u
∂ 2u
ρu
≈η 2
∂x
∂y
dimensionless values → O(1)
u
x
ū =
x̄ =
u∞
L
y
ȳ =
δ
u∞
u∞
→ ρu∞
≈η 2
L
δ
L
=p
→δ∼q
u∞ρL
ReL
L
η
√
δ∼ L
Boundary layer thickness
Introduction of dimensionless variables in the Navier-Stokes equations for 2-d steady, incompressibles flows. Dimension varaiables are
O(1).
1
Neglect all terms with the factor
or smaller.
Re
→ Boundary layer equations are valid for Re ≫ 1 without strong
curvatures.
∂u ∂v
continuity:
+
=0
∂x ∂y
x-mom:
y-mom:
∂u
∂ 2u
∂u
∂p
ρu + ρv
+η 2
=−
∂x
∂y
∂x
∂y
|{z}
dp
=
dx
∂p
=0
∂y
y-mom.: The pressure is constant normal to the main stream
direction. It is impressed from the frictionless outer
flow.
Boundary layer edge for a flat plate (y = δ)
∂u
∂ 2u
u=U →
=0→ 2 =0
∂y
∂y
x-mom:
∂p
∂U
=−
ρU
∂x
∂x
frictionless outer flow field
Euler equation for y = δ
3 cases depending on the pressure gradient
∂p
∂U
1
=0→
= 0 → U = const
∂x
∂x
flat plate, boundary layer (Blasius)
U= const
∂p
∂U
∧
2
<0→
> 0 = accelerated flow
∂x
∂x
convergent channel (nozzle)
U1< U 2
1
2
∂p
∂U
∧
3
>0→
< 0 = decelerated flow
∂x
∂x
divergent channel (diffusor)
U1> U 2
1
2
Separation
At the wall (y = 0)
No-slip condition: u = v = 0
→
x-mom.:
→
∂ 2u
∂p
= η 2 |y=0
∂x
∂y
∂u
τw = η
∂y
∂p ∂τ
= |y=0 =
∂x ∂y
∂p
flat plate:
= 0 (no pressure gradient, U = const.)
∂x
∂ 2u
∂u
|y=0 = 0
( |y=0 = const)
2
∂y
∂y
→ no curvature of the velocity profile at the wall
Displacement-/Momentum thickness
δ1 (displacemant thickness): characteristic measure for the displacement of an undisturbed streamline
y
y
U
δ
δ1
U
u
Areas are the same
δ1
Mass flux
width
u
= const
Displacement-/Momentum thickness
δ
δ1
δ1 from ṁ = const.
U (δ − δ1) =
→
δ1 =
Zδ
0
Zδ
0
u dy →
Zδ
0
U − u dy = U δ1
u(y)
(1 −
) dy
U
δ1
in dimensionless form =
δ
Z1
0
u(y) y
)d
(1 −
U
δ
(δ, δ1 6= f (y))
Displacement-/Momentum thickness
Due to friction some momentum losses occur.
From the momentum balance
δ2 =
Z∞
u
u
(1 − ) dy
U
U
Z1
u y
u
(1 − ) d
U
U δ
0
δ2
=
δ
0
To compute the drag, these two measures +
the von Kàrmàn integral equation are used.
Displacement-/Momentum thickness
Integration of the x-momentum equation
d 2
dU τw
(U δ2) + δ1U
=
dx
dx
ρ
dδ2 1 dU
τw
or
+
(2δ2 + δ1) =
dx U dx
ρU 2
• Assume a function (polynomial, . . . ) for the velocity
• Use the boundary conditions to compute the coefficients
• Compute δ1 and δ2
• Use the von Kàrmàn integral equation to compute τw (x) or δ(x).
Ansatz: polynom for the velocity profile
n
y i
X
y
u(x, y)
ai
= f (x, )
=
U (x)
δ
δ
i=0
selfsimilar profile ai(x), δ(x)
boundary conditions
y
1. no-slip condition (Stokes) for = 0 → u = v = 0
δ
y
2. boundary layer edge = 1 → u = U
δ
(uB = uw )
3. from x-momentum
∂ 2u
∂p
η 2 |y=0 =
(= 0 for a flat plate)
∂x
∂y
∂p
from Euler equation (Bernoulli)
∂x
Only: if the degree of the polynomial is > 2, other boundary conditions are necessary
from the continuity at the boundary layer edge
∂u
y
=0
4. > 1 →
δ
∂y
continous transition from the boundary layer to the outer flow
y
∂ 2u
5. = 1 → 2 = 0
δ
∂y
frictionless flow
Example 15.8
In the stagnation point of a flat plate that is flown against normally
to the outer flow ua(x) is accelerated in such a way that a constant
boundary layer thickness δ0 is generated. The velocity profile is assumed to be linear as a first approximation.
u(x, y)
y
= a0 + a1
ua(x)
δ0
Determine:
a) the constants a0, a1
b) the distribution of the outer velocity ua(x) using the von Kármán
integral equation.
c) the tangential force that is applied between x = 0 and x = L on
the plate with the width B.
Given: δ0, L, η, ρ, B
Example 15.8
von Kármán integral equation
1 dua
τw
dδ2
+
(2δ2 + δ1) = 2
dx ua dx
ρua
u a (x)
d0
y
x
x=0
x=L
Example 15.8
Remark: ua = Ua = U = ue = . . .
1) a0, a1 = ?
different notations
no-slip condition u(y = 0) = 0 → a0 = 0
u
at the boundary layer edge: | y =1 = 1 → a1 = 1
U δ0
u(x, y)
y
→
=
U (x)
δ0
Example 15.8
δ1
=
δ0
2) displacement thickness:
Z1
0
u(y) y
(1 −
)d
U
δ
2#1
y
1
1 y
=
=
−
δ0 2 δ0
2
"
0
δ2
momentum thickness: =
δ0
Z1
0
→
1
δ1 = δ0
2
u y
u
(1 − ) d
U
U δ
2
3 1
1
1 y
1 y
1
−
=
= → δ2 = δ0
2 δ0
3 δ0
6
6
0
∂δ2 1
∂δ2
→
= = const →
=0
∂δ0 6
∂x
Example 15.8
∂u
Wall shear stress: τw = η
∂y y/δ=0
U
U ∂( Uu )
→ τw = η
y =η
δ0 ∂( δ )
δ0
0
Example 15.8
von Kàrmàn integral equation
1
U
dU
=U
η 2
2
dx
ρU δ0
η 6
U (x) = 2 x
ρδ0 5
3) F =
ZL
0
6 η η
B
τ (x) B dx =
2
5 ρδ0 δ0
1
2 16 + 21
!
η 6
= 2
ρδ0 5
(usually δ = δ(x))
ZL
x dx
0
3 η 2 BL2
F =
5 ρ δ03
15.3
From exercise 15.2 the drag of a flate plate, wetted on both sides, of
length x and width B can be determined with
W =
Z x
0
τw (x)Bdx = ρ
Z δ(x)
0
u(u∞ − u)Bdy.
Using this equation and with the approximation of the velocity profile
y 2
y
u(x, y)
= a0 + a1 + a2
u∞
δ
δ
the boundary layer thickness δ(x) isrto be determined and compared
νx
.
with the Blasius solution δ(x) = 5.2
u∞
15.3
Given: ν, u∞
Determination of the coefficients a0, a1, a2 by using the boundary
conditions:
B.C.1: no slip u(x, y = 0) = 0
B.C.2: boundary layer edge u(x, y = δ) = u∞
∂ 2u(x, y = 0) ∂p
=
R.B 3: at the wall (from x-momentum) η
=0
2
∂x
∂y
15.3
If additional boundary conditions are necessary, a steady transition
y
at = 1 can be assumed, i.e.
δ
n
∂ u
= 0 with n ≥ 1
n
∂y y=δ
=⇒ from B.C.1 follows a0 = 0
from B.C.2 follows a1 + a2 = 1
from B.C.3 follows
∂ 2u
1 ∂ 2(u/u∞)
1
= u∞ 2
= 2u∞ 2 a2 = 0
2
2
∂y
δ ∂(y/δ)
δ
15.3
=⇒ a2 = 0, a1 = 1
u(x, y) y
linear distribution.
=
=⇒
u∞
δ
Permutation of the order of boundary conditions, e,g,:
B.C.1 u(x, y = 0) = 0
B.C.2 u(x, y = δ) = u∞
∂u
B.C.3
=0
∂y y=δ
=⇒ a0 = 0; a1 = 2; a2 = −1
parabolic distribution
15.3
Since the approximation of the velocity profiles is not an exact solution of the boundary layer equations, the boundary conditions are
not satisfiable in all cases. Hence, it has to be paid attention, that
the physical boundary conditionsa re satisfied first.
Computation of the boundary layer thickness δ(x):
∂u
u∞ ∂u/u∞
u∞
τw = η
=η
=η
∂y y=0
δ ∂y/δ y=0
δ
2
y
y
y
y
2
u(u∞ − u) = u∞ u∞ − u∞
= u∞
−
δ
δ
δ
δ
15.3
with the equation
Z x
Z δ(x)
B
τw (x)dx = Bρ
u(u∞ − u)dy
0
0
Z 1 Z x
y
y
y 2
u∞
2
δ
d
dx = ρu∞
−
=⇒
η
δ(x)
δ
δ
δ
0
0
1
Z x
2
3
1 y
1 y
dx
2
−
= ρu∞δ(x)
ηu∞
2 δ
3 δ
0 δ(x)
0
1
2
= ρu∞δ(x)
6
15.3
Differentiating gives
1
1ρ
dδ(x)
= u∞
δ(x) 6 η
dx
1ρ
=⇒ dx = u∞δ(x)dδ(x)
6η
Integration:
1ρ
x=
u∞δ 2(x)
12 η
r
12νx
=⇒ δ(x) =
u∞
r
r
√
νx
νx
≈ 3, 5
=⇒ δ(x) = 12
u∞
u∞
15.3
Compare with Blasius solution: δ(x) = 5.2
r
νx
u∞
15.4
The velocity profile of a laminar incompressible boundar layer with
constant viscosity η can be described with a polynomial:
y 3
y y 2
u(x, y)
+ a3(x)
= a0 + a1(x)
+ a2(x)
ua(x)
δ
δ
δ
The outer velocity ua(x) is given by the following approach:
ua(x) = ua1 − C · (x − x1)2.
ua1 is the outer velocity at x1 and C is a positive constant. The boundary layer thickness at x2 is δ(x2).
15.4
Given: ρ, η, x1, ua1, δ(x2), C, mit: C > 0
Determine:
a) the pressure gradient∂p/∂x in the flow as a function of x.
b) the coefficient a0 and the coefficients a1(x), a2(x), a3(x).
15.4
∂p
a)
in the flow:
∂x
∂p
∂ua
=−
frictionless outer flow: ρua
∂x
∂x
∂p
∂ua
=⇒
= −ρua
= −ρ(ua1 − C(x − x1)2) · (−2C(x − x1))
∂x
∂x
∂p
= +2ρC(x − x1)(ua1 − C(x − x1)2)
∂x
∂p
=0
∂x x1
∂p
= 2ρC(x2 − x1)(ua1 − C(x2 − x1)2)
∂x x2
15.4
b) 4 B.C:
y
I = 0 u = 0 H.B. =⇒ a0 = 0
δ
y
II = 1 u = ua
δ
y
∂p
∂ 2u
III = 0
= η 2 at the wall
δ
∂x
∂y
y
∂u
IV = 1
=0
δ
∂y
15.4
2
3
y
y
y
u = a1
+ a3
ua
+ a2
δ
δ
δ
1
y
y2
a1 + 2a2 2 + 3a3 3
δ
δ
δ
∂u
=
∂y
∂ 2u
∂y 2
=
!
1
y
+ 2a2 2 + 6a3 3
δ
δ
ua
ua
15.4
II:
1 = a1 + a2 + a3
IV:
0 = a1 + 2a2 + 3a3
2a2
∂p
= η ua 2
=⇒
∂x
δ
III:
=⇒
1 δ 2 ∂p
a2 =
2 η ua ∂x
δ 2 2 ρ C (x − x1)(ua1 − C (x − x1)2)
a2(x) =
2η
(ua1 − C · (x − x1)2)
δ2 ρ C
(x − x1)
a2(x) =
η
15.4
II / IV:
1
1
a3 = −
− a2
2
2
3
1
a1 =
− a2
2
2
therefore:
a1(x) =
1 δ2 ρ C
3
−
(x − x1)
2
2 η
1 δ2 ρ C
1
−
(x − x1)
a3(x) = −
2
2 η
15.5
In the boundary layer of a flat plate the velocity profiles and the
pressure distribution are measured. Thre pressure distribution on
x 2
p(x)
, with k = const < 1
= 1−k
the surface is described with
p0
l
and the velocity profiles are presented with
1
u(x, y)
y 2
=
ua(x)
δ0
with a constant boundary layer thickness δ0.
15.5
Determine the wall shear stress τw using the Kármán integral equation
1 dua
τ (y = 0)
dδ2
+
(2δ2 + δ1) +
=0
2
dx ua dx
ρua
Given: p0, k, δ0, l
Boundary layer equation (x-momentum):
∂u
∂u
1 ∂p η ∂ 2u
u +v
=−
+
∂x
∂y
ρ ∂x ρ ∂y 2
15.5
a)
δ1
•
δ0
1 dua
τ (y = 0)
dδ2
+
(2δ2 + δ1) +
=0
2
dx ua dx
ρua


Z 1
1
Z 1
y 2
y
u
y
d
1 −
d
1−
=
=
=
ua
δ
δ0
δ
0
0
y 3
2
y 2
−
δ 3 δ
!1
0
1
=⇒ δ1 = δ0
3
R
R
1/2
3/2
2
• δδ2 = 01 uua 1 − uua d yδ = 01 yδ
− yδ d yδ = 23 yδ
− 12 yδ
0
1
=⇒ δ2 = δ0
6
dδ2
•
= 0, since δ0 = const.
dx
15.5
from the x-momentum equation for y = δ0:
dp
dua
=−
ρua
dx
dx
dp
=⇒ τw = − (2δ2 + δ1)
dx x 2
2x
d
dp
1−k
= −p0 k 2
= p0
with
dx
dx
l
l
2x 2
4
δ0 x
=⇒ τw = p0k 2 δ0 = p0k 2
3
l 3
l
15.7
The velocity profile in the laminar boundary layer of a flat plate
(length L) is described by a polynomial of fourth order
y y 3
y 4
y 2
u
= a0 + a1
+ a3
+ a4
.
+ a2
ua
δ
δ
δ
δ
15.7
a) Determine the coefficient of the polynomial!
b) Proof the following relationships:
δ1
= 3/10
δ
δ2
= 37/315
δ
p
δ
= 5.84/ Rex
x
p
cw = 1.371/ ReL
15.7
a) Boundary conditions:
v
u
y
= 0,
=0
=0:
δ
ua
ua
y
u
=1
=1:
δ
ua
from boundary layer equation
∂ 2u
∂u
∂u
=η 2 :
ρ u +v
∂x
∂y
∂y
y
=0:
δ
y
=1:
δ
∂ 2(u/ua)
u=v=0:
=0
2
∂(y/δ)
∂u ∂u
∂ 2(u/ua)
=0
=
=0:
2
∂x ∂y
∂(y/δ)
15.7
frictionless outer flow
∂(u/ua)
y
=1: τ ∼
=0
δ
∂(y/δ)
y 3 y 4
y u
+
−2
=2
ua
δ
δ
δ
b)
Z 1
y
u
δ1
3
d
1−
=
=
δ
ua
δ
10
0
Z 1 u
y
37
u
δ2
1−
d
=
=
δ
ua
δ
315
0 ua
15.7
von K ármánsche integral equation:
dδ2 τ (y = 0)
=0
+
2
dx
ρua
ηua d(u/ua)
ηua
τ (y = 0) = −
= −2
δ d(y/δ) y/δ = 0
δ
δ
5.84
Integration: = √
x
Rex
Z L
Z L
τw
τ (y = 0)
2
1.371
2
dx = −
dx = √
cw =
2
2
L 0 ρua
L 0
ReL
ρua