Boundary Layers As a fluid flows over a body, body, the no-slip condition ensures that the fluid next to the boundary is subject su bject to large shear. shear. A pipe is enclosed, so the fluid is fully bounded, b ounded, but in an open flow at what distance away from the boundary bound ary can we begin to ignore this shear? There are three main definitions of boundary layer thickness: 1. 99% thickness 2. Displacement thickness 3. Momentum thickness
99% Thickness
y
U
x
H(x)
U is the free-stream velocity
H(x) is the boundary layer thickness when u(y) ==0.99U
Displacement thickness y
There is a reduction in the flow rate due to the presence of the boundary layer U u
y This is equivalent to having a theoretical boundary layer with zero flow
Hd Uu
Displacement thickness The areas under each curve are defined as being equal: g
q!
´ U u dy
and
q ! d U
0
Equating these gives the equation for the displacement thickness: g
d
¨1 u ¸dy ! ´© ¹ Uº 0ª
Momentum thickness In the boundary layer, the fluid loses momentum, so imagining an equivalent layer of lost momentum: y
g
´
m ! u U u dy
y
and
m ! U 2 m
0
Equating these gives the equation for the momentum thickness: g
m
¨1 u ¸dy !´ © ¹ Uª Uº 0 u
Laminar boundary layer growth dX X + dX
H(x)
y
dy X
x
Boundary layer => Inertia is of the same magnitude as Viscosity a) Inertia Force: a particle entering the b.l. will be slowed from a velocity U to near zero in time, t. giving force FI w VU/t. But u=x/t => w t l /U /U where U is the characteristic velocity and l the characteristic length in the x direction. Hence FI w VU2/l b) Viscous force: FQ w xX/ xX/xy w Qx2u/x u/xy2 w QU/H2 since U is the characteristic velocity and y direction
H the characteristic length in the
Laminar boundary layer growth Comparing these gives: VU2/l w QU/H2
w
l
!5
l ( Blasius )
l
So the boundary layer grows according to
Alternatively, Alternatively, dividing div iding through by l , the non-dimensionalised boundary layer growth is given by:
w l
1 Rl
Note the new Reynolds number characteristic velocity and characteristic length
!
l
Ul Ul
!
Boundary layer growth
Length Reynolds Number U
l Rl !
l
Flow at a pipe entry
d
U
l
If the b.l. meet while the flow is still laminar the flow in the pipe will be laminar If the b.l. goes turbulent before they meet, then the flow in the pipe will be turbulent
Length Reynolds number and Pipe Reynolds number The critical Reynolds number for flow along a surface is Rl =3.2*105 In a pipe, the Reynolds number is given by
Re !
Vum d Q
Considering a pipe as two boundary layers meeting, meeting, d=2a=2 H and from above
!5
l
The mean velocity in the pipe, u m, is comparable to the free-stream velocity, U
Re !
U
.10
l U
! 10
Ul
! 10 Rl
If Rl =3.2*105 then Re=5657
Boundary layer equations for laminar flow These may be derived by solving the Navier-Stokes equations in 2d. ¨ x 2 u x 2 u ¸ du 1 xp xu xv ©© 2 2 ¹¹ ! !0 xy º dt xx ª xx xx xy Continuity
Momentum U
Assume: 1. The b.l. is very thin compared to the length 2. Steady state
Boundary layer equations for laminar flow This gives Prandtl¶s Prandtl¶s b.l. equation:
1 xp xx
x 2u xy 2
!u
xu xx
v
xu xy
rate of change of u with x is small compared to y
Blasius produced a perfect solution of these equations valid for 0
Blasius Solution y' 0 1 2 3 4 5 6 7
5 ' y
f' (or u/U) /U) 0 0. 330 0. 630 0. 846 0. 956 0. 992 0. 999 1. 000
0 0
1
u/U
y' ! y
U l
f'' 0. 332 0. 0. 332 0. 0. 323 0. 0. 267 0. 0. 161 0. 0. 064 0. 0. 002 0. 0. 000
Laminar skin friction The shear stress at the surface can be found fou nd by evaluating the velocity gradient at the surface 0 !
xu xy
0
The friction drag force along the surface is then found by integrating over the length
¨ xu ¸ ´0 ©©ª xy ¹¹º y!0dx l
Ff ! b
where b is the breadth of the surface su rface
Laminar skin friction From the Balsius solution, the gradient of the velocity profile at y=0 yields the result:
¨ ¸ ¹ ªxº
0 ! 0.332 ©
0.5 x
The shear force can be obtained by integration along the surface l
´
Ff ! b 0 dx ! 0.66 .664 b
0.5 l
0
The frictional drag coefficient can then be calculated C f !
Ff 1 2
AU
12
2
! 1.33R l
Force and momentum momentum in fluid mechanics - refresher Newton¶s laws still apply. Consider a stream tube: u1,A1
u2,A2
q1=u1A1
q2=u2A2
mass entering in time, t, t, is u 1A1t momentum entering in time, t, is m 1 = (u1A1t)u 1 momentum leaving in time, t, is m 2 = (um 2A2t)u2 Impulse = momentum change, F = (m 2 ± m1)/ t = (u22A2-u12A1)
The von Karman Integral Equation (VKI) C
U
H2 - H1
B
H1
H2
u2(y)
u1(y) A
(x
D
Flow enters on AB and BC, and leaves leav es on CD
VKI The momentum change between entering and leaving the control volume is equal to the shear force on the surface: 2
1
´
2 0 (x ! u 2 dy
Force on fluid
0
´
2
0
(CD)
2
u 1 dy
2
1
(BC)
(AB)
By conservation of fluid mass, any fluid entering the control volume must also leave, therefore H 2
(H U 2 H 1) !
H 1
y ´ u d y ´ u d 2
1
0
2
0
1
0 (x ! ´ u 22 Uu 2 dy ´ u12 Uu1 dy 0
0
VKI As (x p0, the two integrals on the right become beco me closer and the equation may be written as a differential:
¨ 2 ¸ 0 dx ! d ©© ´ u u dy ¹¹ ª0 º ¨ d u¨ u¸ ¸ 2 © 1 ¹dy ¹ 0 ! © ´ dx ©ª 0 ª º ¹º
The integral is the definition d efinition of the momentum thickness, so 0 !
2
d m dx
d
d dx
if
(x)
Turbulent boundary layers The assumption is made that the flat plate approximates to the behaviour in a pipe. p ipe. The free stream velocity, velocity, U, corresponds to the velocity at the centre, and the boundary layer thickness, H, corresponds to the radius, R. 1/7 Power Law
From experiments, one possibility for the shape of the boundary layer profile is u
y¸ ¨ !© ¹ ªº
1 7
and measurements of the shear profile give
¨ ¸ ¹ ª U º
0 ! 0.0225 U 2 ©
1 4
Turbulent boundary layers Putting the expression for the 1/7 power law into the equations for displacement and momentum thickness
d !
8
, m !
7 72
H=99% Hd Hm
Turbulent boundary layers 0 !
2
d m
becomes
dx
0 !
7 72
2
d dx
Equating this to the experimental value of shear stress: 7 dH ¨ ¸ ! 0.0225© ¹ 72 dx ª º
1 4
Integrating gives:
¨ Ux ¸ ¹ ª º
! 0.37x ©
15
The turbulent boundary grows as x 4/5, faster than the laminar boundary layer.
Turbulent boundary layers Momentum thickness
m
!
7 72
¨ Ux ¸ ¹ ª º
15
! 0.036x ©
2 To find the total force, first find the shear stress 0 ! U
d m dx
then integrate over the plate length l
l
f ! ´ 0 dx ! ´ U 0
0
2
d m dx
dx ! U 2 m
For a plate of length, l , and width b, 15
Ul ¨ ¸ Ff ! 0.036 U l b© ¹ ª º 2
f ! 0.074
15 l
(5 *10
5
7
R l 10 )
Logarithmic boundary layer From the mixing length hypothesis it can be shown that the profile is logarithmic, but the experimental values are different from those in a pipe
u V
*
! 5.56 5.85ln !
and the friction coefficient
f
yV* 0.455
logR
2.58
l
¨ 0.455 1.328 ¸¹ © A! © logR rit 2.58 R rit ¹R rit ª º
A R l
9 (0 R 10 ) l
c
c
c
(A is a correction co rrection constant if part of the b.l. is laminar)
Quadratic approximation to the laminar boundary layer Blasius (exact) (exact) Quadratic
1 H
/ y
0 0
1 u/U
Quadratic approximation to the laminar boundary layer Remember - boundary layer theory is only applicable inside the boundary layer. u
!2
¨y¸ ¹ ªº y
2
©
This is sometimes written with L=y/H and F(L)=u/ as
F ! 2 2
It provides a good approximation to the shape of the laminar boundary layer and to the shear stress at the surface
Turbulent Boundary Layer
Sub-Layer Laminar
