boundary layer theory_p2
DESCRIPTION
Chapter presentation on Boundary Layer TheoremsTRANSCRIPT
CVEN 311 Fluid Dynamics 1
� Further the equation can be re-written as
MOMENTUM INTEGRAL EQUATION
( ) 0
0 0
h hd dU dU
u u U dy udy h Udx dx dx
ρ − − ρ = ρ − τ
∫ ∫
( ) ( ) 0
0 0
h hd dU
u U u dy U u dydx dx
ρ − + ρ − = τ
∫ ∫
or
( ) ( )2 *
0
d dUU U
dx dxρ θ + ρ δ = τ
� For a flat plate at zero incidence (i.e. no imposed
pressure, dp/dx = 0, then
� Hence,
( ) ( )2 * 0d dUU U
dx dx
τθ + δ =
ρ
MOMENTUM INTEGRAL EQUATIONMOMENTUM INTEGRAL EQUATION of
the boundary layer which forms the
basis for approximate methods of
solving boundary layer problems
0=dx
dUU
0
2
d
U dx
τ θ
ρ=
CVEN 311 Fluid Dynamics 2
� In order to use this equation to estimate the boundarylayer thickness as a function of x, we must first:� Obtain a first approximation to the freestream velocity
distribution, U(x). This is determined from inviscid flow theory(the velocity that would exist in the absence of boundarylayer) and depends on body shape.
� Assume a reasonable velocity-profile shape inside theboundary layer
� Derive an expression for τ0 using the result obtained from theprevious item.
� Reviewing the assumptions made in the derivation, it can be seen thatthe equation is:
� Restricted to steady, incompressible, two-dimensional flow with nobody forces parallel to the surface.
� Valid for either a laminar or turbulent boundary layer flow.
� Step 1. Assume a suitable velocity profile for u(y)
inside the BL
� For example, assume a linear velocity
distribution inside a BL
� Determine constants A and B with the boundary
conditions as
� u(0) = 0 → B = 0
� u(δ) = U → A = U/ δ
VON KARMAN’S MOMENTUM INTEGRAL APPROACH
BAyyu +=)(
δ
y
U
u=
CVEN 311 Fluid Dynamics 3
� Thus δ
µµτU
y
u
y
=
∂
∂=
=0
0
VON KARMAN’S MOMENTUM INTEGRAL APPROACH
*
0 0
1 12
u ydy dy
U
δ δδ
δ = − = − = δ
∫ ∫
0 0
1 16
u u y ydy dy
U U
δ δδ
θ = − = − = δ δ
∫ ∫
dx
d
U
θ
ρ
τ=
2
01
6
d
dx U U
δ µ υ= =
ρ δ δ
2( ) 12d
dx U
δ υ=
� Integrating
� Local drag coefficient
� Skin friction drag
VON KARMAN’S MOMENTUM INTEGRAL APPROACH
2 12 xC
U
υδ = +
At x=0, δ = 0, hence C = 0
2 12 x
U
υδ =
2
2
12 12
Rexx Ux
δ υ= = 12 3.464
Re Rex xx
δ= =
0
2
2 0.5774
Re
2
f
x
cU U
τ µ= = =
ρ ρ δ
3 2 1 2 1 2
0
0
0.5774
L
DF Bdx U L B= τ = ρ υ∫
CVEN 311 Fluid Dynamics 4
VON KARMAN’S MOMENTUM INTEGRAL APPROACH� Average drag coeffcient
� Displacement thickness
� Momentum thickness
2
1.1548
2 Re
Df
L
F BLC
U= =
ρ where µ
ρULL =Re
* 1.732
2 Re x
xδδ = =
0.5774
6 Rex
xδθ = =
This problem dealt with linear velocity profile as an approximate solution. The results
obtained are rough. However the exercise illustrates the use of the momentum
integral method. Practice this method with other types of approximated velocity
profile, such as parabolic, sinusoidal, … etc.
� One key point to remember
� Be careful not to confuse the calculation for cf and Cf.
cf is a local calculation at a particular x location
(including x=L) and can only be used to calculate local
shear stress, NOT drag force. Cf is an integrated
average over a specified length (including any x ≤ L)
and can only be used to calculate average shear
stress and the integrated force over the length
CVEN 311 Fluid Dynamics 5
EXAMPLE PROBLEM
� Water at 15°C flows over a flat plate at a speed of 1 m/s. Theplate is 0.4 m long and 1 m wide. The boundary layer on eachsurface of the plate is laminar. Assume that the velocity profilemay be approximated as linear. Determine the drag force on theplate.
� Given
� Working fluid is water at T = 15 °C ⇒ ρ = 999 kg/m3 & µ = 1.14 × 10-3
N⋅s/m2
� U = 1 m/s
� L = 0.4 m
� W = 1 m
� The boundary layer on each surface of the plate is laminar
� Velocity profile is linear (assuming approximately)
� Assumptions
� Steady state condition
� Incompressible fluid flow
� Laminar boundary layer
� System diagramU = 1 m/s
L = 0.4 m
CVEN 311 Fluid Dynamics 6
� Governing Equations
� Skin friction coefficient definition:
� Reynolds number definition for a flat plate:
2
2
1U
C w
f
ρ
τ=
µ
ρUxx =Re
LAMINAR BOUNDARY LAYER ON A FLAT PLATE:
APPROXIMATE SOLUTION USING PARABOLIC
VELOCITY PROFILE
� Consider two-dimensional laminar boundary
layer flow along a flat plate. Assume the
boundary layer as parabolic.
� Find expressions for:
�The rate of growth of δ as a function of x.
�The displacement thickness, δ*, as a function of x.
�The total friction force on a plate of length L and width b.
CVEN 311 Fluid Dynamics 7
� In 1908 Blasius, a student of Prandtl, obtained an exact solution of thefollowing BL equations for a flat plate and demonstrated the shape ofthe boundary layer profile.
� With the following boundary conditions� u(y = 0) = 0
� v(y = 0) = 0
� u→ U as y → ∞
� Blasius’ exact solution is valid only for laminar BL flow with no pressuregradient.
LAMINAR BOUNDARY LAYER (BLASIUS’ EQN.)
Parallel flow along a
plate with zero
pressure gradient
LAMINAR BOUNDARY LAYER � Inside the boundary layer since the viscous forces
are predominant� Reasonable to assume: inertial and viscous forces are of
the same order of magnitude in a laminar boundary layer
Inertial forces/unit volumex
U
x
uu
2
ρρ ≈∂
∂For a flat plate
Viscous forces/unit volume 22
2
δ
µµµ
τ U
y
u
y
u
yy∝
∂
∂=
∂
∂
∂
∂=
∂
∂
If these forces are proportional to each other, then
x
k
Ux
k
x Re2
2 ′=
′=
ρ
µδ2
2
δµ
ρ Uk
x
U′=
x
Uk
υδ= A non-dimensional
parameter
Rex
kk
x Ux
δ µ
ρ= =
CVEN 311 Fluid Dynamics 8
RESULTS OF BLASIUS’EXACT SOLUTION
� u approaches to 99 % of U at k = 5.
� In other words when k becomes 5, y becomes 5.
� Therefore using the definition of k, BL thickness atany x becomes
� Using local Reynolds number definition in theabove equation we get
U
xυδ 5=
δ grows with
υ
Uxx =Re
x
x
Re
5=δ
� Shear stress
� Since in the boundary layer
� Replacing δ in the above equation
RESULTS OF BLASIUS’EXACT SOLUTION
δµµτ
δ
U
y
uU
y
u
yy
≈
∂
∂=∴≈
∂
∂
== 0
0
0
;
x
U ρµτ
3
0 constant =
xx
fUx
cU Re
0.664
Re
constantconstant
2
2
0 ====
µ
ρρ
τ Blasius’exact analytical solution
cf = local drag coefficient
CVEN 311 Fluid Dynamics 9
� Total horizontal force (or skin friction drag)
� Average drag coeffcient
� Displacement thickness
� Momentum thickness
RESULTS OF BLASIUS’EXACT SOLUTION
BLUBdxF
L
D
21
0
2123
0 664.0∫ == υρτ
L
Df
U
BLFC
Re
328.1
22==
ρwhere
µ
ρULL =Re
x
x
Re
729.1* =δ
x
x
Re
664.0=θ
VON KARMAN’S MOMENTUM INTEGRAL APPROACH
Blasius’exact solution
• laminar BL
• over a flat plate
• With zero pressure gradient (dp/dx=0)
Momentum Integral Approach (MIA)
• both laminar and turbulent BLs
• over flat and curved surfaces
• for any known U(x) and poutside (x) distributions
CVEN 311 Fluid Dynamics 10
COMPARISON OF DIFFERENT SOLUTIONS
Source: Munson, Yong, Okiishi, “Fundamentals of Fluid
Mechanics”, 3rd ed. Willey, 1998
PROBLEM
� Air flows over a sharp edged flat plate 1 m long,
3 m width with a velocity of 2 m/s. For one side
of the plate, determine δ at the end of the plate,
τ0 at the middle of the plate, FD. [ρ = 1.23 kg/m3;
υ =1.46 × 10-5 m2/s]
CVEN 311 Fluid Dynamics 11
TURBULENT BOUNDARY LAYER
� Turbulent Boundary layers are usually thicker
than laminar ones.
� Velocity distribution in a turbulent boundary
layer is much more uniform than that in a
laminar boundary layer
� Large velocity change occur in a relatively small
vertical distance
� Velocity gradient (dv/dy) is steeper in a turbulent
boundary layer than in laminar boundary layer
CVEN 311 Fluid Dynamics 12
� From experiments,
� Velocity distribution in a turbulent boundary profilefollows 1/7th power law i.e.
� Satisfactorily describes velocity distribution for mostof the region of turbulent boundary layer but givesinfinite slope at the wall,
� Therefore it can not be used to predict �0
TURBULENT BOUNDARY LAYER
17u y
U δ
=
( )61
7 7u
1 7 U at y = 0yy
− −∂= δ = ∞
∂
� Instead experimentally obtained measurements of the
shear profile are used such as
� Putting the expression for the 1/7 power law into the
equations for displacement and momentum thickness
14
2
0 0.0225 UU
υ τ = ρ
δ
* 7,
8 72
δδ = θ = δ
δ=99%
δd
δm
TURBULENT BOUNDARY LAYER
CVEN 311 Fluid Dynamics 13
� Using momentum thickness obtained and the above �0
relation in the integral momentum equation we can
obtain
� Equating this to the experimental value of shear stress:
2
0
7
72
dU
dx
δτ = ρ
147
0.022572
d
dx U
δ υ =
δ
Integrating gives:
The turbulent boundary grows as x4/5, faster than the laminar
boundary layer where δ increases as x1/2.
2
0
dU
dx
θτ = ρ
TURBULENT BOUNDARY LAYER
51
376.0
−
=
υδ
Uxx ( )5
1
Re
059.0
x
fc =
� Momentum thickness
� To find the total force, integrate over the plate
length
TURBULENT BOUNDARY LAYER
157
0.03672
Uxx
−
θ = δ = υ
2 2
0
0 0
L L
D
dF dx U dx U
dx
θ= τ = ρ = ρ θ∫ ∫
For a plate of length, L, and width B,
15
20.036D
ULF U LB
−
= ρ υ
15C 0.074Re
f L
−= 5 7(5 10 Re 10 )
L× < <
CVEN 311 Fluid Dynamics 14
� For , H. Schlichting assumed alogarithmic velocity distribution for the boundary layerflow and obtained a semi-empirical relation as follows
� The previously obtained relations of average dragcoefficient (Cf) including the above, (for laminar as wellas turbulent flows) is applicable for the entire length ofthe plate.
� However, when the plate is of such length as bothlaminar and turbulent boundary layer exist then,
TURBULENT BOUNDARY LAYER7 9(10 Re 10 )
L< <
( )2.58
10
0.455C
log Ref
L
=
15
0.074C
ReRef
LL
A= −
( ) 58.2Relog
370.0
x
fc =61
Re
22.0
xx
=δ
Critical ReCritical ReCritical ReCritical Rexxxx 3333××××101010105555 5555××××101010105555 101010106666 3333××××101010106666
Constant A 1050 1700 3300 8700
TURBULENT BOUNDARY LAYER
15
0.074 1700C
ReRef
LL
= −
Turbulent spots and the transition from laminar to
turbulent boundary layer flow on a flat plate. Flow
from left to right.
(Photograph courtesy of B. Cantwell, Stanford
University.)
CVEN 311 Fluid Dynamics 15
PRESSURE GRADIENT INSIDE A BOUNDARY
LAYER
� dp/dx=0: Fluid particles inside the BL slow down due to shear stressonly.� No flow separation can occur.
� dp/dx<0: Pressure decreases in the flow direction (favorable pressuregradient).� Pressure force is in the flow direction. It helps the flow to attach to the
surface even stronger.
� No flow separation can occur.
� dp/dx>0: Pressure increases in the flow direction (adverse pressuregradient).� Pressure force is in opposite direction of the flow.
� Fluid particles close to the wall with low momentum may come to a stop oreven move in opposite direction of the main flow, called backflow.
� Therefore, adverse pressure gradient is the necessary, but notsufficient, condition for separation.
� Boundary layer theory is no longer applicable after the separationpoint.
PRESSURE GRADIENT INSIDE A BOUNDARY
LAYER
CVEN 311 Fluid Dynamics 16
FLOW SEPARATION
� Flow separation is generally undesired. It reduces liftforce on an airfoil or increases drag force on a bluntbody.
� Turbulent BLs are more resistive to separationbecause, compared to a laminar one, in a turbulentBL velocities close to the wall are higher.
Separation over an airfoil Separation over a tennis ball
SOLVED IN CLASS
� Air enters a square duct through a 1-ft opening as is shown inFig. Because the boundary layer displacement thicknessincreases in the direction of flow, it is necessary to increase thecross-sectional size of the duct if a constant velocity (U=2 ft/s)is to be maintained outside the boundary layer. Plot a graph ofthe duct size, d, as a function of x for 0 ≤x≤10 ft if U is toremain constant. Assume laminar flow. [ν= 1.57× 10-4 ft2/s]
CVEN 311 Fluid Dynamics 17
� Assuming a velocity distribution defined by u/U =
sin (πy/2δ ) determine the general expressions
for growth of the laminar boundary layer and for
the surface shear stress for a smooth flat plate.
PROBLEM 1
� A flat plate is drawn submerged through still
water at a velocity of 9 m s−1. If the plate is 3 m
wide and 20 m long determine the position of
the laminar to turbulent transition and the total
drag force acting on the plate. Take water
temperature as 20 °C. [Ans: 0.01 m, 9.5 kN]
CVEN 311 Fluid Dynamics 18
� Estimate the skin friction drag on an airship 92 m
long, average diameter 18 m, being propelled at
130 km h−1 through air at 90 kN m−2 absolute
pressure and 27 °C. [Ans:6.7 kN]
PROBLEM 2
� Air at 20 °C and 760 mm Hg absolute pressure flows past asmooth wind tunnel wall, with a free stream velocity of 160 kmh−1. Determine the position along the wall, in the flowdirection, at which the boundary layer becomes turbulent andthe distance to a boundary layer thickness of 25 mm. All wallmeasurements may be assumed to be taken from the workingsection entrance and edge corner effects may be ignored.
[Ans: 33.6 mm, 1.4 m]
PROBLEM 3
CVEN 311 Fluid Dynamics 19
� Show that, if a flat plate, sides a, b in length, is towedthrough a fluid so that the boundary layer is entirelylaminar, the ratio of towing speeds so that the drag forceremains constant regardless of whether a or b is in theflow direction is given by
where Ua is the free stream velocity if side a is in the flowdirection and Ub is the corresponding fluid velocity if b isin the flow direction.
� Repeat Problem above if the boundary layer is consideredfully turbulent.
3 / baUU ba =
9 / baUU ba =
PROBLEM 4