boundary layer theorytoufiquehasan.buet.ac.bd/me 323-(jan 2020)-lec-4.pdf©dr. a.b.m. toufique hasan...

© Dr. A.B.M. Toufique Hasan (BUET) 1L-3 T-2, Dept. of ME ME 323: Fluid Mechanics-II (Jan. 2020)

ME 323: FLUID MECHANICS-IIDr. A.B.M. Toufique Hasan

Professor Department of Mechanical Engineering

Bangladesh University of Engineering and Technology (BUET), Dhaka

Lecture-0404/03/2020

[email protected]

Boundary Layer Theory


Recall RTT (ME 321 Fluid Mechanics‐I)

sysCV CS

ˆdB d ρ dV dAdt dt

V n

RTT (Reynolds Transport Theorem) relates between the system approach with finite controlvolume (CV) approach for a system property:

B = any extensive property (such as mass, momentum, energy etc.)β = any intensive property per unit mass (such as mass per mass,momentum per mass, etc.)

0ˆCS

dAnV

Conservation of mass for steady incompressible flow in integralformulation is (covered in ME 321)

0ˆCSCV

dAVddtd nV

0

const.

syst

systsys

dtdm

mB

1massmass


Recall linear momentum equation (ME 321 Fluid Mechanics‐I)

Newton’s second law of motion for a system is

contents of the CV CScontrol volume ˆs BdF F F ρ dV dAdt

V V V n

** Vector equation

sys

sys contents of thecontrol volume

d mF F

dt

V

For fluid dynamics:

: External forces acting on the content of the control volume (CV)(such as pressure force, viscous shear force, gravity etc.)

F

Conservation of linear momentum(momentum)(momentum),

(mass)mB m

m VV V

RTT takes the form of

sys

CV CSˆ

d m d dV dAdt dt

V

V V V n


The integrand in the mass flow rate integral represents the product of the component ofvelocity, V perpendicular to the small portion of the control surface and the differential area,dA. As shown in figure (dot product)

veˆ nV ; +ve for flow out from the control volume veˆ nV ; -ve for flow in to the control volume


Momentum Integration Equation

U

U

①

②

③

④

0ˆCS

dAnV

Conservation of mass for steady incompressibleflow in integral formulation is (covered in ME 321)

000 )(00

xh

bdyubdyU

)(

0

xdyuhU

① ② ③ ④

… … … . …… (1)

Consider a flat plate boundary layer and bound thisregion by a finite control volume with 4 controlsurfaces as shown in Fig.


Momentum Integration EquationConservation of linear momentum for steadyincompressible flow in integral formulation is(covered in ME 321)

① ② ③ ④

CV CS ˆs BdF F F ρ dV dAdt

V V V n

CS

ˆsF dA V V n

( )

0 00 0

h x

DF U U bdy u u bdy

( )2 2

0

x

DF U bh b u dy

( ) ( ) 2

0 0

x x

DF U b udy b u dy

)(

0:)1(Eq.

xdyuhU

( )

0( )

x

DF b u U u dy

FD

Pressure is uniform, so there is nonet pressure force on the controlvolume



( )2

0( ) 1

x

Du uF x bU dyU U

2( ) ... ... ... ... ... ... (2)DF x bU

( )

01

x u u dyU U

Momentum thickness is thus a measure of total plate drag.

Drag also equals the integrated wall shear stress along the plate:

0( ) ( )

( ) .... .... ... ... ....(3)

x

D w

Dw

F x b x dx

dF b xdx

FD



2 ... ... ... ... ... ... (4)DdF dbUdx dx

Constant for 0 (ZPG flow)dpUdx

Combining Eq. (3) and (4):

The derivative of Eq. (2):

2Dw

dF dbU bdx dx

2w

dUdx

Momentum integral equation applied to a general boundary layerflows (laminar/turbulent).And is of general use in deriving further important relations in theboundary layer flows.

θ = f(x)

FD


Application of Momentum Integral EquationLaminar flowsVelocity profile inside a laminar boundary layer can be approximated by:

y

xyyy

Uyxu

for1

)(0for2),(2

U∞U∞

δ(x)

x

y

(0,0)

Laminar

( )

0

2 2( )

0

1

2 21

... ... ... ... ... ... ...

... ... ... ... ... ... ...

x

x

u u dyU U

y y y y dy

Momentum thickness for this given velocity profile:

215lam lam


Application of Momentum Integral Equation

Again the shear stress can be calculated as:

0

2

0

2

y

w

yw

yyUy

yu

U∞U∞

δ(x)

x

y

(0,0)

Laminar

Uw

2

From momentum integral equation:

;15

1522 2

2

dxU

d

dxdUUdxdUw


Application of Momentum Integral EquationOn integrating the above equation; and considering

U∞U∞

δ(x)

x

y

(0,0)

Laminar

Ux 15

2

2

)platetheonlocation -at(;)platetheofedgeleadingat(0;0

xxxx

21

5.5

xUx

x

lam

xx

Re5.5)(

x

lam

xx

Re0.5)(

(1908))solutionBlasius(Exact

δ(x) = f(x)

Growth of laminar boundary layer thickness with distance.Laminar boundary layer thickness along a flat plate varies inverselywith the square root of the length Reynolds number.

xU

xRe



Displacement thickness, δ*

x

x

d

dUu

dyUu

Re5.5

31

31

3

21

1

1

*

*

1

0

32*

1

0

2*

1

0

*

0

*

; Change of variable from y/δ to η

ddyy

x

lam

xx

Re83.1)(*

U∞U∞

δ(x)

x

y

(0,0)

Laminar

y

xyUyxu

for1)(0for2),( 2

x

lam

xx

Re721.1)(*


Growth of laminar boundary layer displacementthickness with distance.



Momentum thickness, θ

xlam

xRe5.5

152

U∞U∞

δ(x)

x

y

(0,0)

Laminar

x

lam

xx

Re664.0)(


215lam

x

lam

xx

Re733.0)(

Growth of laminar boundary layer momentumthickness with distance.

boundary layer theorytoufiquehasan.buet.ac.bd/me 323-(jan 2020)-lec-4.pdf©dr. a.b.m. toufique hasan...

Documents