BOUNDARY LAYER THEORY

HIGH RENOLDS NUMBER FLOW BOUNDARY LAYERS (Re ∞)

BOUNDARY LAYER Thin region adjacent to surface of a body where viscous forces dominate over inertia forces

⎛ inertia forces ⎞ Re = ⎜ ⎟ Re >> 1 ⎝viscous forces ⎠

Boundary layer separation Wake: viscous effects not Outer flow Flow fieldimportant Viscouseffect aroundvorticityan not s negligible Inner flowarbitraryzero Vorticity zero Strongshape (Inviscid viscous potential flow) effects Steady ,incompressible 2-D flow with no body

ddθ * 1 Uτ 0 ∂u n ++(2δθ) = forces. Valid for laminar flow τ 0 ∼ () dx θρdx U 2 dy

O.D.E for θ () x To solve eq. we first ”assume” an approximate velocity profile inside the B.L Relate the wall shear stress to the velocity field Typically the velocity profile is taken to be a polynomial in y, and the degree of fluid this polynominal determines the number of boundary conditions which may be satisfied u EXAMPLE: =+abη +cηη2 =f () LAMINAR FLOW OVER A FLAT PLATE: U

U∞

U ≈ 0,99U∞ • Laminar boundary layer predictable • Turbulent boundary layer poor predictability UL • Controlling parameter Re = ν

• To get two boundary layer flows identical match Re (dynamic similarity) • Although boundary layer’s and prediction are complicated,simplify the N-S equations to make job easier

High Reynolds Number Flow

2-D , planar flow uv, xy, u* = v * = , x*, y*= U∞ L → Dimensionless gov. eqs. ∇.0V =

**** 2*2* ∂∂uu**∂u∂P1 ∂u∂u X ; ++u ν =−+()22+ ∂∂tx∂y ∂xxRe∂∂y viscous terms * P P = 2 22 ∂∂ν νν∂ ∂P 1 ∂ν∂ν ρU∞ Y; ++u ν =−+()+ ∂∂tx∂y∂yRe ∂x22∂y

“Naïve” way of solving problem for

1 Re →∞ → 0 Re

If you drop the viscous term Euler’s eqs. (inviscid fluid) • We can not satisfy all the boundary B.C.s because order of eqs. Reduces by 1

Inside B-L can not get rid of viscous terms

U∞ U∞

U (x,y) y

δ δ 1 δ * =〈 L 100 L

Derivation of B-L eqs. From the N-S eqs

• Physically based argument :determine the order of terms in N-S

• Limiting procedure as Re ∞ eqs. and throw out small terms Assumption 1

δ δ * =〈〈1 L Term Order

* (1) ∂u =1 (1) ∂x* δ * ∂ν * =1 δ * ∂y* v* δ *

* * ∂ν δ * = δ ∂x* 1

∂2*u 1 2 ∂y*2 δ *

* du ∂u* u* =1 dt* ∂x* ∂∂uu**∂u*∂P*1 ∂2u*∂2u* ++u**ν =−+()+ ∂∂tx∂y∂xRe ∂x22∂y

(1) (1) * 1 (1) (1) δ =1 (1) *2 * =1 δ 2 *2 δ (1) (1) ()δ Neglect since of order (1) *2 >>>1 Also for y –direction ()δ

* * * * 2* 2* ∂∂ν **νν∂∂P 1 ∂ν∂ν ++u ν =−+()22+ ∂∂tx**∂y*∂y*Re ∂∂xy**

* * ()δ * ()δ ** (1) ()δ *2 δδ * (1) * ()δ { + } ()δ (1) 2*(δ ) 2 * ()δ * * * ()δ ()δ ()δ ∂P* ∂P* ()δ * small relative to (1) ∂y* ∂x*

To good approximation P ≅ Px() pressure at the edge of B- L. is equal to pressure on boundary layer.

• Time – dependant P ≅ Px (, t ) known from the other flow

• Pressure at all points is the same

• Only need to consider x-direction B-L. eqs. Prandtl (1904) Outer flow (inviscid)

y 2-D planar x Governing ∂∂uv +=0 1) ∂∂xy eqs.for B.L B-L.eqs. ∂∂uu∂u1 ∂P∂2u 2) ++=uv− +ν still non-linear ∂∂tx∂yρ ∂x∂y2 but parabolic type

unknows u,v (x,y,t)

PP≅ (,xt) known from the potential flow Need B.C.s & I.C.(time dependant)

• 2-D, steady

BCs

• u=ν =0 at y=0

• u=u(y) at x=0

• u= U ∞ (x) y ∞ (y δ ) marching condition

• B-L. eqs. can be solved exactly for several cases • Can approximate solution for other cases Limitation of B.L egs.: where they fail? (1) Abrupt chances (2) Eqs. are not applicable near the leading edge

δ L is small δ * = 〈〈 1 invalid L

(3) Where the flow separates not valid beyond the separation point

Separation point

Bernouilli eqs. ρ =constant

PV2 11dP dV +=constant + 20U = ρ 2 ρ dx 2 dx Valid along the streamlines

1 dP dU −=U substitute the B.L eqs u,v can be found ρ dx dx

known

SIMILARITY SOLUTION TO B.L. EQS

Example 1 Flow over a semi-infinite flat plate dp Zero pressure gradient = 0 p = constant dx

dp Steady ,laminar & U=constant ( = 0 ) dx U y

x

1 • Bernouilli eqs. outsideB.L pU+=ρ 2 cons. dp 2 U=constant , = 0 dx

Governing (B.L. eqs.) become

∂∂u ν +=0 (1) ∂∂xy

∂∂uu∂2u uv+=ν (2) ∂∂x yy∂2 B.C. • y=0 u= v =0 (no-slip) & y ∞ , u U • x=0 u=U

Blasuis(1908) :

1.Introduce the stream function ψ (x,y)

• Recall ; ∂ψ ∂ψ u = ν =− ∂y ∂x

note that ψ satisfies cont. eqs. substitute intoB.L. mom. Eqs

∂∂ψ 22ψψ∂∂ψ ∂3ψ ..−=ν (2’) ∂∂y x∂y ∂x ∂yy23∂ • Now, assume that we have a similarity “stretching” variable, which has all velocity profiles on plate scaling on . δ i.e uy = f () y U∞ δ

δ x

dimensional analysis δ = gU(,∞ x,ν )

δ Ux 1 ==gg()∞ (Re) ∼()δ 2 δ ∼ ν x ν Re

ν x 2 δ ∼ mm δ 1 U ..sm= [] ∼ both ()δ ∞ sm x Rex

Viscous dif. Depth Ux ν x Re = ∞ δ ≈ 5 ν U∞

y Let η = [-] similarity variable δ

U u η = y ∞ = f ()η ν x U

Use similarity profile assumption to turn 2 P.D.E 1 O.D.E

∂ ψ yy η ν x u = ψ ==udy Uf ()ηηdy =Uf () dη ∂ y x = fixed ∫∫ ∫ 00 0 U 2

η 2 η d dF = '' F () F

νη η Ux d dF = = ' dx ψ F ∞ x ψ U

ν ∂ ψ y r (x,y) to get O.D.E for F( ) y =

η ψ yx ∂∂ ∂ =+

η x ψ dd ' F

η to P.D.E fo () x n

F ν ∞

ν ψ Ux y ud FU y 0 ∫ η ν () ∞ d η () F U 0 () F ψ f 2 νη 1 −= 0 η ∫ Ux =+ ψ νη ψ xx Ux = ∂∂ ∂∂

== ψ •Now, substitute i ψ ∂η 11U∞ 1 ∂ψ 1 U ν ' ∂ψ U =− y =− η =−∞ ()FFη ==Uxν F''∞ UF ∂xx22ν xx ∂xx2 ∂yx∞∞ν

∂2ψ U 2 3 2 =− ∞ ηF '' ∂ ψ U∞ '' ∂ ψ U∞ ''' = UF∞ = F ∂∂xy 2x ∂y2 ν x ∂yx3 ν

Substituting into eq. (2’)

2 11 UU∞∞⎡⎤1 ν 22⎡U∞⎤U∞ UF∞∞'(−−ηηF''') ⎢⎥( ) (F−F') ⎢U( ) F''⎥=νF''' 22xx⎣⎦⎣ννx⎦x or

U 2 1 U 2 1 U 2 U 2 − ∞ η F '' F '− ∞ FF'' + ∞ ηFF'' ' = ∞ F ''' 2x 2 x 2 x x 1 FF'''+=F'' 0 blasius eq. 3rd order , non linear ODE 2 U Note: FF ''' += F '' 0 for η = y ∞ BVP 2ν x

BC’s are

At y=0 u=v=0 η = 0

∂ψ F’(0)=0 BC 1) u ==0 UF∞ '0= y=0 η =0 ∂y y=0

1 U ν F(0)=0 BC 2) ν = 0 − ∞ ('FF−=η )0 y=0 2 x

BC 3) (x,y ∞ ) U ∞

∂ψ →U UF' = U F '(η →∞) 1 F '(∞) = 1 ∞ ∞∞η→∞ ∂y y→∞ F( η ) dimensionless function

Or At x=0 uU= UF∞∞' x=0 = U ∞ η→∞

F’( ∞ )=1 same with BC 3) Matching B.C

• Solution to blasius eg a)power series b)runge-kutta

• results tabulated form for F,F’,F’’,etc

p.g 121 U u η = y ∞ F F ' = F '' ν x U∞

0 0 0 0.33206

5.0 3.28329 0.99155 0.01591

F’’= 0.33206 From the solution • Velocity profile

U = y ∞ γ x η 5 5

F’= u ν U∞ Re U x 0.8 ∞

1 U ν ∂ψ 1 U ν ην→∞ = ∞ (5x1−3.28) ν =− = ∞ ('ηFF− ) ∞ 2 x ∂xx2 ν ∞ 1 ν 1 − 1 = 0.86 =−Re 2 []ηFF' x U ∞ Rex U∞ 2 Shear stress distribution along the flat plate

∂∂u ν τµ=+( ) τ(x, y) 4 ν ∞ 1 ∂∂yx For Rex =⇒10 =0.00865 ≈ U∞ 100 ∂∂uu**ν ∂ τµ≅ 6 ν ∞ 1 ** For Re =⇒10 =0.000865 ≈ ∂∂yx ∂y x U∞ 1000

∂u At the wall (y=0) τµ0 ()x = ∂y y=0

τ w ()x

2 3 ∂ ψ U∞ U τµ()xU==µ F'' ∞ 0 2 ∞ η=0 τµ0 (xF) = ''(0) ∂yxy=0 ν ν x

Distribution along the wall 0.332 Non dimensionalize :

τ 2F ''(0) 0.664 Ux. C ==0 = Re = ν fxC f = 0.664 1 2 ν ρU RexxRe Ux 2 ∞

Friction coef.

Note : x →⇒0 τ 0 →∞ y ν →∞

B.L eqs.are not valid near the leading edge x x

Up to the point we are considering

Drag force acting on the flat plate We have to integrate shear stress x = τ ()ζζd τ FD ∫ 0 0 ↓ per unit width

x 3 21FbD = .328()U∞µρx x

dimensionless drag coef.( CD )

we have 2 wetted sides

2F C = D A=2bx D 1 ρUA2 2 ∞ Width normal to the blackboard

1.328 56 CDx=< valid for laminar flow i.e for Re 5.10 to 10 Rex 6 for Rex >10 → turbulent drag becomes considerably greater

Boundary Layer Thickness : δ

U u ηη==yy∞ at 5 ⇒ =0.99 → =δ (Table) ν xU

UU∞∞5x x 5 ≅≅δδ Rex = ννx Rex

δ:defined as the distance from the wall for which u=0.99U∞

Boundary Layer Parameter (thicknesses)

Most widely used is δ but is rather arbitrary y=δ when u=0.99 U∞ hard to establish more physical parameters are needed

Displacement thickness: δ

U ∞ U∞

* δ δ δ * an imaginary displacement of fluid from the surface to account for “lost” mass flow in boundary layer

* . ∞∞ ∞ δ mutot ==ρρdyUdy=ρUdy−ρUdy or ∫∫∞∞∫ ∫∞ 00y=δ * 0 * −ρδU∞ ∞∞u ρδUU**=−(ρ ρu)dy δ=(1−)dy ∞∞∫∫ 00U∞ if ρδ=>cons. δ* always by definition

Momentum thickness: θ

U ∞ θ

an imaginary displacement of fluid of velocity U ∞ to account for “lost” momentum due to the formation of a boundary layer velocity profile

∞∞ ρθU 2 =−()ρudy U ()ρudy u ∞∞∫∫ 00 Mass flow in B.L Possible momentum actual momentum "lost" momentum ∞ uu θ =−∫ (1 )dy will occur in B.L eqs. 0 UU∞∞

notes(remaks) ∗ Various thinknesses defined above are,to some extend,an indication of the distance over which viscous effects extend. ∗ δθ*,(x) only ∗ δδ>* >θ (always) ∗ Definition is same for ZPG,APG,FPG,turbulance 5x From flat plate analysis δ ≅ δ u and δ x =−(1 )dy Re x ∫ 0 u∞ uu remember η =⇒y ∞ d =d ∞ ν xxη y ν ηη==55uxννx δ *'=−(1 ) dF= (1 −)d ∫∫η η 00uu∞∞ u∞

ν xx 5 x 1.72x η −=F 53−.283= []0 [] ux∞ Rex Re x F(5) = 3.283 1.72x δ * = Re x δ uu 0.664x Similarly, θ =−(1 )d = ∫ y 0 uu∞∞ Re x δ δ δ * δ * θ θ

FALKNER-SKAN SIMILARITY SOLUTIONS

Stagnation-point flow (Hiemenz flow) Similarity methods Flow over a flat plate (Blasius flow) (,xy)⇒η Falkner & Skan (1931) Ægeneral similarity solution of the B-L eqs. Family of similarity solutions to the 2-D,steady B-L egs. Look for general similarity solutions of the form

ux(,y)= U(x)f'(η) where (1) y ζ (x) - unspecified function of x which will be determined later η = ζ ()x

∂ψ 1 (2) ψζ(x, y) ==Ux( ) (x) f(η) check : u= Ux( )ζ(x) f '(η) ∂y ζ ()x ∂∂uudU∂2u B.L eqs. uv+=U+ν (3) ∂∂xydx∂y2 ∂∂ψψ22∂ψ∂ψ dU ∂3ψ or in terms of ψν(xy, ) −=U + (3') ∂∂yx∂y ∂x∂y23dx ∂y B.C.s no-slip, smooth matching y Substitute eq .(2) into (3’) η = ζ ()x

df d 2 f ψ ==Ux( )ζη(x) f( ) ψ(x, y) f' =, f" = ddη η 2 ∂ψ ==Uf ' ( u) ∂y ∂ψζdU d df dη =+ζζfU f+U ∂xdx dx dη dx dyη dζζ1 d =− =−η dx ζζ2 dx dx ∂ψζdU d dζ =+ζηfU f−U f' ∂xdx dx dx ∂∂2ψψ⎛⎞∂∂ dU ∂f ' ==⎜⎟ []Uf ''=f +U ∂∂xy ∂x⎝⎠∂y ∂x dx ∂x dU df '1∂ηζdU ⎡ d ⎤ = fU'+=fU'+f"⎢−η ⎥ dx dηζ∂xdx ⎣ dx⎦ ∂2ψζdU U d = f '"− η f ∂∂xy dxdζ x ∂2ψ ∂∂η U = []Uf '"==Uf f " ∂y2 ∂∂yyζ ∂3ψ U = f "' ∂y32ζ Substitute above results into (3') ⎡⎤dU Udζζ⎡⎤dU dUζ d dUU Uf ''f −−η fU" ζνf +U f −η f' f "=U +f "' ⎢⎥⎢⎥2 ⎣⎦dx ζ dx ⎣⎦dx dc dx ζζdx dU dU 1 dζ dU U Uf(')22−−Uff"U ff"=+Uν f"' dx dx ζζdx dx 2 dU Ud dU U uf( ')2 −=()Ufζνf"U + f"' dx ζζdx dx 2 ζ 2 To put the eq. into standard form, multiply by νU

Transformed gov. Eq. 2 ⎡⎤ζζd ⎡⎤dU 2 fU"'++()ζ ff" ⎡⎤1−()f' =0 (4) ⎢⎥ννdx ⎢⎥dx ⎣⎦ ⎣⎦ ⎣ ⎦ α β If a similarity solution exists, eq.(4) must be an ODE for the function f in terms of η. So, coefficieuts α & β must be constant for a similarity solution

ff"'++αβf" ⎡⎤1−f' 2 =0 Falker-Skan eq. (5) ⎣⎦()

B.C same as for flat plate ff(0) = '(0) = 0 f '(η →∞) →1 remark : BCs don't depend on αβ, Exact solutions to the B-L. Eqs. May be obtained by pursuing the following PROCEDURE Step 1: Select αβ& . (a particular flow configuration is considered this will not be known a priori but will be exident when step 2 is completed). Step 2: Determine U(x) , ζ (x) ζζdd2 U αζ==(U ) , β (6a-b) ν dx ν dx Step 3 : Determine the function f (η)which is the solution of the following problem

'' ' 2 ff'''++αβf⎣⎦⎡⎤1 −( f) =0 with BCs ff(0) ==''(0) 0 , f(ηη) →1 as →∞ Step 4 : Calculate the stream functionin physical coord. ⎛⎞ ⎜⎟y ψζ(,xy)= U(x) (x)f⎜⎟ ζ ()x ⎜⎟ ⎝⎠η Remark in step #2 , instead of working with eqs. 6 a-b) ζ 2 dU d ==βζ (6a)' ()U 2 ν()2α−β (6b)' ν dx dx Example #1 Flate Plate (ZPG) 1 step #1 αβ= , =0 2 d step #2 ()Uζν2 = (6a)' dx ζ dU = 0 (6b)' ν dx

' dU ζ (xb) ≠⇒0 (6 ) leads to =0⇒ U=const. dx this means that flat plate at ZPG dxζν2 ν (6a)' =→ζ 2 = dx U U ν x ζ ()x = U 1 Step #3 : ff'''+=f'' 0 f(0) =f'(0) =0 2 y ηη=→ ∞; f ' →1 compare with Blasius solution ν x U ⎛⎞ ⎛⎞yxν ⎜⎟y Step #4 ψζ(xy, ) ==U f⎜⎟U f⎜⎟ ⎝⎠ζ U ⎜⎟ν x ⎜⎟ ⎝⎠U ⎛⎞U ψ ()xy, = Uν xf⎜⎟y ← same as Blasius solution ⎝⎠ν x Example #2 FLOW OVER WEDGE Step #1 αβ==1, arbitrary constant d (6a' ) ()UUζν22=−()22β⇒ζ=ν()−βx (7) dx dU (6b') ζν2 = β dx Divide eq. (6b') by (7) 11dU β = Udx 2 − β x

β β lnU = ln xc+⇒ln Ux()= cx 2−β outer flow is that over a wedge of angle πβ (Fig.) 2 − β

dU β −−2(1 β ) ζν22==β ζcx2−β νβ dx 2 − β νβ(2− ) 1−β ζ ()xx= 2−β (9) c Step #3 Solve the BVP

2 ff'''++f'' β ⎣⎦⎡⎤1−( f') =0 ff(0) =='(0) 0 as ηη→∞ f ' →1 Solve numerically to get f ( ), f '(η), f ''(η) Step 4: Go back to the physical coordinate

1 ⎛⎞ ⎛⎞yy()21−−ββ()−(2−β) ψζ(,xy)=−U(x) (x)f⎜⎟ = c()2 βν x f⎜⎟x ζ ()x ⎜⎟ ⎝⎠ ⎝⎠()2/− βν c

STAGNATION-POINT FLOW; β = 1 α =1

Flow over a wedge →= Let β 1

Eq. (8) gives, Ux()= cx ν (9) →=ζ (xf) "'+ff"+1−( f')2 =0 c ff(0) =='(0) 0 as ηη→∞ f'( ) →1 ⎛⎞y ψν(xy, ) = c x f⎜⎟ ⎝⎠ν / c Note: See Hiemenz flow Exact solution to the full Navier-Stokes equations obtained by Hiemenz for a stagnation point. FLOW IN A CONVERGENT CHANNEL α = 0 , β = 1

Boundary layer flow on the wall of a convergent channel. Exercise: pg. 132. Solve the BVP (F-S. eq.)

More on similarity solutions to the B.L. Evans (1968) “Laminar Boundary Layers” Numerical Solutions Finite differences H.B. Keller (1978) Ann. Rev. of Fluid Mech. Vol.10.pp. 417-433 Finite Element Methods, Finite Volume Methods Spectral (Element) Methods

APPROXIMATE SOLUTIONS: Solve exact eq. approximately Von Karman Momentum Integral Eqn (General Momentum Integral Equation for Boundary Layer) Idea: Develop an eqn. which can accept "approximate" vel. profiles as input & yield accurate (close, but approximate) shear stress δδ, * ,θ as output.

Approach: Integrate the differential B-L. eqs. across the B-L. 0 ≤≤y δ

Start with B-L. eqs. ∂∂uudU∂2u uv+=U+ν ∂∂xydx∂y2 ∂∂uv +=0 ∂∂xy BC. y==0 u,v 0 yu==δ U ∂∂uu()v∂v∂ ∂u ∂∂vu First note v =−u =()uv +u =− (continuity) ∂∂yy∂y∂y ∂x ∂y ∂x Substitute into B.L eq. & integrate from y=0 to y= δ. δδ∂∂ud∂ ()uv δUδ 2u 2udy+=dyU dy+ν dy ∫∫ ∫ ∫ 2 00∂∂xy0dx0 ∂y (1) (2) (3) (4) δ ∂()uv δ Consider term (2) dy ==uv U vx(,δ )−0 ∫ ∂y 0 0 ? δ Integrate cont. eq. ∫ dy 0 δδ∂∂uv δ∂u δ∂u ∫∫dy +=dy 0 ⇒ ∫dy +v(x,δ ) −0 =0 Uv(,x δ )=−U ∫dy 00∂∂xy 0∂x 0∂x

δ δδ∂∂2uu⎛⎞∂∂u∂u∂u Integrate term (4) dy ==dy =− ∫∫2 ⎜⎟ 00∂∂yy⎝⎠∂y∂y00∂yyy==δ ∂y

=0

du ττ δδ∂∂uu()2 τµ=⇒ (4) ⇒− 00ν =− Term(1) ⇒ 2udy= dy 0 ∫∫ dy y=0 µρ 00∂∂xx

δδ∂()ud2 δ ∂u Uτ B-L. eq. becomes ∫∫dy −−U ∫ dy = U dy 0 0 ∂xd0 ∂x 0 xρ

δ ∂u δ ∂udδ ⎛⎞∂ ()uU U U ∫ dy ==∫Udy∫⎜⎟−u dy 0 ∂x 0 ∂∂xx0 ⎝ dx⎠ δδ∂()ud2 ∂ (uU ) δUδdUτ Thus, get dy − dy +−u dy U dy =−0 ∫∫∂∂x x ∫dx ∫dx ρ 00 00

δδ∂ dU τ ∫∫()uu2 −+Udy ()u−U dy=−0 00∂xd xρ

∂ δ ∫()u2 − uU dy ∂x 0 Using Leibnitz’s rule permits the order of integ. & dif. to be interchanged

∂ δδ⎛⎞uu2 ⎛⎞u dU τ Ud2 −+y−1 Udy=−0 ∫∫⎜⎟2 ⎜⎟ ∂xU00⎝⎠U ⎝⎠U dx ρ

Multiply by -1 & factor U terms out of integrals, ⎡⎤ ⎢⎥δδ ∂ 2 uu⎛⎞ dU⎛⎞u τ 0 ⎢⎥Ud⎜⎟11− y+−U⎜⎟dy= ∂xU∫∫U dx U ρ ⎢⎥00⎝⎠ ⎝⎠ ⎣⎦⎢⎥θ ()x δ *( x) ∂ dU τ ()UU2*θδ+=0 ∂xdxρ ↓ ∂∂θθdU dθ UU2 +→θθ2 , (x) only ∂xdx ∂xdx ddθ 1 Uτ Divide eq. by U2 & get ++(δθ* 2) = 0 dxUdxρU2 ddθθUC δ* τ or +=(H + 2) f H = C =0 H= shape factor f 1 dx U dx 2 θ ρU 2 2 Ordinary Differential eq. for θ(x) & is called von Karman Momentum Integral eqn. or Generalized momentum integral equation To solve the integral eq. we first “assume” an approximate velocity profile, i.e. one that “fits” & has proper “shape” and satisfies the proper B.C we do thin by using similarity concept again y & writing potential similarity velocity profiles in terms of the variable , η = & apply δ ()x B.C & get particular form.

evaluate θ(x), δ*(x) and τ 0 from their definitions.

integral equation can be solved for the B.L. thickness, δ(x) n u dy ∂ () ∼ 0 τ

δ = and turbulent flow y ree of this polynomial AT PLATE: laminar and the deg η U Steady ,incompressible 2-D flow with no body forces. Valid for as an example polynomial in y, imate velocity profile inside the B.L later LAMINAR FLOW OVER A FL 2 example 0 onditions which may be satisfied x τ U ()

θ () f = an approx = 2 2 U

η dx ηη c c ss to the velocity field 1 + + θρ of boundary c ) η

η 2 O.D.E for y c ab + =+ * ab y δθ b u (2 U =+ ++ u U θ the velocity profile is taken to be a y =+ l dx dd ua mines the number cal i An approximate velocity profile, for

To solve eq. we first ”assume” Relate the wall shear stre Typ deter EXAMPLE: laminar profile or B.C 1-)u=0 at y=0 ( η =0) a=0 b=2

2-)u=U at y= δ ( η =1) 1=b+c c=-1 ∂u 3-) = 0 at y= δ ( η =1) 0=b+2c ∂y

uyy =−2ηη22=2()−() U δ δ

Now use the approximate velocity profile to obtain terms in the momentum integral eq. NOTE: Using the approximate velocity profile across the B.L will reduce the momentum integral to an O.D.E for the B.L thickness, δ (x).

δ * u y dy δ =−∫ (1 )dy η = dη = 0 U δ ()x δ

η=1 1 1 *2u 1 δ δ =−∫∫(1 )δηdd=δ (1 −2η+η) η δδ*2=−()ηη+η3= 00U 330 δ uu η =1 uu θ =−∫∫(1 )dy =δη(1 −)d 00UU UU

1 22 2 θ =−δη∫ (2 η)(1−2η+η)dη θ = δ 0 15

du du dη 1 du U τ 0 ==µµ =µ =2µ dy y==00dη dy η δηd η =0 δ or U (2− 2η ) η=0

∂ ⎡⎤yy2 U τ 0 =−22ηµU ⎢⎥()()=2 ∂y ⎣⎦δ δδy=0

Momentum Integral eq. becomes ddθ 1 Uτ ++(2δθ* ) =0 dx U dx ρU 2 dd24δ δδ1U2µU2ν ()++( ) = = dx 15 3 15 U dx δρδU 2 U

dU For a flow over a flat plate = 0 U=const. dx

22dδ ν * = ODE for δ (x).solve δ (x)first then δ ,,θτ0 15 dx δU

δ 15ν x δν2 15 x Solving for δ , ∫∫δδdd=⇒x= 00UU2 δ ν x ννxx5.477x Ux δ * ==1.826 δ ==30 5.477 = , Re = , UU x 3 U Rex ν

τ 0.73 2δ ν x C ==0 U θ ==0.73 , f 1 2 Re τ = 2µ 15 U ρU x 0 2 ν Comparing to (exact) blasius solution

δ 5, 477 note:2nd order profile ∂2uU2 ==1.095 (,x 0)= −≠0 ∂y22δ δblasius 5 but it should be zero

δ * 1,826 * ==1.061 ∼ 10% δ blasius 1, 72

θ 0.73 ==1.099 θB 0.664

u =+A BCη +ηη23+D+Eη4 U

Additional BCs need to be imposed ∂∂2uP1 dU ν ==−U (=0 for flat plate) ∂∂y2 ρ xdx ∂u ∂∂uP1 ∂2u y=0 u + ν =− +ν 2 ∂x y=0 ∂∂yxy=0 ρ ∂yy=0 BC#5 at y= δ ∂2u all higher derivates should also be zero at y= δ = 0 ∂y2 for a smooth transition from the B-L. to the outer flow

u yy nd Note: =−2( ) ( )2 2 order profile U δ δ

∂2uU2 BC#4 (,x 0)=− ≠0 ∂y22δ

u by employing 3rd order profile , i.e =+abη +cηη23+d the above condt. may be imposed U More accurate results are obtained

Flat Plate at zero incidence

Vel. Dist. uy ==ff() (η) U δ

f ()η =η 2 , f ()η =−2ηη

0.647x 4.64 C = 313 δ = f f ()η =−ηη Re Rex 22 x

34 5.84 0.685x f ()η =−2ηη2 +η δ = C f = Rex Rex π f ()η = sin( η) δ = 4.80 C = 0.65 2 f

Note 1: Once the variation of τ 0 is known, viscous drag on the surface can be evaluated by integration over the area of the flat plate.

Note 2: B-L thickness at transition 5 Rex = 5.10

U = 30m δ 5.48 s xm= 0.24 air ( ν ) ==0.00775 x Rex

δ ==0.00775xm1.86 m←less than 1% of development length,x. viscous effects are confined to a very thin layer near surface of body Boundary layer seperation

Separation wake formation increase in drag

Wake total force exerted on body in direction of fluid motion

Boundary layers have a tendency to separate and form wake Wake leads to large streamwise pressure differentials across the body

results in substantial pressure drag (form drag)

For large Re ( 10 4 or higher) bluff bodies (e.g circular cylinder) pressure drag constitutes almost all the total drag

Total drag = pressure drag + viscous drag

due to shear stress along the surface

due to pressure differences caused by separation of flow 2) AIRFOILS-LIFT drops sharply “STALL” due to separation

force normal to flow direction Shape of streamlines near point of separation

3) FLAT PLATE S No separation ~ No separation t

chord

REMARK: After separation point ,external (decelerating) stream ceases to flow nearly parallel to the boundary surface Condition for separation dP Pressure gradient , dx

dP >0 adverse pressure gradient (decelerating external stream) increasing dx pressure in the flow direction dP dP <0 favourable P.G and =0 (zero pressure gradient) dx dx NOTE: pressure gradient along a B.L is determined by the outer flow

dU 1 dP U =− (Bern. Eq.) dx ρ dx

Separation occurs only for APG condition o Momentum contained in the fluid layers adjacent to surface will be insufficient to overcome the force exerted by the pressure gradient , so that a region of reverse flow occurs. i.e at some point downstream, the APG will cause the fluid layers adjacent to the surface to flow in a direction opposite to that of the outer flow B.L separation

velocity profiles in a

B.L near separation

δ

∂u 0 ∂u ∂u > sep. point = 0 < 0 ∂y 0 Note :shear stress changes∂ itsy 0 sign after separation∂y 0

Definition of separation point = point at which the shear (or velocity gradient) vanishes

∂u (x,0) = 0, for separation ∂y • Question show that separation can occur only in region of adverse pressure gradient ! Steady state B.L eqs.

2 ∂u ∂∂uP1 ∂u 2 3 u + v =− +ν ∂ udP ∂ u 2 µ = ∂x ∂∂yxρ ∂y 2 3 = 0 y=0 y=0 y=0 ∂ydx y=0 ∂y y=0

If dP <0 dx

y U ?

∂u 2 ∂ u ∂y u 2 ∂y 2 ∂∂uP ∂u u = 0 y = 0 2 = > 0 ∂∂yx ∂y

2 ∂ u ∂u the same y = δ 2 = 0 = 0 uU= ∂y ∂y ∂P case = 0 ∂x

yyy constant slope

∂2u ∂u u 2 = 0 ∂y ∂y

∂P Case > 0 APG ∂x y y y U PI PI

u ∂2u ∂u = 0 ∂y2 ∂y 2 ∂2udP ∂ u µ = > 0 PI= point of inflection where 2 = 0 2 ∂y ∂ydwall x Control of separation by suction

Control of separation by variable geometry and by blowing

How to calculate the separation point ?

Goldstein

Stewartson

The Karman – Pohlhausen Approximate Method

Fourth order polynomial for u (y). Pohlhausen (1921)

Step #1 :coefs. a,b,c,d,e, in general, will be functions

of x, so that solutions which are not similar may be obtained. u =+abη +cηη23+d +eη4 U y η = δ )

u η ∂ ∂ U dx d (

ν

ηδ f ∂ ∂ =− e o 11 δ r 0 = u

η ow uU y 2 ∂ fl ∂ () r 22 e ∂∂ 6 a meas 1 Λ ∂∂

δη ; e == l 1- out 0 = n = ab y i r 2 e uu a nt i 2 yy e v 2 ∂ ∂ = i Λ s d + 2 les n gra o i c s d=- 2 sure en s = m - i = dx :d dU Λ pre 2 c 0

0 δ 62 = ΛΛ = −

δ δ

δ = = = y y 2 =U Λ u = y 2 u y u y ∂ ∂ ∂ ∂ =− 0 b=2+ +3d+4e = x

U 2 U b+c+d+e d a +6d+12e dx dp

u ην + () C.s x ∂ . 1 2 µ → () ν ∂ y=0 B = u=0 1=a tion 1 =− u pose = 2 uU =0 0=a =0 =1 0=b+2c =1 0=2c 2 yd

im sol ∂ ∂ η η η η η u δ 2 dU = FG(η) + Λ (η) (1) Λ(x)= -12 ≤ Λ≤12 U ν dx where F(ηη)=1-(1+ )(1-η)3 Pohlhausen parameter G(ηη)= (1-η 3 ) 6

Note : for Λ= 0 velocity profile corresponds to a flat plate Plot function F( η ) & G( η ) u

G( η ) F( η ) U Λ >12 1 1 0.016 Λ > 0

Λ = 0

Λ < 0 Λ <−12

η η 0.25 1 1 u Λ=0 : =F(η) Flat surface in which the represantation is a 4th order polynominal U u Λ>>12 1 vel. in B.L. is not expected to exceed that of the outer flow locally. U So Λ must be less than 12 Λ⇒<-12 negative velocity ∴ reverse flow.B.L. theory is not applicable after separation

Step#2 Displacement thickness δ * δ uu1 δδ* ()xd=−∫∫(1 )y=(1−)dη 00UU 1 ⎡⎤33ΛΛ3 =δη∫ ⎢⎥(1+ )(1-ηη) −−(1 η) dη=δ( −) (2) 0 ⎣⎦610120 momentum thickness 1 uu 37 ΛΛ2 θδ(x)= ∫ (1−=)dηδ( −−) (3) 0 UU 315 945 9072

wall shear stress : τ 0

Uu∂Λ()U U τµ== τµ(2 +) 00δη∂ δ 6 η =0 b Uθ Step #3 Plug into the general momentum eq. Multiply the mom. Eq. by ν Udθ θθdU τ θ ++(2θδ* ) =0 or ννdx dx µU 1 ddθδ2*θ2Uτθ U ()++(2 ) =0 (5) 2 dx νθνdx µU δ 2 dU ΛΛ(x)= evaluate each term in terms of (x) ν dx θθ22dU 37 ΛΛ2 θ 2 dU =Λ=( − − )2 Λ=Kx( ) = K()x ν dx δ 2 315 945 9072 ν dx 3 Λ * ( − ) δ = 10 120 = fK( ) (6) θ 37 ΛΛ2 ()−− 315 945 9072 f(Λ→) f(x) but K=K(x) ⇒ f(K) τθ ΛΛ37 Λ2 0 =−gK( ) , g(K)=(2+ )( −) µU 6 315 945 9072 U Λ τµ=+(2 ) 0 δ 6 1 d θ 2 Uf( ) ++[]2 (K) K=g(K) (7) 2 dx ν θ 2 dU where K= = Kx( ) ν dx θ 2 dU Now , let us take Z= as the new dependent variable so that K=Z and the mom,int. becomes ν dx dZ dZ U2=−{}gK()[]2+f(K)K =H(K) or U = HK( ) (8) dx dx

H(K) is known (1st order nonlinear , ODE for Z , solve numericallay , start x=0 →Λstop =-12 []separation ) but complex H(Λ) ODE for Z(x) - mom. int. reduces to above form IVP for ODE for any Λ→(x) K & H(K) may be evaluated H(K)

H(K)=0.47-6K (9) approximation 0.47 Linear in K over the range of interest K 0.0783 Mom. Int. eq. becomes dZ dU U =−0.47 6KH=(K) =0.47 −6Z or dx dx 1 d (ZU 6 ) = 0.47 U5 dx

0.47 x Zx()= U5 (ςς)d Mom. int. may be expressed interms of this quadrature 6 ∫ U ()x 0

2 then , since Z=θ , the value of θ will be ν 0.47ν x θς25(xU) = ( )dς (10) 6 ∫ Ux()0

Procedure: Potential flow problem should be solved to yield the outer velocity U(x) (for a given boundary shape) Use eq. (10) to evaluate the momentum thickness θ ()x Pr essure parameter Λ(x) may be evaluated from the relation

θ 22dU 37 ΛΛ ==Kx() ( −−)2 (11) difficult to find Λ(x) ν dx 315 945 9072 having found Λ(x), δ (x) is evaluated from eq. (3)

37 Λ Λ2 θδ(x)=( −− ) and δ * eq. (2) 315 945 9072 3 Λ δδ* =−( ) 10 120

u =+FG(ηη) Λ( ) ← vel. distribution eq (1) U shear stress at the surface is given by eq. (4)

U Λ τµ=+(2 ) 0 u 6 In practice it is difficult to evaluate the quality Λ () x from eq (11) unless Λ is a constant

Instead : choose specific functions Λ () x and use foregoing eqs. to determine the outer-flow vel. & hence the nature of the boundary shape

EXAMPLE Karman-Pohlhausen approx. applied to the case of flow over a flat plate

ννx xx Uc=→onstan t eq. (10) θθ2 =0.47 → =0.686 θ=0.686 U U Rex dU =⇒0 eq. (11) dx ⇓ δ 2 dU Λ=0 ( ) ν dx 37 ν x 5.84x From eq. (3) θδ(x)= →= δ=5.84 315 U Rex

31.75x eq.(2) ⇒= δδ** → δ= 10 Re x 3.5% error

U τ 0 0.686 eq.(4) ⇒= τµ0 2 ⇒ = Θ δ 1 2 Exact 0.664 ρU Rex 2 4th order vel. pr 0.686 2nd order vel. pr 0.73

STABILITY OF STEADY FLOWS

Boundary – Layers Instabilities Usually laminar flow becomes turbulent flow EXAMPLE: Flow over a circular cylinder

CD ~82 ~108

Re Laminar B.L Turbulent B.L

C • Significant drop in the drag coefficent D • Due to vel. profile difference between lam. & turb. flow

y parallel flow V(y) is known⎫ ⎬ undisturbed flow "base flow" v=0 ⎭

x Linear Stability Analysis: The Method of Small Perturbations

Introduce arbitrary small (infinitesimal) disturbance into the flow eqs. & determine whether this disturbance grows or decays with time if the disturbance grows with time, the flow (the B.L) will be classified as unstable if the disturbance decays with time, the flow (the B.L) will be classified as stable marginal stability (neutral): the disturbance neither grows nor decays Non linear stability analysis: no restriction on disturbance size

A1 Introduce small disturbance to the velocity profile u(x,y,t) = V(y) + u’(x,y,t)

u(,xyt,) = V(y) + u'(x,yt,) v(,xyt,)=+0 v'(,xyt,)

p(,xyt,)=+p0 (x) p'(x,yt,) u' v' p' where <<<1 ; <1 ; <<1 VVp0 A2 Substitute A1 into the N-S eqs. & continuity

∂∂uv'' +=0 ∂∂xy ∂∂uu''dV∂u'1dp dp'∂22u'∂2V ∂u' xV ; ++( u') +v'( + ) =−( 0 + ) +ν ( + + ) ∂∂txdy∂yρ dx dx∂x22∂y2 ∂y ∂∂vv''∂v'1dp'∂22v'∂v' y ; ++(Vu') +v' =− +ν ( + ) ∂∂tx∂∂xρ dyx22∂y

A3 When the perturbation is zero, the above eqs. reduce to

2 1 dp0 dV Undisturbed flow 0 =− +ν 2 ρ dx dy (parallel)

A4 Drop term A3 in x-mom. Eq.

Since the perturbation is assumed to be small, products of all primed quantities may be

neglected as being small Thus , Linearized eqs. governing the motion of the disturbances are

∂∂uv'' +=0 ∂∂xy ∂∂u ''u dV 1dp '∂22u '∂u ' XV ; ++v' =− +ν ( +) ∂∂txdy ρ dx ∂x22∂y ∂∂vv''1dp'∂22v'∂v' YV ; +=− +ν ( +) ∂∂txρ dy∂x22∂y A5 Introduce a perturbation stream - function ψ (to reduce number or eqs. by one) ∂ψ ∂ψ uv' =− , '= ∂y ∂x

In terms of this stream function the governing eqs. become

∂∂22ψ ψψ∂dV 1'dp ∂3ψ∂3ψ +−V =−+ν ()+ ∂∂yt ∂x∂y ∂x dy ρ dx ∂x23∂y ∂y ∂∂22ψ ψψ1'dp ∂3∂3ψ −−V =− −ν ( + ) ∂∂xt ∂x23ρ dy ∂x ∂∂xy2 ∂2 p ' A6 Eliminate the pressure term by forming ∂xy∂ mixed derivative, above two eqs. above two eqs. may be reduced to one ,

∂∂∂22ψ ∂ψψdV2∂ ∂4ψ∂4ψ∂4ψ ()++V ( )−=ν (+2+) ∂∂tx∂∂y22xdy2∂x∂y4∂y2∂x2∂4x

Stream function for the disturbance must satisfy this linear , 4th order , PDE

A7 Since the disturbance under consideration is arbitrary in form, Perturbation stream function may be represented by the following Fourier – lntegral:

∞ ψ (xy, ,t) = ∫φα(y)eixα ()−ctd c:time coefficient 0 α : real & positive (inverse wavelength) 2π λ= []m α wave length of the disturbances note : time variation e-iαct -iαct cc=+rici → if ci > 0 → e → ∞ as t →∞ disturbance will grow → unstable -iαct in general complex number: if ci < 0 →→ e 0 as t →∞ disturbance will decay → stable

ci = 0 → neutrally stable (c=0)

Plug in A6 yields the integro – differential equation:

∞ 2iα (x-ct) ∫ ⎣⎦⎡⎤()−+icααiV("φ−αφ)−iαφVd" e α 0 ∞ 2 4iα (x-ct) 24 =−∫νφ⎣⎦⎡⎤( ''' 2αφφ"+α ) e dα , i = −1 i =1 0 dd24φφ φ " = , φ""= ,... ddy24y Above equation should be valid for arbitrary α. Thus, the integrand should vanish (because eq. should be valid for arbitrary disturbance) ν (V-c)(φα"- 22φ)-Vφ= (φ''''−+2αφ" α4φ) (A) iα Orr-Sommerfield equation

B.C disturbance should vanish at the surface y=0 and at the edge of the Boundary Layer

ux' ( , y= 0,t) ==0 , v'(x, y0,t) =0 ux' ( , y,t) = v'(x, y,t) =→0 as y ∞

in terms of the stream function ψ (y) ∂ψ u' ==0 →φ '(0) = 0 ∂y y=0 ∂ψ (B) v' =− =0 → φ(0) = 0 ∂x y=0 φφ'(yy) =→( ) 0 as y → ∞ Solution of the Orr – Sommerfeld Equation

Undisturbed vel. profile V(y) and disturbance wavelength α is specified Vy( ) & α known

Eq. (A) with BC.(B)represent an eigenvalue problem for the time coefficient , c cc =+ i c , c<0 ⇒ flow stable rii ψ = φ(y) ei(α x-ct) ci >⇒0 flow unstable

ixα( −=ct) , ci 0 ⇒ neutral stablity

stable , c < 0 α δ * i Ra = Gr . Pr

TL

unstable TH

Steady laminar flow Uα * can become another stable Re = ν steady lam. flow Recritical Stability Diagram: Orszag (1971):

α

ci < 0

ci > 0

ci < 0 V

typical stability – calculation result for fixed V , α is varied. Then, by considering all possible values of the undisturbed B.L vel. (which less than the outer –flow vel.) a stability diagram is constructed All possible values of V (y) in the range 0 ≤≤Vy( ) U(x)

Flow over a flat surface Uα * Re ==cr 420 cr ν stability boundary α δ * Schlichting 575 = Recr 0.34 unstable

Uα * Re = ν 420

Re >420 arbitrary disturbance will be unstable. manifest themselves in the form of turbulence FREE – SHEAR FLOWS (LAYERS)

Unaffected by walls Develop and spread in an open ambient fluid Possess vel. gradient created upstream mechanism viscous diffusion ⇔ convective deceleration

EXAMPLE: 1) The free-shear layer between parallel moving streams:

U1

ρ1 U1 u y µ1 S - shaped free-shear layer

u0 is due to viscous diffusion ρ U interface U 2 2 2 µ2

At x=0 , upper free stream U ⎫ 1 meets as x=0 lower free stream ⎬ U12 & U uniform U2 ⎭ For each stream , can define a Blasius – type similarity variable Lock(1951) – two different fluids with physical parameters

(,ρ11µρ) & (2,µ2)

u U1 j η jj==yf , ′ , j=1,2 2xUν j 1

ψ j1= 2νηjjUx f(j)

Following the same procedure as in derivation of Blasius equation, one can obtain Blasius-type eq. for each layer

ffjj'''+=fj'' 0 j=1,2 BC..s 1) f '(+∞) =1 asymptotic approach to the two stream velocitei s 1

U2 y →∞(- ) → η →−∞ ⇒ uf22 → U →2' = U1

uU11→→ as η +∞

BC..s 2) Kinematics equality , uu12= and v1= v2 at the interface

η j1=→0 ff'(0) = 2'(0) ≠0 =u0 u1=u2 ∂ψ ∂ψ ff(0) == (0) 0 v=v ⇒ 12= 12 12∂xx∂

BC..s 3) Equality of shear stress at the interface

∂∂uu12 yU1 µµ12(0) ==(0) or ηj ∂∂yy 2xν i ∂∂uf' ∂η U µµ11==UU1µf'' 1 (1) 11∂∂yy1η ∂111 y=0 1 0 2xν1 ∂u2 U1 µµ22= Uf12'' (2) ∂y y=0 2xν 2

11 ρµ22 (1)=(2) ⇒= ff11''(0)µµ2''(0) 2 → f1''(0) =f2''(0) νν12 ρµ11

ρ22µ fk12''(0) = f''(0) k= ρ11µ

Most practical cases

Case 1 : k=1 (identical fluids) ρ12= ρµ ; 12= µ Case 2 : a gas flowing over a liquid k>>1 ex. air-water interface k ≈⇒60000 k ≈245

TURBULENCE

INTRODUCTION

LAMINAR FLOW : Smooth , orderly flow limited to finite values of critical parameters: Re, Gr, Ta, Ri

Beyond the critical parameter, Laminar flow is unstable a new flow regime turbulent flow

Turbulent Transition

Laminar

x Characteristics

1) Disorder : not merely white noise but has spatial structure (Random variations)

2) Eddies : (or fluid packets of many sizes) Large & small varies continuously from shear – 3 νδ 1 layer thickness δ down to the Kolmogorov length scale , L = ()4 U 3 3) Enhanced mixing in laminar flow molecular action

mixing in turbulent flow turbulent eddies actively about in 3-D and cause rapid diffusion of mass, momentum & energy

Heat transfer & friction are greatly enhanced compared to Lam. Flow

4) Fluctuations : (in pressure, vel. & temp. )

Velocity fluctuates in all three directions

5) Self-sustaining motion: Once trigged turbulent flow can maintain. Itself by producing new eddies to replace those lost by viscous dissipation Experimental measurement : Hot-wire anemometer measure fluctuations in velocity via heat transfer Examine change in resistance assoc. with temp. (use wire ~ 0.0001” dia.)

u Laminar B.L t u

Shedding cylinder

t Turbulent B.L u

t Mathematical Description

N-S eqs. do apply to turbulent flow Direct Numerical Simulation :Solve the N-S eqs. directly using computers

Problem: wide range of flow scales involved solutions requires supercomputers and even then are limited to very low Reynolds numbers

Mesh points : beyond the capacity of present computers (trillions) Eq. Turbulent flow in a pipe

722 At Red =→10 requires 10 numerical operatious ⇒ computation would take thousand years to complete (for the fine details of the turbulent flow) Direct numerical simulation DNS

Because of complexity of the fluctuations, a purely numerical computation of turbulent flow has only been possible in a few special cases.

Therefore, consider time average of turbulent motion

Difficulties in setting up eqs. of motion for mean motion

Turbulent fluctuations coupled with mean motion Time averaging N-S additional terms (determined by turbulent fluctuations)

Additional unknowns in computation of mean motion We have more unknowns than eqs. To close system of eqs. of motion ⇒ need additional eqs

These eqs. can no longer be set up purely from the balances of mass momentum & energy

But, they are model eqs. which model relation between the fluctuations & mean motion called turbulence modelling central problem in computing the mean motion of turbulent flows Mean Motion & Fluctuations

u ' u , time average value

Decompose the motion into a mean motion & a fluctuating motion

uu=+u' vv=+v' In compressible turbulent flows ww=+w' ρρ = +=ρ' ; TT+T' pp=+p'

Average is formed as the time average at a fixed point in space

1 tT0 + uu=← dt integral is to be taken over a sufficently large time interval T so that u≠f (t) T ∫ t0 Characterization of fluctuation ⇒ RMS 1 2 2 ⎪⎪⎧ 1 T ⎫ ug'(= t) uu=−⎨⎬()udt T ∫ uu=+u'(=f t) ⎩⎪⎪0 ⎭

By definition time average of fluctuating quautities are zero i.e. uv' ==0 , ' 0 , w' =0 , p' =0 First assume that mean motion indep. of time ⇒ steady turbulent flow

steady unsteady u u steady unsteady

Turb. flow u Lam . flow t t Fluctuations u ' ,v' , w' influence the progrees of mean motion u ,v , w , so that mean motion exhibit an apparent increase in resistance aganist deformation.Increased apparent viscosity is cenral of all theoretical considerations on turbulent flow

Rules of computation

uu=+ , uv=u +v , u .v=u . v ∂∂u u == , udx udx ; uv =uv +u'v' ; u'v=0 ∂∂x x ∫∫

∂u ττ=+τ =µ−ρuv'' xy xy lam xy tur ∂y Additional shear stress (Reynolds stress)

x Ex : uv =(u+u ') (v+v ')=uv +++uv ' vu ' u 'v ' uv uv =+uv u'v' u'v'0≠

Physical Interpretation of ρuv' ' as a stress a)Consider fluid particle moving up from 1 to 2 u ' vu' ><0 ' 0 (since particle has velocity deficit i.e uu12< ) 2 uv' ' <⇒0 τ >⇒0 decel. of flow at 2 v ' turb 1 b)if particle moves down from 2 to 1 vu' <>0 ' 0 (particle has excess vel.)

∴< uv' ' 0 ⇒ τ turb >0 ⇒ accel. of flow at 1

2

Momentum Turbulent shear stress is higher exchange 1 Basic Eqs. for Mean Motion of Turbulent Flows

Consider flows with constant properties

Continuity equation

∂∂uv∂w (1) ++ uu=+u' ∂∂xy∂z ∂∂uu∂u' Time −=averaging of (1) + ∂∂xx∂x ∂∂uv∂w (2) ++ =0 ∂∂xy∂z ∂∂uv''∂w' (3) Also , using (1) ++ =0 ∂∂xy∂z Both time average values &fluctuations satisfy laminar flow continuity eq

Momentum Eqs.(Reynolds eqs.) ∂V Incomp. N-S eqs. ρµ( +∇(VV. ) ) =-∇p+∇2V (4) ∂t 1) Substitute uu=+u' v=v+v' w=w+w' p=p+p' into N-S egs

2) Time average the equations 3) Drop-out terms which `average` to zero . Use “Rules of Computation”

∂∂uu''2 ==0 0 ←terms which are linear in fluctuating quantities ⇒ 0 ∂∂tx2

uu'2 ≠≠0 'v' 0 ←terms which are quadratic in fluctuating quantities ⇒ 0

Resultant eqs. (called Reynolds eqs.) ∂∂uu∂u∂p ∂∂uu''2 v'∂u'w' ρµ()uv++w=−+∇−2 uρ()++ ∂∂xy∂z ∂x ∂xy∂∂z ∂∂vv∂v∂p ∂∂uv'' v'2 ∂v'w' ρµ()uv++w=−+∇2 v−ρ()+ + ∂∂xy∂z ∂y ∂∂xy∂ z ∂∂ww∂w∂p ∂∂uw'' v''w ∂w'2 ρ()uv++w=−+µ∇2 w−ρ ( ++) ∂∂x yz∂ ∂z ∂∂xy∂z ∴treat unsteady "fluctuations" additional terms due to turbulent as added stresses ⇒ called fluctuating motion ⇒ momentum Reynolds stresses(turbulent stresses) exchange due to fluctuations ⇒ "stresses"

Complete stresses consist of ∂u σµ=−pu+2 −ρ'2 → fluctuatios xx ∂x ∂∂uv τµ=+( ) −ρuv' ' ,...... xy ∂∂yx Re ynolds stress apparent turbulent viscous stresses laminar stresses

In general , Reynolds stresses dominate over viscous stresses, except for regions directly at the wall Closure problem too few eqs : 4 too many unknowns : 10 Figure some way to approximate Reynolds stresses

Objective : Establish relationship between Reynolds stresses & mean motions, i.e uv, , w

⇒ model eqs. must be developed ∴ turbulence models or turbulence modeling. model equations contain empirical elements A. Eddy vis cos ity − Attempt to approximate a "turbulent" viscosity ∂∂uu idea : Since τµ==ρν lam ∂∂yy ∂u Let τρ==∈ −ρu ''v turb ∂y Eddy viscosity ⇒ ∈>>ν Problem : how to model ∈ ? For some situations ⇒∈ ≈ const. ∂u In general ∈≠ const. ⇒ ∈ = f (u, y, ,etc.) ∂y In general, many wild guesses are made, not many work

Energy Equation

Consider the energy equation for incompressible flow with constant properties DT ρck= ∇+2TΦ p Dt Taking the time-average of the energy eq. , we obtain following eq. for the average temp. field T = (xy, , z) ∂∂TT∂T⎫ ρcup ( ++v w )⎬ convection ∂∂xy∂z⎭ ∂∂∂22TTT2⎫ =k( 22+ + 2) ⎬ molecular heat transport ∂∂xy∂z⎭ ∂∂uT'' v''T ∂w''T ⎫ -ρcp ( ++)⎬ turbulent heat transport("apparent" heat conduction) ∂∂xy∂z⎭

⎡⎤∂uv22∂ ∂∂w2u∂v2∂u∂∂w2v∂w 2 ⎪⎫ +2µ ⎢⎥( )++2()2( )+( + )+( + )+( + ) ⎬ direct dissipation ⎣⎦∂∂xy∂z∂y∂x∂z∂x∂z∂y ⎭⎪

The same eq.holds for the average temp. fields as for laminar temp. fields, apart from two additional terms "apparent" heat conduction ⇒ div(VT' ') "turbulent" dissipation , ρ ∈

⎡⎤⎫ ∂∂uv22∂w2∂u∂v2∂u∂w2∂v∂w2⎪ ρµ∈= ⎢⎥2( )+2()+2( )(+ +)+( + )(+ + )⎬ ∂∂xy∂z∂y∂x∂z∂x∂z∂y ⎣⎦⎢⎥⎪⎭ In turbulent flows mechanical energy is transformed into internal energy in two different ways:

a) Direct dissipation : transfer is due to the viscosity (as in laminar flow)

b) Turbulent dissipation : transfer is due to the turbulent fluctuations

The Turbulence Kinetic Energy Equation (K-equation)

Many attemps have been made to add “turbulence conservation” relations to the time-averaged continuity, momentum and energy equations derived. A relation for the turbulence kinetic energy K of fluctuations. 11 Ku≡+()′′uv′v′+w′w′=uii′u′ 22 Einstein summation notation,

uui ==(,12u,u3) (u,v,w) A conservation relation for K can be derived by forming the mechanical th energy equation i.e., dot product of ui ve i momentum equation subtract instantaneous mechanical energy equation from its time averaged value. Result: Turbulence kinetic energy relation for an incompressible fluid. ⎡⎤∂u ′ DK ∂ ′′⎛⎞1 ′p′ ′′j =− ⎢⎥uuij⎜⎟uj+ −uiuj + Dtx∂∂2 ρ x ii⎣⎦⎢⎥⎝⎠ I III II ⎡⎤ ∂ ⎛⎞∂∂uu′′∂∂uu′ ′′⎛⎞∂u ⎢⎥ννu ′ ⎜⎟ii+−jj⎜⎟+j ⎢⎥j ∂xi ⎜⎟∂∂xx′ ′′∂x⎜⎟∂x′∂x′ ⎢⎥⎝⎠ji i⎝⎠ji ⎣⎦ IV V I. Rate of change of turbulent (kinetic) energy II. Convective diffusion of turbulence energy III. Production of turbulent energy IV. Viscous diffusion (work done by turbulence viscous stresses) V. Turbulent viscous dissipation Reynolds stress equation: conservation equations for Reynolds stresses see F. White pg. 406 2-D Turbulent Boundary Layer Equations

Just as laminar flows, turbulent flows at high Re also have boundary layer character, i.e. large lateral changes and small longitudinal changes in flow properties.

Ex.: Pipe flow, channel flow, wakes and jets.

y

δ(x)<

Same approximations as in laminar boundary layer analysis, ∂∂Assume that mean flow structure is 2D vu<< << ∂∂xy

∂ 2 ww==0 0 but ′ ≠0 ∂z Basic turbulent equations (Reynolds equations) reduce to

∂∂uv Continuity: +=0 (1) ∂∂xy ∂∂uu dU 1 ∂τ x-momentum: uv+≈Ue + (2) ∂∂xye dxρ ∂y

Ue: free stream velocity ⎛⎞∂∂TT∂q∂u Thermal energy: ρτcup ⎜⎟+≈v + (3) ⎝⎠∂∂xy∂y∂y ∂u where τ = µρ− uv′′ ∂y ∂T qk=−ρcv′′T (4) ∂y p turbulent flux molecular flux Above equations closely resemble the laminar flow equations except that τ and q contain turbulent shear stress and turbulent heat flux (Reynolds Stress) must be modelled. y-momentum equation reduces to

∂∂pv′2 ≈−ρ (5) ∂∂yy Integrating over the boundary layer yields: 2 pp≈−e ()xρv′ Unlike laminar flow, p varies slightly across the boundary layer due to velocity fluctuations normal to the the wall

p +≈ρvc′2 onst.

Note: pw : wall pressure

no-slip ⇒≡ v′ 0 ⇒ ppw =e (x)

Bernoulli equation in the (inviscid) free stream dpee≈ −ρU dUe Boundary Conditions: Free stream conditions Ue(x) and Te(x) are known.

No-slip, no jump: ux( ,0) ==v(x,0) 0 , T(x,0) =Tw (x)

Free stream matching: ux( ,δδ) ==UeT , T(x, ) Te(x)

The velocity and thermal boundary layer thicknesses (δ, δT) are not necessarily equal but depend upon the Pr, as in laminar flow. Eqs. 1 and 2 can be solved for u v if a suitable correlation for total shear τ is known.

Turbulent Boundary Layer Integral Relations: The integral momentum equation has the identical form as laminar flow dθθdUewτ c f ++()2 H = 2 = dx Ueedx ρU 2 ∞ uu⎛⎞ δ * θ =−∫ ⎜⎟1 dy , H= (momentum shape factor) 0 UUee⎝⎠ θ Turbulent velocity profile is more complicated in shape * ⎛⎞u δ =−⎜⎟1dy and many different correlations have been proposed. ⎝⎠Ue Example: Turbulent pipe flow Often used correlation is the empirical power-law velocity profile r x R 1/n ur⎛⎞ =−⎜⎟1 VRc ⎝⎠

n=f(Re) for many practical flows n = 7 n=10 n 1.0 n=8 8 r/R n=6 7 laminar Turbulent 6 profile 5 0 104 105 106 Re=ρVD/µ 0 u 1.0

Vc ‰ Turbulent profiles are much “flatter” than laminar profile ‰ Flatness increases with Reynolds number (i.e., with n)

Turbulent velocity profile(s): The inner, outer, and overlap layers. Key profile shape consist of 3 layers Inner layer: very narrow region near the wall (viscous sublayer) viscous (molecular) shear dominates laminar shear stress is dominant, random eddying nature of flow is absent Outer layer: turbulent (eddy) shear (stress) dominates Overlap layer: both types of shear important; profile smoothly connects inner and outer regions.

Example: Structure of turbulent flow in a pipe Viscous R pipe wall R sublayer τ overlap layer lam τtur r r τ outer layer 0 0 τ(r) τw Vc Shear Average velocity stress Inner law: uf= (τρw , , µ, y) (1) Velocity profile would not depend on free stream parameters. Outer law: dp Uu−=g(τρ, , y,δ, e ) (2) ewdx Wall acts as a source of retardation, independent of µ.

Overlap law: uuinner = outer (3) We specify inner and outer functions merge together smoothly. Dimensionless Profiles: The functional forms in Eqs.(1)-(3) are determined from experiment after use of dimensional analysis. Primary Dimensions: (mass, length, time) : 3 Eq.(1) : 5 variables Π groups : 5-3 = 2 (dimensionless parameters) Proper dimensionless inner law:

* 1/2 uy⎛⎞v * ⎛⎞τ w * = fv⎜⎟ ; =⎜⎟ v ⎝⎠νρ⎝⎠ Variable v* [m/s] called wall friction velocity. v* is used a lot in turbulent flow analyses. Outer law using Π -theorem:

Uuee− ⎛⎞y δ dp * = g ⎜⎟,ξξ ; = vd⎝⎠δτw x

Often called velocity defect law, with Uue − being “defect” or retardation of flow due to wall effects. At any given position x, defect g(y/δ) will depend on local pressure gradient ξ. Let ξ have some particular value. Then overlap function requires

Overlap law:

* uv⎛⎞δ yUe ⎛⎞y **==fg⎜⎟ -⎜⎟ vv⎝⎠ν δδ⎝⎠ From functional analysis: both f and g must be logarithmic functions. Thus, in overlap layer: uy1 v* Inner variables: =+ln B vk* ν Uu− 1 y Outer variables: e = −+ln A vk* δ

Where K and B are near-universal constants for turbulent flow past smooth, impermeable walls. K≈0.41 , B≈5.0 pipe flow measurements, data correlations A varies with pressure gradient ξ (perhaps with other parameters also).

uyv* Let u++== , and y v* ν Inner layer details, Law of the wall. At very small y, velocity profile is linear. u y++≤=5: τµ or u =y+ w y Example: Thickness of viscous sublayer 5ν ν δ = : viscous length scale of a turbulent boundary layer sub vv**

Flat plate airfoil data: v*=1.24 m/s , νair≈1.51x10-5 m2/s Between 5 ≤y+≤30 buffer layer. Velocity profile is neither linear nor logarithmic but is a smooth merge between two. Spalding (1961) single composite formula.

⎡⎤++23 + ()Ku ()Ku yu++=+e−KB ⎢⎥eKu −1−Ku+− − ⎢⎥26 ⎣⎦ Notes: 1 ur⎛⎞n =−⎜⎟1 VRc ⎝⎠ 1 Power law profile cannot be valid near the wall. −1 n du Vc ⎛⎞r ⎛1 ⎞Power law profile cannot be precisely valid =−⎜⎟1 ⎜−⎟ dr n ⎝⎠R ⎝R ⎠near the centreline. du rR== ∞ However, it does provide a reasonable dr approximation to measured velocity profiles du r =≠0 0 across most of the pipe. dr Example: Water at 20 °C (ρ=998 kg/m3), ν=1.004x10-6 m2/s

Q=0.04 m3/s D=0.1m dp = 2.59kPa / m dx

δs = ? thickness of viscous sublayer? centreline velocity, Vc = ? ratio of turbulent to laminar shear stress, τturb/τlam = ? at a point midway between the centreline and pipe wall i.e., at r = 0.025 m. Lawof thewallvalid y± ≤ 5 viscous sublayer yv* y± =≤5 ν δ v* 5ν yy==δδ ± 5 ⇒ s =5 = ssν v* τ v* = w ρ Pressure drop and wall shear stres in a fully developed pipe flow is related by

4lτ ∆=p w (Valid for both laminar & turbulent flow) D (Exercise: Obtain the above equation considering the force balance of a fluid element)

Dp∆ (0,1)(2,59.103 ) τ == Pa =64,8N / m2 w 44lm(1) 64,8Nm/ 2 So, v* ==0,255ms/ 998kg / m3 5.1,004.10−6 δ ==1, 97.10 −5 mm≅0, 02 m s 0, 255

Imperfections on pipe wall will protrude into this sublayer and affect some of the characteristics of flow(i.e.,wall shear stres & pressure drop) Qm0,04 3 / s Vm== =5, 09 / s Amπ (0,1)22/ 4 VD 5,09.(0,1) Re == =5,07.105 ν 1,004.10−6 Re =⇒5, 07.105 n =8, 4

Power-law profile

ur ≅−(1 )1/8,4 VRc R r Q ==A.(V udA =V 1−)1/n (2π r)dr ∫∫c 0 R 2 2 n QR= 2π Vc (1nn++)(21) Recall that Vc=2V for laminar pipe flow: V2n2 QR=∴π 2V = Vc (nn++1)(2 1) nV==8,4 : Vc 1,186 =1,186(5,09) =6,04m/ s τ turb = ? Shear stres distribution throughout the pipe τ lam rm=0,025 2τ r τ = w (Valid for laminar or turbulent flow) D

R=D/2 2(64,8).0,025 τ (0rN==,025) =32,4/m2 r 0,1

ττ=+lam τturb =32,4 du r du V r τµ=− ; uV= (1 − )1/nn⇒ =− c (1 − ) (1− )/n lam dr c R dr nR R du 6,04 0,025 =− (1 − ) (1−8,4) /8,4 =−26, 5 dr 8,4(0,05) 0,05 r=0,025

du du τµ=− =−()νρ As expected lam dr dr Thus =−(1,004.10−62).(998).(−26,5) =0,0266Nm/ τ >>τ τ 32,4 − 0,0266 turb lam turb ==1220 τ lam 0,0266 Turbulent Boundary Layer on a Flat Plate

Problem of flow past a sharp flat plate at high Re has been studied extensively, numerous formulas have been proposed for friction factor. -curve fits of data -use of Momentum Integral Equation and/or law of the wall -numerical computation using models of turbulent shear Momentum Integral Analysis dp dθ C τ ==0 (U const.) =f =w dx dx 2 ρU 2 Momentum Interal Equation valid for either laminar or turbulent flow. For turbulent flow u a reasonable approximation to the velocity profile = fy(/δ ) U Functional relationship describing the wall shear stress Need to use some empirical relationship ∂u For laminar flow τµw = ∂y y=0 Example: Turbulent flow of an incompressible fluid past a flat plate Boundary layer velocity profile is assumed to be uy =←()1/7 power law profile suggested by Prandtl U δ (taken From pipe data!) Reasonable approximation of experimentally observed profiles, except very near the plate, ∂u =∞! ∂y y=0 uy 1 = ()1/7 U δ y Laminar Turbulent η = δ Assume shear stress aggrees with experimentally determined formula

−1/4 ⎧ 2 ν 1/4 ⎫ 0 CUfw==0,045Reδ ⎨or τρ0,0225 ( ) ⎬ 0 1 ⎩⎭Uδ Determine; * Uδ δδ, ,θ and τw as a function of x. Re = ν What is the friction drag coefficient CD,f=? Momentum Integral Equation (with U=constant) dyθ C τ uy ==f w ηη=; =( )1/7 =1/7 dx 2 ρδU 2 U δ ∞ uu 11uu 7δ θδ=−∫∫(1 )dy =(1 −)dη=δ∫η1/7 (1 −η1/7 )dη= 00UU o UU 72 7 dδν ==0,0225Re−1/4 0,0225( )1/4 72 dx δ Uδ δ ν x ∫∫δδ1/4dd= 0, 231( )1/4 x 00U ν δ 0,370 δ = 0,370( )1/5 x4/5 or in dimensionless form = U 1/5 x Rex Boundary layer at leading edge of plate is laminar but in practice,laminar boundary layer often exists over a relatively short portion of plate. ∴ error associated with starting turbulent boundary layer with δ =0 at x=0 can be negligible. ∞ uu11δ δδ*1=−∫∫(1 )dy =(1 −)dη=δ∫(1 −η/7)dη= 00UU0 8 δ * 0,0463 = 1/5 x Rex 7 ν θδ==0,0360( )1/5 x4/5 72 U

θ 0,036 * =<1/5 θδ<δ x Rex

1/4 2 2 ⎡⎤νρ0,0288 U τρw ==0,0225 U ⎢⎥1/5 4/5 1/5 ⎣⎦UU(0,37)(ν / ) x Rex 0,058 C f = 1/5 Rex

Friction drag on one side of plate,Df llν Db==τρdxb(0,0288 U21) ( ) /5dx fw∫∫ o 0 Ux

2 A DUf = 0,0360ρ 1/5 where A=b.l area of plate Rel

Df 0,0720 Note:Results presented in this example are CDf ==1/5 1 2 Re valid only in the range of validity of original ρUA l 2 data, assumed velocity profile & shear stres. 4/5 −1/5 The range covers smooth flat plates Turbulent flow:δτ(xx) ~ ; (xx) ~ 5 7 w with 5x10

∴= ab110, =4 / 3 u 4 δ /2 =≤ηη : 0 ≤1/ 2 U 3 u 12 1 Similarly, =+ηη for ≤<1 u U 33 2 2U/3 U 11uu /244 112 12 θ =−∫∫(1 )δηdd=δ η(1 −η) η+δ∫( +η)(1 −−)dη 00UU 33 1/233 33 = 0,1574δ ∂∂u uU4 τµw = ==µ µ ∂∂y y=0 ηδη =0 3 dδν4 0,1574 = dx 3 δU δ 4 ν ν x 0,1574∫δδdd=⇒x δ ()x = 4,12 0 3 U U

τ w 0,648 C f == 1 2 ρU Rex 2

Example 2 : Viscous drag in thin plate 1 a) FU= ρ 2 AC Da,,2 Da

l 1, 328 1, 328 2 CADa, == & =4l Rel Ul ν 4l U U ρ,UA, is the same 4l l 1,328 1,328 CD,b == Re4l U 4l ν a) b) FC Da,,==Da 2 FCDb,,Db The shear stres decreases with distance from the leading edge of the plate. Thus, even though the plate area is the same for case (a) or (b), the average shear stress (and the drag) is greater for case (a). Example 3: Thin flat plate in water tunnel

Parabolic velocity profile: uyy =−2( ) ( )22=2ηη− U δδ δ(x) Ul 1, 6.(0, 3) Re == =4,8.1055<5.10 l ν 10−6 ∴ Flow is laminar L U Viscos drag ==F 2 τ bdx (2 sides of plate) b=1 Dw∫ m 0 ∂∂uu∂ηµ τµ==µ.(=U 2−2η) w η =0 x ∂∂yyy==00ηδ∂η L=0, 2µU 3m = δ 5, 48x δ = Rex LL24µµUUdx8bUUL F ==2 bdx bµU = D ∫∫ 00δ 5, 48 ννx 5, 48

FND =1, 62 Continuity eq. for incompressible flow, 23 Qdinlet ==0 U(0.3*0.3)*0.7 =0.063 m/ s *2 QQinlet ==()xUA=U(d−2δ ) A: effective area of the duct (allowing for the decreased flowrate in the b.l.) Thus, 2*2 ** dd0 =−(2δ )=0.09⇒=dd0 +20δδ=.3+2 []m

ν xx1.5*10−5 δ * ==1.72 1.72 =0.00796 x []m U 0.7 dx= 0.3+ 0.0159 []m dx(3= m)≅ 0.328 []m