II. Ideal fluid flow

II. Ideal fluid flow Ideal fluids are Inviscid Incompressible The only ones decently understood mathematically Governing equations u=0 Continuity u 1 +( u ) u= ρ p+ f t Euler

Boundary conditions Normal to surface u n=u n Free-slip (velocity is parallel to surface) Velocity of surface Potential flow (special case) u = (u = / x, v = / y, w = / z) Potential flow is irrotational Continuity equation for potential flow =0 Continuity equation (with boundary conditions) can be solved alone for velocity

Then plug into momentum equation (Bernoulli form) to solve for pressure

4. D potential flows 4.1. Stream function D ideal continuity equation u v + =0 x y Velocity potential φ φ u=, v= x y Introduce streamfunction (counterpart of potential) so that ψ ψ u=, v= y x

Streamfunction satisfies continuity equation by construction ψ ψ =0 x y y x Streamfunction exists for any ideal D flow Before going further, consider vorticity in D flow [ i ω= u=det x u j y v k z w ]

Streamfunction satisfies continuity equation by construction ψ ψ =0 x y y x Streamfunction exists for any ideal D flow Before going further, consider vorticity in D flow [ i ω= u=det x u j y v k 0 0 ]

Vorticity in D flow ( ) v u ω=k =k ω x y For D, effectively a scalar Now consider an irrotational D flow v u ω= =0 x y Express velocity in terms of streamfunction ( ) ( ) ψ ψ ω= =0 x x y y ψ=0

Properties of streamfunction Streamlines are lines of = const Difference in the value of between two streamlines equals the volume of fluid flowing between them Streamlines = const and potential lines = const are orthogonal at every point in the flow

dy Why = const is a streamline ds dx ( ) dψ ψ x ψ y d ψ= ds= + ds= v dx+u dy ds x s y s dy dx = d ψ=0 means v dx=u dy ; v u Streamline equation!

Flow rate between two streamlines Direction along AB: ds = (dx,dy) = 1 B v ds u dn u = Direction normal to AB: dn = (dy,-dx) Volume flow rate A B B B B Q= A u n ds= A u d n= A u dy A v dx B Q= A d ψ=ψ1 ψ

Orthogonality between streamlines and potential lines Along a streamline d ψ= v dx+u dy=0 Along an isopotential line ( = const)... φ φ d φ= dx+ dy=u dx+v dy=0 x y Normal to streamline: (-v, u) Normal to isopotential line: (u, v) They are orthogonal: (-v, u) (u, v) 0

4.. Complex potential and velocity Complex variable z = x+iy Function of a complex variable F(z) = (x,y) + i (x,y) Cauchy-Riemann condition for function of a complex variable to be holomorphic* φ ψ φ ψ = ; = ; x y y x Holomorphic function complex-valued function of a complex variable which is differentiable in a neighborhood of every point within its domain

Complex potential constructed from velocity potential and streamfunction F(z) = (x,y) + i (x,y) Cauchy-Riemann condition satisfied by construction Advantages of using complex potential If and are the real and imaginary parts of any holomorphic function, = 0 and = 0 automatically Complex velocity w = df/dz = u - iv directly related to flow velocity

Magnitude of complex velocity w*w = (u + iv)(u iv) = u+v = u u = Polar coordinates in complex plane x + iy = r (cos + i sin ) = rei y eq u = ur cos - u sin er v = ur sin + u cos r x q w = (ur - iu ) e-i

4.3. Uniform flow i α F (z )=C e z df i α w (z )= =C e =C cos α ic sin α dz u=c cos α, v=c sin α y a x

4.4. Source, sink, and vortex iθ F (z )=C log z=c log (r e )=C (log r+i θ) First, let C be real and positive φ=c log r, ψ=c θ df C C i θ w (z )= = = e dz z r y x C u r=, uθ =0 r Source at z = 0

Source strength (discharge rate) π π m= 0 u r r d θ= 0 C d θ= πc Complex potential of a source of strength m at z = z0 m F ( z )= log(z z 0) π Complex potential of a sink of strength m at z = z0 m F ( z )= log ( z z 0 ) π y x

Now consider a purely imaginary constant in the logarithmic potential: iθ F (z )= ic log z = ic log (r e )= ic log r+c θ y φ=c θ, ψ= C log r df C C i θ w (z )= = i = i e dz z r C u r=0, u θ= r x Point vortex

Vortex strength (circulation) π Γ= u d l = 0 u θ r d θ= π C L Complex potential of a vortex with circulation G at z = z0 Γ F ( z )= i log (z z 0 ) π Note 1. z = z0 is a singularity (uq ) Note. This flow field is called a free vortex: Γ L' = u d l 0 L' Any contour not including z0

4.5. Flow in a sector n F (z )=U z, n 1 Abraham de Moivre's formula Abraham de Moivre 1667-1754 Author of The Doctrine of Chances

4.5. Flow in a sector n F (z )=U z, n 1 Abraham de Moivre's formula n inθ e =( cos θ+i sin θ ) =cos (n θ)+i sin(n θ) Use polar coordinates iθ z=r e n n F (z )=U r cos(nθ)+i U r sin (n θ) Potential and stream function n n φ=u r cos (n θ), ψ=u r sin( n θ)

Complex velocity n 1 n 1 i(n 1)θ w (z )=nuz =n U r e = n 1 n 1 i θ =( n U r cos n θ+i nu r sin n θ ) e Velocity components u r=n U r n 1 u θ= n U r cos n θ n 1 sin n θ n = 1: uniform flow n = : flow in a right-angle corner q=p/n n = 3: shown

4.6. Flow around a sharp edge 1 / 1 / i θ / F (z )=C z =C r e Potential and streamfunction φ=c r 1 / 1/ θ θ cos, ψ=c r sin Complex velocity df 1 C iθ / 1 / w (z )= = C z = 1 / e = dz r C iθ i θ/ C i θ θ θ = 1 / e e = 1 / cos +i sin e r r C C θ θ u r= 1/ cos, u θ= 1 / sin r r ( )

y Singularity = 0, q = 0 x = 0, q = p C C θ θ u r= 1/ cos, u θ= 1 / sin r r

4.7. Doublet y Source at x = -e Sink at x = +e x Now let e 0

Complex potential of source and sink m m F (z )= log(z +ε) log ( z ε) π π m z+ε m 1+ε/ z F (z )= log = log π z ε π 1 ε/ z For small /z, expand denominator into series: 1 (1 ε/ z) =1+ε/ z+ Plug that into F(z) m F (z )= log ( ( 1+ε /z )( 1+ε/ z+ ) ) π m ε F (z )= log 1+ + π z ( )

Use series expansion for logarithm near 1 m m ε ε F (z )= log 1+ + = + π z π z ( ) ( ) If we take the limit of this as 0, the result will be trivial: F(z) = 0 For a non-trivial result, let lim mε=πμ ε 0 Then μ μ x iy x iy lim F (z)= = =μ =μ z x+iy (x+iy)(x iy) ε 0 x +y x y φ=μ, ψ= μ x +y x +y

Consider a streamline = const y ψ= μ x +y ψ( x + y )= μ y μ x + y + ψ y=0 μ μ μ x + y + ψ y+ = ψ ψ ( ) ( ) μ μ x + y+ = ( ψ) ( ψ) Circle of radius m/(y) and center at x = 0, y = -m/(y)

y x

μ μ i θ μ iθ w (z )= = e = e (cos θ i sin θ) z r r μ u r= cos θ y r μ u θ= sin θ r x

Doublet of strength m at z = z0 μ F (z )= z z 0

4.8. Circular cylinder flow Let uniform flow go past a doublet μ F ( z )=Uz+ z Potential and stream function μ μ μ F (z )=Ure + i θ = Ur+ cos θ+i Ur sin θ r r re iθ ( ) Potential ( ) Stream function Consider streamline y = 0 Ur = m/r means that this streamline is a circle of 1/ radius a = (m/u)

Can rewrite complex potential as ( ) a F ( z )=U z+ z z, F (z) U z Uniform flow dominates the far field a z 0, F ( z ) U z Doublet dominates the flow near the origin

Singularity at origin Velocity=0 (forward stagnation point) Flow symmetry: F(-z) = -F(z) Velocity=0 (rear stagnation point)

4.9. Cylinder with circulation Take cylinder flow, add rotation around the origin ( ) a iγ F (z )=U z+ + log z+c z π Constant to Vortex at origin keep y = 0 at r = a Pretty easy to find C, tuck it into the logarithm ( ) a iγ z F (z )=U z+ + log z π a Complex velocity ( ) df a iγ 1 w= =U 1 + dz π z z

( ) ( ) a iγ 1 a i θ i Γ 1 iθ w=u 1 + =U 1 e + e π z π r z r [( ] ) a i θ i Γ 1 i θ w= U e e + e π r r iθ [( ) (( ) a a Γ w= U 1 cos θ+i U 1+ sin θ+ πr r r )] i θ e Remember that w = (ur-iuq)e-iq ( ) ( ) a a Γ u r=u 1 cos θ, uθ = U 1+ sin θ π r r r

On the surface (r = a), Γ u r=0, u θ= U sin θ πa Boundary! Find stagnation points (velocity = 0, r = a) Γ sin θs = 4 πu a Possibilities: stagnation points on the cylinder 1 stagnation point on the cylinder 0 stagnation points on the cylinder (but maybe somewhere else in the flow?)

Two stagnation points Γ 0< <1 4 πua One stagnation point Γ =1 4 π Ua

No stagnation points on the cylinder Γ >1 4 π Ua Look for stagnation point (rs, qs) elsewhere (for rs > a) Cannot be 0 Must be 0! ( ) ( ) a u r=u 1 cos θs =0, rs a Γ u θ= U 1+ sin θ s =0 πrs rs

cos qs = 0 means qs = p/ or qs = 3p/ ( ) a Γ U 1+ sin θ s= π rs rs negative positive Must be -1, so qs = 3p/ ( ) a Γ U 1+ = π rs rs Solve this for rs

r s= Γ ± 4πU Two stagnation points ( Γ a 4πU ) - inside the cylinder (so who cares?) + outside the cylinder (good stuff)

4.10. Blasius integral laws Find potential Find velocity components Plug velocity into Bernoulli equation to find pressure on body surface Integrate to find Hydrodynamic force on the body Hydrodynamic moment on the body MUCH simpler with complex potential!

Any contour fully enclosing the body C0 Body of an arbitrary shape Surface: streamline y = 0 Complex force: X - iy cece r o Y FFor c.g. X M y x Blasius first law ρ X iy =i w dz C 0 Blasius second law ρ M= ℜ ( zw C0 dz )

Evaluating complex integrals Taylor series (real variable) f ( x x 0)= a n ( x x 0 ), a n= n=0 n f (n) ( x 0) n! This expansion is valid in an interval x - x0 < dx

Evaluating complex integrals Laurent series (complex variable) f ( z z 0 )= a n ( z z 0 ), n n= 1 n 1 a n= f (ζ )( ζ z 0) dζ π i C This expansion is valid in an annulus where f is holomorphic: R1 < z - z0 < R If R1 = 0, z0 isolated singularity Coefficient a-1 of Laurent series: residue of f at z0 C z0

Cauchy theorem If complex function f(z) is holomorphic everywhere inside contour C, f (z)dz=0 C Cauchy residue theorem If complex function f(z) is holomorphic everywhere inside contour C, except isolated singularities, f (z)dz=iπ a 1, k C k

Example ez holomorphic everywhere in a disk of radius r with center at z = 0 3 z z e =1+z+ + + 6 z ez/z holomorphic everywhere in a disk of radius r with center at z = 0, except at it center z e 1 z z = +1+ + + z z 6 a-1=1 Note. a-m 0, a-m-k 0, k = 1,,... at z = z0 z0 is a pole of order m

4.11. Force and moment on a circular cylinder Complex potential ( ) a iγ z F (z )=U z+ + log z π a Complex velocity ( ) df a iγ w= =U 1 + dz πz z Blasius first law ρ X iy =i w dz C 0

4 U a U a i U Γ i U Γ a Γ w =U + 4 + 3 πz z z πz 4π z 0 - -4 Term order in z -1-3 - iu Γ a 1= π z = 0 sole isolated singularity of w, thus iu Γ X iy =iπ a 1,k =iπ π = iρ U Γ k X = 0 (D'Alembert's paradox) Y = rug (Zhukovsky-Kutta law) Similar analysis for zw produces M = 0

4.1. Conformal transformations Helps deal with boundaries z = x+iy z = x+ih y x h z = f(z) It's only good if the Laplace equation is also transformed into something nice... x

Consider f holomorphic function mapping (x,y) into (x,h) In (x,y) plane, let (x,y) = 0 Then in (x,h) plane, (x,h) = 0 (proof: p. 93) Laplace equation is preserved by conformal mapping What happens with complex velocity? df df ( ζ) d ζ d ζ w (z )= = = w( ζ) dz d ζ dz dz Velocity scales during conformal mapping

Let's prove that conformal mapping preserves sources, sinks, etc. Γ= u dl = ( u dx+v dy ) C C C Circulation of all point vortices inside C dy dl dx m= u dn= ( u dy v dx ) C C Strength of all sources/sinks inside C

w( z ) dz= C = ( u iv )( dx+idy )= ( u dx+v dy )+i ( udy vdx )= C C C =Γ+i m Could have proven the same with residue theorem... Now consider a conformal mapping (x,y) (x,h) ( Γ+i m ) z = w(z )dz= C z dζ = w(ζ) dz = dz C = w(ζ) d ζ=( Γ+i m ) ζ z C ζ

Conformal mapping preserves strength of sources, sinks, and vortices

4.13. Zhukovsky transformation c z=ζ+ ζ ζ, z ζ dz c =1 dζ ζ z = 0: singularity (let's contain it inside the body) dz ζ=±c, =0 dζ Nikolai Egorovich Zhukovsky (1847-191) Man will fly using the power of his intellect rather than the strength of his arms. z = c: critical points (angle not preserved)

Critical points of Zhukovsky transform z = x+iy y q -c z0 q1 z = x+ih h z0 -c +c n1 n x +c ζ=±c c z=±c+ =±c ±c Can prove: q1 - q = (n1 - n) A smooth curve passing through z = c will correspond to a curve with a cusp in z-plane x

Example: z = cein z = x+iy z = x+ih y h x x -c +c -c +c c iν i ν z=ce + i ν =c ( e +e ) =c cos ν ce iν Zhukovsky transform recipe. Start with flow around a cylinder in z-plane, map to something

4.14. Flow around ellipses Circle in z-plane, radius a > c, center at origin iν ζ=a e ( ) ( ) c i ν c c z=a e + e = a+ cos ν+i a sin ν a a a iν Major semiaxis Minor semiaxis Parametric equation of an ellipse

z = x+iy z = x+ih y h x x -c +c -c +c

Flow past a cylinder ( ) a F ( z )=U z+ z

Now consider freestream flow at an angle Can get this by conformal mapping too (in plane z': z = eiaz' - rotation) x' y' a Correspondingly, z' = e-iaz

In plane z' ( a F (z ')=U z ' + z' ( ) ( a a iα i α + i α =U z e + e z ze i α F =U z e ) Let's have this flow in z-plane: ( i α F (ζ)=u ζ e Now recall that ) a iα + e ζ c z=ζ+ ζ )

Express z in terms of z: ζ +c ζ z =0 z ζ= ± ( ) z c Recall that for z, z z. Thus select z ζ= + ( ) z c Plug this into F(z) to get F(z)... (skip derivation)

[ i α F ( z )=U ze ( a i α i α + e e c )( ( ) )] z z c Uniform flow at angle a approaching an ellipse with major semiaxis a + c/a and minor semiaxis a - c/a

a Stagnation points: z = aeia

Stagnation points in z-plane... c i α z=±ae ± e a iα a

( ) ( ) ( ) ( ) c c z=± a+ cos α±i a sin α a a c x=± a+ cos α a c y=± a sin α a - forward stagnation point + downstream stagnation point a = 0: horizontal flow approaching horizontal ellipse a = p/: vertical flow, horizontal ellipse (or horizontal flow, vertical ellipse)

4.15. Kutta condition and the flat-plate airfoil Martin Wilhelm Kutta (1867-1944)

4.15. Zhukovsky-Chaplygin postulate and the flat-plate airfoil Sergey Chaplygin (1869-194), Hero of Socialist Labour (1 February 1941)

4.15. Zhukovsky-Chaplygin postulate and the flat-plate airfoil Flow around a sharp edge (section 4.6)... 1 / F (z )=C z df C w (z )= = 1/ dz z z = 0: singularity At a sharp edge, velocity goes to infinity

4.15. Zhukovsky-Chaplygin postulate and the flat-plate airfoil Flow around a sharp edge (section 4.6)... 1 / F (z )=C z df C w (z )= = 1/ dz z z = 0: singularity At a sharp edge, velocity goes to infinity This is not the case in experiment, luckily Need a fix for theory near sharp edges That's not the only problem though...

c z=ζ+ ζ z = x+iy z = x+ih r=c a a Herein lies the problem!

Stagnation point is ALWAYS at the trailing sharp edge! Smoke visualization of wind tunnel flow past a lifting surface Alexander Lippisch, 1953

Zhukovsky-Chaplygin postulate: For bodies with sharp trailing edges at modest angles of attack to the freestream, the rear stagnation point will stay at the trailing edge Dealing with trailing-edge singularity In modeling real lifting surfaces, trailing edge has sharp but finite curvature

How to fix the flat-plate flow? z = x+ih a Angle of attack

Add circulation... z = x+iy z = x+ih a...to move the stagnation point to the trailing edge!

We want to move the rear stagnation point to z = c That would correspond to z = c in the z-plane Need to move it there from z = ceia For cylinder flow with circulation... Γ sin θs = 4 πu a If sin qs = - sin a, Γ=4 π U a sin α

Recipe for constructing a complex potential for corrected flat-plate flow (Eq. 4.b) Cylinder flow Add circulation G = 4p a U sin a Rotate the plane a degrees counterclockwise Zhukovsky transform??? Profit!

Lift on a flat-plate airfoil extending from -a to a Blasius law for cylinder flow: Y =ρ U Γ In our case Y =4 π ρu a sin α

Introduce dimensionless lift coefficient C L= wing chord Y 1 ρu l Characteristic length scale (for wings chord length) For our flat plate, l = 4a and C L =π sin α At small angles of attack, lift coefficient on a flat plate increases with angle of attack!

4.16. Symmetrical Zhukovsky airfoil Goal: airfoil with sharp trailing edge and blunt leading edge z = x+iy z = x+ih y h r = a = c(1 + e) -c t x x l -c +c +c Center: -m = -ec -(c + m) small

Leading edge in z-plane: -(c + m) In z-plane, the leading edge is... c z= c (1+ε) = c+o(ε ) c 1+ ε Chord length l = 4c Similarly (more series expansions, linearization) thickness t 3 3 t =3 3c ε, l = 4 ε Thickness ratio Maximum thickness occurs at x = -c

Extra Flugzeugbau EA300, 1987, Walter Extra design, Zhukovsky wing profile

Can find e in z-plane from desired l and t in zplane: 4 t t ε= 0.77 l 3 3 l Equation for symmetric Zhukovsky profile in zplane ( y x =± 1 l l 3 3 ) ( ) x 1 l At zero angle of attack, stagnation point is at trailing edge, lift = 0 Add angle of attack a...

To satisfy the Zhukovsky/Kutta/whatever condition... z = x+ih Need to move this stagnation point... r=a x a...here! For a cylinder of radius a, the needed amount of circulation is (same as for flat plate...) G = 4p a U sin a

Express radius a in terms of l and t... ( l 4 t a=c+m=c (1+ε)= 1+ 4 3 3 l ) For an angle of attack a, circulation we need to add is... ( ) 4 t Γ=4 π U a sin α=π U l 1+ sin α 3 3 l Lift coefficient for symmetrical Zhukovsky airfoil ( ) t C L π 1+0.77 sin α l t 0, this reduces to lift coefficient of flat plate Zhukovsky symmetrical profile has better lift!

4.17. Arc airfoil Airfoil of zero thickness but finite curvature ) h a r ( z = x+ih m Use cosine theorem to get r π a =r +m r m cos ν In z-plane, -c c i ν z=r e + e = r x y c c = r+ cos ν+i r sin ν r r iν ( ) ( ) x n +c

( 4 ) ( 4 ) c c x = r +c + cos ν, y = r c + sin ν r r sin n ( ( cos n 4 ) ) c r cos ν sin ν=x sin ν c + cos ν sin ν r = 4 c r cos ν sin ν= y cos ν+ c cos ν sin ν r x sin ν y cos ν=4c cos ν sin ν Use cosine theorem: ( ) r c c 1 y sin ν= = r = rm r m msin ν

y y sin ν=, cos ν=1 m m y cannot be negative!!! x sin ν y cos ν=4c cos ν sin ν ( ) ( ) y y y y x y 1 =4 c 1 m m m m x y c y y+ = c m m m m y x m y+ y =4 c c m c m c m x + y +c =c 4+ m c m c [ ( )] [ ( ) ]

[ ( )] [ ( ) ] c m x + y +c m c c m =c 4+ m c y 0 Equation of an arc in the z-plane z = x+iy z = x+ih y h -c +c m h will find r a x -c x n +c

Otto Lilienthal and his glider, 1895

[ ( )] [ ( ) ] c m x + y +c m c c m =c 4+ m c Recall that m/c = e, linearize (not essential here but nice) ( ) ( ) c c x + y+ =c 4+ m m Find arc height h Since y = m sinn, ymax = h =m Next have to add circulation to put stagnation point at the trailing edge (trickier, because cylinder is moved upward in the z-plane)

Stagnation point needs to rotate by a + tan-1(m/c) Angle of attack Vertical shift Linearize: tan-1(m/c) m/c = e, a c Amount of circulation to be added: ( ) ( ) ( ) ( ) m m Γ=4 π U a sin α+ 4 πu c sin α+ c c Lift coefficient: m h C L =π U c sin α+ = π U c sin α+ c l Again, more lift than flat plate!

4. 18. Zhukovsky airfoil Know how to create lifting surfaces with: Straight chord, finite thickness Zero thickness, small finite curvature (camber) Both improve lift, compared with flat plate Create a lifting surface with both thickness and camber (Zhukovsky profile)

z = x+ih h h/ r a -c x +c 0.77 tc/l l chord t max. thickness h max. camber

z = x+iy y x t h -c +c l

Maxim Gorky (ANT-0, PS-14) plane, 1935

Circulation ( ) ( t h Γ=π U l 1+0.77 sin α+ l l Lift coefficient ( thick ness ) ( cam ber t h C L =π 1+0.77 sin α+ l l ) )

Kalinin K-7 (Russia, 1930)

Dornier X flying boat (Germany, 199)