IDEAL FLOWS.! n. " u! n = u w. $ u = u i

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1 IDEAL FLOWS Generalities Inviscid, incompressible flow with conservative body forces g =!G : with: D! Dt = "!# u = 0! Du = "#p + # +!#G Dt * p and u specified at some point in the flow velocity BCs on the flow boundaries: Notice: # u = 0 +u, +t dx u u + p! " G = B t " u! n = u w! n # u = u i on solid surfaces walls on internal surfaces! " constant! the energy equation is decoupled can be dropped Kelvin theorem! " # u = 0 if the flow originates from an irrotational stream Thus, for incompressible and irrotational flows:! " u = 0! u = " #! " vector stream function! " u = 0! u = "#! " scalar velocity potential In particular: goverrning eq s in 2D flows planar, polar, axisymmetric, etc.! reduces to a single component e k!

2 IDEAL FLOWS Velocity Potential Formulation With u =!" :! " u = 0 #u #t " dx u "u + p G = B t * +! 2, = 0 * #, #t + 1 2!, "!, + p G = B t Notice:! is harmonic! 2 " = 0 # linear! superposition holds for! and u p! nonlinear in u,!! superposition does not hold for p d! = "! # dx = u # dx! equipotential surfaces d! = 0 are are orthogonal to u Closed-form solution approaches: superposition of elementary solutions boundary integral methods conformal mapping planar 2D separation of variables 3D flows in ON co-ordinates and BCs Numerical solution approaches: discretized boundary integral methods collocation, finite differences, finite elements, etc.

3 2D IDEAL FLOWS Scalar Stream Function In 2D flows planar, polar, axisymmetric, etc.:! = e k! orthogonal to the 2D coordinate plane e i,e j! " # = x k = 0 As the flow is irrotational and u =! " # :! " u =! "! " # =!! #! 2 # = e 3! 2 # = 0! 2 # = 0! " harmonic In rectangular coordinates planar Cartesian coordinates x, y, z : # "# = e x x + e # y u = # * y x = * y! = e z!! u =! * " = e x y e * y v = # + * x y = + * x Also: u! "# = "!"# = x # x + y # y = # y # x # x # y = 0 Thus, in general: lines surfaces at constant! are parallel to the velocity lines surfaces at constant! and! are mutually orthogonal and w = 0 The infinitesimal volume flux through a line element dx is: dq =!v dx + u dy = " x dx + " y dy = d"!! " constant along streamlines dq = 0 dx dx dy u u v

4 2D IDEAL FLOWS Problem Show that in: polar coordinates planar cylindrical coordinates r,!, z : u = "# "r = 1 " r "! = e z!! v = 1 "# r " = " and w = 0 * "r axisymmetric cylindrical coordinates r,!, z, where dx = e r dr + e! r d! + e z dz : u = "# "r = " "z! = e "!! w = "# "z = 1 and v = 0 " r "r r axisymmetric spherical coordinates r,!,!, where dx = e r dr + e! r d! + e " r sin! d# : u = "#!! = r sin" e! "r = 1 " r 2 sin " # v = 1 and w = 0 "# r " = 1 " * r sin "r Derive the expression of the Laplacian in the above coordinate systems

5 PLANAR IDEAL FLOWS Separable Solutions Assuming separable solutions:! = R rcos m" # m of Laplace s equation in polar coordinates r,! :! 2 "!r + 1!" 2 r!r + 1! 2 " r 2!# = 0! R "" + 1 R " # m2 lnr for m = 0 R = 0! R " 2 2 r r r ± m for m # 0 where m is integer for single-valued 2! m -periodic solutions in!, and therefore: Using:! = a 0 ln r + a m r m cos m" # m m0 u =!"!r = 1!# r!, v = 1 r it is readily shown that: d! = "! "! dr + "r "# d# = 1 r and, integrating:!"!# =!!r! = a 0 " + a m r m sin m" # m m0 and! 2 "!r r!"!r + 1! 2 " r 2!# = 0 2 " " dr + r "# "r d# is perfect since!!#!"!r =!!r!#!"

6 PLANAR IDEAL FLOWS Complex Velocity Potential From earlier results:! = a 0 ln r + a m r m cos m" # m and! = a 0 " + a m r m sin m" # m obtain: Also: f z m0 =! + i" = a 0 ln z + a m e #i m z m m0 u =!"! x =!#! y ; v =!"! y = #!! x are the Cauchy-Riemann equations for f z! analytic " f # z = f z f! z = w z = m0 complex velocity potential + i * * + i + * and u =!"!r = 1!# r! ; v = 1 r = f x + iy = f re i! x e i r = + i i y = + i ire i -"# " x + i " = u iv for z = x + iy / " x. / 1 "# e i "r + i " * "r +, = u iv for z = re i 0/ e i!"!# =!!r and therefore: for z = x + iy for z = re i complex velocity

7 ELEMENTARY SOLUTIONS Uniform Flow Recall: =! + i" = a 0 ln z + a m e #i m z m f z m0 Thus, choosing a 1 = U,! 1 =! and a m = 0 for m! 1, obtain: f z =! + i" = Ue #i z! Differentiating f z: # = u! iv = Ue!i"! u = U cos" w z Therefore f z " = U x cos# + ysin# = U x sin# + ycos# v = U sin" = Ue!i" z represents a uniform flow with: absolute value of the velocity U orientation at an angle! w.r.t. the positive x -axis U From Bernoulli s equation with p z o p! = p o! + G " G o = p o and G z o = G o : Show that u! r 0 " the kinetic energy of the whole flow is divergent

8 ELEMENTARY SOLUTIONS Source-Vortex Flow Q or Γ Choosing a 0 = Q! i" 2# and a m = 0 for m! 0, obtain: f z =! + i" = Q # i 2 log z Using polar representation z = re i! " ln z = lnr + i! : Qln r + "# Q" # lnr! = and! = 2 2 Differentiating f z with z = re i! : w z = u! iv = Q! i# u = Q and v = # e i" 2re i" 2r 2r On any oriented contour C of radius r centered at z = 0 : 2!" u! n ds = ur d# C " = Q and u! t ds 0!" = vr d# C " = 0 i.e. flow with source-vortex intensities Q and! From Bernoulli s equation with p! p! = p "! + G # G # Q2 + 2 " 8 2 r 2 2 = p! and G! = G! : Show that u! r "1 # the kinetic energy of the far field flow is divergent streamlines equipotential lines for source vortex flows Γ or Q streamlines equipotential lines for vortex source flows

9 ELEMENTARY SOLUTIONS Doublet Flow Choosing a!1 = µe i" and a m = 0 for m! "1, obtain: f z =! + i" = µei# z! " = µ r cos # = µ * r sin # streamlines and equipotential lines are circles through the center of the dipole Differentiating f z with z = re i! : w z = u! iv e i" Notice: d Q! i" log z 2# = Q! i" 2#z =! µei# r 2 e! u = " µ i2" r cos # " and v = µ 2 r sin! " # 2 dz Hence the flow can be visualized as the superposition of: a source doublet of strength ±Q! " at z o + dz and z o with Q dz = 2!µe i" a vortex doublet of strength ±! " # at z o + dz and z o with! dz = 2"µe i # +" 2 a doublet is characterized by its orientation which is not true for source-vortices Show that u! r "2 # the kinetic energy of the far field flow is convergent

10 Cylinder in a Uniform Stream SUPERPOSITION SOLUTIONS Consider the superposition of a uniform stream U and a doublet parallel to U : f z =! + i" = Uz + µ z : Using polar coordinates z = re i! = r cos! + isin! Hence: "! = Ur + µ # r cos and! = # Ur " µ r! = 0 " constant on: Differentiating f z Problem w z = u! iv with z = re i! : sin " the circle of radius R = # the x-axis! =0 µ U = U! µ # u = U 1! R2 R2 e i" z 2 r 2 cos" and v =!U 1+ r 2 sin" Compute the sensitivity! p 1 " p 2!# of the Pitot tube of the figure in the measurement of the freestream direction from the pressures at the taps 1 and 2. U

11 Blasius Integral Formulas Assumptions: FLOW FORCES fluid force F on a body in a steady force-free flow For the fluid inside a simple countour C around the body:!#!uu " ds + p ds = F where from Bernoullis equation p = B! 1 C 2 " u2 + v 2 Hence, noting that! Bds = 0 on closed contours and that ds = e x dy! e y dx " dx : C F x =!! # "u udy! v dx + pdy C = "! # uv dx! 1 2 u2! v 2 dy C F y =!! # "v udy! v dx! pdx C = "! #! 1 2 u2! v 2 dx! uv dy C F body dx ds C In complex notations: " w 2 = u! iv 2 = u 2! v 2! 2iuv # dz = dx + i dy! F " if = i # x y 2! zdz 1st Blasius formula w 2 C Problem Show that for the fluid moment M on the body, using:! x! "uu # ds + pds = M! M = " # 2 Re zw2 zdz C { } 2nd Blasius formula! C

12 Source-Vortex Forces Assumptions: FLOW FORCES source-vortex pair Q! i" in a uniform stream U Here: f z = Uz + Q! i" 2# ln z! z o Therefore: = f! z = U + Q " i# 1! w 2 z = U 2 + Q " i# 2 z " z o w z and, by residue integration of Blasius formulas: F x! if y = i " 2#i Res w2 z 2 { } z = zo =!"U Q! i M = M Q,! = " 1 # Re 2i Res zw2 z 2 { { } } = "#Q! z = zo Therefore: F Q =!"UQ F # =!"U# M Q,# =!"Q# M Q + M # = 0 no superposition y U U + Q " i# z " z o 2 *! F x = F Q = "#UQ F y = F = "#U Q! i" 2 + F 1 2 z " z o Kutta-Joukowski Law x

13 SUPERPOSITON SOLUTIONS Flow around a Cylinder Consider a flow with: two opposite conjugate source-vortex pairs: Q ± i! at A, A! z =!a,! 1 a, reals at B,!! Q ± i" B z = b, 1 b, reals a vortex! o in the origin O z = 0 y z # plane Q + i" Q # i" #Q + i" #Q # i" " o + + # # A A! O B! 1 B x The complex velocity potential is therefore: = Q 2! log z + a z + 1 a z " b z " 1 b " i # 2! log z + a z " 1 b z + 1 a z " b " i # o = Q 2! log " b z + a 1 z + a " i # a z " b 1 z " b 2! log " a z + a 1 z " b b 1 z + a z " b f z 2! log z = " i # o 2! log z Problem Recalling that: log z = ln z + i arg z products of complex conjugates are real numbers ratios of complex conjugates have unit modulus show that! " constant on z = e i! 1 z = e!i" = z, i.e. that the unit circle is a streamline

14 Some Special Flows around a Cylinder SUPERPOSITON SOLUTIONS y Ideal Flows z # plane Recall: = Q 2! log z + a z + 1 a z " b z " 1 b + z! 1 b z + 1 a z! b! i " o f z!i " 2# log z + a 2# log z Q + i" Q # i" #Q + i" #Q # i" " o + + # # A A! O B! 1 B x For a single source-vortex pair in A b! +", neglecting the constant term in log b : f z! Q 2" log z + a z + 1 a # i z 2" log z + az z + 1 a 2" log z For a circle in a uniform stream a = b! +" : z + 1 a z! 1 b z! 1 b z + 1 a z! b log z + a z! b log z + a = log!1 = log!1 # i o " 1+ z a + log 1+ 1 az # 1! z a 1! 1 az + log " 1 + z a 1! 1 az # az 1! z a : and therefore, neglecting the constant term in log!1 f z! Ue "i# z + U e "i# z " i o 2 log z where Ue!i" = Q! i# a 2z a + 2 az + constant 2z a! 2 az + constant

15 COORDINATE TRANSFORMATIONS Invertible 2D Transformations by Real Variables x and y For globally invertible mappings from a set D to its image! : #! =! x, y " = " x, y # x = x!," y = y!," for any x, y in D By hypothesis: # x = x! x, y," x, y on D y = y! x, y," x, y Then, if!,! are differentiable on D and x, y on! : # x!! x + x " " x = 1 # y!! x + y " " x = 0 and x!! y + x " " y = 0 y!! y + y " " y = 1 and for unique solutions in the unknowns x!, x! and y!, y! : J x, y =! x " x # 0! x = # " y J! y " y x # = " y J Then, if! x, y,! x, y and J x o, y o N x o, y o of x o, y o! x, y differentiable on the image N! o," o and y! = "# x J y # =! x J are continuously differentiable i.e.! x,! y,! x,! y! continuous on D! 0, then from the above equations by continuity J x, y! 0 on a neighborhood, x!, x!, y!, y! are continuous and therefore the transformation! x, y, is one-to-one locally invertible on N x o, y o, with x!," and y!," continuously of N x o, y o under the transformation. y " # D x o, y o N x o, y o! o," o N! o," o x!

16 Invertible 2D Transformations The complex transformation:! =! z MAPPING WITH ANALYTIC FUNCTIONS for any z in a complex set D is invertible, i.e. if:! # z 2! z " = " #1 " maps D into its image! if! z z 1! z 2 " # z 1 Let! = " + i# and z = x + iy, then it follows that: If! z #! =! x, y " = " x, y is also analytic:! z in D and "! z = # x + i x with Since! z! x = " y! y = #" x # x = x!," y = y!,"! " in! C-R eq ns for! z is invertible, from earlier results: =! x " y #! y " x =! 2 x + " 2 x = z 0! #" z 0 J x, y x! = " y J =! x J = y " x " = #! y J = " x J = #y! = z : "! z = 1! z "# = 1 #" z Besides, differentiating z! z z! " C-R eq ns for z!! also z " # analytic y " D N z o N # o z plane z o = x o + iy o x # plane # o =! o + i o!

17 Conformality of Analytic Mappings MAPPING WITH ANALYTIC FUNCTIONS Conformal 2D transformations! x, y, " x, y magnitude locally preserve: orientation of the angle between any two regular curves in the x, y! plane Consider the analytic transformation:! =! z on D at a point z where: "! z # 0 i.e.! z " invertible Then for any line element dz from z and its image d! : d! =!" zdz! d" = dz " # z argd" = argdz + arg "# z Hence, the element dz, regardless of its orientation, is: elongated/shortened by the same factor "! z rotated by the same angle arg "! z It follows that complex analytic transformations! =! z : represent conformal mappings in the z! plane at any point z where "! z y D C z! C " z " d" # 0 dz Argdz z - plane x " # plane Argd"

18 Flow Solutions by Conformal Mapping In the physical plane! = " + i# :! 2 " #, CONFORMAL MAPPING = 0 f = "#, + i #, analytic Given an analytic transformation! z = f! z f z, define: by assigning equal values to the complex velocity potential f at corresponding points z and! z in the two planes As f! and! z are both analytic, then by construction: =! x, y + i" x, y # analytic 2! x, y = 0 f z Hence: Laplace s equation is invariant under conformal mappings the flow in the! -plane maps in another simpler flow in z -plane Idea: map the physical! -plane and BCs into a simpler geometry in the transformed plane z = x + iy with the inverse transformation z =! "1! obtain if possible the flow solution in the transformed plane obtain the physical solution in the! -plane with the direct transformation! =! z! y # C # z # C z w # w z # plane z plane S # Q # i # Q z i z " S z x

19 CONFORMAL MAPPING Complex Velocity under Conformal Mappings Given the analytic mapping:! =! z at any point where the mapping is invertible, i.e.: 1 "! z = # 0 z! " the velocity potential in the z plane transformed plane is: f z = f! z Hence differentiating: f! z = f! " where: f! " f! "! z = f!" z!" # f! " = f! z "! z = w " is the complex velocity in the physical plane z = w z is the complex velocity in the transformed plane and therefore the complex velocity transformation formulas are: = w! z "! w z and viceversa w! = w z!" z! y # C # z # C z w # w z # plane S # Q # i # z plane Q z i z " S z x

20 CONFORMAL MAPPING Source-Vortex Transformations under Conformal Mappings For a source-vortex Q! " i#! in the! plane: Q! =!# u " n ds = u d v d C!!# { = Im w! d! C!!# } C!! " =! u # t ds = u d + v d C "! { = Re w " d" C "! } C " where: C! is a simple closed contour surrounding Q! " i#! = u " iv is the complex velocity in the! plane ds = d! + i d" is the line element of C! n, t are the normal/tangential unit vectors along C! w! and similarly for the source-vortex intensities Q z and! z in the z -plane from the! plane to z plane, using: = f "!, f! = f z and w z = f! z Under the conformal mapping z = z! w! obtain:! " + iq " = f #" d"! = f " C " " 2 z "1 = f z 2 z1 =! f # zdz =! z + iq z Hence source-vortex intensities are not affected by conformal mappings C z! y # C # z # C z w # w z # plane S # Q # i # z plane Q z i z " S z x

21 CONFORMAL MAPPING Doublet Transformations under Conformal Mappings Recall that for doublet generated by a source Q! in the! plane: where: 2!µ " e i# " = Q " d" µ! e i"! = Q! d! 2# is the complex doublet intensity Under the conformal mapping from the! plane to z plane: z = z! "! =! z from earlier results: the source intensity is unaffected: Q z = Q! the line element d! transforms into: dz = z!" d" Hence the doublet intensity in the z plane is: µ z e i! z = Q z dz 2" = Q d# # z # = µ # e i! # z # 2" Hence doublet intensities are affected by conformal mappings proportionally to z! "! y # C # z # C z w # w z # plane z plane S # Q # i # Q z i z " S z x

22 CONFORMAL MAPPINGS Linear Transformations LTs In this case:! z = az + b with a! 0, b " complex constants y, # z - and - planes a LTs are globally invertible since: = a # 0! z " locally invertible everywhere "! z z! =! " b a globally valid, also a LT D 1 b! Arga Special cases: x," a = 1! " z = z + b translation by a displacement b a = a > 0! " z = a z change of scale by a factor a a = e i!! " z = e i# z rotation through an angle! = Arg a mod 2" For every LT:! z = " + b with! = a µ and µ = e i Arg a z translation + scale change + rotation For every analytic mapping:! z "! z o +!# z o z z o as z! z o locally approximated by a LT

23 CONFORMAL MAPPINGS The Inversion Transformation In this case:! z = 1 z for z! 0 z =1 y z - plane z z For z! 0 the inversion is globally invertible since: Arg z x = # 1 z 0! z " locally invertible for z! 0 2 "! z Problem z! = 1! globally valid for! " 0, also an inversion Show that the inversion transformation maps: the z = 1 into! = 1 the exterior of z = 1 into the interior of! = 1, and viceversa lines and circles into lines or circles, and in particular: lines through the origin z = 0 into lines through the origin! = 0 lines through the origin z = 0 into lines through the origin! = 0 lines not through the origin z = 0 into circles through the origin! = 0 circles not through the origin z = 0 into circles not through the origin! = 0 # =1! Argz 1 z 1 z # plane "

24 CONFORMAL MAPPINGS Linear Fractional or Möbius Transformations LFTs In this case:! z = az + b cz + d for z! " d and c For z! " d c the LFT is globally invertible since: "! z = z! = d! " b a " c! For every LFT:! z = a c ad # bc cz + d 0! z 2 + bc " ad c a,b,c! 0,d " complex ad # bc! 0 " locally invertible globally valid for! " a c, also LFT # with! = 1 µ Hence, recalling the properties of LTs and inversions: LFTs map lines and circles onto lines or circles Problem and µ = cz + d LT + inversion + LT Show that the LFT that maps the upper half z -plane Im z > 0 onto! < 1 see Figure is:! z = " z " i z + i A y 1 0 1! # =1 i B z - plane C x i # plane C B 1 " A

25 CONFORMAL MAPPINGS The Joukowski Transformation JT In this case:! z = z + 1 z for z! 0 For z > 1 z < 1 the JT is globally invertible since: = 1# 1 z 0! z " locally invertible for z! ±1 2 "! z A 1 y r! z plane B 1 x and, choosing the root such that z! " or z! 0 as! " # : Problem z! = #! +! 2 " 4 2! "! 2 " 4 2 on z > 1 on z < 1 globally valid for any! # plane A B 2 2 " Show that the JT maps: the unit circle z = e i! twice on the segment!2 " # " 2,! = 0 circles z = re i! onto the ellipses of semi-axes r ± 1 r the interior and the exterior of the unit circle z = e i! on the entire! -plane more generally, circles in the z -plane into ellipses in the! -plane

26 CONFORMAL MAPPINGS The Schwarz-Christoffel Transformation SCT In this case, using the principal branch for the complex power function: 1 2 n1 z! = #!" zdz = a z x 1 z x 2... z x # n1 dz + b with a, b constants and! z For z = x real and dx > 0 Argdx = 0 : z 0 defined and locally invertible for z! x 1, x 2,...x n"1 n1 d! =!" xdx # Argd! = Arg a + Arg x x k constant for z! x 1, x 2,...x n"1 x! x k! " k # = e Log x! x k!" k # k =1! " k = e! " k # ln x! x k +i Arg x! x k Arg x! x k Therefore: Argd! changes by an angle! k when x crosses x k from the left if! ! n = 2", with! n " turning angle for x n =!, then:! describes a closed polygon! 1,! 2,...! n in the! -plane # =!" k for x < x k * + 0 for x > x k x 1 x 2 y z! plane x n!1 x x n " Besides: the SCT maps the upper z -plane inside! 1,! 2,...! n a and b introduce a change-of-scale/rotation and a traslation the n sides of! 1,...! n are linked by n! 1 similarity conditions Hence the n!1 real unknowns x 1, x 2...x n!1 can be uniquely determined # n! 2 #! plane 1 # n!1 # 2 # n # 1

27 CONFORMAL MAPPINGS The Unstaggered Cascade Transformation UCT In this case:! z = s 2" log e# + z e # z + log e# + 1 z e # 1 z * for z! ±e±" For z! ±e ±" the UCT is locally invertible since: "! z = s 1 2# z + e 1 z e + 1 z + e 1 z e * +, 0 Obtain the UCT by consecutively mapping: z > 1 outside a single flat plate in the w -plane by the JT: w z = c! z " # z where w ±e! = ±1 " c = e! + e #! the w-plane onto the strip! < s 2 by the SCT with: a! 1 "1 =! 2 1 = # w = w + 1 w " 1 with the conditions on the central blade: = b = 0! w = # a = #i a 2 = i s 2 2 log1+ w 1# w + b integrating "! w! 0! " Since log! = Log! + i2k" with k! integer,! =! w is s -periodic in! blade cascade A 1 B #e e v A O B #1 1 A y u! " - plane O c z - plane s x w - plane B

28 CONFORMAL MAPPINGS The Staggered Cascade Transformation SCT In this case stagger angle! " 0 :! z = s For z! ±e ±" "! z = s 2" e#i log e + z e # z + ei log e + 1 z e # 1 z * + 2# Obtain the CT from the UCT:! z = s the CT is locally invertible since: e i z + e ei z e + e i z + e e i z e 2" log e# + z e # z + log e# + 1 z e # 1 z * for z! ±e±" * +, - 0 by considering that! must: tend to i s 2e!i" instead of i s 2 for large z :! the 1st log-term must be proportional to e!i" be real on the central blade ln z = Log z for z = e i! " 1 z = z :! the 2nd log-term must be the conjugate of the 1st y z - plane A O 1 B e e x v w - plane A O B 1 1 u " # - plane O s B A c!

29 The Staggered Cascade Chord and Solidity CONFORMAL MAPPINGS The CT fails to be analytic at the leading and trailing edges, L T : "! z # ei z + e ei z e + e i z + e e i z e = 0 Solving for z = z o after some algebra obtain: and: z o = ± e! +i" #! #i" + e z e! #i" + e #! +i" o 2 = 1 z T = e i T * z L = #z T = e i T + z o! 1 z o = ei"! e!i" e #! e!# = T tan!1 tan" tanh# z o + 1 z o e i" + e!i" e i# + e!# * L = T + Substituting in the transformation! z c =! z L "! z T = 2s : # sin sin tan"1 + + sinh 2 + cos 2 * + 2s # cos ln sinh 2 + cos 2 + cos + sinh * which implicitly determines the parameter:! =! ", c s as a function of the stagger angle! and the cascade solidity! = c s A O 1 B e e A O B 1 1 A y v u " # - plane O c z - plane w - plane s! x B

30 . Ideal Flows The Radial Cascade Transformation RCT CONFORMAL MAPPINGS +! - plane Map an axial cascade onto a radial logarithmic cascade by: a rotation through an angle! : = "e i# + iµ = + i cos# + isin#! " L A " O c s T B * a change of scale 1 c and a logarithmic transformation: ln! " = " # ln + i = + iµ c O c Then:! " 0, + # as! "!# far fields A, B i!" =!# c i 2 N B = i s c N B = 2c s No. of blades A " L µ O c plane s T B # On the central blade!c 2 " # " c 2 : ln = "! " = "e i# O c cos# = " ln L,T = ± 1 * c sin# O 2 cos# Therefore, eliminating! : ln! = " cot# d! = cot# logarithmic spiral! O! d" A L, plane T O T O L B

31 Flow Velocity Assumptions: flat plate, chord length c freestream velocity U angle of attack! FLAT PLATE FLOW U! " plane A O B c # V A y O o z plane B x Map the plate on the unit circle by the Joukowski transformation: = c 4 z + 1 # z! z " In the z -plane:! #" z = c z 2 f z = Ve!i" z + V e!i" z! i # o 2 log z = f! z = Ve "i# " V w z e "i# z " i o 2 2z where V and! are determined from the far-field condition: w z lim w! = lim Ue i = Vei! "# z"# * V = Uc 4! z c 4 + = and! o by the Kutta condition at the trailing edge B z = 1: w z z =1 = V e!i"! e i"! i # o * 2 = 0! " o = 2V sin 2#

32 FLAT PLATE FLOW Flow Forces U # " plane! From Blasius integral formula in the! -plane: Fe!i" = i # w 2 d 2! = i # 2 w z * A O B C, zdz 2! C z z + c where C! is any simple closed contour surrounding the profile and: w z = Ve!i"! V e!i" z! i # o and "! z = c 2 2z 4 1 # 1 z 2 V y In particular, for C! "! # C z " z # : z " plane w 2 z "! z # Vei V e i z i o * 2 2z, - c * z , - A O o B from which: # Res w z x = i2 * o 4 "! z 2+ Vei, c = iu 2 ce i- sin- Thus, by residue integration of Blasius formula: Fe i! = 1 "U 2 2 c2# sin e i +# 2! F " U lift force! C L = 2" sin# Notice: the suction due to the LE vortex rotates F by an angle! in the upstream direction Problem Show that the momentum w.r.t. the origin is M O =! 1 4 "#U 2 c 2 sin cos

33 SOLUTION SUPERPOSITION Line-Distributed Singularities Assumptions: line-distributed singularities curvilinear abscissa s linear source-vortex intensity q s! i" s Integrating on a contour surrounding a line element ds : ds u u # + u # [ q s! i" s ]ds C u q sds =! u " n ds = 2u # ds! 1 2 q s C! sds = " u #t ds = 2u ds! 1 2 " s C = u " = u normal velocity component tangential velocity component Integrating the elementary contributions: q s! i" s f z = log z! sds w z = L 2# q s! i" s ds L 2# z! s The superposition of line-distributed singularities is often used to satisfy the velocity BCs on solid surfaces.

34 FLAT PLATE FLOW Flow Solution by Line-Distributed Singularities Assumptions: blade chord c, freestream velocity U, angle of attack! vortex sheet distribution! " on the blade surface z =! zero vertical normal velocity on the blade kinematic BC Complex velocity: w z = u! iv = Ue!i" + L d!i# 2 z! Hence, in particular the vertical velocity on the blade at z = x is: c 2 " # v x = U sin! + = 0 for!c 2 " x " c 2 kinematic BC c 2 2 x # d# The general solution of this homogeneous integral equation for! "! " = #2U sin c 2 # " c 2 + " and, using the Kutta-Joukowski law, the blade force lift, in this case is: c 2 F = F x! if y = i"u # d =!i"u 2 sin!c 2 U! can be shown to be: y A O B c z-plane " # x

35 AXIAL CASCADE FLOW Flow Geometry and Far-Field Velocities Assumptions: infinite cascade of flat plates chord c, spacing s, stagger angle! upstream velocity U 1, angle of attack! 1 downstream velocity U 2, angle of attack! 2 blade circulation! By continuity in the direction normal to the cascade: Let: = U 2 cos! 2 + " U 1 cos! 1 + " plane c s! A B " U 1 # U 12 = 1 2 U + U 2 1 and!u = U 2 " U 1 Then, on the contour C closing in the far fields! A,B =!" :! = su 2 sin " 2 + # su 1 sin " 1 + # = su and, from the velocity triangles: w!" = U 12 e i# ! 2 Uei = U 12 e!i# 12! i 1 2 Uei w +! = U 12 e i" 12 # 1 2# 2 Uei = U 12 e #i" 12 + i 1 2 Uei U 12 U 2 12 U

36 Transformed Velocity Potential AXIAL CASCADE FLOW Map the blades periodically on z = e i! by means of:! z = s 2" e#i log e + z e # z + ei log e + 1 z e # 1 z * + Then:! =!" # z =!e the uniform flows at!! map into source- vortex singularities Q A,B! i" A,B at!e! lim w! = lim! "!# z"!e Hence in the z -plane: f z = Q A! i" A 2# + Q A + i" A 2# + Q B! i" B 2# + Q B + i" B 2# w z! z + log z + e log z + e! + log z! e + log z! e! with: Q A! i" A Q A + i" A Q B + i" B Q B! i" B A + A # O! B# 1! B!e!e! e! e * y! plane c s A B C z! plane x

37 AXIAL CASCADE FLOW Transformed Flow Velocity Differentiating: y z! plane = Q A! i" A w z "! z = s + Q + i" A A 2# z + e! + 2# z + e + Q B! i" B 2# z! e 2# Then, from w! + Q B + i" B 2# z! e! e i z + e ei z e + e i z + e e i z e = w z!" z in the limits for! "!# z "!e : U 12 e!i" 12! i 1 2 #Uei = Q! i A A! Q se!i A " i# A = su 12 e "i 12 + " i 1 su 2 U 12 e!i" 12 + i 1 2 #Uei =! Q! i B B! Q se!i B " i# B = "su 12 e "i 12 + " 1 isu 2 Thus, using! = "s#u, the complex velocity potential is: w z = su 12 2! +i, 4! * +, e "i # 12 + " e"i # ei #12 + z + e z " e z + e " ei #12 + " z " e " + * + 1 z + e + 1 z " e " 1 z + e " 1 " Q A! i" A Q A + i" A Q B + i" B Q B! i" B z " e " * + A + A # O! B# 1! B!e!e! e! e x

38 AXIAL CASCADE FLOW Blade Circulation and Flow Turning U 1 For unstalled flow: = w z T " = 0 # w z T = 0 w! T! z T Kutta condition where z T = e i! T! T = tan "1 tan# tanh and: After some heavy, uninteresting algebra: 2sU! = "s#u = " 12 sin 12 sinh 2 + cos 2 L azimuthal direction c s!! axial direction U 12 U 2 #U " 12 For given U 12 and! 12, determine:! 1,! 2 from: tan! 1 + " # tan! 12 + " = tan! 12 + " # tan! 2 + " = U 1, U 2 from the continuity equation: U 1 cos! 1 + " = U 2 cos! 2 + " = U 12 cos! 12 + " U 2 U 12 cos! 12 + " F a, F! axial azimuthal blade forces from the momentum balance of a blade channel: =!F a s p 2! p 1 * s"u 1 cos # 1 + U =!F where p 2 = p ! U " U 2 Bernoulli s equation

39 AXIAL CASCADE FLOW Flow Deviation Angle The deviation angle! 2 : is a measure of the guiding effectiveness of the blades because of the decreasing aspect ratio of the blade channels: increases with the blade stagger angle! decreases rapidly with the solidity! = c s for! < 1 for! = c s > 1 is relatively: small a few degrees insensitive to the angle of attack! 1 " well guided flow γ = π 4 c s = 0.5 c s = 1 γ = π 6 γ = 0 γ = π 4 γ = π 6 γ = 0 For the flow from hub to tip r H 2! r 2! r T 2 in typical axial bladings: Hence:!h t! r 2 u " 2 # constant head rise cos # u " 2 1 u z2 r 2 c a # ccos # constant axial chord c s # c cos a 0! r 2 2r 2 N B γ = π 4 the solidity c s is nearly constant! small effect on! 2 the stagger angle! increases!! 2 increases with r 2

40 Flow Velocity Assumptions: axial cascade: azimuthal velocity V = dµ dt radial cascade: rotational velocity! = d" dt ROTATING CASCADES Using the radial cascade transformation RCT: ln! = " # ln O + i t = + iµ t c # = V c Thus the translational velocity V in the! -plane transforms in the rotational speed! = V c in the!, " -plane Solution approach: decompose the absolute velocity in the! -plane: u = u! + V relative + dragging components solve for u! on stationary blades with previous approach add V to the solution for u! A use the RCT to obtain the velocity past a rotating radial cascade in the!, " -plane L µ O c A L s " # plane T T O L O B T!, # plane B

41 Flow Velocity on a Moving Cylinder Assumptions: cylinder of radius R in a resting fluid density!, far field pressure p! variable position x o t UNSTEADY FLOWS in the x -direction! = 0 variable velocity U t x Decompose the absolute velocity at the time t in the dragging and relative components: U # uniform stream u = U + u! " u! # flow on a stationary cylinder at x o With no loss of generality x o = 0 at the observation time t, then: = Uz! U z + R2 f z " # z =!U R2 z + f z 2, 2Re f z -2 = w z = u! iv { } = Re,!U R2 t in a uniform stream U + - = U R2 e i* r 2 e i2* re i*. / = 1 0 Therefore the velocity potential and the radial and azimuthal flow velocity components are:! = " UR2 r cos#, u = U R2 R2 cos! and v = U 2 r r sin! 2 y z o z " plane U!

42 UNSTEADY FLOWS Flow Forces and Apparent Mass of a Moving Cylinder The force components on the cylinder are: 2# F x =! p z = Re i" Rcos" d" 0 2# F y =! p z = Re i" Rsin" d" 0 From earlier results and the unsteady Bernoulli s equation:!"!t ww + p # = p p # z = Re i = p + # du dt Rcos 1 2 #U 2 = # du x dt Rcos + c Hence, carrying out the integrations: F x =!" du 2 du dt R2 cos 2 # d# =!"R 2 0 dt F y =!" du * F =! R 2 " du 2 dt dt R2 sin# cos# d# = 0 0 Hence F has the typical form of an inertial force F =!m a du dt where: m a =!"R 2 added mass, here equal to the mass of the displaced fluid Added masses can be quite important in determinng the critical speeds of turbopumps Problem Show that F =!m a du dt also holds with m a =!"R 2 for the force acting on a cylinder whirling on a circular orbit with constant radius a and angular velocity!. y z o z " plane U!

43 PROBLEM Simplified Cascade Flow Model In the simplest 2D approximation, the flow through an axial cascade can be represented by an infinite sequence of identical vortices! equally spaced at a distance s along the imaginary axis in a uniform stream of velocity U o and angle! o w.r.t. the positive real axis. s U 1 " 1 y i2 i 0 # # # z! plane U 2 " 2 Using the SCT show that the complex velocity # x potential for such flow is: f z = U o e!i" o z! i # 2 z log sinh s *!i Determine the upstream and downstream flow velocities U 1, U 2 and angles! 1,! 2 at!!. What is the force acting on a blade i.e. vortex? Check that the result is consistent with the application of the momentum equation to a control volume bounded by two vertical planes in the upstream and downstream far fields.

44 AXISYMMETRIC IDEAL FLOWS Governing Equations In spherical coordinates r,!,! : dx = e r dr + e! r d! + e " r sin! d" with:! = 0 axial symmetry!" u =!" u =! " e # r sin irrotationality incompressibility obtain: u =!"!r = 1! r 2 sin#!#! "u = 1 # ur 2 + r 2 #r 1 r sin v = 1 r!"!# = 1! r sin#!r # vsin # = 0! and w = Also notice that: d! = "! "! dr + d# = v r sin# dr + u r sin# "r "# r d# = dq 2 where dq is the volumetric flux through a surface element dr and r d! O! r 1 "# r sin! " = 0 - " "r r "# 2 "r + 1 " "# sin* sin* "* "* = 0 /. / " "+ "r "*, " "+ "* "r = 0 0/ " z

45 AXISYMMETRIC IDEAL FLOWS Separable Solutions Looking for separable solutions! = R r" # of:! 2 " = 1 # r 2 #r r #" 2 1 # #" #r + r 2 sin* #* #* sin* = 0 with separation constant n n + 1, obtain: R = 0 r 2 R!! + 2r R! " n n + 1 #!! + #! cot + n n + 1# = 0 Substituting R = r! where:! 2 +! " n n + 1 1! x 2 X ""! 2x " P n x : and x = cos! " X x = #! x = 0! R = A n r n + B n r "n"1 X + n n + 1 X = 0! " = C n P n cos# + D n Q n cos# and Q n x are the Legendre functions of the 1st and 2nd kind diverges! even integer for single-valued regular solutions in 0! "! # C n and D n are constants with D n = 0 for physical solutions as Q n ±1 n n + 1 and therefore: " 1! = A n r n * + B # n P n cos where P n x n=0 r n+1 is the Legendre polynomial of degree n O = 1 2 n n!! r d n x 2! 1 n dx n " z

46 AXISYMMETRIC IDEAL FLOWS Uniform Flow Setting:! n = 1 " = Ur cos velocity potential # A 1 = U r and differentiating: u =!" = U cos# radial velocity!r v = 1!" = U sin# azimuthal velocity r!# O! which clearly represents a uniform flow with velocity U in the axial direction " z U Substituting: d! = "vr sin# dr + ur 2 sin# d# = Ur sin 2 # dr + Ur 2 sin# cos# d# where:!!" Ur sin2 " =!!r Ur 2 sin" cos" Thus, integrating with! = 0 on the negative z -axis! = " :! = 1 2 Ur 2 sin 2 " stream function = 2Ur sin" cos"! d" # perfect differential

47 AXISYMMETRIC IDEAL FLOWS Source-Sink Flow Setting: # n = 0 =! Q velocity potential B 0 =! Q 4" 4"r r and differentiating:! u =!"!r = Q Q radial velocity 4#r 2 v = 1!" = 0 azimuthal velocity r! which clearly represents a source-sink flow with volumetric flux:!" = r 2 d# Q ds!u S Substituting:!" = Q S 4r 2 d! = "vr sin# dr + ur 2 sin# d# = Q sin# d# perfect differential 4 Thus, integrating with! " = 0 on the negative z -axis! = " :! = " Q 1+ cos stream function 4# " u z

48 AXISYMMETRIC IDEAL FLOWS Source-Sink Doublet Flow The velocity potential of a source-sink doublet:! = µ 4"r cos# corresponding to " n = 1 # 2 B 1 = µ 4! is obtained by superposing two singularities ±Q at z! a :! = Q # Q = Q r 1 # r 2 4"r 2 4"r 1 4" r 1 r 2 and letting: # a! 0 2aQ = µ " finite # r r! 2acos 1 2 r 1 r 2! r 2 Hence the radial and meridional velocity components are: u =!"!r = # µ 2r cos ; v = 1!" 3 r!# = µ 4r sin# 3 Substituting:! µ 4*r 2 cos d! = "vr sin# dr + ur 2 sin# d# = µ 4r 2 sin2 # dr " µ sin# cos# d# 2r again a perfect differential, and integrating with! "! = " µ 4#r sin2 stream function = 0 on the negative z -axis! = " :

49 AXISYMMETRIC IDEAL FLOWS Flow around a Sphere By superposing a uniform stream and a source-sink doublet:! = Ur cos" + µ 4#r 2 cos"! = 1 2 Ur 2 sin 2 " # µ 4r sin2 " Thus, without loss of generality:! = 0 " constant! 1 2 Ur 2 " µ 4#r sin 2 * = 0 and therefore the streamsurface! = 0 is determined by the conditions: r = " # µ 2!U 1 3! the sphere of radius a =! = 0, "! the z -axis " # µ 2!U 1 3 The corresponding radial and meridional velocity components are: + u =!" a3 = U 1#!r r 3 cos* radial velocity -, - v = 1!" a3 = #U 1+ r!* 2r 3 sin* azimuthal velocity -. µ = 2!Ua 3

50 PROBLEM Flow near a Blunt Nose Show that the superposition of: a uniform stream with: velocity U far field pressure p! a source Q at the origin represents the flow around a semi-infinite Rankine body of radius: R 0 = Q!U with a blunt nose located at a distance: Q 4!U = R 0 2 upstream of the source location. By suitable mass and momentum balances show that the fluid force on the cylinder is: F = Q U p!

51 AXISYMMETRIC IDEAL FLOWS D Alambert Paradox Assumptions: axisymmetric finite body uniform stream U with zero angle of attack velocity perturbation u! due to the presence of the body U + u! F r S " z From the momentum equation for the flow inside the surface S! :! ds! "uu + p1 = F where p = B! 1 "u # u Bernoulli s equation S # 2 Therefore: F =!! ds 1 2 u " u! ds "uu S #! where u = U + u! S # For finite bodies the net source-sink strength is zero B 0 = 0 and: + "! = B n P n cos# r n+1! u " = # " = O 1 r 3 n=1 Therefore, on a surface S! in the far field ds = O r 2 d! and: F! "! ds 1 2 U #U " ds #UU S! = 0 D Alambert paradox S since the integrand is constant

52 AXISYMMETRIC IDEAL FLOWS Source-Sink Forces Assumptions: source-sink singularity Q in a flow with velocity U velocity u! due to the presence of the singularity Recall: F =!!# ds 1 2 u "u! ds "uu S!# S where: # ds = O r 2 d! and, by hypothesis u! = "#! = " Q u = U + u" 4r * + = O 1 r 2 On a vanishingly small sphere centered at Q, to order 1 r 2 : u!u " U!U + 2U! u # and uu! UU + U u " + u" U where U! constant Neglecting noncontributing constants and using the identity a! b! c = a " cb # a " bc : F =!! dsu " u # ds "U u # ds " u # U =! U ds u# S!! ds " u# U S! S and finally: ds! u " = 0 and!# ds! u" = Q! F = "#QU Joukowski Law in 3D S U u! Q F S o z

53 FREE SURFACE WAVES Governing Equations Assumptions: laterally unbounded ideal fluid u =!" depth h and unperturbed velocity U = 0 density!, surface tension S free surface f x,t = 0 constant ambient pressure p a and gravity g =!" gz Field equation for the velocity potential and the pressure:! 2 " = 0 Laplaces equation #" #t + p + 1 2!"!" + gz = B t Bernoullis equation Boundary conditions: Df Dt =! f + "# "f = 0 free surface points remain on the interface!t p p a = S 1 R free surface curvature * R!" = 0 on solid impermeable surfaces of normal n!n on f x,t = 0

54 2D FREE SURFACE WAVES 2D Linearized Equations Assumptions: 1D plane surface waves, wave length! free surface in normal form no breaking waves: = z! " x,t = 0 f x,t small-amplitude waves! << " : both! and! are of perturbation order To the 1st perturbation order:!f = " # # x e x + e z * + e z ; Therefore, subtracting the unperturbed flow:!"!t + p # + gz = B t and 1 p a! = B 0 and including B t! B 0 in the definition of! R! " 2 # " x "# " x! " 2 # 2 " x 2 # B 0 " = " + B t dt the linearized perturbation equations and BCs are:! 2 #"! x +! 2 #" 2!z = 0! "# 2 "t + " "z = 0 and p! p a =!S " 2 # on z = # " x!#"!t + p p with 2 a + gz = 0 " = 0 on z =!h "z

55 2D FREE SURFACE WAVES Planar Harmonic Waves Assuming a complex separable solution:!" = "e ˆ i kx #t!! = ˆ! i kx "#t ze! from the Laplace s equation obtain: i kx #t!p # p a = ˆpe d 2 ˆ! dz 2 " k 2 ˆ! = 0! ˆ" = Ae kz + Be #kz with A and B integration constants and BCs: i! ˆ " + d ˆ# dz = 0 and ˆp = Sk 2 ˆ! on z =! where!i" ˆ# + ˆp + g ˆ = 0 d ˆ! = 0 on z =!h dz Thus, eliminating ˆp and! ˆ and evaluating the free surface conditions at z = 0 : " S k 2 #! + g d ˆ dz * # 2 ˆ = 0 at z = 0! k 2 S " + g k A B d ˆ! dz = 0 at z =!h! k Ae"kh " Be kh = 0 * 2 A + B = 0 which represents a linear homogeneous problem for the integration constants A and B

56 2D FREE SURFACE WAVES Dispersion Relation and Wave Speed For nontrivial solutions for A and B: k 2 S! + gk " # 2 " k 2 S! + gk " # 2 = 0 ke "kh "ke kh Thus the dispersion relation for the wave speed is: a =! k = ± # 1 + k S 2 "g Special cases: k 2 g k tanhkh S!g >> 1! a = ± S k tanh kh capillary waves " a 2 gh capillary waves deep-fluid waves shallow-fluid waves 1 kh S = 0! a = ± g k tanhkh waves in finite-depth fluids kh = 2!h " >> 1! a = ± g waves in deep fluids k kh = 2!h << 1! a = ± gh waves in shallow fluids " Notice: a = a! " dispersive except for waves in shallow fluids

57 Velocity Potential From earlier results: 2D FREE SURFACE WAVES Be kh = Ae!kh! ˆ" = Ae kz + Be #kz = 2Ae #kh coshk z + h i! " ˆ + d ˆ# dz = 0 on z = 0! 2Ae"kh = " i# ˆ k sinhkh and, assuming! ˆ > 0 wave amplitude: i kx "#t!! = ˆ!e = "ia ˆ coshk z + h i e kx "#t sinh kh Ideal Flows If x! is the displacement of a fluid particle of unpertubed position x, in complex notations: d!! x " #! dt r = x! x!" = # ˆ + cosh k z + h i e x e kx t 2 * sinhk z + h i + e z e kx t., - sinh kh sinh kh / 0 Thus the particle oscillates in the x and z directions with amplitudes:!x = " ˆ cosh k z + h ˆ " tanhkh for z = 0 = with 90 phase delay w.r.t.! sinh kh " ˆ sinh kh for z = #h!z=" ˆ sinh k z + h ˆ " for z = 0 = in phase with! sinh kh 0 for z = #h Problem Work out the solution for surface waves in a fluid of infinite depth 2ˆ! 2! ˆ tanh kh x " # x 2 ˆ! sinhkh z = 0 h

58 Internal Waves between Two Fluids Assumptions: INTERNAL WAVES two semi-infinite, ideal, irrotational fluids densities! 1,! 2 and velocities U 1, U 2 constant gravity g =!" gz waves, length!, small-amplitude! << " free interface f x,t = 0 Field equations for the velocity potentials in the two fluids and the pressure:! 2 " 1,2 = 0 Laplaces equation #" 1,2 + p 1,2 + 1 #t 1,2 2!"!" + gz = B t 1,2 1,2 1,2 Bernoullis equation Boundary conditions: Df Dt =! f!t + "# 1,2 "f = 0 free surface points remain on the interface p 1 = p 2 negligible surface tension! 1,2 " finite as z! ±" unperturbed far fields on f x,t = 0

59 Linear Internal Waves between Two Fluids Additional assumptions: small-amplitude plane waves! << " 2D INTERNAL WAVES free surface in normal form no breaking waves: = z! " x,t = 0 f x,t linear perturbation approach both! and! are of perturbation order To the 1st perturbation order:!f = " # # x e x + e z * + e z and!" 1,2!t Thus, subtracting the unperturbed flow and including B t dt " 1,2! = " 1,2 + # B 1,2 t with u 1,2 = U 1,2 +! " " 1,2 + p 1,2! + U 1,2 # 1,2! x + gz = B t 1,2 # 1,2 in the perturbation potential: the linearized perturbation equations and BCs are:! 2 # 1,2 " +! 2 # 1,2 " p 1 = p 2 = 0! x 2!z 2 + Df!# 1,2 " + p" and 1,2!#" + U 1,2 1,2!t 1,2! x + gz = 0 Dt! " # #t " U # 1,2 # x + # 1,2 * #z = 0 +, + 1,2 finite as z ± on z =!

60 Linear Internal Waves between Two Fluids Assuming a complex separable solution:!! 1,2 = ˆ! i kx "#t 1,2 ze! from the Laplace s equation obtain: d 2 ˆ!1,2 2D INTERNAL WAVES!" = "e ˆ i kx #t i kx #t!p 1,2 = ˆp 1,2 e dz 2 " k 2 ˆ!1,2 = 0! ˆ" 1,2 = A 1,2 e kz + B 1,2 e #kz with BCs: Df Dt = 0! i" # ˆ iku ˆ 1# k ˆ1 = 0 i" # ˆ iku ˆ 2# + k ˆ2 = 0 ˆp 1 = ˆp 2! * 1 i" ˆ 1 iku 1 ˆ1 g ˆ #! = * 2 i" ˆ 2 iku 2 ˆ2 g# ˆ Eliminating ˆ! 1 and ˆ! 2 with the first two equations: ˆ" 1 # e kz 0 as z + * ˆ" 2 # e kz 0 as z on z =! 0 *! 1 "# " ku 1 # k " U -, 1 " g/ +. "! * 2 # " ku 2 # k " U -3 1, 2 " g/ 46 ˆ = Thus, for nontrivial solutions! ˆ = 0, the dispersion relation for the wave speed is: a =! k = " 1U 1 + " 2 U 2 ± " 2 # " 1 g " 1 + " 2 " 1 + " 2 k # " 1 " 2 " 1 + " 2 U 2 1 # U 2 2

61 2D INTERNAL WAVES Instabilities of the Interface between Two Fluids Recall:!! =!e ˆ ik x "at where: a =! k = " 1 U 1 + " 2 U 2 " 1 + " 2 ± is generally complex " 2 # " 1 " 1 + " 2 g k # " U 1 # U 2 1" 2 " 1 + " 2 2 Special cases:! 2 >! 1 and U 1 = U 2 = 0 stationary ocean, no wind: a = ±! 2 "! 1! 1 +! 2 g k # ± g k for! 1 " 0! 2 <! 1 and U 1 = U 2 = 0 inversely stratified fluids: a = ±i! 1 "! 2 g! 1 +! 2 k Taylor instability! 2 =! 1 and U 1! U 2 shear layer, no density gradient: stable, as expected a = U 1 + U 2 2 ± i U 1! U 2 2 Rayleigh instability

62 PROBLEM Boundary Layer Stability Assumptions: infinite flat wall laterally unbounded boundary layer: thickness! zero velocity semi-infinite free stream: uniform velocity U ideal, irrotationa fluid, density! Under the above assumptions: write the appropriate governing equations and boundary conditions formulate the linearized problem for small-amplitude harmonic flow perturbations formulate the eigenvalue problem for the flow oscillations derive the dispersion relation for the propagation speed of the flow oscillations is the flow unstable? if so, determine the wavelength of the most unstable instability