0 = p. 2 x + 2 w. z +ν w

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Eugene Foster
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1 Solution (Elliptical pipe flow (a Using the Navier Stokes equations in three dimensional cartesian coordinates, given that u =, v = and w = w(x,y only, and assuming no body force, we are left with = p x, = p y, = p ( z +ν w x + w y where note that u w = as w is independent of z. Since we are given p = Gz, the first two equations are consistent and substituting this form for p into the final equation gives the required result w x + w y = G ν., (b Substituting w = Ax + By + C in to the partial differential equation in part (a, we immediately deduce A+B = G ν. (c Using the no-slip boundary condition, i.e. that w = on x a + y b = y = b ( x a, generates the equation Ax +Bb ( x a +C = (A B b a x +Bb +C = which must hold for all x [ a,a]. Hence equating coefficients of x and x, we arrive at the following three equations (including the result from part (b above for the three unknowns A, B and C: A B b a =, Bb +C = and A+B = G ν. Solving the first equation for A in terms of B and substituing this into the third equation we find B (+ b a = G B = G ν ν Substituting this into the first equation reveals a a +b. A+ G ν a a +b b a = A = G ν b a +b. Finally substituting these two answers for A and B into the second equation = G ν a a +b b +C = C = G ν a b a +b.

2 (d For a small patch of area ds of an elliptical cross section of the pipe, the volume of fluid passing through ds per unit time is equivalent to the volume of the cylinder of cross sectional area ds and length w (the orthogonal flow rate through ds, i.e. w ds. Summing over all such small patches of areas to make up the complete cross section generates the integral wds over the whole elliptical cross sectional area. Computing the integral using the substitutions x = arcosθ and y = brsinθ we see: wds = Ax +By +Cdxdy = π = G ν a 3 b 3 a +b = πga3 b 3 ν(a +b = πa3 b 3 G 4ν(a +b. ( Aa r cos θ +Bb r sin θ +C abrdrdθ π r r 3 dr ( r rdrdθ

3 3 Solution (Bernoulli s Theorem and Venturi tube (a (i Since the flow is incompressible and homogeneous, ρ is uniform and constant. The flow is also irrotational so that u = ϕ. Euler s equations = Using the given identity we get ( ϕ t t ( ϕ+u u = ( p ρ V. + ( u u ( u = ( ϕ ( p V ρ t + u + p ρ +V = u ( u H = u H = for the quantity H given in the question. (ii Hence H = f(t for some function f = f(t. Using the suggested redefined potential for u given by Φ = ϕ and substituting this into we get which gives the result. t f(τdτ ϕ = Φ+ H t = f (t t f(τdτ ( ϕ t t + u + p ρ +V = f (t ( Φ t t +f(t+ u + p ρ +V = f (t ( Φ t t + u + p ρ +V = (b (i Fluid incompressible, so volume flux (= cross-sectional area uniform velocity through wide region equals that in narrow Since A < A = A u = A u. u = A A u > u. (ii Part (a = Bernoulli quantity H same in wide and narrow regions on same streamline passing through tube u + p ρ = u + p ρ. (Assume potential difference along any streamline negligible.

4 4 (iii Bernoulli result p p = ρ(u u > since u > u. Pressure in narrow region less! (iv Examples are: device for measuring flow speeds (measure p p, A /A known = u, u ; carburetor, pressure drop draws in another fluid from side channel; or lift of an aircraft wing.

5 5 Solution (Viscous drag on sphere (a Assume the set-up given: stationary Stokes flow around solid sphere with no swirl and axisymmetric with respect to the line through the sphere centre aligned with the uniform far-field flow of speed U. The stream function ψ is given by ψ u r = r and u θ = ψ sinθ θ rsinθ r. First let us consider the boundary conditions as r. Decomposing the far-field axial directed velocity field of speed U into components along ˆr and ˆθ we get U cosθ and U sinθ, respectively. Hence in the far-field limit we have ψ r sinθ θ U cosθ and ψ rsinθ r U sinθ. The solution to this pair of first order partial differential equations is ψ Ur sin θ, generating the far-field boundary condition in terms of ψ. Second consider the boundary conditions on the surface of the sphere. The no-slip condition on r = a = ψ a sinθ θ = and ψ asinθ r =. Hence ψ is independent of r and θ along the boundary r = a; we can therefore take ψ = and ψ/ r = to be the boundary conditions on r = a. (b Take as given that for the stationary Stokes flow under consideration D (D ψ =, where D is the second order partial differential operator D := r + sinθ ( r. θ sinθ θ Look for a solution of the form ψ = Uf(rsin θ. First compute D ψ = D ψ = U (f (r r f(r sin θ, in which we set Now compute D (D ψ = = F(r := f (r r f(r. D (D ψ = D (UF(rsin θ = U Hence D (D ψ = if and only if (F (r r F(r sin θ. F (r rf(r =. This is a linear second order ordinary differential equation whose two independent solutions are r and /r; thus F(r is a linear combination of these two solutions. However we require f(r which satisfies f (r rf(r = F(r.

6 6 This is a non-homogeneous linear second order ordinary differential equation, again whose two independent homogeneous solutions are r and /r; while the particular integral for the non-homogeneous component F(r, which is a linear combination of r and /r, has the form Ar 4 +D/r for some constants A and D. Hence f = f(r necessarily has the form f(r = Ar 4 +Br +Cr + D r, where B and C are two further constants. (c Use the boundary conditions from part (a. First we see that as r we must have ( U Ar 4 +Br +Cr + D sin θ = r Ur sin θ so that A = and B =. Second we see that on r = a we must have ( a U +Ca+ D sin θ = and U (a+c D a a sin θ =. Hence we get two simultaneous equations for C and D, namely a /+Ca+D/a = and a+c D/a =, whose solution is C = 3a/4 and D = a 3 /4. We thus have ( r ψ = 4 Ua a 3r a + a sin θ. r (d Using the stationary Stokes equations p = µ u we first compute u, i.e. the components u r and u θ from the stream function ψ. Using the relations given ( u r = 4 Ua a 3 ar + a r 3 cosθ and u θ = Ua ( 4r 4 r a 3 a a r sinθ. Then second, using the hint given, compute ω = u and then ω. Note ω has only one non-zero component, namely ω ϕ = ( ( ruθ ur r r r θ = 3Ua r sinθ, with a lot of terms cancelling. Then we have ω = 3Ua cosθ sinθ. r 3 Thus to find the pressure p we must solve the pair of first order partial differential equations: = 3Uaµ ( cosθ r 3 sinθ ( p r p r θ These give p = p 3Uaµ r cosθ. (e We are given that the total force on the sphere is π F = πa ( pˆr +µdrrˆr +µd rθˆθ sinθdθ,

7 7 The axial component of the force F ax (i.e. along the direction of the far-field flow, since the components of ˆr and ˆθ in the axial direction are given by ˆrcosθ and ˆθsinθ, respectively, is thus given by π F ax = πa ( ( p+µdrr cosθ µd rθ sinθ sinθdθ. From the formulae sheet we find D rr = u ( r and D rθ = uθ r r u θ r + u r. r θ The no-slip boundary conditions on the sphere surface r = a = u r = u θ =. Further from our expressions for u r and u θ in part (d above we see that u r / r = u r / θ = and u θ / r = (3U/asinθ on r = a. Substituting these into the expressions for the total force above we get π ( 3Uµ (cos F ax = πa θsinθ +sin 3 θ dθ a = 3πUµa = 6πUµa. π sinθdθ

8 8 Solution (Shallow layer equations with applied surface stress (a (i Assume x and y have typical scale L and z has typical scale H L. Further suppose u and v have the same typical scale U while w has typical scale W. Then the continuity equation (incompressibility condition reveals the scalings u x U/L + v y U/L + w =. z W/H Hence we deduce W = O ( U(H/L. Since H L = W U. (ii Note that the time scale T = O(L/U. Substituting for the modified pressure P := p + ρgz into the Navier Stokes equations and examining the scalings of the individual terms reveals: t u+u x u+v y u+w z u U /L t v +u x v +v y v +w z v U /L = ν( xx u+ yy u νu/l = ν( xx v + yy v νu/l + zzu νu/h x P/ρ, + zzv νu/h y P/ρ, for the first and second components. In the first equation, comparing the viscous terms xx u and yy u to zz u we see that the ratio of their scalings is νu/l νu/h = H L. Thus we omit the xx u and yy u terms. Now compare the inertia terms (all those on the left-hand side to the viscosity term ν zz u, the ratio of their scalings is U /L νu/h = UH νl = UH ν H L, and hence we omit all the inertia terms under the assumption that the modified Reynolds number R m = UH/ν is small or order one. Finally to keep the pressure term it must have the scaling x P/ρ = O ( νu/h = P = O ( µul/h. An exactly analogous scaling argument applies to the second equation above and thus we see that the final equations for these two components are (with µ = ρν P x = u µ z, P y = v µ z. Now conisder the third equation, the scalings of the individual terms reveals: t w+u x w+v y w+w z w = ν( xxw + yyw UW/L=U H/L νw/l =νuh/l 3 + zzw νw/h =νu/lh z P/ρ. νul/h 3

9 9 The largest scaling term appears to be the pressure term so we compare the other terms to that. The ratio of the inertia terms to the pressure term in terms of their scalings is U H/L νul/h 3 = UH4 νl 3 = UH ν ( H L 3, again under the assumption the modified Reynolds number R m = UH/ν is small or order one. The ratio of the viscous terms ν xx w and ν yy w to the pressure term in terms of their scalings is νuh/l 3 ( H 4 νul/h 3 =. L Finally the ratio of the viscous term ν zz w to the pressure term is νu/lh ( H νul/h 3 =. L Hence the third equation reduces to P/ z =, completing the shallow layer equations. (b Note from the shallow layer equations that the third equation = P z = p = ρg p = ρg(h z P = ρgh. z where we assume that p = at z = h(x,y,t (the pressure is constant on the free surface and we take it to be zero there. Substituting this expression for the pressure into the first two shallow layer equations = and similarly µ u z = ρg h x µ u z = ρg h Γ (z h+ x x µu = ρg h z(z h+z Γ x x, µv = ρg h z(z h+z Γ y y, where in the first integration from z to h, we used the given conditions for the applied surfaces stresses at z = h(x,y,t and in the second integration from to z, we used the no-slip boundary conditions at z =. Let us now restrict ourselves to the surface z = h(x,y,t. The acceleration of any particle at the surface is given by h t +u(x,y,h,t h x +v(x,y,h,t h y.

10 From the incompressibility condition we have h t +u(x,y,h,t h x +v(x,y,h,t h y = h = x ( u x + v dz y h udz y +u(x,y,h,t h x +v(x,y,h,t h y. Cancelling like terms on each side and subsituting our expressions for u and v above into this last expression reveals that µ h t = x ( 3 ρgh3 h x h Γ x + y h vdz ( 3 ρgh3 h y h Γ. y Simplifying the terms under the partial derivarives on the right-hand side reveals µ h t + ( h ( Γ x x 3 ρgh + ( h Γ y y( 3 ρgh =.

11 Solution (Navier Stokes regularity in three dimensions (a (i To show that the Navier Stokes equations can be expressed in the form u t +ω u = ν u ( p+ u +f, we simply substitute the identity u u = ( u u ( u into the standard formulation of the Navier Stokes equations. (ii Directly compute (d/dt u L as follows: d dt u L = t (u udx = u ( t udx = u ( ω u+ν u ( p+ u +fdx = ν u dx+ u f dx. In the last step we have used that u (ω u, and that u ( ( p+ u dx = u ( p+ u dx ( u ( p+ u dx and u ( udx = (( u u dx u dx. Note that in these last two integral identites we simply use the product formulae for the vector fields indicated and thus these are also just integration by parts formulae. The divergence terms on the right, by the divergence theorem, generate surface integrals on, the bounding surface of. Since u = on due to the viscous boundary conditions assumed, these terms are zero. Of course, we are also assuming incompressibility u = everywhere. Further note, to be clear, u dx = ( xj u i ( xj u i dx. i,j= Now using the Hölder and Young inequalities we see that u,f L u L f L δ u L + δ f L, for some δ >. Combining these results gives us the estimate d dt u L +ν u L δ u L + δ f L. (iii To derive a uniform upper bound for u L in time, we can set δ = ν and use Poincaré s inequality u L c u L

12 for some constant c (note that if u(x, has mean zero then u(x,t also has mean zero, for t >. Using this in the evolution inequality above, we get d dt u L + ν c u L ν f L. Now we can integrate this time differential inequality (more rigorously we apply the Gronwall lemma to get: u(,t L u(, L exp( νt/c ( c + f(,t ν L ( exp ( νt/c. This establishes that u L is uniformly bounded in time. Now note if we simply time integrate the time differential inequality (before we applied the Poincaré inequality we get u(,t L + ν T T c u(,τ L dτ ν u(,τ L dτ + ν From this we can deduce for any time T >, we have T u(,τ L dτ <. T f(,τ L dτ. (b (i We consider the L -inner product of u with the Navier Stokes formulation quoted above, this generates: d dt u L +ν u L = u,ω u L u,f L. Note that for the time derivative term, we have used the product formula again (integration by parts and that u = on the boundary so that t u = on the boundary as well. Now consider the term generated from the nonlinear term in the Navier Stokes equations. Since u ( u for divergence free fields u we see that u,ω u L = ω,ω u L = ω, (ω u L = ω,u ω L + ω,ω u L = ω,dω L, where D is the deformation matrix (the symmetric part of u. In this sequence of equalities, we have used the product formula ω,v L = (ω v dx ω, v L with v = ω u together with the divergence theorem and that u = on the boundary. We have also used that for divergence free fields u, with ω = u, we have (ω u = u ω +ω u, and that ω u = Dω as Rω (where R is the antisymmetric part of u and that ω,u ω L = ( u ω dx =.

13 3 Putting all this together, we arrive at the equality: d dt u L +ν u L = ω,dω L u,f L. (ii Assume d = 3. The vorticity stretching term Dω is now an important mechanism of the flow. Using the Hölder inequality, followed by the Sobolev Gagliardo Nirenberg inequality u L 4 c u 3/4 L u /4 L, and then the Young inequality, we see that ω,dω L c u L u L ( 4 3/4 c u L u L u /4 L = c u 3/ L u 3/ L 3 4 ν u L + c ( u 3. 4ν 3 L Inserting this into the equality we have above and using similar estimates for the forcing term we see for some constant c, we have d dt u L + ν 8 u L c ( u 3 ν 3 L + ν f L. For simplicity assume f. Then using the Gronwall lemma we see, as a function of time, the best we can show is u L has an upper bound that blows up in finite time.

F11AE1 1. C = ρν r r. r u z r

F11AE1 1. C = ρν r r. r u z r F11AE1 1 Question 1 20 Marks) Consider an infinite horizontal pipe with circular cross-section of radius a, whose centre line is aligned along the z-axis; see Figure 1. Assume no-slip boundary conditions