Introductory fluid mechanics: solutions

Transcription

1 Detailed solutions Introductory fluid mechanics: solutions Simon J.A. Malham Simon J.A. Malham 22nd October 214 Maxwell Institute for Mathematical Sciences and School of Mathematical and Computer Sciences Heriot-Watt University, Edinburgh EH14 4AS, UK Tel.: Fax:

2 2 Simon J.A. Malham Solution Trajectories and streamlines: expanding jet By definition the equations for the particle trajectories are dx dt = xe2t z, dy dt = ye2t z, and dz dt = 2e2t z. We solve the third equation first. Separating variables we see it is equivalent to z t e ζ dζ = 2e 2τ dτ z e z e z = e 2t 1 e z = e z 1 + e 2t. We substitute this expression for e z into the second trajectory equation, this gives Similarly we can show that dy dt = y e 2t e z 1 + e 2t y t dη y η = e 2τ dτ e z 1 + e2τ y log = 1 e 2t de 2τ y 2 1 e z 1 + e 2τ y log = 1 e z y 2 log 1 + e 2t e z y = y e z 1 + e 2t e z 1/2. x = x e z 1 + e 2t e z 1/2. If x = 1, y = 1 and z =, then the parametric equations for the particle trajectory are x = e t, y = e t and z = 2t. To find the streamlines, we fix t, and solve the system of equations dx ds = e2t xe z, dy ds = e2t ye z, and dz ds = 2e2t e z. Solving the third equation we get e z = e z + 2e 2t s. Substituting this into the second streamline equation implies dy ds = y s dη y η = e2t y log y e 2t e z + 2e 2t s y 1 e z + 2e 2t σ dσ = 1 e z 2 log + 2e 2t s e z y = y 1 + 2e 2t z s 1/2. Similarly we find x = x 1 + 2e 2t z s 1/2.

3 Introductory fluid mechanics: solutions 3 Solution Trajectories and streamlines: three dimensions The velocity field ux, t = u, v, w T is given for t > 1 by u = x 1 + t, v = y t and w = z. To find the particle paths or trajectories, we must solve the system of equations dx dt = u, dy dt = v and dz dt = w, and then eliminate the time variable t between them. Hence for particle paths we have dx dt = x 1 + t, dy dt = y t and dz dt = z. Using the method of separation of variables and integrating in time from t to t, in each of the three equations, we get x 1 + t y ln = ln, ln = 2 ln 2 t z x 1 + t y t and ln = t t, z where we have assumed that at time t the particle is at position x, y, z T. Exponentiating the first two equations and solving the last one for t, we get x = 1 + t, x 1 + t y y = t t 2 and t = t + lnz/z. We can use the last equation to eliminate t so the particle path/trajectory through x, y, z T is the curve in three dimensional space given by 1 + t + lnz/z x = x, and y = y 2 t lnz/z t t 2. To find the streamlines, we fix time t. We must then solve the system of equations dx ds = u, dy ds = v and dz ds = w, with t fixed, and then eliminate s between them. Hence for streamlines we have dx ds = x 1 + t, dy ds = y t and dz ds = z. Assuming that we are interested in the streamline that passes through the point x, y, z T, we again use the method of separation of variables and integrate with respect to s from s to s, for each of the three equations. This gives x ln = s s y x 1 + t, ln = s s z y t and ln = s s. z Using the last equation, we can substitute for s s into the first equations. If we then multiply the first equation by 1 + t and the second by t, and use the usual log law ln a b = b ln a, then exponentiation reveals that 1+t 1+ 1 x y 2 t = = z, z x y which are the equations for the streamline through x, y, z T.

4 4 Simon J.A. Malham Solution Streamlines: plane/cylindrical polar coordinates We are concerned with describing the streamlines associated with the velocity field u, v, w T = αt x y, x + y, T, where α = αt is an arbitrary function of t. Using the hint given we change to cylindrical polar coordinates for which x = r cos θ, y = r sin θ and z = z. Hence if the coordinates x = xs, y = ys and z = zs represent a streamline for a given fixed t for a particle starting at x, y, z T, then in cylindrical coordinates the same streamline is represented by r = rs, θ = θs and z = zs. By direct computation we observe that dx ds = dr dθ cos θ r sin θ ds ds, dy ds = dr dθ sin θ + r cos θ ds ds. Multiply the first equation above by r sin θ and the second equation by r cos θ. Adding the resulting two equations reveals that r 2 dθ dx = y ds ds + x dy ds. Now multiply the first equation above by r cos θ and the second equation by r sin θ. Adding these resulting two equations reveals that The equations for the streamlines are dx ds = αt x y, r dr ds = x dx ds + y dy ds. dy ds = αt x + y and Hence zs = z for all s. The equation for r = rs is given by r dr ds = x dx ds + y dy ds = x αt x y + y αt x + y = αt x 2 + y 2 = αtr 2. dz ds =. Hence we see r = rs satisfies r = αtr which implies rs = r exp αts. Similarly the equation for θ = θs is r 2 dθ ds = y dx ds + x dy ds = y αt x y + x αt x + y = αt x 2 + y 2 = αtr 2. Hence we find θ = θs satisfies θ = αt, i.e. θs = θ + αts. Hence for any particle, the θ = θs coordinate increases linearly in time, while the r = rs coordinate exponentially expands or contracts depending on the sign of αt and the z = zs is constant. The streamlines thus look like horizontal exponential spirals.

5 Introductory fluid mechanics: solutions 5 Solution Steady oscillating channel flow The streamlines for this problem by definition are given by dx ds = 1 + x sin y and dy = cos y. ds Using the chain rule, in other words using that dy dx = dy/ds dx/ds or dx dy = dx/ds dy/ds we can solve for the streamlines directly as follows. dx dy = 1 + x sin y cos y dx dy = 1 cos y + sin y cos y x. We observe this is a first order linear differential equation for x = xy, i.e. we are treating y as the independent variable and x as the dependent variable and solution. We can solve first order linear differential equations using the integrating factor technique, where in this case the integrating factor is y exp sin η cos η dη = exp logcos y = cos y. Hence the integrating factor is cos y and we multiply the first order linear differential equation above through by it. Hence we have cos y dx = 1 + sin y x dy d cos y x = 1 dy cos y x = y y + cos y x x = y y + cos y x. cos y This gives the equation for the streamline passing through x, y T. Note that when x = 1 and y = π/2 as well as when x = 1 and y = π/2, one can check by direct inspection that dx/ds = and dy/ds = and thus both points are stagnation points of the flow.

6 6 Simon J.A. Malham Solution Channel shear flow As stated, we consider the flow to be two-dimensional with u =, v = U1 x 2 /a 2, and we ignore the z coordinate as well as the velocity component w which is zero. In the hint we are told that we should assume the flow is incompressible, which in two-dimensions corresponds to the condition u x + v y =. Hence there exists a stream function ψ = ψx, y such that u = ψ y and v = ψ x. This choice is consistent with the incompressibility condition above. To find the stream function we set ψ y = and ψ 1 x = U x2 a 2. This is a pair of partial differential equations for the stream function ψ = ψx, y. The first equation implies ψ = fx for some arbitrary function f = fx i.e. it is a function of x only. Substituting this into the second equation we find f x = U 1 x2 a 2 fx = U x x3 3a 2 + C ψx, y = U x x3 3a 2 + C, where C is an arbitrary constant. The total volume flux across y = y, x a, is +a [ ] x=+a U 1 x2 a a 2 dx = U x x3 3a 2 x= a = 4 3 Ua. The volume flux across y = y, x 4 3 a, is +3a/4 [ U 1 x2 3a/4 a 2 dx = U x x3 3a 2 3 = 2Ua The ratio of the two volume fluxes is = Ua Ua = Ua ] x=+3a/4 x= 3a/4

7 Introductory fluid mechanics: solutions 7 Solution Flow inside and around a disc We are given a two dimensional flow in plane polar coordinates as follows u r = U cos θ 1 a2 r 2 and u θ = U sin θ 1 + a2 r 2 Γ 2πr. Here U, a and Γ are constants. Our goal is to compute the stream function for this flow. First we check the incompressibility condition, which in polar coordinates is 1 r r r ur + 1 r u θ θ =. Since by direct computation we find r r ur = U cos θ r a2 r r = U cos θ 1 + a2 r 2 and u θ θ = U sin θ 1 + a2 θ r 2 = U cos θ 1 + a2 r 2, we see that the incompressibility condition is satisfied. Second, the stream function ψ = ψr, θ is the solution to the pair of partial differential equations 1 ψ ψ = ur and r θ r = u θ. Substituting the given forms for u r and u θ, these partial differential equations are 1 ψ r θ = U cos θ 1 a2 r 2 and Solving the first partial differential equation we find ψ = U sin θ r a2 + F r, r ψ r = U sin θ 1 + a2 r 2 Γ 2πr. where F = F r is an arbitrary function of r. Substituting this expression into the second partial differential equation we find U sin θ r a2 + F r = U sin θ 1 + a2 r r r 2 Γ 2πr U sin θ 1 + a2 r 2 F r = U sin θ 1 + a2 r 2 Γ 2πr F r = Γ 2πr F r = Γ log r + C, 2π where C is an arbitrary constant which we can chose to set to. Hence the stream function is ψ = U sin θ r a2 + Γ log r. r 2π

8 8 Simon J.A. Malham Solution Couette flow a We know that u r = u z =, / t stationary flow, / θ for any of the velocity components the component u θ is independent of θ and f r = f θ = f z =. We are also told ρ 1. We have to show that for the given velocity field, we can indeed find a pressure field such that the incompressible Euler equations hold. Substituting all the above conditions into Euler s equations in cylindrical polar coordinates implies u2 θ r = p r, = 1 r p θ and = p z. Note u = 1 ru r + 1 u θ r r r θ + uz z = trivially, and that u, which simplifies to u θ /r / θ, also acts trivially on the velocity field which independent of θ. The second and third equations in the system above imply p = pr. The first equation implies p r = 1 2 A r r + Br p r = A2 r 3 + 2AB + B 2 r r p = A2 2r 2 + 2AB log r B2 r 2 + K, where K is an arbitrary constant. Hence there is a pressure field for which the Euler equations are satisfied by the given velocity field. b The angular velocity is given in terms of the velocity component u θ = u θ r by u θ /r. To compute the angular velocity at r = R 1 and r = R 2 we first directly compute u θ R 1 and u θ R 2 as follows. First we have, using the formulae for A and B given, Second, similarly we have u θ R 1 = A R 1 + BR 1 = R 1R 2 2ω 2 ω 1 R 2 2 R2 1 = R 1R 2 2ω 1 R 3 1ω 1 R 2 2 R2 1 = R 1 ω 1. R2 1ω 1 R 2 2ω 2 R 2 2 R2 1 R 1 u θ R 2 = A R 2 + BR 2 = R2 1R 2 ω 2 ω 1 R 2 2 R2 1 = R 2 ω 2. R2 1ω 1 R 2 2ω 2 R 2 2 R2 1 Hence we deduce that u θ R 1 /R 1 = ω 1 and u θ R 2 /R 2 = ω 2. R 2 c Directly computing the vorticity we get ω = u =,, 1 r r ru T θ =,, 2B T.

9 Introductory fluid mechanics: solutions 9 Solution Venturi tube a The fluid is incompressible, so the volume flux = cross-sectional area uniform velocity through wide region equals that in narrow, i.e. we have A 1 u 1 = A 2 u 2. Since A 2 < A 1 this implies u 2 = A 1 A 2 u 1 > u 1. b Bernoulli s Theorem implies the quantity H is the same in wide and narrow regions on same streamline passing through the tube, i.e. we have 1 2 u2 1 + p 1 ρ = 1 2 u2 2 + p 2 ρ. We assume the potential difference along any streamline is negligible. c The Bernoulli result above holds if and only if p 1 p 2 = 1 2 ρu2 2 u 2 1. The quantity on the right is positive since u 2 > u 1. Hence the pressure in the narrow region is less! d Examples are: a device for measuring flow speeds measure p 1 p 2, A 1 /A 2 known = u 1, u 2 ; a carburetor the pressure drop draws in another fluid from a side channel; or the lift of an aircraft wing.

10 1 Simon J.A. Malham Solution Clepsydra or water clock a Use Bernoulli s Theorem for a streamline starting at the free surface and going through the orifice. The potential energy of any fluid particle at the orifice outlet compared to that at the free surface is ρg. This we have 1 dz 2 P 2 + dt ρ = 1 2 U 2 + P ρ ρg 1 dz 2 2 = 1 dt 2 U 2 ρg. Note that inside the container at the level at which water is pushed through the orifice, the pressure is P = P + ρgz. The difference P P accelerates the water through the orifice. b Using incompressibility, we equate the volume flux through the orifice to the rate of volume drop at the free surface as it descends. We know S is the cross-sectional area of the orifice and U is the velocity of the fluid through the orifice. The volume flux through the orifice is equal to the cross section area S times the fluid velocity and is thus SU. Similarly the cross-sectional area of the free surface at height z is say A, while its velocity is dz/dt. Hence the drop in volume per unit time at the free surface i.e. the volume flux at the free surface is A dz/dt. Hence incompressibility implies SU = A dz dt. c Rearranging the result from part b we see that 1 dz U dt = S A. Thus using the result from part a we find that gz = 1 2 U 2 dz 2 dt = 2 1 U 2 1 dz 2 1 U dt = 2 1 U 2 S 2 1. A We assume S A so that S/A 1. Hence with an error of order S/A 2 we have U 2gz. d We want dz/dt to be constant, so combining the results of parts b and c, and using that A = πr 2 as the Clepsydra is a volume of revolution, implies as required. S 2gz = πr 2 dz dt π dz 2r 4 z =, 2g S dt

11 Introductory fluid mechanics: solutions 11 Solution Coffee mug a We are given u r = u z =, / t, / θ and f r = f θ = with f z = g. Substituting these into Euler s equations in cylindrical polar coordinates implies Note we have u = 1 r u2 θ r = 1 p ρ r, = 1 p ρr θ, = 1 p ρ z g. ru r r + 1 r u θ θ + uz z = trivially. And u = u θ /r / θ thus also acts trivially on the velocity field which is independent of θ. The second equation in the system implies p = pr, z. b Using that u θ = Ωr this is the rigid body rotation ansatz implies p r = ρω2 r pr, z = 1 2 ρω2 r 2 + Gz, where G an arbitrary function. Substitute this into the other equation 1 p ρ z = g G z = ρg. Hence Gz = ρgz + C where C is an arbitrary constant. This implies we have At the free surface p = P : pr, z = 1 2 ρω2 r 2 ρgz + C. P = 1 2 ρω2 r 2 ρgz + C z = Ω 2 /2g r 2 C P /ρg. c Choose C = P z = Ω 2 /2g r 2. Relative to the chosen origin O, the bottom of the mug is at z = z. i Initially the coffee depth is d. This implies the total volume is πa 2 d. The incompressibility volume constraint implies a πa 2 Ω 2 r 2 z + 2πr dr = πa 2 d. 2g The term πa 2 z represents the cylindrical volume of coffee lying under the origin O. The integral term represents the volume of coffee above the origin O but under the free surface. ii The constraint in part i is equivalent to the condition z + Ω2 a 2 4g = d.

12 12 Simon J.A. Malham The coffee at the edge of the mug is at height z = z + Ω2 a 2 2g = d Ω2 a 2 4g Spillage occurs when this exceeds h, i.e. when + Ω2 a 2 2g = d + Ω2 a 2 4g. z h Ω 2 4gh d a 2. If the mug is initially less than half full then d < 1 2 h. Spillage occurs when Ω 2 Thus we observe that 4gh d a 2 Ω2 a 2 h d Ω2 a 2 h d. 4g 4g z = d Ω2 a 2 d h d = 2d h <. 4g In other words the centre part of the free surface the dip in the middle will have hit the bottom of the mug before spillage.

13 Introductory fluid mechanics: solutions 13 Solution Channel flow: Froude number Recall the general theory from the notes. We saw there that in the case when the actual maximum value of the river bed undulation y max does not reach the maximum possible bound y, y max < y, that when the Froude number F was such that while when F < 1 F > 1 then then d h + y <, dy d h + y >. dy In the case when y max = y, these last two properties still hold upstream until y reaches it maximum y max = y all the arguments in the text for the case y max < y apply until y actually reaches its maximum. Recall the upper bound y occurred when h = h where h := UH2/3 g 1/3. Recall that due to incompressibility UH = uh and in particular at the maximum y = y we have UH = uh. Hence we see that h = uh 2/3 g 1/3 h 3 = uh 2 g h = u2 g at y = y. Since we have see the general theory in the notes 1 dh 1 dy = u2, gh we see that as y y then dh/dy. u2 gh = 1 We know dy/dx = at y max = y so what happens to the free surface at this point, and in particular, what is dh/dx? To answer this we need to expand locally and look at higher order terms. Suppose that x = at the maximum value for y = yx so that y = y max = y. We expand y = yx and h = hx about x = as follows: y = y Kx 2 +, h = h 1 + a1 x + a 2 x 2 + where we will ignore terms of degree x 3 and higher. The quantities K, a 1 and a 2 are all constants. Naturally since x = is a local maximum for y = yx we know K >. We substitute these expansions into the Bernoulli constraint again see the general theory in the notes y = U 2 2g + H h UH2 2gh 2. Note to expand h 2 we use the Binomial expansion with X := a 1 x + a 2 x 2 + so h 2 = h a1 x + a 2 x = h X = h X X 2 + 2! = h 2 1 2a1 x + a 2 x a 1 x + a 2 x = h 2 1 2a1 x + 3a 2 1 2a 2 x 2 +.

14 14 Simon J.A. Malham F<1 sub a > F>1 1 sub a > 1 sub a 1 < super a < a 1> super super a < 1 1 y y x= x x= x Fig. 1 The cases for the two incident upstream Froude numbers are shown. The flow is subcritical if a 1 x > and supercritical if a 1 x <. For a real flow friction will force the supercritical solution upstream in each case. Thus for example, in the case F > 1 upstream, there will be a discontinuity in the gradient of h = hx at x =. Substituting this into the Bernoulli constraint we find y Kx 2 + = U 2 2g +H h 1+a1 x+a 2 x 2 UH a1 2gh 2 x+3a 2 1 2a 2 x 2 +. Thus equation coefficients of powers of x we have x : y = U 2 x 1 : x 2 : 2g + H h UH2 2gh 2 = h a 1 UH2 2gh 2 2a 1, K = h a 2 UH2 2gh 2 3a 2 1 2a 2. Note that the relation between the coefficients at x simply reproduces the definition for the maximum value y. Then for the relation between the coefficients at x 1, since UH = uh and u 2 = gh we see that the expression on the right is a 1 h UH2 gh 2 = a 1 h u2 h 2 gh 2, = a 1 h u2 =, g generating no new information. The relation between the coefficients at x implies K = h a 2 + u2 h 2 2gh 2 3a 2 1 2a 2 K = h a 2 + h 2 3a2 1 2a 2 2K 1/2 a 1 = ±. 3h Around x = where y = y = y max, using the Binomial expansion we see that u 2 gh = UH2 gh 3 = UH2 gh 3 1 3a 1 +. Hence locally the flow is subcritical so F < 1 if a 1 x >, and supercritical so F > 1 if a 1 x < ; see Fig. 1. Friction forces the supercritical solution downstream in each case.

15 Introductory fluid mechanics: solutions 15 Solution Bernoulli s Theorem for irrotational unsteady flow a Since the flow is incompressible and homogeneous, ρ is uniform and constant. The flow is also irrotational so that u = ϕ. The Euler equations imply p t ϕ + u u = Φ. ρ Using the given identity we get ϕ t u 2 u u = p Φ ρ ϕ t u 2 + p ρ + Φ = u u H = u H = for the quantity H given in the question. b Hence H = ft for some function f = ft. Using the suggested redefined potential for u given by t t V = ϕ fτ dτ ϕ = V + fτ dτ and substituting this into we get which gives the result. H t = f t ϕ t t u 2 + p ρ + Φ = f t V t t + ft u 2 + p ρ + Φ = f t V t t u 2 + p ρ + Φ =

16 16 Simon J.A. Malham Solution rigid body rotation Need to show that the rigid body rotation ansatz u, v T = Ωy, Ωx T, where Ω = Ωt represents an angular velocity, is a solution to the incompressible Euler equations when a two-dimensional body force given by f = αx + βy, γx + δy T is applied. We assume that the third velocity component w =. We first note that the incompressibility condition u = is trivially satisfied by the rigid body rotation ansatz. Second we note that u u = u x + v u y v = = Ωy Ω 2 x Ω 2 y x + Ωx y. Ωy Ωx Hence if we substitute the rigid body rotation ansatz u, v T = Ωy, Ωx T, into the Euler equations we find Ωy Ω 2 x + Ωx Ω 2 = 1 p/ x αx + βy +. y ρ p/ y γx + δy Consider the curl of these equations the curl of a vector u = u, v T which only has two components is the scalar u/ y + v/ x. By direct computation this gives Ω = 1 γ β. 2 Now let us separate the equations above, i.e. those resulting from substituting the rigid body rotation ansatz into the Euler equations. We consider them as two partial differential equations for the pressure p as follows 1 p ρ x = Ωy + Ω 2 x + αx + βy, 1 p ρ y = Ωx + Ω 2 y + γx + δy. Recall to show that the rigid body rotation ansatz is a solution to the incompressible Euler equations we must find a pressure field p such that u, v T = Ωy, Ωx T and p satisfy them. We integrate the first equation above with respect to x. This generates the relation p ρ = Ωyx Ω2 x αx2 + βxy + F y, where F = F y is an arbitrary function. If we now substitute this expression into the second equation above and use that Ω = 1 2 γ β we find F y = Ω 2 y + δy. Hence we deduce F y = 1 2 Ω2 + δy 2 + C, where C is an arbitrary constant. Thus a solution to the two partial differential equations is p ρ = 1 2 Ω2 + αx Ω2 + δy β + γxy + C. 2 Hence there exists a pressure p such that the rigid body rotation ansatz represents a solution to the incompressible Euler equations.

17 Introductory fluid mechanics: solutions 17 Solution Vorticity and streamlines Starting with the steady incompressible two-dimensional Navier Stokes equations for the velocity field u = u, v T our goal is to show they are equivalent to the system of equations v H = ω u where ω is the scalar vorticity and H := p/ρ u2 + v 2. Using the identity 1 2 u 2 = u u + u u, we can write the steady Euler equations as ρ is constant and we assume no body force 1 2 u 2 p u u =. ρ In our two-dimensional context u 2 = u 2 + v 2, while by direct computation u u u = v ω vω = uω. Substituting these expressions back into the equation above we find ignoring the last component 1 2 u 2 + v 2 p vω + =, ρ uω thus demonstrating the equivalence. To show the scalar vorticity ω is constant along streamlines we observe that H y vω x uω v y + u ω + u ω x x + v ω y. Incompressibility implies the first term on the left is zero. Hence the relation above is equivalent to u ω and thus ω is constant along streamlines.

18 18 Simon J.A. Malham Solution Poiseuille flow a Use the incompressible Navier Stokes equations in cylindrical polar coordinates. Assume that u r =, u θ = and u z = u zr and ignore any possible body forces f r = f θ = f z =. Note we have implicitly assumed the flow is steady. Only the third equation for u z = u zr generates a non-trivial condition the first two equations tell us that the pressure is independent of r and θ. The third equation is u z t + u uz = 1 p ρ z + ν 1 r r r uz r + f z. Since the flow is steady u z/ t =, and as u r = u θ = and u z = u zr only then we see that u u z =. Further we assume there are no body forces, including gravity, so f z =. Since we assume p = Cz for some constant C, this equation becomes C = ρν r r r uz r b We integrate the equation above with respect to r twice to find u z = u zr as follows. Rearranging the equation and then integrating twice we find r uz = C r r ρν r r uz r = C 2ρν r2 + A u z r. = C 2ρν r + A r u z = C 4ρν r2 + A log r + B where A and B are arbitrary constants. We want the solution to be bounded for r a and therefore we must insist A = because log r as r. The no-slip boundary condition at the pipe wall r = a implies Hence we see that = C 4ρν a2 + B B = Ca2 4ρν. u z = C 4ρν r2 + Ca2 4ρν = C 4ρν a2 r 2. c Let S represent a disc cross-section of the pipe. Let ds represent a small patch of area on S. Since ρ is the mass per unit volume and u z ds represents the volume of fluid passing through ds per unit time, the mass flow rate across ds is ρu z ds. The total mass flow rate across S is thus a 2π ρu z ds = ρu zr rdθdr S a C = 2πρ 4ρν a2 r 2 rdr = πc 2ν = πca4 8ν. [ 1 2 a2 r r4] r=a r=

19 Introductory fluid mechanics: solutions 19 Solution Elliptical pipe flow a Using the Navier Stokes equations in three dimensional Cartesian coordinates, given u =, v = and w = wx, y only, and assuming no body force, we are left with = p x, = p y, = p 2 z + ν w 2 x + 2 w 2 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 2 w 2 x + 2 w 2 y = G ν., b Substituting w = Ax 2 + By 2 + C into the partial differential equation above, we immediately deduce A + B = G 2ν. c Using the no-slip boundary condition, i.e. that w = on generates the equation x 2 a 2 + y2 b 2 = 1 y2 = b 2 1 x2 a 2, Ax 2 + Bb 2 1 x2 a 2 + C = A B b2 a 2 x 2 + Bb 2 + C = which must hold for all x [ a, a]. Hence equating coefficients of x and x 2, we arrive at the following three equations including the result from part b above for the three unknowns A, B and C: A B b2 a 2 =, Bb2 + C = and A + B = G 2ν. Solving the first equation for A in terms of B and substituting this into the third equation we find B 1 + b2 a 2 = G 2ν B = G 2ν a 2 a 2 + b 2. Substituting this into the first equation reveals A + G 2ν a 2 a 2 + b 2 b2 a 2 = A = G 2ν b 2 a 2 + b 2.

20 2 Simon J.A. Malham Finally substituting these two answers for A and B into the second equation implies G 2ν a 2 a 2 + b 2 b2 + C = C = G 2ν a 2 b 2 a 2 + 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 w ds over the whole elliptical cross sectional area. Computing the integral using the substitutions x = ar cos θ and y = br sin θ we see: w ds = Ax 2 + By 2 + C dxdy 2π 1 = Aa 2 r 2 cos 2 θ + Bb 2 r 2 sin 2 θ + C abr dr dθ = G 2ν a 3 b 3 2π 1 a 2 + b 2 1 r 2 r dr dθ = 2πGa3 b 3 1 2νa 2 + b 2 r r 3 dr = πa3 b 3 G 4νa 2 + b 2. e Denote the flow rate from part d above by Q := πa3 b 3 G 4νa 2 + b 2. We wish to find the maximum of Q subject to the constraint that the cross sectional area of the elliptic pipe is given, say by K, i.e. πab = K with K fixed. Simply substitute that b = K/πa into Q and differentiate with respect to a implies Q a = πk/π 3 G a 4νa 2 + K 2 π 2 a 2 = πk/π3 G 4ν Thus Q/ a is zero and Q maximized if and only if Note that when a = K/π, then 2a 2K2 π 2 a 3 a 2 + K 2 π 2 a a 2K 2 π 2 a 3 = a = K/π. b = K/πa = K/π K 1/2 /π 1/2 = K 1/2 /π 1/2 = a. Hence the flow rate Q is maximized when a = b.

21 Introductory fluid mechanics: solutions 21 Solution Stokes flow: between hinged plates For the first part of the problem we can follow the Stokes flow section in the notes together with the viscous corner flow example given in that section. Since we assume we have an incompressible Stokes flow and it is stationary, we have p = µ u and u = where p is the pressure field, u the velocity field and µ is the viscosity. Since the flow is uniform in the direction aligned with the imaginary line of intersection between the plates denote this the z-axis there exists a stream function ψ = ψr, θ such that in cylindrical polar coordinates ψ =. To deduce this we simply follow the argument in the viscous corner flow example in the Stokes flow section. Note the velocity components u r and u θ and stream function are related by u r = 1 r ψ θ and u θ = ψ r. For the second part of the problem let us determine the appropriate boundary conditions. We will assume a lower plate that is stationary and Ω is the angular velocity of the upper plate relative to the lower plate. Using no-slip boundary conditions on the plate along θ = which is stationary we see that 1 ψ r θ = and ψ = on θ =. r Also using the no-slip boundary conditions on the moving plate which at time t lies along θ = βt >, we have 1 ψ r θ = and ψ = Ωr on θ = β. r Hence our goal now is to solve the biharmonic equation ψ = subject to the four boundary conditions above. We thus look for a solution of the form ψ = Ωr 2 fθ, for some arbitrary function f = fθ. This solution ansatz is suggested by the boundary condition on the moving upper plate. We solve the biharmonic equation here in two stages. First we observe that Ωr 2 fθ 2 = Ω r r r r 2 fθ r 2 θ 2 = Ω 4fθ + f θ. If we set F := f + 4f then, second, we observe that Ωr 2 fθ = ΩF θ 2 = Ω r r = ΩF θ. r F θ r 2 θ 2

22 22 Simon J.A. Malham Hence our solution guess ψ = Ωr 2 fθ satisfies the biharmonic equation if and only if F θ =. Hence we find F = 4A + 4Bθ for some constants A and B the prefactor 4 is chosen for convenience. Hence we find, combining the complementary function and particular integral for f + 4f = F, we have f = A + Bθ + C cos2θ + D sin2θ, where C and D are two further constants. For our solution ansatz ψ = Ωr 2 fθ the four boundary conditions become f = f = f β = and fβ = 1 2. The four constants are determined by the four boundary conditions above, and indeed we find sin θ cosβ θ θ cos β fθ =. 2β tan β cos β We finally focus on the issue of the distance within O within which this solution is self-consistent. Our solution ansatz was inferred from the boundary condition u θ = Ωr on the upper plate. Hence the inertia terms u u scale like Ω 2 r 2 /r while the viscous terms ν u scale like νωr/r 2. Our Stokes flow assumption was Ω 2 r 2 /r νωr/r 2 1 Ωr 2 1. ν Hence a necessary condition for the solution to be valid is r ν/ω 1/2. What happens when tan β = β?

23 Introductory fluid mechanics: solutions 23 Solution Stokes flow: rotating sphere We consider a Stokes flow around a rotating sphere, of radius a and angular velocity Ω. Our goal is to compute the couple exerted on the fluid due to the sphere; the fluid is at rest at infinity. Using the hint provided we use spherical polar coordinates r, θ, ϕ and assume that the axis to which the latitude angle θ is measured is aligned with, and in the same direction as, the angular velocity vector Ω. Given the geometry of the set up, we assume a stationary flow and that u r = and u θ = and u ϕ = u ϕr, θ only. Hence we have u = u ϕ and u ϕ as r. On the surface of the sphere r = a the velocity is u ϕ = Ωa sin θ where Ω = Ω the magnitude of the angular velocity Ω. This is because any point on the surface of the sphere at latitude angle θ is a distance a sin θ from the axis of rotation of the sphere. The form of the velocity field u ϕ = u ϕr, θ at the sphere surface r = a suggests we look for a solution of the form u ϕ = Ωafr sin θ. The steady Stokes flow equations involving the velocity field u and pressure p are µ u = p, where µ is the first coefficient of viscosity. Since there is only one non-zero component of velocity u ϕ, the Stokes flow equations in spherical polar coordinates reduce to µ u ϕ = 1 p r sin θ ϕ. Since u ϕ = u ϕr, θ only, we deduce that p = pr, θ can only be linear in ϕ. However p must be periodic in ϕ hence p must be constant, i.e. invariant to ϕ. Hence we have p ϕ =. Hence the Stokes flow equations reduce to u ϕ = or u =. We employ the strategy suggested in the notes for such Stokes flow equations and write where u = = u ϕ = u ϕ r sin 1 θ D2 =, r sin θ u ϕ D 2 := 2 r 2 + sin θ r 2 1 θ sin θ. θ

24 24 Simon J.A. Malham Substituting the solution ansatz u ϕ = Ωafr sin θ into the equation above we find 1 r sin θ D2 r sin θ u ϕ = D 2 r sin θ Ωafr sin θ = D 2 rfr sin 2 θ = sin 2 θ 2 fr 1 rfr + sin θ r 2 r θ sin θ θ sin2 θ = sin 2 θ fr + rf fr r + sin θ 2 cos θ = r r θ sin 2 θ 2f r + rf r 2 sin 2 θ fr r = rf r + 2f r 2 fr = r r 2 f r + 2rf r 2fr =. This is a linear second order Cauchy Euler type of equation. If we set ρ := log r and suppose F ρ = fr then we see using the chain rule that and f r = df dr = dρ df dr dρ = 1 r F ρ f r = d 1 dr r F ρ = 1 r 2 F ρ + 1 dρ r dr d dρ F ρ = 1 r 2 F ρ + 1 r 2 F ρ. Substituting these expressions into the Cauchy Euler equation above we find F + F 2F =, for F = F ρ. This is a second order constant coefficient differential equation for F whose independent solutions are expρ = r and exp 2ρ = 1. Hence the general r 2 solution to the Cauchy Euler equation for f = fr is f = Ar + B r 2, for arbitrary constants A and B. Using the boundary condition that u ϕ as r implies A =. Then on the boundary r = a of the sphere we must have u ϕa, θ = Ωa sin θ Ωafa sin θ = Ωa sin θ fa = 1 B = a 2. Hence the solution to the Stokes flow equations and boundary conditions is u ϕ = Ωa3 sin θ r 2.

25 Introductory fluid mechanics: solutions 25 Having computed the velocity field, we can now compute the coupling exerted on the fluid due to the sphere. Recall that for an incompressible Newtonian fluid the deviatoric stress ˆσ is related to the deformation matrix D through the formula ˆσ = 2µD. The components of the deformation matrix D are given the appendix of the notes. Since the only non-zero component of the velocity field is u ϕ = Ωa 3 sin θ/r 2 the only non-zero component of the deformation matrix D is D rϕ = r uϕ 2 r r = 3Ωa3 sin θ 2r 3. The total couple exerted on the fluid at the sphere boundary is π a sin θ ˆR ˆσn 2πa 2 sin θ dθ. By analogy with the way we computed the force on a sphere in a uniform Stokes flow, we split the sphere here into concentric rings about the axis of rotation and compute the coupling at the sphere surface r = a. Hence in the above integral 2πa 2 sin θ dθ represents the area of the concentric ring at angle θ to the axis of rotation. Note we have n = ˆr and so ˆσn = ˆσˆr represents the stress at the angle θ on the surface r = a of the sphere. The vector ˆR is a vector orthogonal to the axis of rotation and a sin θ represents the distance from that axis of rotation to a point on the sphere. Thus the coupling on the sphere at angle θ is a sin θ ˆR ˆσˆr. Now we compute the stress: ˆσˆr = 2µDˆr D rr D rθ D rϕ 1 = 2µ D rθ D θθ D θϕ D rϕ D θϕ D ϕϕ = 2µ D rϕ = 2µD rϕ ˆϕ. Hence the coupling is 2µa sin θd rϕ ˆR ˆϕ. We observe that ˆR ˆϕ = ˆΩ as it should be and 2µa sin θd rϕ is the component we are after. We can now compute the integral to obtain the total coupling π π 2µa sin θd rϕ2πa 2 sin θ dθ = 4µπa 3 3Ωa 3 sin θ 2a 3 sin 2 θ dθ π = 6µπΩa 3 sin 3 θ dθ π π = 6µπΩa 3 sin θ dθ cos 2 θ sin θ dθ π = 6µπΩa cos 2 θ dcos θ = 6µπΩa = 8µπΩa 3.

26 26 Simon J.A. Malham Solution Lubrication theory: shear stress We assume lubrication theory holds for this scenario. Hence the flow in the narrow layer is governed by the shallow layer equations. We also assume the flow is uniform in one of the horizontal directions the y-direction from the set up given and so the only relevant shallow layer equation is p x = µ 2 u z 2, where z represents the vertical coordinate, u = ux, z is the velocity in the x-direction and µ is the viscosity. Integrating this equation with respect to z we find u = 1 p zh z. 2µ x The volume flux through the layer is constant say Q, i.e. we have h u dz = Q. Computing the integral on the left we find 1 p 2µ x as we required so the constant A 12µQ. Now suppose h = Ce Bx ; this implies h 3 6 = Q p x = 12µQ h 3, p x = 12µQ C 3 e3bx px = p 4µQ e 3Bx 1. BC 3 In particular we see that Hence we see that pl p = 4µQ BC 3 1 e 3BL ˆp = 4µQ BC 3 1 e 3BL. p 3B ˆp e3bx = x 1 e 3BL. The shear stress on z = is µ u z = 1 p z= 2 x h = 3B ˆp 2 1 e 3BL Ce2Bx. The maximum shear stress is at x = L and has the value 3B ˆp 2 e 3BL 1 Ce2BL.

27 Introductory fluid mechanics: solutions 27 Solution Lubrication theory: shallow layer a This whole part follows directly from the derivation of the shallow layer equations in the lubrication theory section of the notes once the replacement p = P ρgz is made, where P is the modified pressure. The explanation is as follows. With this replacement since p x = P x, and p y = P y, the first and second component Navier Stokes equations are unchanged the body force ρgẑ does not act on either of these components anyway. The body force ρgẑ does act on the third component Navier Stokes equation and the pressure term and body force term in combination are 1 p ρ z ρg = 1 P ρgz ρg ρ z = 1 P ρ z. Hence making the replacement for the pressure in terms of the modified pressure in the Navier Stokes equations with a body force ρgẑ, we obtain the Navier Stokes equations in terms of the modified pressure P without any body force. We can thus proceed with the derivation of the shallow layer equations in the lubrication theory section of the notes. b Note from the shallow layer equations that the third equation implies P z = p z = ρg p = ρgh z P = ρgh. where we assume that p = at z = hx, 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 implies and similarly µ 2 u h = ρg z2 x µ u z h Γ = ρg z h + x x µu = 1 2 µv = 1 2 h Γ ρg zz 2h + z x x, h Γ ρg zz 2h + z y y, where in the first integration from z to h, we used the given conditions for the applied surfaces stresses at z = hx, 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 = hx, y, t. The acceleration of any particle at the surface is given by h h h + ux, y, h, t + vx, y, h, t t x y.

28 28 Simon J.A. Malham From the incompressibility condition we have h h h h u + ux, y, h, t + vx, y, h, t t x y = x + v dz y = h u dz h v dz x y + ux, y, h, t h h + vx, y, h, t x y. Cancelling like terms on each side and subsituting our expressions for u and v above into this last expression reveals that 2µ h t = x 23 ρgh 3 h x h2 Γ x + 23 ρgh 3 h y y Γ h2. y Simplifying the terms under the partial derivatives on the right-hand side reveals 2µ h t + h 2 Γ 1 x x 3 ρgh 2 + h 2 Γ 1 y y 3 ρgh 2 =.

29 Introductory fluid mechanics: solutions 29 Solution Lubrication theory: Hele Shaw cell Part a is straightforward and part b follows from the standard theory underlying the shallow layer equations in the notes. c We are asked to show that the vertically averaged velocity field u = u, v T is incompressible and irrotational. Recall that by definition T ux, y, vx, y := 1 h T ux, y, vx, y dz. h By direct computation we find note that h is constant u = h 1 u dz + h 1 v dz x h y h = 1 h u h x + v y dz = 1 h w dz h z [ = w ] z=h h z= =, since w = on z = and z = h. Integrating the shallow layer equations with respect to z twice we find u = 1 2µ p x zh z and v = 1 2µ p y zh z, using the no-slip boundary conditions at z = and z = h. Hence the scalar quantity representing the curl of u = u, v T is given by u = h 1 u dz y h = 1 h u h y v x dz = 1 h zh z h 2µ =. 1 x h h v dz 2 p x y 2 p dz y x Hence the vertically averaged velocity field is both incompressible and irrotational. d Within a distance Oh of the cylinder the solution above is no longer representative as no-slip boundary conditions must hold on the cylinder boundary. The solution further away from the cylinder is least disturbed if we suppose the boundary condition for u on the cylinder to be u n =.

30 3 Simon J.A. Malham Solution Boundary layer theory: axisymmetric flow We consider an axisymmetric flow in cylindrical polar coordinates r, θ, z where u r = αr/2, u θ = u θ r, t only and u z = αz, where α is a constant. First we show this flow is incompressible. In cylindrical polar coordinates we have u = 1 r r rur + 1 r = 1 r r =. u θ θ + uz z α 2 r2 + + z αz Second we compute the form of the vorticity for the flow, we have 1 u z r θ u θ z u = u r z uz r 1 r r ru θ 1 r ur θ = 1 r r ru θ =, ω where we set ω = ωr, t to be ω = 1 r r ru θ. The equation for the evolution of vorticity ω = ωx, t is given by ω t + u ω = ν ω + ω u. For the vorticity form above, in cylindrical polar coordinates, using for any vector v, we have v = v r r + v θ r θ + vz z, and using the form for the Laplacian in cylindrical polar coordinates from the formulae in the Appendix, the equations above for the evolution of vorticity collapse to ω ω + ur t r = ν 1 r ω + ω uz r r r z. a For the moment we assume the flow is inviscid and at time t = the initial data has the form ω = ω fr for some function f = fr and some constant ω. We are required to verify that ω = ω e αt f re αt/2 is a solution for all t >. Note that this ansatz satisfies the initial condition at t =. Looking at the evolution equation for ω = ωr, t above with ν =, we see that we need to compute ω t = αω e αt f re αt/2 + ω e αt f re αt/2 rα 2 eαt/2, ω r = ω e αt f re αt/2 e αt/2.

31 Introductory fluid mechanics: solutions 31 Subsituting the ansatz for ω above and these two expressions for its partial derivatives with respect to t and r into the acceleration terms in evolution equation for ω = ωr, t, we find ω + ur ω t r = αω e αt f re αt/2 + ω e αt f re αt/2 rα 2 eαt/2 αr ω e αt f re αt/2 e αt/2 2 = αω e αt f re αt/2 = ω uz z. Hence we deduce that the ansatz does indeed give a solution to the evolutionary vorticity equation in the inviscid case when ν =. We can interpret this solution ansatz as follows. Consider a fluid particle whose radial position is prescribed by r = rt. Suppose that at t = its radial position was r = r. Its radial trajectory is thus given by dr dt = α 2 r r = r e αt/2. Thus a particle that at time t has radial coordinate r started with radial coordinate re αt/2 at time t =. Thus in the ansatz ω = ω e αt f re αt/2 we see that as time evolves the vorticity intensifies exponentially fast due to: i Stretching via the exponential prefactor e αt, and ii Inwards advection via the term re αt/2 in the argument of f. b Now consider steady viscous flow. The equation for the vorticity above becomes αr ω 2 r = ν 1 r ω + αω. r r r We look for a solution ω = ωr with the ansatz ω = ω e αr2 /4ν, where ω is a constant. We observe that ω r = ω αr /4ν 2ν e αr2, and r ω = ω α r 2 e αr 2 /4ν r r 2ν r = ω α 2r αr3 e αr2 /4ν. 2ν 2ν Substituting these expressions into the vorticity equation above we find ν 1 r ω + αω = ω α 2 αr2 e αr2 /4ν + αω e αr2 /4ν r r r 2 2ν = ω α 2 r 2 e αr2 /4ν 4ν = αr ω αr /4ν 2 2ν e αr2 = αr ω 2 r. Hence ω = ω e αr2 /4ν is a solution to the steady viscous vorticity equation. A steady solution is possible as there is a dominant balance between outwards diffusion, due to the viscous term, and inwards advection and stretching.

32 32 Simon J.A. Malham Solution Boundary layer theory: rigid wall Our first goal is to seek a similarity solution in the form ψ = Uxδxfη, where η = y/δx, to the boundary layer equation ψ 2 ψ y x y ψ 2 ψ x y 2 = U U x + ν 3 ψ y 3, and find the corresponding equation satisfied by f. Using the chain rule, we compute the following partial derivatives: as well as ψ y = Uf, ψ x = δ Uδf U x x ηf 2 ψ y 2 = U δ f and and 2 ψ x y = U x f U δ Substituting these into the boundary layer equations above we find Uf U x f U δ δ x ηf 3 ψ y 3 = U δ 2 f, δ x ηf. δ Uδf U x x ηf U δ f = U U x + ν U δ 2 f U U x f 2 x Uδ U δ ff = U U x + ν U δ 2 f. Rearranging this equation we find f + αff + β 1 f 2 =, where α := δ / xuδ ν and β := U/ xδ2. ν Since f is independent of x, this last relation implies α and β are independent of x also and are thus constants. The product rule implies x δ2 U = δ x δu + δ x δu U = δ Uδ δ + να x x = να β + να = ν2α β, where we have used the definitions of α and β above. Hence if 2α = β then δ 2 U = K a constant. Then from the definition of β we have U/ x = νβ/δ 2 = νβ/ku. Thus we have U = C expνβx/k, for some constant C, and δ = K/C 1/2 exp νβx/2δ. However if 2α β then δ 2 U = K + 2α βνx, where K is a constant. Again using the definition of β we find U x = νβ δ 2 = νβ K + 2α βνx U. Solving this first order differential equation implies U = C K +2α βνx β/2α β, where C is another constant. Hence U is algebraic generically and exponential in the special case 2α = β. If there is a rigid wall at y = then no-slip boundary conditions imply u = and v = at y =. Since ψ/ y = u and ψ/ y = v we observe that these no-slip boundary conditions translate to f = f =. Also, since u Ux as y and u = ψ/ y = Uf, we necessarily have f 1 as η.

33 Introductory fluid mechanics: solutions 33 Acknowledgements The lecture notes were to a large extent grown out of a merging of, lectures on Ideal Fluid Mechanics given by Dr. Frank Berkshire [1] in the Spring of 1989, lectures on Viscous Fluid Mechanics given by Prof. Trevor Stuart [5] in the Autumn of 1989 both at Imperial College and the style and content of the excellent text by Chorin and Marsden [2]. They also benefited from lecture notes by Prof. Frank Leppington [3] on Electromagnetism. Some of these solutions are derived from their solutions to the problems in their lecture notes. References 1. Berkshire, F Lecture notes on ideal fluid dynamics, Imperial College Mathematics Department. 2. Chorin, A.J. and Marsden, J.E. 199 A mathematical introduction to fluid mechanics, Third edition, Springer Verlag, New York. 3. Leppington, F Electromagnetism, Imperial College Mathematics Department. 4. Majda, A.J. and Bertozzi, A.L. 22 Vorticity and incompressible flow, Cambridge Texts in Applied Mathematics, Cambridge University Press. 5. Stuart, J.T Lecture notes on Viscous Fluid Mechanics, Imperial College Mathematics Department.