Candidates must show on each answer book the type of calculator used. Only calculators permitted under UEA Regulations may be used.

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1 UNIVERSITY OF EAST ANGLIA School of Mathematics May/June UG Examination FLUID DYNAMICS MTH-3D41 Time allowed: 3 hours Attempt FIVE questions. Candidates must show on each answer book the type of calculator used. Only calculators permitted under UEA Regulations may be used. Notes are not permitted in this examination. Do not turn over until you are told to do so by the Invigilator. MTH-3D41 Module Contact: Dr M.J. Cooker, MTH Copyright of the University of East Anglia Version: 1

2 - 2 - Note: Below are the Navier-Stokes equations, for an incompressible fluid of constant density ϱ and constant kinematic viscosity ν, in which the velocity u = u r e r +u θ e θ +u z k and the pressure is p, with zero body force, expressed in cylindrical polar coordinates r, θ, z : where the operator u r t + u )u r u θ r u θ t + u )u θ + u ru θ r and the Laplace operator is 2 = 1 p ϱ r + ν 2 u r u r = 1 ϱr p θ + ν u z t + u )u z = 1 p ϱ z + ν 2 u z u ) = u r r + u θ r 2 = 1 r r 2 ) u θ 2 r 2 θ 2 u θ u θ r + 2 ) u r 2 r 2 θ θ + u z z r ) + 1r 2 r r 2 θ z. 2 Also u = 1 r r ru r) + 1 u θ r θ + u z z. In spherical polar coordinates r, θ, α F = F r e r + 1 F r θ e θ + 1 F r sin θ α e α and u = 1 ) r 2 1 u θ sin θ) u r 2 r u α ) r r sin θ θ r sin θ α. MTH-3D41 Version: 1

3 A straight rigid pipe has uniform cross-section, D, which is a finite simply-connected region of the x, y plane. A fluid of dynamic viscosity µ and constant density ϱ flows steadily along the pipe with velocity u = wx, y)k, where k is a unit vector directed along the axis of the pipe. There is zero gravity. The cartesian coordinates x, y have axes which are orthogonal to k. i) Starting from the Navier-Stokes equations deduce the following relations in D: dp dz = constant, 1) 2 1 dp w = µ dz. 2) [8 marks] ii) For a pipe of circular cross-section w = wr) and p = pr) where r is the cylindrical polar coordinate measured from the axis of the pipe. Suppose µ 1 dp/dz = G a constant. Show that the general solution of equation 2) is wr) = G 4 r2 + B logr) + C, where B and C are constants. [4 marks] iii) If the pipe wall is fixed at r = a, show that wr) = G 4 a2 r 2 ). 3) [4 marks] iv) Suppose G < 0. Draw a graph of w as a function of r for 0 r a, and mark clearly the position of the maximum value of w. [4 marks] MTH-3D41 PLEASE TURN OVER Version: 1

4 i) A certain flow has the velocity field v. Define concisely its vorticity. [3 marks] ii) Fluid of kinematic viscosity ν occupies a channel y : 0 y H, in the x, y plane, where H > 0 is constant. The wall y = H is at rest. The wall y = 0 moves with velocity U 0 t)i. There are no body forces acting and you may assume that the pressure is independent of x. Assuming that the fluid velocity is u = uy, t) i, in 0 y H, show that the two-dimensional Navier-Stokes equations simplify to u t = νu yy 1) and p = pt). 2) [5 marks] iii) Write down the boundary conditions on ux, t) for equation 1). [3 marks] iv) Suppose U 0 t) = constant. equation 1) is Confirm that the corresponding steady solution of u = U 0 1 y ). 3) [4 marks] H v) Up until the time t = 0 the fluid flows with the velocity field in equation 3). Then, at t = 0 the motion of the wall at y = 0 is halted so that U 0 t) = 0 for t 0. The subsequent flow for t 0 in the channel is unsteady and governed by equation 1) with initial condition 3). You are given that the form of solution of 1) is: uy, t) = a n sinnπy/h) exp νn 2 π 2 t/h 2 ). n=1 Write an integral for the coefficient a n, but do not evaluate it. vi) Comment on what happens to the flow as t increases to infinity. [3 marks] [2 marks] MTH-3D41 Version: 1

5 i) Define the Reynolds number, defining the symbols you use. [3 marks] ii) For steady flow at high Reynolds number the two-dimensional Navier-Stokes equations, for the velocity u = ui+vj, simplify to the following dimensionless boundarylayer equations for fluid flow above a fixed impermeable surface at y = 0 for x 0: uu x + vu y = U du dx + u yy 1) u x + v y = 0, 2) where, as y, u Ux) i. Suppose there exists a stream function ψx, y) such that u = ψ y and v = ψ x. Show that equations 1) and 2) are satisfied provided ψ satisfies the partial differential equation ψ y ψ xy ψ x ψ yy = U du dx + ψ yyy. 3) [6 marks] iii) Let U be constant. Suppose ψ = x m fη) where x > 0 and η = yx m. Find the constant m such that this is a similarity solution of equation 3). [6 marks] iv) From the boundary conditions for f and from the expression ψ = x m fyx m ), show that in this situation the boundary layer thickens as x increases. [5 marks] MTH-3D41 PLEASE TURN OVER Version: 1

6 i) A certain two-dimensional fluid velocity field, parallel to the r, θ plane is v = u θ r, t) e θ, where e r, e θ are unit vectors with respect to plane polar coordinates r, θ. Assume that p = pr, t). Show that the velocity field is compatible with the flow of an incompressible fluid. ii) Show also that u 2 θ r = 1 p ϱ r 1 u θ ν t = 1 r r r u ) θ r 1) [3 marks] u θ, 2) [6 marks] r2 where ϱ is the constant density and ν is the constant kinematic viscosity. iii) Now assume that the flow is steady so that u θ is independent of time. Show that r λ is a solution of equation 2) for certain values of the constant λ that you must find. [4 marks] iv) Hence find the general solution u θ r) of the steady form of equation 2). [2 marks] v) Honey flows steadily in a circular jar of radius a. The jar rotates about its axis, which is the z -axis, in such a way that on the jar s surface r = a, u θ = U 0, where U 0 is a constant. Use your solution of equation 2) to find the field of steady velocity v = u θ r)e θ of the honey. [5 marks] MTH-3D41 Version: 1

7 A viscous liquid lubricates the narrow gap between a moving flat plate at z = 0 and a fixed curved surface z = hx), where hx) > 0 is a differentiable function for x : 0 x 1. The plate has constant velocity U 0 i, where U 0 > 0. At x = 0 and at x = 1 the pressure p is zero. The flow is two-dimensional in the x, z plane, and such that the pressure, p = px), and the velocity u = ux, z)i + wx, z)k, are governed by 2 u z = 1 dp 2 µ dx 1) and w z = u x. 2) i) From equation 1) and the boundary conditions, show that ux, z) = 1 dp 2µ dx z 2 hz ) + U 0 1 z ). 3) [4 marks] h ii) State why it is reasonable that the quantity Q = hx) 0 ux, z) dz, is independent of x. iii) Use this fact about Q to deduce from equation 3) that [2 marks] x px) = 6µ U 0 h 2 dx 12µ Q x 0 0 h 3 dx. [7 marks] iv) Suppose hx) = h 0 x 1, where h 0 px). is a positive constant. Find an expression for [4 marks] v) For what choice of Q does p vanish at x = 1? [3 marks] MTH-3D41 PLEASE TURN OVER Version: 1

8 A layer of oil is flowing steadily under the force of gravity downhill. The oil moves in a layer of constant thickness h supported by a sheet of glass which is at a constant downward inclination angle α to the horizontal. The x -axis and the unit vector i are parallel to the glass such that x increases downhill, and the y -axis is directed normal to the glass such that y = 0 is the glass surface. The oil flows in the region y : 0 y h. The pressure on the free surface is constant and equal to p 0 for all x. The velocity of the oil is u = uy)i. i) Show that the two-dimensional Navier-Stokes equations reduce to the following: 1 p ϱ x + g sinα) + ν 2 u y = 0 1) 2 1 p and + g cosα) = 0, 2) [6 marks] ϱ y where g is gravitational acceleration, ϱ is density and ν is kinematic viscosity. ii) Show that the boundary conditions are u = 0 on y = 0, and u/y = 0 on y = h. [2 marks] iii) From equation 2) derive the following expression for the pressure distribution in the oil: py) = p 0 + ϱgh y) cosα). [5 marks] iv) From equation 1) show that the velocity in the direction i is uy) = g y 2h y) sinα). 2ν v) Sketch the velocity profile for the case α = π/2. [5 marks] [2 marks] END OF PAPER MTH-3D41 Version: 1

UNIVERSITY OF EAST ANGLIA

UNIVERSITY OF EAST ANGLIA School of Mathematics May/June UG Examination 2007 2008 FLUIDS DYNAMICS WITH ADVANCED TOPICS Time allowed: 3 hours Attempt question ONE and FOUR other questions. Candidates must