inviscid fluid dynamics

Upload: messy-uzunov

Post on 12-Oct-2015

139 views

Category:

Documents


5 download

DESCRIPTION

Inviscid Fluid Dynamics

TRANSCRIPT

  • Inviscid Fluid Dynamics MAT2050

    Semester 2

    Dr Matt Turner, Spring 2014

    i

  • Inviscid Fluid Dynamics MAT2050

    Semester 2

    Dr Matt Turner, Spring 2014

    LECTURER: Dr Matt Turner, Room 26aAA04

    [email protected]

    OFFICE HOURS: Tuesday 3-4pmThursday 3-4pm

    UNASSESSED COURSEWORKS: Week 5 (13/03/14 4pm)Week 9 (08/05/14 4pm)Week 11 (22/05/14 4pm)

    CLASS TEST: Week 8 (03/04/14 10am)

    PREFERED CONTACT METHOD: Please use my office hours above, and the discussionforums on SurreyLearn when seeking help for problems. Emailed problems may be uploaded toSurreyLearn if they are generic and would benefit the whole class. For personal problems eithercome and see me or email me.

    Books:

    1. Acheson: Elementary Fluid Dynamics

    2. Paterson: A first course in fluid dynamics

    3. Batchelor: An introduction to Fluid Dynamics

    4. Lighthill: An informal introduction to theoretical fluid dynamics

    None of these books follow this course exactly. Each book includes most of the material covered inthis course, but also includes other topics from the field of fluid mechanics. The books by Achesonand Paterson are probably the most useful, with Acheson the best one to purchase if you are goingto purchase one. On the reading list for this course there are also some e-books which you mightfind useful.

    ii

  • Aims of the Course

    1. Understand what we mean by a fluid (includes both liquids and gases)

    2. What equations govern the motion of fluid particles?

    3. How is pressure related to flow field?

    4. How quickly will a barrel empty from a hole in side?

    5. What causes the Severn Bore a tidal wave moving up River Severn?

    6. How can we model flows around obstacles?

    7. What equations describe waves on the surface of water? (eg behind duck or ship)

    Expectation of each student

    1. You are expected to attend the lectures and prepare accordingly by reading the notes fromprevious lectures.

    2. You are expected to revise material covered in previous courses which is applied in thiscourse.

    3. You are expected to work though the lecture notes filling in the missing in examples steps,and completing all examples which are left as exercises.

    4. You are expected to use the library resources (e-books and books) to research module contentwhich you are having trouble to understand.

    5. You are expected to submit as complete a set of solutions as possible to all the unassessedpieces of coursework.

    6. You are expected to attempt all the coursework questions which are not handed in asunassessed coursework, in the lead up to class tests and the exam.

    7. The solutions given to the coursework questions will only be sketch solutions and you areexpected to fill in the missing algebraic steps for yourself.

    8. If are having problems with any of the above items then you are expected to be proactiveand use my office hours and the discussion forums on SurreyLearn to help you resolve theseissues.

    iii

  • Notation for module

    Here is a list of some of the more common notation you will meet throughout this course. Someletters have multiple definitions, but their context will always be clear. You can add more elementsto this list if you find it helpful.

    Variable Usage(x, y, z) Cartesian coordinates

    u = (u, v, w) Cartesian velocity components

    x, y, z Unit vectors in (x, y, z) directions

    (r, , z) Cylindrical polar coordinates

    u = (ur, u, uz) Cylindrical polar velocity components

    r, , z Unit vectors in (r, , z) directions

    r position vector

    s unit position vector

    t time

    p pressure

    density

    w(z) = w(x+ iy) complex potential

    U Velocity

    g = 9.81ms2 gravity vorticity

    vorticity in z direction for 2D flows/frequency

    velocity potential

    log(x) = loge(x) = ln(x) Natural logarithm

    Circulation

    Stream function

    n, N normal vector

    F Force

    S Surface area element

    Free surface/height

    T Temperature, Kinetic energy

    V Volume, Potential energy

    S Surface

    iv

  • The Greek Alphabet

    Small Capital Name A alpha

    B beta

    gamma

    delta

    E epsilon

    Z zeta

    H eta

    theta

    I iota

    K kappa

    M mu

    N nu

    xi

    o O omicron

    pi pi

    P rho

    sigma

    T tau

    upsilon

    phi

    X chi

    lambda

    psi

    omega

    v

  • Contents

    1 Vector Calculus 11.1 Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

    1.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.3 Normals to Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.4 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

    1.2 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.1 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.2 Important Corollary to Divergence Theorem . . . . . . . . . . . . . . . . . . 61.2.3 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

    2 Introduction 72.1 Density and the Continuum Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.4 Statics & Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 Buoyancy and Archimedes Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.6 One-dimensional flow in tubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

    3 Kinematics 183.1 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Describing flow fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

    3.2.1 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2.2 Particle Pathlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2.3 Streaklines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

    3.3 Convective Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.4 Mass Conservation and Compressibility . . . . . . . . . . . . . . . . . . . . . . . . . 273.5 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.6 Velocity Potential for Irrotational Flows . . . . . . . . . . . . . . . . . . . . . . . . 29

    4 Dynamics 324.1 Eulers Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.2 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.3 Bernoullis Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

    4.3.1 Irrotational Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.3.2 Steady Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

    4.4 Momentum Integral Theorem for Steady Flows . . . . . . . . . . . . . . . . . . . . . 374.5 Calculating Velocity Potential for general flows . . . . . . . . . . . . . . . . . . . . . 39

    4.5.1 Uniqueness of velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . 39

    5 Channel Flows 425.1 Steady flow under a Sluice Gate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425.2 Hydraulic Jumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.3 Flow over obstacle on river bed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

    vi

  • 6 Two-dimensional Flows 476.1 The stream function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

    6.1.1 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.1.2 Governing Equations and Boundary Conditions . . . . . . . . . . . . . . . . 496.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

    6.2 Line Vortices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516.2.1 Velocity potential of line vortex . . . . . . . . . . . . . . . . . . . . . . . . . 516.2.2 Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.2.3 Relation between vorticity and circulation . . . . . . . . . . . . . . . . . . . 526.2.4 Kelvins Circulation Theorem Inviscid Flows . . . . . . . . . . . . . . . . . 536.2.5 Significance of Irrotational Flows . . . . . . . . . . . . . . . . . . . . . . . . 53

    6.3 Combining simple examples of 2D velocity potentials . . . . . . . . . . . . . . . . . 536.4 General solution of velocity potential in plane polar coordinates . . . . . . . . . . . 556.5 Complex Potentials for incompressible, irrotational 2-D flows . . . . . . . . . . . . . 56

    6.5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566.5.2 Boundary Conditions on w(z) . . . . . . . . . . . . . . . . . . . . . . . . . . 576.5.3 Velocity in terms of Complex Potential . . . . . . . . . . . . . . . . . . . . . 576.5.4 Simple Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576.5.5 Combining Simple Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

    6.6 Calculating Forces on Body using Pressure . . . . . . . . . . . . . . . . . . . . . . . 646.7 Induced Velocity and Image Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 69

    6.7.1 Induced Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696.7.2 Image Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

    7 Surface Waves 757.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757.2 Basic wave theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757.3 Potential Flow with a Free Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . 767.4 Small amplitude disturbances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.5 Small-amplitude travelling wave solutions . . . . . . . . . . . . . . . . . . . . . . . . 787.6 Linear Superposition and Standing Waves . . . . . . . . . . . . . . . . . . . . . . . 797.7 Phase Velocity and Group Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 807.8 Particle paths for water waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

    A Cylindrical Polar coordinates 86A.1 Vector operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

    B Divergence theorem 87

    C Stokes theorem 89

    D Proof of Kelvins Circulation Theorem 91

    E Flow around a cylinder 92

    F Propagation of energy for deep water waves 93

    vii

  • 1 Vector Calculus

    1.1 Vector Calculus

    1.1.1 Definitions

    In this course I will be using x, y, z for the unit vectors in the 3 Cartesian directions, rather than i, j, k.

    The position vector of a point P relative to a fixed origin O is written as

    OP = r = xx + yy + zz.We now consider scalars and vectors which are functions of position.

    1. A single-valued function of the three variables x, y, z written as (x, y, z) can be considered as a scalar function of position (r). This is known as a scalar field.

    2. A single-valued vector function of x, y, z, u(x, y, z), can be written as

    (r) = u(x, y, z)x + v(x, y, z)y + w(x, y, z)z, and this is known as a vector field.

    Therefore given a scalar field (x, y, z) and a vector field

    u(x, y, z) = u(x, y, z)x + v(x, y, z)y + w(x, y, z)z,we define

    = xx+ y y +

    z

    z, (This is a vector!)

    2= 2

    x2+2

    y2+2

    z2,

    u= ux

    +v

    y+w

    z, (Pronounced divergence)

    u=

    x

    y z

    x

    y

    z

    u v w

    , (Pronounced curl)

    u

    =2u 2ux + 2vy + 2wz.

    The form of these definitions in polar coordinates can be found in Appendix A. You should know them in Cartesian coordinates along with in polar coordinates, but the other polar expressions

    will be given in an exam.

    1

  • 1.1.2 Directional DerivativesNote that s is the component of a vector in the direction of the unit vector s. We now show that

    s = s = Rate of change of in the s direction.ProofConsider a point P with position vector r. At this point is given by (r) = (x, y, z).Now consider a second point Q close to P, with position vector r + r. The value of at Q is

    (r + r) = (x + x, y + y, z + z).

    This can be expanded using Taylors Theorem,

    (x + x, y + y, z + z) = (x, y, z) + xx + yy + zz + smaller terms, so

    the change in is given by

    =(x + x, y + y, z + z) (x, y, z)

    x

    =xx + yy + zz=(xx + yy + zz) (x + yy + zz) =r

    Now let r be in the direction of unit vector s, sor = (s)s, where s = |r| = distance PQ

    and hence

    s= s. = s s, =

    Taking the limit s 0 gives required result.

    x

    Example 1Calculate the rate of change of = xyz at the point (1, 1, 1) in the direction parallel to x+2y+2z. The unit vector parallel to + 2y + 2z is

    s = 11 + 22 + 22

    (x + 2y + 2z) =13

    (x + 2y + 2z).Then

    =yzx + xzy + xyzs= 13(yz + 2xz + 2xy)

    =1

    3(1 + 2 + 2) =

    5

    3, when x = y = z = 1.

    2

  • 1.1.3 Normals to Surfaces

    We know that in two-dimensional space, f(x, y) =constant defines a line, and similarly the equation (x, y, z) =constant defines a surface in 3-D space.Examples:

    Set of parallel planes for different values of kx + 2y z =k x2 + y2 + z2 =k Set of concentric spheres

    In 1.1.2 it was shown that the rate of change of in a direction s iss = s .

    Now if s is chosen to be a tangent to the level surface, so s = t then in this direction

    s= 0, = t = 0.

    Thus is orthogonal to any vector tangent to the surface =constant, and so is normal to the surface. (Another proof will be seen later in the course).The unit normal to the surface =constant is given by

    n= || ,the two possibilities corresponding to normals on either side of the surface.

    Example 2Calculate the normal to the surface x2y + 2xz = 4 at the point A = (2, 2, 1).Calculate the equation of the tangent plane to the surface at A and find the perpendicular distance of this plane from the origin.

    at A

    =x2y + 2xz,=(2xy + 2z)x + x2y + 2xz, =6x + 4y + 4z = 2(3x + 2y + 2z).Hence the unit normal to the plane isn= 1 (3x + 2y + 2z).17

    The equation of a plane with unit normal n is rn = p, where |p| is the perpendicular distance of the plane from the origin.

    rn = (xx + yy + zz)(3x + 2y + 2z)117

    =117

    (3x + 2y + 2z) = p.

    Since the plane passes through (2, 2, 1),117

    (6 4 + 2) = p.

    3

  • Hence the perpendicular distance of the plane from the origin is

    817

    |p| = ,

    and the equation of the tangent plane is

    3x + 2y + 2z = 8.

    1.1.4 Identities

    () = 0, ( u) = 0.A further identity we will use, but which you have probably not seen before is

    u ( u) = 2u2(1 ) (u)u.It is important to understand how to calculate each term in this identity. Note that u2 = u u. If u = ux + vy + wz, then

    (12u

    2)= 1

    2

    x

    (u2 + v2 + w2

    ) x+ 12 y (u2 + v2 + w2)y + 12 z (u2 + v2 + w2) ,(u)u=

    (u

    x+ v

    y+ w

    z

    )(ux + vy + wz),

    =(uux + vuy + wuz)x + (uvx + vvy + wvz)y + (uwx + vwy + wwz)z.(Proof not included here show by components)

    Example 3Show that the identity is valid when u = yx 2zy + xz.

    (+ u=2x y z,

    u ( u)=(2z x)x + (y + 2x)y + (4z y)z, 2u

    2 = 12 y2 + 4z2 + x2(1 ) ) ,

    x(y

    x

    y(u)u= 2z + x

    z

    = x + yy + 4zz, )(yx 2zy + xz),

    (12u

    2) =2zx 2xy + yz, (u)u=(x + 2z)x + (y + 2x)y + (4z y)z, =u ( u).

    Another useful identity that we shall use is

    (u) = u + u . This identity is analogous to the product rule for differentiation.

    4

  • 1.2 Integral Theorems

    1.2.1 Divergence Theorem

    Let w be a vector field, V be a volume of space with surface S and outward unit normal n.Thenwn dS = w dV,

    S V

    where dS is an element of surface area. For the proof of this see appendix B.

    Example of verifying Divergence Theorem (Example 4)Verify the divergence theorem for w = 4xx 2y2y + z2z taken over the region bounded by x2 + y2 = 4, z = 0 and z = 3.First the volume integral:

    V

    w dV = ( V x

    (4x) +

    y(2y2) +

    z(z2)

    )dV,

    =

    (4 4y + 2z) dV,

    = 2x=2 y=

    4x2

    V 4x2 3z=0

    (4 4y + 2z) dxdydz,=84pi.

    The surface S of the cylinder consists of a base S1 (z = 0), the top S2 (z = 3) and the convex portion S3 (x2 + y2 = 4). Thus the surface integral is

    S

    wn dS =S1

    wn dS1 +S2

    wn dS2 + S3

    wn dS3.On S1 (z = 0)

    n= z, w = 4xx 2y2y and w n = 0, S1

    wn dS1 = 0.On S2 (z = 3) n= z, w = 4xx 2y2y + 9z and w n = 9,therefore

    wn dS2 = 9 dS2 = 36pi.S2 S2

    On S3 (x2 + y2 = 4), the unit normal to this surface is

    u =1

    2(xx + yy) ,

    thus

    w n =(4xx 2y2y + z2z) 12

    (xx + yy) = 2x2 y3.5

  • On the surface of the cylinder

    x = 2 cos and y = 2 sin ,

    thusdS3 = 2ddz.

    Therefore S3

    wn dS3 = 2pi 3z=0

    [2 (2 cos )2 (2 sin )3

    ]2dzd,

    =

    =0 2pi0

    [ ]d,

    =48

    48 cos2 48 sin3 2pi0

    cos2 d,

    =48pi.

    Then the surface integral = 0 + 36pi + 48pi = 84pi, agreeing with the volume integral and thus verifying the divergence theorem.

    1.2.2 Important Corollary to Divergence Theorem

    By writing w = pa where p(r) is a scalar field of the position vector r and a is an arbitrary constant,

    w = (pa) = (p) a,

    pan dS =the divergence theorem

    gives a p dV, = a

    ( pn dS) = a(

    V

    p dV),

    S V S

    taking the constant vector a out of the integrals. Since a is

    arbitrary,

    S

    pn dS = V

    p dV.

    1.2.3 Stokes Theorem

    Let C be a simple closed curve in space, spanned by surface S with unit normal n,thenS

    ( w)n dS = C

    wdr.

    For the proof of this see appendix C.

    6

  • Example of verifying Stokes Theorem (Example 5) Verify Stokes Theorem for the vector field

    w = z2x + y2y + xzwhere the closed curve C is the triangle with vertices A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1). We first calculate the line integral going from A to B to C and back to A.

    AB z =0 x=1 y dr=dy(x + y) wdr= 2y + xz)(x + y)dy = y2 dy0

  • For larger a, the average density will change due to general variations of the fluid (eg temperature of gas, salinity of seawater etc). As a gets very small and approaches the molecular separation distance there will only be a small number of molecules in the cube and the average density will change significantly for small changes of a.For a small, but much larger than the molecular separation distance, the average desnsity is approximately independent of a and we define this as the density at point P. This is known as the Continuum Hypothesis. Other quantities such as fluid velocity are defined in a similar way. Typical valuesFor most liquids the density depends on temperature (but is almost independent of pressure see later). Typical values at room temperature and atmospheric pressure:

    WaterMercuryAir

    = 1000kg m3

    = 13, 500kg m3

    = 1.29kg m3 sensitive to temperature and pressure

    Nlying on the boundary of a region occupied2.2 PressureConsider a small element of area S with normal by

    a fluid.

    8

  • The fluid exerts a force, F on the area element perpendicular to the element,

    F = F N , F = |F |.The pressure, p on the area element is defined by

    p = limS0

    F

    S,

    where again the continuum hypothesis is applied (S is not taken so small that motion of individual molecules becomes important).For a general area S, the total force on S due to the fluid is

    F =

    SpN dS, Nis normal out of fluid.

    (Note that if the surface is curved, Nis in different directions at different points on S.)

    Example 6A water tank consists of a cube occupying the region

    0 x 1, 0 y 1, 0 z 1,and the pressure of the water inside the tank is given by p = 1 z (we will look later at how p can be calculated).

    Calculate the force on each face of the cube.Let n the normal out of the fluid.Face 1 z = 0 = z p = 1 F1 = 1

    0

    10

    (z)dx dy = z,Face 2 z = 1

    NN= z p = 0 F2 =0,

    Face 3 x = 0 N= x p = 1 z F3 = 10

    10

    (1 z)(x)dy dz,=x 10 (1 z)dz = x [z 12z2]10 = 12 ,x

    Face 4 x = 1 N= x p = 1 z F4 = 12 ,xFace 5 y = 0 = y p = 1 z F5 =21 y,Face 6 y = 1

    NN= y p = 1 z F6 = 12 y.

    9

  • 2.3 Equations of State

    The equation of state relates the pressure p and the density .For a liquid the density is constant, independent of pressure.Ideal Gas Laws:For an ideal gas, the pressure p, density and temperature T are related by

    p

    T= constant.

    If the temperature is constant then p . Other special cases also apply.

    2.4 Statics & Hydrostatics

    Statics: Consider a body of mass m suspended on a string.

    m

    mg

    g

    T z

    There is a an upward force due to tension in the string, T = T z and a force mg = mgz due togravity.If the body is in static equilibrium the total force on the body is zero,

    T +mg = 0, = T = mgz.

    Hydrostatics: Now consider a large volume of stationary fluid and a smaller region R of fluidwithin this volume. The surface of this region is taken to be S with n the unit normal out of R.

    10

  • Since the fluid is stationary, the weight of the fluid in the volume must be balanced by a force on R due to the pressure p of the surrounding fluid.From 2.2 the total force on R is

    F =

    S

    p N dS,

    where N is the unit vector normal to S and out of the surrounding fluid. Since N = n, applying the corollary to the divergence theorem (see 1.2) gives

    F = S

    p n dS = R

    p dV.

    If the density of the fluid is , the mass of fluid in region R is

    m =

    dV.R

    Since the fluid is stationary, the total force on the fluid inside R is zero and so

    g dV

    R

    p dV = 0.R

    This result must be true for all volumes and so

    p = g.Since g = gz, this vector equation is equivalent to the three scalar equations

    p

    x= 0,

    p

    y= 0,

    p

    z= g,

    and so p is a function of z only.If is known as a function of p from the equation of state (see 2.3) then p can be found as a function of z.

    Example 7A diver is distance h below the surface of water with constant density, = 1000kg m3.At the surface the air pressure is pa = 105N m

    2. Calculate the pressure experienced by a diver 10m below the surface. At what depth is the pressure three times atmospheric pressure?Defining the z-axis as being vertically up, with the surface of the water at z = 0, the pressure is given by

    dp= g, p(0) = pa.

    dzSince the density is constant, we can integrate to get p(z),

    p = gz + constant = pa gz.Thus for a diver at z = h, the pressure is pa + gh. When h = 10m,

    p = (105 + 1000 9.8 10)N m2 = 198000N m2 = 1.98 pa.When p = 3pa,

    3pa = pa + gh, = h =2pag 20.4m.

    11

  • Example 8In an isothermal atmosphere,

    p =p0

    0, p0, 0 pressure and density at sea level.

    If the 0 = 1.3kg m3 and p0 = 1000N m2, calculate the pressure at 10000m above sea-level.

    dp

    dz= g = kp, k = 0g

    p0 1.27 104m1.

    Solving by separation of variables,

    dp= k

    dz, = log p = kz + constant, = p = p0ekz.p

    So when z = 10000m, (approximately summit of Mt Everest),

    p = p0e1.27 0.28p0,

    the pressure is approximately a quarter of that at sea level.

    Measuring Pressures

    1. Simple Barometer

    Pressure depends only on z so

    pA = pB = pAtm.

    Vacuum at top of tube pC = 0.

    Hence choosing z = 0 at C, p

    = gz, pAtm = gh.

    If liquid is mercury, = 13, 500kg m3, pAtm = 105N m2 and so

    h =105 0.756m,

    9.8 1.35 104

    or 756 mmHg (millimetres of mercury).

    12

  • 2. Pressure differences using a manometer

    Consider a U-shaped tube attached to two vessels containing gases at different pressures. It is

    assumed that the density of the gases in the two vessels is much lower than the density of the

    fluid in the tube.

    pA = pC , pB = pD (ignoring density of gas)

    pE = pF , (same vertical level & no flow)

    pE = pC + g(EC), pF = pD + g(FD),

    pA = pE g(EC) = (pD + g(FD)) g(EC),= pB + gh.

    2.5 Buoyancy and Archimedes Principle

    DefinitionFor a body in a static fluid the buoyancy is the force due to the pressure of the surrounding fluid. Consider a body, volume V, completely submerged in a fluid of density . The buoyancy force isgiven by

    B =

    pn dS,S

    where S is the surface of the body, n is the unit normal out of the body and into the fluid, and p is the fluid pressure at the surface of the body.

    13

  • The fluid pressure is a function of z only, with the pressure at A the same as at B. Thus the pressure on S is the same if the body is replaced by fluid (figure (b)). So we can then apply the divergence theorem

    B=

    S

    pn dS = V

    p dV = zg V

    dV = V gz,=mW gz,

    where mW is the mass of fluid displaced by the body.This is Archimedes Principle the buoyancy force is equal to the weight of displaced fluid. The same result holds for a partially submerged body.

    V1 = Volume of water displaced W

    = Water density

    mB = Mass of body

    The buoyancy force is then B = W V1gz. Since it is floating, the buoyancy force is equal and opposite to the weight of the body, and so

    mB = W V1.

    So if the volume of displaced water can be measured and the density of water is known, the mass

    of the body can be found.

    2.6 One-dimensional flow in tubes

    We now consider flow through a tube (which may have varying cross-sectional area, and may not be level).We define s to be the distance along the tube from some fixed point, A(s) to be the cross-sectional area, and z(s) the height of the mid-point of the tube above some fixed level surface.

    14

  • We assume the density is constant, that the flow is parallel to the walls of the tube and that the average velocity across the tube is given by u(s). (NB As long as A(s) changes slowly, the assumption of parallel one-dimensional flow is valid.)We now consider a section of tube.

    In time t, the mass flow of fluid through A1 is tu1A1, while the flow out of A2 is tu2A2. Since the density of the fluid is constant, the total mass in = total mass out, so

    u1A1 = u2A2.

    A more general form of this equation will be derived in 3.4 and is known as the Mass Conservation Equation or Continuity Equation.Energy EquationFor a particle of mass m in a gravitational field, with velocity u and vertical coordinate z, we define:

    Kinetic Energy (due to motion) T = 12m|u|

    2 Potential Energy (due to position) V =mgz

    15

  • In MAT1036 (Classical Dynamics) last year it was shown that the total energy is conserved:

    2T + V = 1m|u|2 + mgz = constant.

    A similar relation can be derived for fluid motion along a pipe12|u|2 +gz + p =constant.

    KE PE PressureEffect

    This is a form of Bernoullis Equation.We do not derive this here a general derivation is given in 4.NB If gravity is ignored, then Bernoullis Equation says that at higher flow speeds the pressure is lower.

    Example 9: Canvas sided truck inside the truck the fluid velocity is zero, while outside the fluid velocity is high, so pressure is lower and sides bulge out.

    Returning to our example of a pipe, Bernoullis equation becomes

    12u

    21 + gz1 + p1 =

    12u

    22 + gz2 + p2,

    or

    p2 p1 = 12(u21 u22) + g(z1 z2).This is supplemented by the conservation of mass equation

    u1A1 = u2A2.

    Example 10 Venturi TubeFor a tube of cross-sectional area A1, a constriction of cross-section area A2 is inserted. Vertical tubes are attached to the pipe in the uniform section and at the constriction. If the difference in height between the fluid levels in the pipe is H we can find the fluid velocity in the tube using Bernoullis equation and the equation of mass conservation.

    16

  • Applying Bernoullis equation at the points P and Q, where the pressure is p1 and p2 respectively

    2 2p1 + gr1 + 1 u21 = p2 + gr2 +

    1 u22.

    Using the hydrostatic balance in the two vertical tubes and noting that the pressure at R and S isatmospheric pressure, pa,

    p2 = pa + gh2,p1 = pa + gh1,

    Eliminating p1 and p2 from Bernoullis Equation,

    12(u

    22 u21) = g ((h1 + r1) (h2 + r2)) = gH.

    Conservation of mass gives A1u1 = A2u2 and so(A21

    A22 1)u21 = 2gH, = u21 =

    2gHA22

    A21 A22 .

    Note:

    1. Bernoullis equation holds along a stream tube (or streamline) for steady flows and in theabsence of viscous (frictional) forces.

    2. For many flows, the whole space occupied by the fluid may be regarded as a stream tube, e.g.the wall of a pipe is a surface of a stream tube. In this case the solution is only approximate aswe have neglected viscous effects which will be important near the walls of the pipe, althoughtheir overall effect on these integral properties may remain small.

    17

  • 3 Kinematics

    Kinematics is the study of velocity and acceleration of fluid particles.

    3.1 VelocityIn Cartesian coordinates, we write the velocity of the fluid particle at a point with coordinates (x, y,

    z) asu(x, y, z, t) = ux + vy + wz.

    If the fluid velocity depends on time, t, we say the flow is unsteady, while if

    u= 0 = flow is steady. t

    The fluid speed is given by

    |u| = (u2 + v2 + w2)12 .Definitions

    1. A point at which u = 0 is known as a stagnation point.

    2. A flow is two-dimensional if

    u

    z=v

    z= 0, and w = 0.

    Examples

    11- u = xx yy + 2zz steady flow with stagnation point at (0, 0, 0).12- u = a(y t)x axy, a two-dimensional unsteady flow with stagnation point at x = 0, y = t.Polar Coordinates for 2-D flowsFor a general point P with Cartesian coordinates (x, y), the polar coordinates are (r, ) with

    x = r cos ,

    y = r sin ,

    r =cos x + sin y, = sin x + cos y.

    18

  • The expression for r comes from r/r, while is obtained by noting that r = 0.3.2 Describing flow fields

    If the fluid velocity u is known everywhere, there are a number of ways in which we can visualise theflow.In 2-D we can draw an arrow at various points illustrating the direction and magnitude of the flow.

    Example 13u = ayx + axy.

    In polars this isu = ar ( sin x + cos y) =

    ar. Hence the flow can be visualised by

    An alternative way of visualising the flow is by considering particle pathlines, streaklines andstreamlines. We provide a general definition of each line but only illustrate the ideas for 2-D flows.

    3.2.1 Streamlines

    A streamline is a curve which, at an instant of time, is tangent to the flow velocity at all points alongthe curve.Note that for an unsteady flow the pattern of streamlines changes with t.In simple cases the pattern of streamlines can be deduced from the direction of flow at each point.

    Example 13a see previous example.

    u = ayx + axy = ar.Streamlines are concentric circles.

    19

  • Mathematical FormulationConsider the streamline passing through the point x = xx + yy + zz = x0, and choose s to be a coordinate along the streamline with s = 0 when x = x0.NB s is any coordinate along the curve, not necessarily the distance from x0.

    Since the tangent to the curve is parallel to the velocity field,

    dx= u, for some .

    ds

    Choosing the scaling on s correctly we can set = 1 and so we must solve

    dx

    ds= u(x), x(0) = x0.

    Writing u = ux + vy + wz, an alternative way of writing this differential equation isdx

    u=dy

    v=dz

    w(= ds),

    from which it follows thatdy

    dx=v

    u,

    dz

    dx=w

    u, etc.

    Example 13b see previous example.

    u = ayx + axy.Hence,

    dy

    dx=v

    u= ax

    ay.

    Solving by separation of variables

    y dy =

    x dx, = y2

    2= x

    2

    2+ C, = x2 + y2 = constant.

    So the streamlines are concentric circles centred on the origin.

    20

  • Example 14u = x +

    ty. The streamlines at time t = t0 are given bydy

    dx= t0

    ie straight lines with gradient t0.

    t < 1 t > 1

    3.2.2 Particle Pathlines

    Consider a fluid particle at point P (x0, y0, z0) (position vector x0) at some fixed time t = 0 (say).As t increases, the fluid particle moves through space the path taken is known as the pathline. Byplotting a set of pathlines, the flow pattern can be visualised.NB As t increases, the picture only changes in that the pathlines get longer.

    Example 15u = ayx,

    21

  • t = t1 t = t2 > t1

    Mathematical Formulation

    dx= u (x, t) , x(t = 0) = x0.

    dt

    For each x0 chosen, there is a particle

    path.

    Example 16u = ayx axy.

    The equation governing the particle path passing through the point (x0, y0) at t = 0 is

    dx

    dt= ay,

    dy

    dt= ax, x(0) = x0, y(0) = y0.

    Solving these two linked differential equations is easiest by differentiating the first equation andeliminating y using the second equation,

    d2x

    dt2= a

    dy

    dt= a2x.

    Solvingd2x

    + a2x = 0 = x = A cos(at) + B sin(at).dt2

    Then using the initial conditions we obtain

    x=x0 cos(at) + y0 sin(at),

    y=y0 cos(at) x0 sin(at),

    Note thatx2 + y2 = x20 + y0

    2,

    and so the pathlines are circles.

    22

  • Example 17u = yx (x bt)y. Prove that particle path through (x0, y0) are given by

    x=(y0 b) sin t + x0 cos t + bt,y=(y0 b) cos t x0 sin t + b.

    Results from MATLAB for b = 1 for the particle paths through (0, 0.5) and (0, 0).

    0 2 4 6 8 10 12 140

    0.5

    1

    1.5

    2

    Example 18u = x + ty.

    Calculate the particle path for a particle at position (x0, y0) at time t = .

    dx

    dt=1, = x = t + C1 = t + x0, (using x() = x0)

    dy

    dt 2=t, = y = 1 t2 + C2 = 1 2(t2 2) + y0, (using y() = y0).

    Hence the particle path for a particle at the origin at t = 0 is

    x = t, y = 12t2, or equivalently y = 12x

    2.

    3.2.3 Streaklines

    A common technique for visualising flow in experiments is to release smoke or dye at a point and seehow it is carried by the flow (eg smoke from the top of a chimney). The position of these markedparticles at a particular time is known as the streakline. For an unsteady flow, the streakline changesposition as well as lengthening as t increases. For a steady flow the streakline will lengthen as tincreases but not change shape.Mathematical FormulationSolving

    dx

    dt= u(x, t), x(t = ) = x0.

    23

  • This gives the position of a particle x as a function of t, (release time) and x0 (release position) x(x0, t, ).The streakline at time t0 is then given by the set of points

    x(x0, t0, ), 0 t0.

    Example 19From example 18 of previous section, if

    u = x + ty,

    then the particle pathlines for a particle released from the origin at time t = are

    x = t , y = 12(t2 2).

    Thus the streakline at time t0, originating from (0,0) is given by the parametric curve

    x = t0 , y = 12(t20 2), 0 t0.

    The parameter can be eliminated to give

    y = xt0 12x2, 0 x t0.

    The streaklines at t0 = 1 and t0 = 2 are plotted on the handout.

    24

  • Streaklines and Pathlines

    u = x + ty,

    0 0.2 0.4 0.6 0.8 10

    0.1

    0.2

    0.3

    0.4

    0.5

    Streakline at t0 = 1, path lines for particles released at = 0, 0.5.

    0 0.5 1 1.5 20

    0.5

    1

    1.5

    2

    Streakline at t0 = 2, path lines for particles released at = 0, 0.5, 1.0, 1.5.

    25

  • Example 20u = sin(t)x + y.

    Slope of streamlines is given bydy

    dx=

    1

    sin(t).

    Pathlines Consider particle released from origin at time t = 0.

    dxdt

    =sin(t), x(0) = 0, = x(t)= 1cos(t)

    ,dydt

    =1, y(0) = 0, = y(t)=t.This is plotted as a parametric curve below.Streaklines Consider the position of a particle at time t which was released from origin attime t = .

    dxdt

    =sin(t), x() = 0, = x(t)= cos() cos(t)

    ,dydt

    =1, y() = 0, = y(t)=t .The streakline at t = 2pi/ is then given as a parametric equation,

    x() =cos() 1

    , y() =

    2pi

    .

    0 0.05 0.1 0.15 0.2 0.25 0.30

    0.5

    1

    1.5

    2

    0.35 0.3 0.25 0.2 0.15 0.1 0.050

    0.2

    0.4

    0.6

    0.8

    1

    Pathline for particle at origin at t = 0 with = 2pi. Streakline at t = 2pi/ for particlesreleased from origin at 0 < < 1.

    26

  • 3.3 Convective Derivatives

    Consider a fluid in motion with density and velocity u depending on time and position. If weconsider a small blob of fluid, we examine how the density of this blob changes as it moves throughthe fluid. We first consider a 1-D flow field. Then a blob centred at P x at time t, is at point Q x+ u(x, t)t at time t + t. Hence the density of the blob at time t + t can be written as

    Q = (x + u(x, t)t, t + t) (x, t) + (t)

    t+ (u(x, t)t)

    x+ . . . ,

    as t 0, using a Taylor expansion about (x, t). Hence the rate of change of as we move withthe fluid is

    Q Pt

    t

    x= + u(x, t) + Small Terms.

    This is known as the convective derivative of (also known as the material derivative),written as

    D

    Dt=

    t+ u(x, t)

    x.

    This is readily extended to a 3-D velocity field,

    D

    Dt=

    t+ u , (3.3.1)

    where the convective derivative is the operator

    D

    Dt=

    t+ u .

    3.4 Mass Conservation and CompressibilityBy considering flow in and out of a fixed volume, we can derive the Conservation of Mass

    Equation

    t+ (u) = 0,

    or equivalently,D

    Dt+ u = 0.

    Proof

    Consider a volume of fluid V fixed in space, with surface S. In time t, the mass flow through a smallelement of surface S with normal n is given by unS t, since un is the component of velocity normal to the surface.Then the total mass flow into V through the surface S is given by

    tS

    un dS,the minus sign arising because n is the normal out of the

    volume. The change in mass contained in volume V is then

    m =V

    dV = t

    V

    tdV,

    27

  • and so V

    tdV =

    S

    un dS,=

    V

    (u) dV, (using divergence theorem).

    This is true for all volumes hence result.For an incompressible fluid, the density is constant as we move with the fluid, so

    D

    Dt= 0, = u = 0.

    Throughout this course, all fluids are assumed to be incompressible, so

    u = 0.

    This is a good approximation for water and also for air at low speeds (M = u/c 1, where M isthe Mach number and c is the speed of sound).

    Example 21a,b constants.u = (2x + y)x + (ay + 2yz)y +

    bz2z, Find a and b such that the flow is incompressible.u = 2 + (a + 2z) + 2bz = (2 + a) + (2 + 2b)z.

    Need u = 0 for all z and so need2 + a = 0 = a = 2 AND 2 + 2b = 0 = b = 1.

    3.5 VorticityDefinition Vorticity

    The vorticity of a flow is given by = u.

    Note that this is a vector quantity.

    Example 22Calculate the vorticity of the flow with velocity

    u = 2yx + zy + 4xz. u =

    x

    y z

    x

    y

    z2y z 4x

    = (0 1)x 4y 2z.

    28

  • Vorticity of 2-D flowConsider

    u = u(x, y)x + v(x, y)y,then

    = u =x uy

    (v )

    z = z.So for any 2-D flow, the vorticity is normal to the plane of the flow.

    Definition Irrotational

    A flow is irrotational = 0, u = 0.

    Example 23If u = yz2x + xz2y + 2xyzz, show that the flow is irrotational.

    u =

    x

    y z

    x

    y

    z

    yz2 xz2 2xyz

    = (2xz 2xz)x (2yz 2yz)y + (z2 z2)z = 0.

    3.6 Velocity Potential for Irrotational FlowsTheorem from Vector Calculus

    For any vector field f,

    f = 0 There exists a scalar field such that f = .

    Note that it is easy to show that f = implies f = 0.The reverse implication (ie that a can be found) is proved by constructing in terms of thecomponents of u.

    Hence for an irrotational flow, u, there exists a scalar field (x, y, z) such that

    u = ,and is known as the velocity potential.

    Incompressible and Irrotational FlowsIf the flow is irrotational, then u = . If the flow is also incompressible

    u = 0, = () = 2 = 0,

    so satisfies Laplaces Equation.

    29

  • Example 24If = xy + yz3, calculate u and show directly that flow is irrotational.

    = x x+ yy + zz = yx + (x + z3)y + 3yz2z.

    To show that flow is irrotational

    u =

    x

    y z

    x

    y

    zy x + z3 3yz2

    = (3z2 3z2)x (0 0)y + (1 1)z = 0.

    Example 25If = 3x2yz y3z calculate u and show that the flow is incompressible.

    u = =6xyzx + (3x2z 3y2z)y + (3x2y y3)z,u=6yz 6yz + 0 = 0.

    To show that flow is incompressible without first calculating u,

    2 = 6yz 6yz = 0.

    Example 25aGiven

    u = 6xyzx + (3x2z 3y2z)y + (3x2y y3)z,calculate .We have

    u =

    x x+ yy + zz = 6xyzx + (3x2z 3y2z)y + (3x2y y3)z,and so

    x

    y= 6xyz, = 3x2z 3y2z,

    z= 3x2y y3.

    Integrating the first of these equations with respect to x we have

    = 3x2yz + f(y, z),

    where f(y, z) is an arbitrary function of y and z at this stage. Substituting into the secondequation,

    y

    (3x2yz + f(y, z)

    )= 3x2z 3y2z, = f

    y= 3y2z, = f(y, z) = y3z + g(z).

    The function g(z) is determined by substituting = 3x2yz y3z + g(z) into the third equation,

    z3x yz y z + g(z)( 2 3 )

    3x2y y3 + gz

    =3x2y y3,

    =3x2y y3,

    30

  • and so g(z) is a constant, giving

    = 3x2yz y3z + C, (compare with previous example)(NB The velocity potential is only ever defined up to an additive constant. Whatever the value of

    the constant chosen, u = is unchanged.)

    31

  • 4 Dynamics

    4.1 Eulers Equation of Motion

    Consider a blob of fluid occupying a small volume centred on the point P .

    DuAcceleration of Fluid Blob = .

    Dt

    We then relate this acceleration to the forces acting on the blob using Newtons Second

    Law Mass x Acceleration=Total Force Applied.

    We assume that the only forces applied are due to gravity and the pressure of the surrounding fluid, and so

    V

    Du

    DtdV =

    g dV,

    =

    pn dS + S

    p dV + V V

    g dV, using divergence theorem.V

    This is true for all volumes V and so

    Du

    =p + g, g = gzu

    t

    Dt

    + (u)u= 1p + g.

    This is known as Eulers Equation (or Eulers Equation of Motion).

    Notes:

    1. We have ignored other forces acting on the fluid particles, in particular neglecting the effect of viscosity (which can be thought of as friction between fluid particles due to their relative motion). The full equation of motion including this effect and more general externally appliedforces is

    (u + (u)u

    )= p + F + 2u,

    t

    where F is the body force per unit mass and is the coefficient of viscosity.

    2. In many fluid flows viscosity is small. If the viscosity of the fluid is set to zero ( = 0), the flow is said to be INVISCID, in which case Eulers equation applies. In this course we only consider inviscid flows.

    3. For an incompressible, inviscid flow we have four scalar equations

    ut + (u)u = 1p + g, (1 vector equation = 3 scalar equations)u = 0, (1 scalar equation)

    with four unknowns the three components of the velocity vector, and the pressure p.

    32

  • 4.2 Boundary ConditionsA fixed, impermeable surface is surface fixed in space through which no fluid can pass.

    At a fixed, impermeable surface the normal component of fluid velocity on the surface is zero.

    un = 0, nis normal to surface.Hence flow must be tangential to a fixed, impermeable surface.

    Example 26If u = yx + xy show that the flow is incompressible and irrotational and satisfies the boundary conditions on the impermeable surfaces y = x.

    u = (1 1)z = 0.u = 0 + 0, The normal to the line y = x is

    n= 12(x + y),xsince the normal must be perpendicular to + y (the tangent to the line).

    Then on y = x,

    2

    ( 1 )un = x(x +

    y) (x + y) = 0.xSimilarly can show that normal to other surface is parallel to + y.

    4.3 Bernoullis EquationIn an earlier section we suggested that fluid flow must obey some form of energy equation. We wrote

    down

    2p + 1 |u|2 + gz = constant,

    without fully explaining how the contribution to the energy from pressure arises, and without explaining when the equation is valid.

    33

  • We now derive this result for an inviscid flow starting from Eulers equation. We first note that from 1, for any vector field f,

    f ( f) = 2f2(1 ) (f)f.Hence setting f = u (fluid velocity) we have f = (vorticity) and

    (u)u = 2u2(1 ) u ,and Eulers equation becomes

    u u = (p

    + 12 |u|2 + gz).

    t

    We now consider two separate

    cases:4.3.1 Irrotational Flow

    In this case = 0 and we can write the velocity field in terms of the velocity potential u = ,()t

    = p

    + 12()2 + gzso we obtain ( )

    .

    Or equivalently

    (t

    +p

    + 12()2 + gz

    )= 0,

    which implies that

    p + (t

    + 12()2 + gz)

    = constant,

    throughout the fluid.If instead we consider a general conservative body force, this is one such that F = then

    p + (t

    + 12()2 )

    = constant. (4.3.2)

    Gravity is a conservative body force F = gz, = gz.

    34

  • Example 27If u = (etx, ety, 2etz) and p = p0 at the point (0, 0, 0), show that the flow is irrotational and incompressible, and find the pressure everywhere, in the absence of a body force.

    u=et

    x

    y z

    x

    y

    z

    x y 2z

    = 0,

    u=et(1 + 1 2) = 0.

    Hence the flow is irrotational and incompressible. Next we must construct the velocity potential,

    xetx=

    = = 12 x

    2et + f(y, z, t),

    ety= y

    = f + 0=ety,y

    f(y, z, t)= 12 y

    2et + g(z,

    t),= 1

    2 x2et + 12y

    2et + g(z, t),

    2etz = z =g

    + 0=2etz, z

    g(z, t)=z2et + h(t).Combining these results gives

    =1

    2

    (x2 + y2 2z2 ) et + h(t).

    (NB. We can show that 2 = 0) The unsteady pressure equation gives

    p=

    ( 1

    ()2 2 t)

    + F (t),

    =

    (0 1

    2(x2 + y2 + 4z2)e2t +

    1

    2(x2 + y2 2z2)et

    )+ F (t) h(t).

    And p = p0 at (0, 0, 0) which fixes F (t) h(t) = p0.

    35

  • 4.3.2 Steady Flow

    If the flow is steady we have

    u = (p

    + 12 |u|2 + gz).

    Taking the dot product of this equation with u,

    u(p

    + 12 |u|2 + gz)

    = 0.

    Recalling that is normal to the line =constant, we see that vector u is tangential to the line =constant where

    =p

    + 12 |u|2 + gz,

    and so this line is a streamline. Hence

    p + (12|u|2 + gz) = Constant along a Streamline,

    for any steady flow. However if the vorticity is non-zero, then the constant is different along different streamlines.

    36

  • 4.4 Momentum Integral Theorem for Steady Flows

    Starting from Eulers equation for a fluid in a gravitational field

    (ut

    + (u)u)

    = p gz.We can integrate over a fixed control volume V to give

    V

    u

    tdV =

    V

    (p + (u)u + gz) dV=

    pn dS (u)u dV Mgz, M is mass of fluid in volume.

    S V

    If we now consider a steady flow with a constant everywhere,

    pn dS = (u)u dV Mgz.S V

    For any scalar field and any vector field f such that f = 0,

    V

    f dV =

    V

    ((f) f) dV,

    =

    fn dS, using divergence theorem.

    S

    From this it follows that for any g and f such that f = 0,

    V

    (f)g dV =

    S

    g(fn) dS,(taking the x, y and z components separately).

    Thus for a steady flow with constant density, for any fixed volume

    F = pn dS = S

    u(un) dS Mgz,S

    where F is the total force on the fixed volume.

    NB The total force on a volume due to a constant pressure is zero since

    pan dS = pa dV = 0, for constant pa.S V

    Hence we can set atmospheric pressure to be zero when using this result for real problems.

    37

  • Example 28Consider a jet of water directed at right angles to a flat plate.If the flow rate and velocity of the incoming jet are Q and U respectively, calculate the force on the plate, ignoring the effect of gravity.Let area of incoming jet be A and assume that the fluid velocity through this incoming jet is constant, so Q = UA and the pressure in the jet is approximately constant and equal to atmospheric pressure (which we take to be zero).

    We split the surface of the control volume into 4 parts:

    1. Circle with area A through incoming jet far from plate

    2. The curved surface of the jet

    3. A disc of radius R on the plate

    4. The rim of the disc, radius R through the fluid spreading over the plate.

    We take the radius of the disc large enough that the flow is virtually parallel to the plate as it passes out of the control volume through surface (4).Considering each of the surfaces in turn,

    On 1pn dS =0 (since p = pa = 0)

    uu( n) dS =AU2x (since u = Ux, n = x)On 2 pn dS =0uu( n) dS =0On 3 pn dS =F x

    (since p = pa = 0)(since un = 0)(F is force on disc, tends to total force on plate as R .)

    u(un) dS =0 (since un = 0)On surface (4), the contributions of both integrals are zero by symmetry. Hence using the momentum integral theorem we have

    F = AU2 = QU.

    38

  • 4.5 Calculating Velocity Potential for general flows

    If the flow is irrotational and incompressible, then we obtain Laplaces Equation,

    2 = 0,

    this follows since 0 = u = () = 2. So to determine we must solve this equation with the appropriate boundary conditions.

    Boundary conditions on For a solid, fixed boundary un = 0, and so

    n= 0 on S,

    using the identity n = /n.For example, if the surface is a cylinder, radius a, centred at the origin, then the normal to the surface is in the radial direction and the boundary condition becomes

    r= 0 on r = a.

    4.5.1 Uniqueness of velocity potential

    1. Uniqueness theorem

    There cannot be two different forms of irrotational motion for a confined mass of fluid with

    prescribed velocities on the boundaries.

    Proof

    Assume there are two different solutions 1 and 2. On the boundary surface S,

    1n

    =2n

    ,

    as is prescribed.nLet = 1 2, then

    2 = 21 22 = 0 0 = 0.i.e. 2 = 0 in V and

    n=1n 2

    n= 0,

    on S. Now noteV

    () dV =S

    () n dS by Divergence theorem,=

    S

    ( n) dS,=

    S n

    dS.

    Note that () = () + = 2 + ()2,

    39

  • and so V

    2 dV +

    ()2 dV = dS,

    i.e.

    V

    ()2 dV =S

    ndS

    V

    V S n

    2 dV, Greens theorem.

    Thus, V

    ()2 dV =S

    ndS

    V

    2 dV,=0.

    As ()2 0 this integral implies that = 0. = constant,

    1 = 2 + constant.

    i.e. solutions 1 and 2 are identical (to within an arbitrary constant) and hence the solution for u is unique.

    Corollary

    If the boundaries S are at rest then irrotational motion is possible.

    Proof

    So

    =0 on S

    n2=0 in V.

    Therefore a solution is = constant, but by the above theorem this is the unique solution.

    If = constant = u = = 0 so no flow.

    2. Theorem

    If there is a given distribution of vorticity in the fluid and the normal boundary velocities are

    given then the flow is uniquely determined.

    Proof

    Assume there are two possible flows u1 and u2 (No potential as 6= 0). Let u = u1 u2 then

    u = u1 u2 = 0, by conservation of mass for an incompressible fluid.

    u = u1 u2 = 1 2 = 0,

    40

  • as prescribed vorticity. Therefore u = and then 2 = u = 0. On the boundary u n = u1 n u2 n = 0 as given on the

    boundary. n

    = 0 on S.

    Also 2 = 0 in V . Therefore represents an irrotational motion with boundaries at rest. Hence by the above corollary, = constant in region V .

    u = = 0,

    and sou1 = u2. i.e. unique.

    41

  • 5 Channel Flows

    5.1 Steady flow under a Sluice Gate

    Consider steady flow of depth h1 and flow speed u1 (independent of depth) far upstream of a sluice gate.We assume that downstream of the sluice gate the flow has depth h2 and uniform flow u2.

    By mass conservationu1h1 = u2h2.

    Applying Bernoulli along the surface streamline we have

    12u

    21 + gh1 =

    12u

    22 + gh2,

    since the pressure is atmospheric on the free surface.We will use the momentum integral theorem, on the volume indicated so need to find pressure on the surfaces (1) and (5). On surface (1) the flow is uniform and hence irrotational, so we can apply the unsteady Bernoulli equation (4.3.2),

    p + (12u

    21 + gz

    )= pa +

    (12u

    21 + gh1

    ), = p = g(h1 z),

    taking pa = 0 as explained earlier. Similarly on (5), p = g(h2 z). Considering each of the surfaces in turn, we consider the component of

    S

    (u(un) + pn) dS,in the x direction.

    42

  • On 1pn dS =x h10 g(h1 z) dz = 12 gh21 x

    u(un) dS =h1u21x (since u = u1 = x)On 2 and 4

    pn dS =0 (since p = pa = 0)

    u(un) dS =0 (since un = 0)On 3

    pn dS =T x (T is force on gate.)

    u(un) dS =0 (since un = 0)On 5

    pn dS =x h20 g(h2 z) dz = 12gh22 x

    u(un) dS =h2u22x (since u = u2

    n

    n

    x,

    x,

    = x)On 6 x pn dS =0 n(since = z)

    u(un) dS =0 (since un = 0)Substituting these results into the momentum integral theorem we find that the total force on thegate is

    T = 12g(h21 h22) +

    (h1u

    21 h2u22

    ).

    43

  • 5.2 Hydraulic Jumps

    In many flows we see a rapid change in the surface accompanied by turbulent flow. For example,water from a tap falling onto a horizontal plate forms a disc of shallow fast flow with a jump in thesurface at a particular radius. A similar effect is often seen at the bottom of weirs or downstreamof sluice gates.

    These are known as hydraulic jumps and typically stay in a fixed position. When a strong tidalsurge enters a narrowing river estuary a similar jump occurs, but moves upstream.

    These are usually known as hydraulic bores, but much of the theory is the same for the two cases.Working out the position of such a jump depends on the physical problem and is typically verycomplicated. Here we only consider the relationship between the fluid velocity and water deptheither side of the jump, using methods similar to the example of flow underneath a sluice gate.

    44

  • The equation of mass conservation gives u1h1 = u2h2.Applying the momentum integral theorem, there is no external force acting on the water (compareto sluice gate problem, where the gate exerted a force on the fluid). Hence

    12g(h21 h22) +

    (h1u

    21 h2u22

    )= 0.

    However, in this case we can not apply Bernoullis equation along the surface streamline, since weknow that the flow is turbulent in the region of the jump, and so energy is lost by the fluid (lostin form of sound or heat).Combining these equations we can obtain u2 and h2 in terms of u1 and h1.

    5.3 Flow over obstacle on river bed

    Consider a flow of depth h and speed U encountering a small obstacle on the bed given by z = hb(x),with 0 b(x) 1 and b(x) 0 as |x| .Let the surface of the water be given by z = h+hs(x), with |s(x)| 1 and s(x) 0 as x .

    Letting the velocity at x be u(x), the depth at x is h+hs(x)hb(x), so conservation of mass givesUh = u(x)h (1 + s(x) b(x)) , = u = U(1 + s b)1.

    45

  • Applying Bernoulli on the surface streamline,

    pa + (

    12U2 + gh

    )= pa +

    (12u2 + gh(1 + s)

    ),

    since the pressure is atmospheric at the surface. Eliminating u,

    U2 + 2gh = U2(1 + s b)2 + 2gh(1 + s), = F 2 = F 2(1 + s b)2 + 2s,

    where

    F =Ugh, is known as the Froude Number.

    Since s and b are both small, we can use the binomial expansion,

    (1 + s b)2 = 1 2(s b) + . . . .

    to finally giveF 2 = F 2(1 2s+ 2b+ . . .) + 2s, = (F 2 1)s = F 2b.

    Since b > 0, if F > 1 corresponding to a fast, shallow incoming flow, then s > 0 and the flow risesover the bump. If F < 1 (a slower deep flow), the flow dips over the bump.

    F > 1 F < 1Supercritical Subcritical

    46

  • 6 Two-dimensional Flows

    When solving two-dimensional flow fields,

    u = u(x, y)x + v(x, y)y,simplifications can be made to the analysis. In particular, we can find an easier way of calculating streamlines.

    6.1 The stream function

    For any 2-D incompressible flow, we can write the velocity in the form

    u =

    y, v =

    x,

    where (x, y) is known as the Stream Function. NB

    1. This definition is equivalent to u = ((x, y)z).2. The property of incompressibility is immediately satisfied,

    u = x

    (y

    )+

    y

    (x

    )= 0.

    6.1.1 Properties

    1. =Constant is a streamline.

    = (x,y, 0) = (v, u, 0).Hence,

    u= 0. (6.1.3)Now, is always perpendicular to the line=constant (see MAT1005). Also, the vectors , u and the line=constant all lie on the x, y plane.

    47

  • From (6.1.3), u is perpendicular to and henceu is tangent to =constant.

    2.|| = (y2 +x2)

    12 = (u2 + v2)

    12 = |u|.

    3. The magnitude of u is inversely proportional to the distance between streamlines.

    = c

    = c

    u2

    d du1

    1

    1

    2

    2

    1Conservation of mass = u1d1 = u2d2, i.e. u . d

    This can be stated more precisely. For any line, , joining two points x and y,

    Flux of volume across =(y) (x).Proof

    Flux=

    un dl, n=

    (y,

    x

    ) (dy, dx) =

    dl = (dy, dx),

    (ydy +

    xdx

    ),

    =

    d=(y) (x).

    4. In plane polars,(r, ),

    ru =

    ur + u, ur = 1r

    , u =

    r.

    48

  • 6.1.2 Governing Equations and Boundary Conditions

    v

    x = 0, 0, u

    y

    Recall, for a two-dimensional flow, the vorticity is given by( ),

    so we can write = z, where =

    v

    x uy.

    Substituting the expressions for u and v in terms of from (6.1) we have

    = 2

    x2

    2

    y2,

    and hence, the stream functionsatisfies,

    2= .

    Boundary Condition on

    Solid fixed boundary ===

    No flow normal to surface, Surface is a streamline, =Constant on surface.

    6.1.3 Examples

    Given the velocity potential , find the stream functionand plot the streamlines.

    (29) =1

    2(x2 y2), (30) = K

    2pilog r.

    Example 29

    =1

    2(x2 y2), u = (u, v) = = (x, y).

    Also, u =(y,

    x

    )and so

    u =

    y= x =xy + f(x),

    v = x

    = y =xy + g(y),

    where f(x) and g(y) are undetermined functions. However, for both these results to be true f(x) = g(y) =constant,

    = xy + constant.

    An alternative method is to show that= xy + f(x) from the first equality, then substitute this into the second equation, hence proving that f(x) = 0.Streamlines are given by =Constant. Hence the streamlines are rectangular hyperbola, xy =Constant.

    49

  • For the direction of flow along the streamlines, use the expression for u. In this case u = x, so u > 0 when x > 0 and u < 0 when x < 0. Always try to mark direction of flow on streamlines with arrows.NB Same flow is possible if any streamline is replaced by a solid, fixed boundary. In this case, x = 0 is a streamline, so the flow in the upper half plane is like that due to a jet directed normal to a flat wall (x = 0).

    Example 30A 2-D source, using polar coordinates (r, ).

    =K

    2pilog r,

    u = = K2pirr.

    Writing the velocity in terms of the stream function,

    u =1

    r

    r r .Hence

    1

    r

    =

    K

    2pir

    = 0

    =K

    2pi(+Constant).

    r

    The streamlines,=Constant, correspond to the lines =Constant, which are straight lines directed out from the origin.

    50

  • Flow directions on the streamlines are marked for K > 0, ie outward flow from a source. The flux across the circular contour r = a is given by

    0 2piaFlux = u n dl = r ra d = K

    2pi

    2pi ( K ) 2pi0

    d = K.

    So the velocity potential represents a source of strength K at the origin.

    6.2 Line Vortices

    6.2.1 Velocity potential of line vortex

    Consider a flow with velocity potential

    =K

    2pi, K = constant.

    Then

    u=

    r r+ 1r ,=K

    2pir.

    Hence the velocity is all in the direction and |u| as r 0. Calculating the stream function, since

    u =1

    r

    r r ,we have

    1

    r

    = 0,

    r=

    K

    2pir,

    and hence

    = K2pi

    log r.

    The streamlines are given by=constant which corresponds to r =constant, i.e. concentric circles centred on the origin. This flow is described as being due to a line vortex at the origin. For the case K > 0 the streamlines are:

    51

  • 6.2.2 Circulation

    =

    Definition

    Given a velocity field u and a curve C in space, the circulation of the flow around the curve C is defined to be

    C

    udl.

    Example 31For the line vortex with u =

    K ,

    2pir

    calculate the circulation around the curve C given by r = a (ie a circle radius a centred on the

    origin).On C we have dl = (a d), and so

    = 2pi

    0

    ( K

    2pia

    )(a) d = K.

    Thus the circulation round a circle centred on the origin is independent of the radius of the circle. The velocity potential

    = , = constant.

    2pirepresents flow due to a vortex of strength (or circulation) at the origin.

    6.2.3 Relation between vorticity and circulation

    For a 2-D flow, consider a closed curve C on the x, y plane, with A the area enclosed by C, using the definition of circulation and Stokes Theorem

    =C

    udl =A

    ( u) zdS = A

    dS,

    since for a 2-D flow = u = z.52

  • Example 32Consider line vortex strength 1 at origin. Flow is irrotational except at the origin where |u| .Hence by previous result the circulation round any curve not including the origin is zero.We showed in previous example that circulation round circle of radius a is 1.Hence we can consider the line vortex to be the limit of a circle of radius a and vorticity 1 witha 0 and 1 , with pia21 = 1.This serves as a model for a concentrated patch of vorticity as may be seen in a river flow.

    NB

    1. It can easily be shown that the circulation round any simple closed curve surrounding the avortex of strength 1 at the origin is 1.

    2. For a system of line vortices of strength i located at (xi, yi), then the circulation round acurve C is

    =

    i, sum over vortices inside curve C.

    The analogue of circulation in 3D is, given curve C in 3-D space, spanned by any surface S, theCirculation round C is

    =

    udl =

    n dS.

    6.2.4 Kelvins Circulation Theorem Inviscid Flows

    Consider a flow with constant density and let C be a closed simple curve at time t = 0. As timeincreases the particles lying on the curve will move we let the C(t) be the curve described bythese particles at time t.Kelvins Circulation Theorem states that the circulation round the curve C(t) is independent oftime.For the proof of this theorem see appendix D

    6.2.5 Significance of Irrotational Flows

    If at some time a flow is irrotational, and the subsequent flow is inviscid, then the flow will beirrotational at later times.ProofConsider a small closed curve C at time t = 0. Since the flow is irrotational at t = 0, then thecirculation round C is zero. Since the circulation is constant, the circulation round the curvecarried with the fluid is zero at later times and so the vorticity inside the small curve is zero.NoteIncluding viscosity can lead to generation of vorticity in a flow. For example consider a flat platein a uniform flow initially in line with flow, then rotated. Vortices shed due to viscous effects atsharp edges. Same principle for spoon impulsively moved in liquid.

    6.3 Combining simple examples of 2D velocity potentials

    As the equation of motion (ie Laplaces Equation) is linear, we can superpose solutions. We canmake more complex flow fields by combining some of velocity potentials which we have alreadyseen. Below are just a few examples, and more will be seen during the course.

    53

  • 1. A source at the origin in a uniform flow

    = Ux +k

    2pilog((x2 + y2)1/2).

    2. A source at (x0, y0) in a uniform flow

    = Ux +k

    2pilog(((x x0)2 + (y y0)2)1/2).

    i.e. shifted the origin.

    3. Source at r1, strength k and a sink at r2, strength k.

    =k

    2pilog R1 (Source at r1) R1 = |r r1|

    k2pi

    log R2 (Sink at r2) R2 = |r r2|.

    Here r is a general position vector, e.g. r = xx + yy in Cartesians.Example 33Consider a source, strength K at (x0, 0) and a sink strength K at (x0, 0). Write down the velocity potential (x, y).Find the velocity at (0, 2x0).Find the approximate value of (x, y) as x0 0.Velocity potential

    =K

    4pilog((x x0)2 + y2

    ) K4pi

    log((x + x0)

    2 + y2).

    Velocity is given by u = so

    u=( K2pi

    x x0(x x0)2 + y2

    K2pi

    x + x0(x + x0)2 + y2

    )x

    +( K2pi

    y

    (x x0)2 + y2 K

    2pi

    y

    (x + x0)2 + y2

    )y.

    So at (0, 2x0)

    u=K

    2pi

    ( x0

    5x20 x05x20

    )x+ K2pi (2x0

    5x20

    2x05x20

    )y,

    = K5pix0 .x

    As x0 0

    ((x x0)2 + y2 =(x2 + y2) 2x0 + x

    20,

    =(x2 + y2) 1 2x0xx2 + y2

    +x20

    x2 + y2

    ).

    54

  • Similarly

    (x + x0)2 + y2 = (x2 + y2)

    (1 +

    2x0x

    x2 + y2+

    x20x2 + y2

    ).

    So,

    =K

    4pi

    ( (1 2x0x

    x2 + y2+

    x20x2 + y2

    ))K

    4pi

    log(x2 + y2) + log(log(x2 + y2) + log

    (1 +

    2x0x

    x2 + y2+

    x20x2 + y2

    )).

    Using log(1 + X) X + ...... for |X| 1 this simplifies to

    K4pi

    ( 2x0xx2 + y2

    2x0xx2 + y2

    )+ ....

    Kx0pi

    x

    x2 + y2.

    6.4 General solution of velocity potential in plane polar coordinates

    Consider a velocity potential (r, ),

    r

    r

    (r

    r

    )1

    r22

    22 = 1 + = 0.This can be solved using separation of variables. Try = f(r)g(),

    (rf)gr

    +gfr2

    =0,

    r(rf)

    f+g

    g=0.

    The first term depends only on r, the second only on and hence each term must be equal to a constant (independent of r and ). So

    g

    g= 2, g = A cos + B sin , constant

    but g() must be 2pi-periodic and so = n, an integer. [NB If g/g is taken as equal to a positive constant, g() would not be 2pi-periodic.]The r equation then becomes

    r(rf) n2f = 0.Trying f rm we obtain m2 = n2, ie m = n. The general solution is then given by

    (r, )=n=1

    [rn (An cos n + Bn sin n) + r

    n (Cn cos n + Dn sin n) ] ,

    +A0 + B0 log r,

    where the final two terms correspond to an additional flow due to a line vortex and a point source.

    55

  • Example 34Find the velocity potential describing flow past a cylinder r = a, with zero circulation, subject to uniform flow speed, U, far from the cylinder.

    Boundary conditionsUx = Ur cos ,r

    =0 r = a.

    rUsing the boundary condition at r , together with the zero circulation condition, the only term with a positive power of r is Ur cos and so

    (r, ) = Ur cos + rn (Cn cos n + Dn sin n) .n=1

    Applying the boundary condition at r = a gives

    = U r +a2

    r

    C1 = Ua2, D1 = 0, Ci = Di = 0 (i 2). Hence (

    )cos . (6.4.4)

    6.5 Complex Potentials for incompressible, irrotational 2-D flows

    6.5.1 Definition

    For a two-dimensional flow which is both incompressible and irrotational, we have seen that

    x= u =

    y,

    y= v =

    x. (6.5.5)

    These equations are known as the Cauchy-Riemann equations (see MAT3034, Functions of a complex variable, if selected next year).As andsatisfy these equations, we can construct a complex function

    w(z) = + i, z = x + iy (6.5.6)

    known as the complex potential. This function is an analytic function of the complex variable z, which means it has a derivative at every value of z. Some complex functions are not analytic and some points in the complex plane have a different derivative depending upon how you approach that point.NB

    1. Any analytic function satisfies the Cauchy-Riemann equations.

    2. Throughout this section

    z = x + iy NOT spatial coordinate in third direction, w = + iNOT velocity component in third direction.

    3. Any analytic function represents some two-dimensional, irrotational, incompressible flow, but this flow is often physically unrealistic.

    56

  • 6.5.2 Boundary Conditions on w(z)

    On a solid boundary, =Constant.

    Hence, Im(w) = Constant on solid boundary. (6.5.7)

    6.5.3 Velocity in terms of Complex Potential

    It is possible to show that the derivative of a complex function can be written in terms of its real

    and imaginary parts,dw

    dz

    x

    x= + i .

    As the function is analytic we can take the derivative along the x-axis. Thus

    dw

    dz= u iv, (6.5.8)

    is the key result to remember. Hence, if w(z) is given, the velocity components u and v can be obtained immediately, without explicitly calculating or in terms of x and y. The magnitude of velocity at any point is given by dwdz

    = u2 + v2 = |u|. (6.5.9)6.5.4 Simple Examples

    Here we choose some elementary functions for w(z) and consider the resulting flow field. Hopefully this provides us with the necessary insight in order to work out the required complex potential for more complicated flows.

    Example 35 w = Az, A = Uei, U, real.

    u iv = dwdz

    = A = U(cos + i sin ).

    Hence u = U cos and v = U sin and w(z) represents uniform flow, speed U, inclined at angle to the positive x direction.Alternatively we can obtain the velocity potential and stream function from the complex potential,

    + i=w = Ueiz,

    =U(cos + i sin )(x + iy),

    =U ((x cos y sin ) + i(y cos + x sin )) ,

    =U(x cos y sin ),=U(y cos + x sin ).

    The streamlines given by=Constant, have equation

    y = x tan + C.

    57

  • The directions on the streamlines are for U > 0, in which case the velocity in the x-direction is positive everywhere.

    Example 36 w = Bz2, B is real and positive.

    + i= w = B(x + iy)2 = B(x2 + 2ixy y2).Hence

    = B(x2 y2),= 2Bxy,and the streamlines are given by xy = c (c a constant). ie rectangular hyperbola as in 6.1.3, Example 29.

    Example 37 w = Bzn, B is real.This time it is easier to calculate the flow field using polar coordinates, and hence we write z = rei.

    + i= w = Brnein = Brn(cos n + i sin n),

    58

  • which gives dw = Brn sin n, |u| = = Bnrn1.dz

    The streamlines are given by rn sin n =Constant. If we now consider the streamline= 0, then

    pi

    n

    2pi

    n= 0 = sin n = 0 = = 0, , . . . ,

    and hence these streamlines are straight lines. We now let = pi/n and restrict ourselves to the region of the plane 0 .Since = 0 and = are streamlines, we can replace them by solid boundaries and flow is unchanged. The resulting flow field depends on the value of n chosen.

    (a) n > 2 0 < < pi2

    This gives the streamlines for flow into an acute corner. The direction of the streamlines

    corresponds to B < 0 (since u < 0 on = 0).

    As r 0, |u| = Bnrn1 0. Hence, O is a stagnation point.(b) 1 < n < 2 pi

    2< < pi

    59

  • This corresponds to flow in the vicinity of an obtuse angle. As above, O is a stagnation

    2

    point.(c) 1 < n < 1 pi < <

    2pi

    This corresponds to flow outside of a corner.

    2

    Since n < 1, rn1 as r 0, hence this flow has infinite velocity at the point O.(d) n = 1 = 2pi

    This corresponds to flow round the edge of a semi-infinite plate. As before, infinite velocities would occur at the tip of the plate.

    The appearance of infinite velocities in these last two examples is clearly physically unrealistic. Indeed, using Bernouillis equation, we see that if |u| , then p and so we have infinite suction. However, the solutions obtained are still useful away from the vertex. Close to the vertex, we can not ignore viscous effects.

    60

  • Flow (d) is particularly useful, corresponding to the limiting case of flow round a rounded edge. This is important in aerofoil theory.Also the solutions above are useful as local solutions near corners in more complicated flows.

    Example 38 w = B log z, B is real and positive.

    + i= B log(rei) = B(log r + i).

    [NOTE: This is a general result log(z) = log |z| + i arg(z).]Hence = B log r,= B and the streamlines are given by =Constant, ie lines radially outwards from the origin. also,

    |u| =dw = B|z| .dz

    Comparing with that in Example 30 in 6.1.3,

    w =K

    2pilog z, is a source, strength K at O.

    Example 39 w = iB log z, B is real and positive.Following example 38, = B,= B log r and the streamlines are r =Constant, ie concentric circles centred at the origin. Comparing with the Example in 6.2.1,

    w = i 2pi

    log z, is a line vortex at O, circulation .

    NB All these examples can be generalised, eg.

    K w = 2pi

    log(z z0),is a source at z = z0.

    6.5.5 Combining Simple Examples

    As with the velocity potentials in 6.3, we can superpose solutions.

    Example 40 Source in Uniform Stream

    w=Uz +K

    2pilog z,

    =Uy +K

    2pi,

    =Uy +K

    2pitan1

    ( yx

    ), Cartesian

    =Ur sin +K

    2pi, Polars

    u iv = dwdz

    =U +K

    2piz.

    61

  • At a stagnation point the velcity is zero,

    u = v = 0 = z = K2piU

    .

    Hence the stagnation point is located at ( K

    2piU, 0

    ).

    The streamlines are given by

    y =K

    2piU( ), = Constant.

    Plotting the streamline passing through the stagnation point is usually a good idea for visualising the flow. In this case, the stagnation point is at = pi, y = 0 and so = pi. The easiest way to sketch this streamline is to tabulate y for various ,

    0 pi 4pi2

    3pi4 pi

    3pi2

    y 12KU

    38KU

    14KU

    18KU 0 14

    K U

    Construction of further streamlines gives a flow field like,

    62

  • Since streamlines can be replaced by solid surfaces, the same flow is obtained if a solid body is inserted inside the streamline= K/2U.

    Example 41 Dipole or Doublet

    Consider a sink, strength K at z = 0, and a source of strength K at z = ei,

    K

    2pi 2pi

    K

    2pi

    ei

    z

    ( )w = log(z ei) K log z = log 1 .

    For small ,

    log

    (1 e

    i

    z

    ) e

    i

    z,

    and so, if 0 and K in such a way that K is kept fixed, then

    w = ei

    z, =

    K

    2pi.

    This field is known as a dipole (or doublet) of strength , orientation

    ei

    rei

    ( ) r

    = Im(w) = Im = sin( ).

    The streamlines are then given by

    r = sin( ), = Constant.

    Exercise: Show that when = 0, the streamlines are circles, centred on the y-axis. We can then show that the pattern of streamlines is given by,

    63

  • Example 42 Uniform Flow, Dipole and Line Vortex.

    w = Uz +Ua2

    z i

    2pilog z, real

    Uniform DipoleFlow at Origin

    Line Vortex at Origin

    This corresponds to flow around a cylinder with circulation . See 6.6.

    6.6 Calculating Forces on Body using Pressure

    Recall from 2.2 that the force on a body is given by

    F =p S (n),F = pn dS.

    For two-dimensional flows this simplifies to

    Force = pn dl. (6.6.1)

    For irrotational flows, the pressure is then calculated using the unsteady Bernoulli equation (4.3.2) or by using Bernoullis equation along the streamlines which lie on the surface of the body.

    Example 43

    Consider a cylinder, radius a, in uniform flow (zero circulation), in a gravitational field. If the cylinder is centred on the origin, and p p0 on y = 0 as |x| , find the pressure on the cylinder.

    64

  • From (6.4.4), the flow is given by

    = U

    (r +

    a2

    r

    )cos ,

    or in terms of the complex potential

    w = Uz +Ua2

    z.

    The streamlines look like

    The speed anywhere in the fluid is given by

    |u| =dwdz

    = U 1 a2z2.

    Far from the cylinder, |z| and |u| U. On the cylinder, z = aei and |u| = U 1 e2i = U (1 cos 2)2 + sin2 2( )1 2 = 2U sin .

    Using Bernoulli on the streamline which passes along y = 0 and the surface of the cylinder,

    p0

    +1

    2U2 = p

    +1

    2|u|2 + gy,

    Conditions at y = 0, x , Conditions on surface of cylinder.Hence, the pressure on the surface of the cylinder is given by

    p = p0 + (

    12U2 2U2 sin2 ga sin

    ).

    nOn the surface of the cylinder, the normal is in the radial direction, = cos x + sin y and the element of arc length is dl = ad. Hence the force on the body is

    F =pn dl = 2pi

    0

    p(cos x + sin y)a d,=(p0 +

    1

    2U2)a

    2pi(cos x + sin y)d

    +2U2a 2pi

    0

    0

    sin2 (cos x + sin y)d+ga2 2pi

    0

    sin (cos x + sin y)d.65

  • 2pi0 sin

    2 d = pi andEvaluating all these integrals, the only non-zero contribution comes from hence

    F = ga2piy,which is the buoyancy force (independent of fluid flow).Important: There is no force in the x-direction ie No Drag.NBIn this case we could guess that there is no drag on a cylinder in inviscid flow by fore-aft symmetry. BUT a more general result exists:

    Any body placed in a uniform, inviscid flow with zero circulation experiences no drag.

    This result is known as DAlemberts Paradox.

    Example 44Find the force on a half cylinder attached to a plane wall, normal to a uniform flow.

    See Problem Sheet.

    66

  • Example 45 Cylinder in Uniform Flow with circulation

    Ua2

    z

    i

    2piw = Uz + log z.

    Sketch the streamlines for various values of and calculate the force on the cylinder. I

    Show that z = aei is a streamline.

    w=Ua(ei + ei) i2)pi(log a + i),=

    (2Ua cos +

    2pi i log a

    2pi.

    Hence Im(w) =Constant, and the circle |z| = a is a streamline. II Show that the flow is uniform far from the cylinder.

    u iv = dwdz

    = U Ua2

    z2 i

    2pi

    1

    z,

    U as z .III Find the Stagnation points in the flow. At the

    stagnation points, u = v = 0 and hence dwdz

    = 0. Solving for z,

    z2 2aiz a2, = 0, z = a(

    i

    1 2),

    where = /4piUa. Thus there are two stagnation points z1 and z2. From the quadratic equation for z, the product of the roots is a2,

    z1z2 = a2 = |z1||z2| = a2,and so, either one stagnation point is inside the cylinder and the other is outside, or both lie on the surface |z| = a.We now consider the separate cases:

    (i) = 0

    There is no circulation and the stagnation points are at a.

    2 1 0 1 22

    1

    0

    1

    2

    x

    y

    67

  • [NB These plots are obtained using MATLAB. See Appendix E. The streamlines inside the cylinder are also plotted, though they are not relevant to flow considered here.]

    (ii) 4piUa < < 0In this case 1 < < 0 and (|z1,2| = a 2 + (1 2) )12 = a,and both stagnation points lie on the surface of the cylinder.

    2 1 0 1 22

    1.5

    1

    0.5

    0

    0.5

    1

    1.5

    x

    y

    (iii) = 4piUa ( = 1)In this case there is only one stagnation point, z1 = z2 = ai.

    2 1 0 1 22

    1.5

    1

    0.5

    0

    0.5

    1

    1.5

    x

    y

    (iv) < 4piUaIn this case, < 1, ( )

    z = a 2 1 i,so both stagnation points lie on the imaginary axis, but only one is outside the cylinder.

    2 1 0 1 22

    1.5

    1

    0.5

    0

    0.5

    1

    1.5

    x

    y

    68

  • IV Find the force on the cylinderOn the cylinder z = aei, so

    |u|2 =U Ue2i 2Uiei2 ,=U2 |(1 cos 2 2 sin ) + i (sin 2 2 cos )|2 ,=U2 (1 cos 2 2 sin )2 + (sin 2 2 cos )2 ,=4U2 sin2 2 sin + 2( ) .

    Using Bernoulli along a streamline from which encircles r = a (neglecting gravity)

    2( )p0 + 1U2 =p + 2

    1

    U2

    2 (sin2

    2

    2 sin 2

    + 2 ,

    2p=p0 + U 1 4 4 sin + 8 sin ).

    Therefore,

    F =pn dl,

    = 2pi

    0

    p (cos x + sin y) a d,=4U2piay, =Uy.

    So(i) Positive circulation = Downward Force(ii) Negative circulation = Upward Force (Lift)A physical illustration of these effects is seen when balls are spinning (eg Table Tennis). Top Spin balls drop shorter case (i)Back Spin balls carry further case (ii).In football, side spin allows balls to be bent round a defensive wall.Swing of cricket balls is considerably more complicated.

    6.7 Induced Velocity and Image Systems

    6.7.1 Induced Velocity

    Consider two vortices, one of strength 1 at z = 0, and one of strength 2 at z = 2,

    w = i12pi

    log z i22pi

    log (z 2),

    u iv = dwdz

    = i12pi

    1

    z i2

    2pi

    1

    z 2 .

    Velocity due to Velocity due to vortex at z = 0 vortex at z = 2

    = (u1 iv1) + (u2 iv2).As z 0, |u1| and u2 (0, 2/4pi).

    69

  • The velocity u2 is known as induced velocity of the vortex at z = 0 due to the vortex at z = 2. Ifthe vortices are free to move through the fluid, then each vortex moves with the velocity inducedby the other vortex.For example, if 1,2 > 0,

    For pairs of vortices, the induced velocity is always perpendicular to the line between the vortices.If 1 = 2 = , the two vortices move on a circular path.

    If 1 = 2 = > 0,

    70

  • In this case the separation of the vortices stays constant, and the vortices move as a pair with speed /2pia, where a is the distance between the two vortices. This explains why two vortices generated by a spoon in a cup of tea move across the mug.These ideas can be extended to systems with more than two vortices. If n vortices are present, the induced velocity at vortex 1 is due to the effect of vortices 2, 3, . . . n.

    6.7.2 Image Systems

    Consider a vortex of strength at z = ai and a vortex of strength at z = ai.

    w= i2pi

    i

    2pilog(z + ai),

    = i2pi

    logz + ai

    log(z ai) + (z ai), (6.7.2)

    u iv = dwdz

    = i2pi

    (1

    z ai 1

    z + ai

    )=

    2pi

    2a

    z2 + a2.

    On the line y = 0, z = x and 2a

    u iv = 2 2 .

    2pi x + aHence v = 0 which proves that y = 0 is a streamline (no normal velocity component). As discussed previously, we can replace the line y = 0 with a solid boundary.

    71

  • The induced velocity at vortex 1 is

    u iv = 4pia

    .

    Hence the flow due to a single vortex in the presence of a plane wall y = 0 is given by (6.7.2). The vortex at z = ai is known as the image vortex. It is located at the mirror image of the original vortex.Two more general results extend the concept of image systems.

    A Schwartz Reflection Principle

    Let w = f(z) describe a flow everywhere in the complex plane, with no singularities (points where f(z) ) in y < 0. If a solid wall is now introduced at y = 0, the new complex potential is given by

    w = f(z) + f(z),

    where (.) denotes the complex conjugate.This can be proved by showing that the line y = 0 is a streamline. Setting y = 0, z = z and w = f(z) + f(z) = 2Re(f). Thus w is real, and so = Im(w) = 0, a constant on y = 0.

    Example 46From the example introduced above, without the wall, the complex potential for a vortex at z = z0 is

    iw = f(z) = log(z z0).2pi

    With the boundary included, the complex potential is given by

    w = f(z) + f(z)= i2pi

    log(z z0)

    2pii log(z z0),

    = i2pi

    log(z z0) +i

    2pilog(z z0),

    where we have used the identity, log(Z) = log(Z).Hence for a vortex at z0, the image vortex must be located at z0, ie at the mirror image location.

    72

  • This method can also be used to find the flow due to a source near a wall, or for more than one vortex near a wall.

    Example 47 Tea-cup Vortices

    The same motion will occur for a single vortex in a right-angled corner. In this case there are three image vortices. See problem sheet.

    B Images for CircleThere is an equivalent principle to the Schwartz reflection principle which defines an image system for a two-dimensional flow in the presence of a circle, radius a, centred at the origin.If w = f(z) describes a flow everywhere in the complex plane, with no singularities in |z| < a, then

    w = f(z) + f(a2z

    ),

    73

  • is the complex potential for the flow in the presence of the circle.Proof:Put z = aei and show that Im(w) = 0, to demonstrate that the surface of the circle is a streamline. Hence for a source or vortex at z = z0, (|z0| > a), an image is located at z = z1, where z1 = a2/z0.

    Clearly |z1| < a and the image lies inside the circle.

    Example 48If the flow is uniform in the absence of the circle, at angle to the positive x-axis, w = Ueiz, where U is real. With the circle present,

    w(z) = Ueiz + Ueia2

    z= Ueiz +

    Ueia2

    z.

    This can be compared with earlier examples.

    Example 49If a vortex is located at z = 2a, with a circle |z| = a, then

    w=

    2pi

    {i log(z 2a) i log(a2

    z 2a

    )},

    = i2pi

    {log(z 2a) log

    (a2z 2a

    )},

    u iv= i2pi

    {1

    z 2a +1

    z+

    2a

    a2 2az}.

    Hence the induced velocity at the vortex located at z = 2a is

    u1 iv1 = i

    2pi

    { 12a

    +2a

    a2 4a2}

    =i

    12pia.

    74

  • 7 Surface Waves

    7.1 Introduction

    We now consider a fluid layer of average depth h with surface given by z = (x, y, t) and fixed impermeable boundary z = h. We wish to consider how the position of the surface can change with time eg move surface with a paddle (or hand)

    7.2 Basic wave theory

    We consider a 2-D problem with surface given

    by k known as wave number.z = (x, t) = A cos(k(x ct)),

    At t = 0, (x, 0) = A cos(kx)

    Wave crests located at

    x = 0,2pi

    k,

    4pi

    k, . . .

    75

  • so distance between crests is

    =2pi

    k.

    This is known as the wavelength.The maximum height of the wa