Fluid dynamics 1


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Fluid dynamics 1

  1. 1. Introduction to Fluid Dynamics
  2. 2. Applications of Fluid Mechanics • Fluids are the principle transport media and hence play a central role in nature (winds, rivers, ocean currents, blood etc.) • Fluids are a source of energy generation (power dams) • They have several engineering applications – Mechanical engineering (hydraulic brakes, hydraulic press etc.) – Electrical engineering (semi-conductor industries) – Chemical Engineering (centrifuges) – Aerospace engineering (aerodynamics)
  3. 3. List the fluid systems in … Typical Home Car Aircraft
  4. 4. Leaking crude oil from the grounded tanker Argo Merchant (Nantucket Shoals 1976) ctsy: An album of fluid motion by Van Dyke
  5. 5. Smoke plumes
  6. 6. Turbulent Jet impinging into fresh water ctsy: An album of fluid motion by Van Dyke
  7. 7. Trapped plume in a stratified ambient flow ctsy: CORMIX picture gallery
  8. 8. A salt wedge propagating into fresh water ctsy: Gravity currents in the environment and the laboratory, author : John E. Simpson
  9. 9. What is a fluid ? • Fluids (liquids and gases) cannot resist shearing forces (tangential stresses) and will continue to deform under applied stress no matter how small. • Solids can resist tangential stress and will deform only by the amount required to reach static equilibrium. This class will concentrate on the dynamics of fluid motion Note – There are several examples where the distinction between solids and fluids blur (e.g. silly putty). – Individual fluid types have distinct characteristics that will play a critical role in their behavior (e.g. water, oil, air)
  10. 10. What is a Fluid? … a substance which deforms continuously under the action of shearing forces however small. … unable to retain any unsupported shape; it takes up the shape of any enclosing container. ... we assume it behaves as a continuum
  11. 11. Continuum Hypothesis Properties of fluids result from inter-molecular interactions. Individual interactions are very difficult to quantify. Hence fluid properties are studied by their lumped effects Continuum hypothesis states that “macroscopic behavior of fluid is perfectly continuous and smooth, and flow properties inside small volumes will be regarded as being uniform” Is this hypothesis valid ?
  12. 12. Continuum hypothesis breaks down at molecular scales Example: density is defined as the mass / unit volume Measure of density is determined by the volume over which it is calculated. To study fluctuations in density (e.g. variation in m air density with altitude) we need a local estimate v v 0 How small should δv be to get a true local estimate ? ( Number of molecules) * (mass / molecule) At the molecular scale v Density will depend upon the number of molecules in the sample volume
  13. 13. (adapted from Batchelor) For continuum hypothesis to hold macroscopic properties should not depend on microscopic fluctuations True for most fluid states (exception : gas flows at very low pressure have mean free path of molecular motion approaching length scales of physical problems) Scales of fluid processes (1mm ~ 1000 km) >> molecular scales (10-8 cm) Thus fluid flow and its various properties will be considered continuous
  14. 14. Properties of fluids Common fluid properties are described below – Mass • Denoted by the density of the fluid (ρ). It is a scalar quantity • Density varies with temperature, pressure (described below) and soluble compounds (e.g. sea water as opposed to fresh water) – Velocity • Is a vector quantity, and together with the density determine the momentum of the flow – Stresses • Are the forces per unit area acting on the fluid particles. They are of two types – Normal stress – Shear stress • Normal stresses in liquids are generally compressive and are referred by a scalar quantity “pressure”
  15. 15. – Viscosity • It is the ability of the fluid to flow freely • Mathematically it is the property of fluid that relates applied shear stress to rate of deformation (shall be studied in detail later) • Viscosity usually varies with temperature (to a greater extent) and pressure (to a lesser extent). Note, viscosity in a liquid decreases with increase in temperature but in a gas increases with increase in temperature. – Thermal conductivity • It is the ability of the fluid to transfer heat through the system • Mathematically it is similar to viscosity (viscosity is the ability of the fluid to transfer momentum).
  16. 16. – Bulk modulus of elasticity and compressibility • Compressibility is the change in density due to change in normal pressure dp Ev • Reciprocal of compressibility is known as the bulk modulus of d elasticity • Liquids have very high values of Ev in comparison to gases. (Thus, for most practical problems liquids are considered incompressible. This is a major difference between liquids and gases) – Coefficient of thermal expansion • Thermal expansion is the change in fluid density due to change in d temperature dT • Liquids in general are less sensitive to thermal expansion than gas • In some cases coefficient of thermal expansion may be negative (e.g. water inversion near freezing point)
  17. 17. • Fluid properties are inter-related by equations of state. • In gases these equations of state are determined by the collision of molecules and are given by the kinetic theory of gases (the perfect gas law) p RT • In liquids these equations of state are very difficult to achieve due to forces of inter-molecular attraction and are thus determined empirically. • Response of a fluid to external forcings is to a large extent determined by its properties. External Forcings + Fluid properties + laws of physics = Fluid motion
  18. 18. Equation of state for an ideal gas 2 1 p pV mRT or RT Sometimes written … pV = nRuT n= number of moles of gas (kmol/kg) Ru= Universal gas constant (kJ/kmol K)
  19. 19. Liquids Gases Almost incompressible Relatively easy to compress Forms a free surface Completely fills any vessel in which it is placed
  20. 20. Hydrostatic Equation p g z z h IF = a p gz gh constant
  21. 21. Shear stress in moving fluids y Newtonian fluid U 3 U y = viscosity (or dynamic viscosity) kg/ms = kinematic viscosity m2/s 4
  22. 22. Continuity . AU = constant = m Mass is conserved What flows in = what flows out ! 5 1A1U1= 2A2U2
  23. 23. Bernoulli 6 p 2 U2 1 gz … cons tan t if no losses 7 2 p1 U1 p2 U2 2 z1 z2 1H 2 g 2g g 2g Where H2 1 is the head loss (m)
  24. 24. Reynolds Number UL UL Re inertia f orce rate of change of momentum U2L2 v iscousf orce shear stress x area L2
  25. 25. Venturi2 A 2U2 1A 1U1 p 2 U2 1 gz cons tan t p1 p2   
  26. 26. Discharge Coefficients Q' CD Q 8 Actual Flowrate = CD x Predicted flowrate
  27. 27. Orifice Plate D D/2 D d    Vena contracta Orifice plate
  28. 28. Discharge Coefficients Q' CD Q Venturi meter Orifice plate CD CD 0.65 Increasing 0.98 values of d/D 0.6 0.94 104 106 Re 10 10 106 Re D 4 5 D
  29. 29. A Mathematical Review
  30. 30. Taylor series Taylor stated that in the neighborhood of x a the function f(x) can be given by Primes denote differentials 2 3 x a x a f x f a f a x a f a f a  2 3! p0 x p1 x p2 x A Taylor series expansion replaces a complex function with a series of simple polynomials. This works as long as the function is smooth (continuous) Note: p0 x A constant value at x a p1 x A straight line fit at x a p2 x A parabolic fit at x a For x a 1 each additional term is smaller than the previous term
  31. 31. Example x Find the Taylor series expansion for f x e about x 1 1 p0 x f a e 1 1 p1 x f a f a x a e e x 1 2 2 x a 1 1 1 x 1 p2 x f a f a x a f a e e x 1 e 2 2 Each additional term provides a more accurate fit to the true solution for x 1.2 for x 1.5 1.2 1.5 f (1.2) e 0.3012 f (1.5) e 0.2231 1 1 p0 1.2 e 0.3679 p0 1.5 e 0.3679 1 1 1 1 p1 1.2 e e 0.2 0.2943 p1 1.5 e e 0.5 0.1839 2 2 1 1 1 0.2 1 1 1 0.5 p2 1.2 e e 0.2 e 0.3017 p2 1.5 e e 0.5 e 0.2299 2 2
  32. 32. Scalar Field Variables that have magnitude, but no direction (e.g. Temperature) Vector Field Variables that have both magnitude and direction (e.g. velocity). A vector field is denoted by the coordinate system used. There are two ways to denote a vector field Mathematically a vector field is denoted by the magnitude along each orthogonal coordinate axis used to describe the system  V ui vˆ wk ˆ j ˆ (Vector notation for Cartesian coordinates) ˆ j ˆ i , ˆ, k unit vecto along the x, y, z directionrespective rs ly Geometrically a vector field is denoted by an arrow along the direction of the field, with the magnitude given by the length of the arrow z w y   v m agV V u2 v2 w2 u x
  33. 33. Dot Product  V     V U V U cos  U   For θ = 90 (perpendicular vectors) V U 0     For θ = 0 (collinear vectors) V U VU For a Cartesian coordinate system, if  V j ˆ v xi v y ˆ vz k ˆ iˆ ˆ j ˆ j ˆ ˆ iˆ iˆ k k iˆ j ˆ ˆ j ˆ k k ˆ 0  iˆ iˆ ˆ j j ˆ ˆ ˆ k k 1 U ˆ j ˆ u x i u y ˆ uz k Then   V U j ˆ ˆ j ˆ vxi v y ˆ vz k uxi u y ˆ uz k ˆ vxux v yuy vzuz Note : Dot product of two vectors is a scalar
  34. 34. Deconstruction of a vector into orthogonal components    Consider a pair of vectors U and V U θ   V  Aim is to separate U into two components such that one of them is collinear with V     U U1 U2  Represents a unit vector along the direction of V V Define a vector v ˆ   V Component of U along V    Now U ˆ v ˆ U v cos U cos   Therefore U1 ˆ ˆ U vv provides the direction of the vector    and U2 U U1             2 also U 2 U1 U U1 U1 U U1 U1 U1 U U1 cos U1   But U1 U cos   2 2 2 2 Substituting we get U2 U1 U cos U cos 0 Thus, a vector can be split into two orthogonal components, one of which is at an arbitrary angle to the vector.
  35. 35. Divergence (Example of dot product)   Consider a vectorfield U in an arbitraril y U enclosed volumeV V ˆ n Unit vector normal to surface dA ˆUnit vector tangential to surface  ˆ U n Component of flow normal to surface dA  Therefore ˆ U ndA T ot alflow normalto thesurface of V A T ot alflow into/outof thesyst em Then Represents the total flow  U ndAˆ across a closed system per  div U A unit volume V limitV 0
  36. 36. For a Cartesian system     U ui vj wk x0 , y0 , z0 dz V dxdydz dx dy At the front face: ˆ ˆ n i  ˆ U ndA u( x0 dx / 2, y0 , z0 )dydz A At the back face: ˆ n ˆ i  ˆ U ndA u( x0 dx / 2, y0 , z0 )dydz A Thus, contribution from these two faces to the divergence is u x0 dx / 2, y0 , z0 u x0 dx / 2, y0 , z0 u dx limdx 0 x
  37. 37. Similar exercise can be carried out for the remaining faces as well to yield  u v w div U x y z This can be written in a more concise form by introducing an operator  ˆ i ˆ j ˆ k x y z known as the “dell” operator Thus    div U U Note : A vector operator is different from a vector field
  38. 38. Gradient Consider the differential of a scalar T along the path PC, which we shall denote as s C Let the unit vector along this path be ˆ ˆ sy ˆ ˆ P s sxi j sz k   R R ˆ ss Then by definition   dT T R ˆ ss T R ds s lim s 0 Using differentiation by parts we get dT T dx T dy T dz ds x ds y ds z ds T T T sx sy sz x y z Gradient of a scalar  ˆ T s
  39. 39.  Thus ˆ ˆ T s represents the change in scalar T along the vector s  T Represents a vector of the scalar quantity T. What is its direction? angle between the gradient and direction of change dT    T ˆ s ˆ T s cos T cos ds dT Maximum value of occurs when cos 1 or 0 ds  Thus T lies along direction of maximum change
  40. 40. Now consider that the path s lies along a surface of constant T dT  ˆ T s 0 ds A vector along a surface is denoted by the tangent to the surface dT  T ˆ 0 ds From the definition of dot product, this means that  T is perpendicular to surface of constant T In summary, the gradient of a scalar represents the direction of maximum change of the scalar, and is perpendicular to surfaces of constant values of the scalar
  41. 41. Cross Product A cross product of two vectors is defined as     U V U V sin e ˆ ˆ e is a unit vector perpendicular to the plane of the two vectors and its direction is determined by the right hand rule as shown in the figure     For θ = 90 (perpendicular vectors) U V UV   For θ = 0 (collinear vectors) U V 0 For a Cartesian coordinate system, if  V ˆ v xi vy ˆ j ˆ vz k ˆ ˆ ˆ ˆ ˆ ˆ i i j j k k 0  U ˆ uxi uy ˆ j ˆ uz k i ˆ ˆ j ˆ k; ˆ i j ˆ ˆ ˆ ˆ k;i k ˆ j ˆ ˆ k i j j ˆ ˆ; ˆ k ˆ ˆ i;k ˆ j ˆ i T hen   U V ˆ i u y vz uz v y ˆ ux vz j v x uz ˆ k ux v y vxu y
  42. 42. A convenient rule is ˆ j ˆ i ˆ k   U V u x u y uz ˆ i u y vz uz v y ˆ ux vz j v x uz ˆ k ux v y vxu y v x v y vz Curl of a vector (an example of the cross product)  Consider a velocity field given by V ˆ j ˆ ui vˆ wk    Then curl V V iˆ j ˆ ˆ k   w v w u v u V iˆ ˆ j ˆ k x y z y z x z x y u v w What does this represent?
  43. 43. Consider a 2-dimensional flow field (x,y)   Then ˆ v V k u x y y A fluid particle in motion has both translation B and rotation Δy A rigid body rotates without change of shape, O Δx A but a deformable body can get sheared x Thus for a fluid element, angular velocity is defined as the average angular velocity of two initially perpendicular elements v A vO v ˆ Angular velocity of edge OA = k x lim x 0 x uO u B u ˆ Angular velocity of edge OB = k y lim x 0 y 1 v u ˆ Angular velocity of fluid element = k 2 x y Thus, the curl at a point represents twice the angular velocity of the fluid at that point
  44. 44. Recap • For continuous functions, properties in a neighborhood can be represented by a Taylor series expansion • Fluid flow properties are represented as either scalars (magnitude) or vectors (magnitude and direction)  • ˆ The dot product of a vector and a unit vector V s represents the magnitude of the vector along the direction of the unit vector   • The divergence V represents the net flow out / in to a closed system per unit volume  • The gradient of a scalar T represents the direction and magnitude of maximum change of the scalar. It is oriented perpendicular to surfaces of constant values  • The dot product of a gradient and a unit vector T s represents the change of ˆ the scalar along the direction of the unit vector   • The curl of a velocity vector V represents twice the angular velocity of the fluid
  45. 45. Kinematics of Fluid Motion • Kinematics refers to the study of describing fluid motion • Traditionally two methods employed to describe fluid motion – Lagrangian – Eularian
  46. 46. Lagrangian Method In the Lagrangian method, fluid flow is described by following fluid particles. A “fluid particle” is defined by its initial position and time    X F X 0 , t0 , t For example : Releasing drifters in the ocean to study currents Velocity and acceleration of each particle can be obtained by taking time derivatives of the particle trajectory     dX dF X 0 , t0 , t u dt dt     du d 2 F X 0 , t0 , t a dt dt2 Lagrangian framework is useful because fundamental laws of mechanics are formulated for particles of fixed identity. However, due to the number of particles that would be required to accurately describe fluid flow, this method has limited use and we have to rely on an alternative method
  47. 47. Eularian Method Alternatively fluid velocity can be described as functions of space and time   in domain u f x, t Describing the velocity as a function of space and time is convenient as it precludes the need to follow hundreds of thousands of fluid particles. Hence the Eularian method is the preferred method to describe fluid motion. Note : Particle trajectories can be obtained from the Eularian velocity field The trick is to apply Lagrangian principles of conservation in an Eularian framework
  48. 48. Acceleration in an Eularian coordinate system Consider steady (not varying with time) 1-D flow in a narrowing channel U0 d0 U(x) d(x) In a Eularian framework the flow field is given by   u U ( x)i Where, according to the conservation of mass (to be discussed later) U0d0 U x d x Standing at one point and taking the time derivative we find that the acceleration is zero However a fluid particle released into the channel would move through the narrow channel and its velocity would increase. In other words its acceleration would be non zero What are we missing in the Eularian description ?
  49. 49. Consider a particle at x0 inside the channel with a velocity u U x0 After time Δt the particle moves to x0 + Δx where x u t The new velocity is now given by u u U x0 x, t U x0 u t, t Applying Taylor series expansion we get U U u u U x0 u t t x t u U U or u t x t In the limit t 0 u Du U U accelerati on u t t 0 Dt t x Material derivative
  50. 50. We can derive a 3D form of the material derivative using the chain rule of differentiation Consider a property f of a fluid particle f g x t , y t , z t ,t Rate of change of f with respect to time is given by df dt According to the chain rule of differentiation df f f dx f dy f dz dt t x dt y dt z dt where dx velocit yalong t he x direct ion u dt dy velocit yalong t he x direct ion v dt dz velocit yalong t he x direct ion w dt Therefore Material derivative df f f f f Df u v w dt t x y z Dt
  51. 51. Stream lines Streamlines are imaginary lines in the flow field which, at any instant in time, are tangential to the velocity vectors. Streamlines are a very useful way of denoting the flow field in a Eularian description In steady flows (not changing with time) stream lines remain unchanged Consider an element of the stream line curve given by dx  And the flow field at this element by U Then the required condition for the element to be tangential (parallel) to the flow is  U dx 0  Substituting U uiˆ vˆ j ˆ wk and dx ˆ dxi dyˆ j ˆ dzk We get dx dy dz Differential equation that needs to be solved to u v w determine stream lines
  52. 52. Example:  Consider a flow field given by U ˆ xi yˆ j Substituting in our equation for stream lines we get dx dy x y Integrating we get ln xy constant or xy constant Graphically this represents flow around a corner
  53. 53. Path lines Path lines are the lagrangian trajectories of fluid particles in the flow. They represent the locus of coordinates over time for an identified particle In steady flows, path lines = stream lines  Returning to our previous example U ˆ xi yˆ j Path lines can be obtained by integrating the flow field in time dx dy u x; v y dt dt dx dy or dt; dt x y Constants of integration Which yields x a0et ; y b0e t Note that xy a0b0 constant Which is the same as the stream line curves. The initial position of the particle determines the constants of integration, and thus which stream line the particle is going to be on
  54. 54. Relative motion of a fluid particle When introducing the concept of a cross product we showed that for a deforming fluid particle the angular velocity of the rotating fluid particle is given ˆ 1   i w v ˆ j w u ˆ k v u V 2 2 y z 2 x z 2 x y    V Referred to as vorticity Apart from rotation the fluid particle is also stretched and distorted u dydt y v dy dydt dy y v dxdt x dx u dx dxdt x
  55. 55. u dx dxdt dx The stretching (or extensional strain) is given by dt x xx dx u or xx x v w Similarly we get yy and zz y z The shear strain for a fluid particle is defined by the average rate at which the two initially perpendicular edges are deviating away from right angles 1 u v From the figure this yields xy 2 y x 1 v w 1 u w And similarly yz and xz 2 z y 2 z x Also note that the shear strain is symmetric yx xy
  56. 56. The total rate of deformation of a moving fluid particle can be written in matrix form u u u x y z v v v D x y z Deformation tensor w w w x y z Which can be split up to yield u 1 u v 1 u w 1 u v 1 u w 0 0 0 0 x 2 y x 2 z x 2 y x 2 z x v 1 u v 1 v w 1 u v 1 v w D 0 0 0 0 y 2 y x 2 z y 2 y x 2 z y w 1 u w 1 v w 1 u w 1 v w 0 0 0 0 z 2 z x 2 z y 2 z x 2 z y stretching shearing rotating stretching + shearing = strain tensor
  57. 57. Acceleration in a rotating reference frame Reference frames that are stationary in space are referred to as inertial reference frames In many situations the reference frames themselves are moving. For newtonian dynamics it is important to take the motion due to the moving reference frame into account A very common example is the acceleration due to rotation of the earth ˆ e Effect of rotation on a vector B ˆ er Unit vectors along Consider a point B on the  ˆ ˆ ˆ ez , er , e R orthogonal coordinate surface of a cylinder rotating axes about its vertical axis with O angular velocity Ω  ˆ ˆ The position vector is given by R zez rer   dR ˆ der d Velocity vector is then given by V r r ˆ e ˆ re dt dt dt   At the same time R ˆ ez ˆ zez ˆ rer ˆ re  dR   Thus R dt
  58. 58. Now consider a fluid particle in an arbitrarily moving coordinate system   R is the position vector of the origin of the r moving coordinate system with respect to  the inertial coordinate system R  r is the position vector of the fluid particle in the moving coordinate system Any arbitrary motion can be reduced to a translation and a rotation. Thus the velocity of the fluid particle in the inertial coordinate system can be reduced to three components   dr a) Motion relative to moving frame uR dt   b) Angular velocity r  dR c) Motion of moving reference frame dt Thus the velocity is given by   dR    uI uR r dt
  59. 59. Differentiating once again yields acceleration   2 d R d        aI uR r uR r dt2 dt Simplifying we get     2 d R duR d       aI r 2 uR r dt2 dt dt (1) (2) (3) (4) (5) (1) Acceleration of moving coordinate system (2) Acceleration of fluid particle in moving coordinate system (3) Tangential acceleration (acceleration due to change in rotation rate) (4) Coriolis acceleration (5) Centripetal acceleration
  60. 60. Example: Acceleration in a cartesian coordinate system fixed on the surface of the rotating earth  E  Angular velocity of the earth is 7.29x 10-5 sec-1 y z Using cartesian coordinates we have x into the paper  ˆ ˆ r xi yˆ zk j  R Rkˆ  ˆ uR ui vˆ wk ˆ j   cos ˆ j ˆ sin k E  Since change in R Is occurring purely due to rotation, we have  dR   E R dt  2 d R    E E R dt2
  61. 61. We thus get ˆ i j ˆ ˆ k   2 uR 2 j ˆ 2 cos w 2 sin v i 2 sin uˆ 2 cos uk ˆ x y z u v w          r r r x 2iˆ 2 sin z cos y sin ˆ j 2 cos y sin ˆ z cos k Where we have used the vector identity          A B C B CA A BC Similarly we get    R 2 R sin cos ˆ j 2 ˆ R cos2 k E E Assuming a constant rotation rate  d 0 dt
  62. 62. Individual components of acceleration can thus be given by du 2 ax 2 cos w 2 sin v x dt dv 2 ay 2 sin u sin R cos z cos y sin dt dw 2 az 2 cos u cos R cos z cos y sin dt Note that for problems relating to earth’s rotation 2 Thus, centripetal accelerations can usually be ignored. Also for most problems involving ocean circulation vertical velocities and accelerations are much smaller than horizontal motion (to be covered later) In their simplest form the accelerations are thus given by du ax fv Note : f has opposite signs in the northern and dt southern hemispheres. This is why hurricanes are dv counter clockwise in the northern hemisphere and ay fu dt clockwise in the southern where f 2 sin
  63. 63. Equations of motion for fluids • Conservation of mass (continuity equation) – Net mass loss or gain is zero • Conservation of momentum (Navier Stokes equations) – In each direction   ma F mass Sum of all forces acting on fluid acceleration
  64. 64. Conservation of mass Consider a control volume of infinitesimal size Let density = x, y, z, t dz  ˆ j ˆ Let velocity = V u x, y, z, t i v x, y, z, t ˆ w x, y, z, t k dy dx Mass inside volume = dxdydz Mass flux into the control volume = udydz vdxdz wdxdy Mass flux out of the control volume = u u dx dydz v v dy dxdz x y w w dz dxdy z Conservation of mass states that Mass flux out of the system – Mass flux into the system = Rate of change of mass inside the system Thus u v w x y z t
  65. 65. Or, using differentiation by parts, we get u v w u v w 0 t x x y y z z Rearranging, we get u v w u v w 0 t x y z x y z D   V Dt Using vector notation 1 D   V 0 Conservation of mass equation, Dt valid for both water and air Physically it means that the divergence at any point (net flow out/in) is balanced by the rate of change in density through that point
  66. 66. Conservation of momentum Law of conservation of momentum states that for a fluid particle accelerating in water   ma F Let us look at the forces acting on a fluid particle Forces Two types of forces acting on fluid particles •Body forces – occur through the volume of the fluid particle (e.g. gravity) •Surface forces – occur at the surfaces of the fluid particle •Two types of surface forces •Normal stress – acting either in tension or compression •Shear stress – acting to deform the particle Shear stress Normal stress
  67. 67. Simple case : static fluid In static fluid, there is no shear force, since the particles are stationary and not deforming. Hence all the forces occur due to normal compressive forces Consider a fluid particle as shown Since there is no net motion forces in the x and z direction must balance each nn other out dz Force balance in the x direction xx dx θ xx dz nn dl sin nn dz or xx nn zz Force balance in the z direction 1 1 zz dx nn dl cos gdxdz ; or zz nn gdz 2 2 Reducing particle to zero size we get zz nn Thus, in a static fluid, the normal stresses are isotropic and are given by nn ˆ pn Hydrostatic pressure (scalar) Negative sign indicating compressive forces (convention)
  68. 68. What is the expression for pressure ? p z dz / 2 Consider a finite sized particle cube in dy stationary fluid dz p x dx / 2 p x dx / 2 Force balance in the x direction dx p z dz / 2 p x dx / 2 p x dx / 2 0 p dx p dx p p 0 x 2 x 2 p 0 x p Similarly 0 y Force balance in the z direction p z dz / 2 p z dz / 2 dxdy gdxdydz p g z Integrating yields p gz C
  69. 69. Surface forces in a moving fluid Viscous forces between the different fluid particles will lead to 1. Non – isotropic normal stresses 2. Development of shear stresses zz At any given plane of a fluid particle zx there are three forces acting on the fluid xz particle (one normal and two shear yz stresses) zy xx yx (Notation : First subscript refers to direction xy of force and second subscript to direction of yy perpendicular to plane) xx xy xz For each axis there are three ij yx yy yz contributions from the stress tensor Referred to as stress tensor zx zy zz
  70. 70. How are the shear stresses related ? yxy yx y 2 Consider the moments acting on a fluid particles xy x xy x xy xy Counter clockwise moments x 2 x 2 xxy x xy x x y xy y xy y yx x 2 2 x 2 2 yx y 2 xy y x clockwise moments yx y y yx y y yx x yx x y 2 2 y 2 2 yx y x
  71. 71. Conservation of moments Rotational acceleration moments I z  For a rectangular element Moment of inertia Iz x2 y2 x y 12 Therefore xy yx x2 y2  12 if xy yx then as x, y 0;  Therefore xy yx Similarly xz zx ; yz zy
  72. 72. Thus the stress tensor xx xy xz is said to be symmetric ij yx yy yz zx zy zz An important property of symmetric tensors is that the sum of diagonals is independent of the coordinate system used p0 0 For a hydrostatic fluid ij 0 p 0 Sum of diagonals = -3p 0 0 p Which is independent of coordinate system Using the analogy of hydrostatic fluid we define xx p xx Separates normal stress into a compressive part + yy p yy deviations zz p zz Note: This is a mechanical definition of pressure for xx yy zz moving fluids and is not equal to that of hydrostatic p fluids 3
  73. 73. The stress tensor can thus be written as p0 0 xx xy xz ij 0 p 0 yx yy yz ij 0 0 p zx zy zz (deviatoric stress tensor) What is the expression for deviatoric stress tensor ? For a newtonian fluid, Stokes (1845) hypothesized that •The stress tensor is at most a linear function of strain rates (rate of deformation)in a fluid •The stresses are isotropic (independent of direction of coordinate system) •When the strain rates are zero (no motion), the stresses should reduce to hydrostatic conditions This leads to u 1 u v 1 w u x 2 y x 2 x z 1 u v v 1 w v ij 2 2 y x y 2 y z Coefficient of viscosity 1 w u 1 w v w 2 x z 2 y z z
  74. 74. Conservation of momentum equations xx xx dx For x direction: xx x Net normal surface force = xx dx dydz dydz xx xx x xx dxdydz x p xx dxdydz dxdydz x x xy xz Similarly, net shear surface flow = dxdydz y z Let the body force in x direction = Xdxdydz Conservation of momentum states that ma F Du p xx xy xz dxdydz X dxdydz Dt x x y z
  75. 75. or u u u u 1 p 1 xx xy xz u v w X t x y z x x y z Similarly in y and z direction v v v v 1 p 1 yx yy yz u v w Y t x y z y x y z w w w w 1 p 1 zx zy zz u v w Z t x y z z x y z What about body forces ? Navier Stokes equations For gravity : X Y 0; Z g Also, for Newtonian fluid u u v u w xx 2 ; xy ; xz x y x z x Assuming that μ does not vary we get 2 1 xx xy xz u u v u w 2 x y z x2 y y x z z x =0 2 2 2 u u u u v w x2 y2 z2 x x y z
  76. 76. Thus 2 2 2 u u u u 1 p u u u u v w t x y z x x2 y2 z2 2 2 2 v v v v 1 p v v v u v w t x y z y x2 y2 z2 2 2 2 w w w w 1 p w w w u v w g t x y z z x2 y2 z2 The equations of motion, can thus be written in vector form as   U 0  U    1   U U ˆ gk p 2 U t where Kinematic coefficient of viscosity Laplace’s operator (or Laplacian) 2   2 2 2 x2 y2 z2