Fluid Mechanics Chapter 4. Differential relations for a fluid flow
- 1. 1 Diffrential Relations For A Fluid Flow Chapter 4 Fluid Mechanics (MEng 2113) Mechanical Engineering Department Prepared by: Addisu Dagne April, 2017
- 2. Contents of the Chapter ❖ Introduction ❖ Acceleration Field ❖ Differential equation Conservation of mass ❖ Stream function ❖ Differential equation of Linear momentum ❖ Inviscid flow ❖ Vorticity and Irrotationality ❖ Velocity Potential ❖ Basic Plane Potential Flows ❖ Superposition of Basic Plane Potential Flows ❖ Solved Problems 2
- 3. Introduction In analyzing fluid motion, we might take one of two paths: 1. Seeking an estimate of gross effects (mass flow, induced force, energy change) over a finite region or control volume or 2. Seeking the point-by- point details of a flow pattern by analyzing an infinitesimal region of the flow. 3
- 4. The control volume technique is useful when we are interested in the overall features of a flow, such as mass flow rate into and out of the control volume or net forces applied to bodies. Differential analysis, on the other hand, involves application of differential equations of fluid motion to any and every point in the flow field over a region called the flow domain. When solved, these differential equations yield details about the velocity, density, pressure, etc., at every point throughout the entire flow domain. 4 Introduction
- 5. The Acceleration Field of a Fluid Velocity is a vector function of position and time and thus has three components u, v, and w, each a scalar field in itself. This is the most important variable in fluid mechanics: Knowledge of the velocity vector field is nearly equivalent to solving a fluid flow problem. The acceleration vector field a of the flow is derived from Newton’s second law by computing the total time derivative of the velocity vector: 5
- 6. Since each scalar component (u, v , w) is a function of the four variables (x, y, z, t), we use the chain rule to obtain each scalar time derivative. For example, But, by definition, dx/dt is the local velocity component u, and dy/dt =v , and dz/dt = w. The total time derivative of u may thus be written as follows, with exactly similar expressions for the time derivatives of v and w: 6 The Acceleration Field of a Fluid
- 7. Summing these into a vector, we obtain the total acceleration: 7 The Acceleration Field of a Fluid
- 8. The term δV/δt is called the local acceleration, which vanishes if the flow is steady-that is, independent of time. The three terms in parentheses are called the convective acceleration, which arises when the particle moves through regions of spatially varying velocity, as in a nozzle or diffuser. The gradient operator is given by: 8 The Acceleration Field of a Fluid
- 9. The total time derivative—sometimes called the substantial or material derivative— concept may be applied to any variable, such as the pressure: Wherever convective effects occur in the basic laws involving mass, momentum, or energy, the basic differential equations become nonlinear and are usually more complicated than flows that do not involve convective changes. 9 The Acceleration Field of a Fluid
- 10. Example 1. Acceleration field Given the eulerian velocity vector field find the total acceleration of a particle. 10 Solution step 2: In a similar manner, the convective acceleration terms, are
- 11. 11 Solution step 2: In a similar manner, the convective acceleration terms, are
- 12. An idealized velocity field is given by the formula Is this flow field steady or unsteady? Is it two- or three dimensional? At the point (x, y, z) = (1, 1, 0), compute the acceleration vector. Solution The flow is unsteady because time t appears explicitly in the components. The flow is three-dimensional because all three velocity components are nonzero. Evaluate, by differentiation, the acceleration vector at (x, y, z) = (−1, +1, 0). 12 Example 2. Acceleration field
- 13. 13 Example 2. Acceleration field
- 14. Exercise 1 The velocity in a certain two-dimensional flow field is given by the equation where the velocity is in m/s when x, y, and t are in meter and seconds, respectively. 1. Determine expressions for the local and convective components of acceleration in the x and y directions. 2. What is the magnitude and direction of the velocity and the acceleration at the point x = y = 2 m at the time t = 0? 14
- 15. The Differential Equation of Mass Conservation Conservation of mass, often called the continuity relation, states that the fluid mass cannot change. We apply this concept to a very small region. All the basic differential equations can be derived by considering either an elemental control volume or an elemental system. We choose an infinitesimal fixed control volume (dx, dy, dz), as in shown in fig below, and use basic control volume relations. The flow through each side of the element is approximately one-dimensional, and so the appropriate mass conservation relation to use here is 15
- 16. The element is so small that the volume integral simply reduces to a differential term: 16 The Differential Equation of Mass Conservation
- 17. The mass flow terms occur on all six faces, three inlets and three outlets. Using the field or continuum concept where all fluid properties are considered to be uniformly varying functions of time and position, such as ρ= ρ (x, y, z, t). Thus, if T is the temperature on the left face of the element, the right face will have a slightly different temperature For mass conservation, if ρu is known on the left face, the value of this product on the right face is 17 The Differential Equation of Mass Conservation
- 18. Introducing these terms into the main relation Simplifying gives 18 The Differential Equation of Mass Conservation
- 19. The vector gradient operator enables us to rewrite the equation of continuity in a compact form so that the compact form of the continuity relation is 19 The Differential Equation of Mass Conservation
- 20. Continuity Equation in Cylindrical Coordinates Many problems in fluid mechanics are more conveniently solved in cylindrical coordinates (r, θ, z) (often called cylindrical polar coordinates), rather than in Cartesian coordinates. 20 The Differential Equation of Mass Conservation
- 21. Continuity equation in cylindrical coordinates is given by Steady Compressible Flow If the flow is steady , and all properties are functions of position only and the continuity equation reduces to 21 The Differential Equation of Mass Conservation
- 22. Incompressible Flow A special case that affords great simplification is incompressible flow, where the density changes are negligible. Then regardless of whether the flow is steady or unsteady, The result is valid for steady or unsteady incompressible flow. The two coordinate forms are 22 The Differential Equation of Mass Conservation
- 23. The criterion for incompressible flow is where Ma = V/a is the dimensionless Mach number of the flow. For air at standard conditions, a flow can thus be considered incompressible if the velocity is less than about 100 m/s. 23 The Differential Equation of Mass Conservation
- 24. Example 3 Consider the steady, two-dimensional velocity field given by Verify that this flow field is incompressible. Solution Analysis. The flow is two-dimensional, implying no z component of velocity and no variation of u or v with z. The components of velocity in the x and y directions respectively are To check if the flow is incompressible, we see if the incompressible continuity equation is satisfied: So we see that the incompressible continuity equation is indeed satisfied. Hence the flow field is incompressible.24
- 25. Example 4 Consider the following steady, three-dimensional velocity field in Cartesian coordinates: where a, b, c, and d are constants. Under what conditions is this flow field incompressible? Solution Condition for incompressibility: Thus to guarantee incompressibility, constants a and c must satisfy the following relationship: a = −3c 25
- 26. Example 5 An idealized incompressible flow has the proposed three- dimensional velocity distribution Find the appropriate form of the function f(y) which satisfies the continuity relation. Solution: Simply substitute the given velocity components into the incompressible continuity equation: 26
- 27. Example 6 For a certain incompressible flow field it is suggested that the velocity components are given by the equations Is this a physically possible flow field? Explain. 27
- 28. Example 7 For a certain incompressible, two-dimensional flow field the velocity component in the y direction is given by the equation Determine the velocity in the x direction so that the continuity equation is satisfied. 28
- 29. 29 Example 7 - solution
- 30. Example 8 The radial velocity component in an incompressible, two dimensional flow field is Determine the corresponding tangential velocity component, required to satisfy conservation of mass. Solution. The continuity equation for incompressible steady flow in cylindrical coordinates is given by 30
- 31. 31 Example 8
- 32. The Stream Function Consider the simple case of incompressible, two-dimensional flow in the xy-plane. The continuity equation in Cartesian coordinates reduces to (1) A clever variable transformation enables us to rewrite this equation (Eq. 1) in terms of one dependent variable (ψ) instead of two dependent variables (u and v). We define the stream function ψ as (2) 32
- 33. Substitution of Eq. 2 into Eq. 1 yields which is identically satisfied for any smooth function ψ(x, y). What have we gained by this transformation? First, as already mentioned, a single variable (ψ) replaces two variables (u and v)—once ψ is known, we can generate both u and v via Eq. 2 and we are guaranteed that the solution satisfies continuity, Eq. 1. Second, it turns out that the stream function has useful physical significance . Namely, Curves of constant ψ are streamlines of the flow. 33 The Stream Function
- 34. This is easily proven by considering a streamline in the xy-plane 34 The Stream Function Curves of constant stream function represent streamlines of the flow
- 35. The change in the value of ψ as we move from one point (x, y) to a nearby point (x + dx, y + dy) is given by the relationship: Along a line of constant ψ we have dψ = 0 so that and, therefore, along a line of constant ψ 35 The Stream Function
- 36. Along a streamline: where we have applied Eq. 2, the definition of ψ. Thus along a streamline: But for any smooth function ψ of two variables x and y, we know by the chain rule of mathematics that the total change of ψ from point (x, y) to another point (x + dx, y + dy) some infinitesimal distance away is 36 The Stream Function
- 37. Total change of ψ: By comparing the above two equations we see that dψ = 0 along a streamline; 37 The Stream Function
- 38. In cylindrical coordinates the continuity equation for incompressible, plane, two dimensional flow reduces to and the velocity components, and can be related to the stream function, through the equations 38 The Stream Function
- 39. The Differential Equation of Linear Momentum Using the same elemental control volume as in mass conservation, for which the appropriate form of the linear momentum relation is 39
- 40. Again the element is so small that the volume integral simply reduces to a derivative term: The momentum fluxes occur on all six faces, three inlets and three outlets. 40 The Differential Equation of Linear Momentum
- 41. Introducing these terms A simplification occurs if we split up the term in brackets as follows: The term in brackets on the right-hand side is seen to be the equation of continuity, which vanishes identically 41 The Differential Equation of Linear Momentum
- 42. The long term in parentheses on the right-hand side is the total acceleration of a particle that instantaneously occupies the control volume: Thus now we have This equation points out that the net force on the control volume must be of differential size and proportional to the element volume. 42 The Differential Equation of Linear Momentum
- 43. These forces are of two types, body forces and surface forces. Body forces are due to external fields (gravity, magnetism, electric potential) that act on the entire mass within the element. The only body force we shall consider is gravity. The gravity force on the differential mass ρ dx dy dz within the control volume is The surface forces are due to the stresses on the sides of the control surface. These stresses are the sum of hydrostatic pressure plus viscous stresses τij that arise from motion with velocity gradients 43 The Differential Equation of Linear Momentum
- 44. 44 The Differential Equation of Linear Momentum
- 45. 45 The Differential Equation of Linear Momentum Fig. Elemental Cartesian fixed control volume showing the surface forces in the x direction only.
- 46. The net surface force in the x direction is given by Splitting into pressure plus viscous stresses where dv = dx dy dz. Similarly we can derive the y and z forces per unit volume on the control surface 46 The Differential Equation of Linear Momentum
- 47. The net vector surface force can be written as 47 The Differential Equation of Linear Momentum
- 48. In divergence form is the viscous stress tensor acting on the element The surface force is thus the sum of the pressure gradient vector and the divergence of the viscous stress tensor 48 The Differential Equation of Linear Momentum
- 49. The basic differential momentum equation for an infinitesimal element is thus In words 49 The Differential Equation of Linear Momentum
- 50. the component equations are This is the differential momentum equation in its full glory, and it is valid for any fluid in any general motion, particular fluids being characterized by particular viscous stress terms. 50 The Differential Equation of Linear Momentum
- 51. Inviscid Flow: Eulers’ Equation For Frictionless flow τij =0, for which This is Eulers’equation for inviscid flow 51
- 52. Newtonian Fluid: Navier-Stokes Equations For a newtonian fluid, the viscous stresses are proportional to the element strain rates and the coefficient of viscosity. where μ is the viscosity coefficient Substitution gives the differential momentum equation for a newtonian fluid with constant density and viscosity: 52
- 53. These are the incompressible flow Navier-Stokes equations named after C. L. M. H. Navier (1785–1836) and Sir George G. Stokes (1819–1903), who are credited with their derivation. 53 Newtonian Fluid: Navier-Stokes Equations
- 54. Inviscid Flow Shearing stresses develop in a moving fluid because of the viscosity of the fluid. We know that for some common fluids, such as air and water, the viscosity is small, therefore it seems reasonable to assume that under some circumstances we may be able to simply neglect the effect of viscosity (and thus shearing stresses). Flow fields in which the shearing stresses are assumed to be negligible are said to be inviscid, nonviscous, or frictionless. For fluids in which there are no shearing stresses the normal stress at a point is independent of direction—that is σxx = σyy = σzz. 54
- 55. Euler’s Equations of Motion For an inviscid flow in which all the shearing stresses are zero and the Euler’s equation of motion is written as In vector notation Euler’s equations can be expressed as 55 Inviscid Flow
- 56. Vorticity and Irrotationality The assumption of zero fluid angular velocity, or irrotationality, is a very useful simplification. Here we show that angular velocity is associated with the curl of the local velocity vector. The differential relations for deformation of a fluid element can be derived by examining the Fig. below. Two fluid lines AB and BC, initially perpendicular at time t, move and deform so that at t + dt they have slightly different lengths A’B’ and B’C’ and are slightly off the perpendicular by angles dα and dβ. 56
- 58. We define the angular velocity ωz about the z axis as the average rate of counterclockwise turning of the two lines: But from the fig. dα and dβ are each directly related to velocity derivatives in the limit of small dt: Substitution results 58 Vorticity and Irrotationality
- 59. The vector is thus one-half the curl of the velocity vector A vector twice as large is called the vorticity 59 Vorticity and Irrotationality
- 60. Many flows have negligible or zero vorticity and are called irrotational. Example. For a certain two-dimensional flow field the velocity is given by the equation Is this flow irrotational? Solution. For the prescribed velocity field 60 Vorticity and Irrotationality
- 62. Velocity Potential The velocity components of irrotational flow can be expressed in terms of a scalar function ϕ(x, y, z, t) as where ϕ is called the velocity potential. In vector form, it can be written as so that for an irrotational flow the velocity is expressible as the gradient of a scalar function ϕ. The velocity potential is a consequence of the irrotationality of the flow field, whereas the stream function is a consequence of conservation of mass 62
- 63. It is to be noted, however, that the velocity potential can be defined for a general three-dimensional flow, whereas the stream function is restricted to two-dimensional flows. For an incompressible fluid we know from conservation of mass that and therefore for incompressible, irrotational flow (with ) it follows that 63 Velocity Potential
- 64. Velocity Potential This differential equation arises in many different areas of engineering and physics and is called Laplace’s equation. Thus, inviscid, incompressible, irrotational flow fields are governed by Laplace’s equation. This type of flow is commonly called a potential flow. Potential flows are irrotational flows. That is, the vorticity is zero throughout. If vorticity is present (e.g., boundary layer, wake), then the flow cannot be described by Laplace’s equation.64
- 65. For some problems it will be convenient to use cylindrical coordinates, r,θ, and z. In this coordinate system the gradient operator is 65 Velocity Potential
- 67. Example 1 The two-dimensional flow of a nonviscous, incompressible fluid in the vicinity of the corner of Fig. is described by the stream function where ψ has units of m2/s when r is in meters. Assume the fluid density is 103 kg/m3 and the x–y plane is horizontal that is, there is no difference in elevation between points (1) and (2). FIND a) Determine, if possible, the corresponding velocity potential. b) If the pressure at point (1) on the wall is 30 kPa, what is the pressure at point (2)? 67
- 68. Example 1 Solution The radial and tangential velocity components can be obtained from the stream function as 68
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- 73. Basic Plane Potential Flows For simplicity, only plane (two-dimensional) flows will be considered. In this case, by using Cartesian coordinates Since we can define a stream function for plane flow, we can also let 73
- 74. Uniform Flow The simplest plane flow is one for which the streamlines are all straight and parallel, and the magnitude of the velocity is constant. This type of flow is called a uniform flow. For example, consider a uniform flow in the positive x direction as is illustrated in Fig a. 74
- 75. In this instance, u = U and v = 0, and in terms of the velocity potential These two equations can be integrated to yield where C is an arbitrary constant, which can be set equal to zero. Thus, for a uniform flow in the positive x direction 75 Uniform Flow
- 76. The corresponding stream function can be obtained in a similar manner, since and, therefore, These results can be generalized to provide the velocity potential and stream function for a uniform flow at an angle α with the x axis, as in Fig. b. For this case and 76 Uniform Flow
- 77. Source and Sink Consider a fluid flowing radially outward from a line through the origin perpendicular to the x–y plane as is shown in Fig. Let m be the volume rate of flow emanating from the line (per unit length), and therefore to satisfy conservation of mass or 77
- 78. A source or sink represents a purely radial flow. Since the flow is a purely radial flow, , the corresponding velocity potential can be obtained by integrating the equations It follows that If m is positive, the flow is radially outward, and the flow is considered to be a source flow. If m is negative, the flow is toward the origin, and the flow is considered to be a sink flow. The flowrate, m, is the strength of the source or sink. 78 Source and Sink
- 79. The stream function for the source can be obtained by integrating the relationships To yield The streamlines (lines of ψ = constant ) are radial lines, and the equipotential lines (lines of ϕ = constant) are concentric circles centered at the origin. 79 Source and Sink
- 80. Example 2 A nonviscous, incompressible fluid flows between wedge- shaped walls into a small opening as shown in Fig. The velocity potential (in ft/s2), which approximately describes this flow is Determine the volume rate of flow (per unit length) into the opening. 80
- 81. 81 The negative sign indicates that the flow is toward the opening, that is, in the negative radial direction
- 82. Vortex We next consider a flow field in which the streamlines are concentric circles—that is, we interchange the velocity potential and stream function for the source. Thus, let and 82 where K is a constant. In this case the streamlines are concentric circles with and This result indicates that the tangential velocity varies inversely with the distance from the origin
- 83. A mathematical concept commonly associated with vortex motion is that of circulation. The circulation, Γ, is defined as the line integral of the tangential component of the velocity taken around a closed curve in the flow field. In equation form, Γ, can be expressed as 83 Circulation where the integral sign means that the integration is taken around a closed curve, C, in the counterclockwise direction, and ds is a differential length along the curve
- 84. For an irrotational flow This result indicates that for an irrotational flow the circulation will generally be zero. However, for the free vortex with , , the circulation around the circular path of radius r is which shows that the circulation is nonzero. However, for irrotational flows the circulation around any path that does not include a singular point will be zero.84 Circulation
- 85. The velocity potential and stream function for the free vortex are commonly expressed in terms of the circulation as and 85 Circulation
- 86. Doublet Doublet is formed by combining a source and sink in a special way. Consider the equal strength, source–sink pair shown. The combined stream function for the pair is 86
- 87. which can be rewritten as From the Fig. above it follows that And These results substitution gives 87 Doublet
- 88. So that For small values of the distance a 88 Doublet A doublet is formed by letting a source and sink approach one another. since the tangent of an angle approaches the value of the angle for small angles
- 89. The so-called doublet is formed by letting the source and sink approach one another while increasing the strength so that the product remains constant. In this case, since where K, a constant equal to is called the strength of the doublet. The corresponding velocity potential for the doublet is 89 Doublet
- 90. Plots of lines of constant ψ reveal that the streamlines for a doublet are circles through the origin tangent to the x axis as shown in fig below. 90 Doublet
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- 92. Rankine Ovals To study the flow around a closed body, a source and a sink of equal strength can be combined with a uniform flow as shown in Fig. below. 92
- 93. The stream function for this combination is 93 Rankine Ovals
- 94. Using the stream function for the source–sink pair, the stream function for Rankine Ovals can be written as Or The corresponding streamlines for this flow field are obtained by setting ψ = constant. If several of these streamlines are plotted, it will be discovered that the streamline ψ =0 forms a closed body as shown in fig. above. 94 Rankine Ovals
- 95. Stagnation points occur at the upstream and downstream ends of the body. These points can be located by determining where along the x axis the velocity is zero. The stagnation points correspond to the points where the uniform velocity, the source velocity, and the sink velocity all combine to give a zero velocity. The locations of the stagnation points depend on the value of a, m, and U. The body half-length, When y= 0), can be expressed as or 95 Rankine Ovals
- 96. The body half-width, h, can be obtained by determining the value of y where the y axis intersects the ψ = 0 streamline. Or both are functions of the dimensionless parameter, . A large variety of body shapes with different length to width ratios can be obtained by using different values of Ua/m, 96 Rankine Ovals
- 97. As this parameter becomes large, flow around a long slender body is described, whereas for small values of the parameter, flow around a more blunt shape is obtained 97 Rankine Ovals
- 98. Flow around a Circular Cylinder When the distance between the source–sink pair approaches zero, the shape of the Rankine oval becomes more blunt and in fact approaches a circular shape. Since the Doublet was developed by letting a source–sink pair approach one another, it might be expected that a uniform flow in the positive x direction combined with a doublet could be used to represent flow around a circular cylinder. This combination gives for the stream function and for the velocity potential 98
- 99. In order for the stream function to represent flow around a circular cylinder it is necessary that where a is the radius of the cylinder. which indicates that the doublet strength, K, must be equal to Thus, the stream function for flow around a circular cylinder can be expressed as 99 Flow around a Circular Cylinder
- 100. and the corresponding velocity potential is The velocity components are On the surface of the cylinder it follows 100 Flow around a Circular Cylinder
- 101. Fig. The flow around a circular cylinder. 101 Flow around a Circular Cylinder
- 102. Solved Problems 1. The velocity potential for a certain flow field is ϕ = 4xy. Determine the corresponding stream function. 102
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- 104. 2. The stream function for an incompressible, two dimensional flow field is Where a and b are constants. Is this an irrotational flow? Explain. 104 bxay 2
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- 106. 3. The stream function for a given two dimensional flow filed is Determine the corresponding velocity potential. 106 32 )3/5(5 yyx
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- 108. 4. Determine the stream function corresponding to the velocity potential ϕ = x3 – 3xy2. Sketch the streamline ψ = 0, which passes through the origin. 108
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- 110. 5. The velocity potential for a certain inviscid, incompressible flow field is given by the equation Where ϕ has the units of m2/s when x and y are in meters. Determine the pressure at the point x = 2 m, y = 2m if the pressure at x = 1 m, y = 1 m is 200 kPa. Elevation changes can be neglected and the fluid is water. 110 32 ) 3 2 (2 yyx
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- 112. 6. Water is flowing between wedge-shaped walls into a small opening as shown in the Fig. below. The velocity potential with units m2/s for this flow is ϕ = -2ln r with r in meters. Determine the pressure differential between points A and B. 112
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- 114. 7. An ideal fluid flows between the inclined walls of a two dimensional channel into a sink located at origin. The velocity potential for this flow field is 114
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- 117. End of Chapter 4 Next Lecture Chapter 5: Dimensional Analysis And Similitude 117