fluid06 - FUNDAMENTALS OF FLUID MECHANICS Chapter 6 Flow...

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Unformatted text preview: FUNDAMENTALS OF FLUID MECHANICS Chapter 6 Flow Analysis Using Differential Methods Jyh-Cherng Shieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 12/24/2007 1 MAIN TOPICS Fluid Element Kinematics Fluid Conservation of Mass Conservation Conservation of Linear Momentum Conservation Inviscid Flow Inviscid Some Basic, Plane Potential Flow Some 2 Motion of a Fluid Element Fluid Translation: The element moves from one point to another. Fluid Fluid Rotation: The element rotates about any or all of the x,y,z axes. Fluid Fluid Deformation: Fluid Angular Deformation:The element’s angles between the sides Angular change. Linear Deformation:The element’s sides stretch or contract. Linear 3 velocity Fluid Translation velocity and acceleration The velocity of a fluid particle can be expressed The r r r r v V = V ( x , y, z , t ) = u i + v j + wk Velocity field The total acceleration of the particle is given by The r r r r r r DV ∂V ∂V dx ∂V dy ∂V dz a= = + + + Dt ∂t ∂x dt ∂y dt ∂z dt dx dy dz = u, = v, = w dt dt dt r r r r r r DV ∂V ∂V ∂V ∂V +v +w ⇒a= = +u Dt r ∂t ∂x ∂y ∂z Acceleration field r DV a= is called the material , or substantial derivative. Dt 4 Physical Significance r r r r r r DV ∂V ∂V ∂V ∂V a= +v +w + =u Dt ∂x ∂y ∂z ∂t Total Acceleration Of a particle Local Acceleration Convective Acceleration r r r ∂V v r DV a= = (V ⋅ ∇)V + Dt ∂t 5 Scalar Component ∂u ∂u ∂u ∂u +u +v +w ∂z ∂t ∂x ∂y ∂v ∂v ∂v ∂v ay = +u +v +w ∂t ∂x ∂y ∂z ∂w ∂w ∂w ∂w az = +u +v +w ∂t ∂x ∂y ∂z ax = Rectangular coordinates system ∂Vr ∂Vr Vθ ∂Vr V 2 θ ∂V ar = + Vr + − + Vz r ∂t ∂r r ∂θ r ∂z ∂Vθ ∂Vθ Vθ ∂Vθ Vr Vθ ∂Vθ aθ = + Vr + + + Vz ∂t ∂r r ∂θ r ∂z ∂Vz Vθ ∂Vz ∂Vz ∂Vz az = + Vr + + Vz ∂r ∂t r ∂θ ∂z Cylindrical coordinates system 6 Linear Translation All points in the element have All the same velocity (which is only true if there are o velocity gradients), then the element will simply translate from one position to another. 7 1/2 Linear Deformation 1/2 The shape of the fluid element, described by the angles at The its vertices, remains unchanged, since all right angles continue to be right angles. A change in the x dimension requires a nonzero value of change ∂u / ∂x A ……………… y ∂ v / ∂ y ……………… A ……………… z ∂ w / ∂ z ……………… 8 2/2 Linear Deformation 2/2 The change in length of the sides may produce change in The volume of the element. The change in ⎛ ∂u ⎞ δV = ⎜ δx ⎟( δyδz )( δt ) ⎝ ∂x ⎠ The rate at which the δV is changing per unit volume due to gradient ∂u/ ∂x 1 d (δV ) ∂u = δV dt ∂x If ∂v/ ∂y and ∂w/ ∂z are involved Volumetric dilatation rate r 1 d (δV ) ∂u ∂v ∂w = + + = ∇⋅V δV dt ∂x ∂y ∂z 9 1/4 Angular Rotation 1/4 δα ωOA = lim The angular velocity of line OA δt → 0 δ t ⎛ ∂v ⎞ ⎜ ⎟δxδt ∂v For small angles tan δα = δα = ⎝ ∂x ⎠ = δt & δx ∂x ωOA ∂v = CCW ∂x ωOB ∂u = ∂y CW “-” for CCW 10 2/4 Angular Rotation 2/4 The rotation of the element about the z-axis is defined as the average of the angular velocities ωOA and ωOB of the two mutually perpendicular lines OA and OB. 1 1 ⎛ ∂v ∂u ⎞ ωz = (ωOA + ωOB ) = ⎜ ⎟ ⎜ ∂x − ∂y ⎟ 2 2⎝ ⎠ 1 ⎛ ∂w ∂v ⎞ ωx = ⎜ ⎜ ∂y − ∂z ⎟ ⎟ 2⎝ ⎠ In vector form 1 ⎛ ∂u ∂w ⎞ ωy = ⎜ − ⎟ 2 ⎝ ∂z ∂x ⎠ r v v r ω = ωx i + ω y j + ωz k 11 3/4 Angular Rotation 3/4 1 ⎛ ∂u ∂w ⎞ 1 ⎛ ∂w ∂v ⎞ 1 ⎛ ∂v ∂u ⎞ ⎟ ωy = ⎜ − ωx = ⎜ − ⎟ ωz = ⎜ − ⎟ ⎜ ∂y ∂z ⎟ 2 ⎝ ∂z ∂x ⎠ 2⎝ 2 ⎜ ∂x ∂y ⎟ ⎠ ⎝ ⎠ r 1 ⎡ ⎛ ∂w ∂v ⎞ r ⎛ ∂u ∂w ⎞ r ⎛ ∂v ∂u ⎞ r ⎤ ω = ⎢⎜ ⎜ ∂y − ∂z ⎟ i + ⎜ ∂z − ∂x ⎟ j + ⎜ ∂x − ∂y ⎟ k ⎥ ⎜ ⎟ ⎟ 2 ⎢⎝ ⎝ ⎠ ⎝ ⎠ ⎠ ⎣ ⎦ r1 r 1 ⎡ ∂w ∂v ⎤ r 1 ⎡ ∂u ∂w ⎤ r 1 ⎡ ∂v ∂u ⎤ r r1 ω = curlV = ∇ × V = ⎢ − ⎥i + ⎢ − ⎥ j + 2 ⎢ ∂x − ∂y ⎥ k 2 2 2 ⎣ ∂y ∂z ⎦ 2 ⎣ ∂z ∂x ⎦ ⎣ ⎦ Defining vorticity r r ζ = 2ω = ∇ × V r Defining irrotation ∇ × V = 0 12 4/4 Angular Rotation 4/4 r i r1 r 1∂ r1 ω = curlV = ∇ × V = 2 2 2 ∂x u r j ∂ ∂y v r k ∂ ∂z w 1 ⎡ ∂w ∂v ⎤ r 1 ⎡ ∂u ∂w ⎤ r 1 ⎡ ∂v ∂u ⎤ r =⎢ − ⎥i + ⎢ − ⎥ j + 2 ⎢ ∂x − ∂y ⎥ k 2 ⎣ ∂y ∂z ⎦ 2 ⎣ ∂z ∂x ⎦ ⎣ ⎦ 13 Vorticity Defining Vorticity ζ which is a measurement of the rotation of a Defining fluid element as it moves in the flow field: r r r r ζ = 2 ω = curl V = ∇ × V r r 1 ⎡ ⎛ ∂w ∂v ⎞ r ⎛ ∂u ∂w ⎞ r ⎛ ∂v ∂u ⎞ r ⎤ 1 ω = ⎢⎜ ⎜ ⎟ ⎜ ∂y − ∂z ⎟ i + ⎜ ∂z − ∂x ⎟ j + ⎜ ∂x − ∂y ⎟ k ⎥ = 2 ∇ × V ⎟ 2 ⎢⎝ ⎝ ⎠ ⎝ ⎠⎥ ⎠ ⎣ ⎦ In cylindrical coordinates system: In r r ⎛ 1 ∂ Vz ∂ Vθ ⎞ r ⎛ ∂ Vr ∂ Vz ⎞ r ⎛ 1 ∂ rVθ 1 ∂ Vr ⎞ − − − ∇ × V = er ⎜ ⎟ ⎟ + eθ ⎜ ⎟ + ez ⎜ ∂r ⎠ ∂z ⎠ r ∂θ ⎠ ⎝ ∂z ⎝ r ∂r ⎝ r ∂θ 14 1/2 Angular Deformation 1/2 Angular deformation of a particle is given by the sum of the two Angular angular deformation δγ = δα + δβ ⎛ ∂u ⎞ ∂u δξ = ⎜ u + δy ⎟ δt − uδt = δ yδ t ⎜ ⎟ ∂y ⎠ ∂y ⎝ δα = δη / δ x δβ = δξ / δ y ∂v ⎞ ∂v ⎛ δxδt δη = ⎜ v + δx ⎟ δt − vδt = ∂x ∂x ⎠ ⎝ ξ(Xi)η(Eta) Rate of shearing strain or the rate of angular deformation ⎛ ∂v δx ∂u δy ⎞ δt + δt ⎟ ⎜ ∂x δx ∂y δy ⎟ δα + δβ ⎝ ⎠ = ... = ⎛ ∂ v + ∂ u ⎞ & γ = lim = lim ⎜ ⎜ ∂x ∂y ⎟ ⎟ δt → 0 δt → 0 δt δt ⎝ ⎠ 15 2/2 Angular Deformation 2/2 The rate of angular deformation in xy plane ⎛ ∂v ∂u ⎞ ⎜ ⎜ ∂x + ∂y ⎟ ⎟ ⎝ ⎠ The rate of angular deformation in yz plane ⎛ ∂w ∂v ⎞ ⎜ ⎜ ∂y + ∂z ⎟ ⎟ ⎝ ⎠ The rate of angular deformation in zx plane ⎛ ∂w ∂u ⎞ + ⎜ ⎟ ∂x ∂z ⎠ ⎝ 16 Example 6.1 Vorticity For a certain two-dimensional flow field th evelocity is given by For r r r 2 2 V = 4 xy i + 2 ( x − y ) j Is this flow irrotational? 17 Example 6.1 Solution u = 4 xy v = x2 − y2 1 ⎛ ∂w ∂v ⎞ ⎜ ⎜ ∂y − ∂z ⎟ = 0 ⎟ 2⎝ ⎠ 1 ⎛ ∂u ∂w ⎞ ωy = ⎜ − ⎟=0 2 ⎝ ∂z ∂x ⎠ w=0 ωx = This flow is irrotational 1 ⎛ ∂v ∂u ⎞ ωz = ⎜ − ⎟ = 0 2 ⎜ ∂x ∂y ⎟ ⎝ ⎠ 18 1/5 Conservation of Mass 1/5 With field representation, the property fields are defined With by continuous functions of the space coordinates and time. To derive the differential equation for conservation of To mass in rectangular and in cylindrical coordinate system. The derivation is carried out by applying conservation of The mass to a differential control volume. With the control volume representation of the conservation of mass vr ∂ ∫CV d V + ∫CS V ⋅ n dA = 0 ∂t The differential form of continuity equation??? 19 2/5 Conservation of Mass 2/5 The CV chosen is an infinitesimal cube with sides of length δx, δ y, The and δ z. ∂ ∂ρ ∫ CV ρdV = ∂t δxδyδz ∂t ∂ (ρu ) δx ρu | ⎛ dx ⎞ = ρu + x +⎜ ⎟ ∂x 2 ⎝2⎠ Net rate of mass Outflow in x-direction ρu | = ρu − ⎛ δx ⎞ x −⎜ ⎟ ⎝2⎠ ∂ (ρu ) δx ∂x 2 20 3/5 Conservation of Mass 3/5 Net rate of mass Outflow in x-direction ∂ (ρu ) δx ⎤ ∂ (ρu ) δx ⎤ ∂ (ρu ) ⎡ ⎡ = ⎢ρu + ⎥δyδz − ⎢ρu − ∂x 2 ⎥δyδz = ∂x δxδyδz ∂x 2 ⎦ ⎣ ⎦ ⎣ Net rate of mass Outflow in y-direction Net rate of mass Outflow in z-direction ∂ (ρv ) δxδyδz = ⋅⋅⋅⋅⋅ = ∂y ∂ (ρw ) = ⋅⋅⋅⋅⋅ = δxδyδz ∂z 21 4/5 Conservation of Mass 4/5 Net rate of mass Outflow ⎡ ∂ (ρu ) ∂ (ρv ) ∂ (ρw ) ⎤ ⎢ ∂x + ∂y + ∂z ⎥δxδyδz ⎣ ⎦ The differential equation for conservation of mass r ∂ρ ∂ (ρu ) ∂ (ρv ) ∂ (ρw ) ∂ρ + + + = + ∇ ⋅ ρV = 0 Continuity equation ∂t ∂x ∂y ∂z ∂t 22 5/5 Conservation of Mass 5/5 Incompressible fluid Incompressible r ∂u ∂v ∂w + + = ∇⋅V = 0 ∂x ∂y ∂z Steady flow Steady r ∂ (ρu ) ∂ (ρv ) ∂ (ρw ) + + = ∇ ⋅ ρV = 0 ∂x ∂y ∂z 23 Example 6.2 Continuity Equation The velocity components for a certain incompressible, steady flow The field are u = x 2 + y2 + z2 v = xy + yz + z w=? Determine the form of the z component, w, required to satisfy the continuity equation. 24 Example 6.2 Solution The continuity equation ∂u ∂v ∂w + =0 + ∂x ∂y ∂z ∂u = 2z ∂x ∂v =x+z ∂y ∂w = − 2 x − ( x + z ) = − 3x − z ∂z z2 ⇒ w = − 3xz − + f (x, y) 2 25 Conservation of Mass 1/3 Cylindrical Coordinate System 1/3 The CV chosen is an infinitesimal cube with sides of The length dr, rdθ, and dz. dr dz The net rate of mass flux out through the control surface The ∂ ρ Vr ∂ ρ Vθ ∂ ρ Vz ⎤ ⎡ ⎢ρ Vr + r ∂ r + ∂ θ + r ∂ z ⎥ δ rδθδ z ⎣ ⎦ The rate of change of mass The inside the control volume ∂ρ rd θ drdz ∂t 26 Conservation of Mass 2/3 Cylindrical Coordinate System 2/3 The continuity equation ∂ ρ 1 ∂ ( rρ Vr ) 1 ∂ (ρ Vθ ) ∂ (ρ Vz ) + + + =0 ∂t r ∂r r ∂θ ∂z By “Del” operator r ∂ r1∂ r∂ ∇ = er + eθ +k ∂r r ∂θ ∂z The continuity equation becomes r ∂ρ + ∇ ⋅ ρV = 0 ∂t 27 Conservation of Mass Cylindrical Coordinate System 3/3 Incompressible fluid Incompressible r 1 ∂ ( rVr ) 1 ∂ ( Vθ ) ∂ ( Vz ) + + = ∇⋅V = 0 r ∂r r ∂θ ∂z Steady flow Steady r 1 ∂ ( rρ Vr ) 1 ∂ (ρ Vθ ) ∂ (ρ Vz ) + + = ∇ ⋅ ρV = 0 r ∂r r ∂θ ∂z 28 1/6 Stream Function 1/6 Streamlines ? Lines tangent to the instantaneous velocity vectors at Streamlines every point. Stream function Ψ(x,y) [Psi] ? Used to represent the velocity Stream component u(x,y,t) and v(x,y,t) of a two-dimensional incompressible flow. Define a function Ψ(x,y), called the stream function, which relates Define the velocities shown by the figure in the margin as ∂ψ u= ∂y ∂ψ v=− ∂x 29 2/6 Stream Function 2/6 The stream function Ψ(x,y) satisfies the two-dimensional form of The the incompressible continuity equation ∂u ∂v ∂ 2ψ ∂ 2ψ + =0⇒ − =0 ∂x ∂y ∂ x ∂ y ∂ y∂ x Ψ(x,y) ? Still unknown for a particular problem, but at least we have simplify the analysis by having to determine only one unknown, Ψ(x,y) , rather than the two function u(x,y) and v(x,y). 30 3/6 Stream Function 3/6 Another advantage of using stream function is related to the fact that Another line along which Ψ(x,y) =constant are streamlines. How to prove ? From the definition of the streamline that the slope How at any point along a streamline is given by dy ⎞ v = ⎟ dx ⎠ streamline u Velocity and velocity component along a streamline 31 4/6 Stream Function 4/6 The change of Ψ(x,y) as we move from one point (x,y) The to a nearly point (x+dx,y+dy) is given by ∂ψ ∂ψ dψ = dx + dy = − vdx + udy ∂x ∂y >> d ψ = 0 >> − vdx + udy = 0 Along a line of constant Ψ dy ⎞ v = ⎟ dx ⎠ streamline u This is the definition for a streamline. Thus, if we know the function Ψ(x,y) we can plot lines of constant Ψto provide the family of streamlines that are helpful in visualizing the pattern of flow. There are an infinite number of streamlines that make up a particular flow field, since for each constant value assigned to Ψa streamline can be drawn. 32 5/6 Stream Function 5/6 The actual numerical value associated with a particular streamline is The not of particular significance, but the change in the value of Ψ is related to the volume rate of flow. For a unit depth, the flow rate across AB is For y2 y 2 ∂ψ ψ2 q = ∫ udy = ∫ dy = ∫ d ψ = ψ 2 − ψ 1 y1 y1 ∂ y ψ1 For a unit depth, the flow rate across BC is For ψ x x ∂ψ q = ∫ vdx = − ∫ dx = − ∫ d ψ = ψ 2 ψ 1 2 x1 2 x1 1 ∂x ψ2 33 6/6 Stream Function 6/6 Thus the volume flow rate between any two streamlines can be Thus written as the difference between the constant values of Ψ defining two streamlines. The velocity will be relatively high wherever the streamlines are The close together, and relatively low wherever the streamlines are far apart. 34 Stream Function Cylindrical Coordinate System Cylindrical For a two-dimensional, incompressible flow in the rθ For plane, conservation of mass can be written as: ∂ ( rv r ) ∂v θ + =0 ∂r ∂θ The velocity components can be related to the stream The function, Ψ(r,θ) through the equation 1 ∂ψ vr = r ∂θ and ∂ψ vθ = − ∂r 35 Example 6.3 Stream Function The velocity component in a steady, incompressible, two The dimensional flow field are u = 2y v = 4x Determine the corresponding stream function and show on a sketch several streamlines. Indicate the direction of glow along the streamlines. 36 Example 6.3 Solution From the definition of the stream function ∂ψ u= = 2y ∂y ∂ψ v=− = 4x ∂x ψ = y 2 + f1 (x) ψ = −2 x + y 2 2 ψ = − 2 x 2 + f 2 (y) Ψ=0 ψ = −2 x + y + C 2 2 For simplicity, we set C=0 2 2 y x − =1 ψ ψ/2 Ψ≠0 37 Conservation of Linear Momentum Applying Newton’s second law to control volume Applying r r DP r r r F= Psystem = ∫ Vdm = ∫ Vρd V M ( system ) V ( system ) Dt SYS ( ) r r r r r r D V δm ⎛ ∂V ∂V ∂V ∂V ⎞ δF = = δm ⎜ ⎟ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ Dt ⎠ ⎝ r r DV = δm = δm a Dt For a infinitesimal system of mass dm, what’s the The infinitesimal The differential form of linear momentum equation? 38 1/2 Forces Acting on Element 1/2 The forces acting on a fluid element may be classified as body forces and surface forces; surface forces include normal forces and normal tangential (shear) forces. r r r δF = δFS + δFB r r r = δFsx i + δFsy j + δFsz k r r r + δFbx i + δFby j + δFbz k Surface forces acting on a fluid element can be described in terms of normal and shearing stresses. δF1 δFn δF2 τ 1 = lim τ 2 = lim σ n = lim δt → 0 δA δt → 0 δA δt → 0 δA 39 2/2 Forces Acting on Element 2/2 ⎛ ∂σ xx ∂τ yx ∂τzx ⎞ δFsx = ⎜ ⎟ ⎜ ∂x + ∂y + ∂z ⎟δxδyδz ⎠ ⎝ ⎛ ∂τ xy ∂σ yy ∂τ zy ⎞ δFsy = ⎜ ⎜ ∂x + ∂y + ∂z ⎟δxδyδz ⎟ ⎝ ⎠ ⎛ ∂τ xz ∂τ yz ∂σzz ⎞ δFsz = ⎜ ⎟ ⎜ ∂x + ∂y + ∂z ⎟δxδyδz ⎠ ⎝ δFbx = ρg x δxδyδz δFby = ρg y δxδyδz Equation of Motion δFbz = ρg z δxδyδz 40 Equation of Motion δFx = δma x δFy = δma y δFz = δma z ⎛ ∂u ∂σ xx ∂τ yx ∂τ zx ∂u ∂u ∂u ⎞ + + = ρ⎜ + u +v +w ⎟ ρg x + ⎜ ∂t ∂x ∂y ∂z ∂x ∂y ∂z ⎟ ⎝ ⎠ ∂τ xy ∂σ yy ∂τzy ⎛ ∂v ∂v ∂v ∂v ⎞ ρg y + + + = ρ⎜ + u +v +w ⎟ ⎜ ∂t ∂x ∂y ∂z ∂x ∂y ∂z ⎟ ⎝ ⎠ ⎛ ∂w ∂τ xz ∂τ yz ∂σzz ∂w ∂w ∂w ⎞ ρg z + + + = ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ ⎟ ∂x ∂y ∂z ⎝ ⎠ These are the differential equations of motion for any fluid satisfying the continuum assumption. How to solve u,v,w ? 41 Double Subscript Notation for Stresses τ xy The direction of the stress The direction of the normal to the plane on which the stress acts 42 Inviscid Flow Shearing stresses develop in a moving fluid because of the viscosity Shearing of the fluid. For some common fluid, such as air and water, the viscosity is small, For and therefore it seems reasonable to assume that under some circumstances we may be able to simply neglect the effect of viscosity. Flow fields in which the shearing stresses are assumed to be Flow negligible are said to be inviscid, nonviscous, or frictionless. Define the pressure, p, as the negative of the normal stress − p = σ xx = σ yy = σzz 43 Euler’s Equation of Motion Under frictionless condition, the equations of motion are reduced to Euler’s Equation: ⎛ ∂u ∂p ∂u ∂u ∂u ⎞ ρg x − = ρ⎜ ⎜ t + u ∂x + v ∂y + w ∂z ⎟ ⎟ ∂x ⎝ ⎠ ⎛ ∂v ∂p ∂v ∂v ∂v ⎞ ρg y − = ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ ⎟ ∂y ⎝ ⎠ ρg z − ⎛ ∂w ∂w ∂w ⎞ ∂p ∂w = ρ⎜ +u +v +w ⎟ ⎜ ∂t ∂x ∂y ∂z ⎟ ∂z ⎝ ⎠ r r r⎞ ⎛ ∂V r ρg − ∇p = ρ ⎜ + ( V ⋅ ∇) V ⎟ ⎜ ∂t ⎟ ⎝ ⎠ 44 1/3 Bernoulli Equation 1/3 Euler’s equation for steady flow along a streamline is Euler r r r ρg − ∇ p = ρ ( V ⋅ ∇ ) V Selecting the coordinate system with the z-axis vertical so that the acceleration of gravity vector can be expressed as r g = − g∇z ( ) ( ) ( r r 1 rr r r V⋅∇ V = ∇ V⋅V − V × ∇× V 2 − ρg∇ z − ∇ p = ρ 2 ) Vector identity …. rr r r ∇(V ⋅ V ) − ρ (V × ∇ × V ) 45 2/3 Bernoulli Equation 2/3 ( r r ∇p 1 2 + ∇(V ) + g∇z = V × ∇ × V ρ2 ) r r r V × ∇ × V perpendicular to V ( ) r r ⋅ d s ∇p r 1 r r vr 2 ⋅ d s + ∇(V )⋅ d s + g∇z ⋅ d s = V × ∇ × V ⋅ d s ρ 2 [( With )] s r r r d s = dx i + dy j + dzk r ∂p ∂p ∂p ∇p ⋅ d s = dx + dy + dz = dp ∂x ∂y ∂z 46 3/3 Bernoulli Equation 3/3 r r ∇p r 1 2 ⋅ d s + ∇(V ) ⋅ d s + g∇z ⋅ d s = 0 ρ 2 dp 1 + d (V 2 ) + gdz = 0 ρ2 dp V 2 + + gz = cons tan t Integrating … ∫ ρ 2 For steady inviscid, incompressible fluid ( commonly called ideal fluids) along a streamline along p V2 + + gz = cons tan t ρ2 Bernoulli equation 47 1/4 Irrotational Flow 1/4 Irrotation ? The irrotational condition is r ∇ ×V = 0 In rectangular coordinates system ∂v ∂u ∂w ∂v ∂u ∂w − = − = − =0 ∂x ∂y ∂y ∂z ∂z ∂x In cylindrical coordinates system 1 ∂ Vz ∂ Vθ ∂ Vr ∂ Vz 1 ∂ rVθ 1 ∂ Vr = − = =0 − − r ∂θ r ∂r r ∂θ ∂z ∂z ∂r 48 2/4 Irrotational Flow 2/4 A general flow field would not be irrotational flow. general A special uniform flow field is an example of an irrotation special flow 49 3/4 Irrotational Flow 3/4 A general flow field A solid body is placed in a uniform stream of fluid. Far away from solid the body remain uniform, and in this far region the flow is irrotational. The flow around the body remains irrotational except very near the The boundary. Near the boundary the Near velocity changes rapidly from zero at the boundary (no-slip condition) to some relatively large value in a short distance from the boundary. Chapter 9 50 4/4 Irrotational Flow 4/4 A general flow field Flow from a large reservoir enters a pipe through a streamlined Flow entrance where the velocity distribution is essentially uniform. Thus, at entrance the flow is irrotational. (b) In the central core of the pipe the flow remains irrotational for some In distance. The boundary layer will develop along the wall and grow in The thickness until it fills the pipe. Viscous forces are dominant Chapter 8 51 1/3 Bernoulli Equation for Irrotational Flow 1/3 The Bernoulli equation for steady, incompressible, and inviscid The flow is p V2 + + gz = cons tan t ρ2 The equation can be applied between any two points on the same The streamline. In general, the value of the constant will vary from streamline streamline to streamline. streamline Under additional irrotational condition, the Bernoulli equation ? Under the Starting with Euler’s equation in vector form r 1 rr r r r r 1 ( V ⋅ ∇ ) V = − ∇ p − gk = ∇ V ⋅ V − V × ∇ × V ρ 2 ( ) ( ) ZERO Regardless of the direction of ds 52 2/3 Bernoulli Equation for Irrotational Flow 2/3 r ∇ ×V = 0 With irrotaional condition irrotaional r 1 rr r r r r 1 ( V ⋅ ∇) V = − ∇p − gk = ∇ V ⋅ V − V × ∇ × V ρ 2 r 1 1 1 rr 2 ∇ V ⋅ V = ∇ V = − ∇ p − gk 2 ρ 2 ( ( ) () ) ( ) r ⋅ dr rv 1 1 v v 2 ∇ (V ) ⋅ d r = − ∇ p ⋅ d r − gk ⋅ d r 2 ρ 1 dp dp 1 2 >> d (V ) = − − gdz >> + d (V 2 ) + gdz = 0 2 ρ ρ2 53 3/3 Bernoulli Equation for Irrotational Flow 3/3 Integrating for incompressible flow 2 dp V ∫ ρ + 2 + gz = con tan t p V2 + + gz = cons tan t ρ2 This equation is valid between any two points in a steady, incompressible, inviscid, and irrotational flow. 2 2 p1 V1 p 2 V2 + + z1 = + + z2 2g γ 2g γ 54 Velocity Potential Φ(x,y,z,t) 1/4 The stream function for two-dimensional incompressible The flow is Ψ(x,y) which satisfies the continuous condtions For an irrotational flow, the velocity components can be For expressed in terms of a scalar function ψ(x,y,z,t) as ∂φ u= ∂x ∂φ v= ∂y ∂φ w= ∂z where ψ(x,y,z,t) is called the velocity potential. 55 Velocity Potential Φ(x,y,z,t) 2/4 In vector form In r V = ∇φ For an incompressible flow For r ∇⋅V = 0 called a potential flow For incompressible, irrotational flow r r ∂ 2φ ∂ 2φ ∂ 2φ V = ∇φ ⇒ ∇ ⋅ V = ∇ 2φ = 2 + 2 + 2 = 0 ∂z ∂y ∂x Laplace’s equation Laplacian operator 56 Velocity Potential Φ(x,y,z,t) 3/4 Inviscid, incompressible, irrotational fields are governed Inviscid by Laplace’s equation. This type flow is commonly called a potential flow. This To complete the mathematical formulation of a given To problem, boundary conditions have to be specified. These are usually velocities specified on the boundaries of the flow field of interest. 57 Velocity Potential Φ(x,y,z,t) 4/4 In cylindrical coordinate, r, θ, and z In 1 ∂φ ∂φ ∂φ vz = vr = vθ = r ∂θ ∂z ∂r 1 ∂ ⎛ ∂φ ⎞ 1 ∂ 2 φ ∂ 2 φ + 2 =0 ⎜r ⎟+ 2 2 r ∂r ⎝ ∂r ⎠ r ∂θ ∂z 58 Potential Flow Theory Velocity potential ψ (x,y,z,t) exists only for irrotational flow. Irrotationality may be a valid assumption for those regions of a flow in which viscous forces are negligible. In an irrotational flow, the velocity field may be defined by the potential function, ψ(x,y,z,t), the theory is often referred to as potential flow theory. 59 1/2 Laplace’s Equation 1/2 For two-dimensional, incompressible flow u= ∂ψ ∂y v=− ∂ψ ∂x (1) For two-dimensional, irrotational flow u= ∂φ ∂x v= ∂φ ∂y (2) ∂u ∂v ∂ 2ψ ∂ 2ψ (1) + irrotational condition … ∂ y − ∂ x = 0 ⇒ ∂ x 2 + ∂ y 2 = 0 (2) + continuity equation … ∂u ∂v ∂ 2φ ∂ 2φ + =0⇒ + 2 =0 2 ∂x ∂y ∂x ∂y 60 2/2 Laplace’s Equation 2/2 For a two-dimensional incompressible flow, we can define a stream function Ψ; if the flow is also irrotational, Ψ will satisfy Laplace’s equation. For an irrotational flow, we can define a velocity potential Φ; if the flow is also incompressible, Φ will satisfy Laplace’s equation. Any function ψ or Ψ that satisfies Laplace’s equation represents a possible two-dimensional, incompressible, irrotational flow field. 61 Φ and Ψ 1/2 For Ψ=constant, dΨ =0 and For ∂ψ ∂ψ dψ = dx + dy = 0 ∂x ∂y The slope of a streamline – a line of constant Ψ The dy ⎞ ∂ψ / ∂x v = ⎟ =− dx ⎠ ψ ∂ψ / ∂y u Along a line of constant ψ, dψ =0 and d φ = Along ∂φ ∂φ dx + dy = 0 ∂x ∂y The slope of a potential line – a line of constant ψ The dy ⎞ ∂φ / ∂x u =− ⎟ =− dx ⎠ φ ∂φ / ∂y v Line of constant Ψ and constant ψ are orthogonal. 62 Φ and Ψ 2/2 The lines of constant ψ (called The equipotential lines) are orthogonal to lines of constant Ψ (streamlines) at all points where they intersect.. For any potential flow a “flow net” can be For drawn that consists of a family of streamlines and equipotential lines. Velocities can be estimated from the flow Velocities net, since the velocity is inversely proportional to the streamline spacing Flow net for 90° bend 63 Example 6.4 Velocity Potential and 1/2 Inviscid Flow Pressure 1/2 The two-dimensional flow of a nonviscous, incompressible fluid in The the vicinity of the 90° corner of Figure E6.4a is described by the stream function ψ = 2 r 2 sin 2 θ Where Ψ has units of m2/s when r is in meters. (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)? 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). 64 Example 6.4 Velocity Potential and 2/2 Inviscid Flow Pressure 2/2 65 Example 6.4 Solution1/2 1 ∂ψ = 4r cos 2θ r ∂θ ∂ψ vθ = − = −4 r sin 2θ ∂r vr = vr = ∂φ = 4 r cos 2 θ ⇒ φ = 2 r 2 cos 2 θ + f1 ( θ ) ∂r vθ = 1 ∂φ = − 4 r sin 2 θ ⇒ φ = 2 r 2 cos 2 θ + f 2 ( r ) r ∂θ φ = 2 r 2 cos 2 θ + C Let C=0 2 φ = 2 r cos 2 θ 66 Example 6.4 Solution2/2 Bernoulli equation between points (1) and (2) with no elevation change p1 V12 p 2 V2 2 ρ + = + ⇒ p 2 = p1 + ( V12 − V2 2 ) γ 2g γ 2g 2 V 2 = v r 2 + v θ2 V12 = ... = 16m 2 / s 2 p 2 = ... = 36kPa V2 2 = ... = 4 m 2 / s 2 φ = 2 r 2 cos 2 θ = 4 r 2 cos θ sin θ = 4 xy 67 Some Basic, Plane Potential Flow Since Laplace’s equation is linear, various solutions can be added to Since obtain other solution – that is , if ψ1(x,y,z) and ψ2(x,y,z) are two solutions to Laplace’s equation, then ψ= ψ1 + ψ2 is also solution. The practical implication of this result is that if we have certain The basic solution we can combine them to obtain more complicated and interesting solutions. Several basic velocity potentials, which describe some relatively Several simple flows, will be determined. These basic velocity potential will be combined to represent These complicate flows. 68 1/2 Uniform Flow 1/2 A uniform flow is a simplest plane flow for which the uniform streamlines are all straight and parallel, and the magnitude of the velocity is constant. u=U and v=0 ∂φ = U, ∂x ∂ψ = U, ∂y ∂φ =0 ∂y ∂ψ =0 ∂x ⇒ φ = Ux + C ⇒ ψ = Uy 69 2/2 Uniform Flow 2/2 For a uniform flow of constant velocity V, inclined to an For angle α to the x-axis. ψ = ( V cos α ) y − ( V sin α ) x φ = − ( V sin α ) y − ( V cos α ) x 70 1/2 Source and Sink 1/2 For a source flow ( from origin radially) with volume flow For rate per unit depth m (m=2π r vr ) (m=2 m vr = 2 πr m θ vθ= 0 ⇒ ψ = 2π m φ= ln r 2π 1 ∂ψ ∂ψ vr = and vθ = − r ∂θ ∂r ∂φ 1 ∂φ vr = and vθ = ∂r r ∂θ 71 2/2 Source and Sink 2/2 For a sink flow (toward origin radially) with volume flow For rate per unit depth m m vr = − 2πr m vθ = 0 ⇒ ψ = − θ 2π m φ = − ln r 2π 1 ∂ψ ∂ψ vr = and vθ = − r ∂θ ∂r ∂φ 1 ∂φ vr = and vθ = ∂r r ∂θ 72 Example 6.5 Potential Flow - Sink A nonviscous, incompressible fluid flows between wedge-shaped nonviscous walls into a small opening as shown in Figure E6.5. The velocity potential (in ft2/s), which approximately describes this flow is φ = − 2 ln r Determine the volume rate of flow (per unit length) into the opening. 73 Example 6.5 Solution The components of velocity ∂φ 2 vr = =− ∂r r 1 ∂φ =0 vθ = r ∂θ The flowrate per unit width π/6 π q = v r Rdθ = ... = − = −1.05ft 2 / s 3 0 ∫ 74 Vortex A vortex represents a flow in which the streamlines are vortex concentric circles. Vortex motion can be either rotational or irrotational. Vortex For an irrotational vortex (ccw, center at origin) with For vortex strength K At r=0, the velocity At r=0, the velocity becomes infinite. singularity K vr= 0 vθ = r ⇒ ψ = − K ln r φ = Kθ 75 1/2 Free Vortex 1/2 Free (Irrotational) vortex (a) is that rotation refers to the Free orientation of a fluid element and not the path followed by the element. A pair of small sticks were pair placed in the flow field at location A, the sticks would rotate as they as they move to location B. 76 2/2 Free Vortex 2/2 One of the sticks, the one that is aligned the streamline, One would follow a circular path and rotate in a counterclockwise direction The other rotates in a clockwise The direction due to the nature of the flow field – that is, the part of the stick nearest the origin moves faster than the opposite end. The average velocity of the two The sticks is zero. 77 Forced Vortex If the flow were rotating as a If rigid body, such that vθ=K1r where K1 is a constant. Force vortex is rotational and Force cannot be described with a velocity potential. Force vortex is commonly Force called a rotational vortex. 78 Combined Vortex A combine vortex is one with a forced vortex as a central combine core and a velocity distribution corresponding to that of a free vortex outside the core. K vθ = r v θ = rω r > r0 r ≤ r0 79 1/3 Circulation 1/3 Circulation Γ is defined as the line integral of the Circulation tangential velocity component about any closed curve fixed in the flow: rr r Γ = V ⋅ d s = 2 ω Z dA = ( ∇ × V ) Z dA ∫ ∫ c A ∫ A r ds where the is an element vector tangent to the curve and having length ds of the element of arc. It’s positive ds corresponds to a c.c.w. direction of integration around the curve. 80 2/3 Circulation 2/3 For irrotational flow , Γ =0 For rr Γ = ∫ V ⋅ ds = c ∫ A r r ( ∇ × V ) dA = ∫ ∇ φ ⋅ d s = ∫ d φ = 0 c c For irrotational flow , Γ =0 The circulation around any path that does not The circulation around any path that does not include the singular point at the origin will be include the singular point at the origin will be zero. zero. 81 3/3 Circulation 3/3 For free vortex For K vθ = r K ( rdθ) = 2πK Γ=∫ 0 r K = Γ/2 π 2π Γ ψ = − ln r 2π Γ φ= θ 2π The circulation around any path that encloses The circulation around any path that encloses singularities will be nozero. singularities will be nozero. 82 Example 6.6 Potential Flow – Free Vortex A liquid drains from a large tank through a small opening as liquid illustrated in Figure E6.6. A vortex forms whose velocity distribution away from the tank opening can be approximated as that of a free vortex having a velocity potential Γ φ= θ 2π Determine an expression relating the surface shape to the strength of the vortex as specified by the circulation Γ. 83 Example 6.6 Solution Since the free vortex represents an irrotational flow field, the Bernoulli equation p1 V12 p 2 V2 2 + + z1 = + + z2 γ γ 2g 2g 2 2 V1 V2 = + zs 2g 2g 1 ∂φ Γ vθ = = r ∂θ 2πr At free surface p 1 = p 2 = 0 Far from the origin at point (1), V1=vθ=0 Γ2 zs = − 2 2 2π r g 84 1/2 Doublet 1/2 For a doublet ( produced mathematically by allowing a For source and a sink of numerically equal strength to merge) with a strength m The combined stream function for the pair is m ψ = − (θ1 − θ2 ) 2π tan θ1 − tan θ2 ⎛ 2πψ ⎞ tan ⎜ − ⎟ = tan (θ1 − θ2 ) = 1 + tan θ1 tan θ 2 ⎝ m⎠ tan θ1 = r sin θ (r cos θ − a ) and tan θ2 = r sin θ (r cos θ + a ) m ⎛ 2πψ ⎞ 2ar sin θ ⎛ 2ar sin θ ⎞ ⇒ tan ⎜ − ⇒ ψ = − tan −1 ⎜ 2 ⎟= 2 ⎟ m ⎠ (r − a 2 ) 2π r − a2 ⎠ ⎝ ⎝ 85 2/2 Doublet 2/2 a →0 m 2ar sin θ mar sin θ ψ=− =− 2 2 2 2 2π r − a π(r − a ) The so-called doublet is formed by letting a→0,m→∞ r 1 → 2 2 r −a r K sin θ ψ=− r ma K= π K cos θ φ= r 86 Streamlines for a Doublet Plots of lines of constant Ψ reveal that the streamlines for Plots a doublet are circles through the origin tangent to the x axis. 87 Summary 88 Superposition of Elementary Plane 1/2 Flows 1/2 Potential flows are governed by Laplace’s equation, which is a linear partial differential equation. Various basic velocity potentials and stream function, ψ and Ψ, can be combined to form new potentials and stream functions. ψ 3 = ψ1+ ψ 2 φ 3 = φ1 + φ 2 89 Superposition of Elementary Plane 2/2 Flows 2/2 Any streamline in an inviscid flow field can be considered as a solid boundary, since the conditions along a solid boundary and as streamline are the same – that is, there is no flow through the boundary or the streamline. We can combine some of the basic velocity potentials or stream functions to yield a streamline that corresponds to a particular body shape of interest, that combination can be used to describe in detail the flow around that body. Methods of superposition 90 1/4 Half-Body : Uniform Stream + Source 1/4 ψ = ψ uniform− flow + ψ source = Ur sin θ + m θ 2π m φ = φuniform− flow + φsource = Ur cos+ ln r 2π 1 ∂ψ m vr = = U cos θ + r ∂θ 2πr ∂ψ vθ = − = − U sin θ ∂r The stagnation point occurs at x=-b m m vr = U + =0⇒b= − 2πb 2πU The combination of a uniform flow and a source can be used to describe the flow around a streamlined body placed in a uniform stream. 91 2/4 Half-Body : Uniform Stream + Source 2/4 The value of the stream function at the stagnation point can be obtained by evaluating Ψ at r=b θ=π m m ψ stagnation = = πbU 2 2 The equation of the streamline passing through the stagnation point is πbU = Ur sin θ + bUθ b( π − θ) The streamline can be replaced by a solid boundary. r= The body is open at the downstream end, and thus is sin θ called a HALF-BODY. The combination of a uniform called a HALF-BODY. The combination of a uniform flow and a source can be used to describe the flow around a streamlined body placed in a uniform stream. 92 3/4 Half-Body : Uniform Stream + Source 3/4 b ( π − θ) r= ⇒ y = b ( π − θ) sin θ As θ 0 or θ= π the half-width approaches ±bπ. The width or of the half-body asymptotically approach 2πb. The velocity components at any point m 1 ∂ψ vr = = U cos θ + r ∂θ 2 πr ∂ψ vθ = − = − U sin θ ∂r b= m 2 πU 2 Um cos θ ⎛ m ⎞ b b2 ⎞ 2 2 2 2 2⎛ V = v r + vθ = U + +⎜ ⎟ = U ⎜1 + 2 cos θ + 2 ⎟ ⎜ r r⎟ πr ⎝ 2πr ⎠ ⎝ ⎠ 93 4/4 Half-Body : Uniform Stream + Source 4/4 With the velocity known, the pressure at any point can be determined from the Bernoulli equation 1 12 2 p 0 + ρU = p + ρV 2 2 Far from the body Where elevation change have been neglected. 94 Example 6.7 Potential Flow – Half-Body The shape of a hill arising from a plain can be approximated with The the top section of a half-body as is illustrated in Figure E6.7a. The height of the hill approaches 200 ft as shown. (a) When a 40 mi/hr wind blows toward the hill, what is the magnitude of the air velocity at a point on the hill directly above the origin [point (2)]? (b) What is the elevation of point (2) above the plain and what is the difference in pressure between point (1) on the plain far from the hill and point (2)? Assume an air density of 0.00238 slugs/ft3? 95 Example 6.7 Solution1/2 The velocity is ⎛ b b2 ⎞ V = U 1 + 2 cos θ + 2 ⎟ ⎜ r r⎟ ⎝ ⎠ 2 2⎜ At point (2), θ=π/2 b ( π − θ) π b r= = sin θ 2 ⎛ b2 ⎞ ⎟ = U 2 ⎛1 + 4 ⎞ = ... = 47.4mi / hr V = U 1+ ⎟ ⎜ 2 ⎜ ( πb / 2) 2 ⎟ ⎝ π⎠ ⎝ ⎠ 2 2⎜ πb 200ft = = 100ft The elevation at (2) above the plain is y 2 = 2 2 96 Example 6.7 Solution2/2 From the Bernoulli equation p1 V12 p 2 V2 2 + + z1 = + + z2 γ 2g γ 2g ρ2 p1 − p 2 = ( V2 − V12 ) + γ ( y 2 − y1 ) = ... = 9.31lb / ft 2 = 0.0647 psi 2 ⎛ 5280ft / mi ⎞ V1 = ( 40mi / hr )⎜ ⎟ = 58.7ft / s 3600s / hr ⎠ ⎝ ⎛ 5280ft / mi ⎞ V2 = ( 47.4 mi / hr )⎜ ⎟ = 69.5ft / s 3600s / hr ⎠ ⎝ 97 1/3 Rankine Oval: Uniform Stream + Doublet 1/3 m ψ = Ur sin θ − (θ1 − θ2 ) 2π m φ = Ur cos θ − (ln r1 − ln r2 ) 2π ψ = Ur sin θ − m ⎛ 2ar sin θ ⎞ tan −1 ⎜ 2 ⎟ 2 2π ⎝ r −a ⎠ ⎞ m 2ay −1 ⎛ ⎟ ⎜ tan ψ = Uy − ⎜ x 2 + y2 − a 2 ⎟ 2π ⎠ ⎝ The corresponding streamlines for this flow field are obtained by setting Ψ=constant. It is discovered that the streamline forms a closed body of length 2l and width 2h. Rankine ovals 98 2/3 Rankine Oval: Uniform Stream + Doublet 2/3 The stagnation points occur at the upstream and downstream ends of The 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 The 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, The and U. Dimensionless The body half-length 2l 1/ 2 ⎛ ma 2⎞ l=⎜ +a ⎟ ⎠ ⎝ πU 1/ 2 l ⎛m ⎞ ⇒ =⎜ + 1⎟ a ⎝ πUa ⎠ 99 3/3 Rankine Oval: Uniform Stream + Doublet 3/3 The body half-width, h, can be obtained by determining the value of The y where the y axis intersects the Ψ=0 streamline. Thus, with Ψ=0, x=0, and y=h. The body half-width 2h 2 ⎤ ⎡ ⎛ πUa ⎞ h ⎤ h2 − a2 2πUh h 1 ⎡⎛ h ⎞ h= tan ⇒ = ⎢⎜ ⎟ − 1⎥ tan ⎢2⎜ ⎟⎥ 2a m a 2 ⎢⎝ a ⎠ ⎥ ⎣ ⎝ m ⎠a⎦ ⎣ ⎦ Dimensionless 100 1/4 Flow around a Circular Cylinder 1/4 When the distance between the source-sink pair approaches zero, the When shape of the rankine oval becomes more blunt and in fact approaches a circular shape. ⎛ a2 ⎞ K sin θ ψ = Ur ⎜1 − 2 ⎟ sin θ ψ = Ur sin θ − ⎜ r⎟ ⎝ ⎠ r K cos θ ⎛ a2 ⎞ φ = Ur cos θ + φ = Ur ⎜1 + 2 ⎟ cos θ ⎜ r⎟ r ⎝ ⎠ In order for the stream Ψ=constant for r=a function to represent flow around a circular cylinder Ψ=0 for r=a K=Ua2 101 2/4 Flow around a Circular Cylinder 2/4 The velocity components On the surface of the cylinder (r=a) ⎛ a2 ⎞ ∂φ 1 ∂ψ = U⎜1 − 2 ⎟ cos θ vr = = ⎜ r⎟ ∂r r ∂θ ⎝ ⎠ vr = 0 ⎛ a2 ⎞ ∂ψ 1 ∂φ = − U⎜1 + 2 ⎟ sin θ =− vθ = ⎜ r⎟ ∂r r ∂θ ⎝ ⎠ v θ = −2 U sin θ The pressure distribution on the cylinder surface can be obtained from the Bernoulli equation 12 1 2 p 0 + ρU = p s + ρv θs 2 2 2 Far from the body 1 p s = p 0 + ρU 2 (1 − 4 sin 2 θ) 2 v θ = −2 U sin θ 102 3/4 Flow around a Circular Cylinder 3/4 On the upstream part of the cylinder, there is approximate agreement between the potential flow and the experimental results. Because of the viscous boundary layer that develops on the cylinder, the main flow separates from the surface of the cylinder, leading to the large difference between the theoretical, frictionless solution and the experimental results on the downstream side of the cylinder. 103 4/4 Flow around a Circular Cylinder 4/4 2π ∫ = − p sin θadθ ∫ Fx = − ps cos θadθ Fy 0 2π 0 s 104 Flow around a Circular Cylinder + Free 1/4 Vortex 1/4 Adding a free vortex to the stream function or velocity Adding potential for the flow around a cylinder. ⎛ a2 ⎞ Γ ψ = Ur ⎜1 − 2 ⎟ sin θ − ln r ⎜ r⎟ 2π ⎝ ⎠ ⎛ a2 ⎞ Γ θ φ = Ur ⎜1 + 2 ⎟ cos θ + ⎜ r⎟ 2π ⎝ ⎠ The circle r=a will still be a streamline, since the streamlines for the added free vortex are all circular. The tangential velocity on the surface of the cylinder ∂ψ vθ = − ∂r r =a Γ = −2 U sin θ + 2πa 105 Flow around a Circular Cylinder + Free 2/4 Vortex 2/4 This type of flow could be approximately created by This placing a rotating cylinder in a uniform stream. Because of the presence of viscosity in any real fluid, the Because fluid in contacting with the rotating cylinder would rotate with the same velocity as the cylinder, and the resulting flow field would resemble that developed by the combination of a uniform flow past a cylinder and a free vortex. 106 Flow around a Circular Cylinder + Free 3/4 Vortex 3/4 A variety of streamline patterns variety can be developed, depending on the vortex strength Г. The location of stagnation The points on a circular cylinder (a) without circulation; (b, c, d) with circulation. sin θstag vθ = 0 Γ = 4πUa θ = θstag 107 Flow around a Circular Cylinder + Free 4/4 Vortex 4/4 For the cylinder with circulation, the surface pressure, ps, is For obtained from the Bernoulli equation 12 1⎛ Γ⎞ p 0 + ρU = ps + ρ⎜ − 2U sin θ + ⎟ 2 2⎝ 2πa ⎠ 2 Γ2 ⎞ 1 2⎛ 2Γ sin θ 2 − 2 2 2⎟ ps = p 0 + ρU ⎜1 − 4 sin θ + ⎜ πaU 2 4π a U ⎟ ⎝ ⎠ 2π Drag Lift Fx = − ∫ ps cos θadθ = 0 0 2π Fy = − ∫ ps sin θadθ = −ρUT 0 108 1/2 Example 6.8 Potential Flow – Cylinder 1/2 When a circular cylinder is placed in a uniform stream, a stagnation When point is created on the cylinder as is shown in Figure E6.8a. If a small hole is located at this point, the stagnation pressure, pstag, can be measured and used to determine the approach velocity, U. (a) Show how pstag and U are related. (b) If the cylinder is misaligned by an angle α (Figure E6.8b), but the measured pressure still interpreted as the stagnation pressure, determine an expression for the ratio of the true velocity, U, to the predicted velocity, U’. Plot this ratio as a function of α for the range -20 °≦α≦20°. °≦α≦ 109 2/2 Example 6.8 Potential Flow – Cylinder 2/2 110 Example 6.7 Solution1/2 The Bernoulli equation between a point on the stagnation streamline upstream from the cylinder and the stagnation point p 0 U 2 pstag + = γ 2g γ The difference between the pressure at the stagnation point and the upstream pressure ⎡2 ⎤ ⇒ U = ⎢ ( pstag − p 0 )⎥ ⎣ρ ⎦ 1/ 2 If the cylinder is misaligned by an angle, α, the pressure actually measured, pa, will be different from the stagnation pressure. ⎡2 ⎤ U ' = ⎢ ( p a − p 0 )⎥ ⎣ρ ⎦ 1/ 2 1/2 ⎛ pstag − p 0 ⎞ U(true) ⇒ =⎜ ⎜ p −p ⎟ ⎟ U' (predicted ) ⎝ a 0⎠ 111 Example 6.7 Solution2/2 On the surface of the cylinder (r=a) v θ = −2U sin θ The Bernoulli equation between a point upstream if the cylinder and the point on the cylinder where r=a, θ=α. 12 1 p 0 + ρU = p a + ρ( −2U sin α ) 2 2 2 1 p a − p 0 = ρU 2 (1 − 4 sin 2 α) 2 1 p stag − p 0 = ρU 2 2 U = (1 − 4 sin 2 α)1 / 2 U' 112 Viscous Flow To incorporate viscous effects into the differential analysis of fluid motion General equation of motion ⎛ ∂u ∂σ xx ∂τ yx ∂τ zx ∂u ∂u ∂u ⎞ + + = ρ⎜ ρg x + ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ ⎟ ∂x ∂y ∂z ⎝ ⎠ ∂τ xy ∂σ yy ∂τ zy ⎛ ∂v ∂v ∂v ∂v ⎞ ρg y + + + = ρ⎜ +u +v +w ⎟ ⎜ ∂t ∂x ∂y ∂z ∂x ∂y ∂z ⎟ ⎠ ⎝ ⎛ ∂w ∂τ xz ∂τ yz ∂σ zz ∂w ∂w ∂w ⎞ ρg z + + + = ρ⎜ ⎟ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ ∂x ∂y ∂z ⎠ ⎝ Stress-Deformation Relationship Stress 113 1/2 Stress-Deformation Relationship 1/2 The stresses must be The expressed in terms of the velocity and pressure field. Cartesian coordinates r ∂u 2 σ xx = − p − µ ∇ ⋅ V + 2µ ∂x 3 r ∂v 2 σ yy = − p − µ ∇ ⋅ V + 2µ ∂y 3 r ∂w 2 σ zz = − p − µ ∇ ⋅ V + 2µ ∂z 3 ⎛ ∂v ∂u ⎞ τ xy = τ yx = µ ⎜ ⎜ ∂x + ∂y ⎟ ⎟ ⎝ ⎠ τ xz = τ zx ⎛ ∂w ∂u ⎞ = µ⎜ + ⎟ ⎝ ∂x ∂z ⎠ τ yz = τ zy ⎛ ∂w ∂v ⎞ = µ⎜ ⎜ ∂y + ∂z ⎟ ⎟ ⎝ ⎠ 114 2/2 Stress-Deformation Relationship 2/2 ∂v r σ rr = − p + 2µ ∂r ⎛ 1 ∂v θ v r ⎞ + σ θθ = − p + 2µ ⎜ ⎟ r⎠ ⎝ r ∂θ ∂v z σ zz = − p + 2µ ∂z ⎛ ∂ ⎛ v ⎞ 1 ∂v r τ rθ = τ θ r = µ ⎜ r ⎜ θ ⎟ + ⎜ ∂r r ⎝ ⎠ r ∂θ ⎝ ⎛ ∂v θ 1 ∂v z ⎞ + τ θz = τ zθ = µ ⎜ ⎟ r ∂θ ⎠ ⎝ ∂z ⎛ ∂v r ∂v z ⎞ τ rz = τ zr = µ ⎜ + ⎟ ∂r ⎠ ⎝ ∂z ⎞ ⎟ ⎟ ⎠ Cylindrical polar coordinates Introduced into the differential equation of motion…. 115 1/5 The Navier-Stokes Equations 1/5 These obtained equations of motion are called the NavierThese Stokes Equations. Cartesian coordinates r ⎞ ⎤ ∂ ⎡ ⎛ ∂u ∂v ⎞ ⎤ Du ∂p ∂ ⎡ ⎛ ∂u 2 − ∇ ⋅ V ⎟⎥ + ρ = ρg x − + ⎢µ ⎜ ⎢µ ⎜ 2 ⎜ ∂y + ∂x ⎟ ⎥ + ⎟ Dt ∂x ∂x ⎣ ⎝ ∂x 3 ⎠ ⎦ ∂y ⎢ ⎝ ⎠⎥ ⎣ ⎦ r ⎞⎤ ∂p ∂ ⎡ ⎛ ∂u ∂v ⎞ ⎤ ∂ ⎡ ⎛ ∂v 2 Dv ρ = ρg y − + ⎢µ ⎜ ⎜ ∂ y + ∂ x ⎟ ⎥ + ∂ y ⎢µ ⎜ 2 ∂ y − 3 ∇ ⋅ V ⎟ ⎥ + ⎟ ⎜ ⎟ ∂y ∂x ⎢ ⎝ Dt ⎢⎝ ⎠⎥ ⎠⎥ ⎦ ⎣ ⎦ ⎣ ∂ ∂z ⎡ ⎛ ∂w ∂u ⎞ ⎤ + ⎟⎥ ⎢µ ⎜ ⎣ ⎝ ∂x ∂z ⎠ ⎦ ⎡ ⎛ ∂v ∂w ⎞ ⎤ ⎢µ ⎜ ⎜ ∂z + ∂y ⎟ ⎥ ⎟ ⎢⎝ ⎠⎥ ⎦ ⎣ r ⎞⎤ ∂ ⎡ ⎛ ∂w ∂u ⎞ ⎤ ∂ ⎡ ⎛ ∂v ∂w ⎞ ⎤ ∂ ⎡ ⎛ ∂w 2 ∂p Dw ⎟⎥ + ρ = ρg z − + + + − ∇ ⋅ V ⎟⎥ ⎟⎥ + ⎢µ ⎜ ⎢µ ⎜ ⎢µ ⎜ 2 ∂ z ∂ x ⎣ ⎝ ∂ x ∂ zx ⎠ ⎦ ∂ y ⎢ ⎜ ∂ z ∂ y ⎟ ⎥ ∂ z ⎣ ⎝ ∂ z Dt 3 ⎠⎦ ⎠⎦ ⎣⎝ ∂ ∂z 116 2/5 The Navier-Stokes Equations 2/5 Cylindrical polar coordinates 2 ⎛ ∂v r ∂v r v θ ∂v r v θ ∂v ⎞ ρ⎜ + vr + − + vz r ⎟ ⎜ ∂t r ∂θ r ∂r ∂z ⎟ ⎝ ⎠ ⎡ 1 ∂ ⎛ ∂v r ⎞ v r 1 ∂2v r 2 ∂v θ ∂ 2 v r ⎤ ∂p =− + ρg r + µ ⎢ −2 + ⎜r ⎟− 2 + 2 ⎥ 2 r ∂r ⎝ ∂r ⎠ r ∂r r ∂θ r ∂θ ∂z 2 ⎥ ⎢ ⎦ ⎣ v ∂v vv ∂v ∂v ⎞ ⎛ ∂v ρ⎜ θ + v r θ + θ θ + r θ + v z θ ⎟ r ∂θ r ∂r ∂z ⎠ ⎝ ∂t ⎡ 1 ∂ ⎛ ∂v θ ⎞ v θ 1 ∂ 2 v θ 2 ∂v r ∂ 2 v θ ⎤ 1 ∂p +2 + =− + ρg θ + µ ⎢ ⎜r ⎟− 2 + 2 ⎥ 2 r ∂θ r ∂r ⎝ ∂r ⎠ r r ∂θ r ∂θ ∂z 2 ⎥ ⎢ ⎣ ⎦ v ∂v ∂v ∂v ⎞ ⎛ ∂v ρ⎜ z + v r z + + θ z + v z z ⎟ r ∂θ ∂r ∂z ⎠ ⎝ ∂t ⎡ 1 ∂ ⎛ ∂v z ∂P =− + ρg z + µ ⎢ ⎜r ∂z ⎢ r ∂r ⎝ ∂r ⎣ 2 2 ⎞ 1 ∂ vz ∂ vz ⎤ + ⎟+ 2 2 2⎥ ∂z ⎥ ⎠ r ∂θ ⎦ 117 3/5 The Navier-Stokes Equations 3/5 Under incompressible flow with constant viscosity Under conditions, the Navier-Stokes equations are reduced to: ⎛ ∂2u ∂2u ∂2u ⎞ ⎛ ∂u ∂u ∂u ∂u ⎞ ∂p ⎜ ρ⎜ + 2⎟ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂x + ρ g x + µ ⎜ 2 + ⎟ 2 ∂z ⎟ ∂y ⎝ ⎠ ⎝ ∂x ⎠ ⎛ ∂2v ∂2v ∂2v ⎞ ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p ⎜ ⎟ + ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂y + ρ g y + µ ⎜ 2 + ⎟ 2 2⎟ ∂z ⎠ ∂y ⎝ ⎠ ⎝ ∂x ⎛ ∂2w ∂2w ∂2w ⎞ ⎛ ∂w ∂w ∂w ∂w ⎞ ∂p ⎜ ⎟ + + ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂z + ρ g z + µ ⎜ ⎟ 2 2 2⎟ ∂z ⎠ ∂y ⎝ ⎠ ⎝ ∂x 118 4/5 The Navier-Stokes Equations 4/5 Under frictionless condition, the equations of motion are Unde frictionless reduced to Euler’s Equation: ⎛ ∂u ∂p ∂u ∂u ∂u ⎞ ⎜ ⎟=− + ρg x ρ⎜ +u +v +w ⎟ ∂x ∂x ∂y ∂z ⎠ ⎝ ∂t ⎛ ∂v ∂v ∂v ∂v ⎞ ∂p ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂y + ρ g y ⎟ ⎝ ⎠ ⎛ ∂w ∂w ∂w ∂w ⎞ ∂p ρ⎜ ⎜ ∂t + u ∂x + v ∂y + w ∂z ⎟ = − ∂z + ρ g z ⎟ ⎝ ⎠ r r DV ρ = ρg − ∇p Dt 119 5/5 The Navier-Stokes Equations 5/5 The Navier-Stokes equations apply to both laminar and The turbulent flow, but for turbulent flow each velocity component fluctuates randomly with respect to time and this added complication makes an analytical solution intractable. The exact solutions referred to are for laminar flows in The which the velocity is either independent of time (steady flow) or dependent on time (unsteady flow) in a welldefined manner. 120 Some Simple Solutions for Viscous, Incompressible Fluids A principal difficulty in solving the Navier-Stokes principal equations is because of their nonlinearity arising from the convective acceleration terms. convective There are no general analytical schemes for solving There nonlinear partial differential equations. There are a few special cases for which the convective There acceleration vanishes. In these cases exact solution are often possible. 121 Steady, Laminar Flow between Fixed 1/6 Parallel Plates 1/6 Consider flow between the two horizontal, infinite Consider parallel plate. For this geometry the fluid particle move in the x For direction parallel to the pates, and there is no velocity in the y or z direction – that is, v=0 and w=0. 122 Steady, Laminar Flow between Fixed 2/6 Parallel Plates 2/6 From the continuity equation that ∂u/∂x=0. From There would be no variation of u in the z direction for There infinite plates, and for steady flow so that u=u(y). The Navier-Stokes equations reduce to The ⎛ ∂ 2u ⎞ ∂p 0=− + µ⎜ 2 ⎟ ⎜ ∂y ⎟ ∂x ⎝ ⎠ ∂p ∂p − ρg 0=− 0=− ∂z ∂y 123 Steady, Laminar Flow between Fixed 3/6 Parallel Plates 3/6 ∂p − ρg ∂y ∂p 0=− ∂z ⎛ ∂2u ⎞ ∂p + µ⎜ 2 ⎟ 0=− ⎜ ∂y ⎟ ∂x ⎝ ⎠ 0=− Integrating p = − ρ gy + f1 (x ) Integrating 1 ⎛ ∂p ⎞ 2 u= ⎜ ⎟ y + c1 y + c 2 2µ ⎝ ∂ x ⎠ c1 ? c2 ? 124 Steady, Laminar Flow between Fixed 4/6 Parallel Plates 4/6 With the boundary conditions u=0 at y=-h u=0 at y=h With 1 ⎛ ∂p ⎞ 2 c 2 = 0, c1 = − ⎟h ⎜ 2µ ⎝ ∂ x ⎠ Velocity distribution ( 1 ⎛ ∂p ⎞ 2 2 u= ⎜ ⎟y −h 2µ ⎝ ∂ x ⎠ ) 125 Steady, Laminar Flow between Fixed 5/6 Parallel Plates 5/6 Shear stress distribution Shear ⎛ ∂p ⎞ =⎜ ⎟y ⎝ ∂x ⎠ τ yx Volume flow rate Volume q= ∫ h udy = −h ∫ h −h 1 ⎛ ∂p ⎞ 2 2h 3 2 ⎜ ⎟ ( y − h ) dy = − 2µ ⎝ ∂ x ⎠ 3µ ⎛ ∂p ⎞ ⎜ ⎟ ∂x ⎠ ⎝ p − p1 ∆p ∂p = cons tan t = 2 =− ∂x l l 2 h 3∆p >> q = 3µ l 126 Steady, Laminar Flow between Fixed 6/6 Parallel Plates 6/6 Average velocity Average Vaverage q h 2 ∆p = = 2h 3µ l Point of maximum velocity Point du =0 dy at y=0 u = u max h 2 ⎛ ∂p ⎞ 3 =U=− ⎜ ⎟ = Vaverage 2µ ⎝ ∂ x ⎠ 2 127 1/3 Couette Flow 1/3 Since only the boundary conditions have changed, there Since is no need to repeat the entire analysis of the “both plates stationary” case. 1 ⎛ ∂p ⎞ 2 u= ⎟ y + c1 y + c 2 ⎜ 2µ ⎝ ∂ x ⎠ c1 ? c2 ? 128 2/3 Couette Flow 2/3 The boundary conditions for the moving plate case are The u=0 at y=0 u=U at y=b U 1 ⎛ ∂p ⎞ ⇒ c1 = − ⎜ ⎟b b 2µ ⎝ ∂ x ⎠ Velocity distribution b2 P= 2µ U ⎛ ∂p ⎞ ⎜ ⎟ ⎝ ∂x ⎠ c2 = 0 Uy 1 ⎛ ∂p ⎞ 2 1 ⎛ ∂p ⎞ + u= ⎜ ⎟ by ⎜ ⎟y − b 2µ ⎝ ∂ x ⎠ 2µ ⎝ ∂ x ⎠ u y b2 =− U b 2µ U ⎛ ∂p ⎞⎛ y ⎞ ⎡ ⎛ b ⎞ ⎤ ⎜ ⎟ ⎜ ⎟ ⎢1 − ⎜ ⎟ ⎥ ∂x ⎠⎝ b ⎠ ⎢ ⎜ y ⎟ ⎥ ⎝ ⎣ ⎝ ⎠⎦ 129 3/3 Couette Flow 3/3 Simplest type of Couette flow U = ri ω b = ro − ri τ = µ ri ω /( ro − r ) y ∂p =0⇒ u=U ∂x b This flow can be approximated by the flow between closely spaced concentric cylinder is fixed and the other cylinder rotates with a constant angular velocity. Flow in the narrow gap of a journal bearing. 130 Example 6.9 Plane Couette Flow A wide moving belt passes through a wide container of a viscous liquid. The belt moves vertically upward with a constant velocity, V0, as illustrated in Figure E6.9a. Because of viscous forces the belt picks up a film of fluid of thickness h. Gravity tends to make the fluid drain down the belt. Use the Navier Stokes equations to determine an expression for the average velocity of the fluid film as it is dragged up the belt. Assume that the flow is laminar, steady, and fully developed. 131 Example 6.9 Solution1/2 Since the flow is assumed to be fully developed, the only velocity component is in the y direction so that u=w=0. From the continuity equation ∂v = 0 , and for steady flow, so that v=v(x) ∂y ∂p ∂p =0 =0 dv γ ∂x ∂z = x + c1 2 0 = −ρg + µ dv dx 2 Integrating dx τ xy µ ⎛ dv ⎞ = µ⎜ ⎟ ⎝ dx ⎠ 132 Example 6.9 Solution2/2 τ xy = 0 at x = h c1 = − γh µ v = V0 at x = 0 Integrating γ 2 γh v= x− x + c2 2µ 2µ γ 2 γh v= x− x + V0 2µ 2µ 133 1/5 Steady, Laminar Flow in Circular Tubes 1/5 Consider the flow through a horizontal circular tube of Consider radius R. ∂v z v r = 0, vθ = 0 ⇒ = 0 v z = v z (r ) ∂z 134 Steady, Laminar Flow in Circular Tubes 2/5 Navier – Stokes equation reduced to ∂p 0 = −ρg sin θ − ∂r 1 ∂p 0 = −ρg sin θ − r ∂θ ⎛ 1 ∂ ⎛ ∂v z ∂p 0=− + µ⎜ ⎜ r ∂r ⎜ r ∂r ∂z ⎝ ⎝ p = −ρg (r sin θ) + f1 (z ) Integrating ⎞⎞ ⎟⎟ ⎟ ⎠⎠ p = −ρgy + f1 (z ) Integrating 1 ⎛ ∂p ⎞ 2 vz = ⎜ ⎟ r + c1 ln r + c 2 4µ ⎝ ∂z ⎠ c1 ? c2 ? 135 Steady, Laminar Flow in Circular Tubes 3/5 At r=0, the velocity vz is finite. At r=R, the velocity vz is zero. 1 ⎛ ∂p ⎞ 2 c1 = 0, c 2 = − ⎜ ⎟R 4µ ⎝ ∂z ⎠ Velocity distribution ( 1 ⎛ ∂p ⎞ 2 2 vz = ⎜ ⎟r −R 4µ ⎝ ∂z ⎠ ) 136 4/5 Steady, Laminar Flow in Circular Tubes 4/5 The shear stress distribution The τ rx du r ⎛ ∂p ⎞ =µ =⎜ ⎟ dr 2 ⎝ ∂x ⎠ Volume flow rate Volume π R 4 ⎛ ∂p ⎞ Q = u z 2 π rdr = ..... = − ⎜ ⎟ 8µ ⎝ ∂ z ⎠ 0 ∫ R p 2 − p1 ∂p = cons tan t = = − ∆p / l l ∂z π R 4 ⎛ ∂p ⎞ π R 4 ⎛ ∆ p ⎞ π ∆ pD 4 >> Q = − ⎜ ⎟= ⎜ ⎟= 8µ ⎝ ∂ z ⎠ 8µ ⎝ l ⎠ 128 µ l 137 5/5 Steady, Laminar Flow in Circular Tubes 5/5 Average velocity Average Vaverage Q Q R 2 ∆p = = = 2 A πR 8µ l Point of maximum velocity Point dv z =0 dr at r=0 2 v max vz R ∆p ⎛r⎞ =− = 2 Vaverage ⇒ =1− ⎜ ⎟ 4µ l v max ⎝R⎠ 2 138 1/2 Steady, Axial, Laminar Flow in an Annulus 1/2 For steady, laminar flow in circular tubes 1 ⎛ ∂p ⎞ 2 vz = ⎜ ⎟ r + c1 ln r + c 2 4µ ⎝ ∂z ⎠ c1 ? c2 ? Boundary conditions vz = 0 , at r = ro vz = 0 , at r = ri 139 2/2 Steady, Axial, Laminar Flow in an Annulus 2/2 The velocity distribution 2 ri2 − ro 1 ⎛ ∂p ⎞ ⎡ 2 2 r⎤ vz = ln ⎥ ⎜ ⎟ ⎢ r − ro + 4µ ⎝ ∂z ⎠ ⎢ ln( ro / ri ) ro ⎥ ⎣ ⎦ The volume rate of flow 2 π ⎛ ∂p ⎞ ⎡ 4 4 ( ro − ri2 ) 2 ⎤ Q = v z ( 2πr )dr = − ⎜ ⎟ ⎢ ro − ri − ⎥ 8µ ⎝ ∂z ⎠ ⎢ ln( ro / ri ) ⎥ ri ⎣ ⎦ ∫ ro The maximum velocity occurs at r=rm ∂v z =0 ∂r 1/ 2 ⎡ ⎤ rm = ⎢ ⎥ ⎢ 2 ln( ro / ri ) ⎥ ⎣ ⎦ 2 ro − ri2 140 ...
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This note was uploaded on 11/17/2011 for the course MECHANICAL 303 taught by Professor Shieh during the Spring '11 term at California Coast University.

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