Fluids Lecture 1 Notes

Transcription

1 Fluids Lecture Notes. Introductory Concepts and Definitions. Properties of Fluids Reading: Anderson. (optional),.,.3,.4 Introductory Concepts and Definitions Fluid Mechanics and Fluid Dynamics encompass a huge range of topics which deal with the behavior of gasses and liquids. In UE we will focus mainly on the topic subset called Aerodynamics, with a bit af Aerostatics in the beginning. Merriam Webster s definitions: Aerostatics: a branch of statics that deals with the equilibrium of gaseous fluids and of solid bodies immersed in them Aerodynamics: a branch of dynamics that deals with the motion of air and other gaseous fluids and with the forces acting on bodies in motion relative to such fluids Related older terms are Hydrostatics and Hydrodynamics, usually used for situtations involving liquids. There is surprisingly little fundamental difference between the Aero and Hydro disciplines. They differ mainly in the applications (e.g. airplanes vs. ships). Difference between a Solid and a Fluid (Liquid or Gas): Solid: Applied tengential force/area (or shear stress) τ produces a proportional deformation angle (or strain) θ. τ = Gθ The constant of proportionality G is called the elastic modulus, and has the units of force/area. Fluid: Applied shear stress τ produces a proportional continuously-increasing deformation (or strain rate) θ. τ = µθ The constant of proportionality µ is called the viscosity, and has the units of force time/area. τ stress τ stress θ strain. θ strain rate Solid Fluid

2 Properties of Fluids Continuum vs molecular description of fluid Liquids and gases are made up of molecules. Is this discrete nature of the fluid important for us? In a liquid, the answer is clearly NO. The molecules are in contact as they slide past each other, and overall act like a uniform fluid material at macroscopic scales. In a gas, the molecules are not in immediate contact. So we must look at the mean free path, which is the distance the average molecule travels before colliding with another. Some known data for air: Mean free path at 0 km (sea level) : mm Mean free path at 0 km (U flight) : 0.00 mm Mean free path at 50 km (balloons) : 0. mm Mean free path at 50 km (low orbit) : 000 mm = m The mean free path is vastly smaller than the typical dimension of any atmospheric vehicle. So even though the lift on a wing is due to the impingement of discrete molecules, we can assume the air is a continuum for the purpose of computing this lift. In contrast, computing the slight air drag on an orbiting satellite requires treating the air as discrete isolated particles. Pressure Pressure p is defined as the force/area acting normal to a surface. A solid surface doesn t actually have to be present. The pressure can be defined at any point x, y, z in the fluid, if we assume that a infinitesimally small surface ΔA could be placed there at whim, giving a resulting normal force ΔF n. ΔF n p = lim ΔA 0 ΔA The pressure can vary in space and possibly also time, so the pressure p(x, y, z, t) in general is a time-varying scalar field. ΔΑ Δ F n Normal force on area element due to pressure Density Density ρ is defined as the mass/volume, for an infinitesimally small volume. Δm ρ = lim Δ 0 Δ Like the pressure, this is a point quantity, and can also change in time. So ρ(x, y, z, t) is also a scalar field.

3 elocity We are interested in motion of fluids, so velocity is obviously important. Two ways to look at this: Body is moving in stationary fluid e.g. airplane in flight Fluid is moving past a stationary body e.g. airplane in wind tunnel The pressure fields and aerodynamics forces in these two cases will be the same if all else is equal. The governing equations we will develop are unchanged by a Galilean Transformation, such as the switch from a fixed to a moving frame of reference. Consider a fluid element (or tiny blob of fluid) as it moves along. As it passes some point B, its instantaneous velocity is defined as the velocity at point. B at a point = velocity of fluid element as it passes that point This velocity is a vector, with three separate components, and will in general vary between different points and different times. (x, y, z, t) = u(x, y, z, t) î + v(x, y, z, t) j ˆ + w(x, y, z, t) k ˆ So is a time-varying vector field, whose components are three separate time-varying scalar fields u, v, w. fluid element B streamline elocity at point B A useful quantity to define is the speed, which is the magnitude of the velocity vector. (x, y, z, t) = = u + v + w In general this is a time-varying scalar field. Steady and Unsteady Flows If the flow is steady, then p, ρ, don t change in time for any point, and hence can be given as p(x, y, z), ρ(x, y, z), (x, y, z). If the flow is unsteady, then these quantities do change in time at some or all points. For a steady flow, we can define a streamline, which is the path followed by some chosen fluid element. The figure above shows three particular streamlines. 3

4 Fluids Lecture Notes. Hydrostatic Equation. Manometer 3. Buoyancy Force Reading: Anderson.9 Hydrostatic Equation Consider a fluid element in a pressure gradient in the vertical y direction. Gravity is also present. y dp dy dx dy dz x z g If the fluid element is at rest, the net force on it must be zero. For the vertical y-force in particular, we have Pressure force + Gravity force = 0 ( ) dp p da p + dy da ρ g d = 0 dy dp dy da ρ g d = 0 dy The area on which the pressures act is da = dx dz, and the volume is d = dx dy dz, so that dp dx dy dz ρ g dx dy dz = 0 dy dp = ρg dy () which is the differential form of the Hydrostatic Equation. If we make the further assumption that the density is constant, this equation can be integrated to the equivalent integral form. p(y) = p 0 ρgy () The constant of integration p 0 is the pressure at the particular location y = 0. Note that this integral form is valid provided the density is constant within the region of interest.

5 Application to a Manometer A manometer is a U-shaped tube partially filled with a liquid, as shown in the figure. Two different pressures p and p are applied to the two legs of the tube, causing the two liquid columns to have different heights h and h. y p p h h p 0 We now pick p 0 to be the pressure at some point of the tube (at the bottom for instance), and apply equation () to each leg of the tube. p = p 0 ρgh p = p 0 ρgh Subtracting these two equations then gives the difference of the pressures in terms of the liquid height difference. p p = ρg(h h ) (3) If tube is left open to the atmosphere, so that p = p atm, then p can be measured simply by applying it to tube, measuring the height difference Δh = h h, and applying equation (3) above. p = p atm + ρg Δh This requires knowing the density ρ of the fluid to sufficient accuracy. Buoyancy Now consider an object of arbitrary shape immersed in the pressure gradient. The object s volume can be divided into vertical matchstick volumes, each of infinitesimal crosssectional area da = dx dz, and finite height Δh. The vertical y-direction pressure force on each volume is ( ) dp df = p da p + Δh da dy dp df = Δh da dy df = ρg d

6 y da = dx dz dp dy Δ h x z g where dp/dy has been replaced by ρg using the Hydrostatic Equation (), and the volume of the infinitesimal volume is Δh da = d. Integrating the last equation above then gives the total buoyancy force on the object. F = ρg It is important to note that is the overall volume of the object, while ρ is the density of the fluid. The product ρ is recognized as the mass of the fluid displaced by the object, and ρg is the corresponding weight, giving the well known Archimedes Principle: Buoyancy force on body = Weight of fluid displaced by body 3

7 Fluids Lecture 3 Notes. -D Aerodynamic Forces and Moments. Center of Pressure 3. Nondimensional Coefficients Reading: Anderson.5.6 Aerodynamics Forces and Moments Surface force distribution The fluid flowing about a body exerts a local force/area (or stress) f on each point of the body. Its normal and tangential components are the pressure p and the shear stress τ. r f r τ ds f p force/area distribution on airfoil M R local pressure and shear stress components ( magnitude greatly exaggerated) τ N L R α α D A resultant force, and moment about ref. point alternative components of resultant force The figure above greatly exaggerates the magnitude of the τ stress component just to make it visible. In typical aerodynamic situations, the pressure p (or even the relative pressure p p ) is typically greater than τ by at least two orders of magnitude, and so f is very nearly perpendicular to the surface. But the small τ often significantly contributes to drag, so it cannot be neglected entirely. The stress distribution f integrated over the surface produces a resultant force R, and also a moment M about some chosen moment-reference point. In -D cases, the sign convention for M is positive nose up, as shown in the figure. Force components The resultant force R has perpendicular components along any chosen axes. These axes are arbitrary, but two particular choices are most useful in practice.

8 Freestream Axes: The R components are the drag D and the lift L, parallel and perpendicular to. Body Axes: The R components are the axial force A and normal force N, parallel and perpendicular to the airfoil chord line. If one set of components is computed, the other set can then be obtained by a simple axis transformation using the angle of attack α. Specifically, L and D are obtained from N and A as follows. L = N cos α A sin α D = N sin α + A cos α Force and moment calculation A cylindrical wing section of chord c and span b has force components A and N, and moment M. In -D it s more convenient to work with the unit-span quantities, with the span dimension divided out. A A/b N N/b M M/b α y p l (s l ) s l θ ds l τ l (s ) l s u ds u θ p u (s u) θ τ u (s u) On the upper surface, the unit-span force components acting on an elemental area of width ds u are dn da u u And on the lower surface they are = ( p u cos θ τ u sin θ) ds u = ( p u sin θ + τ u cos θ) ds u dnl = (p l cos θ τ l sin θ) ds l dal = (p l sin θ + τ l cos θ) ds l Integration from the leading edge to the trailing edge points produces the total unit-span forces. N A = = TE TE dn u + dnl LE LE TE TE da u + dal LE LE x ds u p u τ u b c

9 The moment about the origin (leading edge in this case) is the integral of these forces, weighted by their moment arms x and y, with appropriate signs. TE TE TE TE M = x dn + x dnl + y da + y da LE u u l From the geometry, we have LE LE LE LE dy ds cos θ = dx ds sin θ = dy = dx dx which allows all the above integrals to be performed in x, using the upper and lower shapes of the airfoil y u (x) and y l (x). Anderson.5 has the complete expressions. Simplifications In practice, the shear stress τ has negligible contributions to the lift and moment, giving the following simplified forms. ( ) c c dy l dy u L = cos α (p l p u ) dx + sin α p l p u dx 0 0 dx dx [ ( ) ( )] c dy u dy l M = p u x + y u p l x + y l dx LE 0 dx dx A somewhat less accurate but still common simplification is to neglect the sin α term in L, and the dy/dx terms in M. L c (p l p u ) dx M LE (p l p u ) x dx 0 The shear stress τ cannot be neglected when computing the drag D on streamline bodies such as airfoils. This is because for such bodies the integrated contributions of p toward D tend to mostly cancel, leaving the small contribution of τ quite significant. Center of Pressure Definition The value of the moment M depends on the choice of reference point. Using the simplified form of the M LE integral, the moment M ref for an arbitrary reference point x ref is c M ref = (p l p u ) (x x ref ) dx 0 = M LE + L x ref This can be positive, zero, or negative, depending on where x ref is chosen, as illustrated in the figure. At one particular reference location x cp, called the center of pressure, the moment is defined to be zero. M cp = M + L x cp 0 LE x cp = M LE /L 3

10 p l p u L L L x cp or or M < 0 M = 0 M < 0 The center of pressure asymptotes to + or as the lift tends to zero. This awkward situation can easily occur in practice, so the center of pressure is rarely used in aerodynamics work. For reasons which will become apparent when airfoil theory is studied, it is advantageous to define the standard location for the moment reference point of an airfoil to be at its quarter-chord location, or x ref = c/4. The corresponding standard moment is usually written without any subscripts. c M c/4 M = (p l p u ) (x c/4) dx 0 Aerodynamic Conventions As implied above, the aerodynamicist has the option of picking any reference point for the moment. The lift and the moment then represent the integrated p l p u distribution. Consider two possible representations:. A resultant lift L acts at the center of pressure x = x cp. The moment about this point is zero by definition: M cp = 0. The x cp location moves with angle of attack in a complicated manner.. A resultant lift L acts at the fixed quarter-chord point x = c/4. The moment about this point is in general nonzero: M c/4 0. The figure shows how the L, M, and x cp change with angle of attack for a typical cambered airfoil. Note that with representation, the x cp location moves off the airfoil and tends to + as L approaches zero. Fixing the moment reference point, as in representation, is a simpler and preferable approach. Choosing the quarter-chord location for this is especially attractive, since M then shows little or no dependence on the angle of attack. This surprising fact will come from a more detailed airfoil analysis later in the course. 4

11 p l p u α = 5 α = 0 α = 5. L (M =0) x cp x cp x cp. L M c/4 c/4 c/4 Nondimensional Coefficients The forces and moment depend on a large number of geometric and flow parameters. It is often advantageous to work with nondimensionalized forces and moment, for which most of these parameter dependencies are scaled out. For this purpose we define the following reference parameters: Reference area: S Reference length: l Dynamic pressure: q = ρ The choices for S and l are arbitrary, and depend on the type of body involved. For aircraft, traditional choices are the wing area for S, and the wing chord or wing span for l. The nondimensional force and moment coefficients are then defined as follows: L Lift coefficient: C L q S D Drag coefficient: C D q S M Moment coefficient: C M q Sl For -D bodies such as airfoils, the appropriate reference area/span is simply the chord c, and the reference length is the chord as well. The local coefficients are then defined as follows. Local Lift coefficient: c l q c D Local Drag coefficient: c d q c M Local Moment coefficient: c m q c These local coefficients are defined for each spanwise location on a wing, and may vary across the span. In contrast, the C L, C D, C M are single numbers which apply to the whole wing. 5 L

12 Fluids Lecture 4 Notes. Dimensional Analysis Buckingham Pi Theorem. Dynamic Similarity Mach and Reynolds Numbers Reading: Anderson.7 Dimensional Analysis Physical parameters A large number of physical parameters determine aerodynamic forces and moments. Specifically, the following parameters are involved in the production of lift. Parameter Symbol Units Lift per span L mt Angle of attack α Freestream velocity lt Freestream density ρ ml 3 Freestream viscosity µ ml t Freestream speed of sound a lt Size of body (e.g. chord) c l For an airfoil of a given shape, the lift per span in general will be a function of the remaining parameters in the above list. L = f(α, ρ,, c, µ, a ) () In this particular example, the functional statement has N = 7 parameters, expressed in a total of K = 3 units (mass m, length l, and time t). Dimensionless Forms The Buckingham Pi Theorem states that this functional statement can be rescaled into an equivalent dimensionless statement Π = f( Π, Π 3... Π N K ) having only N K = 4 dimensionless parameters. These are called Pi products, since they are suitable products of the dimensional parameters. In the particular case of statement (), suitable Pi products are: L Π = = c l lift coefficient ρ c Π = α = α angle of attack ρ c Π3 = = Re Reynolds number µ Π 4 = = M Mach number a The dimensionless form of statement () then becomes c l = f(α, Re, M ) ()

13 We see that the original 6 dimensional parameters which influence L has been reduced to only 3 dimensionless parameters which influence c l. Benefits of non-dimensionalization The reduction of parameter count is potentially a huge simplification. Consider an exaustive lift-measurement experiment where the effect of all parameters is to be determined. Let s assume that in this experiment we need to give each parameter 5 distinct values in order to adequately ascertain its effect on the lift. If we work with the 6 dimensional parameters in statement (), then the number of possible parameter combinations and experimental runs required is 5 6 = 565 (!). But if we work with the 3 dimensionless parameters in statement (), the number of parameter combinations and experimental runs is only 5 3 = 5, which is more than a hundredfold reduction in effort. Nondimensionalization is clearly a powerful technique for minimizing experimental effort. The benefits of non-dimensionalization also extend to theoretical work. Deducing a statement such as () at the outset can be useful to guide subsequent detailed analysis. Theoretical results are also usually more concise and clear when presented in dimensionless form. Derivation of dimensionless forms Anderson.7 has details on how the Pi product combinations can be derived for any complex situation using linear algebra. In many cases, however, the products can be obtained by physical insight, or perhaps by inspection. Several rules can be applied here: Any parameter which is already dimensionless, such as α, is automatically one of the Pi products. If two parameters have the same units, such as and a, then their ratio (M in this case) will be one of the Pi products. A power or simple multiple of a Pi product is an acceptable alternative Pi product. For example, ( /a ) is an acceptable alternative to /a, and ρ is an acceptable alternative to ρ. Which particular forms are used is a matter of convention. Combinations of Pi products can replace the originals. For example, the 3rd and 4th products in the example could have been defined as ρ a c Π3 = = Re/M µ Π 4 = = M a which is workable alternative, but perhaps less practical, and certainly less traditional. Dynamic Similarity It is quite possible for two differently-sized physical situations, with different dimensional parameters, to nevertheless reduce to the same dimensionless description. The only requirement is that the corresponding Pi products have the same numerical values.

14 Airfoil flow example Consider two airfoils which have the same shape and angle of attack, but have different sizes and are operating in two different fluids. Let s omit the () subscript for clarity. Airfoil (sea level) Airfoil (cryogenic tunnel) α = 5 α = 5 = 0m/s = 40m/s ρ =.kg/m 3 ρ = 3.0kg/m 3 µ = kg/m-s µ = kg/m-s a = 300m/s a = 00m/s c =.0m c = 0.5m Airfoil Sea level air Airfoil Cryogenic tunnel The Pi products evaluate to the following values. Airfoil Airfoil α = 5 α = 5 Re = Re = M = 0.7 M = 0.7 Since these are also the arguments to the f function, we conclude that the c l values will be the same as well. f(α, Re, M ) = f(α, Re, M ) c l = c l When the nondimensionalized parameters are equal like this, the two situations are said to have dynamic similarity. One can then conclude that any other dimensionless quantity must also match between the two situations. This is the basis of wind tunnel testing, where the flow about a model object duplicates and can be used to predict the flow about the fullsize object. The prediction is correct only if the model and full-size objects have dynamic similarity. 3

15 Fluids Lecture 5 Notes. Effects of Reynolds Number. Effects of Mach Number Reading: Anderson.0,. Dimensional Analysis has identified the important dimensionless parameters (or Π products), which determine the nature of a given aerodynamic flow situation. The main parameters are the Reynolds number and the Mach number, and their values in any given flow situation will determine the nature of the flow (strongly viscous vs. nearly inviscid, low speed vs. high speed) Effects of Reynolds Number Interpretation The Reynolds number can be interpreted as a comparison of the pressure forces and viscous shear forces which act on the body, done by forming their ratio. pressure forces p p ρu ρ ρ c = Re shear forces τ µ du/dn µ /c µ Note that p p is considered rather than p itself, since p is just a large constant bias which produces no net force on a body. The Reynolds number is therefore a measure of the importance of pressure forces relative to viscous shear forces. Hence, if the Reynolds number increases, the viscous effects on the flow get progressively less important. The viscosity of air and water is quite small when expressed in common units. STP 5 C µ kg/m-s kg/m-s µ/ρ = ν m /s m /s The ratio µ/ρ = ν is called the kinematic viscosity. This is more relevant than µ itself, since the ratio is what actually appears in the Reynolds number. Re = From the small values of ν in the table above, it is clear that typical aerodynamic and hydrodynamic flows will have very large Reynolds numbers. This can be seen in the following table, which gives the Reynolds numbers based on the chord length of common winged objects. c ν Object Butterfly Pigeon RC glider Sailplane Business jet Boeing 777 Re 5 K 50 K 00 K M 0 M 50 M

16 The Reynolds number is large even for insects, which means that the flow can be assumed to be inviscid (i.e. µ = 0 and τ = 0) almost everywhere. The only place where the viscous shear is significant is in boundary layers which form adjacent to solid surfaces and become a wake trailing downstream. boundary layer boundary layer Re = 0 K Re = M c d c d The larger the Reynolds number is, the thinner the boundary layers are relative to the size of the body, and the more the flow behaves as though it was inviscid. This has a number of consequences of great practical significance:. Neglecting the viscous shear stress greatly simplifies the equations of fluid motion, allowing them to be manipulated and solved much more easily. Simple solutions then give better insight and understanding of aerodynamic behavior of bodies such as airfoils, wings, propellers, turbine blades, etc.. If the Reynolds number is large enough (i.e. viscosity is small enough), then its effects on some aerodynamic forces and moments can be neglected. For example, if Re > 0 6 or so, then it has almost no influence on the aerodynamic lift and moment, which means that the general dependencies for c l and c m simplify as follows. c l = c l (α, M, Re ) c l = c l (α, M ) (Re large) c m = c m (α, M, Re ) c m = c m (α, M ) Such a simplification is not appropriate for c d (α, M, Re ), however. Streamlined vs. Bluff Bodies The above assumption that viscosity has negligible effects on lift and moment is valid only for streamline bodies which gradually taper to a point or sharp edge, and are roughly aligned with the freestream flow direction. Examples are slender bodies, airfoils, and wings, operating at small angles of attack. For bluff bodies which have blunt rear faces, or for streamline bodies at large angles of attack, viscosity always plays an important role for all the forces and moments. Such flows exhibit flow separation, which is the sudden thickening or breakaway of the boundary layer from the surface, resulting in a thick trailing wake. flow separation flow separation Stalled Streamline body Bluff body c d c d 0.6. Bodies with separation exhibit much larger drag forces than comparably-sized streamline bodies without separation. When an airfoil stalls, its drag will typically increase by a factor

17 of 0 or more. Most of this drag increase is due to a sharp rise of pressure drag, defined as the integrated pressure p p normal to the body. The friction drag, which is the integrated tangential shear stress τ, is not significantly different between streamline and bluff bodies of similar size. Drag reference length The reference length used to define c d is arbitrary. Traditionally, for streamline bodies such as an airfoil we use the streamwise length c. For bluff bodies we use the transverse height, typically denoted by h. For a circular cylinder they are of course the same, and equal to the diameter d. Effects of Mach Number Interpretation The Mach number is defined as the ratio of the freestream speed to the speed of sound. a M The speed of sound a is the velocity at which pressure disturbances travel in the fluid, although what this implies for the flow is not immediately apparent. Another interpretation of the Mach number arises when we estimate the fractional variation of pressure in the flowfield. ρ Δp q γ ρ γ γ M = = = = p p p γp a where γ = c p /c v is the thermodynamic ratio of specific heats. So the Mach number squared is a measure of the fractional pressure variations in the flowfield, as shown in the figure. p / p Δ p / p ~ M s s 3

18 Low-Speed flows A low-speed flow is defined as one for which M. From the above derivation this implies that fractional pressure changes are small, and from thermodynamics we then conclude that that density and temperature changes must also be small. Δp Δρ ΔT,, p ρ T So an important feature of a low-speed flow is nearly constant fluid properties throughout the flow field. All hydrodynamic flows can be classified as low-speed flows, since liquids feature an essentially constant density. One practical feature of low-speed flows is that they are extremely insensitive to Mach number variations, as long as M. So a flow with M = 0.0 is almost indistinguishible from a flow with M = 0. when compared using dimensionless variables. So for low speed flows we can remove the influence of M from the dependency list. c l = c l (α, M ) c l = c l (α) ( ) c m = c m (α, M ) c m = c m (α) M c d = c d (α, M, Re ) c d = c d (α, Re ) This is a considerable simplification. 4

19 Fluids Lecture 6 Notes. Control olumes. Mass Conservation 3. Control olume Applications Reading: Anderson.3,.4 Control olumes In developing the equations of aerodynamics we will invoke the firmly established and timetested physical laws: Conservation of Mass Conservation of Momentum Conservation of Energy Because we are not dealing with isolated point masses, but rather a continuous deformable medium, we will require new conceptual and mathematical techniques to apply these laws correctly. One concept is the control volume, which can be either finite or infinitesimal. Two types of control volumes can be employed: ) olume is fixed in space (Eulerian type). Fluid can freely pass through the volume s boundary. ) olume is attached to the fluid (Lagrangian type). olume is freely carried along with the fluid, and no fluid passes through its boundary. This is essentially the same as the free-body concept employed in solid mechanics. Eulerian Control olume fixed in space Lagrangian Control olume moving with fluid Both approaches are valid. Here we will focus on the fixed control volume. Mass Conservation Mass flow Consider a small patch of the surface of the fixed, permeable control volume. The patch has

20 area A, and normal unit vector ˆn. The plane of fluid particles which are on the surface at time t will move off the surface at time t + Δt, sweeping out a volume given by Δ = n A Δt n is the component of the veleocity vector normal to the area. where n = ˆ ^ n n swept volume ^ n A Δ t n Δt The mass of fluid in this swept volume, which evidently passed through the area during the Δt interval, is Δm = ρ Δ = ρ n A Δt The mass flow is defined as the time rate of this mass passing though the area. mass flow = m = lim Δm Δt 0 Δt The mass flux is defined simply as mass flow per area. mass flux = ṁ A = ρ n = ρ n A A Δ t Mass Conservation Application The conservation of mass principle can now be applied to the finite fixed control volume, but now it must allow for the possibility of mass flow across the volume boundary. d (Mass in volume) = Mass flow into volume dt Using the previous relations we have d ρ d = ρ n ˆ da () dt where the negative sign is necessary because ˆn is defined to point outwards, so an inflow is where n ˆ is positive. Using Gauss s Theorem and bringing the time derivative inside the integral we have the result [ ] ρ ( ) + ρ d = 0 t

21 This relation must hold for any control volume whatsoever. If we place an infinitesimal control volume at every point in the flow and apply the above equation, we can see that the whole quantity in the brackets must be zero at every point. This results in the Continuity Equation ρ ( ) + ρ = 0 () t which is the embodiment of the Mass Conservation principle for fluid flow. The steady flow version is: ( ) ρ = 0 (steady flow) For low-speed flow, steady or unsteady, the density ρ is essentially constant, which gives the very great simplification that the velocity vector field has zero divergence. = 0 (low-speed flow) Control olume Applications Steady Surface-Integral Form All of the above forms of the continuity equation are used in practice. The surface-integral form () with the steady assumption, ρ n ˆ da = 0 (3) is particularly useful in many engineering applications. Use of equation (3) first requires construction of the fixed control volume. The boundaries are typically placed where the normal velocity and the density are known, so the surface integral can be evaluated piecewise. Placing the boundary along a solid surface is particularly useful, since here n ˆ = 0 (at solid surface) and so that part of the boundary doesn t contribute to the integral. In the figure, we note that n ˆ da = cos θ da = n da = da s where da s is the area of the surface element projected onto a plane normal to. n ^ n θ or A A A s The form of the integrand using da s is most useful on a portion of the boundary which has a constant value of ρ (in magnitude and direction). The contribution of this boundary portion to the overall integral (3) is then ρ n ˆ da = ρ da s = ρ A s (portion with constant ρ ) 3

22 boundary portion with constant flow A s Note that the boundary portion does not need to be flat for A s to be well defined. Only ρ must be a constant. Channel Flow Application Placing the control volume inside a pipe or channel of slowly-varying area, we now evaluate equation (3). ρ n ˆ da = ρ A + ρ A = 0 ρ A = ρ A The negative sign for station is due to n ˆ = at that location. The control volume faces adjacent to walls do not contribute to the integral, since their normal vectors are perpendicular to the local flow and therefore have n ˆ = 0. ^ n.n ^ = 0 A ^ n A ^ n Placing plane, say, at any other location, gives the general result that ρ A = constant inside a channel. The product ρ A is also recognized as the constant channel mass flow. 4

23 Fluids Lecture 7 Notes. Momentum Flow. Momentum Conservation Reading: Anderson.5 Momentum Flow Before we can apply the principle of momentum conservation to a fixed permeable control volume, we must first examine the effect of flow through its surface. When material flows through the surface, it carries not only mass, but momentum as well. The momentum flow can be described as momentum flow = (mass flow) (momentum /mass) where the mass flow was defined earlier, and the momentum/mass is simply the velocity vector. Therefore ( ) momentum flow = m = ρ nˆ A = ρ n A n as before. Note that while mass flow is a scalar, the momentum flow is a where n = ˆ vector, and points in the same direction as. The momentum flux vector is defined simply as the momentum flow per area. momentum flux = ρ n ^ n. m A. m ρ Momentum Conservation Newton s second law states that during a short time interval dt, the impulse of a force F applied to some affected mass, will produce a momentum change dp a in that affected mass. When applied to a fixed control volume, this principle becomes. P in P(t) F. P out dp a dt = F () dp + P out P in = F () dt

24 In the second equation (), P is defined as the instantaneous momentum inside the control volume. P(t) ρ d The P out is added because mass leaving the control volume carries away momentum provided by F, which P alone doesn t account for. The P in is subtracted because mass flowing into the control volume is incorrectly accounted in P, and hence must be discounted. Both terms are evaluated by a surface integral of the momentum flux over the entire boundary. ( ) P out P in = ρ n ˆ da The sign of nˆ automatically accounts for both inlow and outflow. Applied forces The force F consists of two types. Body forces. These act on fluid inside the volume. The most common example is the gravity force, along the gravitational acceleration vector g. F gravity = ρ g d Surface forces. These act on the surface of the volume, and can be separated into pressure and viscous forces. F pressure = p nˆ da The viscous force is complicated to write out, and for now will simply be called Fviscous. ρ g pn F viscous pn F viscous Integral Momentum Equation Substituting all the momentum, momentum flow, and force definitions into Newton s second law () gives the Integral Momentum Equation. ( ) d ρ d + ρ nˆ da = p nˆ da + ρ g d + F viscous (3) dt

25 Along with the Integral Mass Equation, this equation can be applied to solve many problems involving finite control volumes. Differential Momentum Equation The pressure surface integral in equation (3) can be converted to a volume integral using the Gradient Theorem. p ˆn da = p d The momentum-flow surface integral is also similarly converted using Gauss s Theorem. This integral is a vector quantity, and for clarity the conversion is best done on each component separately. After substituting = u î + v ĵ + w k, ˆ we have ( ) ( ) ( ) ρ n ˆ u î + v ĵ + w k ˆ da = î ρ u d ( ) + ĵ ρ v d ( ) + k ˆ ρ w d The x-component of the integral momentum equation (3) can now be written strictly in terms of volume integrals. [ ] (ρu) ( + ρu ) p + ρg x (Fx) viscous d = 0 (4) t x This relation must hold for any control volume whatsoever. If we place an infinitesimal control volume at every point in the flow and apply equation (4), we can see that the whole quantity in the brackets must be zero at every point. This results in the x-momentum Equation (ρu) ( + ρu ) = p + ρg x + (Fx) viscous (5) t x and the y- and z-momentum Equations follow by the same process. (ρv) ( + ρv ) = p + ρg y + (Fy) viscous (6) t y (ρw) ( + ρw ) = p + ρg z + (Fz) viscous (7) t z These three equations are the embodiment of the Newton s second law of motion, applied at every point in the flowfield. The steady flow version has the / t terms omitted. 3

26 Fluids Lecture 8 Notes. Streamlines. Pathlines 3. Streaklines Reading: Anderson. Three types of fluid element trajectories are defined: Streamlines, Pathlines, and Streaklines. They are all equivalent for steady flows, but differ conceptually for unsteady flows. Streamlines Streamline equations A streamline is defined as a line which is everywhere parallel to the local velocity vector (x, y, z, t) = u î + v ĵ + w k. ˆ Define d s = dx î + dy ĵ + dz ˆk as an infinitesimal arc-length vector along the streamline. Since this is parallel to, we must have d s = 0 (w dy v dz) î + (u dz w dx) j ˆ + (v dx u dy) ˆ k = 0 Separately setting each component to zero gives three differential equations which define the streamline. The three velocity components u, v, w, must be given as functions of x, y, z before these equations can be integrated. To set the constants of integration, it is sufficient to specify some point x o, y o, z o through which the streamline passes, y y ds x o y o z o ds x v dy dx x o y o u x z 3 D streamline D streamline In -D we have dz = 0 and w = 0, and only the ˆk component of the equation above is non-trivial. It can be written as an Ordinary Differential Equation for the streamline shape y(x). dy v = dx u Again, u(x, y) and v(x, y) must be given to allow integration, and x o, y o must be given to set the integration constants. In a numerical integration, x o, y o would serve as the initial values.

27 Streamtubes Consider a set of x o, y o, z o points arranged in a closed loop. The streamlines passing through all these points form the surface of a streamtube. Because there is no flow across the surface, each cross-section of the streamtube carries the same mass flow. So the streamtube is equivalent to a channel flow embedded in the rest of the flowfield. y y x o y o z o x x o y o x z 3 D streamtube D streamtube In -D, a streamtube is defined by two streamlines passing through two specified x o, y o points. The flow between these two streamlines carries the same mass flow/span at each cross-section, and can be considered as a -D channel flow embedded in the rest of the flowfield. Pathlines The pathline of a fluid element A is simply the path it takes through space as a function of time. An example of a pathline is the trajectory taken by one puff of smoke which is carried by the steady or unsteady wind. This path is fully described by the three position functions x A (t), y A (t), z A (t), which can be computed by integrating the three velocity-field components u(x, y, z, t), v(x, y, z, t), w(x, y, z, t) along the path. The integration is started at time t o, from the element s initial position x o, y o, z o (e.g. the smoke release point), and proceeds to some later time t. t ( ) x A (t) = x o + u x A (τ), y A (τ), z A (τ), τ dτ t o t ( ) y A (t) = y o + v x A (τ), y A (τ), z A (τ), τ dτ t o t ( ) z A (t) = z o + w x A (τ), y A (τ), z A (τ), τ dτ t o The dummy variable of integration τ runs from t o to t. Streaklines A streakline is associated with a particular point P in space which has the fluid moving past it. All points which pass through this point are said to form the streakline of point P. An example of a streakline is the continuous line of smoke emitted by a chimney at point P, which will have some curved shape if the wind has a time-varying direction.

28 Unlike a pathline, which involves the motion of only one fluid element A in time, a streakline involves the motion of all the fluid elements along its length. Hence, the trajectory equations for a pathline are applied to all the fluid elements defining the streakline. The figure below illustrates streamlines, pathlines, and streaklines for the case of a smoke being continuously emitted by a chimney at point P, in the presence of a shifting wind. One particular smoke puff A is also identified. The figure corresponds to a snapshot when the wind everywhere is along one particular direction. streakline from point P (smoke line) instantaneous streamlines instantaneous wind velocity P A streakline at successive times pathline of fluid element A (smoke puff) In a steady flow, streamlines, pathlines, and streaklines all cooincide. In this example they would all be marked by the smoke line. 3

29 Fluids Lecture 9 Notes. Momentum-Integral Simplifications. Applications Reading: Anderson.6 Simplifications For steady flow, the momentum integral equation reduces to the following. ( ) ρ n ˆ da = p n ˆ da + ρ g d + Fviscous () Defining h as the height above ground, we note that h is a unit vector which points up, so that the gravity acceleration vector can be written as a gradient. g = g h Using the Gradient Theorem, this then allows the gravity-force volume integral to be converted to a surface integral, provided we make the additional assumption that ρ is nearly constant throughout the flow. ρ g d = ρ g h d = ρgh ˆn da We can now combine the pressure and gravity contributions into one surface integral. p n ˆ da + ρ g d (p + ρgh) n ˆ da Defining a corrected pressure p c = p+ρgh, the Integral Momentum Equation finally becomes ( ) ρ n ˆ da = pc n ˆ da + F viscous () Aerodynamic analyses using () do not have to concern themselves with the effects of gravity, since it does not appear explicitly in this equation. In particular, the velocity field will not be affected by gravity. Gravity enters the problem only in a secondary step, when the true pressure field p is constructed from p c by adding the tilting bias ρgh. h h h h = + g p p c ρg h net aerodynamic aerostatic (gravity neglected) (gravity alone)

30 For clarity, we will from now on refer to p c simply as p. The understanding is that the forces and moments computed using this p will not include the contributions of buoyancy. However, buoyancy is relatively easy to compute, and can be added as a secondary step if appropriate. It should be noted that buoyancy is rarely significant in aerodynamic flow situations, one exception being lighter-than-air vehicles like blimps. One situation where gravity is crucial is a flow with a free surface, like water waves past a ship. Here, gravity enters not in equation (), but in the free-surface boundary condition. Free-surface flows are not relevant to most aeronautical problems. Applications Basic Procedures Application of the integral momentum equation () uses the same basic techniques as for the integral continuity equation. Both can use the same control volume, and both demand that the integrals are evaluated for the entire surface of the control volume. There are three significant differences, however: ) Momentum is a vector. Each of the three x, y, and z components of equation () is independent, and must be treated separately. ) A volume boundary lying along a streamline now has a nonzero pˆn contribution. 3) A volume boundary lying along a solid surface now has a nonzero pˆn contribution, and possibly also a viscous shear τ contribution if viscous effects are deemed significant. Effect of body in flow Equation () assumes that there is no internal force acting on the fluid other than gravity. If there is a solid body in the flow, it is therefore necessary to construct the control volume such that the body is on the volume s exterior. This is easily done by placing the body in an indentation cdefg of the volume surface, as shown in the figure. u p = p a streamline far away e p = p dṁ f d b c g p = p u i streamline far away h The surface integrals are now broken up into the individual pieces: the outermost part abhi, the body surface def, and the cut surfaces cd and fg. ( ) da = ( ) da + ( ) da + ( ) da + ( ) da abhi def cd fg

31 Surfaces cd and fg have the same flow variables, but opposite ˆn vectors, so that all their contributions are equal and opposite, and hence cancel. ( ) da + ( ) da = 0 cd fg The resultant force/span R on the body is the integrated pressure and shear force distribution on the body surface. Because the body surface and volume surface have opposite normals n, ˆ this R is precisely equal and opposite to all the def surface integrals for the control volume. ( ) da = R def R Forces on airfoil p τ e d e d f f τ p R Forces on control volume An intuitive explanation is that the aerodynamic force exerted by the fluid on the body is exactly equal and opposite to the force which the body exerts on the fluid. Collecting the remaining integrals produces the very general result ( ) ρ ˆn da + p ˆn da = R (3) abhi abhi which gives the resultant force on the body only in terms of the surface integrals on the outer surface of the volume. Drag relations To obtain the drag/span, we define the x direction to be aligned with, and take the x-component of equation (3). ( ) ρ n ˆ u da + p n ˆ î da = D (4) abhi abhi The pressure on the outer surface is uniform and equal to p, and hence the pressure integral is zero. The momentum-flux integral is zero on the top and bottom boundaries, since 3

32 these are defined to be along streamlines, and hence have zero momentum flux. Only the momentum flux on the inflow and outflow planes remain. ρ u da ρ u da = D bh ia (5) Using the concept of a streamtube, we see that planes and have the same mass flows, with ρ u da = ρ u da = d ṁ as shown in the figure. Therefore both the plane and plane integrals can be lumped into one single plane integral. b D = ρ u (u u ) da (6) h Equation (6) gives the drag only in terms of the freestream speed u, and the downstream wake profiles ρ and u. These downstream profiles can be measured in a wind tunnel, and numerically integrated in equation (6) to obtain the drag. A physical interpretation of (6) can be obtained by writing it as D = b (u u ) d ṁ h This is simply the summation of the momentum flow lost by all the infinitesimal streamtubes in the flow. The lost momentum appears as drag on the body. 4

33 Fluids Lecture 0 Notes. Substantial Derivative. Recast Governing Equations Reading: Anderson.9,.0 Substantial Derivative Sensed rates of change The rate of change reported by a flow sensor clearly depends on the motion of the sensor. For example, the pressure reported by a static-pressure sensor mounted on an airplane in level flight shows zero rate of change. But a ground pressure sensor reports a nonzero rate as the airplane rapidly flies by a few meters overhead. The figure illustrates the situation. wing location at t = t o p (t) p (t) Note that although the two sensors measure the same instantaneous static pressure at the same point (at time t = to), the measured time rates are different. p p (t) p (t) dp dp p (to) = p (to) but (to) (to) dt dt Drifting sensor We will now imagine a sensor drifting with a fluid element. In effect, the sensor follows the element s pathline coordinates xs(t), y s (t), zs(t), whose time rates of change are just the local flow velocity components dxs dy s dzs = u(xs, y s, zs, t), = v(xs, y s, zs, t), = w(xs, y s, zs, t) dt dt dt pressure field p s t o t pathline sensor drifting with local velocity Consider a flow field quantity to be observed by the drifting sensor, such as the static pressure p(x, y, z, t). As the sensor moves through this field, the instantaneous pressure value reported by the sensor is then simply p (t) s Dp Dt dp dt p s (t) = p (xs(t), y s (t), zs(t), t) () s t

34 This p s (t) signal is similar to p (t) in the example above, but not quite the same, since the p sensor moves in a straight line relative to the wing rather than following a pathline like the p s sensor. Substantial derivative definition The time rate of change of p s (t) can be computed from () using the chain rule. dp s p dxs p dy s p dzs p = dt x dt y dt z dt t But since dxs/dt etc. are simply the local fluid velocity components, this rate can be expressed using the flowfield properties alone. dp s p p p p Dp = + u + v + w dt t x y z Dt The middle expression, conveniently denoted as Dp/Dt in shorthand, is called the substantial derivative of p. Note that in order to compute Dp/Dt, we must know not only the p(x, y, z, t) field, but also the velocity component fields u, v, w(x, y, z, t). Although we used the pressure in this example, the substantial derivative can be computed for any flowfield quantity (density, temperature, even velocity) which is a function of x, y, z, t. D( ) ( ) ( ) ( ) ( ) ( ) + u + v + w = + ( ) Dt t x y z t The rightmost compact D/Dt definition contains two terms. The first / t term is called the local derivative. The second term is called the convective derivative. In steady flows, / t = 0, and only the convective derivative contributes. Recast Governing Equations All the governing equations of fluid motion which were derived using control volume concepts can be recast in terms of the substantial derivative. We will employ the following general vector identity (a v) = v a + a v which is valid for any scalar a and any vector v. Continuity equation Applying the above vector identity to the divergence form continuity equation gives ρ t ρ t + ( ρ ) = 0 + ρ + ρ = 0 Dρ Dt + ρ = 0 () The final result above is called the convective form of the continuity equation. A physical interpretation can be made if it s written as follows.

35 or... Dρ = ρ Dt fractional density rate = velocity divergence fractional volume rate = velocity divergence For a fluid element of given mass, the volume must vary as /density, which gives the second interpretation above. Both interpretations are illustrated in the left figure below, where the fluid element expands when it flows through a flowfield region where > 0. In low speed flows and in liquid flows the density is essentially constant, so that Dρ/Dt = 0 and by implication = 0. Momentum equation The divergence form of the x-momentum equation is (ρu) ( + ρu ) = p + ρg x + (Fx) viscous t x Applying the vector identity again, and also cancelling some terms by use of the continuity equation (), produces the convective form of the momentum equation. The y- and z- momentum equations are also derived the same way. Du p ρ = + ρg x + (Fx) viscous Dt x (3) Dv p ρ = + ρg y + (Fy) viscous Dt y (4) Dw p ρ = + ρg z + (Fz) viscous Dt z (5) The Du/Dt etc. substantial derivatives are recognized as the acceleration components experienced by a fluid element. This leads to a simple physical interpretation or these equations as Newton s law applied to a fluid element of unit volume. mass/volume acceleration = total force/volume The element s mass/volume is simply the density ρ, and the total force/volume consists of the buoyancy-like pressure gradient force, the gravity force, and the viscous force. Δ ρ = D ρ Δ t D t Δ = D D t Δ t t Δ. t + Δt expanding fluid element t Δ t + Δt p + ρ g +F viscous accelerating fluid element 3

36 Fluids Lecture Notes. orticity and Strain Rate. Circulation Reading: Anderson.,.3 orticity and Strain Rate Fluid element behavior When previously examining fluid motion, we considered only the changing position and velocity of a fluid element. Now we will take a closer look, and examine the element s changing shape and orientation. Consider a moving fluid element which is initially rectangular, as shown in the figure. If the velocity varies significantly across the extent of the element, its corners will not move in unison, and the element will rotate and become distorted. y (y+dy) z x element at time t (y) element at time t + Δt In general, the edges of the element can undergo some combination of tilting and stretching. For now we will consider only the tilting motions, because this has by far the greatest implications for aerodynamics. The figure below on the right shows two particular types of element-side tilting motions. If adjacent sides tilt equally and in the same direction, we have pure rotation. If the adjacent sides tilt equally and in opposite directions, we have pure shearing motion. Both of these motions have strong implications. The absense of rotation will lead to a great simplification in the equations of fluid motion. Shearing together with fluid viscosity produce shear stresses, which are responsible for phenomena like drag and flow separation. tilting stretching General edge movement Rotation (vorticity) Shearing motion (strain rate) Tilting edge movements