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Low Speed Aerodynamics 3: Conservation Equations in Integral Form Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Deriving Conservation Equations From the Laws of Physics y Laws Physical Fluids, being matter, must obey the laws of Physics. The ones we need are: 1. Conservation of Mass: Mass is neither created nor destroyed (nor changed to anything else) 2. Newton's Laws of Motion – •1st Law: concept of equilibrium •2nd 2nd Law: force, force acceleration, acceleration momentum •3rd Law: action & reaction Conservation of Energy -- 1st Law of Thermodynamics These must be valid for all flow problems that we consider. http://abyss.uoregon.edu/~js/images/relativistic.jpg They might need modification if the flow is inside a star or a Black Hole, where there may be substantial conversion between matter and energy Also, for problems involving spacecraft (or atoms or subatomic particles) accelerating to speeds close to that of light. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. http://www.dailygalaxy.com/photos/unc ategorized/2007/07/05/black_hole_3_3 .jpg Constitutive Relations In addition, we can get relations which are specific to the kind of fluid with which we are dealing. These equations are called the "Constitutive Relations". They include the equations of state. Equations. q of State [relations [ between different properties] p p ] •Perfect Gas Law: Thermal eqn. of state •Caloric eqn. of state Energy contained per unit mass of a substance = (specific heat)*(temperature) The independent variables are: velocity, density, pressure and temperature. To solve for these in a given problem, we have the 3 conservation equations of mass, momentum and energy, and the thermal state equation. In addition, we may be interested in other variables, like the speed of sound. Here the properties of the fluid are taken into account further using the caloric l i equation ti off state. t t Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Conservation of Mass g of interest,, ((a "control volume"), ), If we have a fluid ggoingg in and comingg out of a ggiven region we can say for sure that what goes in per unit time = what comes out per unit time + what accumulates inside per unit time. Mass going in per second= { Sum of masses going out per second + mass accumulated inside in 1 second. } In general, mass may be going in and/or out everywhere across an (imaginary ?) surface enclosing the space in which you are interested. Also, the velocity of the inflow/outflow may be nonuniform and in some arbitrary direction. nonuniform, direction Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Mass flowing out of the surface of the control volume per unit time = ds u n S where h dS is i a smallll element l t off the th ttotal t l surface f area S S. An "integral" is a neat way of saying, “collect all the little bits and add them up." Law of conservation of mass for a control volume becomes u n dS S dV 0 t V “Conservation" of anything can be expressed this way. as we will see. Here the r.h.s. is zero: mass can't get converted to anything. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Example: Closed circuit wind tunnel Settling Chamber velocity = 15 fps, fps uniform; steady. Chamber diameter: 20 feet. Test section is a 9-foot diameter duct. Our "control volume" has one face in the settling chamber, volume chamber other face in the test section. Assume that no flow can escape out the sides of the control volume. volume "Steady" means that at any given point in the flow, the properties don't change as time changes. g (anything thi ) 0 t u n 0 d s The “integral form of the mass conservation equation” becomes: S Subscript "1" refers to settling chamber face of control volume, subscript "2" refers to test section. Air cannot escape out of the sides of the "control volume, so the integral reduces to just ρ1A1U1 + ρ2 A2U2 = 0 -ρ Negative sign: Look at direction of vector normal to each face, with respect to the direction of U. Thus U2 = (ρ1A1U1) / (ρ2A2) Since the density change can be neglected, and each area is a circle, U2 = (20/9)2*15 = 74.07 ft/s. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Conservation of Momentum Newton's 2nd law of Motion: "Rate of Change of Momentum = Force" or "Net force acting on a system = time rate of change of momentum of system" Note units: M Momentum t = mass x velocity l it = density x volume x velocity = density x area x distance x velocity Momentum per unit time = density x velocity x velocity x area What is being conserved? "momentum per unit volume". So the conservation equation has 2 terms on the left hand side: 1. "rate of change of the quantity, integrated over the whole control volume". This is the "unsteady term". 2 "net 2. net outflow per unit time of the quantity across the surfaces of the control volume volume, integrated over the whole control surface". This is the convection term. On the right hand side, we have surface and volume integrals of the quantities to which our "conserved quantity" may have changed. In this case, when momentum disappears, the rate of change of quantity momentum produces forces of various kinds. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Forces In general, general we consider two kinds of forces: a) "Body forces" : forces that act per unit mass, such as gravity or electromagnetic forces. b) "Surface Surface forces forces":: forces that act per unit area, area acting normal to the area and parallel to the area. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Types of Forces to be included in the momentum equation a) Body forces Gravitational force per unit volume = g (Important for liquid flows, and buoyant gas flows). Electromagnetic force: it would be something like B Important in ion propulsion, spark plugs, plasmas. Generally we neglect body forces in aerodynamics Generally, aerodynamics. b) Surface forces Pressure is the result of molecules, flying about at random, crossing [ or colliding with ] the surface, thus transferring momentum. So it acts normal to the surface. It must be related to # of molecules/volume [thus related to density], and speed of random motion of molecules [thus related to temperature]. p ] Thus,, ppressure is related to the pproduct of densityy and temperature. p This relation is a simple proportionality (multiply by a constant) where R is the gas constant. This is called the "perfect gas law." dS because pressure acts on the surface , opposite to n p n So force due to pressure is S Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Viscous Forces Fluids, like other forms of matter, resist movement. Part of the resistance is explained p byy ppressure: this part is reversible. The rest is irreversible: when fluids move, there is some loss. This is attributed to "viscosity." There must be some relation between stress and strain. In solids, stress is a function of strain. In the simplest case, stress is directly proportional to strain Fluids don’t resist shear strain if applied slowly. The resistance is to the ) in fluids, rate of strain. Thus, ideallyy ((this is called “Newtonian Fluid”), stress is proportional to Rate of Strain. The proportionality constant is called Absolute Viscosity . Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Shear Stress vs. Rate-of-Strain Relations u v xy The direction is indicated by the second subscript. The units of stress y x are Force/Area, such as Newtons/m2, psi, or psf. v w yz z y u w zx z x These can be written in compact form form. Here u is used to represent any of the velocity components components, and x is used to represent any of the spatial coordinates. i and j representing x,y,z in turn. u j u i ij x x j i The normal stress, on the other hand, is simply the pressure. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Normal Stress Here “normal” normal is a fashionable term for “perpendicular perpendicular to a surface surface”, as opposed to “tangential” tangential , not as opposed to “abnormal”. . The normal stress is simply due to the pressure. xx pi yy pj pj zz pk The unit vectors are included to show the direction of each of these stress components. Don’t fluids resist the strain rate due to compression and expansion? They do, and a relation between normal stress and rate of normal strain can be written, just like the shear strain relations on the previous slide, slide and added on to the above expressions. expressions However, in gases, the coefficient of resistance to normal rate of strain is very small, so that this “bulk viscosity” does not become significant unless the strain rate is extremely high. This occurs, for instance, across a shock wave, where velocity decreases by a large factor in the space needed for a few (under 10) collisions between molecules. In most other situations,, it is a usual assumption p that “Bulk Viscosityy is Zero” in a ggas so that the normal stress is attributed to just the pressure. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Energy Equation In addition to mass and momentum, we can use the fact that “energy is neither created nor destroyed, but can change form” y is energy gy pper unit volume. The lhs has the terms describingg time rate of The “conserved qquantity” change, and “flux” across the surfaces of the control volume. The rhs has the terms describing what changes the energy in the control volume: work and heat transfer. We use the first law of thermodynamics, y which says y that if yyou do work or release heat, yyour energy gy is exhausted. We will leave detailed study of this equation to the course on compressible flow and thermodynamics. The measure of total energy in a flow is the “stagnation enthalpy”, which is related to the stagnation value of temperature. So the energy equation reduces to: "Stagnation temperature goes up if work or heat are added to the flowing fluid". In steady flow, with no work being added, and no heat being added (adiabatic flow), the rhs is zero, so that the energy equation reduces to “Stagnation Temperature is Constant”: u2 T0 T cons tant 2c p The cp here is not pressure coefficient, but “specific R heat at constant pressure. cp 1 For diatomic gases such as air, at usual temperatures that we encounter in low speed aerodynamics aerodynamics, the “gamma” gamma has the value 1.4. Thus cp = 3.5R Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Why we don’t worry about the energy equation too much here In low speed aerodynamics, density is usually assumed constant (changes due to velocity changes are negligibly small). Then you need one less equation because density is no longer an unknown variable. Thus the Th th energy equation ti is i rarely l needed, d d exceptt when h one worries i about b t what h t happens when work is added to the flow, as by a propeller or rotor. In such cases, cases we will relate the stagnation pressure to the stagnation enthalpy, enthalpy and use that instead as a measure of work added or removed. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Fluid Dynamics Summary 1 11. 2. Assumed that fluid flow is a continuum phenomenon. phenomenon Fluids, being composed of matter, obey the laws of physics. To calculate the lift and drag forces, forces we would like to be able to calculate the pressure acting on the surface. This depends on the velocity, and also the state of the fluid, which requires knowledge of the temperature and density. Thus we would like to be able to calculate the velocity vector Thus, vector, density density, temperature and pressure everywhere, as functions of space (x,y,z) and time (t). For this we use 3 laws of physics: Mass is neither created nor destroyed.......... (1) Rate of change of momentum = Net force .....(2) Energy can change form, but is neither created nor destroyed ......(3) In addition addition, the state of the gas is specified by the “State State Equations” Equations . The Thermal Equation of State for a gas reduces to the Perfect Gas Law, which relates density, temperature and pressure. To solve very any flow problem, we can specify the boundary conditions and/or initial conditions, and solve all these equations simultaneously all over the flow field. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. In the next sections, we will see how to reduce these conservation equations to forms which can be usedd tto calculate l l t changes h iin velocity l it andd pressure over space andd titime, andd thus th to t calculate l l t flow properties at given points, or over entire vehicles. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Reducing the Integral Conservation Equations to Differential Form In this section we will first convert the integral conservation equations to a form� suitable to apply pp y at each ppoint,, so that we can track changes g from one point to another. So we have to go from the "integral form" over a control volume, to a "differential form" which deals with small changes from point to point. f z f t f(x,y,z,t) Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. f y f x Relations between integrals g over lines, surfaces and volumes Why? Because we would like to bring everything under the same integral sign, set the whole integral to zero, and relate the quantities inside, thus getting rid of all the integrals dS S dV V Stokes' Theorem Relates line integral over a closed contour to surface integral over the surface enclosed by the contour u dl u dS c S This can also be written with the area element dS as a scalar, and the unit normal vector indicating its orientation u dl u n dS c Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. S Divergence Theorem Relates surface integral over a control surface to the volume integral over the volume enclosed byy the control surface. u n dS u dV S V Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Gradient Theorem Relates surface integral of a scalar a control surface to the volume integral of the ggradient of the scalar over the volume enclosed byy the control surface. pdS p dV S V Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Identityy on the Gradient of a Product of a Vector and a Scalar u u u u Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Substantial Derivative D u Dt t The first term on the rhs is the "local" local or "unsteady" unsteady term. term The second is the "convective" convective term. term D u v w Dt t x y z "local" or " "unsteady" d " “convective" The rate of change D()/Dt is for two reasons: 1. Things are changing at the point through which the element is moving (unsteady, local) 2. The element is movingg into regions g with different properties.� Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. The Substantial Derivative: Consider the rate of change of atmospheric pressure inside a TV News Copter flying towards an approaching storm. If the helicopter hovers over a given place, the occupants feel the pressure dropping there: that’s the local rate of change. If the helicopter flies towards the storm, the occupants feel the pressure dropping even faster as they approach the storm: this is the rate of change due to the helicopter moving through a pressure gradient, i.e., the convective rate of change. visibleearth.nasa.gov/view_rec.php?id=18224 rotored.arc.nasa.gov/activity/ranger.html Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Differential Form of the Continuity Equation u n dS dV 0 t V S Use the Divergence Theorem u n dS u dV S V u dV 0 V t We can take the limiting case as V tends to 0 – this gives a point, where Use the vector identity u u u u u 0 t D u 0 Dt D u v w ( ) 0 Dt x y z u 0 t t Note: This means that if there is a variation of density in time and/or space, it must result in dilatation. One consequence q is: If density is constant, then dilatation is zero. Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Momentum Conservation: Differential Form By arguments similar to those used with the continuity equation, the integral form of the momentum equation is reduced to the differential form (See textbook for derivation). Writing in terms of components, components Du p f x Fx,viscous Dt x p Dv f y Fy ,viscous Dt y Where f is body force per unit mass, and F is viscous force per unit volume. Dw p f z Fz ,viscous Dt z In vector form, Du p f Fviscous Dt Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Energy Equation The energy equation, reduced to differential form, is: u 2 v 2 w2 D e 2 q pu f u Qviscous Wviscous Dt Here, e is internal energy per unit mass, q is heat transfer rate (per unit time), per unit mass, Q is heat transfer per unit volume, f is body force per unit mass, W is work per unit volume. The equation says that th t the th rate t off change h off energy per unitit volume, l which hi h consists i t off iinternal t l energy off th the molecules and kinetic energy of the flow, is equal to the sum of the heat transfer rate per unit volume, the work done by pressure on the flow (which is negative of work done by the flow against pressure), the work done by body forces and viscous forces, forces and the rate of heat transfer due to viscous effects effects, into the flow. We have not included “potential energy” due to a gravitational field or an electromagnetic field in the above. Gravitational potential is important when dealing with water flows, or with balloon flight (aerostatics). (aerostatics) Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Simplifying the Momentum Equation Reynolds number: Re UL Ratio of “inertial forces” to “viscous forces”. For typical speeds and sizes encountered in flows over aircraft, the Reynolds number is very high (on the order of several millions). This allows us to neglect some terms involving viscous stresses, compared to other terms in the momentum equation. Du p f Fviscous Dt Neglecting g g the viscous terms simplifies p the momentum equation q down to the “Euler equation”. q The Euler equation Du p f Dt Where body forces can be neglected (i.e., except flows with very high radial acceleration) Du p Dt Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Example 1 1. A spinning cylinder is oriented such that the rotation about its axis is 68i + 47j radians per second, it is in a freestream 40k m/s, with the fluid density being 1.2kg/m3. Find the lift vector. L' U i j U 68 47 0 0 k 0 40 40 47i 68 j = 2256i – 3264j Newtons Hey, is this perpendicular to the other two vectors? Do the directions make sense? Please verify for yourself. What is the magnitude of the lift? L' 22562 32642 ? Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. Example Derive the Bernoulli equation for steady incompressible flow, along a streamline. Du p Dt Euler equation u u u p t 0 t Steady flow: u ui vj wk i j k x y x u u u v w u y x x Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved. u u u 1 p u v w y z x x v v v 1 p u v w z y y x w w w 1 p v w u y z z x Multiply the first equation by dx dx, second by dy dy, and third by dz. dz Consider the first equation: u u u u 1 p dx dx v dx w dx x z x y Along a streamline, v/u=dy/dx etc. So, vdx = udy, etc. u u u 1 p 1 p u dx u dy u dz dx u du dx x y z x x Adding the 3 equations together, we get 1 1 2 d u 2 v 2 w2 dp d dp d U 0 2 2 p 2 U 2 2 dp p 1 U 2 1 2 d U 0 2 1 1 p2 U 2 2 p1 U12 const p0 2 2 Copyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved.