Low Speed Aerodynamics 3: Conservation Equations … · 1. Assumed that fluid flow is a continuum...

Low Speed AerodynamicsLow Speed Aerodynamics

3: Conservation Equations in Integral Form3: 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

Physical Laws y

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 Law: force acceleration momentum2nd Law: force, 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 energyconversion between matter and energy

Also, for problems involving spacecraft (or atoms or subatomic particles) accelerating to speeds


close to that of light. http://www.dailygalaxy.com/photos/uncategorized/2007/07/05/black_hole_3_3.jpg

Constitutive RelationsIn 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. of State [relations between different properties]q [ 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

l i ti f t tCopyright 2007 N. Komerath. Other rights may be specified with individual items. All rights reserved.

caloric equation of state.

Conservation of Mass

If we have a fluid going in and coming out of a given region of interest, (a "control volume"), g g g g g , ( ),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 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, and in some arbitrary direction.

h dS i ll l t f th t t l f S

S

nusd Mass flowing out of the surface of the control volume per unit time =

where dS is a small element of the total surface area 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

dVdS 0

VS

dVt

dSnu 0

“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.


Example: Closed circuit wind tunnelSettling Chamber velocity = 15 fps uniform; Settling Chamber velocity = 15 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 has one face in the settling 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. 0)( thig 0)(

anything

t

Snusd 0The “integral form of the mass conservation equation” becomes:

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 = 0ρ1 1 1 ρ2 2 2

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.

Conservation of MomentumNewton'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 t l it Momentum = mass x velocity

= density x volume x velocity = density x area x distance x velocity

Momentum per unit time = density x velocity x velocity x area 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: 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 outflow per unit time of the quantity across the surfaces of the control volume integrated over 2. net outflow per unit time of the quantity across the surfaces of the control 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 may have changed. In this case, when momentum disappears, the rate of change of momentum produces forces of various kinds.


In general we consider two kinds of forces:

ForcesIn general, we consider two kinds of forces:

a) "Body forces" : forces that act per unit mass, such as gravity or electromagnetic forces.

b) "Surface forces": forces that act per unit area acting normal to the area and parallel to b) Surface forces : forces that act per unit area, acting normal to the area and parallel to the area.


Types of Forces to be included in the momentum equation

a) Body forces

Gravitational force per unit volume = Gravitational force per unit volume =

(Important for liquid flows, and buoyant gas flows).

Electromagnetic force: it would be something like

g

BElectromagnetic force: it would be something like

Important in ion propulsion, spark plugs, plasmas.

Generally we neglect body forces in aerodynamics

B

Generally, we neglect body forces in aerodynamics.

b) Surface forces

Pressure is the result of molecules, flying about at random, crossing [ or colliding with ] the surface, 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]. Thus, pressure is related to the product of density and temperature.p ] , p p y p

This relation is a simple proportionality (multiply by a constant) where R is the gas constant. This is called the "perfect gas law."


So force due to pressure is S

dSnp because pressure acts on the surface , opposite to n

Fluids, like other forms of matter, resist movement. Part of the resistance is explained by pressure: Viscous Forces

p y pthis 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 therate of strain. Thus, ideally (this is called “Newtonian Fluid”), in fluids,


y ( )stress is proportional to Rate of Strain. The proportionality constant is called Absolute Viscosity .

Shear Stress vs. Rate-of-Strain Relations

The direction is indicated by the second subscript. The units of stressare Force/Area, such as Newtons/m2, psi, or psf.

xy uy

vx

yz

vz

wy

These can be written in compact form Here u is used to represent any of the velocity components and

zx uz

wx

These can be written in compact form. Here u is used to represent any of the velocity components, andx is used to represent any of the spatial coordinates. i and j representing x,y,z in turn.

uj ui

ij xi

x j

The normal stress, on the other hand, is simply the pressure. The normal stress, on the other hand, is simply the pressure.


Here “normal” is a fashionable term for “perpendicular to a surface” as opposed to “tangential”Normal Stress

Here normal is a fashionable term for perpendicular to a surface , as opposed to tangential ,not as opposed to “abnormal”. . The normal stress is simply due to the pressure.

xx pi

pj

The unit vectors are included to show the direction of each of thesestress components.

yy pj

zz pk

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 and added on to the above expressions just like the shear strain relations on the previous slide, and added on to the above 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 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 that “Bulk Viscosity is Zero” in a gas so that the


, p y gnormal stress is attributed to just the pressure.

In addition to mass and momentum, we can use the fact that Energy Equation

In addition to mass and momentum, we can use the fact that “energy is neither created nor destroyed, but can change form”

The “conserved quantity” is energy per unit volume. The lhs has the terms describing time rate of q y gy p gchange, 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, which says that if you do work or release heat, your energy is exhausted. y y y y gyWe 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”:

T0 T u2

2cp cons tant For diatomic gases such as air, at usual

temperatures that we encounter in low speed aerodynamics the “gamma” p

The cp here is not pressure coefficient, but “specificheat at constant pressure. c R

speed aerodynamics, the gamma has the value 1.4. Thus cp = 3.5R


cp 1

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.

Th th ti i l d d t h i b t h t Thus the energy equation is rarely needed, except when one worries about what happens when work is added to the flow, as by a propeller or rotor.

In such cases we will relate the stagnation pressure to the stagnation enthalpy In such cases, we will relate the stagnation pressure to the stagnation enthalpy, and use that instead as a measure of work added or removed.


Fluid Dynamics Summary 11 Assumed that fluid flow is a continuum phenomenon 1. Assumed that fluid flow is a continuum phenomenon. 2. Fluids, being composed of matter, obey the laws of physics.

To calculate the lift and drag forces we would like to be able to calculate the pressure acting on the To calculate the lift and drag 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 density temperature and pressure Thus, we would like to be able to calculate the velocity vector, 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 the state of the gas is specified by the “State Equations” The Thermal Equation of StateIn addition, the state of the gas is specified by the State Equations . The Thermal Equation of Statefor 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.

In the next sections, we will see how to reduce these conservation equations to forms which can bed t l l t h i l it d d ti d th t l l t used to calculate changes in velocity and pressure over space and time, and thus to calculate

flow properties at given points, or over entire vehicles.


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 at each point, so that we can track changes pp y p , gfrom 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

f

fyz

f fx

ftf(x,y,z,t)


Relations between integrals over lines, surfaces and volumesWhy? 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

g

S

dS V

dV

Stokes' TheoremRelates line integral over a closed contour to surface integral over the Relates line integral over a closed contour to surface integral over the surface enclosed by the contour

Sdld Sduldu

Sc

dSnulduSc

This can also be written with thearea element dS as a scalar, and theunit normal vector indicating its orientation


Divergence Theorem

Relates surface integral over a control surface to the volume integral overthe volume enclosed by the control surface. y

dVudSnu VS

dVudSnu


Gradient Theorem

Relates surface integral of a scalar a control surface to the volume integral of the gradient of the scalar over the volume enclosed by the control surface. g y

dVpSpd

VS

dVpSpd


Identity on the Gradient of a Product of a Vector and a Scalary

u u u u u u


Substantial Derivative

D

utDt

D

The first term on the rhs is the "local" or "unsteady" term The second is the "convective" term

wvutDt

D

The first term on the rhs is the local or unsteady term. The second is the convective term.

zyxtDt

"local" or" d " “convective"

The rate of change D()/Dt is for two reasons:

"unsteady" “convective"

1. Things are changing at the point through which the element is moving (unsteady, local)

2. The element is moving into regions with different properties.�g g


The Substantial Derivative: Consider the rate of change of atmospheric pressure inside a TV News Copterflying 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


Differential Form of the Continuity Equation

VS

dVt

dSnu 0

dVudSnu Use the Divergence Theorem

dVu 0

VS

Use the Divergence Theorem

V t

We can take the limiting case as V tends to 0 – this gives a point, where 0 u

t

t uuu

Use the vector identity

0 Note: This means that if there is a 0 uu

t

0 uD

Note: This means that if there is avariation of density in time and/or space, it must result in dilatation.One consequence is: 0 u

Dt q

If density is constant, then dilatation is zero.

0)(

wvuD


0)(

zyxDt

Momentum Conservation: Differential Form

By arguments similar to those used with the continuity equation, the integral form of themomentum equation is reduced to the differential form (See textbook for derivation).Writing in terms of componentsWriting in terms of components,

viscousxx Ffxp

DtDu

,

Where f is body force per unit

viscousyy Ffyp

DtDv

,

mass, and F is viscous force per unit volume.

y

viscouszz Ffzp

DtDw

,

In vector form,

viscousFfpDt

uD


Energy Equation

The energy equation, reduced to differential form, is:

WQufupq

wvueD

2

222

viscousviscous WQufupqDt

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

th t th t f h f it l hi h i t f i t l f th says that the rate of change of energy per unit volume, which consists of internal energy of 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 and the rate of heat transfer due to viscous effects the work done by body forces and viscous forces, and the rate of heat transfer due to viscous 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)


or with balloon flight (aerostatics).

Simplifying the Momentum EquationULReynolds number:

ULRe Ratio of “inertial forces” to “viscous forces”.

For typical speeds and sizes encountered in flows over aircraft, the Reynolds number is very 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 viscousstresses, compared to other terms in the momentum equation.

uD

viscousFfpDt

uD

Neglecting the viscous terms simplifies the momentum equation down to the “Euler equation”. The Euler equation

fpuD

g g p q q

fpDt

Where body forces can be neglected (i.e., except flows with very high radial acceleration)

pDt

uD


1 A spinning cylinder is oriented such that the rotation about its axis is 68i + 47jExample

1. A spinning cylinder is oriented such that the rotation about its axis is 68i + 47jradians per second, it is in a freestream 40k m/s, with the fluid density being 1.2kg/m3. Find the lift vector.

UL 'kji

400004768kji

U

ji 684740

= 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? ?32642256' 22 L


ExampleDerive the Bernoulli equation for steady incompressible flow, along a streamline.

pDt

uD

Euler equation

Dt

puuut

Steady flow:

0t

kx

jy

ix

kwjviuu

ux

wy

vx

uuu


xp

zuw

yuv

xuu

1

xzyx

yp

zvw

yvv

xvu

1

zp

zww

ywv

xwu

1

Multiply the first equation by dx second by dy and third by dz Consider the first equation:Multiply the first equation by dx, second by dy, and third by dz. Consider the first equation:

dxxpdx

zuwdx

yuvdx

xuu

1

y

Along a streamline, v/u=dy/dx etc. So, vdx = udy, etc.

dxpdzuudyuudxuu

1 dxpduu 1dxxpdz

zudy

yudx

xu

dx

xpduu

dd 11 222 2

Adding the 3 equations together, we get

dpwvud1

21 222 0

22

Uddp

02

2 22

UpUddp

02

112

2211 pconstUpUp


022

11

UpUddp

01122 22pconstUpUp

Low Speed Aerodynamics 3: Conservation Equations … · 1. Assumed that fluid flow is a continuum...

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