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Low Speed Aerodynamics
3: Conservation Equations in Integral Form
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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.
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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
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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
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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.
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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.
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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.
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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.
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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
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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 .
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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.
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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.
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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
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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.
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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.
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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.
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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)
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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
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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
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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
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Identityy on the Gradient of a Product of a Vector and a Scalar

 
  u    u
 u  u 
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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.�
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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
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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.
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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
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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)
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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
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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  ?
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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
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 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
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