AOE 2104--Aerospace and Ocean Engineering
Fall 2009
AOE 2104
Introduction to Aerospace
Engineering
Lecture 2
Basic Aerodynamics
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Reminder: The first homework assignment
(paper copy) is due AT THE BEGINNING OF
NEXT CLASS!!
Also I would appreciate any feedback on the
class that you have. You are welcome to see me
after class, tell me during class, or send me an
email.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Standard Atmosphere
Any questions ?
2 equations used to construct the standard atmosphere
model ?
2 types of regions found in the temperature variations
with altitude and their characteristics ?
3 steps to determine p, r, and T at any altitude ?
Name and define the different types of altitudes.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Basic Aerodynamics
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Basic Aero – Why? How? What do we have so far?
Why are we looking into aerodynamics?
To determine the forces acting on a vehicle in flight
Remember aerodynamic forces arise from two natural phenomena
How are we going to proceed ?
 Using Laws of Physics to quantify the interaction between the vehicle
and the environment it is evolving in.
What do we have so far ?
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Our Aerodynamic Tool Box
Four aerodynamic quantities that define a flow field
Steady vs unsteady flow
Streamlines
Sources of aerodynamic forces
Equation of state for perfect gases
Hydrostatic Equation
Standard Atmosphere Model
6 different altitudes
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Aerodynamic Tools Needed: Governing Laws
We are going to need the 3 following physical principles to describe the interaction
between the vehicle and its associated flow field:
Conservation of Mass
Continuity Equation (§§ 4.1-4.2)
Newton’s 2nd Law (and
Conservation of Momentum)
Euler’s and Bernoulli’s
Equations (§§ 4.3-4.4)
Conservation of Energy
Energy Equation (§§ 4.5-4.7)
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Conservation of Mass – The Continuity Equation
Physical Principle:
Mass can neither be created nor destroyed (in other words, input = output).
Eq.(4.2)
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Streamline

v

v

v

v

v
A streamline is a line that is tangent to the local velocity vector.
If the flow is steady, the streamline is the path that a particle follows.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Remarks on Continuity
The equation we just derived assumes that both
velocities and densities are uniform across
areas 1 and 2.
In reality, both velocities and densities will vary
across the area
Continuity Equation is extensively used in the design and
operation of wind tunnels and rocket nozzles (we will see
how later).
A stream tube is delimited by 2 streamlines
and does not have to be bounded by
a solid wall.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2007
2009
Compressible Versus Incompressible Flows
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Continuity for Incompressible Flows
•
All fluids are compressible in reality.
•
However, many flows are “incompressible enough” so that the incompressibility
assumption holds.
•
Incompressibility is an excellent model for
Flows of liquids (e.g. water and oil)
Air at low speed (V < 100 m/s or 225 mi/h)
•
Equation of Continuity for Incompressible Flows reduces to
•
So that if A2 < A1 then V2 > V1.
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Lecture 2
A1 V2

A2 V1
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Continuity – Sample Problem 1
A convergent duct was found in the basement of Randolph. The
inlet and exit areas are measured to be Ai = 5m2 and Ae = 2m2.
Assuming we use this duct with an inlet velocity of Vi = 9 mi/h,
find the exit velocity.
First, we need to be consistent with the unit system. Let’s work
in SI units.
Vi = 9 mi/h = 9x1609/3600 m/s  Vi = 4 m/s.
Vi << 100 m/s so the flow is considered incompressible.
A
5
From Incompressible Continuity, Ve  i Vi   4  Ve  10m/s
Ae
2
Therefore, the exit velocity will be 10 m/s.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Continuity – Sample Problem 2
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum Equation
Continuity is a great addition to our toolbox, however it says nothing about
pressure.
Why is pressure important? Let’s look at Newton’s 2nd Law:
Sum of the forces
=
Time rate of change of momentum
F
=
d(mv)/dt
F
=
m dV/dt assuming m = const.
F
=
ma
The pressure is going to translate into force, which by Newton’s 2nd Law results
in change of momentum. Assuming incompressibility (m = const), this will result
in change of velocity (thus impacting performance for example).
To find momentum, simply apply F = ma to an infinitesimally small fluid element
moving along a streamline.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum Equation – Free Body Diagram
v
O
p
F = ma
dx
dy
p
dz
dp
dx
dx
Assume fluid element is moving in the x-direction.
3 types of force act on the element:
• Pressure force (normal to the surface) p
• Shear stress (friction, parallel to the surface) tw
• Gravity r dxdydz g
Ignore gravity (smaller than other forces) and assume inviscid flow (non-viscous
i.e. no friction), balance of the forces on x.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum Equation – Force Balance
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum for Incompressible Flows – Bernoulli’s Equation
•
For incompressible flows, r = const.
•
Integrating Euler’s equation between 2 points along a streamline gives:

1
2
2
V2  V1
2
which can be rewritten as
p2  p1   ρ

1
1
2
2
ρV2  p1  ρV1
2
2
in other words
p2 
p
•
1
ρV 2  const  p0
2
along a streamline
This equation is known as Bernoulli’s Equation.
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Description of Bernoulli’s Equation
1
2
p  ρV  p0
2
Static Pressure
• Pressure felt by an object or person suspended in the fluid and moving with it.
• Can be thought of as internal energy.
Dynamic Pressure
• Pressure due to the fluid motion.
• Can be thought of as kinetic energy.
Total (stagnation) Pressure
• Pressure that would be felt if the fluid was brought isentropically to a stop.
• Can be thought of as total energy.
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
3 New Tools – Continuity, Euler, and Bernoulli’s Equations
•
Continuity Equation
r A V = const
Assumptions: steady flow.
•
Euler’s Equation
dp = - r V dV
Assumptions: steady, inviscid flow.
•
Bernoulli’s Equation
1
p  ρV 2  p0
2
Assumptions: steady, inviscid, incompressible flow along a streamline.
Euler and Bernoulli’s equations are essentially applications of Newton’s
2nd Law to fluid dynamics.
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum Equations - Sample Problem 1
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Momentum Equations - Sample Problem 2
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Practical Applications
By combining Continuity, Euler, and Bernoulli’s equation, one
can obtain the velocity at any point on an aircraft assuming
surrounding conditions are known (either through
measurements or using Standard Atmosphere).
Two major applications for this:
Low-Speed Subsonic Wind Tunnel testing/designing
Flight measurements of velocity
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Low-Speed Subsonic Wind Tunnels (§4.10)
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Wind Tunnel Calculations
•
From Bernoulli, between points 1 and 2:
2
V2 
•
Using Continuity:
V1 
•
•
A2
V2
A1
Combining the two, we get:
V2 
•
2
p1  p2   V12
ρ
2p1  p2 
2
  A2


ρ 1  
A1  



Since the ratio of throat to reservoir area (A2/A1) is fixed for wind tunnel and r is
constant for low-speed (incompressible) flows, the quantity driving the tunnel is
p1-p2.
But how can we determine p1-p2 ???
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Manometer
p1 A  p2 A  wAh, where w  ρf g is the specific weight (weight per unit volum e) of the reference fluid.
 p1  p2  wh since w is constant for the reference fluid, this means that the pressure difference is
directly proportion al to the height of fluid.
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Wind Tunnels – Sample Problem 1
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Wind Tunnels – Sample Problem 1 Solution
•
•
•
•
Height of liquid: h = 10cm = 0.1m
Specific weight of liquid mercury: w = (1.36x104)x9.8 = 1.33x105 N/m2
Actual pressure difference: p1-p2 = w h = 1.33x104 N/m2.
To find V2 from Bernoulli, use
V2 
•
•
2p1  p2 
2
  A2


ρ 1  
A1  



We computed p1-p2, A1/A2 = 15 is given, so we need to find r.
Since we are in a low-speed wind tunnel, flow is incompressible, so r = const, which
means we can compute it at any point in the tunnel. Since p1 and T1 are given, use
Equation of State to find r = r 1:
pp11 1.1
1.1(1.01
(1.0110
1055))
ρρ11


ρρ111.29kg/m
1.29kg/m33
RT
RT11
287
287300
300
•
Combining all the results we get V2 = 144 m/s (slightly over the incompressible velocity
limit, which means compressibility effects should be taken into account).
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Measurement of Airspeed (§4.11)
Bernoulli’s equation provides an easy method for determining the velocity of any
fluid
V
2 p0  p 
ρ
Therefore, we need to know p and p0
ρ
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p
RT
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Total (stagnation) Pressure (p0 ) Measurement
•
The total pressure is easy to measure if the flow direction is known. An openedend tube aligned with the flow direction is enough. This type of tube is called
"Pitot probe”
(named after Henri Pitot who invented it in 1732; see §4.3 for historical
background)
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Static Pressure (P) Measurement
The static pressure is also easy to measure using a tube with a close end and
pressure taps around its circumference.
“Static probe”
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Dynamic Pressure Measurement
Finally, it is possible to measure directly the difference between stagnation and
static pressure by combining the Pitot and static probes into a Pitot-static probe (!).
“Pitot-Static probe”
Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Airspeed Indicator
If the only known
density is at sea level,
“Indicated or Equivalent
Airspeed”
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
True Airspeed
and
Therefore, the relationship between true and indicated
airspeed is:
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
www.aeromech.usyd.edu.au/
aero/instruments/
http://www.tech.purdue.edu/at/courses/
aeml/airframeimages/pitottube.jpg
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
www.aeromech.usyd.edu.au/
aero/instruments/
http://home4.highway.ne.jp/tpark/tp/image/seventh/s-port.jpg
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Measurement of Airspeed – Sample Problem
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Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
Measurement of Airspeed – Sample Problem Solution
From Standard Atmosphere (App. B), at 5000ft, p = 1761 lb/ft2.
Pitot tube measures stagnation pressure so p0 = 1818 lb/ft2.
Density is found from measured temperature and tabulated pressure
r= p/(RT) = 1761/(1716*505)  r = 2.03x10-3 slug/ft3.

7.6% difference

Virginia Tech
Lecture 2
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Virginia Tech
Lecture 2
Fall 2009
1 September 2009
AOE 2104--Aerospace and Ocean Engineering
Fall 2009
For Next Class: Review Chapter 4 and let me
know what questions you have
Thursday: HW 1 due. Stay Tuned for HW 2.
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Lecture 2
1 September 2009