MAE 3241: AERODYNAMICS AND
FLIGHT MECHANICS
Overview of Compressible Flows:
Critical Mach Number and Wing Sweep
April 25, 2011
Mechanical and Aerospace Engineering Department
Florida Institute of Technology
D. R. Kirk
EXAMPLES: INFLUENCE OF COMPRESSIBILITY
M∞ ~ 0.85
M∞ < 1
M∞ > 1
WHEN IS FLOW COMPRESSIBLE?
14
Stagnation to Static Density Ratio
Cp/Cv=1.4
12
0    1 2 
 1 
M 
 
2

10
1
 1
8
6
4
2
0
0
0.5
1
1.5
Mach Number
2
2.5
3
WHEN IS FLOW COMPRESSIBLE?
14
Cp/Cv=1.4
1.2
12
Stagnation to Static Density Ratio
Stagnation to Static Density Ratio
Cp/Cv=1.4
10
8
6
4
0    1 2 
 1 
M 
 
2

1.15
1.1
1
 1
1.05
1
0.95
0
0.1
0.2
0.3
0.4
0.5
Mach Number
2
0
0
0.5
1
1.5
Mach Number
2
2.5
3
EXAMPLES: COMPRESSIBLE INTERNAL FLOW

Aexit
A
 e*  77
Athroat A
EXAMPLE: H2 VARIABLE SPECIFIC HEAT, CP
COMPRESSIBILITY SENSITIVITY WITH 
30
Stagnation to Static Density Ratio
Cp/Cv=1.4
Cp/Cv=1.2
25
20
15
10
5
0
0
0.5
1
1.5
Mach Number
2
2.5
3
PRESSURE COEFFICIENT, CP
• Use non-dimensional description instead of plotting actual values of pressure
• Pressure distribution in aerodynamic literature often given as Cp
• So why do we care?
– Distribution of Cp leads to value of cl
– Easy to get pressure data in wind tunnel
– Shows effect of M∞ on cl
p  p
p  p
Cp 

1
q
 V2
2
EXAMPLE: CP CALCULATION
See §4.10
Cp at a point on an airfoil of fixed shape
and fixed angle of attack
COMPRESSIBILITY CORRECTION:
EFFECT OF M∞ ON CP
C p,0
p  p
p  p


1
q
2
 V
2
For M∞ < 0.3,  ~ const
Cp = Cp,0 = 0.5 = const
Flight Mach Number, M∞
COMPRESSIBILITY CORRECTION:
EFFECT OF M∞ ON CP
Cp 
C p,0
1  M 2

0.5
1  M 2
For M∞ < 0.3,  ~ const
Cp = Cp,0 = 0.5 = const
Effect of compressibility
(M∞ > 0.3) is to increase
absolute magnitude of Cp as
M∞ increases
Called: Prandtl-Glauert Rule
M∞
Prandtl-Glauert rule applies for 0.3 < M∞ < 0.7
EXAMPLE: SUPERSONIC WAVE DRAG
F-104 Starfighter
CRITICAL MACH NUMBER, MCR
• As air expands around top surface near leading edge, velocity and M will increase
• Local M > M∞
Flow over airfoil may have
sonic regions even though
freestream M∞ < 1
INCREASED DRAG!
CRITICAL FLOW AND SHOCK WAVES
M CR  M DragDivergence  1.0
MCR
• Sharp increase in cd is combined effect of shock waves and flow separation
• Freestream Mach number at which cd begins to increase rapidly called DragDivergence Mach number
CRITICAL FLOW AND SHOCK WAVES
‘bubble’ of supersonic flow
CRITICAL FLOW AND SHOCK WAVES
MCR
EXAMPLE: IMPACT ON AIRFOIL / WING DRAG
D  D friction  D pressure  Dwave
cd  cd , f  cd , p  cd , w
Profile Drag
Profile Drag coefficient
relatively constant with
M∞ at subsonic speeds
Only at transonic and
supersonic speeds
Dwave= 0 for subsonic speeds
below Mdrag-divergence
AIRFOIL THICKNESS SUMMARY
• Which creates most lift?
– Thicker airfoil
• Which has higher critical Mach number?
– Thinner airfoil
• Which is better?
– Application dependent!
Note: thickness is relative
to chord in all cases
Ex. NACA 0012 → 12 %
CAN WE PREDICT MCR?
A
C p, A
2

M 2
  1 2 
1
M 
p0 
pA
2



p p
 1   1 M 2 
A 
p0 
2

pA
• Pressure coefficient defined in terms of
Mach number (instead of velocity)
PROVE THIS FOR CONCEPT QUIZ
 pA 

 1
 p 

 1
• In an isentropic flow total pressure, p0, is
constant
• May be related to freestream pressure, p∞,
and static pressure at A, pA
CAN WE PREDICT MCR?
C p, A



 1


1

2 
 1

M 
2 
2



 1
2

M   1    1 M 2 


A


2



C P ,CR 
2
2
M CR

2
 1    1 M CR
2

 1

1



2








 1


 1



• Combined result
– Relates local value of CP to local
Mach number
– Can think of this as compressible flow
version of Bernoulli’s equation
• Set MA = 1 (onset of supersonic flow)
• Relates CP,CR to MCR
HOW DO WE USE THIS?
1. Plot curve of CP,CR vs. M∞
2. Obtain incompressible value of CP at minimum pressure point on given airfoil
3. Use any compressibility correction (such as P-G) and plot CP vs. M∞
– Intersection of these two curves represents point corresponding to sonic flow
at minimum pressure location on airfoil
– Value of M∞ at this intersection is MCR
1
3
2
CP ,CR 
2
2
M CR

2
 1    1 M CR
2

 1

1



2

Cp 







 1


 1



C p,0
1  M 2
IMPLICATIONS: AIRFOIL THICKNESS
Note: thickness is relative
to chord in all cases
Ex. NACA 0012 → 12 %
• Thick airfoils have a lower critical Mach number than thin airfoils
• Desirable to have MCR as high as possible
• Implication for design → high speed wings usually design with thin airfoils
– Supercritical airfoil is somewhat thicker
THICKNESS-TO-CHORD RATIO TRENDS
Thickness to chord ratio, %
A-10
Root: NACA 6716
TIP: NACA 6713
F-15
Root: NACA 64A(.055)5.9
TIP: NACA 64A203
Flight Mach Number, M∞
ROOT TO TIP AIRFOIL THICKNESS TRENDS
Boeing 737
Root
Mid-Span
Tip
http://www.nasg.com/afdb/list-airfoil-e.phtml
SWEPT WINGS
• All modern high-speed aircraft have swept wings: WHY?
WHY WING SWEEP?
V∞
V∞
Wing sees component of flow normal to leading edge
WHY WING SWEEP?
V∞
V∞,n
W
W
V∞
V∞,n < V∞
Wing sees component of flow normal to leading edge
SWEPT WINGS: SUBSONIC FLIGHT
• Recall MCR
• If M∞ > MCR large
increase in drag
• Wing sees component
of flow normal to
leading edge
• Can increase M∞
• By sweeping wings of
subsonic aircraft, drag
divergence is delayed
to higher Mach
numbers
SWEPT WINGS: SUBSONIC FLIGHT
• Alternate Explanation:
– Airfoil has same thickness but longer
effective chord
– Effective airfoil section is thinner
– Making airfoil thinner increases
critical Mach number
• Sweeping wing usually reduces lift for
subsonic flight
SWEPT WINGS: SUPERSONIC FLIGHT
 1 

  sin 
 M 
1
• If leading edge of swept wing is outside Mach cone, component of Mach
number normal to leading edge is supersonic → Large Wave Drag
• If leading edge of swept wing is inside Mach cone, component of Mach number
normal to leading edge is subsonic → Reduced Wave Drag
• For supersonic flight, swept wings reduce wave drag
WING SWEEP COMPARISON
F-100D
English Lightning
SWEPT WINGS: SUPERSONIC FLIGHT
M∞ < 1
SU-27

M∞ > 1
 ~ 26º
(M=1.2) ~ 56º
(M=2.2) ~ 27º
WING SWEEP DISADVANTAGE
• At M ~ 0.6, severely
reduced L/D
• Benefit of this design is
at M > 1, to sweep
wings inside Mach
cone
• Wing sweep beneficial in that it increases drag-divergences Mach number
• Increasing wing sweep reduces the lift coefficient
TRANSONIC AREA RULE
• Drag created related to change in cross-sectional area of vehicle from nose to tail
• Shape itself is not as critical in creation of drag, but rate of change in shape
– Wave drag related to 2nd derivative of volume distribution of vehicle
EXAMPLE: YF-102A vs. F-102A
EXAMPLE: YF-102A vs. F-102A
CURRENT EXAMPLES
• No longer as relevant today – more
powerful engines
• F-5 Fighter
• Partial upper deck on 747 tapers off
cross-sectional area of fuselage,
smoothing transition in total crosssectional area as wing starts adding in
• Not as effective as true ‘waisting’ but
does yield some benefit.
• Full double-decker does not glean this
wave drag benefit (no different than
any single-deck airliner with a truly
constant cross-section through entire
cabin area)
EXAMPLE OF SUPERSONIC AIRFOILS
http://odin.prohosting.com/~evgenik1/wing.htm
SUPERSONIC AIRFOIL MODELS
• Supersonic airfoil modeled as a
flat plate
• Combination of oblique shock
waves and expansion fans acting
at leading and trailing edges
– R’=(p3-p2)c
– L’=(p3-p2)c(cosa
– D’=(p3-p2)c(sina
• Supersonic airfoil modeled as
double diamond
• Combination of oblique shock
waves and expansion fans acting
at leading and trailing edge, and
at turning corner
– D’=(p2-p3)t
APPROXIMATE RELATIONS FOR LIFT AND DRAG
COEFFICIENTS
cl 
cd , w 
4a
M 2  1
4a 2
M 2  1
http://www.hasdeu.bz.edu.ro/softuri/fizica/mariana/Mecanica/Supersonic/home.htm
CASE 1: a=0°
Shock waves
Expansion
CASE 1: a=0°
CASE 2: a=4°
Aerodynamic Force Vector
Note large L/D=5.57 at a=4°
CASE 3: a=8°
CASE 5: a=20°
At around a=30°, a detached shock begins
to form before bottom leading edge
CASE 6: a=30°
DESIGN OF ASYMMETRIC AIRFOILS
QUESTION 9.14
• Consider a diamond-wedge airfoil as shown in Figure 9.36, with half angle =10°
• Airfoil is at an angle of attack a=15° in a Mach 3 flow.
• Calculate the lift and wave-drag coefficients for the airfoil.
Compare with your solution
EXAMPLE: MEASUREMENT OF AIRSPEED
• Pitot tubes are used on aircraft as speedometers (point measurement)
Subsonic
M < 0.3
Subsonic
M > 0.3
M < 0.3 and M > 0.3: Flows are qualitatively
similar but quantitatively different
Supersonic
M>1
M < 1 and M > 1: Flows are
qualitatively and quantitatively different
MEASUREMENT OF AIRSPEED:
INCOMPRESSIBLE FLOW (M < 0.3)
1
2
p  V1  p0
2
Static
pressure
V1 
Dynamic
pressure
Total
pressure
2 p 0  p 

Incompressible Flow
• May apply Bernoulli Equation with
relatively small error since compressibility
effects may be neglected
• To find velocity all that is needed is pressure
sensed by Pitot tube (total or stagnation
pressure) and static pressure
Comment: What is value of ?
• If  is measured in actual air around airplane
(difficult to do)
– V is called true airspeed, Vtrue
• Practically easier to use value at standard
seal-level conditions, s
– V is called equivalent airspeed, Ve
MEASUREMENT OF AIRSPEED:
SUBSONIC COMRESSIBLE FLOW (0.3 < M < 1.0)
• If M > 0.3, flow is compressible (density changes are important)
• Need to introduce energy equation and isentropic relations
1 2
c pT1  V1  c pT0
2
2
T0
V1
 1
T1
2c pT1
T0
 1 2
 1
M1
T1
2
p0    1 2 
 1 
M1 
p1 
2

0    1 2 
 1 
M1 
1 
2


 1
1
 1
MEASUREMENT OF AIRSPEED:
SUBSONIC COMRESSIBLE FLOW (0.3 < M < 1.0)
• How do we use these results to measure airspeed?
2  p0 
 
M 
  1  p1 

2
1
 1 
• p0 and p1 give flight Mach number
• Instrument called Mach meter

 1

• M1 = V1/a1
• V1 is actual flight speed
 1 


2 a  p0 
2
 
 1
V1 
  1  p1 


2
1
 1 


2a  p0  p1 
2

 1
 1
V1 
  1  p1



2
1
2
Vcal
2as2  p0  p1 

 1

  1  ps


 1 

 1

• Actual flight speed using pressure
difference
• What are T1 and a1?
• Again use sea-level conditions Ts,
as, ps (a1 = (RT)½ = 340.3 m/s)
• V is called Calibrated Velocity, Vcal
MEASUREMENT OF AIRSPEED:
SUPERSONIC FLOW (M > 1)
p02  p02  p2 
 
 
p1  p2  p1 
p02    1 2 
 1 
M2 
p2 
2

 1 2
1
M1
2
M 22 
 1
M 12 
2

 1
p2
2
 1
M 12  1
p1
 1


2
p02    1 M 12 


p1  4M 12  2  1
  1
1    2M 12
 1
Rayleigh Pitot Tube Formula
EXAMPLE: SUBSONIC AND SUPERSONIC FLIGHT
•
•
Flight at four different speeds, pitot measures p0 = 1.05, 1.2, 3 and 10 atm
What is flight speed if flying in 1 atm static pressure and Tambient = 288 K (a = 340 m/s)?
•
Determine which measurements are in subsonic or supersonic flow
– p0/p = 1.893 is boundary between subsonic and sonic flows
•
1.05 atm → p0/p = 1.05 → subsonic
– Use compressible flow form, M = 0.265, V ~ 90 m/s ~ 200 MPH
– Could use Bernoulli which will provide small error (~ 1%) and give V directly
– Compressible form requires knowledge of speed of sound (temperature)
– Apply Bernoulli safely? p0/p < 1.065
1.2 atm → p0/p = 1.2 → subsonic
– M = 0.52, V ~ 177 m/s ~ 396 MPH
– Use of compressible subsonic form justified (Bernoulli ~ 3% error)
3 atm → p02/p1 = 3 → supersonic
– M1 = 1.39, V ~ 473 m/s ~ 1057 MPH (Bernoulli ~ 22% error)
10 atm → p02/p1 = 10 → supersonic
– M1 = 2.73, V ~ 928 m/s ~ 2076 MPH (MCO → LAX in 1 hour 30 minutes)
•
•
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