introduction to aerospace engineering · 2016. 11. 7. · v= 1208 m/s m1=4 1 bow shock wave p 2=...
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![Page 1: Introduction to Aerospace Engineering · 2016. 11. 7. · V= 1208 m/s M1=4 1 Bow shock wave P 2= 2347 N/m2 T 2 = 953 K P 3= 1240 N/m2 P 4= 704 N/m2 T= ? V= ? M= ? In 3,4 Boundary](https://reader036.vdocuments.site/reader036/viewer/2022071603/613e6f9069193359046d1e95/html5/thumbnails/1.jpg)
1 Challenge the future
Introduction to Aerospace Engineering
Lecture slides
![Page 2: Introduction to Aerospace Engineering · 2016. 11. 7. · V= 1208 m/s M1=4 1 Bow shock wave P 2= 2347 N/m2 T 2 = 953 K P 3= 1240 N/m2 P 4= 704 N/m2 T= ? V= ? M= ? In 3,4 Boundary](https://reader036.vdocuments.site/reader036/viewer/2022071603/613e6f9069193359046d1e95/html5/thumbnails/2.jpg)
23-10-2011
DelftUniversity ofTechnology
Introduction to Aerospace EngineeringAerodynamics 5 & 6
Prof. H. Bijl ir. N. Timmer
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5.Compressibility continuedAnderson 4.13 – 4.14
Ernst Mach
1838-1916
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Summary of equations
Continuity equation A1V1 = A2V2
V22
21
+ p2 = V21
21
+ p1 ρρBernoulli’s equation
For steady, frictionless, incompressible flow:
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Pitot tube to measure velocity
2 21 1 2 2
1 1
2 2p V p Vρ ρ+ = +
21
1 ρ V
2p− =pt
p2=p1
V2=V1
Pt1=Pt2=Pt
12
V=0
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For steady, isentropic compressible flow (adiabatic and frictionless) :
Continuity equation ρ1A1V1 = ρ2A2V2
γγ
=
ρρ
γ
T2
T1 1- 2
1 = p2
p1Isentropic relations
V 21
+ Tc = V 21
+ Tc 222p
211pEnergy equation
Equation of state P = ρRT
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Example : Re-entry Space Shuttle.
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Re-entry Space Shuttle Orbiter
Just info:
h=30,480 m
V= 1208 m/s
M1=4
1 Bow shock wave
P2= 2347 N/m2
T2 = 953 K
P3= 1240 N/m2
P4= 704 N/m2
T= ?
V= ?
M= ?
In 3,4
Boundary layer
4
3
2
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Speed of sound
moving sound wave with speed a
into a stagnant gas
p
ρT
p+dp
ρ +dρT+dT
≡is equal to
motionless sound wave
p
ρT
p+dp
ρ +dρT+dT
a
static observer observer moving with sound wave
a + daa
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Speed of sound
Apply the continuity equation:
ρ1A1V1 = ρ2A2V2 ⇒ ρA1a = (ρ + dρ) A2 (a+da)
1-dimensional flow⇒A1 = A2 = A = constant
Thus: ρa = (ρ + dρ)(a + da) ⇒ ρa = ρa + adρ + ρda + dρda
small, ignoreρρ
dda
- =a ⇒
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ρddp
isentropic =a
Speed of sound
Now apply momentum equation (Euler equation) ⇒
dp = -ρVdV ⇒ 1dp = - ada da = - dp
aρ
ρ⇒
We obtain ⇒1 dp
a = + a d
ρρ ρ ⇒
ρddp
= a2 Going through the sound wave there is
no heat addition, friction is negligible ⇒
daa = -
dρ
ρSubstituting this into
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Speed of sound
Now we may apply the isentropic relations.
We want to rewritedp
dρp
a = γρ
p = RT
ρ⇒
RT =a γ
Use the equation of state:
Speed of sound in a perfect gas depends only on T !
With isentropic relations we find
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a = RTγ
VM
a=
In honor of Ernst Mach the name “Mach number” was introduced in 1929 by Jacob Ackeret
Mach Ackeret
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Equations for a perfect gas
• We defined: h=e + pv
• We found : h=cpT and e= cvT
• With the eq. of state: pv=RT
• we find: cpT=cvT+ RT
• hence: cp=cv+ R => cp-cv= R
p
v
c=
cγ
1p
Rc =
γγ −
We defined:
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Isentropic flow relations, second form
Energy equation: 2p
1T + = constant c V
2
2 21 op 1 p o
1 1 + = + c V c VT T
2 2
Assume index 0 = stagnation point ⇒ Vo = 0 ⇒
2o 12
1 op 1 p1 1p
1 VT + = = 1 + c V cT T2 2cT T
⇒
Substitute: 1p
RC
γγ
=−
(from p vC C R− = )
2
o 1
1 1
-1 VT = 1 + 2T RT
γγ
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Isentropic flow relations, second form
with RT21a γ=
Bring flow isentropically to rest ⇒ Isentropic relation can be used ⇒
γγ
ρρ γ
T1
To 1- = 1
o = p1
po
γ γγ
M21
2
1- + 1 1- =
p1
po
γ γρρ
M21
21-
+ 1 1-1
= 1
o
M21
21-
+ 1 = a21
V21
21-
+ 1 = T1
To γγ
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Second form of isentropic relations
o 21
1
γ-1T = 1 + M2T
γ γρρ
M 21
21-
+ 1 1-1
= 1
o
γ
γ-12o1
1
p γ-1 = 1 + M
2p
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Compressibility
M0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0.6
0.7
0.8
0.9
1 1
2 1-
1
0 M
2
1- + 1 =
−
γ
γρρ
0
ρρ
For M < 0.3 the change in density is less than 5 %
Thus : for M < 0.3 the flow can be treated as incompressible
Derived from energy equation
and isentropic relations
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Supersonic wind tunnel and rocket engine
Rocket motor must be LIGHT
Thus the nozzle is short
Exhaust of Saturn V
that powered the Apollo space missions
M<1 M>1
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Supersonic Wind Tunnel
M=1
Throat
M<1
Supersonic flow
M >1 Pe
Te
p0
T0
V≈0
ReservoirExit
Subsonic flow
For both the wind tunnel and a rocket motor we can derive
an area velocity relation using the continuity and the Euler equation
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Supersonic tunnel and rocket engine
constVAρ = ln ln ln ln( )A V constρ + + =
0d dA dV
A V
ρρ
+ + = dp VdVρ= −
0d VdV dA dV
dp A V
ρ− + + = 2
1 1ddpdp a
d
ρ
ρ= =
20
VdV dA dV
a A V
− + + =
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Area-Velocity Relation
2dA dV = ( -1) M
A V
Conclusions: Case A. Subsonic flow if dV > 0 then dA < 0 and vice versa. Case B. Supersonic flow if dV > 0 then dA > 0 and vice versa. Case C. If the flow is sonic (M=1) then ....
Supersonic wind tunnel: area velocity relation
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Supersonic windtunnel: M=1 in the throat
• 2
dV 1 dA 1 dA = = !
V -1 A 0 AM
• At first glance we see dV
= V
∞ .
But this is not possible on physical basis.
• Then we must have (for finite dV
V): dA
= 0A
⇒dV 0
= = FI NITE!V 0
• If dA
= 0A ⇒
In the THROAT : M = 1 !
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Nozzle flow
o 21
1
γ-1T = 1 + M2T
γ γρρ
M21
21-
+ 1 1-1
= 1
o
flow
ρ/ρ0
1
1 T/T0
1
1
M
P/P0
γ
γ-12o1
1
p γ-1 = 1 + M
2p
X ->
* * for M=1 conditions (T*,p*…)
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Example 4.9: A supersonic wind tunnel is considered.
The air temperature and pressure in the reservoir of the tunnel are:
T0 = 1000 K and p0 = 10 atm.
The static temperatures at the throat and exit are:
T* = 833 K and Te = 300 K.
The mass flow through the tunnel is 0.5 kg/s.
For air, cp = 1008 J/(kg K). Calculate:
a) Velocity at the throat V*
b) Velocity at the exit Ve
c) Area of the throat A*
d) Area of the exit Ae
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6.Viscous flowsAnderson 4.15 - 4.16
Osborne Reynolds Ludwig Prandtl
1842-1912 1874-1953
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Subjects lecture 6
• Viscous flows
• Laminar boundary layers
- calculation of boundary layer thickness
- calculation of skin friction drag
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Viscous flow
Inviscid flow (No friction)NO DRAG
Viscous flow (friction)FINITE DRAG
Up till now we have only dealt with frictionless flow.
What is the effect of friction ? ….
D
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Viscous flow
In real life the flow at the surface adheres to the surface
because of friction between the gas and the solid material:
�Right at the surface the velocity is zero
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Boundary layer
(exaggerated)
Friction forceBoundary layer
V
�In the vicinity of the surface there is a thin region of
retarded flow: the boundary layer
�The pressure through the boundary layer in a direction
perpendicular to the surface is constant
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Viscous flow
Inside the boundary layer
Bernoulli’s law is not valid!!!!!!
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Viscous flows
velocity profile
boundary layerthickness, δ
Shear stress can be written as :
µ = absolute viscosity coefficient or viscosityAir at standard sea level : µ=1.789*10-5 kg/ms)
0yw dy
dU
=
µ=τ
shear stress, τw
⇒ skin friction drag
Boundary layer
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Viscous flow
dy
dU = µτ
Fluids for which the shearing stress is linearly related to the rate of shearing strain
are called: NEWTONIAN FLUIDS.
(Most common fluids (liquids & gases) like are are NEWTONIAN!).
Often viscosity appears as ρµν = = “KINEMATIC VISCOSITY”
viscosity or” dynamic viscosity”
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Reynolds number
Re =⋅V L
ν
Characteristic velocity Characteristic length
(kinematic) viscosity
Sail planes : Re ≈ 1•106, L=wing chordPassenger jets : Re ≈ 50•106 IdemLower leg athlete : Re ≈ 1•105 L=Diameter leg
Osborne Reynolds
1842-1912
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Viscous flows, some definitions
Reynolds number :
Laminar flow : streamlines are smooth and regular and a fluid
element moves smoothly along a streamline
Turbulent flow : streamlines break up and a fluid element moves
in a random irregular way
,µ
xV ρRex
∞
∞∞=
x
δV∞
dimensionless, and varies linearly with x
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Laminar boundary layer, boundary
layer thickness
Consider flat plate flow. What is boundary layer thickness δand skin friction drag Df at location x?
x
δV∞
From laminar boundary layer theory :xRe
x2.5=δ
Thus δ is proportional to : √x (parabolically)
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Total force = total pressure force + total friction force
Total friction force on element dx is: dx (x)w = 1dx(x)w τ⋅⋅τ
Total skin friction drag is: dxw L
o = D f τ∫
τwdx
L
x
1 (unit width)width)
Laminar boundary layer, skin friction
drag
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Laminar boundary layer, skin friction
drag
For the skin friction coefficient we find from laminar boundary layer theory :
Re
0.664 =
q =
V 2
1 = c
x
w
2
wf x
∞∞∞
τ
ρ
τ
Thus xfC and wτ decrease as x .
The skin friction at the beginning of the plate is larger than near the trailing edge.
To calculate the total aerodynamic force we must integrate!
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Laminar boundary layer, skin friction
drag
Re
dx q0.664 = dx q . c = D
x
L
o
f
L
o
f x ∫∫ ∞∞ xdx
L
o
/V
q 0.664 = ∫ν∞
∞
x 2 = x 1/2- = x
dx∫∫
ν∞∞∞
ν∞∞∞
L/V
Lq1.328 = L2
/V
q 0.664 = Df
Sq
Df = Cf∞
Define total skin friction drag coefficient as
1 LReL
1.328.L =
S
L
ReL
1.328 = Cf ⋅ ReL
1.328 = Cf
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ReL
1.328 = Cf
Only for low speed incompressible flow(and reasonably accurate for high speed subsonic flow).
LRe = Reynolds nr. based on length L.
The skin friction coefficient for a
laminar boundary layer
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