Aerothermodynamics of high speed ows AERO 0033{1

Aerothermodynamics of high speed flowsAERO 0033–1

Lecture 6: Prandtl-Glauert theory for thin airfoils (subsonicflow), slender body theory, 1D unsteady flow in straight duct

Thierry Magin, Greg Dimitriadis, and Adrien [email protected]

Aeronautics and Aerospace Departmentvon Karman Institute for Fluid Dynamics

Aerospace and Mechanical Engineering DepartmentFaculty of Applied Sciences, University of Liege

Wednesday 9am – 12:15pmFebruary – May 2021

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Outline

Prandtl-Glauert theory for thin airfoils (subsonic flow)

Exercise: 2D overexpanded jet from nozzle

Slender body theory

1D unsteady flow in straight duct

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Outline



Slender body theory


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2D Linearized velocity potential eq.I Compressible regime (M∞ 6∼ 1 and M∞ 6 1):

(1−M2∞)∂x1 u′1 + ∂x2 u′2 = 0

I Irrotational flow

ω = ∇× v = 0 ⇒ ∂x2 u′1 − ∂x1 u′2 = 0

⇒ v = ∇Φ

with Φ: potential functionI System of partial differential eqs.

∂x1Q + A ∂x2Q = 0

is elliptic for subsonic flow: characteristic theory notapplicable!

⇒ How can we use incompressible results (theory andexperiment) and modify them to take compressibility intoaccount?

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Linearized subsonic flowI Linearized 2D potential eq. for subsonic compressible regime

in non-dimensional form (0 ≤ M∞ < 1 with M∞ 6∼ 1)

β2∂x21φ+ ∂x2

2φ = 0

I Parameter β =√

1−M2∞ (β=1 for incompressible case)

I With x1 = x1/c and x2 = x2/c (c = chord of profile)I Velocity perturbation potential:

u′1 = ∂x1φ/u∞ and u′2 = ∂x2φ/u∞

I Let consider the shape of a family of airfoils in the coordinatesystem attached to the airfoil

X2

c= τ f (

X1

c)

with relative thickness to camber τ 4 / 47

Similarity problem

I Linearized boundary condition: u′2 = dX2dX1− α

(just as for supersonic flow)

with α: angle of attackI For the profile family X2

c = τ f (X1

c )

u′2 = ∂x2φ/u∞ = τ dfdX1− α, for x1 ∈ [0, c]

I Suppose that the flow over an airfoil b with τb known (eithernumerically or experimentally) in free stream Mb at αb

β2b∂x2

1φb + ∂x2

2φb = 0 with ∂x2φb = u∞b(τb

dfdX1− αb)

for x1 ∈ [0, c]

⇒ Is it possible to infer the flow over a geometrical similar airfoila with τa in a free stream Ma at αa?1

1This includes the particular incompressible case (Mb = 0) where τa = τb,αa = αb

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Transformation of variables

I Coordinate transformation:

ξ = x1

η = Ax2Φ = B u∞b

u∞aφa

I Linearized 2D potential eq.

β2a∂x2

1φa + ∂x2

2φa = 0 → β2

a

A2∂ξ2 Φ + ∂η2 Φ = 0

I Linearized B.C.∂x2φa = u∞a(τa

dfdX1− αa)→ ∂ηΦ = u∞b(B

A τadf

dX1− B

Aαa)for x1 ∈ [0, c]

I Transformed system a identical to b

β2b∂x2

1φb + ∂x2

2φb = 0 with ∂x2φb = u∞b(τb

dfdX1− αb)

for x1 ∈ [0, c]

ifβ2a

A2= β2

b (eq. 1),B

Aτa = τb (eq. 2),

B

Aαa = αb (eq. 3)

I Dividing (eq. 2)/(eq. 3) → τa/αa = τb/αbI Geometrical similarity for angle of attack and thickness /

camber ratio : 2 identical airfoils must be compared at thesame incidence (at 6= M)

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Prandtl-Glauert formulaI Providing that this geometrical similarity is satisfied

(eq. 1) → A = βa

βb, (eq. 2) → B = βaτb

βbτaI Keeping only linear terms, the pressure coefficient reads:

Cpa = −2u′1a = −2∂x1φa/u∞a =−2

Bu∞b∂ξΦ =

1

BCpb =

βbτaβaτb

Cpb

I Considering that lift & moment coefficients are in same ratioas pressure coefficients (no drag in subsonic inviscid flow!)

Prandtl-Glauert formula (Ma = M , Mb = 0, τa = τb, αa = αb)

Cp(M) =Cp,inc√1−M2

, Cl(M) =Cl ,inc√1−M2

, Cm(M) =Cm,inc√1−M2

I Compressibility has favorable effect on lift (for given airfoiland angle of attack)

I Since viscous drag depends little on M within subsonic range

→ L / D ratio increases with M (interest of flight at high M)

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Variation of the lift curve slope with Mach numberI The positive effect of Mach number increase ceases when the

transonic regime is entered

I The Prandtl-Glauert formula tends to underestimate slightlythe positive effects of compressibility

I Using the hodograph method, Karman and Tsien proposed animproved formula (not derived here) for the subsonic regime

Cp(M) =Cp,inc

√1−M2 + M2

1+√

1−M2

Cp,inc

2

Summary

I Incompressible flow:Cl = 2π(α− α0)

I Subsonic flow: Cl = 2π(α−α0)√1−M2

∞

I Supersonic flow: Cl = 4α√M2∞−1

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Outline



Slender body theory


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Consider a uniform, parallel, supersonic, two-dimensional jet of air (ratio of specificheats γ = 1.4) which leaves a symmetric nozzle at section 1 with a pressure p1 lessthan that of the exhaust region 0 into which the nozzle discharges (overexpanded jet).Two oblique compression waves AE and CE are generated and the jet boundaries areturned through an angle δ = 10. The flow is assumed to be steady and inviscid.

Jet leaving overexpanded nozzle

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1. Based on the exact oblique shock theory, compute the pressure p0 in the exhaustregion. The Mach number and pressure at section 1 are M1=2 and p1 = 105 Pa.

2. Using as approximation the non-linear isentropic characteristic theory, computethe pressure p0 in the exhaust region. Compare with the value of p0 obtainedpreviously and comment on the validity of this approximation.

3. What is the nature of the waves EB and ED?

4. Compute the pressure p3 in region 3 based on the appropriate theory.

5. Describe qualitatively the nature of the wave between regions 3 and 4.

6. The ambient pressure in region 0 is then decreased to a value slightly lower thanthat of region 1 such that the jet becomes underexpanded. Describequalitatively the nature of the waves AE and CE and EB and ED and commenton the jet deflection following this scenario.

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Solution1. Exact oblique shock theory

The pressure of the jet boundaries must be equal to that of the exhaust region(slip line). Oblique compression waves AE and CD are generated and the jetboundaries are turned through an angle δ such that p2 is equal to the exhaustpressure p0. The interaction of shock and the jet boundary is equivalent to theflow over an inviscid wall.

Notice the jet boundaries of an overexpanded jet are converging.I Deflection angle θ = 10

I Shock-wave angle: β = 39.313931

I Normal Mach: Mn1 = 1.267138I Tangential Mach: Mt1 = 1.547372I Static pressure: p2/p1 = 1.706578⇒ p0 = p2 = 1.706578× 105 Pa

I Static temperature: T2T1

=[1 + 2γ

γ+1(M2

n1 − 1)] [

2+(γ−1)M2n1

(γ+1)M2n1

]= 1.170151

I Normal Mach: M2n2 =

1+ γ−12

M2n1

γM2n1−

γ−12

= 0.803190

I Tangentiel Mach: Mt2 = Mt1

√T1T2

= 1.430453

I Mach number: M2 =√

M2n2 + M2

t2 = 1.640522

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2. Linearized oblique shock theory

On s− emanating from the uniform incoming flow region 1

ν(M2) + θ2 = ν(M1)

ν(M2) = 16.379760 ⇒ M2 = 1.651412

p2

p1=

(1 + γ−1

2M2

1

1 + γ−12

M22

) γγ−1

= 1.705213⇒ p0 = p2 = 1.705213× 105 Pa

3. Because of symmetry conditions, the interaction of the oblique shock waves AEand CE is equivalent to the reflection of a shock from a plane wall. Thestrength of waves EB and ED are determined by the condition that the streamin region 3 has the same direction as in regions 1 and 0. Using an equivalentflow over an inviscid wall, waves EB and ED are found to be shock waves

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4. Exact oblique shock theoryI Deflection angle θ = 10

I Shock-wave angle: β = 49.384050

I Normal Mach: Mn2 = 1.245304I Static pressure: p3/p2 = 1.642579⇒ p3 = 2.803189× 105 Pa

5. The pressure field in region 3 is greater than the exhaust pressure p0. In orderfor the jet boundary to remain at constant pressure, shock waves EB and EDmust be reflected as centred Prandtl-Meyer rarefaction waves where the formerstrike the jet boundaries

6. When p1 < p0, the jet is underexpanded. To reach ambient pressure, theexhaust flow expands via Prandtl-Meyer expansion waves. This expansion occursby the gases turning away from the centerline: the jet boundaries of anunderexpanded jet are diverging. These expansion waves are directed towardsthe centerline of the nozzle. The flow passing through the Prandtl-Meyerexpansion waves turn away from the centerline. The Prandtl-Meyer expansionwaves in turn reflect from the center plane towards the jet boundary (contactdiscontinuity). The gas flow passing through the reflected Prandtl-Meyer wavesis now turned back parallel to the centerline but undergoes a further reductionof pressure. The reflected Prandtl-Meyer waves now meet the contactdiscontinuity and reflect from the contact discontinuity towards the centerline asPrandtl-Meyer compression waves. This allows the flow to pass through thePrandtl-Meyer compression waves and increase its pressure to ambient pressure,but passage through the compression waves turns the flow back towards thecenterline. Notice that when regular reflection is impossible and the Mach typeof reflection is observed. The rest of the process is the same as the processexplained above for the overexpanded nozzle.

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Underexpanded plasma jet in the VKI Plasmatron facility(inlet air mass flow: 4.5 g/s, generator power: 98 kW, chamber pressure: 600 Pa)

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Outline



Slender body theory


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Slender body theory (body of revolution at zero incidence)I Let us consider a body of revolution at zero incidence

I u1 = u∞ + u′1

I u2 = u′2

I u3 = u′3

I The x axis of the coordinate system coincides with the bodyaxis and is aligned with the flow velocity u∞

I The flow is axisymmetric (∂θ = 0)

I In the linear compressible regime, the small disturbanceequation in a cylindrical coordinate system reads

(1−M2∞)

∂2φ

∂x2+∂2φ

∂r 2+

1

r

∂φ

∂r= 0

I The velocity is related to the potential function asI ux = u∞ + u′x with u′x = ∂xφI ur = u′r with u′r = ∂rφ

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Solution for subsonic flow (M∞ < 1, M∞ 6∼ 1)I Laplace eq. for axisymmetric problems (β2 = 1−M2

∞, ρ = βr)∂2φ

∂x2+∂2φ

∂ρ2+

1

ρ

∂φ

∂ρ= 0

I Potential at (x , ρ) due to a source located at position ξI φ = −q

4π[(x−ξ)2+ρ2]1/2

I At distance η fromsource, radial velocity inthe η-direction:uη(η) = q

4πη2

I Source flow rate q = 4π∫ η

0uη(η′)η

′2 dη′

([q] = [u∞]L2 = L3/T = m3/s)I Equipotential: ellipsoid of revolution (x − ξ)2 + β2(y2 + z2) = C

I Superposition principle, the perturbation potential reads

φ(x , r) = −1

4π

∫ l

0

σ(ξ) dξ

[(x − ξ)2 + β2r2]1/2

I Continuous distribution of 3D sources on the axis, betweenx = 0 (leading edge) and x = l (base)

I Source density distribution σ(volume flow/unit length: [σ] = [q]/L)

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Determination of the source density distributionI σ determined from condition: flow tangent to the body of

radius R(x)dR

dx=

u′ru∞ + u′x

∣∣∣∣r=R

∼u′ru∞

∣∣∣∣r=R

I The radial perturbation of velocity is

u′r = ∂rφ = β2

4π

∫ l0

rσ(ξ) dξ

[(x−ξ)2+β2r2]3/2

I This distribution is singular in the vicinityof the axis (r → 0)

I The function r/[(x − ξ)2 + β2r 2]3/2 has apeak O(1/r 2) around ξ = x

I Assume that σ(ξ) ≈ σ(x) around ξ = x

⇒ Using the CVs: ξ = x + βrt and t = tan θ

u′r =β2r

4πσ(x)

∫ l

0

dξ

[(x − ξ)2 + β2r2]3/2=σ(x)

4πr

∫ (l−x)/(βr)

−x/(βr)

dt

(1 + t2)3/2

≈σ(x)

4πr

∫ ∞−∞

dt

(1 + t2)3/2=σ(x)

4πr

∫ π/2

−π/2cos θ dθ =

σ(x)

2πr

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Potential function (M∞ < 1, M∞ 6∼ 1)I The BC reads: dR

dx = u′ru∞

∣∣∣r=R

= σ(x)2πRu∞

⇒ σ(x)u∞

= S ′(x), with

the cross-sectional area distribution S(x) = πR2(x)

I By integrating over r the expression ∂rφ = u′r = u∞S ′(x)2πr , one

getsφ(x , r)

u∞=

S ′(x)

2πln r + g(x)

I The function g(x) in the vicinity of the axis (r → 0) can bedetermined directly from the relation

φ(x , r)

u∞= −

1

4π

∫ l

0

S ′(ξ) dξ

[(x − ξ)2 + β2r2]1/2

⇒

g(x) =S ′(x)

2πlnβ

2−

1

4π

∫ x

0S ′′(ξ) ln(x − ξ) dξ +

1

4π

∫ l

xS ′′(ξ) ln(ξ − x) dξ

provided that S ′(0) = 0 (pointed nose) and S ′(l) = 0 (pointedtip or cylindrical section)

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ProofI Using the CV: ξ − x = βry (− for ξ < x , + for ξ > x)∫

dξ

[(x − ξ)2 + β2r2]1/2= ±

∫dy

(1 + y2)1/2

I Using the CV: y = sinh t = et−e−t

2

(et)2 − 2y et − 1 = 0 → t = arsinh y = ln(y +√

1 + y2)∫dy

(1 + y2)1/2= ln[y + (1 + y2)1/2] + C = ln

|ξ − x |βr

+ [1 +(ξ − x)2

β2r2]1/2+ C

= ln 12|ξ − x |+ 1

2[β2r2 + (ξ − x)2]1/2+ ln

βr

2+ C

= ln |ξ − x |+ constant (independent of ξ)

I Integrating by part, choosing C = − ln βr2

(to obtain the correct asymptoticbehavior in the vicinity of r → 0), and using S ′(0) = 0 and S ′(l) = 0, oneobtains

φ(x , r)

u∞= −

1

4π

[∓(

ln 12|ξ − x |+ 1

2[β2r2 + (ξ − x)2]1/2 − 2 ln

βr

2

)S ′(ξ) (

]x0

+

]lx

)

∓1

4π(

∫ x

0+

∫ l

x) ln(|ξ − x |)S ′′(ξ)dξ

= −1

4π[− ln

βr

2S ′(x)− ln

βr

2S ′(x)]∓

1

4π(

∫ x

0+

∫ l

x) ln(|ξ − x |)S ′′(ξ)dξ

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Solution for supersonic flow (M∞ > 1, M∞ 6∼ 1, M∞ 1)I Laplace eq. for axisymmetric problems (λ2 = M2

∞ − 1, ρ = λri,

i2 = −1)∂2φ

∂x2+∂2φ

∂ρ2+

1

ρ

∂φ

∂ρ= 0

I Potential due to a source located at position ξ

φ =

−q4π[(x−ξ)2−λ2r2]1/2 : x − ξ < λr

0 : otherwise

I Source flow rate q (volume flow: [q] = [u∞]L2 = L3/T = m3/s)I Equipotential: hyperboloid of revolution

(x − ξ)2 − λ2(y2 + z2) = C

I No upstream influence in supersonic flowI The potential function is zero outside of the cone of revolution

defined with the Mach angle µ∞i .e., tanµ∞ = 1/

√M2∞ − 1 = 1/λ

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Solution for supersonic flow (M∞ > 1, M∞ 6∼ 1, M∞ 1)I Superposition principle, the perturbation potential reads

φ(x , r) = − 1

4π

∫ x−λr

0

σ(ξ) dξ

[(x − ξ)2 − λ2r 2]1/2

I Continuous distribution of 3D sources on the axis, betweenx = 0 (leading edge) and x = l (base)

I Only sources at ξ between 0 and x − λr can influence thepoint (x , r)

I Source density distribution σ (volume flow/unit length:[σ] = [q]/L)

I Radial perturbation of velocity u′rI The BC reads: dR

dx =u′ru∞

∣∣∣r=R

= σ(x)4πRu∞

⇒ σ(x)u∞

= 2 S ′(x)

I In subsonic flows, half of the information is sent upstream, asopposed to supersonic flows where the whole information,twice as much as previously, is sent downstream

I By integrating over r the expression ∂rφ = u′r = u∞S′(x)2πr , one

gets φ(x , r)

u∞=

S ′(x)

2πln r + g(x)

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Drag of a slender body at supersonic speedsI In subsonic flow , the small perturbation theory calculation is

only valid when the body closes smoothly in a point, so thatthe inviscid flow assumptions are validI As expected, the total drag is zero in subsonic flow

I In supersonic flow , the calculation can be made irrespective ofthe form of the base, since the base does not exert anyupstream influence on the body surface

I Pressure coefficient non linear in axisymmetric flow (axialvelocity component u′x radial velocity u′r )

Cp = −2u′x − u′r2

I Karman formula for supersonic flow (independent of Machnumber)

D − DB

q∞= − 1

π

∫ l

0S ′′(x)

(∫ x

0ln(x − ξ)S ′′(ξ) dξ

)dx

provided that S ′(0) = 0 (pointed nose) and S ′(l) = 0 (pointedtip or cylindrical section)

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ProofI The function σ(ξ)

[(x−ξ)2−λ2r2]1/2 ∈ L1[(x − λr)] provided that σ(ξ) is regular

enough, whereas the function

σ(ξ) ∂∂x

1[(x−ξ)2−λ2r2]1/2 = (x−ξ)σ(ξ)

[(x−ξ)2−λ2r2]3/2 /∈ L1[(x − λr)]

I To eliminate the singularity, one usesσ(ξ) = σ(x − λr) + [ξ − (x − λr)]σ′(x − λr) + · · ·

⇒ [σ(ξ)− σ(x − λr)] ∂∂x

1[(x−ξ)2−λ2r2]1/2 = ( (x−ξ)σ′(x−λr)

[ξ−(x−λr)]1/2[ξ−x−λr ]3/2 + · · · ) ∈L1[(x − λr)]

I Thus, a more suitable expression for the perturbation function is given by

φ(x, r) = −1

4π

∫ x−λr

0

[σ(ξ)− σ(x − λr)] dξ

[(x − ξ)2 − λ2r2]1/2−σ(x − λr)

4π

∫ x−λr

0

dξ

[(x − ξ)2 − λ2r2]1/2

I Using the CVs: x − ξ = λry and y = cosh t = et+e−t

2, with

(et)2 − 2y et + 1 = 0 → t = arcosh y = ln(y +√

y2 − 1)

∫dξ

[(x − ξ)2 − λ2r2]1/2= −

∫dy

(y2 − 1)1/2= − arcosh[(x − ξ)/(λr)] + C

= − ln(x − ξ)/(λr) + [(x − ξ)2/(λ2r2)− 1]1/2 + C

= − ln 12

(x − ξ) + 12

[(x − ξ)2 − λ2r2]1/2 − lnλr

2+ C

= − ln (x − ξ) + constant (independent of ξ)

I φ(x , r) = − 14π

∫ x−λr0

[σ(ξ)−σ(x−λr)] dξ

[(x−ξ)2−λ2r2]1/2 −σ(x−λr)

4πarcosh

(xλr

)24 / 47

ProofI Leibniz’s formula:

ddr

∫ b(r)a(r)

F (ξ, r) dξ =∫ b(r)a(r)

∂∂r

F (ξ, r) dξ + F [b(r), r ] dbdr− F [a(r), r ] da

dr

∂φ

∂r= −

λ2r

4π

∫ x−λr

0

σ(x − λr) + σ(ξ) dξ

[(x − ξ)2 − λ2r2]3/2−λσ′(x − λr)

4π

∫ x−λr

0

dξ

[(x − ξ)2 − λ2r2]1/2

+λ

4πlim

ξ→x−λr

[σ(ξ)− σ(x − λr)]

[(x − ξ)2 − λ2r2]1/2+ λ

σ′(x − λr)

4πarcosh

(x

λr

)+

x

4πr

σ(x − λr)

[x2 − λ2r2]1/2

I Seeing that∫ x−λr

0dξ

[(x−ξ)2−λ2r2]1/2 = arcosh(

xλr

)and using de l’Hospital’s rule

limξ→x−λr[σ(ξ)−σ(x−λr)]

[(x−ξ)2−λ2r2]1/2 = “ 00

” = σ′(x−λr)

−2λr limξ→x−λr [(x−ξ)2−λ2r2]−1/2 = 0

u′r = ∂rφ = −λ2r

4π

∫ x−λr

0

σ(x − λr)− σ(ξ) dξ

[(x − ξ)2 − λ2r2]3/2+

x

4πr

σ(x − λr)

[x2 − λ2r2]1/2

= O(1) +O(

1

r

)≈

x

4πr

σ(x − λr)

[x2 − λ2r2]1/2≈σ(x)

4πr

I The BC reads:u′ru∞

∣∣∣r=R

= dRdx

= σ(x)4πRu∞

⇒ σ(x)u∞

= 4πR dRdx

= 2 S ′(x)

I The perturbation potential is: φ(x,r)u∞

= − 12π

∫ x−λr0

S′(ξ) dξ

[(x−ξ)2−λ2r2]1/2

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ProofI Integrating by part, choosing C = ln(λr

2), provided that S ′(0) = 0 (pointed

nose),φ(x, r)

u∞=

1

2π[

arcosh

(x − ξλr

)S′(ξ)

]x−λr0

−∫ x−λr

0arcosh

(x − ξλr

)S′′(ξ)dξ

≈1

2π[S′(ξ) ln

(12

(x − ξ) + 12

[(x − ξ)2 − λ2r2]1/2)]x−λr

0−∫ x−λr

0ln(x − ξ)S′′(ξ)dξ

=S′(x)

2πlnλr

2−

1

2π

∫ x−λr


⇒ φ(x,r)u∞

= S′(x)2π

ln r + g(x), with g(x) = S′(x)2π

ln λ2− 1

2π

∫ x−λr0 ln(x − ξ)S ′′(ξ)dξ

I Notice that there is no condition on S ′(l) in supersonic flowI Axial velocity (r → 0):

u′x = ∂xφ

u∞= S′′(x)

2πln λr

2− 1

2πd

dx

∫ x−λr0 ln(x − ξ)S ′′(ξ)dξ

I Pressure coefficient:

Cp = −S′′(x)

πlnλR(x)

2+

1

π

d

dx

∫ x

0ln(x − ξ)S′′(ξ)dξ − R′(x)2

I Wave drag: dD = (p − p∞) dS = 12ρ∞u2

∞ Cp dS , with dS = S ′(x)dx onlateral side

D − DB12ρ∞u2

∞=

∫ l

0Cp S′(x) dx

I Base drag: DB = (p∞ − pB)SB , with the base pressure pB and base surface SB(SB = 0 for pointed tip)

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Proof

I Substituting Cp , integrating by part twice, and using the assumption S ′(0) = 0

D − DB12ρ∞u2

∞= −

∫ l

0

S′′(x)

πS′(x) ln

λR(x)

2dx +

1

π

∫ l

0

(d

dx

∫ x


)S′(x) dx −

∫ l

0R′(x)2S′(x) dx

= −[S′(x)2

2πlnλR(x)

2

]l0

+

∫ l

0

S′(x)2

2π

R′(x)

R(x)dx +

1

π

[∫ x

0[ln(x − ξ)S′′(ξ)dξ] S′(x)

]l0

−1

π

∫ l

0S′′(x)

∫ x

0[ln(x − ξ)S′′(ξ)dξ] dx −

∫ l

0R′(x)2S′(x) dx

= −S′(l)2

2πlnλR(l)

2+ 2π

∫ l

0R(x)R′(x)3 dx +

S′(l)

π

∫ l

0ln(l − ξ)S′′(ξ)dξ

−1

π

∫ l

0S′′(x)

∫ x

0[ln(x − ξ)S′′(ξ)dξ] dx −

∫ l

0R′(x)22πR(x)R′(x) dx

⇒D − DB12ρ∞u2

∞= −

S′(l)2

2πlnλR(l)

2+

S′(l)

π

∫ l

0ln(l − ξ)S′′(ξ) dξ −

1

π

∫ l

0S′′(x)

(∫ x

0ln(x − ξ)S′′(ξ) dξ

)dx

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Outline



Slender body theory


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VKI Longshot gun tunnel

The VKI Longshot gun tunnel allows for partial duplication ofatmospheric entry conditions by providing large Mach and Reynoldsnumbers. This low-enthalpy short duration facility can be operated withnitrogen and carbon dioxide testing gases. It relies on the adiabaticcompression of the testing gas by a light piston to generate largestagnation pressures and temperatures which are then expanded througha contoured nozzle to hypersonic Mach numbers larger than 10

6.1m

Driver tubenitrogen 345 bar Piston

Driven tubenitrogen

27.5m

Check valves Reservoir

Test section Model

Secondary diaphragmPrimary diaphragm Vacuum pump Contoured nozzle: 42cm, Mach 14

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VKI Longshot gun tunnel

Piston starting mechanism

Primary diaphragm Piston + primary diaphragm

I The primary diaphragm separating the driver gas (nitrogen)from the driven gas (nitrogen or carbon dioxide) is subject toa pressure difference eventually causing its ruptureI Driver (left) gas: pL = 345 bar = 3.45×107 PaI Driven (right) gas: pR ∈ [1.4, 3] bar = [1.4, 3] 105 Pa

I When the mass of the piston is low enough, the pistonbehaves as a contact discontinuity in a shock-tube problem

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1D unsteady flow in straight ductI For constant cross-sectional area, the unsteady Euler eqs. are

expressed in conservative form as

∂tU + ∂xF = 0

I with the conservative variables U = (ρ, ρu, ρE )T

I the convective fluxes F = (ρu, ρu2 + p, ρuE + pu)I and the eq. of state p = (γ − 1)ρE − 1

2 (γ − 1)ρu2

I The Riemann problem for this system is the initial valueproblem with the initial conditions

U(x , 0) =

UL if x < 0UR if x > 0

I The Euler eqs. system can be expressed in quasi-linear formas

∂tU + A ∂xU = 0with the Jacobian matrix

A =∂F

∂U=

0 1 012

(γ − 3)u2 (3− γ)u γ − 112

(γ − 1)u3 − uH H − (γ − 1)u2 γu

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Strictly hyperbolic systemI Consider the first order nonlinear system ∂tU + ∂xF = 0,

x ∈ R, t > 0, with U = (U1, . . . ,Up)T and F = (F1, . . . ,Fp)T

Strictly hyperbolic system

This system is strictly hyperbolic if for any state U ∈ Ω, subset ofRp, the p × p Jacobian matrix A(U) = ∂F/∂U has p distinct realeigen values λ1(U) < λ2(U) < · · · < λp(U)

I With each eigen value λk one associates left and right eigenvectors

lk A = λk lk A rk = λk rk

I Since the system is strictly hyperbolic (distinct eigen values)

lj rk = lTj · rk = 0, j 6= k

I Diagonalization: A = RΛL, with R = (r1, · · · , rp), L =

l1...lp

and Λ = diag(λ1, · · · , λp)

I If the eigen vectors are normalized (lk rk = lTk · rk = 1): R−1 = L

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Eigen vectors and eigen values for the Euler eqs.I Right eigen vector r and eigen value λ: A r = λrI Characteristic equation: (A− λI) r = 0 ⇒ |A− λI| = 0

(λ− u)(λ2 − 2uλ+ u2 − a2) = 0

I λ2 = uI λ2 − 2uλ+ u2 − a2 = 0

I Discriminant: ∆ = a2 > 0I Eigen values: λ1 = u − a, λ3 = u + a, with the speed of sound

a =√γRT

⇒ All the eigen values of the unsteady Euler eqs. are real and distinct:

strictly hyperbolic system (for both the subsonic and supersonic

regimes)

I Eigen vectors:

r1 =

1u − a

H − ua

, r2 =

1u

12 u2

, r3 =

1u + a

H + ua

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1D unsteady isentropic flow in straight ductsI The entropy along a particle path in a smooth flow is constant

and the pressure is p = κργ

where κ is a constant

I If the entropy is constant everywhere, the 1D unsteady floweqs. are expressed in conservative form as

∂tU + ∂xF = 0

I with the conservative variables U = (ρ, ρu)T

I the convective fluxes F = (ρu, ρu2 + κργ)

I In quasi-linear form, one gets

∂tU + A ∂xU = 0

with the Jacobian matrix

A =∂F

∂U=

(0 1

−u2 + a2 2u

)

and the speed of sound a =√γp/ρ

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Eigen vectors and eigen values for isentropic flow

I Right eigen vector r and eigen value λ: A r = λr

I Characteristic equation: (A− λI) r = 0 ⇒ |A− λI| = 0

λ2 − 2uλ+ u2 − a2 = 0

I Discriminant: ∆ = a2 > 0I Eigen values: λ1 = u − a, λ2 = u + a

⇒ All the eigen values for 1D unsteady isentropic flow in straightducts are real and distinct: strictly hyperbolic system

I Eigen vectors:

r1 =

(1

u − a

), r2 =

(1

u + a

)

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Characteristic curves

I Multiplying the system of partial differentialeqs. ∂tU + A(U) ∂xU = 0 by the left eigen vector lk

lk(U) ∂tU + λk(U) lk(U) ∂xU = 0

lk(U) [∂t + λk(U) ∂x ]U = 0

I Introducing the characteristic curves Ck as the integral curvesof the differential system

dx

dt= λk [U(x , t)]

I And denoting by sk → [x(sk), t(sk)] a parametricrepresentation of Ck , one gets

lk(U)dU

dsk= 0

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Characteristic curves/Riemann invariants for isentropic flow

I Left eigen vectors: l1 = (a + u − 1) and l2 = (a− u 1)

I Riemann invariants on characteristic curvesl11

dds1ρ+ l12

dds1

(ρu) = 0

l21d

ds2ρ+ l22

dds2

(ρu) = 0⇒

a dρ− ρ du = 0 on C−

a dρ+ ρ du = 0 on C+

∫aρ

dρ =

∫√γκ ρ

γ−32 dρ = 2

γ−1a ⇒

u − 2

γ−1a = J− on C− ≡ dx

dt= u − a

u + 2γ−1

a = J+ on C+ ≡ dxdt

= u + a

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Generalized Riemann invariantsI Using the change of variables ∂W = L ∂U, the system of

partial differential eqs. yieldsL∂tU + ΛL ∂xU = 0

∂tW + Λ ∂xW = 0

∂tWk + λk ∂xWk = 0d

dskWk = 0

Riemann invariant across the k− wave structure

The (smooth) function W : Ω ∈ Rp → R is called a k−Riemanninvariant if it satisfies

∂UW (U) · rk(U) = 0, ∀U ∈ Ω

I A k−Riemann invariant W is constant on a curve v : ξ ∈ R→ v(ξ) ∈ Rp if andonly if

d

dξW [v(ξ)] = ∂UW [v(ξ)] · dv(ξ)

dξ= 0

which holds if v is an integral curve of rk , i .e., dv(ξ)dξ

= rk [v(ξ)]I This means that a k−Riemann invariant W is constant along the trajectories of

rkI There exist (p − 1) k−Riemann invariants whose gradients are linearly

independent dv1rk1

= dv2rk,2

= · · · =dvprk,p

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Isentropic eqs.

I Quantities expressed in conservative variables UI u = ρu

ρI p = κργ

I a = (γκ)12 ρ

γ−12

I s = cv lnκ

I 1-Riemann invariants accros C− ≡ dxdt = u − a

I ∂U(s) = 0⇒ ∂U(s) · r1 = 0I ∂U(u + 2a

γ−1) = 1

ρ(−u + a, 1)T ⇒ ∂U(u + 2a

γ−1) · r1 = 0

I Thus u + 2γ−1

a is constant on C+ ≡ dxdt

= u + a

I 2-Riemann invariants accros C + ≡ dxdt = u + a

I ∂U(s) = 0⇒ ∂U(s) · r2 = 0I ∂U(u − 2a

γ−1) = 1

ρ(−u − a, 1)T ⇒ ∂U(u − 2a

γ−1) · r2 = 0

I Thus u − 2γ−1

a is constant on C− ≡ dxdt

= u − a

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Euler eqs.I Quantities expressed in conservative variables U

I u = ρuρ

I p = (γ − 1)ρE − 12

(γ − 1) (ρu)2

ρ

I a =√γρ

[(γ − 1)ρE − 12

(γ − 1) (ρu)2

ρ]1/2

I s = cvln[(γ − 1)ρE − 12

(γ − 1) (ρu)2

ρ]− γ ln ρ

I 1-Riemann invariants accros dxdt = u − a, if smooth

(isentropic) fieldI ∂U(s) = (γ − 1) cv

p( u2

2− a2

γ−1, − u, 1)T ⇒ ∂U(s) · r1 = 0

I ∂U(u + 2aγ−1

) = 1ρ

(−u− aγ−1

+ γu2

2a, 1− γu

a, γ

a)T ⇒ ∂U(u + 2a

γ−1) · r1 = 0

I 2-Riemann invariants accros dxdt = u

I ∂U(u) = 1ρ

(−u, 1, 0)T ⇒ ∂U(u) · r2 = 0

I ∂U(p) = (γ − 1)( u2

2, − u, 1)T ⇒ ∂U(p) · r2 = 0

I 3-Riemann invariants accros dxdt = u + a, if smooth

(isentropic) fieldI ∂U(s) · r3 = 0

I ∂U(u− 2aγ−1

) = 1ρ

(−u+ aγ−1− γu2

2a, 1+ γu

a, − γ

a)T ⇒ ∂U(u− 2a

γ−1)·r3 = 0

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Riemann problemI The possible types of waves present in the solution of the

Riemann problem depend crucially on closure conditionsI For the non-linear system of Euler eqs. for calorically perfect

gases (such that γ > 1), the only waves present are shocks,contacts and rarefactionsI Shock waves and contact waves are discontinuitiesI Rarefaction waves are smooth transitions

I The eigen values structure for the Euler eqs. is as followsI The λ2-field (contact) is linearly degenerate:

∂Uλ2(U) · r2(U) = 0

I ∂Uλ2(U) · r2(U) = (−u/ρ, 1/ρ, 0)T · (1, u, u2/2)T = 0I λ2 is thus a 2-Riemann invariant

I The λ1- and λ3-fields (rarefaction and shock) are genuinelynon-linear:

∂Uλi (U) · ri (U) 6= 0, i = 1, 3

I Godunov developed the first exact Riemann solver for theEuler eqs.

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Shock-tube problem

I A shock-tube problem is a gas dynamics application of theRiemann problem for the Euler eqs.:I 2 stationary gases (uL = uR = 0) in a tube are separated by a

diaphragm with pL pR (TL = TR , ρL ρR , γL 6= γR)I Rupture of the diaphragm generates a nearly centered wave

system: rarefaction, contact discontinuity and shock wave

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Rarefaction waveI The properties are obtained from the generalised Riemann

invariants valid for smooth transitionI For a 1-rarefaction

2aLγL−1

= u∗L +2a∗LγL−1

pL(ρL)γL

=p∗L

(ρ∗L

)γL⇒ ρ∗L = ρL(

p∗LpL

)1γL

I Using the perfect gas law and definition of sound speed

a∗L = aL(p∗LpL

)γL−12γL ⇒ u∗L = 2aL

γL−1[1− (

p∗LpL

)γL−12γL ]

I Lax’s entropy condition for rarefaction wave: rarefactionsolution admissible only if characteristics on the left and rightof the wave diverge as do characteristics inside the waveI Genuinely non-linear 1-rarefaction

λ(UL) < λ(U∗L )

I Physical meaningρL > ρ∗L , pL > p∗L

I The expansion shock is not physicalI Entropy decreases through expansion shock

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Contact discontinuityI Rankine-Hugoniot relations for 2-contact (see lecture 2):

I σ(U∗L − U∗R) = F ∗L − F ∗R

⇒ σc

ρ∗L − ρ∗R

ρ∗Lu∗L − ρ

∗Ru∗R

ρ∗LE∗L − ρ

∗RE∗R

=

ρ∗Lu∗L − ρ

∗Ru∗R

ρ∗L(u∗L )2 + p∗L − ρ∗R(u∗R)2 − p∗R

ρ∗Lu∗LE∗L + u∗Lp

∗L − ρ

∗Ru∗RE∗R − u∗Rp

∗R

I If σc = u∗L = u∗R = u∗ and p∗L = p∗R = p∗, these relations are satisfiedI The jump in density (temperature and entropy) is non trivialI The jump in density is in the direction of the r2 vector (if γL = γR) ρ∗L − ρ

∗R

ρ∗Lu∗L − ρ

∗Ru∗R

ρ∗LE∗L − ρ

∗RE∗R

= (ρ∗L − ρ∗R)

1u∗

12

(u∗)2

= (ρ∗L − ρ∗R) r2

I Lax’s entropy condition for contact: jump is admissible only ifcharacteristics run parallel to the waveI Linearly degenerate 2-field

λ(U∗L ) = σc = u∗ = λ(U∗R)

I Both gases can have different thermodynamic properties(specific heat ratios): γL 6= γR in general

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Shock waveI Rankine-Hugoniot relations for 3-shock (see lecture 2):

I σ(U∗R − UR) = F ∗R − FR

⇒ σ

ρ∗R − ρRρ∗Ru∗R − ρRuR

ρ∗RE∗R − ρRER

=

ρ∗Ru∗R − ρRuR

ρ∗R(u∗R)2 + p∗R − ρR(uR)2 − pRρ∗Ru∗RE∗R + u∗Rp

∗R − ρRuRER − uRpR

I Introducing the relative velocities v∗R = u∗R − σ and vR = σ (uR = 0),

the relative Mach number MR = σ/aR , one gets (see normal shock

relations in lecture 3)

I p∗RpR

= 1 + 2γRγR+1

(M2R − 1) ⇒ M2

R = γR+12γR

p∗RpR

+ γR−12γR

I ρ∗RρR

= vRv∗R

=(γR+1)M2

R

2+(γR−1)M2R

andT∗RTR

= (p∗RpR

)/(ρ∗RρR

)

I The shock speed is thus given by σ = aR MR = aR [ γR+12γR

p∗RpR

+ γR−12γR

]1/2

I Lax’s entropy condition for shock wave: jump is admissibleonly if characteristics run into the waveI Genuinely non-linear 3-shock

λ(U∗R) > σ > λ(UR)

I Physical meaningρ∗R > ρR , p∗R > pR , T∗R > TR , u∗R − σ < a∗R , s∗R > sR

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Solution for the shock-tube problem

I Rarefaction: u∗L = 2aLγL−1

[1− (p∗LpL

)γL−12γL ]

I Shock:

I σ = aR [ γR+12γR

p∗RpR

+ γR−12γR

]1/2 =

[p∗R+

γR−1γR+1

pR

2γR+1

ρR

]1/2

I p∗R − pR = σρ∗Ru∗R − ρ

∗R(u∗R)2 = ρ∗R(σ − u∗R)u∗R = ρRσu

∗R

⇒ u∗R =p∗R−pRρRσ

= (p∗R − pR)

[2

γR+11ρR

p∗R

+γR−1γR+1

pR

]1/2

I Contact: u∗L = u∗R = u∗ and p∗L = p∗R = p∗

I Combining these relations

I 2aLγL−1

[1− ( p∗

pL)γL−12γL ] = (p∗ − pR)

[2

γR+11ρR

p∗+γR−1γR+1

pR

]1/2

⇒ pLpR

= p∗

pR

1−(γL−1)

aRaL

(p∗

pR−1)

2γR

[2γR+(γR+1)( p∗

pR−1)

]1/2

−2γLγL−1

⇒ The values of p∗ (and then u∗) are obtained by solving thisnonlinear eq.

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VKI Longshot gun tunnel experiment

I Pressure profile at port 3 (14 mfrom the driver tube)

I Events detected

1. Primary shock wave2. Piston path3. Reflected shock wave

[Ilich and Grossir, 2015]

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Exercise

Consider the VKI Longshot gun tunnel. The driver and driven tubeare filled with a nitrogen gas at rest at ambient temperature(uL = uR = 0 m/s, TL = TR = 300 K). The primary diaphragm isruptured for a driver gas pressure pL = 3.45× 107 Pa and drivengas pressure pR = 1.5×105 Pa. The mass of the piston is assumedto be negligible. The specific heat ratio of nitrogen is γ = 1.4.Compute:

1. The pressure p∗ on both sides of the piston, considering that it behaves as acontact discontinuity

2. The velocity u∗ of the piston

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Documents

Aerothermodynamics of high speed ows AERO 0033{1