11.2. LINEARIZEDSUPERSONICFLOW
161
y
x
y=y (x)
+
y-y (x)
-
dy/dx=m (x)
dy/dx=m (x)
+
-
y
x
U
0
α
Figure11.3:Thinairfoilgeometry.
11.2.1 Thinairfoiltheory
Weconsiderfirstthe2Dsupersonicflowoverathinairfoil.
Linearized supersonic c flow results s whenM M > > 1, , linearizedsubsonic flow
whenM<1.ThetransonicregimeM≈1isspecialandneedstobeexamined
asaspecialcase.
Weshowthegeometryofathinairfoilinfigure11.3. Weassumethatthe
slopesdy
±
/dxandtheangleofattackαaresmall. Inthiscase
m
±
(x)≈−α+dy
±
/dx.
(11.40)
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162
CHAPTER11. GASDYNAMICSII
letthechordoftheairfoilbel,soweconsider0<x<l.Wenotethat
1
l
l
0
(m
+
+m
)dx=−2α,
1
l
l
0
(m
2
+
+m
2
)dx=−2α
2
+
1
l
l
0
(y
2
+
+y
2
)dx.
(11.41)
Theanalysisnowmakesusethefollowingfactanalogoustothelinearwave
equationin1D:thelinearoperatorfactorsas
M2−1
∂x
∂y

M2−1
∂x
+
∂y
.
(11.42)
Thusφ=f(x−y
M2−1)+g(x+y
M2−1),wheref,garearbitraryfunc-
tions. Wenowneedtousephysicalreasoningchoosetherightformofsolution.
Inlinearizedsupersonic flowpast anairfoilthe disturbances madebythefoil
propagateoutrelativetothefluidatthespeedofsound,butaresimultaneously
carrieddownstreamwithspeedU
0
. Insupersonicflowthefoilcannottherefore
causedisturbancesofthefluidupstreamofthebody.Consequentlythecharac-
teristiclinesx±y
M2−1=constant,whichcarrydisturbancesawayfromthe
foil,must alwayspointdownstream. . Thusinthehalfspaceabovethefoilthe
correct choiceisφ=f(x−y
M2−1),whileinthespacebelowitthecorrect
choiceisφ=g(x+y
M2−1). Todeterminethesefunctions,wemustmake
theflowtangenttothefoilsurface. Sincewearedealingwiththinairfoilsand
smallangles,theconditionoftangencycanbeapplied,approximately,aty=0.
Thuswehavethetangencyconditions
φ
y
U
0
|
y=0+
=m
+
(x)=−
M2−1U
−1
0
f
(x),
(11.43)
φ
y
U
0
|
y=0−
=m
(x)=
M2−1U
−1
0
g
(x),
(11.44)
Ofinterest toengineers istheliftanddragofafoil. . Tocomputethesewe
firstneedthepressures
p
+
(x)=−U
0
ρ
0
u
(x,0+)=−U
0
ρ
0
f
(x), p
(x)=−U
0
ρ
0
g
(x).
(11.45)
Thisyields
p
±
U
2
0
ρ
0
M2−1
(−α+dy
±
/dx).
(11.46)
Then
Lift=
l
0
(p
−p
+
)dx=
2αρ
0
U2
0
l
M2−1
,
(11.47)
Drag=
l
0
(p
+
m
+
−p
m
)dx=
ρ
0
U2
0
l
M2−1
[2α
2
+
1
l
l
0
[(dy
+
/dx)
2
+(dy
/dx)
2
]dx.
(11.48)
Note that t now inviscid theory gives s a a positive e drag. . We e recall l that for
incompressible potential flowwe obtainedzero drag(D’Alembert’s paradox).
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11.2. LINEARIZEDSUPERSONICFLOW
163
Insupersonicflow,thecharacteristicscarryfinitesignalstoinfinity.Infactthe
disturbancesarebeingcreatedsothattherateofincreaseofkineticenergyper
unittimeisjustequaltothedragtimesU
0
. Thisdragisoftencalledwavedrag
becauseitisassociatedwithcharacteristics,usuallycalledinthiscontextMach
waves,whichpropagatetoinfinity.
Whathappensifwesolveforcompressibleflowpastabodyinthesubsonic
caseM <1? ? Inthecase e of thinairfoiltheory,itiseasytosee thatwemust
get zero o drag. . The e reason n is s that t the equation we e are e now w solvingmay be
writtenφ
xx
¯y¯y
=0where ¯y=
1−M2y. Theboundaryconditionsareat
y= ¯y=0sointhenewvariableswehaveaproblemequivalenttothatofan
incompressiblepotentialflow.
Infactcompressiblepotentialflowpast anyfinitebodywillgivezerodrag
solongastheflowfieldvelocityneverexceedsthelocalspeedofsound,i.e.the
fluidstayslocallysubsoniceverywhere. Inthatcasenoshockwavescanform,
thereisnodissipation,andD’Alembert’sparadoxremains.
11.2.2 Slenderbodytheory
Another case ofinterest is the steady supersonic flowpast aslender body of
revolution.Iftheambientflowisalongthez-axisincylindricalpolarcoordinates
x,r,thebodyweconsideris aslender bodyofrevolutionaboutthez-axis. . It
iseasytoshowthattheappropriatewaveequation(comingfromthelinearized
equationsρ
0
U
0
∂u
∂z
+c
2
0
∂ρ
∂z
=0,U
0
∂ρ
∂z
0
∇·u
=0),is
β
2
φ
zz
−φ
rr
1
r
φ
r
=0, β=
M2−1.
(11.49)
Tofindafundamentalsolutionofthisequation,notthat
a
2
φ
xx
+b
2
φ
yy
+c
2
φ
xx
=0
(11.50)
clearlyhasa“sink-like”solution[(x/a)
2
+(y/b)
2
+(z/c)
2
]
−1
,equivalenttothe
simplesinksolution(−4πtimethefundamentalsolution)1/
(x
2
+y
2
+z
2
)of
Laplace’sequationin3D.This holdsforarbitrarycomplexnumbersa,b,c. . It
followsthatasolutionof(11.49)isgivenby
S(z,r)=
1
z2−β2r2
.
(11.51)
Notethatthisisarealquantityonlyifβr<z,whereSissingular.Wetherefore
wanttocompletethedefinitionofS bysetting
S(z,r)=0, βr>z.
(11.52)
Supposenowthat wesuperimpose thesesolutionsbydistributingthemon
theinterval(0,l)ofthez-axis,
φ=
0
f(ζ)
(z−ζ)2−β2r2
dζ.
(11.53)
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164
CHAPTER11. GASDYNAMICSII
Howevernoticethatifweareinterestedinthesolutiononthesurfacez−βr=C,
thentherecanbenocontributionsfromvaluesofζexceedingC.Wetherefore
proposeapotential
φ(z,r)=
z−βr
0
f(ζ)
(z−ζ)2−β2r2
dζ, if0<z−βr<1,
1
0
f(ζ)
(z−ζ)2−β2r2
dζ,
ifz−βr>1.
(11.54)
wherewenowrequiref(0)=0.
We can n in fact verify that t (11.54) gives s us s a a solution n of (11.49) ) for any
admissiblef,butwillleavethisverificationasaproblem.
Considernowthebehaviorofφnearthebody. Whenr r is smallthemain
contributioncomesfromthevicinityofζ=z,sowemayextractf(z)anduse
thechangeofvariablesζ=z−βrcoshλtoobtain
φ≈f(z)
cosh
−1
(z/βr)
0
dλ=f(z)cosh
−1
(z/βr)∼f(z)log(2z/βr). (11.55)
Nowlet the body be described d by y r r = R(z),0 0 < < z z < l. . The e tangency
conditionisthen
r
φ
r
U
0
r=R
=R
dR
dz
≈−f(z)/U
0
.
(11.56)
IfA(a)denotescross-sectionalarea,thenwehave
f(z)=−
1
U
0
dA
dz
.
(11.57)
Thecalculationofdragisabitmorecomplicatedhere,andwegivetheresult
forthecasewhereR(0)=R(l)=0. Then
Drag=
ρ
0
U2
0
l
0
l
0
A

(z)A

(ζ)
1
log|z−ζ|
dzdζ.
(11.58)
Theform of thisemphasizes the importanceofhaveasmoothdistributionof
area.inordertominimizedrag.
Analternativeformulationandproofofthedragformula
Toprove(11.58)itisconvenienttoreformulatetheproblemintermsofastream-
function. Wegoback k tothebasicequations forsteadyhomentropicpotential
flowincylindricalpolarcoordinates.
∂u
r
∂z
∂u
z
∂r
=0,
∂rρu
z
∂z
+
∂rρu
r
∂r
=0,
(11.59)
u
2
z
+u
2
r
+
2
γ−1
c
2
=constant.
(11.60)
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11.2. LINEARIZEDSUPERSONICFLOW
165
α
z-βr < 0
0< z-βr < 1
z-βr>1
Figure 11.4: : Steady y supersonic flowpast aslender r bodyof f revolution. . Here
tanα=1/
M2−1.
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166
CHAPTER11. GASDYNAMICSII
Fromthesecondof(11.59)weintroducethestreamfunctionψ,
rρu
z
=
∂ψ
∂r
,rρu
r
=−
∂ψ
∂z
.
(11.61)
Wethenexpandtheequationsasfollows:
ψ=
U
0
2
r
2
,ρ=ρ
0
,
(11.62)
andlinearize. Theresultistheequationforψ
,
r
∂r
1
r
∂ψ
∂r
−β
2
2ψ
∂z2
=0.
(11.63)
Nowtheboundaryconditionintermsofthestreamfunctionisthatψequal
zeroontheslenderbodyr=R(z).Approximately,thisgives
ρ
0
U
0
2
r
2
(z,0)≈0.
(11.64)
Wealsowantnodisturbanceupstream,soψ
and
z
shouldvanishonz=
0,r>0.Asolutionofthisproblemisgivenby
ψ
=−
ρ
0
U
0
z−βr
0
(z−ζ)22r2A

dζ.
(11.65)
It is easy to o see that t the equationandupstream m conditions s are satisfied
undertheconditionsthatR(0)=0.Fortheboundaryconditionwehave
ψ
(z,0)=−
ρ
0
U
0
z
0
(z−ζ)A

dζ=−
ρ
0
U
0
z
0
A
dζ=−ρ
0
U
0
2
r
2
. (11.66)
Nowthedragisgivenby
D=
l
0
p
A
(ζ)dζ,
(11.67)
wherethelineartheorygivesp
≈−ρ
0
U
0
u
z
. Howeveritturns s outthattheu
r
velocitycomponentsbecomesufficientlylargenearthebodytomakealeading
ordercontribution. Thuswehave
p
≈−ρ
0
U
0
u
z
1
2
u
2
r
+....
(11.68)
Wenotethat
u
z
=
1
ρr
∂ψ
∂r
1
ρ
0
r
∂ψ
∂r
U
0
ρ
0
c
2
0
p
,
(11.69)
fromwhichwehave
−β
2
u
z
=
1
ρ
0
r
∂ψ
∂r
.
(11.70)
11.2. LINEARIZEDSUPERSONICFLOW
167
Thus
u
z
=−
U
0
z−βr
0
A

(ζ)
(z−ζ)22r2
dζ.
(11.71)
Similarly
u
r
=
U
0
2πr
z−βr
0
(z−ζ)
A

(ζ)
(z−ζ)22r2
dζ.
(11.72)
Weseebylettingr→R≈0in(11.72)that
u
r
U
0
A
(z)/R(z).
(11.73)
Also
u
z
≈−
U
0
A

(z−βr)cosh
−1
z
βr
+
z−βr
0
A

(ζ)−A

(z−βr)
(z−ζ)22r2
(11.74)
≈−
U
0
A

log
2z
βR
z
0
A

(z)−A

(ζ)
z−ζ
.
(11.75)
WearenowinapositiontocomputeD:
D=
ρ
0
U
2
0
l
0
A
(z)
z−βR
0
A

(z)log
2z
βr
z
0
A

(z)−A

(ζ)
z−ζ
dζ−
1
4
A
−1
(z)(A
)
2
(z)
dz.
(11.76)
Afteranintegrationbypartsandacancelationwehave
D=
ρ
0
U2
0
l
0
A
(z)
A

(z)logz−
z
0
A(z)−A(ζ)
z−ζ
dz.
(11.77)
Ourlaststepistoshowthat(11.77)agreeswith(11.58). NowifB(z)=A
(z),
l
0
l
0
B
(z)B
(ζ)log|z−ζ|dζdz=2
l
0
B
(z)
z
0
B
(ζ)log|z−ζ|dζdz
=2
l
0
B
(z)B(z)logzdz+2
l
0
B
(z)
z
0
B(z)−B(ζ)
z−ζ
dζdz
=−2
l
0
B
(z)B(z)logzdz−2
l
0
B(z)
z
0
B
(z)−B
(ζ)
z−ζ
dζdz,
(11.78)
whichprovestheagreementofthetwoexpressions. Herewehaveused
d
dz
z
0
B(z)−B(ζ)
z−ζ
dζ=
B(z)−B(ζ)
z−ζ
ζ
=z+
z
0
(z−ζ)B
(z)−B(z)+B(ζ)
(z−ζ)2
=B
(z)+
(z−ζ)b
(z)−B(z)+B(ζ)
(z−ζ)2
z
0
+
z
0
B
(z)−B(ζ)
z−ζ
=
z
0
B
(z)−B
(ζ)
z−ζ
dζ+B(z)/z.
(11.79)
168
CHAPTER11. GASDYNAMICSII
Chapter12
Shockwaves
12.1 Scalarcase
Wehaveseenthattheequationu
t
+uu
x
=0withainitialconditionu(x,0)=
1−xonthesegment0<x<1producesafamilyofcharacteristics
x=(1−x
0
)t+x
0
.
(12.1)
This familyof linesintersects at(x,t)= (1,1). . Iftheinitialconditionisex-
tendedas
u(x,0)=
1, ifx<0,
0, ifx>1,
(12.2)
we see that t at t t = = 1a discontinuity y develops s in u as a function of x. . We
thusneedtostudyhowsuchdiscontinuitiespropagateforlatertimesasshock
waves. Westudy y first thegeneralscalar waveequationinconservationform,
u
t
+(F(u))
x
=0. Thisequationisassumedtocomefromaconservationlawof
theform
d
dt
b
a
udx=F(u(a,t))−F(u(b,t)).
(12.3)
Suppose now that infact there is a discontinuitypresent at position n ξ(t) ∈
(a,b).Thenwestudytheconservationlawbybreakinguptheintervalsothat
differentiationundertheintegralsignispermitted:
d
dt
ξ
a
udx+
b
ξ
udx=F(u(a,t))−F(u(b,t)).
(12.4)
Nowdifferentiatingundertheintegralandusingthewaveequationtoeliminate
thetimederivativesofuweobtain
dt
[u(ξ+,t)−u(ξ−,t)]=F(u(ξ+,t))−F(u(ξ−,t)).
(12.5)
169
170
CHAPTER12. SHOCKWAVES
Thuswehaveanexpressionforthepropagationvelocityoftheshockwave:
dt
=
[F]
x=ξ
[u]
x=ξ
,
(12.6)
wherehere[·]means“jumpin”.Thedirectionyoutakethejumpisimmaterial
providedthatyoudothesameinnumeratoranddenominator.
Example:Letu
t
+uu
x
=0,
u(x,0)=
0,
ifx<−1,
1+x, if−1<x<0,
1−2x, if0<x<1/2
,0,
ifx>1/2.
(12.7)
The characteristic family associated with the e interval l −1 < x < < 0 0 is s x x =
(1+x
0
)t+x
0
,whilethatof0<x<1/2isx=(1−2x
0
)t+x
0
.Theshockfirst
occurs at(x,t)=(1/2,1/2). . Totherightoftheshocku=0,whiletotheleft
theformerfamilygives
u=1+x
0
(x,t)=1+
x−t
1+t
=
1+x
1+t
.
(12.8)
Then
dt
1
2
[u(ξ−,t)+u(ξ+,t)]=
1
2
ξ+1
1+t
, ξ(1/2)=1/2.
(12.9)
Thus
ξ(t)=
3/2
1+t−1.
(12.10)
Weshowthex,t-diagramforthisinfigure12.1.
12.1.1
Acautionarynote
One peculiarity of shock propagationtheory isthatit isstrongly tiedtothe
physics of f the problem. . Suppose e that u ≥ ≥ 0 solves s u
t
+uu
x
= 0. . Then n it
willalsosolvev
t
+[G(v)]
x
= 0whereG=
2
3
v3/2 andv = = u2. Inthe e firmer
casetheshockwavepropagationspeedis
1
2
(u
+
+u
),whileinthelatteritis
2
3
(u
2
+
+u
+
u
+u
2
)/(u
+
+u
),whichisdifferent. What’sgoingon??
Thepointisthatu
t
+uu
x
=0isbasedfundamentallyonaconservationlaw
involvingF(u)=u
2
/2. Inactualphysicalproblemstheconservationlawswill
beknownandhavetoberespected. Anotherwaytosaythisisthatequivalent
partialdifferentialequationscanarisefromdifferentconservationlaws.Itisthe
conservationlawthatdeterminestherelevantshockvelocityhowever.
Weillustratethiswithasimpleexamplefromthecontinuumtheoryoftraffic
flow. Consider r asingle-lane highwaywithn(x,t)cars per mileasthetraffic
density. The e cars s areassumedtomoveat t a a speed d determined d by y the local
density, equal tou = = U(1−n/n
0
whereU isthe e maximumvelocityand d n
0
isthedensityof fullpackingandzerospeed. . Thefluxof f cars istheF(n)=
nu=Un(1−n/n
0
),andthecorrespondingconservationofcarnumber yields
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