School of Aerospace Engineering
• Previously, we examined
supersonic flow over (sharp)
concave corners/turns
– oblique shock allows flow to
make this (compression) turn
• What happens if:
– turn is convex (expansion)
· already shown expansion
“shock” impossible (entropy
would be destroyed)
– turn is gradual
(concave or convex)
Prandtl Meyer -1
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
M1>1
M1>1
M1>1
M2>M1
M1>1
M2>M1
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M1>1
• Gradual turn is made up of
large number of infinitessimal
turns/corners
µ = sin 1
M
• Each turn has infinitessimal flow
µ = sin
M
change
M2>M1
δ
−1
1
1
−1
a
1
1
Ma
Ma
M
– each turn produced by
infinitessimal wave ÞMach wave
• Flow is uniform and isentropic between
µ = sin 1
each turn/corner
M
M
– length between each is arbitrary
– could be zero length (sharp turn)
and waves collapse to one point
b
µ 2 = sin −1 1
δ
−1
1
M2
1
µ 2 = sin −1 1
1
Prandtl Meyer -2
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
M2
δ
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M2
M2
School of Aerospace Engineering
• Problem
– given upstream conditions (1)
M1
and turning angle (δ)
– find downstream conditions (2)
Prandtl Meyer Fan
δ
M2
• Goal
– Mach number relations (similar to shock relations)
• Equations
– use mass, momentum, energy conservation, Mach
number def’n., state equations
• Assumptions
– steady flow, quasi-1d, reversible+adiabatic (isentropic)
Prandtl Meyer -3
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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School of Aerospace Engineering
• Approach
– begin with single Mach wave that expands supersonic
flow through an infinitessimal (differential) angle of
magnitude dν
vt
vn
– essentially using differential
vt vn+dvn
µ
v
control volume
µ
dν
• Mass/Momentum Conservation
µ+dν
– using same type of approach as for
v+dv
oblique shocks (two momentum components: t, n)
– find lack of pressure gradient tangent to wave gives
vt=constant across wave
Prandtl Meyer -4
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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vt
• Use vt=constant
vn
vt
µ v
v t ,upstream = v t ,downstream
µ
dν
v cos µ = (v + dv )cos(µ + dν )
= (v + dv )(cos µ cos dν − sin µ sin dν )
→1
dν→0
vn+dvn
µ+dν
→dν
v+dv
v cos µ = v cos µ − vdν sin µ + dv cos µ − dvdν sin µ
dv sin µ
=
dν
v cos µ
sinµ=1/M dv
1/ M 2
=
dν
2
sin2µ+cos2µ=1 v
1 − 1/ M
Prandtl Meyer -5
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
→0
dv
1
=
dν
v
M2 − 1
(VIII.1)
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ν
• Relate v and M
v=Ma
tpg/cpg
a2=γRT
dv dM da
=
+
v
M
a
dv dM d T dM 1 dT
=
+
=
+
v
M
M 2 T
T
dv dM [(γ − 1) 2]M 2 dM
=
−
v
M æ γ −1 2 ö M
M ÷
ç1 +
2
è
ø
ù
é
é
(γ − 1) M 2 ù
ú
ú dM ê
dv dM ê
1
2
=
ú
ú=
ê
ê1 −
1
1
γ
−
γ
−
v
M ê æ
M êæ
ö
ö
M2 ÷ ú
M2 ÷ ú
ç1 +
ç1 +
êë è
êë è
2
2
ø úû
ø úû
Prandtl Meyer -6
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
v
µ
dν
v+dv
Energy Conservation
æ γ −1 2 ö
To = Tç1 +
M ÷ = const.
2
è
ø
æ γ −1 2 ö
d ç1 +
M ÷
dTo
dT
2
ø
=0=
+ è
T
æ γ −1 2 ö
To
M ÷
ç1 +
2
è
ø
2
(γ − 1)M dM
dT
=−
æ γ −1 2 ö M
T
M ÷
ç1 +
2
è
ø
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ν
• Relate VIII.1 and last eqn.
ù
é
ú
dv
1
dM ê
1
dν =
=
ú
ê
2
v
M êæ γ − 1 2 ö ú
M −1
M ÷
ç1 +
êë è
2
ø úû
M 2 − 1 dM
dν =
γ −1 2 M
1+
M
2
(VIII.2)
integrate
M2
M − 1 dM
ν 2 − ν1 = ò dν = ò
γ −1 2 M
ν1
M1 1 +
M
2
Prandtl Meyer -7
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
M+dM
dM is change in Mach number
associated with dν turn angle
• Need finite angle, δ=ν2-ν1 and finite ∆M
ν2
µ
dν
M
2
M1
δ
M2
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ν
• Perform Integration
M1
M2
M 2 − 1 dM
ν 2 − ν1 = ò
γ −1 2 M
M1 1 +
M
2
δ
M2
(
)
M2
é γ + 1 −1 γ − 1 2
ù
2
−1
ν 2 − ν1 = ê
tan
M − 1 − tan
M − 1ú
γ +1
ë γ −1
û M1
(VIII.3)
• So, given δ (=ν2-ν1) and M1
– could “solve” VIII.3 for M2
• Can not invert VIII.3 analytically (M2=f(M1, δ))
– either use interative (e.g., numerical or guessing) method
– or find ν as a function of M and tabularize or graph solution
Prandtl Meyer -8
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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School of Aerospace Engineering
(
M
)
é γ + 1 −1 γ − 1 2
ù 2
2
−1
ν 2 − ν1 = ê
tan
M − 1 − tan
M − 1ú
γ
−
1
γ
+
1
ë
û M1
• Want to find ν=ν(M) [really ν2=ν2(M2)] for any M
– need to choose (arbitrary) reference condition,
• analagous to table of h(T)
i.e., pick an M where ν=0
– really h(T)-h(Tref)
– let’s choose ν=0 at M=1
– just chosen h(T )=0
(
)
ref
γ + 1 −1 γ − 1 2
M − 1 − tan −1 M 2 − 1 (VIII.4)
ν=
tan
Appendix D (John)
γ −1
γ +1
for γ=1.4, M≤5
• ν represents angle through which a
sonic flow would
M=1
have to turn
ν1
to reach M
M1
ν1 is turn angle for M=1 to M1
Prandtl Meyer -9
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
δ=ν2-ν1
M2
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• To find M2 given M1 and δ
– find ν1 (for given M1) from table
– get ν2 from δ=ν2-ν1
– look up ν2 in table to find M2
M1
δ=ν2-ν1
M2
• To find T2, p2, ....
– use isentropic flow relations since expansion is
isentropic (no shock)
γ −1 2
– e.g.,
To=const
1+
M
1
T2
γ −1 2
To
2
=
=1+
M Þ
T1 1 + γ − 1 M 2
T
2
2
2
Prandtl Meyer -10
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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• Given: Uniform Mach 2 flow of
nitrogen at 300 K flows over compound
wall corner: two turns, 20° and 6°
M1
20°
6°
• Find:
M and T after final turn
• Assume: N2 is TPG/CPG with γ=1.4, steady, adiabatic, no
work, inviscid,….
Prandtl Meyer -11
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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School of Aerospace Engineering
• Analysis: (class exercise)
• To find M2
given M1 and δ
1. find ν1 (for
given M1)
from table
2. get ν2 from
δ=ν2-ν1
3. look up ν2 in
table to find
M2
Prandtl Meyer -12
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
M1
M2
20°
M3
6°
M
2.00
2.01
2.02
2.03
2.04
2.50
2.51
2.52
2.53
2.54
2.55
2.56
2.57
2.58
2.59
ν
26.38
26.66
26.93
27.20
27.48
39.12
39.36
39.59
39.82
40.05
40.28
40.51
40.74
40.96
41.19
M
2.60
2.61
2.62
2.63
2.64
2.65
2.66
2.67
2.68
2.69
2.70
2.71
2.72
2.73
2.74
ν
41.41
41.64
41.86
42.09
42.31
42.53
42.75
42.97
43.19
43.40
43.62
43.84
44.05
44.27
44.48
M
2.75
2.76
2.77
2.78
2.79
2.80
2.81
2.82
2.83
2.84
2.85
2.86
2.87
2.88
2.89
ν
44.69
44.91
45.12
45.33
45.54
45.75
45.95
46.16
46.37
46.57
46.78
46.98
47.19
47.39
47.59
M
2.90
2.91
2.92
2.93
2.94
2.95
2.96
2.97
2.98
2.99
3.00
3.01
3.02
3.03
3.04
ν
47.79
47.99
48.19
48.39
48.59
48.78
48.98
49.18
49.37
49.56
49.76
49.95
50.14
50.33
50.52
M
3.05
3.06
3.07
3.08
3.09
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
ν
50.71
50.90
51.09
51.28
51.46
51.65
51.83
52.02
52.20
52.39
52.57
52.75
52.93
53.11
53.29
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µ1 = sin−1 1
M1
• Fan angle
– angle between first and last Mach
wave
– useful to determine when
expansion has ended in flowfield
for a given distance away from
wall
Fan Angle
M1
µ2 = sin−1 1
M2
µ1
µ2-δ
δ
M2
• From geometry
Fan Angle = µ1 − (µ 2 − δ )
= (µ1 − µ 2 ) + δ
= (µ1 − µ 2 ) + (ν 2 − ν1 )
Prandtl Meyer -14
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
(VIII.5)
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• Examine plot of ν as function of M
M1
δ=ν2-ν1
150
νmax=130.45°
δmax
M2
ν
100
ν=
50
(
)
γ + 1 −1 γ − 1 2
tan
M − 1 − tan −1 M 2 − 1
γ −1
γ +1
γ=1.4
p→
→0
0
0
10
• Example
20
M
M1 = 2 Þ δ max = 130.45o − 26.38o = 104.1o
M1 = 6 Þ δmax = 130.45o − 84.96o = 45.5o
Prandtl Meyer -15
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• As M increases, reach
maximum turn angle
(νmax~130.5° for γ=1.4)
• So as M1 increases,
max. angle flow can
turn (δmax) decreases
30
M1=2
δPM<104.1°
δ
M2
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• Analytic Expression
(
→ M; tan −1 (M ) → 90o
For M → ∞ :
ν max
ν max
)
γ + 1 −1 γ − 1 2
M − 1 − tan −1 M 2 − 1
tan
γ −1
γ +1
ν=
æ γ +1 ö o
= çç
− 1÷÷90 (VIII.6)
è γ −1 ø
ì 90o
ï
o
ï130.45
=í
o
ï159.2
ïî 208.5o
γ = 5/3
γ = 1.4
γ = 1.3
γ = 1.2
Prandtl Meyer -16
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• As γ decreases (higher
temperatures, bigger
molecules), maximum turn
angle increases
• δmax smaller in real flows
– T and p drop through turn
Þcondensation of gas
Þnonequilibrium flow
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• Already showed that it does
not matter if expansion turn
is sharp or smooth
M1>1
δ
M2>M1
– still get same solution,
P-M fan=infinite set of Mach waves
– unless we exceed the maximum turning angle, final
properties just function of total turn angle
– smooth turn just means expansion process takes
place over longer distance
Prandtl Meyer -17
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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• What happens if we have a smooth
concave turn?
– since flow direction change
is small, can still get set of
weak Mach waves
M2<M1
δ
M1>1
Prandtl-Meyer compression: δ=−−(νν2−ν1), so ν2−ν1<00
– however unlike expansions, compressions merge
– together they coalesce to form oblique shock
– flow that went through PM compression is isentropic,
outer flow has entropy rise (po loss)
– size of PM region depends on M1 and curvature
Prandtl Meyer -18
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
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