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AE3450 Prandtl Meyer -1
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• 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 (entropywould be destroyed)
– turn is gradual
(concave or convex)
M1>1
M1>1
M1>1 M2>M1
M1>1M2>M1
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AE3450 Prandtl Meyer -2
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Gradual turn is made up of
large number of infinitessimal
turns/corners
• Each turn has infinitessimal flowchange
– each turn produced byinfinitessimal wave
• Flow is uniform and isentropic between
each turn/corner – length between each is arbitrary
– could be zero length (sharp turn)
and waves collapse to one point
M1>1M2>M1
δ
1
11 M
1sin−=µ
Ma M b
a
1a M
1sin−=µ
2
12 M
1sin−=µ
M1
M2
δ
1
11 M
1sin−=µ
2
12 M
1sin−=µM1
M2
δ
Mach wave
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AE3450 Prandtl Meyer -3
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Problem
– given upstream conditions (1)
and turning angle (δ) – find downstream conditions (2)
• Goal
– Mach number relations (similar to shock relations)
• Equations
– use mass, momentum, energy conservation, Machnumber def’n., state equations
• Assumptions
– steady flow, quasi-1d, reversible+adiabatic (isentropic)
M1M2δ
Prandtl Meyer Fan
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Prandtl Meyer -4
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Approach
– begin with single Mach wave that expands supersonic
flow through an infinitessimal (differential) angle of
magnitude d ν
• Mass/Momentum Conservation
– using same type of approach as for
oblique shocks (two momentum components: t, n)
d ν
µv
v+dv
µ
vnvt
µ+d ν
vn+dvnvt
– find lack of pressure gradient tangent to wave gives
vt=constant across wave
– essentially using differential
control volume
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Prandtl Meyer -5
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Use vt=constant
( ) ( )
( )( ) νµ− νµ+=
ν+µ+=µ
=
dsinsindcoscosdvv
dcosdvvcosv
vv downstream,tupstream,t d νµv
v+dv
µ
vnvt
µ+d ν
vn+dvnvt
d ν→0 →1 →d ν µ ν−µ+µ ν−µ=µ sindvdcosdvsinvdcosvcosv
ν
µ
µ= d
cos
sin
v
dv→0
sinµ=1/M
sin2µ+cos2µ=1 ν−
= dM/11
M/1
v
dv2
2
ν−
= d1M
1
v
dv
2(VIII.1)
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Prandtl Meyer -6
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
ν νν ν• Relate v and M
d νµv
v+dvtpg/cpg
a2=γ RT
v=Maada
MdM
vdv +=
T
dT
2
1
M
dM
T
Td
M
dM
v
dv+=+=
( )
M
dM
M
2
11
M1
T
dT
M2
11
M
2
11d
TdT0
TdT
.constM2
11TT
2
2
2
2
o
o
2o
−γ +
−γ −=
−γ +
−γ +
+==
=
−+=
( )
−γ
+
=
−γ
+
−γ
−=22
2
M2
1
1
1
M
dM
M2
1
1
M2
1
1M
dM
v
dv
( )[ ]
M
dM
M
2
11
M21
M
dM
v
dv
2
2
−γ +
−γ −=
Energy Conservation
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AE3450 Prandtl Meyer -7
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
ν νν ν• Relate VIII.1 and last eqn.
−γ +
= ν−
=22
M2
11
1
M
dMd
1M
1
v
dv
(VIII.2)M
dM
M2
11
1Md
2
2
−γ +
−= ν dM is change in Mach number
associated with d ν turn angle
d νµM
M+dM
integrate −γ
+
−= ν= ν− ν
ν
ν
2
1
2
1
M
M2
2
12M
dM
M2
1
1
1Md
δ
M1
M2
• Need finite angle, δ= ν2- ν1 and finite ∆M
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AE3450 Prandtl Meyer -8
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
ν νν ν
−γ +
−
= ν− ν
2
1
M
M2
2
12 M
dM
M2
11
1M δ
M1
M2
(VIII.3)
( )
2
1
M
M
2121
121Mtan1M
1
1tan
1
1
−−−
+γ
−γ
−γ
+γ = ν− ν −−
• Perform Integration
• 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
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AE3450 Prandtl Meyer -9
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
( )2
1
M
M
212112 1Mtan1M
1
1tan
1
1
−−−
+γ
−γ
−γ
+γ = ν− ν −−
• Want to find ν= ν(M) [really ν2= ν2(M2)] for any M – need to choose (arbitrary) reference condition,
i.e., pick an M where ν=0
– let’s choose ν=0 at M=1
• analagous to table of h(T)
– really h(T)-h(Tref )
– just chosen h(Tref )=0
( ) 1Mtan1M1
1tan
1
1 2121 −−−+γ
−γ
−γ
+γ = ν −−
(VIII.4)
• ν represents angle through which asonic flow wouldhave to turnto reach M
Appendix D (John)
for γ =1.4, M≤5
δ= ν2- ν1
M1
M2
M=1 ν1
ν1 is turn angle for M=1 to M1
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AE3450 Prandtl Meyer -10
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• 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
• To find T2, p2, ....
– use isentropic flow relations since expansion is
isentropic (no shock)
– e.g.,
δ= ν2- ν1
M1
M2
2o M2
11
T
T −+=
2
2
21
1
2
M2
11
M2
11
T
T
−γ +
−+
=
To=const
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AE3450 Prandtl Meyer -11
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Given: Uniform Mach 2 flow of
nitrogen at 300 K flows over compoundwall corner: two turns, 20° and 6°
• Find:
M and T after final turn
• Assume: N2 is TPG/CPG with γ =1.4, steady, adiabatic, no
work, inviscid,….
20°
M1
6°
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AE3450 Prandtl Meyer -12
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Analysis: (class exercise)
20°
6°
M2
M3
M1
M ν νν ν M ν νν ν M ν νν ν M ν νν ν M ν νν ν
2.00 26.38 2.60 41.41 2.75 44.69 2.90 47.79 3.05 50.71
2.01 26.66 2.61 41.64 2.76 44.91 2.91 47.99 3.06 50.90
2.02 26.93 2.62 41.86 2.77 45.12 2.92 48.19 3.07 51.09
2.03 27.20 2.63 42.09 2.78 45.33 2.93 48.39 3.08 51.28
2.04 27.48 2.64 42.31 2.79 45.54 2.94 48.59 3.09 51.46
2.50 39.12 2.65 42.53 2.80 45.75 2.95 48.78 3.10 51.65
2.51 39.36 2.66 42.75 2.81 45.95 2.96 48.98 3.11 51.83
2.52 39.59 2.67 42.97 2.82 46.16 2.97 49.18 3.12 52.02
2.53 39.82 2.68 43.19 2.83 46.37 2.98 49.37 3.13 52.20
2.54 40.05 2.69 43.40 2.84 46.57 2.99 49.56 3.14 52.39
2.55 40.28 2.70 43.62 2.85 46.78 3.00 49.76 3.15 52.57
2.56 40.51 2.71 43.84 2.86 46.98 3.01 49.95 3.16 52.75
2.57 40.74 2.72 44.05 2.87 47.19 3.02 50.14 3.17 52.93
2.58 40.96 2.73 44.27 2.88 47.39 3.03 50.33 3.18 53.11
2.59 41.19 2.74 44.48 2.89 47.59 3.04 50.52 3.19 53.29
• To find M2
given M1 and δ1. find ν1 (for
given M1)
from table
2. get ν2 from
δ= ν2- ν1
3. look up ν2 in
table to findM2
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AE3450 Prandtl Meyer -14
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
2
12 M
1sin−=µ
1
11 M
1sin−=µ
M1
M2
δ
Fan Angle• Fan angle
– angle between first and last Machwave
– useful to determine whenexpansion has ended in flowfieldfor a given distance away fromwall
• From geometry
µ2-δ
µ1
( )δ−µ−µ= 21AngleFan
( ) δ+µ−µ= 21
( ) ( )1221 ν− ν+µ−µ=
(VIII.5)
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AE3450 Prandtl Meyer -15
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Examine plot of ν νν ν as function of M
• As M increases, reachmaximum turn angle
( νmax~130.5° for γ =1.4)• So as M1 increases,
max. angle flow can
turn (δmax) decreases
δ= ν2- ν1
M1
M2
0
50
100
150
0 10 20 30
M
ν νν ν
νmax=130.45°
γ=1.4
( ) 1Mtan1M1
1tan
1
12121 −−−+γ
−γ
−γ
+γ = ν −−
5.4596.8445.1306M max1 =−=δ
=
• Example
1.10438.2645.1302M max1 =−=δ=
δmax
δ
M1=2
M2δPM<104.1°
p→→→→0
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AE3450 Prandtl Meyer -16
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Analytic Expression
• As γ decreases (highertemperatures, biggermolecules), maximum turnangle increases
• δmax smaller in real flows – T and p drop through turn
condensation of gas
nonequilibrium flow
( ) 90Mtan;M:MFor 1 →→∞→ −
( ) 1Mtan1M11tan
11 2121 −−−
+γ −γ
−γ +γ = ν −−
=γ
=γ
=γ
=γ
= ν
2.15.208
3.1
4.1
2.159
45.130
3/590
max
9011
1max
−
−γ +γ = ν
(VIII.6)
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AE3450 Prandtl Meyer -17
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• Already showed that it does
not matter if expansion turnis sharp or smooth
– still get same solution,
P-M fan=infinite set of Mach waves
M1>1
M2>M1
δ
– 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
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AE3450 Prandtl Meyer -18
School of Aerospace Engineering
Copyright © 2001 by Jerry M. Seitzman. All rights reserved.
• What happens if we have a smooth
concave turn? – since flow direction change
is small, can still get set of
weak Mach waves
M1>1
M2<M1
δ
– 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
δδδδ=−−−−( ν νν ν2−ν−ν−ν−ν1), so ν νν ν2−ν−ν−ν−ν1<0000Prandtl-Meyer compression: