Supersonic Flow Turning

7/24/2019 Supersonic Flow Turning

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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






• 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






• 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



• 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



Prandtl Meyer -5



• 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)



Prandtl Meyer -6



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






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






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






( )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






• 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






• 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°






• 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






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)






• 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






• 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)






• 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






• 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:

Supersonic Flow Turning

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