Normalshocks AE 3450

AE3450Normal Shocks -1

School of Aerospace Engineering

Copyright 2001 by Jerry M. Seitzman. All rights reserved.

Wave propagation is perpendicularto flow direction

Shock is nonequilibrium processinternally, but flow before shock (1) in equilibrium flow after shock (2) in equilibrium

1 2




Start with stationary shock can use steady equations

v1p11T1

v2p22T2

Use control volume analysis only need to consider properties before and after

shock (equilibrium) Equations first studied by Rankine (~1870) and

Hugoniot (~1877)

1 2




Conservation and state equations 1d, steady, inviscid except inside shock

adiabatic, only flow work

Mass

Momentum 112221 vmvmApAp =

v1p11T1h1a1

v2p22T2h2a2

AvAvApAp 21122221 =

2211 vvAm == (VII.1)

( )1221 vvAmpp =

(VII.2a)

2222

2111 vpvp +=+

(VII.2b)Energy o222

211 h2vh2vh =+=+ (VII.3)

Perfect GasState Eqns.

RTp = (a) dTcdh p= (b) RTa2 = (VI.2) 6 equations, 6 unknowns




Only functions of p and Solve for density ratio

Combine energy, momentum, massv1p11T1h1a1

v2p22T2h2a2

(VII.3) ( ) ( )( )2121222112 vvvv21vv21hh +==( )1221 vvAm

pp=

(VII.2a)

2121

vv11Am

+=

+

(VII.1)

TPG/CPG

(VII.5)121

2

1

2

pp

11

pp

111

+++

+

=

( ) ( )211212 11pp21hh +=(VII.4) True for all simple comp. substancesShock Hugoniot Eq.

( ) ( )112212p12 RpRp1RTTchh

==

(b) (a)




For TPG/CPG, entropy state equation p11s1

p22s2

1

2

1

2v12 lnRT

Tlncss

=

( ) ( )

=

=

1

1

2

11

221

1

2

1

2

v

12

ppln

TTln

css

(VII.6)

=

1

2

1

2

v

12

ppln

css

Entropy change as function of pressure and density ratios across shock




Combine with 2/1from (VI.25)

p11s1

p22s2

=

1

2

1

2

v

12

ppln

css

Violates2nd Law

p2




General problem is to find change in all

properties across shock Start by find changes based on M1, M2

then find M2 as function of M1 Velocity ratio, v2/v1

11

22

1

2

aMaM

vv

=

v11T1p1M1

v22T2p2M2

(VII.7)1

2

1

2

1

2

TT

MM

vv

=

Density ratio, 2/ 1

2211 vv =mass2

1

2

1

2

1

1

2

TT

MM

vv

==

(VII.8)




Temperature ratio, T2/T1

(VII.9)

+

+=

22

21

1

2

M2

11

M2

11

TT

2o1o2o1o TThh ==energy

+

+=

21

22

M2

111

M2

111

1o

2o

1o1

2o2

1

2

TT

TTTT

TT

=

1

v11T1p1M1

v22T2p2M2




Pressure ratio, p2/p1momentum

(VII.2b)

2222

2111 vpvp +=+

2222

2111 MppMpp +=+

( )( )

2

2

2

222

Mp

pM

RTM

aMv

=

=

==

( ) ( )222211 M1pM1p +=+

(VII.10)( )( )22

21

1

2

M1M1

pp

++

=

v11T1p1M1

v22T2p2M2




Mach Number, M2 =f(M1) combine all eqns.

212

1

1222

2

2 M2

11M1

MM2

11M1

M +

+=

+

+

1

2

2

1

vv

=mass(VII.1)

Expression of M2 as function of M1- solve

( )( ) 122122

pp

M1M1

=

++mom.

(VII.9)

+

+= 222

11

2 M2

11M2

11TTenergy

(VII.10)

2

1

22

11

vv

RTMRTM

=

M and a

(VI.2,3) 1122

1

2

RTpRTp

=

PG

(a)

v11T1p1M1

v22T2p2M2




Remove square roots by squaring both sides and solve resulting quadratic

( ) ( )

+

+=

+

+2122

1

212

2222

22 M

211

M1

MM2

11M1

M

( ) ( )[ ] ( ) 0MgMg21MMg2

1M 11221

242 =+

g(M1)

No shocksolution

21

21

21

22 Mor

1M1

21

2MM

+

=

(VII.11)

v11T1p1M1

v22T2p2M2




M1 M2

1M1

21

2MM

21

21

22

+

=

In reference frame of normal shock, flow aftershock is always subsonic

M11 Expansion

Shock=1.4

0

1

2

3

4

5

6

7

0 1 2 3

M1

M

2




11lim

1

21pp 12

+=

>>

p increase across normal shock is greatest static property change

Density ratio and velocity ratio approach limit

=1.4

0

5

10

15

20

25

30

35

40

1 2 3 4 5 6 7

M1

T

2

/

T

1

,

p

2

/

p

1

,

2

/

1

p2/p1

T2/T1

2/1,v1/v2

v11T1p1M1

v22T2p2M2

T, p and increase, v decreases

from(VII.5)

1

2

1

2

1

2

pp

11

pp

111

+++

+

=




Stagnation Temperature, To2=To1 Stagnation Pressure

M1To1po1o1

M2To2po2o2

1

21

21

22

1

2

11o

22o

1o

2opp

M2

11

M2

11

pp

pppp

pp

+

+

==

(VII.12)

11M

12

pp 2

11

2

+

+

=

M2 from (VII.11)( )( )22

21

1

2

M1M1

pp

++

=

from (VII.10)

+

+

+

+=

11

21

1

21

21

1o

2o

11M

12

M2

11

M2

1

pp

(VII.13)




Other Useful Expressions M1To1po1o1

M2To2po2o2

1

21

21

22

1o

2opp

M2

11

M2

11

pp

+

+

=

1

2

2

2o

1

2opp

pp

pp

=

1

21

2

1

1o

2opp

TT

pp

=

(VII.14)

1

2122

1

2oppM

211

pp

+= (VII.15)




Stagnation Density, o2/o1

(VII.16)1o2o

1o

2o

pp

=

1

Sonic Area Ratio, A2/A1* *

1o

2o

1o

2o

1o

2oRTRT

pp

=

P.G.

M1To1po1o1A1*

M2To2po2o2 A2*

from(VI.19)

( )( )

=

ff

RTpRTp

AA

mm

1o1o

2o2o*1

*2

1

2

1 1 1

(VII.17)2o

1o*1

*2

pp

AA

=




po , o across normal shock drop (alot)

Entropy increases

M1po1o1A1*

M2po2o2 A2*

=1.4

0

1

2

3

4

1 2 3 4 5 6 7

M1

p

o

2

/

p

0

1

,

(

s

2

-

s

1

)

/

R

0

10

20

30

40

50

60

A*2 /A

*1

po2/po1=o2/o1

(s2-s1)/R

2/A*1 Sonic area ratio

increaseslarger throat required after shockto reach sonic flow (same mass flowrate, less po) o

*pAm




Given: Air flowing through constant area duct encounters stationary shock

1 2v1=500 m/sp1=50 kPaT1=250 K oncoming air 500 m/s,

50 kPa, 250K

Find:M2, T2, p2, v2po2 , To2 (relative to shock/duct)

Assume: Air is TPG, CPG, =1.4




Analysis:

1 2v1=500 m/sp1=50 kPaT1=250 K58.1

sm25020sm500

RTvM 11 ==

=

can either use equations or tables (e.g., Table B.1 for =1.4)M1 M2 p2/p1 T2/T1 po2/po1 2222 ////1111 A*2/A*1 po2/p1

1.50 0.7011 2.458 1.320 0.9298 1.862 1.076 3.4131.51 0.6976 2.493 1.327 0.9266 1.879 1.079 3.4511.52 0.6941 2.529 1.334 0.9233 1.896 1.083 3.4891.53 0.6907 2.564 1.340 0.9200 1.913 1.087 3.5281.54 0.6874 2.600 1.347 0.9166 1.930 1.091 3.5671.55 0.6841 2.636 1.354 0.9132 1.947 1.095 3.6061.56 0.6809 2.673 1.361 0.9097 1.964 1.099 3.6451.57 0.6777 2.709 1.367 0.9062 1.981 1.104 3.6851.58 0.6746 2.746 1.374 0.9026 1.998 1.108 3.7241.59 0.6715 2.783 1.381 0.8989 2.015 1.112 3.765

first calculate M1




Analysis (cont):1 2

v1=500 m/sp1=50 kPaT1=250 K

58.1M1 =

+= 158.114.1)4.1(2

14.1258.1M 2222 M2: B.1/VII.11

T2: B.1/VII.9 374.1675.0214.1158.1

214.11

TT 22

1

2=

+

+=

p2: B.1/VII.12 ( )[ ] ( )[ ] 746.214.14.058.14.04.12pp 212 =+=

v2: B.1/VII.7 ( ) 998.1374.11675.058.1vv 1221 ===

To2: A.1/VI.6 ( )K667.025058.1214.11K250TT 21o2o =

+==

po2: A.1/VII.15 ( ) 724.3746.2675.01pp 5.322 14.112o =+=

M2=0.675

T2=344 K

p2=137 kPa

v2=250 m/s

To2=375 K

po2=186 kPa




Given: Expanding nozzle with exit/inlet area ratio of 4.0 normal shock stands at entrance

Find:1. Mach number at nozzle exit2. (static) pressure at exit

1 2M=2.2po=101.3 kPa

ieAe/Ai=4.0

just before entranceHe, Mach 2.2 withstagnation pressure of 101.3 kPa

Assume: He is TPG/CPG with =5/3

Mepe




Analysis:

pe:

1 2M=2.2po=101.3 kPa

ie

Ae/Ai =4.0

Mepe

but need M2M1=Mi =2.2

( )( ) 0.4AA

AA

AA

2

e

M*

i

M*

e

i

e==

80.420.14 ==

( ) 988.0kPa3.101686.0pe = pe=68.7 kPa118.0Mo

1o1o

2o

oe

eoee

eppp

pp

pppp

=

==

VII.13/B.3 VI.7/A.3

581.0*i

M*e

AA4

AA

e

=

Me: get Me from area ratio

isentropic2e

VI.17,7M A/A* p/po

0.118 4.799 0.98840.581 1.198 0.09063.732 4.799 0.0132VI.17/A.3

VII.11/B.3

M1 M2 po2/po12.20 0.5813 0.6860

Spreadsheet of VII.11,13

M2=0.581

=73.3118.0

Me




Given: Two engine inlets, one astraight tube, the other a converging-diverging diffuser.

both with M=2.0 at their inlets, Ti=300.K, pi=0.50 atm

Compare performance of diffusers: Me, Te, pe, poe/poi

both slow flow down to same MeCD diffuser: isentropicallystraight diffuser: with shock

Assumptions: air is cpg/tpg with =1.4adiabatic diffusersisentropic except at shock

MeTepe

poe/poi

1 2Mi=2.0

pi=0.50 atmTi=300 K

i e

i e




Analysis: Shock Diffuser =1.4, VII.8-11, 13, 17 or

Me=M2=0.577 Te=T2= (T2/T1)T1=(1.688)300K= 506 K pe=p2= (p2/p1)p1=(4.50)0.5atm=2.25 atm; poe/poi = po2/po1= 0.721

CD Diffuser - isentropic (VI.6, 7, 17 or..)

MeTepe

poe/poi

1 2Mi=2.0

pi=0.50 atmTi=300 K

i e

i e

Me=0.577

pe= (Te/Ti)/-1pi=(506/300)3.50.5atm=3.12 atm poe/poi = 1

Te= = 506 K( )( ) K300556.0

938.0TTT

TTi

0.2oi

577.0oe =

same (M, To const)

higher p

no po loss higher mass flowrate or smaller area

(spreadsheet like B.1)

M1 M2 p2/p1 T2/T1 2/1 po2/po1 A*2/A*12.00 0.5774 4.500 1.688 2.667 0.7209 1.387

(spreadsheet like A.1)M A/A* T/To p/po

0.577 1.217 0.9376 0.79802.000 1.688 0.5556 0.1278

Normalshocks AE 3450

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