School of Aerospace Engineering Normal Shock WavesSchool of Aerospace Engineering •Analysis: School of Aerospace Engineering.) –: –: –: : –: –:

1

Normal Shocks -1

School of Aerospace Engineering

Copyright © 2001, 2018, 2020 by Jerry M. Seitzman.

All rights reserved.

AE2010

Normal Shock Waves

• For a normal shock

– wave is perpendicular to

flow (propagation) direction

• Shock is nonequilibrium process

internally, but assume

– flow before shock (1) is in equilibrium

– flow after shock (2) is in equilibrium

1 2

Normal Shocks -2




AE2010

Approach to Finding Shock Properties

• Start with stationary shock (or solve problem in reference frame moving with shock)

can use steady equations

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

shock (equilibrium)

• Equations first studied by Rankine (~1870) and Hugoniot (~1877)

v1

p1

1

T1

v2

p2

2

T2

1 2

2

Normal Shocks -3




AE2010

Governing Equations

• Conservation and state equations

– 1d, steady, inviscid except inside shock, adiabatic, only flow work

Mass

Momentum 112221 vv mmApAp

v1

p1

1

T1

h1

a1

v2

p2

2

T2

h2

a2

AAApAp 2

11

2

2221 vv

2211 vv A

m(V.C1)

1221 vv A

mpp

(V.C2a)

2

222

2

111 vv pp

(V.C2b)Energy

ohhh 2v2v 2

22

2

11(V.C3)

Perfect Gas

State Eqns.

RTp (tpg) dTcdh p from III.6

RTa 2 from V.A3

• 6 equations, 6 unknowns

Normal Shocks -4




AE2010

Shock Density Ratio

• Combine energy, momentum, mass

• Solve for density ratio

v1

p1

1

T1

h1

a1

v2

p2

2

T2

h2

a2

(V.C3) 2121

2

2

2

112 vvvv2

1vv

2

1 hh

1221 vv

Am

pp

(V.C2a)

21

21

vv11

A

m(V.C1)

tpg/cpg

(V.C5)1

2

1

2

1

2

1

1

1

11

p

p

p

p

211212 112

1 pphh

(V.C4) true for all simple comp. substances

Shock Hugoniot Eq.

112212121

RpRp

RTTchh p

(III.6) (tpg)

only functions of p and

3

Normal Shocks -5




AE2010

Entropy Change Across Shock

• For tpg/cpg, entropy state equation p1

1

s1

p2

2

s2

1

2

1

212 lnln

R

T

Tcss v

1

1

2

11

22

1

1

2

1

212 lnln

p

p

T

T

c

ss

v

(V.C6)

1

2

1

212 lnp

p

c

ss

v

entropy change as function of pressure and density ratios

across shock

(tpg)

Normal Shocks -6




AE2010

Entropy Change Across Shock

• Combine s2/s1

result with 2/1

(from V.C5)

p1

1

s1

p2

2

s2

1

2

1

212 lnp

p

c

ss

v

Violates

2nd Law

solution with p2<p1

violates 2nd Law

0.01

0.1

1

10

100

-1 -0.5 0 0.5 1

(s2-s1)/cv

p2/p

1

=1.4

weak shocks

no expansion

“shocks”

1

2

1

2

1

2

1

1

1

11

p

p

p

p

4

Normal Shocks -7




AE2010

Mach Number Relations: v,

• General problem is to find change in

all properties across shock

• Start by finding changes based on

M1, M2

– then find M2 as function of M1

• Velocity ratio, v2/v1

• Density ratio, 2/ 1

11

22

1

2

v

v

aM

aM

v1

1

T1

p1

M1

v2

2

T2

p2

M2

(V.C7)

1

2

1

2

1

2

v

v

T

T

M

M

2211 vv mass2

1

2

1

2

1

1

2

v

v

T

T

M

M

(V.C8)

Normal Shocks -8




AE2010

Mach Number Relations: T

• Temperature ratio, T2/T1

(V.C9)

2

2

2

1

1

2

2

11

2

11

M

M

T

T

2121 oooo TThh energy

2

1

2

2

2

111

2

111

M

M

1

2

11

22

1

2

o

o

o

o

T

T

TT

TT

T

T

1

v1

1

T1

p1

M1

v2

2

T2

p2

M2

5

Normal Shocks -9




AE2010

Mach Number Relations: p

• Pressure ratio, p2/p1

momentum(V.C2b)

2

222

2

111 vv pp

2

222

2

111 MppMpp

2

2

2

222v

Mp

pM

RTM

aM

2

22

2

11 11 MpMp

(V.C10) 2

2

2

1

1

2

1

1

M

M

p

p

v1

1

T1

p1

M1

v2

2

T2

p2

M2

Normal Shocks -10




AE2010

Mach Number Relations: M

• Mach Number, M2 =f(M1)

– combine all eqn’s.

2

12

1

12

22

2

2

2

11

12

11

1M

M

MM

M

M

1

2

2

1

v

v

mass(V.C1)

expression for M2

as function of M1 -can solve

1

2

2

1

2

2

1

1

p

p

M

M

mom.

(V.C9)

2

2

2

1

1

2

2

11

2

11 MM

T

T energy(V.C10)

2

1

22

11

v

v

RTM

RTM

M and a(V.A3)

11

22

1

2

RTp

RTp

tpg

v1

1

T1

p1

M1

v2

2

T2

p2

M2

6

Normal Shocks -11




AE2010

Mach Number Relations: M (con’t)

• Remove square roots by squaring

both sides, solve resulting quadratic

2

122

1

2

12

222

2

2

2

2

11

12

11

1M

M

MM

M

M

0212

111

2

21

24

2

MgMgMMgM

g(M1)

“No shock”

solution

2

12

1

2

12

2 or

11

2

1

2

M

M

M

M

(V.C11)

v1

1

T1

p1

M1

v2

2

T2

p2

M2

Normal Shocks -12




AE2010

Mach Change Across Normal Shock

M1 M2

11

2

1

2

2

1

2

12

2

M

M

M

in shock’s reference

frame, flow after

normal shock is

always subsonic

M1<1,

M2>1

Expansion

Shock

=1.4

0

1

2

3

4

5

6

7

0 1 2 3

M1

M2

7

Normal Shocks -13




AE2010

Property Ratios - Results

• p ratio increase across normal shock is greatest static property change

• and v ratios approach limit

=1.4

0

5

10

15

20

25

30

35

40

1 2 3 4 5 6 7

M1

T2/T

1,p

2/p

1,

2/

1p2/p1

T2/T1

2/1,v1/v2

• T, p and increase across shock and ratios increase withM1

• v decreases

from

(V.C5)

1

2

1

2

1

2

1

1

1

11

p

p

p

p

v1

1

T1

p1

M1

v2

2

T2

p2

M2

1

1lim

1

2

112

pp

Normal Shocks -14




AE2010

Stagnation Properties Across Shock

• Stagnation Temperature, To2=To1

• Stagnation Pressure

M1

To1

po1

o1

M2

To2

po2

o2

1

2

1

2

1

2

2

1

2

11

22

1

2

2

11

2

11

p

p

M

M

p

p

pp

pp

p

p

o

o

o

o

(V.C12)

1

1

1

2 2

1

1

2

M

p

p

M2 from (V.C11)

2

2

2

1

1

2

1

1

M

M

p

p

from (V.C10)

1

1

2

1

1

2

1

2

1

1

2

1

1

1

2

2

11

2

1

M

M

M

p

p

o

o

(V.C13)

8

Normal Shocks -15




AE2010

Stagnation Pressure

• Other Useful Expressions

1

2

1

2

1

2

2

1

2

2

11

2

11

p

p

M

M

p

p

o

o

1

2

2

2

1

2

p

p

p

p

p

p oo

1

21

2

1

1

2

p

p

T

T

p

p

o

o

(V.C14)

1

21

2

2

1

2

2

11

p

pM

p

po

(V.C15)

M1

To1

po1

o1

M2

To2

po2

o2

Normal Shocks -16




AE2010

Stagnation Properties (con’t)

• Stagnation Density, o2/o1

• Sonic Area Ratio, A*2/A

*1

(V.C16)

1

2

1

2

o

o

o

o

p

p

1

1

2

1

2

1

2

o

o

o

o

o

o

RT

RT

p

p

tpg

M1

To1

po1

o1

A1*

M2

To2

po2

o2

A2*

from

(V.B4)

f

f

RTp

RTp

A

A

m

m

oo

oo

11

22

*

1

*

2

1

2

1 1 1

(V.C17)2

1

*

1

*

2

o

o

p

p

A

A

9

Normal Shocks -17




AE2010

Stag. Press., Entropy and A* - Results

• Stagnation pressure and stagnation density drop

• Entropy increases

• Sonic areaincreases

– larger throat required after shock to reach sonic flow (same mass flowrate, lower po)

M1

po1

o1

A1*

M2

po2

o2

A2*

=1.4

0

1

2

3

4

1 2 3 4 5 6 7

M1

po

2/p

01,(

s2-s

1)/

R

0

10

20

30

40

50

60

A*2

/A*1

po2/po1=o2/o1

(s2-s1)/R

A*2/A

*1

opAm *

Normal Shocks -18




AE2010

Examples: Problem 1

• Given: Air flowing through constant area duct encounters stationary shock

– oncoming air 500 m/s,50 kPa, 250K

• Find:

– M2, T2, p2, v2

– po2 , To2 (relative to shock/duct)

• Assume: air is tpg/cpg with =1.4

1 2v1=500 m/sp1=50 kPaT1=250 K

10

Normal Shocks -19




AE2010

Solution: Problem 1• Analysis:

– first calculate M1

– can use equations or tables (e.g., App. B for =1.4)

1 2v1=500 m/sp1=50 kPaT1=250 K

58.125020

500v11

sm

sm

RTM

M1 M2 p2/p1 T2/T1 po2/po1 r2 /r1 A*2/A

*1 po2/p1

1.50 0.7011 2.458 1.320 0.9298 1.862 1.076 3.413

1.51 0.6976 2.493 1.327 0.9266 1.879 1.079 3.451

1.52 0.6941 2.529 1.334 0.9233 1.896 1.083 3.489

1.53 0.6907 2.564 1.340 0.9200 1.913 1.087 3.528

1.54 0.6874 2.600 1.347 0.9166 1.930 1.091 3.567

1.55 0.6841 2.636 1.354 0.9132 1.947 1.095 3.606

1.56 0.6809 2.673 1.361 0.9097 1.964 1.099 3.645

1.57 0.6777 2.709 1.367 0.9062 1.981 1.104 3.685

1.58 0.6746 2.746 1.374 0.9026 1.998 1.108 3.724

1.59 0.6715 2.783 1.381 0.8989 2.015 1.112 3.765

Normal Shocks -20




AE2010

Solution: Problem 1 (con’t.)

• Analysis (con’t):1 2

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

58.11 M

158.1

14.1

)4.1(2

14.1

258.1 222

2M– M2:V.C11

– T2: V.C9 374.1675.02

14.1158.1

2

14.11 22

1

2

T

T

– p2: V.C12 746.214.1

14.158.1

14.1

4.12 2

1

2

p

p

– v2: V.C7 998.1374.11675.058.1v

v

1

2

2

1

– To2: V.A5 KKTT oo 667.025058.12

14.11250 2

12

– po2:V.C15 724.3746.2675.02

14.11

5.3

2

1

2

p

po

M2=0.675

T2=344 K

p2=137 kPa

v2=250 m/s

To2=375 K

po2=186 kPapo1=206 kPaV.A6

11

Normal Shocks -21




AE2010

Examples: Problem 2

• Given: Diverging nozzle with exit/inlet area ratio of 4.0– normal shock stands

at entrance

– just before entranceHe, Mach 2.2 withstagnation pressureof 101.3 kPa

• Find:1. Mach number at nozzle exit

2. (static) pressure at exit

• Assume: He is tpg/cpg, =5/3 for T range (unknown)

1 2M=2.2po=101.3 kPa

i

eAe/Ai=4.0

Mepe

Normal Shocks -22




AE2010

Solution: Problem 2

• Analysis:

– Me

– pe:

1 2M=2.2po=101.3 kPa

i

eAe/Ai =4.0

Mepe

– but need M2

M1=Mi 2.2

0.4

2

*

*

Mi

Me

i

e

AA

AA

A

Ae

80.420.14

988.03.101686.0 kPape pe=68.7 kPa

118.0

1

1

2

eMo

o

o

o

oe

eoee

p

pp

p

p

p

ppp

V.C13 V.A6

581.0

**4

A

A

A

A i

M

e

e

get from area ratio

isentropic2e

V.B2,A6

M A/A* p/po

0.118 4.799 0.9884

0.581 1.198 0.0906

3.732 4.799 0.0132V.B2

V.C11

M1 M2 po2/po1

2.20 0.5813 0.6860

solutions of

V.C11,13M2=0.581

73.3

118.0eM

12

Normal Shocks -23




AE2010

Examples: Problem 3

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

– both with M2.0 at their inlets, Ti300.K, pi0.50 atm

– both slow flow down to same Me

• CD diffuser: reversible flow• straight diffuser: with shock

• Find: compare performance of these 2 diffusers– Me, Te, pe, poe/poi

• Assumptions:

– air is cpg/tpg with =1.4

– adiabatic diffusers

– reversible in straight diffuser except at shock

MeTepe

poe/poi

1 2Mi=2.0

pi=0.50 atm

Ti=300 K

i e

i e

Normal Shocks -24




AE2010

Solution: Problem 3

• Analysis: “Shock Diffuser”– =1.4, solutions of V.C8-11, 13, 17

– 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 (solutions of V.A5-6, B2)

MeTepe

poe/poi

1 2Mi=2.0

pi=0.50 atm

Ti=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

KT

TT

TTi

oi

oe300

556.0

938.0

0.2

577.0

same (M, To const)

higher p

no po loss higher mass flowrate or smaller area

M1 M2 p2/p1 T2/T1 2/1 po2/po1 A*2/A

*1

2.00 0.5774 4.500 1.688 2.667 0.7209 1.387

M A/A* T/To p/po

0.577 1.217 0.9376 0.7980

2.000 1.688 0.5556 0.1278

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School of Aerospace Engineering Normal Shock WavesSchool of Aerospace Engineering •Analysis: School of Aerospace Engineering.) –: –: –: : –: –: