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
AE3450Normal Shocks -2
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -3
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -4
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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)
AE3450Normal Shocks -5
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -6
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
Combine with 2/1from (VI.25)
p11s1
p22s2
=
1
2
1
2
v
12
ppln
css
Violates2nd Law
p2
AE3450Normal Shocks -7
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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)
AE3450Normal Shocks -8
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -9
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -10
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -11
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -12
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -13
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
+++
+
=
AE3450Normal Shocks -14
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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)
AE3450Normal Shocks -15
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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)
AE3450Normal Shocks -16
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
=
AE3450Normal Shocks -17
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -18
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -19
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -20
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -21
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -22
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -23
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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
AE3450Normal Shocks -24
School of Aerospace Engineering
Copyright 2001 by Jerry M. Seitzman. All rights reserved.
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