GDJP - UNIT - III

1. ME 6604 – GAS DYNAMICS AND JET PROPULSION
UNIT – III SHOCK WAVES
A.SANKARA NARAYANA MURTHY
ASSISTANT PROFESSOR
DEPARTMENT OF MECHANICAL ENGG
KAMARAJ COLLEGE OF ENGG & TECH

2. Significance of Compressible Flows
• Isentropic Flow
a. Entropy remains constant
b. All the stagnation parameters remain constant
• Fanno Flow
a. Entropy changes (Non-isentropic process)
b. Stagnation temperature remains constant
• Rayleigh Flow
a. Entropy changes (Non-isentropic process)
b. Stagnation temperature changes
• Flow Through Shock Waves
a. Entropy increases (Non-isentropic process)
b. Stagnation temperature constant
Note: Isentropic flow tables are used for
All the flows. How?

3. Difference between Normal and Oblique Shock Waves
Normal Shock Wave Oblique Shock Wave
Occurs in supersonic flow
(M1 >1)
Occurs in supersonic flow
(M1 >1, No difference)
Normal to the Flow direction Inclined at some other angle
Flow direction does not
change (Flow Deflection
angle = 0)
Flow direction changes (Flow
Deflection angle  0)
The downstream of the shock
wave flow is always subsonic
(M2 <1)
The downstream of the shock
wave flow may be subsonic
or supersonic (M2 <1 or M2
> 1)
Shock strength is high Shock strength is low

4. Applications of Shock Waves
– Supersonic flow over aerodynamic bodies-
Supersonic airfoils, wings, fuselage and other
parts of airplanes
• Supersonic flow through engine components-
Engine intake, Compressors, combustion chambers,
Nozzle
• Propellers running at high speeds

5. Supersonic Flow Turning
• For normal shocks, flow is perpendicular to shock
– no change in flow direction
• How does supersonic flow change direction,
i.e., make a turn
– either slow to subsonic ahead of turn (can
then make gradual turn) =bow shock
M1 M2
M1>1
M1>1
– go through non-normal wave with sudden
angle change , i.e., oblique shock (also
expansions: see later)
• Can have straight/curved, 2-d/3-d oblique shocks
– will examine straight, 2-d oblique shocks

6. Oblique Shock Waves
• Mach wave
– consider infinitely thin body M1>1

• Oblique shock
– consider finite-sized wedge,
half-angle, 
M1>1

– no flow turn required

 1
1
M/1sin 
– infinitessimal wave

– flow must undergo compression
– if attached shock
 oblique shock at angle 
– similar for concave corner

M1>1


7. Equations of Motion
• Governing equations
– same approach as for normal shocks
– use conservation equations and state equations
• Conservation Equations
– mass, energy and momentum
– this time 2 momentum equations - 2 velocity
components for a 2-d oblique shock
• Assumptions
– steady flow (stationary shock), inviscid except
inside shock, adiabatic, no work but flow work

8. Control Volume
• Pick control volume along
shock
p1, h1,
T1, 1
v1
 
p2, h2,
T2, 2
v2

-
• Divide velocity into two
components
– one tangent to shock, vt
– one normal to shock, vn
• Angles from geometry
v1n
v2n
v1n
v2nAn
At
p1
p2
1 2
v1t v2t
v1t v2t
v1t
v2t
– v1n= v1sin; v1t=v1cos
– v2n= v2sin(-); v2t=v2cos(-)
– M1n= M1sin; M1t=M1cos
– M2n= M2sin(-); M2t=M2cos(-)

9. Conservation Equations
• Mass
v1n
v2nAn
At
p1
p2
1 2
v1t v2t
v1t
v2t
  Adnv0 rel
CS

 
2
A
v
2
A
vAv
2
A
v
2
A
vAv
t
t22
t
t11nn22
t
t22
t
t11nn11


• Momentum   
CS
rel
CS
AdnvvAdnp

tangent        nn22t2nn11t1
t
21
t
21 AvvAvv
2
A
pp
2
A
pp 
t2t1 vv 
normal    nn22n2nn11n1n2n1 AvvAvvApAp 
n22n11 vv  (1)
2
n22
2
n1121 vvpp  (2)

10. Conservation Equations (con’t)
• Energy
v1n
v2nAn
At
p1
p2
1 2
v1t v2t
v1t
v2t


















2
v
hAv
2
v
hAv
2
2
2nn22
2
1
1nn11
2
v
2
v
h
2
v
2
v
h
2
t2
2
n2
2
2
t1
2
n1
1 
2
v
h
2
v
h
2
n2
2
2
n1
1  (3)
• Eq’s. (1)-(3) are same equations used to characterize
normal shocks ( with vnv)
• So oblique shock acts like normal shock in direction
normal to wave
– vt constant, but Mt1Mt2 
11t
22t
1t
2t
av
av
M
M
(4)2
1
1t
2t
T
T
M
M

1

11. Oblique Shock Relations
• To find conditions across
shock, use M relations from
normal shocks,
• but replace
M1  M1 sin
M2  M2 sin(-)
• Mach Number
  1sinM
1
2
1
2
sinMsinM 22
1
22
1
22
2 
















1M
1
2
1
2
MM 2
1
2
1
2
2 















from
p1, h1,
T1, 1
v1
 
p2, h2,
T2, 2
v2

-
v1n
v2n
v1t v2t
(5)

12. Oblique Shock Relations (con’t)
• Static Properties
p1, h1,
T1, 1
v1
 
p2, h2,
T2, 2
v2

-
v1n
v2n
v1t v2t
(6)1
1
sinM
1
2
p
p 22
1
1
2






(from pressure ratio of NS wave)
(7)
 
  2sinM1
sinM1
v
v
22
1
22
1
1
2
n2
n1






(from density ratio of NS wave)
   121sinM
1sinM
1
2
sinM
2
1
1
T
T
222
1
22
1
22
1
1
2




















(8)
(from temperature ratio of NS wave)

13. Oblique Shock Relations (con’t)
• Stagnation Properties
1o2o TT 
To (from energy conservation)
   12
1
211o2o ppTTpp 































1
1
22
1
1
22
1
22
1
1o
2o
1
1
sinM
1
2
sinM
2
1
1
sinM
2
1
p
p
(9)

14. Use of Shock Tables
• Since just replacing
M1  M1sin
M2  M2sin(-)
– can also use normal
shock tables
– use M1'=M1sin to look up property ratios
– M2= M2'/sin(-), with M2' from normal shock tables
• Warning
– do not use p1/po2 from tables
– only gives po2 associated with v2n, not v2t
p1, h1,
T1, 1
v1
 
p2, h2,
T2, 2
v2

-
v1n
v2n
v1t v2t

15. Example #1
• Given: Uniform Mach 1.5 air
flow (p=50 kPa, T=345K)
approaching sharp concave
corner. Oblique shock produced
with shock angle of 60°
• Find:
1. To2
2. p2
3.  (turning angle)
• Assume: =1.4, steady, adiabatic, no work, inviscid except
for shock,….
=60°
M1=1.5
T1=345K
p1=50kPa

M2

16. Example #1 (con’t)• Analysis:
Determination of To
30.160sin5.1sinMM 1n1  
191.1
805.1
786.0)(
12
12
2



TT
pp
MTableNS n
– calculate normal component
   K5005.12.01K345
M
2
1
1TTT
2
2
111o2o






 

Determination of p2
    kPa3.90kPa50805.1pppp 1122 
Determination of 



2.11
191.160cos5.1
786.0
tan60 5.0
1







 

  t2n2 MMtan     211n221t1n2 TTcosMMTTMM 

v1t
v1
v1n
-
v2
v2n
v2t
  104.1sinMM n22 NOTE: Supersonic flow okay after oblique shock
=60°
M1=1.5
T1=345K
p1=50kPa

M2

17. Wave/Shock Angle
• Generally, wave angle  is not
given, rather know turning
angle 
• Find relationship between M1,
, and 
  
  22cosM
1sinMtan2
tan 2
1
22
1



 
v2
-
v2n
v1

v1nv1t
v2t
 
  2sinM1
sinM1
v
v
22
1
22
1
n2
n1



(from relation between V1n and V2n and Vel triangle)
 


tanv
tanv
t2
t1
1
(10)

18. Oblique Solution Summary
• If given M1 and turning angle, 
1. Find  from (iteration) or use oblique shock charts
2. Calculate M1n=M1sin
3. Use normal shock tables or Mach relations
4. Get M2 from M2=M2n/sin(-)
M1

M2

19. Example #2
• Given: Uniform Mach 2.4, cool,
nitrogen flow passing over 2-d
wedge with 16° half-angle.
• Find:
, p2/p1 , T2/T1 , po2/po1 , M2
• Assume: =1.4, steady, adiabatic, no work, inviscid except
for shock,….
M1=2.4 
M2

20. Example #2 (con’t)
• Analysis:
Determination of 
52.14.39sin4.2sinMM 1n1  
9227.0;335.1;535.2;6935.0)1.( 1212122  oon ppTTppMNSTable
– use shock relations calculate normal component
  75.1sinMM n22  Supersonic after shock
  
  22cos4.14.2
1sin4.2tan2
16tan 2
22



=16°M1=2.4 
M2
iterate 
4.39

21. Example #2 (con’t)
• Analysis (con’t):
38.21.82sin4.2sinMM 1n1  
5499.0;018.2;425.6;5256.0)( 1212122  oon ppTTppMNStable
– use shock relations calculate normal component
  575.0sinMM n22 
=16°M1=2.4 
M2
– a second solution for 
  
  22cos4.14.2
1sin4.2tan2
16tan 2
22



in addition to 39.4 
1.82
• Eqn generally has 2 solutions for : Strong and Weak oblique shocks
Now subsonic after shock
previous solution 2.535 1.335 0.9227
1.75

22. Strong and Weak Oblique Shocks
• As we have seen, it is possible to get two
solutions to equation (10)
– 2 possible values of  for given (,M1)
– e.g.,
      22cosM1sinMtan2tan 2
1
22
1 
M1=2.4 39.4°
M2=1.75
=16°
M1=2.4 82.1°
M2=0.575
=16°
• Examine graphical solution

23. Graphical Solution
• Weak shocks
–smaller 
–min=
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50
 (deg)
(deg)
0
10
20
30
40
50
60
70
80
90
M1=
1.25
1.5
2 2.5 3
5
10
100
=1.4M2<1
M2>1
Strong
Weak
=sin-1(1/M1)
– usually
M2>1
• Strong shocks
– max=90°
(normal shock)
– always M2<1
• Both for =0
– no turn for
normal shock or
Mach wave

24. Which Solution Will Occur?
• Depends on upstream versus downstream
pressure
– e.g., back pressure
– p2/p1 large for strong shock
small for weak shock
• Internal Flow
– can have p2 much higher
or close to p1
M>1
p1
pb,high
M>1
p1
pb,low
• External Flow
– downstream pressure
usually close to upstream
p (both near patm)
M>1
patm
patm

25. Maximum Turning Angle
• Given M1,
no straight oblique
shock solution for
>max(M)
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50
 (deg)
(deg)
0
10
20
30
40
50
60
70
80
90
M1=
1.25
1.5
2 2.5 3
5
10
100
=1.4
max(M=3)~34°
• Given ,
no solution for
M1<M1,min
• Given fluid (),
no solution for any
M1 beyond max
e.g., ~45.5° (=1.4)
max()

26. Detached Shock
• What does flow look like when no straight
oblique shock solution exists?
– detached shock/bow shock, sits ahead of
body/turn
M1>1
>max
bow shock can
cover whole
range of
oblique shocks
(normal to
Mach wave)
– normal shock at centerline (flow subsonic to negotiate
turn); curves away to weaker shock
asymptotes to Mach wave
M2<1
M2>1
M1>1 >ma
x

27. CRITICAL FLOW AND SHOCK WAVES
0.1 DivergenceDragCR MM
MCR
• Sharp increase in cd is combined effect of shock waves and flow separation
• Freestream Mach number at which cd begins to increase rapidly called Drag-
Divergence Mach number