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ME 6604 – GAS DYNAMICS AND JET PROPULSION
UNIT – III SHOCK WAVES
Mrs. N. PREMALATHAASSOCIATE PROFESSOR
DEPARTMENT OF MECHANICAL ENGG
Significance of Compressible Flows• Isentropic Flow a. Entropy remains constant b. All the stagnation parameters remain constant
• Fanno Flowa. 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?
Difference between Normal and Oblique Shock WavesNormal 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 angleFlow 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
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
Current Research• Design of supersonic combustor for scramjet engine
• Shock interactions
• Shock boundary layer interaction
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
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
11 M/1sin– infinitessimal wave
– flow must undergo compression– if attached shock
oblique shock at angle – similar for concave corner
M1>1
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
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
v1nv2n
v1n
v2nAn
At
p1p2
1 2
v1t v2t
v1t v2t
v1t v2t
– v1n= v1sin; v1t=v1cos– v2n= v2sin(-); v2t=v2cos(-)– M1n= M1sin; M1t=M1cos– M2n= M2sin(-); M2t=M2cos(-)
Conservation Equations• Mass
v1n
v2nAn
At
p1p2
1 2
v1t v2t
v1t v2t Adnv0 rel
CS
2Av
2AvAv
2Av
2AvAv
tt22
tt11nn22
tt22
tt11nn11
• Momentum CS
relCS
AdnvvAdnp
tangent nn22t2nn11t1t
21t
21 AvvAvv2
App2
App
t2t1 vv
normal nn22n2nn11n1n2n1 AvvAvvApAp
n22n11 vv (1)
2n22
2n1121 vvpp (2)
Conservation Equations (con’t)• Energy
v1n
v2nAn
At
p1p2
1 2
v1t v2t
v1t v2t
2vhAv
2vhAv
22
2nn22
21
1nn11
2v
2vh
2v
2vh
2t2
2n2
2
2t1
2n1
1
2vh
2vh
2n2
2
2n1
1 (3)
• Eq’s. (1)-(3) are same equations used to characterize normal shocks ( with vnv)
• So oblique shock acts like normal shock in direction normal to wave– vt constant, but Mt1Mt2
11t
22t
1t
2t
avav
MM
(4)2
1
1t
2t
TT
MM
1
Oblique Shock Relations• To find conditions across
shock, use M relations from normal shocks,
• but replaceM1 M1 sinM2 M2 sin(-)
• Mach Number
1sinM1
21
2sinMsinM 221
221
222
1M1
21
2MM 21
21
22
from
p1, h1,T1, 1
v1
p2, h2,T2, 2
v2
-
v1nv2n
v1t v2t
(5)
Oblique Shock Relations (con’t)• Static Properties
p1, h1,T1, 1
v1
p2, h2,T2, 2
v2
-
v1nv2n
v1t v2t
(6)11sinM
12
pp 22
11
2
(from pressure ratio of NS wave)
(7)
2sinM1sinM1
vv
221
221
1
2
n2
n1
(from density ratio of NS wave)
121sinM
1sinM1
2sinM2
11
TT
2221
221
221
1
2
(8)
(from temperature ratio of NS wave)
Oblique Shock Relations (con’t)• Stagnation Properties
1o2o TT To (from energy conservation)
121
211o2o ppTTpp
1
1
221
1
221
221
1o
2o11sinM
12
sinM2
11
sinM2
1
pp
(9)
Use of Shock Tables• Since just replacing
M1 M1sinM2 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
-
v1nv2n
v1t v2t
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=1.5T1=345Kp1=50kPa
M2
Example #1 (con’t)• Analysis:Determination of To
30.160sin5.1sinMM 1n1
191.1805.1
786.0)(
12
12
2
TTppMTableNS n
– calculate normal component
K5005.12.01K345
M2
11TTT
2
2111o2o
Determination of p2
kPa3.90kPa50805.1pppp 1122
Determination of
2.11191.160cos5.1
786.0tan60 5.01
t2n2 MMtan 211n221t1n2 TTcosMMTTMM
v1t
v1
v1n -v2
v2nv2t
104.1sinMM n22 NOTE: Supersonic flow okay after oblique shock
=60°
M1=1.5T1=345Kp1=50kPa
M2
Wave/Shock Angle• Generally, wave angle is not
given, rather know turning angle
• Find relationship between M1, , and
22cosM
1sinMtan2tan 21
221
v2-
v2n
v1
v1nv1t v2t
2sinM1
sinM1vv
221
221
n2
n1
(from relation between V1n and V2n and Vel triangle)
tanv
tanv
t2
t1
1
(10)
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 relations4. Get M2 from M2=M2n/sin(-)
M1
M2
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
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
1sin4.2tan216tan 2
22
=16°M1=2.4 M2
iterate 4.39
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
22cos4.14.21sin4.2tan216tan 2
22
in addition to 39.4 1.82
• Eqn generally has 2 solutions for : Strong and Weak oblique shocksNow subsonic after shock
previous solution 2.535 1.335 0.9227
1.75
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 21
221
M1=2.4 39.4°
M2=1.75
=16°
M1=2.4 82.1°M2=0.575
=16°
• Examine graphical solution
Graphical Solution• Weak shocks
– smaller –min=
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50 (deg)
(d
eg)
0
10
20
30
40
50
60
70
80
90
M1= 1.25
1.52 2.5 3
510
100
=1.4M2<1
M2>1
Strong
Weak
=sin-1(1/M1)
–usuallyM2>1
• Strong shocks–max=90°
(normal shock)–always M2<1
• Both for =0– no turn for
normal shock or Mach wave
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 higheror close to p1
M>1p1
pb,high
M>1p1
pb,low
• External Flow– downstream pressure
usually close to upstream p (both near patm)
M>1patm
patm
Maximum Turning Angle• Given M1,
no straight obliqueshock solution for >max(M)
0
10
20
30
40
50
60
70
80
90
0 10 20 30 40 50 (deg)
(d
eg)
0
10
20
30
40
50
60
70
80
90
M1= 1.25
1.52 2.5 3
510
100
=1.4
max(M=3)~34°
• Given , no solution forM1<M1,min
• Given fluid (),no solution for any M1 beyond max
e.g., ~45.5° (=1.4)
max()
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 >max
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
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