unit - iii normal & oblique shocks

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