M a = 1.5

supersonic subsonic N O R M A L S H O C K Tape along ceiling and wall of supersonic nozzle. Ma = 1.5

supersonic subsonic N O R M A L S H O C K Ma = 1.5 • can occur if converging, diverging and constant area channels • shock wave involves supersonic to subsonic flow • associated with rapid deceleration, pressure and entropy rise • only To is constant across shock

normal shock (irreversible discontinuity) Ma = 1.5 Ma = 1.7 Can occur in internal and external flow, must be between supersonic to subsonic, irreversible, “abrupt discontinuity” (0.2 microns ~ 10-5 in) Interested in changes across shock rather than what’s happening in shock Important for design of inlets for high performance aircraft and supersonic wind tunnels

p, , T can be very large Decelerations may be of the order of tens of millions of gs Thickness is about 0.2 microns (10-5 inches) “treat” as abrupt discontinuity

Look at fundamental equations: cons. of mass, momentum and energy, 2nd Law of thermodynamics, property relations for an ideal gas with constant specific heats Because shock so thin, A1 = A 2 and Rx = 0; because control volume boundaries are far from shock, no gradients so no heat transfer

Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Cons. Of Mass Cons. of Momentum Cons. of Energy Property relations for ideal gas with cv and cp constant 2nd Law of Thermodynamics

Quasi-One-Dimensional, Steady, FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Conservation of Mass Shock width is extraordinarily thin so A1 = A2 1V1 = 2V2 = (dm/dt)/A

Quasi-One-Dimensional, Steady,FBx = 0, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Momentum Equation 0 Shock width is extraordinarily thin so Rx is negligible; dm/dt = VA p1+ 1V12 = p2+ 2V22

Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant First Law of Thermodynamics Conservation of Energy 0 No temperature gradients at stations 1 and 2 so adiabatic h1+ V12 = h2+ V22 ; h01 = h02

h1+ V12 = h2+ V22 h01 = h02 h = cp T h01 = c0T01 h02 = c0T02 T01 = T02

Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Second Law of Thermodynamics Conservation of Energy No temperature gradients at stations 1 and 2 so adiabatic s22V2A - s11V1A  0; s2 > s1

Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Second Law of Thermodynamics Conservation of Energy s22V2A - s11V1A  0; s2 > s1 2nd Law by itself is little help in calculating entropy. To calculate entropy use 1st and 2nd Laws, ideal gas, constant cp to get: s2 – s1 = cpln(T2/T1) – Rln(p2/p1)

Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0, effects of gravity = 0, ideal gas, cv, cp is constant Equation of State ~ Ideal Gas

(1) (2) (2) (3) (4) (5) (6) To recap: Major simplifying features ~ Because so thin A1 = A 2; Rx = 0 Because boundaries of control volume are far from shock No gradients of V, T, , p there Q/dm = 0

breath

(1) (2) (2) (3) (4) (5) (6) 6 unknowns: p1, 1, T1, s1, h1, and V1 6 equations: and one constraint:

(1) (2) (2) (3) (4) (5) (6) Rayleigh Line heat tranfer IGNORE Fanno Line friction Normal shock must satisfy eqs: 1-6, so must lie on intersection of Fanno and Rayleigh lines.

B R E A T H

s2 > s1 Property Changes Across Shock “explanation” s1 h1 1 s2 h2 2

To = constant Entropy increases for irreversible process 1 2 3 4 5 6 7 8

p02 < p01 T2 > T1 T = To/[1 + (k-1)M2/2] T2/T1 = [1 + (k-1)M12/2]/ [1 + (k-1)M22/2] p02 < p01(prob. 11.2) T2 > T1 1 2 3 4 5 6 7 8

Since T2 > T1, then h2 > h1 ..so V2< V1 1 2 3 4 5 6 7 8

Since V decreases across shock, then  must increase for V to remain constant. 1 2 3 4 5 6 7 8

Since V is constant across shock and V decreases across shock, then p must increase across shock. Can also see from Ts diagram. 1 2 3 4 5 6 7 8

Since V decreases across shock and T increases across shock then Ma = V / (kRT)1/2 must decrease supersonic – to – sonic across shock. 1 2 3 4 5 6 7 8

G I V E N ? Make sure you know what goes in here

Want: p02/p01 = f(M1); T2/T1 = f(M1); p2/p1 = f(M1); V2/V1 = f(M1); M2 = f(M1); 2/ 1 = f (M1)

6 unknowns 6 equations (1) (2) (2) (3) (4) (5) (6) Ma1 Ma2 s2 h2 2 s1 h1 1 Hard to solve, much easier if had ratios (e.g. p1/p2) in terms of M1

Strategy ~ Step #1: Obtain property ratios, p1/p2, T1/T2, etc. in terms of M1 and M2. Step #2: Develop relationship between M1 and M2. Step #3: Recast property ratios, p1/p2, T1/T2, etc. in terms of M1.

Want: p02/p01 = f(M1, M2); T2/T1 = f(M1,M2); p2/p1 = f(M1,M2); V2/V1 = f(M1,M2); M2 = f(M1); 2/ 1 = f (M1, M2)

1 T2/T1 = f(M1, M2) To/T = 1 + {(k – 1)/2}M2 T01 = T02

V2/V1 = f(M1, M2)

2/1 = f(M1, M2)

p2/p1 = f(M1, M2)

T2/T1 = f(M1, M2)

M2 = f(M1) =

M2 = f(M1) Let: L = M12(1 + M12 (k-1)/2) / (1 +k M12)2 after much algebra (M22)2 ((1/2)(k-1) – k2L) + M22(1-2kL) – L = 0 Gas Dynamics - John

ds > 0 M1> 0

Have: T2/T1, V2/V1, 2/1, p2 /p1 = f(M1,M2); M2= f(M1) Still need: p02/p01 = ? po/p = [1 + M2{(k – 1)/2}]k/(k-1)

Have: T2/T1, V2/V1, 2/1, p2 /p1 = f(M1,M2); M2= f(M1) Still need: p02/p01 = ?

And similarly can substitute for M2 = f(M1) for other properties to get:

E X A M P L E

T1 = 0oC p1 = 60 kPa (abs) V1 = 497 m/s T2 = 87oC Find: M2, V2, P02

M a = 1.5

M a = 1.5

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