M a = 1.5

Ma = 1.5

supersonic subsonic

Tape along ceilingand wall of super-sonic nozzle.

NORMAL

SHOCK

Ma = 1.5

supersonic subsonic

NORMAL

SHOCK

• 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

Can occur in internal and external flow, must be between supersonic to subsonic, irreversible, “abrupt discontinuity” (0.2 microns ~ 10-5 in)

Ma = 1.7

Ma

=1.5

normal shock(irreversiblediscontinuity)

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

Thickness is about 0.2 microns (10-5 inches)“treat” as abrupt discontinuity

p, , T can be very largeDecelerations may be of the order

of tens of millions of gs

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

Propertyrelationsfor idealgas withcv and cp

constant

Cons. Of Mass

Cons. of Momentum

Cons. of Energy

2nd Law of Thermodynamics



Conservation of Mass

Shock width is extraordinarily thin so A1 = A2

1V1 = 2V2 = (dm/dt)/A



Momentum Equation

Shock width is extraordinarily thin so Rx is negligible;

dm/dt = VAp1 + 1V1

2 = p2 + 2V22

0

Quasi-One-Dimensional, Steady, dWs/dt = 0, dWshear/dt = 0, dW/dtother = 0,


First Law of ThermodynamicsConservation of Energy

No temperature gradients at stations 1 and 2 so adiabatic

h1 + V12 = h2 + V2

2 ; h01 = h02

0

h1 + V12 = h2 + V2

2 h01 = h02

h = cp T h01 = c0T01

h02 = c0T02

T01 = T02



Second Law of ThermodynamicsConservation of Energy

No temperature gradients at stations 1 and 2 so adiabatic

s22V2A - s11V1A 0; s2 > s1

Second Law of ThermodynamicsConservation of Energy

s22V2A - s11V1A 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)





Equation of State ~ Ideal Gas

(1)

(2)

(2)

(3)

(4)(5)

(6)

To recap:Major simplifying features ~Because so thin

A1 = A 2; Rx = 0Because boundaries of control volume are far from shockNo 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)

Fanno Line friction

Rayleigh Line heat tranfer

Normal shock must satisfy eqs: 1-6, so must lie on intersection of Fanno and Rayleigh lines.

IGNORE

BREATH

s1

h1

1

s2

h2

2

s2 > s1

Property Changes Across Shock

“explanation”

To = constant

Entropy increases forirreversible process

1

23

4

56

7

8

p02 < p01

T2 > T1

T = To/[1 + (k-1)M2/2]T2/T1 = [1 + (k-1)M1

2/2]/[1 + (k-1)M2

2/2]

p02 < p01 (prob. 11.2)

T2 > T1

1

23

4

56

7

8

1

23

4

56

7

8

Since T2 > T1, then h2 > h1

..so V2< V1

1

23

4

56

7

8

Since V decreases across shock,then must increase for V to

remain constant.

1

23

4

56

7

8

Since V is constant across shockand V decreases across shock, then

p must increase across shock.Can also see from Ts diagram.

1

23

4

56

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.

?Make sure you

know what goes in here

GIVEN

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

s1

h1

1

s2

h2

2

(1)

(2)

(2)

(3)

(4)(5)

(6)

6 equations 6 unknowns

Hard to solve, much easier if had ratios (e.g. p1/p2) in terms of M1

Ma1 Ma2

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)

To/T = 1 + {(k – 1)/2}M2

T01 = T02

1T2/T1 = f(M1, M2)