Speed of Propagation

“speed of sound is one of the most important quantities in the study of incompressible flow”

- Anderson : Moderns Compressible Flow

c

SOUND IS A LONGITUDINAL WAVE

dVx

c = ?

Speed of Propagation

• sound wave are propagated by molecular collisions

• sound wave causes very small changes in p, , T

• sound wave by definition is weak(relative to ambient)shock waves are strong(relative to ambient) and travel faster

Speed of Propagation = Isentropic

• changes within wave are small• gradients are negligible particularly for long waves

implies irreversible dissipative effects due to friction and conduction are negligible

• no heat transfer through control volume

implies adiabatic

ISENTROPIC

unsteady

steady

(1) at any position, no properties are changing with time

(2) V and are only functions of x

SOUNDSPEED

(dVxA >> ddVxA)(dVx)A = (d)cA

dVx = (c/) d

cA = (+d)(c-dVx)(A)cA = cA- d(Vx)A +(d)cA - (d)(dVx)A

c

dVx

This terms appears only if CV is accelerating

dRx

dRx represents tangential forces on control volume; because there is no relative motion along wave (wave is on both sides

of top and bottom of control volume), dRx =0. So FSx = -Adp

Total forces = normal surface forces Change in momentum flux

From continuity eq.

Cons. of mass

Cons. of mass

From momentum eq.

From continuity eq.

dp/d = c2

c = [dp/d]1/2

adiabatic? c = [dp/d]s1/2

or isothermal? c = [dp/d]T

1/2

Speed of Propagation

Isentropic & Ideal Gas

Correct Answer Wrong Reasoning

For ideal gas, isentropic, constant cp and cv:

p/k = const p = const k const = p/ k

dp/ds = d(const k)/d = kconstk-1 dp/ds = kp/

dp/ds = k RT/ = kRT

c = (kRT)1/2 ~ 340 m/s~ 1120 ft/s, for air at STP

c = [dp/ds]1/2 = [kRT]1/2

[krT]1/2 ¾ molecular velocity for a perfect gas = [8RT/]1/2

Note: the adiabatic approximation is better at lower frequencies than higher frequencies because the heat

production due to conduction is weaker when the wavelengths are longer (frequencies are lower).

“The often stated explanation, that oscillations in a sound wave are too rapid to allow appreciable conduction of heat,

is wrong.”~ pg 36, Acoustics by Allan Pierce

Newton was the first to predict the velocity of sound waves in air. He used Boyles Law

and assumed constant temperature.

c2 = dp/d = p/|T

FOR IDEAL GAS: p = RT p/ = const if constant temperatureThen: dp/d = d(RT)/d = RT

c = (RT)1/2 ~ isothermal(k)1/2 too small or (1/1.18) (340 m/s) = 288 m/s

Speed of sound (m/s)steel 5050seawater 1540water 1500air (sea level) 340

Moving Sound Source

Shock wave of bullet piercing sheet of Plexiglass bending of shock due to changes in p and T

V = 0; M = 0 V < c; M < 1

V = c; M = 1 V > c; M > 1

.

As measured by the observer the frequency of sound coming from the approaching siren is greater than the frequency of sound from the receding siren.

shock increases pressure

vt

ct

sin = ct/vt = 1/M = sin-1

(1/M)

Mach (1838-1916)

First to make shock waves visible.

First to take photographs of projectiles in flight.

Turned philosopher –“psychophysics”: all knowledge is based on sensations

“I do not believe in atoms.”

(1)What do you put in a toaster?

(2) Say silk 5 times,what do cows drink

(3) What was the first man-made object to break the sound barrier?

POP QUIZ

Tip speed ~ 1400 ft/sM ~ 1400/1100 ~ 1.3

Sound Propagation Problems

Lockheed SR-71 aircraft cruises at around M = 3.3 at an altitude of 85,000 feet (25.9 km). What is flight speed?

Table A.3, pg 71924km T(K) = 220.626km T(K) = 222.525,900m ~ 220.6 + 1900m * (1.9K/2000m) ~ 222 K

PROBLEM 1 (faster than a speeding bullet)

c = {kRT}1/2 = {1.4*287 [(N-m)/(kg-K)] 222 [K]}1/2 = 299 m/s

V = M*c = 3.3 * 299 m/s = 987 m/s

The velocity of a 30-ob rifle bullet is about 700 m/s

Vplane / Vbullet = 987/700 ~ 1.41

PROBLEM 1

Not really linear, althoughnot apparent at the scale

of this plot.

For standard atm. conditionsc= 340 m/s at sea levelc = 295 m/s at 11 km

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(a) What is airspeed of aircraft?(b) What is time between seeing aircraft overhead and hearing it?

PROBLEM 2

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(a) What is airspeed of aircraft?V (airspeed) = Mc = 1.35 * (kRT)1/2 = 1.35 * (1.4*287 [N-m/kg-K] *303 [K])1/2

= 471 m/s (relative to air)

PROBLEM 2

M = V/cV is airspeed

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

* note: if T & not constant, Mach line would not be straight

vt

ct

v is velocityrelative to earth= 471 – 10= 461 m/s

sin = c/v = 1/M

time to travel this distance = distance /velocity of plane relative to earth

3 km

M = 1.35

T = 303 oK

Wind = 10 m/s

(b) What is time between seeing aircraft overhead and hearing it?

= sin-1 (1/M) = sin-1 (1/1.35) = 47.8o

Vearth = 471m/s – 10m/s = 461m/sD = Veartht = 461 [m/s] t = 3000[m]/tan() t = 5.9 s

D = Vearth t

3000m

* note: if T & not constant, Mach line would not be straight

Problem #3

Prove that for an ideal calorically perfect gas that M2 is proportional to:

(Kinetic Energy per unit mass = V2/2)

(Internal Energy per unit mass = u)

Hint: Use ~ u = cvT; cv = R/(k-1); c = (kRT)1/2

show that proportionality constant = k(k-1)/2

Problem #4

Prove that for an ideal calorically perfect gas that M2 is proportional to:

Dynamic Pressure = ½ V2

Static Pressure = p

Hint: Use ~ p = RT; c = (kRT)1/2; M = V/c

show that proportionality constant = k/2

The End