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 dV x 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 ISENTROPICISENTROPIC

unsteady steady

(1) at any position, no properties are changing with time (2) V and  are only functions of x SOUND SPEED

(  dV x A >> d  dV x A)  (dV x )A = (d  )cA dV x = (c/  ) d   cA = (  +d  )(c-dV x )(A)  cA =  cA-  d(V x )A +(d  )cA - (d  )(dV x )A c dV x

This terms appears only if CV is accelerating dR x

dR x 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), dR x =0. So F Sx = -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  = c 2 c = [dp/d  ] 1/2 adiabatic? c = [dp/d  ] s 1/2 or isothermal? c = [dp/d  ] T 1/2

Speed of Propagation Isentropic & Ideal Gas

Correct Answer Wrong Reasoning

For ideal gas, isentropic, constant c p and c v : p/  k = c onst p = c onst  k c onst = p/  k dp/d  s = d(c onst  k )/d  = kc onst  k-1 dp/d  s = kp/  dp/d  s = k  RT/  = kRT

c = (kRT) 1/2 ~ 340 m/s ~ 1120 ft/s, for air at STP c = [dp/d  s ] 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. c 2 = dp/d  =  p/  | T FOR IDEAL GAS: p =  RT p/  = const if constant temperature Then: 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) steel5050 seawater1540 water1500 air (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 = 0V < 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

vtvt ctct  sin  = c  t/v  t = 1/M  = sin -1 (1/M)

Mach ( ) 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/s M ~ 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 km T(K) = kmT(K) = ,900m ~ m * (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 V plane / V bullet = 987/700 ~ 1.41 PROBLEM 1

Not really linear, although not apparent at the scale of this plot. For standard atm. conditions c= 340 m/s at sea level c = 295 m/s at 11 km

3 km M = 1.35 T = 303 o K 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 o K 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/c V is airspeed

3 km M = 1.35 T = 303 o K Wind = 10 m/s * note: if T &  not constant, Mach line would not be straight  vtvt ctct v is velocity relative 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 o K Wind = 10 m/s (b) What is time between seeing aircraft overhead and hearing it?  = sin -1 (1/M) = sin -1 (1/1.35) = 47.8 o V earth = 471m/s – 10m/s = 461m/s D = V earth t = 461 [m/s] t = 3000[m]/tan(  ) t = 5.9 s D = V earth t 3000m  * note: if T &  not constant, Mach line would not be straight 

Problem #3 Prove that for an ideal calorically perfect gas that M 2 is proportional to: (Kinetic Energy per unit mass = V 2 /2) (Internal Energy per unit mass = u) Hint: Use ~ u = c v T; c v = 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 M 2 is proportional to: Dynamic Pressure = ½  V 2 Static Pressure = p Hint: Use ~ p =  RT; c = (kRT) 1/2 ; M = V/c show that proportionality constant = k/2

The End