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# Speed of Propagation

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1. Speed of Propagation “speed of sound is one of the most important quantities in the study of incompressible flow” - Anderson : Moderns Compressible Flow

2. SOUND IS A LONGITUDINAL WAVE dVx c c = ?

3. 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

4. Speed of Propagation = Isentropic I S E N T R O P I C • 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

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

7. cA = (+d)(c-dVx)(A) cA = cA- d(Vx)A +(d)cA - (d)(dVx)A (dVxA >> ddVxA) (dVx)A = (d)cA dVx = (c/) d c dVx

8. dRx This terms appears only if CV isaccelerating

9. 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

10. Total forces = normal surface forces Change in momentum flux From continuity eq.

11. Cons. of mass Cons. of mass

12. From momentum eq. From continuity eq.

13. dp/d = c2 c = [dp/d]1/2 adiabatic? c = [dp/d]s1/2 or isothermal? c = [dp/d]T1/2

14. Speed of Propagation Isentropic & Ideal Gas

16. For ideal gas, isentropic, constant cp and cv: p/k = const p = const k const = p/ k dp/ds = d(const k)/d = kconstk-1 dp/ds = kp/ dp/ds = k RT/ = kRT

17. c = [dp/ds]1/2 = [kRT]1/2 c = (kRT)1/2~ 340 m/s ~ 1120 ft/s, for air at STP [krT]1/2 ¾ molecular velocity for a perfect gas = [8RT/]1/2

18. 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

19. 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 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

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

21. Moving Sound Source Shock wave of bullet piercing sheet of Plexiglass bending of shock due to changes in p and T

22. V = 0; M = 0 V < c; M < 1 V > c; M > 1 V = c; M = 1 .

23. 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.

24. shock increases pressure

25. ct  vt sin  = ct/vt = 1/M  = sin-1(1/M)

26. 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.”

27. POP QUIZ • 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?

28. Tip speed ~ 1400 ft/s M ~ 1400/1100 ~ 1.3

29. Sound Propagation Problems

30. PROBLEM 1 (faster than a speeding bullet) 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 719 24km T(K) = 220.6 26km T(K) = 222.5 25,900m ~ 220.6 + 1900m * (1.9K/2000m) ~ 222 K

31. PROBLEM 1 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

32. 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

33. PROBLEM 2 Wind = 10 m/s M = 1.35 3 km T = 303 oK • What is airspeed of aircraft? • What is time between seeing aircraft overhead and hearing it?

34. PROBLEM 2 M = V/c V is airspeed Wind = 10 m/s M = 1.35 3 km T = 303 oK • 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)

35. * note: if T &  not constant, Mach line would not be straight M = 1.35 Wind = 10 m/s v is velocity relative to earth = 471 – 10 = 461 m/s sin  = c/v = 1/M vt  3 km ct T = 303 oK   time to travel this distance = distance /velocity of plane relative to earth

36. * note: if T &  not constant, Mach line would not be straight D = Vearth t Wind = 10 m/s M = 1.35  3000m 3 km T = 303 oK  (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/s D = Veartht = 461 [m/s]t = 3000[m]/tan() t = 5.9 s

37. 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

38. 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

39. The End