Transcript Document
Speed of Propagation
“speed of sound is one of the most important
quantities in the study of incompressible flow”
- Anderson : Moderns Compressible Flow
SOUND IS A LONGITUDINAL WAVE
dVx
c
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
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
unsteady
steady
SOUND
SPEED
(1) at any position, no properties
are changing with time
(2) V and are only functions
of x
cA = (+d)(c-dVx)(A)
cA = cA- d(Vx)A +(d)cA - (d)(dVx)A
(dVxA >> ddVxA)
(dVx)A = (d)cA
dVx = (c/) d
c
dVx
dRx
This terms appears only
if CV is accelerating
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
From continuity eq.
Change in momentum flux
Cons. of
mass
Cons. of
mass
From momentum eq.
From continuity eq.
dp/d =
2
c
c = [dp/d]1/2
adiabatic? c = [dp/d]s1/2
or
1/2
isothermal? c = [dp/d]T
Speed of Propagation
Isentropic & Ideal Gas
Correct Answer
Wrong Reasoning
For ideal gas, isentropic, constant cp and cv:
k
p/
= const
p = const k
k
const = p/
k
k-1
dp/ds = d(const )/d = kconst
dp/ds = kp/
dp/ds = k RT/ = kRT
c = [dp/ds
1/2
]
=
1/2
[kRT]
1/2
(kRT) ~
c=
340 m/s
~ 1120 ft/s, for air at STP
[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 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
steel
seawater
water
air (sea level)
(m/s)
5050
1540
1500
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
ct
vt
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.
“I do not believe in atoms.”
Turned philosopher –
“psychophysics”: all
knowledge is based on
sensations
POP QUIZ
(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?
Tip speed ~ 1400 ft/s
M ~ 1400/1100 ~ 1.3
Sound Propagation Problems
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
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
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
PROBLEM 2
Wind = 10 m/s
M = 1.35
3 km
T = 303 oK
(a) What is airspeed of aircraft?
(b) What is time between seeing aircraft overhead and hearing it?
PROBLEM 2
Wind = 10 m/s
3 km
M = 1.35
M = V/c
V is airspeed
T = 303 oK
(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)
* note: if T & not constant, Mach line would not be straight
Wind = 10 m/s
M = 1.35
vt
3 km
v is velocity
relative to earth
= 471 – 10
= 461 m/s
ct
T = 303 oK
sin = c/v = 1/M
time to travel this distance =
distance /velocity of plane relative to earth
* 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
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