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From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter Two
WAVE PROPAGATION
IN COMPRESSILBE MEDIA
Problem 1. โ Using the expansion wave and control volume depicted in Figs. 2.8 and 2.9
along with the continuity and momentum equations, rederive Eq. (2.4).
moving wave
dV
p – dp
ฯ – dฯ
dV
a
gas
at
rest
dV
moving wave
p – dp
ฯ – dฯ
dV
a
gas
at
rest
Continuity equation
(ฯ โ dฯ)(a + dV )A โ ฯaA = 0
Expand, neglect products of derivatives and simplify to get
ฯdV โ adฯ = 0
(1)
Momentum equation
pA โ (p โ dp )A = ฯaA[(a + dV ) โ a ]
or
dp = ฯadV
(2)
Combining Eqs. (1) and (2) gives
dp = a 2 dฯ
Since the process is reversible and adiabatic, i.e., isentropic, this can be written as:
โ โp โ
a = โโ โโ
โ โฯ โ s
17
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 2. โ (a) Derive an expression for ks, for a perfect gas, substitute the result into
Eq. (2.10), and thereby demonstrate Eq. (2.7); (b) Derive an expression for kT, for a
perfect gas, substitute the result into Eq. (2.11), and thereby demonstrate Eq. (2.7) and
finally; (c) Derive an expression for ฮฒs, for a perfect gas, substitute your result into Eq.
(2.14), and thereby demonstrate Eq. (2.7).
(a)
ks =
1 โ โฯ โ
โ โ
ฯ โโ โp โโ s
An isotropic process involving a perfect gas is described by P = cฯ ฮณ
โด
dp
ฮณcฯ ฮณ ฮณ p
=
= ฮณcฯ ฮณ โ1 =
dฯ
ฯ
ฯ
Hence,
โ โฯ โ
ฯ
โโ โโ =
โ โp โ s ฮณ p
ks =
1 โ โฯ โ
1
1
โโ โโ =
=
ฯ โ โp โ s ฮณ p ฮณ ฯ RT
So,
a=
(b)
kT =
ฯ=
1
= ฮณRT
ฯk s
1 โ โฯ โ
โ โ
ฯ โโ โp โโ T
p
RT
โ โฯ โ
1
โโ โโ =
โ โp โ T RT
kT =
1
ฯRT
So,
18
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a=
(c)
ฮณ
= ฮณRT
ฯk T
โ ฮณpโ
โ โp โ
ฮฒs = ฯโโ โโ = ฯโโ โโ = ฮณ p = ฮณฯRT
โ ฯ โ
โ โฯ โ s
a=
ฮฒs
= ฮณRT
ฯ
Problem 3. โ Use dimensional analysis to develop an expression for the speed of sound in
terms of the isentropic compressibility, the density and gc.
a = f (k s , ฯ, g e )
a~
L
,
T
ks ~
ML
L2
M
, gc ~
, ฯ~
3
F
FT 2
L
a
b
c
L โโ L2 โโ โ M โ โ ML โ
โโ โโ โโ
โโ = F โa โc M b+c L2a โ3b+c T โ2c
=
3
2
โ
โ
T โ F โ โ L โ โ FT โ
F : โa โ c = 0
M:b+c = 0
L : 2a โ 3b + c = 1
T : โ2c = 1
1
1
a=โ
Hence, c =
2
2
b=โ
1
2
So,
a=
gc
ฯk s
Problem 4. โ Using the data provided in Tables 2-1, 2-2 and 2-3, i.e., the density, and the
isentropic compressibility or the bulk modulus, calculate the velocity of sound at 20ยฐC
and one atmosphere pressure in (a) helium, (b) turpentine, and (c) lead.
(a)
Helium: ฯ = 0.16
kg
m
3
, k s = 5,919
1
GPa
19
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a=
(b) Turpentine: ฯ = 870
kg
m
3
, k s = 0.736
a=
(c) Lead: ฯ = 11,300
kg
m3
1
109
=
= 1027.6 m / s
(0.16)(5,919)
ฯk s
1
GPa
1
=
ฯk s
10 9
= 1249.7 m / s
(870)(0.736)
, ฮฒ s = 16.27 GPa
a=
ฮฒs
=
ฯ
(16.27) 10 9
= 1199.9 m / s
11,300
Problem 5. โ In Example Problem 2.3 the speed of sound of superheated steam was
determined by using a finite difference representation of the compressibility and steam
table data (Table 2-4). Using the same steam table data, determine the speed of sound of
superheated steam for the same pressure and temperature, i.e., at p = 500 kPa and T =
300หC. However, use the following finite differences to obtain two estimates for the
speed of sound:
ฮณ
ฮณ
a2 =
=
โ โฯ โ
โก โ (1 v ) โค
โโ โโ
โข
โฅ
โ โp โ T โฃ โp โฆ T
a2 =
a2 =
1
1
=
โ โฯ โ
โก โ(1 v )โค
โโ โโ
โข
โฅ
โ โp โ s โฃ โp โฆ s
ฮณ
ฮณ
2ฮณโp
=
=
1
1
1
1
โก โ (1 v ) โค
โ
โ
โข โp โฅ
v(p + โp, T ) v(p โ โp, T ) v(p + โp, T ) v(p โ โp, T )
โฃ
โฆT
2โp
From Example 2.3
v(p + โp1T ) = 0.4344
M3
,
kg
v(p โ โp1T ) = 0.6548
M3
, and โp = 100,000 Pa
kg
20
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(2)(1.327 )(100,000) = 342,521.5 m 2
a2 =
1
1
โ
0.4344 0.6548
s2
a = 585.3 m / s
a2 =
1
2โp
=
1
1
โก โ (1 v ) โค
โ
โข โp โฅ
โฃ
โฆ s v(p + โp, s ) v(p โ โp, s )
From Example 2.3
v(p + โp, s ) = 0.4544
M3
M3
, v(p โ โp, s ) = 0.6209
and โp = 100,000 Pa
kg
kg
a2 =
(2)(100,000)
1
1
โ
0.4544 0.6209
= 338,903.2
m2
s2
a = 582.2 m / s
Problem 6. โ Equation (2.16) provides a convenient expression for calculating the speed
of sound in air: a = 20.05 T , where T is the absolute temperature in degrees Kelvin.
Derive the following linear equation for the speed of sound in air:
a = a 0 + 0.6t
where a0 is the speed of sound in air at 0ยฐC and t is ยฐC.
To accomplish this make use of Eq. (2-16) and the expansion
(x + y )n = x n + nx n โ1y + …..
a = ฮณRT = [ฮณR (273 + t )]1 / 2
1/ 2
t โ
โ
= ฮณR (273) โ1 +
โ
โ 273 โ
โ 1 t
โ
= a o โ1 +
+ …..โ
โ 2 273
โ
21
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a o = ฮณR (273) = 20.05 273 = 331
m
,
s
ao
331
=
= 0.6
(2)(273) 546
โด
a = 331 + 0.6t
Problem 7. โ Rather than measure the bulk modulus directly it may be easier to measure
the speed of sound as it propagates though a material and then use it to compute the bulk
modulus. For a Lucite plastic of density 1,200 kg/m3, the speed of sound is measured as
2,327 m/s. Determine the bulk modulus. What is the corresponding isentropic
compressibility?
kg
m
, a = 2,327
Now ฯ = 1,200
3
s
m
a=
ฮฒs
ฯ
2
so,
kg โโ
m โ โ N โ
s2 โ
โ
โโ = 6.498 ร 109 Pa = 6.498 GPa
2
,
327
ฮฒ3 = ฯa 2 = โ1,200
โโ
โ โโ1
ms โ โ
s โ โ kg โ
m โ
โ
ks =
1
1
= 0.1539
GPa
ฮฒs
Problem 8. โ An object of diameter d (m) is rotated in air at a speed of N revolutions per
minute. Draw a plot of the rotational speed required for the velocity at the outer edge of
the object to just reach sonic velocity for a given diameter. Take the speed of sound of
the air to be 331m/s.
The highest speed will occur at R.
1 min
m
โ rev โโ 2ฯrad โ
V = Nโ
= a = 331
โโ
โR (m )
60 s
s
โ min โ โ rev โ
ฯ
m
=
ND,
60
s
22
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The following is a log base 10 plot of N = 6,321.6/D.
Rotational Speed, RPM
5.6
5.4
5.2
5.0
4.8
4.6
4.4
4.2
4.0
3.8
3.6
Supersonic Region
Subsonic Region
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Diameter, m
Problem 9. โ (a) Newton assumed that the sound wave process was isothermal rather
than isentropic. Determine the size of error made in computing the speed of sound by
making this assumption. (b) A flash of lightening occurs in the distance. 20 seconds later
the sound of thunder is heard. The temperature in the area is 23ยฐC. How far away was
the lightening strike?
(a)
as =
1
ฯk s
aT =
โ
a T โ a s โโ 1
=
โ 1โ 100
โ ฮณ
โ
as
โ
โ
(b)
1
ฯk T
โด
aT
=
as
for ฮณ = 1.4
ks
kT
aT โ as
= โ15.5%
as
L = aโt = (344.86 )(20 ) = 6,897.2 m
Problem 10. โ (a) The pressure increase across a compression pulse moving into still air
at 1 atmosphere pressure and 30ยฐC is 100 Pa. Determine the velocity following the pulse.
(b) The velocity changes by 0.1 m/s across a pressure wave that moves into hydrogen gas
that is at rest at a pressure of 100 kPa and temperature 300K. Determine the pressure
behind the wave.
Use Eq (2.2) and write the expression in difference form as
23
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
(a)
air:
โp
,
โp = 100 Pa
ฯa
101,000
kg
ฯ=
= 1.1615
โ 8314 โ
m3
โ
โ(303)
โ 28.97 โ
โV =
a = 20.05 303 = 349.0 m/s
Therefore,
(b)
โV =
100
= 0.247 m/s
(1.1615)(349.0)
โp = ฯaโV ,
hydrogen:
ฯ=
a=
Therefore,
โV = 0.1 m/s
p
100,000
kg
= 0.0808
=
RT โ 8314 โ
m3
โ
โ(300 )
โ 2.016 โ
(1.41)โโ 8314 โโ(300) = 1320.8 m/s
โ 1.016 โ
โp = (0.0808)(1320.8)(0.1) = 10.68 Pa
Problem 11. โ (a) Helium at 35ยฐC is flowing at a Mach number of 1.5. Find the velocity
and determine the local Mach angle. (b) Determine the velocity of air at 40ยฐC to produce
a Mach angle of 38ยฐ
(a) helium: T = 35ยฐC = 308K M = 1.5
V = aM
a = ฮณRT =
(1.667)โโ 8,314 โโ(308) = 1,032.7 m/s
โ 4,003 โ
V = (1,032.7 )(1.5) = 1,549.0 m/s
โ1โ
ยต = sin โ1 โโ โโ = 41.8ยฐ
โยตโ
(b) air: T = 40ยฐC = 313K
a = 20.05 223.3 = 299.6 m/s
24
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
โ1โ
ยต = sin โ1 โ โ
โMโ
M=
1
V
=
sinยต a
V=
a
354.6
=
= 576.0 m/s
sinยต sin (38)
Problem 12. โ (a) A jet plane is traveling at Mach 1.8 at an altitude of 10 km where the
temperature is 223.3K. Determine the speed of the plane. (b) Air at 320 K flows in a
supersonic wind tunnel over a 2-D wedge. From a photograph the Mach angle is
measured to be 45ยฐ. Determine the flow velocity, the local speed of sound and the Mach
number of the tunnel.
(a)
M = 1.8 , T = 223.3 K , a = 20.05 223.3 = 299.6 m/s
V = aM = (299.6 )(1.8) = 539.3 m/s
(b)
air:
T = 320 K, ยต = 45ยฐ , a = 20.05 320 = 358.7 m/s
V=
a
358.7
=
= 507.2 m/s
sinยต sin (45)
M=
V
1
=
= 1.414
a sin ยต
Problem 13. โ A supersonic aircraft, flying horizontally a distance H above the earth,
passes overhead. โt later the sound wave from the aircraft is heard. In this time
increment, the plane has traveled a distance L. Show that the Mach number of the
aircraft can be computed from:
2
2
โ Vโt โ
โLโ
M = โ โ +1 = โ
โ +1
โHโ
โ H โ
Hint: first show that the Mach angle ยต can be expressed as tan โ1 โโ1
โ
connect the Mach angle, ยต, to the geometric parameters H and L.
1
sin ยต =
ยต
M
H
M 2 โ 1 โโ and then
โ
ยต
L
25
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
tan ยต =
M2 โ1
=
H
L
M
1
ยต
2
โLโ
โด M = โ โ +1
โHโ
2
M -1
but L = Vโt
2
โ Vโt โ
= โ
โ +1
โ H โ
Problem 14. โ Given speeds and temperatures, determine the corresponding Mach
numbers of the following (note: 1 mile = 5,280 ft = 1,609.3 m; 1 mi/hr = 1.6093 km/hr =
0.447 m/s):
(a) A cheetah running at top speed of 60 mi/hr; the local temperature is 40ยฐC
(b) A Peregrine falcon in a dive at 217 mi/hr; local temperature of 25ยฐC
(c) In June 1999 in Athens Greece, Maurice Greene became the worldโs fastest human
by running 100 m in 9.79 s; the temperature was 21ยฐC
(d) In June 1999, Alexander Popov became the worldโs fastest swimmer by swimming
50 m in 21.64s; the temperature of the water was 20ยฐC
(a) a = 20.05 313 = 354.7 m/s
mi
m/s
(
60 ) (0.447 )
V
hr
mi/hr = 0.076
M= =
(354.7 ) m/s
a
(b) a = 20.05 298 = 346.1 m/s
V (217 )(0.447 )
=
= 0.28
(346.1)
a
M=
(c) a = 20.05 294 = 343.8 m/s
M=
mi โ
โ 10.21
= 22.9 โ
โ=
hr โ
โ 0.447
50
= 2.31 m/s
21.64
mi โ
โ 2.31
= 5.17 โ
โ=
hr โ
โ 0.447
10.21
= 0.03
343.8
(d) a = 1,481 m/s (from Table 2 – 2) V =
M=
100
= 10.21 m/s
9.79
V=
2.31
= 0.00156
1,481
26
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 15. โ Given speeds and Mach numbers, assuming air is a perfect gas, determine
the corresponding local temperature (note: 1 mi/hr = 0.447 m/s) for the following:
(a) A Boeing 747-400 at a cruise speed of 910 km/hr; M = 0.85.
(b) Concorde at a cruise speed of 1,320 mi/hr; M = 2.0
(c) The fastest airplane, the Lockheed SR-71 Blackbird, flying at 2,200 mi/hr; M =
3.3
(d) The fastest boat, the Spirit of Australia, that averaged 317.6 mi/hr; M = 0.41
(e) The fastest car, the ThrustSSC, averaged 760.035 mi/hr; M = 0.97
(a)
V=
910,000m
m
= 252.8
3600s
s
M = 0.85
2
a=
V 252.8
m
=
= 297.4
M
.85
s
2
โ a โ
โ 297.4 โ
T=โ
โ =โ
โ = 220ยฐK = โ53ยฐC
โ 20.05 โ
โ 20.05 โ
(b)
V = (1320)(0.447 ) = 590.0
m
s
2
M = 2 .0
a=
V
= 295.0 m/s
M
2
โ 295 โ
โ a โ
T=โ
โ = 216.5K = โ56.5ยฐC
โ =โ
โ 20.05 โ
โ 20.05 โ
(3)
V = (2200)(0.447 ) = 983.4
m
s
2
M = 3 .3
a=
983.4
= 298.0 m/s
3.3
2
โ a โ โ 298.0 โ
T=โ
โ =โ
โ = 220.9K = โ52.1ยฐC
โ 20.05 โ โ 20.05 โ
(d)
V = (317.6 )(0.447 ) = 142.0
2
m
s
M = 0.41
a=
142
= 346.3 m/s
.41
2
โ 346.3 โ
โ a โ
T=โ
โ = 298.2K = 25.2ยฐC
โ =โ
โ 20.05 โ
โ 20.05 โ
(e)
V = (760.035)(0.447 ) = 339.7 m/s
2
M = 0.97
a=
339.7
= 350.2 m/s
.97
2
โ 350.2 โ
โ a โ
T=โ
โ = 305.1K = 32.1ยฐC
โ =โ
โ 20.05 โ
โ 20.05 โ
27
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 16. โ A baseball, which has a mass of 145 grams and a diameter of 3.66 cm,
when dropped from a very tall building reaches high speeds. If the building is tall
enough the speed will be controlled by the drag, as the baseball will reach terminal speed.
At this state
W = FD
Where W (weight) = mg, g (acceleration of gravity) = 9.81 m/s2, FD (drag force) =
CDฯairAV2/2, CD (drag coefficient) = 0.5 and A (projected area of sphere) = ฯR2. Find the
terminal speed of the baseball and determine the corresponding Mach number if the
ambient air temperature is 23ยฐC and the ambient air pressure is 101 kPa..
The density of the air is first determined:
ฯ air =
p
101
=
= 1.19kg / m 3
RT (0.287 )(296)
Now
W = mg = FD =
C D Aฯ air V 2
2
Hence,
V=
2mg
=
C D ฯ air A
M=
V
=
a
2(0.145)(9.81)
= 33.76 m / s
(0.5)(1.19)(0.0042)
33.76
(1.4)(287 )(296)
= 0.098
Problem 17. โ Derive the following equation for the speed of sound of a real gas from
Berthelotโs equation of state:
p=
ฯRT ฮฑฯ 2
โ
1 โ ฮฒฯ
T
โก RT
RTฯฮฒ
2ฮฑฯ โค
โ
a = ฮณโข
+
โฅ
T โฅโฆ
โขโฃ1 โ ฮฒฯ (1 โ ฮฒฯ )2
28
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Since T is treated as a constant, we may simply use information from Section 2.6 where
โ โp โ
a = ฮณโโ โโ
โ โฯ โ T
โ โp โ
RT
RTฯฮฒ
โโ โโ =
+
โ 2ฮฑฯ
โ โฯ โ T 1 โ ฮฒฯ (1 โ ฮฒฯ )2
Now replace ฮฑ with ฮฑ/T. Thus, from Eq. (2.24)
โก RT
RTฯฮฒ
2ฮฑฯ โค
a = ฮณโข
+
โ
โฅ
T โฅโฆ
โขโฃ1 โ ฮฒฯ (1 โ ฮฒฯ )2
Problem 18. โ Using the speed of sound expression from the previous problem and the
following constants for nitrogen
R = 296.82 (Nยทm)/(kgยทK)
ฮฑ = 21,972.68 Nยทm4/kg2
ฮฒ = 0.001378 m3/kg
ฮณ = 1.4
determine the speed of sound for the two cases described in Example 2.4.
Case (1) p 0.3 MPa and T = 300K
Iteration
1
2
3
v
0.296823
0.297386
0.297384
f(v)
df /dv v-f/(df/dv)
-4.9286E-05 8.7530E-02 0.297386
1.8796E-07 8.8198E-02 0.297384
2.6975E-12 8.8195E-02 0.297384
ฯ
3.3690
3.3626
3.3627
a
353.7517
353.7505
353.7505
The result differs from the experimental value 353.47 m/s by 0.08%.
Case (2): p 30.0 MPa and T = 300K
29
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Iteration
1
2
3
4
5
6
7
v
0.002968
0.005658
0.004594
0.004088
0.003953
0.003944
0.003944
f(v)
df /dv v-f/(df/dv)
ฯ
-8.2594E-09 3.0708E-06 0.005658 336.9016
5.2436E-08 4.9296E-05 0.004594 176.7430
1.3084E-08 2.5826E-05 0.004088 217.6647
2.2920E-09 1.7035E-05 0.003953 244.6426
1.4088E-10 1.4959E-05 0.003944 252.9695
6.6552E-13 1.4817E-05 0.003944 253.5736
1.5099E-17 1.4817E-05 0.003944 253.5765
a
604.3973
426.1798
457.9898
483.2795
491.8702
492.5088
492.5118
The result differs from the experimental value 483.18 m/s by 1.9%.
Problem 19. โEmploy the finite difference method of Example 2.5 to determine the
speed of sound in nitrogen using the Redlich-Kwong equation of state
p=
a oฯ 2
RTฯ
โ
1 โ ฮฒฯ (1 + ฮฒฯ ) T
where for nitrogen:
R = 296.823 (Nยทm)/(kgยทK)
ao = 1979.453 (Nยทm4ยทโK )/(kg2)
ฮฒ = 0.0009557 m3/kg
ฮณ = 1.4
Compute the speed at a pressure of 30.1 MPa and a temperature of 300 K. Experimental
values of the speed of sound of nitrogen may be found in Ref. (11). For the given
conditions the measured value is 483.730 m/s.
The Redlich-Kwong equation of state is: p =
ao
RT
โ
. Rearrange to obtain:
v โ ฮฒ v (v + ฮฒ ) T
ao โ
a ฮฒ
โ RT โ 2 โ 2 RTฮฒ
โv โ o = 0
โโ v โ โ ฮฒ +
f (v ) = v 3 โ โโ
โ
โ
โ
p
p Tโ
p T
โ p โ
โ
โ
ao โ
โ RT โ
df
RTฮฒ
โ
โโ v โ โ ฮฒ 2 +
โ
= 3v 2 โ 2โโ
โ
โ
dv
p
p
T
โ p โ
โ
โ
Use Newton-Raphson to find v = 0.003279 m3/kg. Thus, ฯ = 304.9917 kg/m3. Use โฯ =
0.1 and compute
30
From Gas Dynamics, Third Edition, by James E. John and Theo G. Keith. ISBN 0-13-120668-0. ยฉ 2006 Pearson Education, Inc.,
Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No Portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
p(ฯ+โฯ,T) = p(305.0917,300) = 30,112,951.62 Pa
p(ฯโโฯ,T) = p(304.8917,300) = 30,087,052.10 Pa
a= ฮณ
โp
m
= 425.79
โฯ
s
The result is 12% too small compared to the experimental value of 483.73m/s. However,
if a more appropriate value of ฮณ at this pressure and temperature is used, i.e., ฮณ = 1.704, a
= 469.75m/s, which is in error by only 2.9%.
31

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