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ocean). (c) Compute the speed of sound at 20C and 9000 atm and compare with the measured
value of 2650 m/s (A. H. Smith and A. W. Lawson, J. Chem. Phys., vol. 22, 1954, p. 351).
Solution 9.12
We may compute these values by differentiating Eq. (1.19) with k 1.0:
Problem 9.13
Consider steam at 500 K and 200 kPa. Estimate its speed of sound by two different methods:
(a) assuming an ideal gas from Table B.4; or (b) using finite differences for isentropic densities
between 210 kPa and 190 kPa.
Solution 9.13
Part (c) is the same as the Spirax result (a), so maybe that’s how the Spirax function works?
Problem 9.14
Benzene, listed in Table A.3, has a measured density of 57.75 lbm/ft3 at a pressure of 700 bar.
Use this data to estimate the speed of sound of benzene.
Solution 9.14
Convert to SI units: 57.7 lbm/ft3 ≈ 924 kg/m3 and 700 bar = 70,000,000 Pa. From Table A.3 at
Problem 9.15
The pressure-density relation for ethanol is approximated by Eq. (1.19) with B = 1600 and n = 7.
Use this relation to estimate the speed of sound of ethanol at a pressure of 2000 atmospheres.
Solution 9.15
Recall that Eq. (1.19) is a curve-fit equation of state for liquids:
Problem 9.16
A weak pressure pulse p propagates through still air. Discuss the type of reflected pulse which
occurs, and the boundary conditions which must be satisfied, when the wave strikes normal to,
and is reflected from, (a) a solid wall; and (b) a free liquid surface.
Solution 9.16
( 1)( ) (1.19)
n
oo
pBB
p
+ −
(a) When reflecting from a solid wall, the velocity to the wall must be zero, so the wall pressure
Problem 9.17
A submarine at a depth of 800 m sends a sonar signal and receives the reflected wave back from
a similar submerged object in 15 s. Using Prob. 9.12 as a guide, estimate the distance to the other
object.
Problem 9.12
Assume that water follows Eq. (1.19) with n 7 and B 3000. Compute the bulk modulus (in
kPa) and the speed of sound (in m/s) at (a) 1 atm; and (b) 1100 atm (the deepest part of the
ocean). (c) Compute the speed of sound at 20C and 9000 atm and compare with the measured
value of 2650 m/s (A. H. Smith and A. W. Lawson, J. Chem. Phys., vol. 22, 1954, p. 351).
Solution 9.17
It probably makes little difference, but estimate a at a depth of 800 m:
Problem 9.18
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the
altitude of Indianapolis, find the Mach number of these cars and estimate whether
compressibility might affect their aerodynamics.
Solution 9.18
Rush to the Almanac and find that Indianapolis is at 220 m altitude, for which Table A.6 predicts
Problem 9.19
In 1976, the SR-71A, flying at 20 km standard altitude, set the jet-powered aircraft speed record
of 3326 km/h. Estimate the temperature, in C, at its front stagnation point. At what Mach
number would it have a front stagnation-point temperature of 500C?
Solution 9.19
At 20 km altitude, from Table A.6, T = 216.66K and a = 295.1 m/s. Convert the velocity from
Problem 9.20
Air flows isentropically in a channel. Properties at section 1 are V1 = 250 m/s, T1 = 330 K, and
p1 = 80 kPa. At section 2 downstream, the temperature has dropped to 0C. Find (a) the
pressure, (b) velocity, and (c) Mach number at section 2.
Solution 9.20
Assume k = 1.4 and, of course, convert T2 = 0C = 273 K. (b) The adiabatic energy equation
will yield the new velocity:
Problem 9.21
N2O expands isentropically through a duct from p1 = 200 kPa and T1 = 250ºC to a downstream
section where p2 = 26 kPa and V2 = 594 m/s. Compute (a) T2 ; (b) Ma2 ; (c) To ; (d) po ; (e) V1 ;
and (f) Ma1.
Solution 9.21
From Table A.4, for N2O, k = 1.31 and R = 189 m2/(s2∙K). Convert T1 = 250+273 = 523 K.
Evaluate cp = kR/(k-1) = (1.31)*189/(1.31-1) = 799 m2/(s2∙K). Proceed systematically through
these various properties, using isentropic and adiabatic relations:
Problem 9.22
Given the pitot stagnation temperature and pressure and the static pressure measurements in
Fig. P9.22, estimate the air velocity V, assuming (a) incompressible flow and (b) compressible flow.
Solution 9.22
Given p = 80 kPa, po = 120 kPa, and T = 100°C = 373 K. Then
Problem 9.23
A gas, assumed ideal, flows isentropically from point 1, where the velocity is negligible, the
pressure is 200 kPa, and the temperature is 300C, to point 2, where the pressure is 40 kPa.
What is the Mach number Ma2 if the gas is (a) air; (b) argon; or (c) CH4? (d) Can you tell,
without calculating, which gas will be the coldest at point 2?
Solution 9.23
This is a standard exercise in using the isentropic-flow formulas. The term “negligible velocity”
is code for stagnation conditions, hence po = 200 kPa and To = 300C = 573 K. Work it out for
the three different gases, using the ideal-gas isentropic-flow formulas:
Problem 9.24
For low-speed (nearly incompressible) gas flow, the stagnation pressure can be computed from
Bernoulli’s equation
2
0
1
2
p p V
=+
(a) For higher subsonic speeds, show that the isentropic relation (9.28a) can be expanded in a
power series as follows:
2 2 4
0
1 1 2
1 Ma Ma
2 4 24
k
p p V
−
+ + + +
(b) Suppose that a pitot-static tube in air measures the pressure difference p0 – p and uses the
Bernoulli relation, with stagnation density, to estimate the gas velocity. At what Mach number
will the error be 4 percent?
Solution 9.24
Expand the isentropic formula into a binomial series:
Problem 9.25
If it is known that the air velocity in the duct is 750 ft/s, use that mercury manometer measurement
in Fig. P9.25 to estimate the static pressure in the duct, in lbf/in2 absolute.
Solution 9.25
Estimate the air specific weight in the manometer to be, say, 0.07 lbf/ft3. Then
Problem 9.26
Show that for isentropic flow of a perfect gas if a pitot-static probe measures p0, p, and T0, the
gas velocity can be calculated from
( 1)/
2
0
0
21
kk
p
p
V c T p
−
=−
What would be a source of error if a shock wave were formed in front of the probe?
Solution 9.26
Assuming isentropic flow past the probe,
Problem 9.27
A pitot tube, mounted on an airplane flying at 8000 m standard altitude, reads a stagnation
pressure of 57 kPa. Estimate (a) the velocity in mi/h, and (b) the Mach number.
Solution 9.27
We assume that the static pressure is the standard atmosphere pressure at 8000 m, which from
Table B.6 is 35,581 Pa. Then the isentropic pressure formula will yield the Mach number:
Problem 9.28
Air flows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa
and 125ºC, and the velocity is 200 m/s. At section 2, the area is 0.25 m2 and the Mach number is
2.0. Determine (a) Ma1 ; (b) T2 ; (c) V2 ; and (d) the mass flow.
Solution 9.28
For air, take k = 1.4. Convert T1 = 125+273 = 398K. (a) Find the speed of sound at section 1:
Problem 9.29
Steam from a large tank, where T = 400C and p = 1 MPa, expands isentropically through a small
nozzle until, at a section of 2-cm diameter, the pressure is 500 kPa. Using the steam tables, estimate
(a) the temperature; (b) the velocity; and (c) the mass flow at this section. Is the flow subsonic?
Solution 9.29
“Large tank” is code for stagnation values, thus To = 400C and po = 1 MPa. This problem
involves dogwork in the tables and well illustrates why we use the ideal-gas law so readily.
Problem 9.30
When does the incompressible-flow assumption begin to fail for pressures? Construct a graph of
po/p for incompressible flow of a perfect gas as compared to Eq. (9.28a). Neglect gravity. Plot
both versus Mach number for 0 Ma 0.6 and decide for yourself where the deviation is too
great.
Solution 9.30
The Bernoulli incompressible equation can be converted to Mach number form:
Problem 9.31
Air flows adiabatically through a duct. At one section, V1 = 400 ft/s, T1 = 200F, and
p1 = 35 psia, while farther downstream V2 = 1100 ft/s and p2 = 18 psia. Compute (a) Ma2;
(b) Umax; and (c) po2/po1.
Solution 9.31
(a) Begin by computing the stagnation temperature, which is constant (adiabatic):
Problem 9.32
The large compressed-air tank in Fig. P9.32 exhausts from a nozzle at an exit velocity of
235 m/s. The mercury manometer reads h = 30 cm. Assuming isentropic flow, compute the
pressure (a) in the tank and (b) in the atmosphere. (c) What is the exit Mach number?
Solution 9.32
The tank temperature = To = 30C = 303 K. Then the exit jet temperature is
Problem 9.33
Air flows isentropically from a reservoir, where p = 300 kPa and T = 500 K, to section 1 in a
duct, where A1 = 0.2 m2 and V1 = 550 m/s. Compute (a) Ma1; (b) T1; (c) p1; (d)
m;
and (e) A*.
Is the flow choked?
Solution 9.33
Use the energy equation to calculate T1 and then get the Mach number:
Problem 9.34
Air in a large tank, at 300ºC and 400 kPa, flows through a converging-diverging nozzle with
throat diameter 2 cm. It exits smoothly at a Mach number of 2.8. According to one-dimensional
isentropic theory, what is (a) the exit diameter, and (b) the mass flow?
Solution 9.34
By “large tank” we infer stagnation conditions, To = 300ºC = 573 K and po = 400 kPa. Then,
from Eq. (9.45) or Table B.1, the exit area ratio is Ae/A* = 3.5001. Exit diameter thus is
Problem 9.35
Helium, at To = 400 K, enters a nozzle isentropically. At section 1, where A1 = 0.1 m2, a pitot-static
arrangement (see Fig. P9.25) measures stagnation pressure of 150 kPa and static pressure of
123 kPa. Estimate (a) Ma1; (b) mass flow
m
; (c) T1; and (d) A*.
Solution 9.35
For helium, from Table A.4, take k = 1.66 and R = 2077 J/kgK. (a) The local pressure ratio is
given, hence we can estimate the Mach number:
1
Problem 9.36
An air tank of volume 1.5 m3 is at 800 kPa and 20C. At t = 0 it begins exhausting through a
converging nozzle to sea-level conditions. The throat area is 0.75 cm2. Estimate (a) the initial
mass flow in kg/s; (b) the time to blow down to 500 kPa; and (c) the time when the nozzle ceases
being choked.
Solution 9.36
For sea level, pambient = 101.35 kPa 0.528ptank, hence the flow is choked until the tank
pressure drops to pambient/0.528 = 192 kPa. (a) We obtain
Problem 9.37
Make an exact control volume analysis of the blowdown process in Fig. P9.37, assuming an insulated
tank with negligible kinetic and potential energy within. Assume critical flow at the exit and
show that both po and To decrease during blowdown. Set up first-order differential equations for
po(t) and To(t) and reduce and solve as far as you can.
