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Problem C2.1
Some manometers are constructed as in Fig. C2.1, where one side is a large reservoir (diameter
D ) and the other side is a small tube of diameter d , open to the atmosphere. In such a case, the
height of manometer liquid on the reservoir side does not change appreciably. This has the
advantage that only one height needs to be measured rather than two. The manometer liquid has
density ρm while the air has density ρa . Ignore the effects of surface tension. When there is no
pressure difference across the manometer, the elevations on both sides are the same, as indicated
by the dashed line. Height h is measured from the zero pressure level as shown.
(a) When a high pressure is applied to the left side, the manometer liquid in the large reservoir
goes down, while that in the tube at the right goes up to conserve mass. Write an exact
expression for p1gage , taking into account the movement of the surface of the reservoir. Your
equation should give p1gage as a function of h , ρm , and the physical parameters in the problem, h,
d , D , and gravity constant g.
(b) Write an approximate expression for p1gage , neglecting the change in elevation of the surface
of the reservoir liquid.
( c ) Suppose h = 0.26 m in a certain application. If pa = 101,000 Pa and the manometer liquid
has a density of 820 kg/m3 , estimate the ratio D/d required to keep the error of the
approximation of part (b) within 1 percent of the exact measurement of part (a). Repeat for an
error within 0.1 percent.
Solution 2.C1
Let H be the downward movement of the reservoir. If we neglect air density, the pressure
difference is p1 − pa =
mg(h + H). But volumes of liquid must balance:
Problem C2.2
A prankster has added oil, of specific gravity SG0 , to the left leg of the manometer in Fig. C2.2.
Nevertheless, the U-tube is still useful as a pressure-measuring device. It is attached to a
pressurized tank as shown in the figure. (a) Find an expression for h as a function of H and other
parameters in the problem. (b) Find the special case of your result in (a) when p tank = pa .
(c) Suppose H = 5.0 cm, pa is 101.2 kPa, ptank is 1.82 kPa higher than pa , and SG0 = 0.85.
Calculate h in cm, ignoring surface tension effects and neglecting air density effects.
Solution 2.C2
Equate pressures at level i in the tube (the right hand water level):
Problem C2.3
Professor F. Dynamics, riding the merry-go-round with his son, has brought along his U-tube
manometer. (You never know when a manometer might come in handy.) As shown in Fig. C2.3,
the merry-go-round spins at constant angular velocity and the manometer legs are 7 cm apart.
The manometer center is 5.8 m from the axis of rotation. Determine the height difference h in
two ways: ( a ) approximately, by assuming rigid-body translation with a equal to the average
manometer acceleration; and ( b ) exactly, using rigid-body rotation theory. How good is the
approximation?
Solution 2.C3
(a) Approximate: The average acceleration of the manometer is
Ravg2 = 5.8[6(2
/60)]2 = 2.29 rad/s toward the center of rotation, as shown. Then
Problem C2.4
A student sneaks a glass of cola onto a roller coaster ride. The glass is cylindrical, twice as tall as
it is wide, and filled to the brim. He wants to know what percent of the cola he should drink
before the ride begins, so that none of it spills during the big drop, in which the roller coaster
achieves 0.55-g acceleration at a 45° angle below the horizontal. Make the calculation for him,
neglecting sloshing and assuming that the glass is vertical at all times.
Solution 2.C4
We have both horizontal and vertical acceleration. Thus the angle of tilt
is
Problem C2.5
Dry adiabatic lapse rate (DALR) is defined as the negative value of atmospheric temperature
gradient, dT/dz, when temperature and pressure vary in an isentropic fashion. Assuming air is an
ideal gas, DALR = − dT/dz when T = T0 (p/p0 )a , where exponent a = ( k − 1)/k, k = cp/cv is the
ratio of specific heats, and T0 and p0 are the temperature and pressure at sea level, respectively.
(a) Assuming that hydrostatic conditions exist in the atmosphere, show that the dry adiabatic
lapse rate is constant and is given by DALR = g (k − 1)/(kR), where R is the ideal gas constant
for air. (b) Calculate the numerical value of DALR for air in units of °C/km.
Solution 2.C5
Write T(p) in the form T/To = (p/po)a and differentiate:
11, But for the hydrostatic condition:
a
o
oo
dT p dp dp
T a g
dz p p dz dz
−
= = −
Problem C2.6
In “soft” liquids (low bulk modulus β ), it may be necessary to account for liquid compressibility
in hydrostatic calculations. An approximate density relation would be
22
00
or ( )dp d a d p p a
= + −
where a is the speed of sound and ( p0 , ρ0 ) are the conditions at the liquid surface z = 0. Use this
approximation to show that the density variation with depth in a soft liquid is
2
/
0
gz a
e
−
=
where g is the acceleration of gravity and z is positive upward. Then consider a vertical wall of
width b, extending from the surface ( z = 0) down to depth z = −h . Find an analytic expression
for the hydrostatic force F on this wall, and compare it with the incompressible result
F = ρ0gh2b/2. Would the center of pressure be below the incompressible position z = − 2h/3?
Solution 2.C6
Introduce this p(
) relation into the hydrostatic relation (2.18) and integrate:
o
z
2
2
0
d g dz
dp a d dz g dz, or: , or: .
aAns
= = − = − = −
2
gz/a
oe−
=
Problem C2.7
Venice, Italy, is slowly sinking, so now, especially in winter, plazas and walkways are flooded
during storms. The proposed solution is the floating levee of Fig. C2.7. When filled with air, it
rises to block off the sea. The levee is 30 m high, 5 m wide, and 20 m deep. Assume a uniform
density of 300 kg/m3 when floating. For the 1-m sea–lagoon difference shown, estimate the angle
at which the levee floats.
Solution 2.C7
The writer thinks this problem is
rather laborious. Assume
seawater = 1025 kg/m3.
B
Problem C2.8
In the U.S. Standard Atmosphere, the lapse rate B may vary from day to day. It is not a
fundamental quantity like, say, Planck’s constant. Suppose that, on a certain day in Rhode Island,
with To = 288 K, the following pressures are measured by weather balloons:
Estimate the best-fit value of B for this data. Explain any difficulties. [ Hint: EES is
recommended.]
Solution 2.C8
If you plot this distribution p(z), it is very smooth, as shown below. But the data are
extraordinarily sensitive to the value of B.
Problem C2.9
The ALVIN submersible vehicle has a passenger compartment which is a titanium sphere of
inside diameter 78.08 in and thickness 1.93 in. If the vehicle is submerged to a depth of 3850 m
in the ocean, estimate (a) the water pressure outside the sphere, (b) the maximum elastic stress in
the sphere, in lbf/in 2 , and (c) the factor of safety of the titanium alloy (6% aluminum, 4%
vanadium).
Solution 2.C9
This problem requires you to know (or read about) some solid mechanics!
(a) The hydrostatic (gage) pressure outside the submerged sphere would be
Problem 2.W1
Consider a hollow cone with a vent hole in the vertex at the top, along with a hollow cylinder,
open at the top, with the same base area as the cone. Fill both with water to the top. The
hydrostatic paradox is that both containers have the same force on the bottom due to the water
pressure, although the cone contains 67 percent less water. Can you explain the paradox?
Solution 2.W1
Problem 2.W2
Can the temperature ever rise with altitude in the real atmosphere? Wouldn’t this cause the air
pressure to increase upward? Explain the physics of this situation.
Solution 2.W2
Problem 2.W3
Consider a submerged curved surface that consists of a two-dimensional circular arc of arbitrary
angle, arbitrary depth, and arbitrary orientation. Show that the resultant hydrostatic pressure
force on this surface must pass through the center of curvature of the arc.
Solution 2.W3
Problem 2.W4
Fill a glass approximately 80 percent with water, and add a large ice cube. Mark the water level.
The ice cube, having SG ≈ 0.9, sticks up out of the water. Let the ice cube melt with negligible
evaporation from the water surface. Will the water level be higher than, lower than, or the same
as before?
Solution 2.W4
Problem 2.W5
A ship, carrying a load of steel, is trapped while floating in a small closed lock. Members of the
crew want to get out, but they can’t quite reach the top wall of the lock. A crew member suggests
throwing the steel overboard in the lock, claiming the ship will then rise and they can climb out.
Will this plan work?
Solution 2.W5
The solution to this word problem is not provided.
Problem 2.W6
Consider a balloon of mass m floating neutrally in the atmosphere, carrying a person/basket of
mass M . m. Discuss the stability of this system to disturbances.
Solution 2.W6
Problem 2.W7
Consider a helium balloon on a string tied to the seat of your stationary car. The windows are
closed, so there is no air motion within the car. The car begins to accelerate forward. Which way
will the balloon lean, forward or backward? Hint: The acceleration sets up a horizontal pressure
gradient in the air within the car.
Solution 2.W7
Problem 2.W8
Repeat your analysis of Prob. W2.7 to let the car move at constant velocity and go around a
curve. Will the balloon lean in, toward the center of curvature, or out?
Solution 2.W8
Problem 2.W9
The deep submersible vehicle ALVIN weighs approximately 36,000 lbf in air. It carries 800 lbm
of steel weights on the sides. After a deep mission and return, two 400-lbm piles of steel are left
on the ocean floor. Can you explain, in terms relevant to this chapter, how these steel weights are
used?
Solution 2.W9
Problem 2.1
For the two-dimensional stress field in Fig. P2.1, let
22
xx yy
3000 lbf/ft 2000 lbf/ft
==
2
xy 500 lbf/ft
=
Find the shear and normal stresses (in lbf/ft2) on plane AA cutting through at a 30 angle as
shown.
Figure P2.1
Solution 2.1
Make cut “AA” so that it just hits the bottom right corner of the element. This gives the freebody
shown at right. Now sum forces normal and tangential to side AA. Denote side length AA as
“L.”
Problem 2.2
For the two-dimensional stress field shown in Fig. P2.1 suppose that σxx = 2000 lbf/ft2
σyy = 3000 lbf/ft2 σn(AA) = 2500 lbf/ft2 Compute ( a ) the shear stress σxy and ( b ) the shear stress
on plane AA .
Solution 2.2
Sum forces normal to and tangential to AA in the element freebody above, with
n(AA) known
and
xy unknown:
Problem 2.3
A vertical clean glass piezometer tube has an inside diameter of 1 mm. When a pressure is
applied, water at 20C rises into the tube to a height of 25 cm. After correcting for surface
tension, estimate the applied pressure in Pa.
Solution 2.3
For water, let Y = 0.073 N/m, contact angle
= 0, and
= 9790 N/m3. The capillary rise in the
tube, from Example 1.9 of the text, is
Problem 2.4
Pressure gages, such as the Bourdon gage in Fig. P2.4, are calibrated with a deadweight piston. If
the Bourdon gage is designed to rotate the pointer 10 degrees for every 2 psig of internal
pressure, how many degrees does the pointer rotate if the piston and weight together total
44 newtons?
Solution 2.4
The deadweight, divided by the piston area, should equal the pressure applied to the Bourdon
gage. Stay in SI units for the moment:
Problem 2.5
Quito, Ecuador has an average altitude of 9,350 ft. On a standard day, pressure gage A in a
laboratory experiment reads 63 kPa and gage B reads 105 kPa. Express these readings in gage
pressure or vacuum pressure, whichever is appropriate.
Solution 2.5
Convert 9,350 ft x 0.3048 = 2,850 m. We can interpolate in the Standard Altitude Table A.6 to a
Problem 2.6
Any pressure reading can be expressed as a length or head, h = p / ρ g . What is standard sea-
level pressure expressed in ( a ) ft of glycerin, ( b ) in Hg, ( c ) m of water, and ( d ) mm of
ethanol? Assume all fluids are at 20°C.
Solution 2.6
Take the specific weights,
=
g, from Table A.3, divide patm by
:
Problem 2.7
La Paz, Bolivia is at an altitude of approximately 12,000 ft. Assume a standard atmosphere.
How high would the liquid rise in a methanol barometer, assumed at 20C? [HINT: Don’t
forget the vapor pressure.]
Solution 2.7
Convert 12,000 ft to 3658 meters, and Table A.6, or Eq. (2.20), give
Problem 2.8
Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the surface of
Mars. Estimate the absolute pressure, in Pa, at the bottom of this speculative lake.
Solution 2.8
Take the specific weights,
=
g, from Table A.3, divide patm by
:
Problem 2.9
A storage tank, 26 ft in diameter and 36 ft high, is filled with SAE 30W oil at 20C. (a) What is
the gage pressure, in lbf/in2, at the bottom of the tank? (b) How does your result in (a) change if
the tank diameter is reduced to 15 ft? (c) Repeat (a) if leakage has caused a layer of 5 ft of water
to rest at the bottom of the (full) tank.
Solution 2.9
This is a straightforward problem in hydrostatic pressure. From Table A.3, the density of SAE
30W oil is 891 kg/m3 515.38 = 1.73 slug/ft3. (a) Thus the bottom pressure is
Problem 2.10
A large open tank is open to sea level atmosphere and filled with liquid, at 20ºC, to a depth of
50 ft. The absolute pressure at the bottom of the tank is approximately 221.5 kPa. From
Table A.3, what might this liquid be?
Solution 2.10
Convert 50 ft to 15.24 m. Use the hydrostatic formula to calculate the bottom pressure:
Problem 2.11
In Fig. P2.11, sensor A reads 1.5 kPa (gage). All fluids are at 20C. Determine the elevations ɀ in
meters of the liquid levels in the open piezometer tubes B and C.
Solution 2.11
(B) Let piezometer tube B be an arbitrary distance H above the gasoline-glycerin interface. The
specific weights are
air 12.0 N/m3,
gasoline = 6670 N/m3, and
glycerin = 12360 N/m3. Then apply
the hydrostatic formula from point A to point B:
Problem 2.12
In Fig. P2.12 the tank contains water and immiscible oil at 20C. What is h in centimeters if the
density of the oil is 898 kg/m3?
Solution 2.12
For water take the density = 998 kg/m3. Apply the hydrostatic relation from the oil surface to the
water surface, skipping the 8-cm part:
Problem 2.13
In Fig. P2.13 the 20C water and gasoline are open to the atmosphere and are at the same
elevation. What is the height h in the third liquid in the right leg?
Solution 2.13
Take water = 9790 N/m3 and gasoline = 6670 N/m3. The bottom pressure must be the same
whether we move down through the water or through the gasoline into the third fluid:
Problem 2.14
For the three-liquid system shown, compute h1 and h2. Neglect the air density.
