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Solution 2.14
The pressures at the three top surfaces must all be atmospheric, or zero gage pressure. Compute
12
Problem 2.15
The air–oil–water system in Fig. P2.15 is at 20°C. Knowing that gage A reads 15 lbf/in2 absolute
and gage B reads 1.25 lbf/in2 less than gage C , compute ( a ) the specific weight of the oil in
lbf/ft3 and ( b ) the actual reading of gage C in lbf/in2 absolute.
Solution 2.15
First evaluate
air = (pA/RT)g = [15 144/(1717 528)](32.2) 0.0767 lbf/ft3. Take
water = 62.4 lbf/ft3. Then apply the hydrostatic formula from point B to point C:
Problem 2.16
If the absolute pressure at the interface between water and mercury in Fig. P2.16 is 93 kPa, what,
in lbf/ft2, is (a) the pressure at the surface, and (b) the pressure at the bottom of the container?
Solution 2.16
Do the whole problem in SI units and then convert to BG at the end. The bottom width and the
slanted 75-degree walls are irrelevant red herrings. Just go up and down:
Problem 2.17
The system in Fig. P2.17 is at 20ºC. Determine the height h of the water in the left side.
Solution 2.17
The bottom pressure must be the same from both left and right viewpoints:
Problem 2.18
The system in Fig. P2.18 is at 20°C. If atmospheric pressure is 101.33 kPa and the pressure at the
bottom of the tank is 242 kPa, what is the specific gravity of fluid X ?
Solution 2.18
Simply apply the hydrostatic formula from top to bottom:
Problem 2.19
The U-tube in Fig. P2.19 has a 1-cm ID and contains mercury as shown. If 20 cm3 of water is
poured into the right-hand leg, what will the free surface height in each leg be after the sloshing
has died down?
Solution 2.19
First figure the height of water added:
Problem 2.20
The hydraulic jack in Fig. P2.20 is filled with oil at 56 lbf/ft3. Neglecting the weight of the
two pistons, what force F on the handle is required to support the 2000-lbf weight for this
design?
Solution 2.20
First sum moments clockwise about the hinge A of the handle:
A
M 0 F(15 1) P(1), = = + −
Problem 2.21
At 20°C gage A reads 350 kPa absolute. What is the height h of the water in cm? What should
gage B read in kPa absolute? See Fig. P2.21.
Solution 2.21
Apply the hydrostatic formula from the air to gage A:
B
Problem 2.22
The fuel gage for a gasoline tank in a car reads proportional to the bottom gage pressure as in
Fig. P2.22. If the tank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how
many centimeters of air remain at the top when the gage erroneously reads “full”?
Solution 2.22
Given
gasoline = 0.68(9790) = 6657 N/m3, compute the gage pressure when “full”:
Problem 2.23
In Fig. P2.23 both fluids are at 20C. If surface tension effects are negligible, what is the density
of the oil, in kg/m3?
Solution 2.23
Move around the U-tube from left atmosphere to right atmosphere:
Problem 2.24
In Prob. 1.2 we made a crude integration of the density distribution ρ(z) in Table A.6 and estimated
the mass of the earth’s atmosphere to be m ≈ 6 E18 kg. Can this result be used to estimate sea-
level pressure on the earth? Conversely, can the actual sea-level pressure of
101.35 kPa be used to make a more accurate estimate of the atmospheric mass?
Solution 2.24
Yes, atmospheric pressure is essentially a result of the weight of the air above. Therefore the air
weight divided by the surface area of the earth equals sea-level pressure:
Problem 2.25*
As measured by NASA’s Viking landers, the atmosphere of Mars, where g ≈ 3.71 m/s2, is almost
entirely carbon dioxide, and the surface pressure averages 700 Pa. The temperature is cold and
drops off exponentially: T To e-Cz, where C 1.3E-5 m-1 and To = 250 K. For example, at
20,000 m altitude, T 193 K. (a) Find an analytic formula for the variation of pressure with
altitude. (b) Find the altitude where pressure on Mars has dropped to 1 pascal.
Solution 2.25*
(a) The analytic formula is found by integrating Eq. (2.17) of the text:
Problem 2.26
For gases that undergo large changes in height, the linear approximation, Eq. (2.14), is
inaccurate. Expand the troposphere power-law, Eq. (2.20), into a power series, and show that the
linear approximation p pa -
a g z is adequate when
Solution 2.26
The power-law term in Eq. (2.20) can be expanded into a series:
2
11 , or : .
2 ( 1)
o
o
T
n B z z Ans
T n B
− −
Problem 2.27
Conduct an experiment to illustrate atmospheric pressure. Note: Do this over a sink or you may
get wet! Find a drinking glass with a very smooth, uniform rim at the top. Fill the glass nearly
full with water. Place a smooth, light, fl at plate on top of the glass such that the entire rim of the
glass is covered. A glossy postcard works best. A small index card or one flap of a greeting card
will also work. See Fig. P2.27 a .
( a ) Hold the card against the rim of the glass and turn the glass upside down. Slowly release
pressure on the card. Does the water fall out of the glass? Record your experimental
observations.
( b ) Find an expression for the pressure at points 1 and 2 in Fig. P2.27 b . Note that the glass is
now inverted, so the original top rim of the glass is at the bottom of the picture, and the original
bottom of the glass is at the top of the picture. The weight of the card can be neglected.
( c ) Estimate the theoretical maximum glass height at which this experiment could still work,
such that the water would not fall out of the glass.
Solution 2.27
This is an experimental problem: Put a card or thick sheet over a glass of water, hold it tight, and
Problem 2.28
A correlation of computational fluid dynamics results indicates that, all other things being equal,
the distance traveled by a well-hit baseball varies inversely as the 0.36 power of the air density.
If a home-run ball hit in Citi Field in New York travels 400 ft, estimate the distance it would
travel in (a) Quito, Ecuador, and (b) Colorado Springs, CO.
Solution 2.28
Citi Field is in the Borough of Queens, NY, essentially at sea level. Hence the standard pressure
is po ≈ 101,350 Pa. Look up the altitude of the other two cities and calculate the pressure:
Problem 2.29
Follow up on Prob. P2.8 by estimating the altitude on Mars where the pressure has dropped to 20 percent
of its surface value. Assume an isothermal atmosphere, not the exponential variation of P2.25.
Solution 2.29
Problem P2.8 we used a surface temperature To = -10ºF = -23ºC = 250 K. Recall that
Problem 2.30
For the traditional equal-level manometer measurement in Fig. E2.3, water at 20C flows
through the plug device from a to b. The manometer fluid is mercury. If L = 12 cm and
h = 24 cm, (a) what is the pressure drop through the device? (b) If the water flows through the
pipe at a velocity V = 18 ft/s, what is the dimensionless loss coefficient of the device, defined by
K = p/(
V2)? We will study loss coefficients in Chap. 6.
Figure E2.3
Solution 2.30
Gather density data:
mercury = 13550 kg/m3,
water = 998 kg/m3. Example 2.3, by going down
from (a) to the mercury level, jumping across, and going up to (b), found the very important
formula for this type of equal-leg manometer:
Problem 2.31
In Fig. P2.31 all fluids are at 20°C. Determine the pressure difference (Pa) between points A and
B .
Solution 2.31
Take the specific weights to be
Benzene: 8640 N/m3
Mercury: 133100 N/m3
Problem 2.32
For the inverted manometer of Fig. P2.32, all fluids are at 20C. If pB − pA= 97 kPa, determine
the height H in centimeters.
Solution 2.32
Gamma = 9790 N/m3 for water and 133100 N/m3 for mercury and (0.827)(9790) = 8096 N/m3 for
Meriam red oil. Work your way around from point A to point B:
Problem 2.33
In Fig. P2.33 the pressure at point A is 25 lbf/in2. All fluids are at 20C. What is the air pressure
in the closed chamber B, in Pa?
Solution 2.33
Take
= 9790 N/m3 for water, 8720 N/m3 for SAE 30 oil, and (1.45)(9790) = 14196 N/m3 for the
Problem 2.34*
Sometimes manometer dimensions have a significant effect. In Fig. P2.34 containers (a) and (b)
are cylindrical and conditions are such that pa = pb . Derive a formula for the pressure difference
pa − pb when the oil–water interface on the right rises a distance Δh < h , for (a) d ≪ D and
(b) d = 0.15 D . What is the percentage change in the value of Δp ?
Solution 2.34
Take
= 9790 N/m3 for water and 8720 N/m3 for SAE 30 oil. Let “H” be the height of the oil in
reservoir (b). For the condition shown, pa = pb, therefore
Problem 2.35
Water flows upward in a pipe slanted at 30, as in Fig. P2.35. The mercury manometer reads
h = 12 cm. Both fluids are at 20°C. What is the pressure difference p1 − p2 in the pipe?
Solution 2.35
The vertical distance between points 1 and 2 equals (2.0 m)tan 30 or 1.155 m. Go around the U-
Problem 2.36
In Fig. P2.36 both the tank and the tube are open to the atmosphere. If L = 2.13 m, what is the
angle of tilt θ
of the tube?
Solution 2.36
Proceed hydrostatically from the oil surface to the slanted tube surface:
Problem 2.37
The inclined manometer in Fig. P2.37 contains Meriam red manometer oil, SG = 0.827. Assume
that the reservoir is very large. If the inclined arm is fitted with graduations 1 inch apart, what
should the angle
be if each graduation corresponds to 1 lbf/ft2 gage pressure pA?
Solution 2.37
The specific weight of the oil is (0.827)(62.4) = 51.6 lbf/ft3. If the reservoir level does not
Problem 2.38
If the pressure in container A in Fig. P2.38 is 200 kPa, compute the pressure in container B.
Solution 2.38
The specific weights are
Problem 2.39
In Fig. P2.39 the right leg of the manometer is open to the atmosphere. Find the gage pressure, in
Pa, in the air gap in the tank.
Solution 2.39
The two 8-cm legs of air are negligible (only 2 Pa). Begin at the right mercury interface and go
to the air gap:
