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Unformatted text preview: 82 mama 3 I FLUID srnncs 61‘ 3.14 An inverted cylindrical container is lowered slowly beneath the surface of a p00 of water. Air trapped in the container is
compressed isothermally as the hydrostatic pressure increases.
Develop an expression for the water height, y, inside the container
in terms of the container height. H. and depth of submersion, h.
Plot y/H versus h/H. 3.15 You close the top of your straw using your thumb and lift it
out of your glass containing Coke. Holding it vertically. the total
length of the straw is I? in., but the Coke held in the straw is in
the bottom 6 in. What is the pressure in the straw just below your
thumb? ignore any surface tension effects. 3.16 A water tank ﬁlled with water to a depth of 5 111 has an in
spection cover (2.5 cm X 2.5 cm square) at its base, held in place
by a plastic bracket. The bracket can hold a load of 40 N. [s the
bracket strong enough? If it is, what would the water depth have
to be to cause the bracket to break? 3.17 A container with two circular vertical tubes of diameters
d. = 39.5 mm and d3 = 12.7 mm is partially ﬁlled with mercury.
The equilibrium level of the liquid is shown in the left diagram. A
cylindrical object made from solid brass is placed in the larger tube
so that it ﬂoats, as shown in the right diagram. The object is D =
37.5 mm in diameter and H = 76.2 mm high. Calculate the pressure
at the lower surface needed to ﬂoat the object. Determine the new
equilibrium level. h. of the mercury with the brass cylinder in place. l‘ﬂ “i Mercury
P3.l7 3.18 A partitioned tank as shown contains water and mercury.
What is the gage pressure in the air trapped in the left chamber? What pressure would the air on the left need to be pumped to in
order to bring the water and mercury free surfaces level? P3. [8. 3.19 P320 3.19 in the tank of Problem 3.18. if the opening to atmosphere
on the right chamber is ﬁrst sealed, what pressure would the air on
the left now need to be pumped to in order to bring the water and
mercury free surfaces level? (Assume the air trapped in the right
chamber behaves isothermally.) 3.20 A manometer is formed from glass tubing with uniform in
side diameter. D = 6.35 mm. as shown. The Utube is partially
ﬁlled with water. Then ti = 3.25 cm3 of Meriam red oil is
added to the left side. Calculate the equilibrium height. H.
when both legs of the Utube are open to the atmosphere. 3.21 Consider the twofluid manometer shown. Calculate the ap
plied pressure difference. P1 P2 Pt P2 air"' 15:
Jﬁvzﬂew‘}. / Carbon
tetrachloride "' Lt.
‘. a P3.2 P322 3.22 The manometer shown contains two liquids. Liquid A has
50 = 0.88 and liquid B has SG = 2.95. Calculate the deﬂection. h. when the applied pressure difference is pl — p2 = 18 lbffftz. ,4 3.23 The manometer shown contains water and kerosene. With
both tubes open to the atmosphere. the freesurface elevations dif
fer by Hu = 20.0 mm. Determine the elevation difference when a
pressure of 98.0 Pa (gage) is applied to the right tube. Liquid .4 ',..e ’ _ e:me ea Kerosene Liquid 8
133.23 P324 3.24 Determine the gage pressure in psig at point a, if liquid A
has SG = 0.75 and liquid 3 has S0 = 1.20. The liquid surrounding
point a is water and the tank on the left is open to the atmosphere. 3.25 The NIH Corporation‘s engineering department is evaluat
ing a sophisticated $80,000 laser system to measure the difference
in water level between two large water storage tanks. It is import
ant that small differences be measured accurately. You suggest
that the job can be done with a $200 manometer arrangement. An
oil less dense than water can be used to give a 10 : l ampliﬁcation
of meniscus movement; a small difference in level between the
tanks will cause 10 times as much deﬂection in the oil levels in
the manometer. Determine the speciﬁc gravity of the oil required
for 10: l ampliﬁcation.  ‘ \
"32x \ ‘ 3.26 Consider a tank containing mercury. water. benzene. and
air as shown. Find the air pressure (gage). If an opening is made
in the top of the tank. ﬁnd the equilibrium level of the mercury in
the manometer. It: 0.25
p__l__m .H 3.27 Water flows downward along a pipe that is inclined at 30‘
below the horizontal. as shown. Pressure difference pA e p3 is
due partly to gravity and partly to friction. Derive an algebraic
expression for the. pressure difference. Evaluate the pressure
difference it'L = 5 ft and It = 6 in. 3.28 A rectangular tank, open to the atmosphere, is ﬁlled with
water to a depth of 2.5 m as shown. A Utube manometer is
connected to the tank at a location 0.7 m above the tank bot
tom. II‘ the zero level of the Meriam blue manometer ﬂuid is P3 .29 PROBLEMS 83 0.2 m below the connection. determine the deﬂection I after
the manometer is connected and all air has been removed
from the connecting leg. 3.29 A reserVoir manometer has vertical tubes of diameter D =
l8 mm and d = 6 mm. The manometer liquid is Meriam red oil.
Develop an algebraic expression for liquid deflection L in the
small tube when gage pressure A]; is applied to the reservoir.
Evaluate the liquid deflection when the applied pressure is equiv
alent to 25 mm of water (gage). 3.30 The manometer ﬂuid of Problem 3.28 is replaced with mer
cury (same zero level). The tank is sealed and the air pressure is in
creased to a gage pressure o‘l't).5 atm. Determine the deﬂection I. 3.3] A reservoir manometer is calibrated for use with a liquid of
speciﬁc gravity 0.827. The reservoir diameter is 5/3 in. and the (verti
cal) tube diameter is 3/16 in. Calculate the required distance between
marks on the vertical scale for l in. of water pressure difference. 3.32 The inclinedtube manometer shown has D = 76 mm and
d = 8 mm. and is ﬁlled with Meriam red oil. Compute the angle.
H. that will give a l5«cm oil deﬂection along the inclined tube for
an applied pressure of 25 mm of water (gage). Determine the sen—
sitivity of this manometer. P332. 3.33 3.33 The inclinedtube manometer shown has D = 96 mm and
d = 8 mm. Determine the angle. 6. required to provide a 5 :  in—
crease in liquid deﬂection. L. compared with the total deﬂection in
a regular Ulube manometer. Evaluate the sensitivity of this
inclinedtube manometer. @6134 A student wishes to design a manometer with better sensi tivity than a waterﬁlled Utube of constant diameter. The stu
dent‘s concept involves using tubes with different diameters
and two liquids. as shown. Evaluate the deﬂection h of
this manometer. if the applied pressure difference is Ap = 250
ij2. Determine the sensitivity of this manometer. Plot the
manometer sensitivity as a function of the diameter ratio dgfdl. f’atm Farm Patm + AP Patrn llﬂil It
Oil \tsc = 0.85) P334 3.35 A barometer accidentally contains 6.5 inches of water on
top of the mercury column (so there is also water vapor instead of 84 CHAPTER 3 I FLUID STATIGS a vacuum at the top of the barometer). On a day when the tem~
perature is 70”F. the mercury column height is 28.35 inches (cor
rected for thermal expansion). Determine the barometric pressure
in psia. [fthe ambient temperature increased to 85F and the baro
metric pressure did not change, would the mercury column be
longer, be shorter, or remain the same length? Justify your
answer. 3.36 If the tank of Problem 3.28 is sealed tightly and water
drains slowly from the bottom of the tank. determine the deﬂec
tion, 1, after the system has attained equilibrium. 3.37 A water column stands 50 mm high in a 2.5mmvdiameter glass
tube. What would be the column height if the surface tension were
zero? What would be the column height in a l.0—mindiameter tube? 3.38 Consider a small diameter openended tube inserted at the
interface between two immiscible fluids of different densities.
Derive an expression for the height difference Ah between the in
terface level inside and outside the tube in terms of tube diameter
D, the two ﬂuid densities, p1 and p2, and the surface tension 0 and
angle 0 for the two ﬁuids‘ interface. [f the two fluids are water and
mercury, ﬁnd the tube diameter such that Ah < 10 mm. 3.39 You have a manometer consisting of a tube that is 1.1cm
ID. On one side the manometer leg contains mercury. 10 cc of an
oil ($0 = 1.67). and 3 cc of air as a bubble in the oil. The other
leg just contains mercury. Both legs are open to the atmosphere
and are in a static condition. An accident occurs in which 3 cc of
the oil and the air bubble are removed from the one leg. How
much do the mercury height levels change? \9‘340 Based on the atmospheric temperature data of the US.
Standard Atmosphere of Fig. 3.3, compute and plot the pressure variation with altitude, and compare with the pressure data of
Table A.3. 6:33.41 Two vertical glass plates 300 mm X 300 mm are placed in
an open tank containing water. At one end the gap between the
plates is 0.1 mm. and at the other it is 2 mm. Plot the curve ofwatcr
height between the plates from one end of the pair to the other. 3.42 Compare the height due to capillary action of water
exposed to air in a circular tube of diameter D = 0.5 mm, and
between two inﬁnite vertical parallel plates of gap a 2 0.5 mm. $3.43 On a certain calm day, a mild inversion causes the atmo—
spheric temperature to remain constant at 85“]3 between sea level
and 16.000 ft altitude. Under these conditions, (a) calculate the
elevation change for which a 2 percent reduction in air pressure
occurs. (b) determine the change of elevation necessary to effect a
[0 percent reduction in density, and (c) plot py’p. and pg/p. as a
function of A2. 6363.44 The Martian atmosphere behaves as an ideal gas with mean
molecular mass of 32.0 and constant tentperature of 200 K. The
atmospheric density at the planet surface is p = 0.015 kg/m'l and
Martian gravity is 3.92 m/sz. Calculate the density of the Martian
atmosphere at height a : 20 km above the surface. Plot the ratio
of density to surface density as a function of elevation. Compare
with that for data on the earth’s atmosphere. 6163.45 At ground level in Denver. Colorado. the atmospheric
pressure and temperature are 83.2 kPa and 25C. Calculate the
pressure on Pike’s Peak at an elevation of 2690 in above the city
assuming (a) an incompressible and (b) an adiabatic atmosphere. Plot the ratio of pressure to ground level pressure in Denver as a
function of elevation for both cases. 3.46 A door 1 m wide and 1.5 m high is located in a plane verti~
cal wall of a water tank. The door is hinged along its upper edge,
which is 1 m below the water surface. Atmospheric pressure acts
on the outer surface of the door. (a) If the pressure at the water
surface is atmospheric. what force must be applied at the lower
edge of the door in order to keep the door from opening? lb) if the
water surface gage pressure is raised to 0.5 atm. what force must
be applied at the lower edge of the door to keep the door from
opening? (c) Find the ratio FIFO as a function of the surface pres
sure ratio p./p,,[,,,. {Fm is the force required when p,, = palm.) .347 A hydropneumatic elevator consists of a pistonrcylinder assembly to lift the elevator cab. Hydraulic oil. stored in an accu
mulator tank pressurized by air, is valved to the piston as needed
to lift the elevator. When the elevator descends, oil is returned to
the accumulator. Design the least expensive accumulator that can
satisfy the system requirements. Assume the lift is 3 ﬂoors, the
maximum load is [0 passengers, and the maximum system pres
sure is 800 kPa (gage). For column bending strength. the piston
diameter must be at least 150 mm. The elevator cab and piston
have a combined mass of 3000 kg, and are to be purchased. Per
fomt the analysis needed to deﬁne. as a function of system operat
ing pressure. the piston diameter, the accumulator volume and
diameter. and the wall thickness. Discuss safety features that your
company should specify for the complete elevator system. Would
it be preferable to use a completely pneumatic design or a com
pletely hydraulic design? Why? ' 6363.48 A door I m wide and 1.5 m high is located in a plane verti cal wall of a water tank. The door is hinged along its upper edge,
which is l m below the water surface. Atmospheric pressure acts
on the outer surface of the door and at the water surface. (a) Deter
mine the magnitude and line of action of the total resultant force
from all ﬂuids acting on the door. (b) If the water surface gage
pressure is raised to 0.3 atm. what is the resultant force and where
is its line of action? (c) Plot the ratios HF” and _v'/_v,. for different
values of the surface pressure ratio pv/pmm. (F0 is the resultant
force when p, = pwm.) 3.49 Find the pressures at points A. B. and C. as shown, and in
the two air cavities. P3 .49 3.50 A triangular access port must he provided in the side ofa
form containing liquid concrete. Using the coordinates and dimen
sions shown, determine the resultant force that acts on the port
and its point of application. u—iasru in it 1.1. trial a H' n hitFt ...
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 Fall '08
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