1–50E

1–50E A pressure gage connected to a tank reads 50 psi at
a location where the barometric reading is 29.1 mm Hg.
Determine the absolute pressure in the tank. Take rHg
848.4 lbm/ft
3
. Answer: 64.3 psia

1–49

1–49 A vacuum gage connected to a tank reads 15 kPa at a
location where the barometric reading is 750 mm Hg. Determine the absolute pressure in the tank. Take rHg 13,590
kg/m
3
. Answer: 85.0 kPa
ANS

1–48

1–48 Consider a 70-kg woman who has a total foot imprint
area of 400 cm2. She wishes to walk on the snow, but the
snow cannot withstand pressures greater than 0.5 kPa. Determine the minimum size of the snowshoes needed (imprint
area per shoe) to enable her to walk on the snow without
sinking.
ANS

1–47E

1–47E A 200-pound man has a total foot imprint area of 72
in2. Determine the pressure this man exerts on the ground if
(a) he stands on both feet and (b) he stands on one foot.
ANS

1–46E

1–46E Show that 1 kgf/cm2 14.223 psi.
ANS

1–45

1–45 The absolute pressure in water at a depth of 5 m is
read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0.85 at the same location.
ANS
1-45The absolute pressure inwater ata specified depth isgiven. The localatmospheric pressure and the
absolutepressure atthe samedepth ina differentliquid are tobe determined.
Assumptions  The liquid and water are incompressible.
Properties The specific gravityof the fluid isgiven tobe SG = 0.85. Wetake the densityof water tobe
1000 kg/m
3
. Thendensity ofthe liquidis obtainedbymultiplying its specific gravity by the density of
water,

1–44

1–44 The gage pressure in a liquid at a depth of 3 m is read
to be 28 kPa. Determine the gage pressure in the same liquid
at a depth of 9 m.
ANS
1-44The gage pressure ina liquid ata certaindepth isgiven. The gage pressure in the same liquid at a
differentdepth istobe determined.
Assumptions  The variation of the densityof the liquid withdepth isnegligible.
Analysis The gage pressure attwo differentdepths of a liquid can be expressed as

1–43 Determine the atmospheric pressure at a location
where the barometric reading is 750 mm Hg. Take the density
of mercury to be 13,600 kg/m3.
ANS
1-43The barometricreading ata location isgiven inheightof mercury column. The atmospheric pressure
is tobedetermined.

1–42

1–42 The water in a tank is pressurized by air, and the
pressure is measured by a multifluid manometer as shown in
Fig. P1–42. Determine the gage pressure of air in the tank if
h1 0.2 m,h2 0.3 m, and h3
0.46 m. Take the densities
of water, oil, and mercury to be 1000 kg/m
3, 850 kg/m3, and13,600 kg/m3, respectively.

ANS

1-42 The pressure ina pressurized water tank ismeasured bya multi-fluid manometer. The gage pressure 

ofair inthe tankis tobedetermined. 

Assumptions The air pressure inthe tankis uniform(i.e., its variationwithelevationis negligible due to its 

low density), and thus we can determine the pressure at the air-water interface. 

Properties  The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kg/m

3

, 

respectively. 

AnalysisStarting with the pressure at point 1 at the air-water interface, and moving along the tube by 

adding (as we go down) or subtracting (as we go up) the  gh ρ terms until we reach point 2, and setting the 

result equal to Patm

since the tube isopen tothe atmosphere gives 

1–41E

1–41E A manometer is used to measure the air pressure in
a tank. The fluid used has a specific gravity of 1.25, and the
differential height between the two arms of the manometer is
28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the
manometer arm with the (a) higher and (b) lower fluid level
being attached to the tank.
ANS
1-41E The pressure ina tank ismeasured witha manometer bymeasuring the differentialheightof the
manometer fluid. The absolutepressure inthe tank isto be determined for the cases of the manometer arm
withthe higher and lower fluid levelbeing attached tothe tank .

1–40

1–40 A vacuum gage connected to a chamber reads 35 kPa
at a location where the atmospheric pressure is 92 kPa.
Determine the absolute pressure in the chamber.
ANS

1–39C

1–39C Consider two identical fans, one at sea level and the
other on top of a high mountain, running at identical speeds.
How would you compare (a) the volume flow rates and
(b) the mass flow rates of these two fans?
ANS
1-39CThe densityof air atsea levelishigher than the densityof air on top of a high mountain. Therefore,
the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of
the fanat sea level will behigher.

1–38C

1–38C Express Pascal’s law, and give a real-world example
of it.
ANS
1-38C  Pascal’s principlestates that the pressure appliedtoaconfinedfluidincreases the pressure
throughout by the same amount. Thisisa consequence of the pressure ina fluid remaining constantinthe
horizontaldirection. An example of Pascal’s principle isthe operation of the hydrauliccar jack.

1–37C

1–37C A tiny steel cube is suspended in water by a string.
If the lengths of the sides of the cube are very small, how
would you compare the magnitudes of the pressures on the
top, bottom, and side surfaces of the cube?
ANS
1-37CIf the lengths of the sides of the tinycube suspended inwater by a string are very small, the
magnitudes ofthe pressures onall sides ofthe cubewill bethe same.

1–37C

1–37C A tiny steel cube is suspended in water by a string.
If the lengths of the sides of the cube are very small, how
would you compare the magnitudes of the pressures on the
top, bottom, and side surfaces of the cube?
ANS
1-37CIf the lengths of the sides of the tinycube suspended inwater by a string are very small, the
magnitudes ofthe pressures onall sides ofthe cubewill bethe same.

1–36C

1–36C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled.
Do you agree? Explain.
ANS
1-36CNo, the absolutepressure ina liquid of constantdensitydoes notdouble when the depth isdoubled.
It is the gage pressure thatdoubles when the depth isdoubled.

1–35C

1–35C Explain why some people experience nose bleeding
and some others experience shortness of breath at high elevations.
ANS
1-35CThe atmospheric pressure, which is the externalpressure exerted on the skin, decreases with
increasing elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference
between the blood pressure inthe veins and the air pressure outside increases. Thispressure imbalance may
cause some thin-walled veins such as the ones inthe nose toburst, causing bleeding. The shortness of
breath iscaused bythe lower air densityathigher elevations, and thus lower amountof oxygen per unit
volume.

1–34C

Pressure, Manometer, and Barometer
1–34C What is the difference between gage pressure and
absolute pressure?
ANS
Pressure, Manometer, and Barometer
1-34CThe pressure relative tothe atmospheric pressure is calledthe gage pressure, and the pressure
relative toanabsolute vacuumis called absolute pressure.

1–33

1–33 Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B
contains 200 kJ of thermal energy at 50°C. Now the systems
are brought into contact with each other. Determine the direction of any heat transfer between the two systems.
ANS
1-33Two systems having differenttemperatures and energy contents are brought in contact. The direction
ofheat transfer is tobedetermined.
Analysis Heat transfer occurs fromwarmer tocooler objects.  Therefore, heat will betransferredfrom
systemB to systemA until both systems reach the sametemperature.

1–32E

1–32E The temperature of a system drops by 45°F during a
cooling process. Express this drop in temperature in K, R,
and °C.
ANS
1-32EA temperature change isgiven in °F. Itistobe expressed in °C, K, andR.
Analysis Thisproblemdeals withtemperature changes, which are identicalin Rankine and Fahrenheit
scales. Thus,
∆T(R) = ∆T(°F) = 45R
The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in
Fahrenheitand Rankine scales by
∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K
and  ∆T(°C) = ∆T(K) = 25°C