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Monday, October 20, 2014

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
1-50E A pressure gage connected toa tank reads 50 psi.The absolutepressure inthe tank istobe  determined.  Properties The densityof mercuryisgiven tobe ρ= 848.4 lbm/ft 3 .  50 psi  Pabs Analysis The atmospheric (or barometric) pressure can be expressed as  psia   14.29 in   144 ft   1 ft/s lbm   32.2 lbf   1 ft)   )(29.1/12 ft/s   )(32.2 lbm/ft   (848.4 2 2 2 2 3 atm =             ⋅ = = h g P ρ Thenthe absolute pressure inthe tankis  psia   64.3 = + = + = 14.29 50 atm gage abs P P P

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
1-49The vacuumpressure reading of a tank isgiven. The absolutepressure inthe tank istobe determined.  Properties The densityof mercuryisgiven tobe ρ= 13,590 kg/m 3 .  Analysis The atmospheric (or barometric) pressure can be expressed as  15kPa Pabs kPa   0 . 100 N/m   1000 kPa   1 m/s kg   1 N   1 m)   )(0.750 m/s   )(9.807 kg/m   (13,590 2 2 2 3 atm =             ⋅ = = h g P ρ Patm=750 mmHg Thenthe absolute pressure inthe tankbecomes  kPa   85.0 = − = − = 15 100.0 vac atm abs P P P

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
1-48 The mass of a woman isgiven. The minimumimprintarea per shoe needed toenableher towalkon  the snow withoutsinking istobe determined.  Assumptions1 The weight of the person isdistributed uniformlyon the imprintarea of the shoes. 2 One  footcarries the entire weightof a person during walking, and the shoe issized for walking conditions  (rather than standing). 3 The weightofthe shoes is negligible.  Analysis The mass of the woman isgiven tobe 70 kg. For a pressure of  0.5 kPa on the snow, the imprintarea of one shoe mustbe  2 m   1.37 =             ⋅ = = = 2 2 2 N/m   1000 kPa   1 m/s kg   1 N   1 kPa   0.5 ) m/s   kg)(9.81   (70 P mg P W A Discussion Thisisa verylarge area for a shoe, and such shoes would be  impracticaltouse. Therefore, somesinking of the snow should be allowed  tohave shoes of reasonable size.

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
1-47E The weight and the foot imprintarea of a person are given. The pressures thisman exertson the  ground when he stands on one and on both feetare tobe determined.  Assumptions  The weightof the person isdistributed uniformlyon footimprintarea. Analysis The weightof the man isgiven tobe 200 lbf. Noting thatpressure is force per  unitarea, the pressure thisman exertson the ground is (a) On both feet:   psi   2.78 = = × = =   lbf/in   78 . 2 in   36 2 lbf   200 2 2 2 A W P (b) On one foot:    psi   5.56 = = = =   lbf/in   56 . 5 in   36 lbf   200 2 2 A W P Discussion Notethatthe pressure exerted on the ground (and on the feet) isreduced by halfwhen the person stands on both feet.

1–46E

1–46E Show that 1 kgf/cm2 14.223 psi.
ANS
1-46EItistobe shown that1 kgf/cm 2 = 14.223 psi.  Analysis Noting that1 kgf = 9.80665 N, 1 N = 0.22481 lbf, and 1 in= 2.54 cm, we have  lbf   20463 . 2 N   1 lbf   0.22481 )   N   9.80665 (   N   9.80665   kgf   1 =       = = and  psi   14.223 = =       = = 2 2 2 2 2 lbf/in   223 . 14 in   1 cm   2.54 )   lbf/cm   20463 . 2 (   lbf/cm   20463 . 2 kgf/cm   1

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