me104hw123 - 1-53 The absolute pressure in water at a...

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Unformatted text preview: 1-53 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be 1000 kg/m3. Then density of the liquid is obtained by multiplying its specific gravity by the density of water, v [3:3pr =(0.85)(1000kg/m3)=850 k m3 H10 Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from Patm =P—pgh 3 2 lkPa =(145kPa)—(1000 kg/m )(9.81m/s )(5m) ‘52 / .1961) kPa . lOOONm (b) The absolute pressure at a depth of 5 m in the other liquid is P = Pam +pgh = (96 OkPa) +(850 kg/m3)(9.8l m/sz)(5 )[ “(Pa 2] :13?“ kPa 1000 N/m Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected. l-60E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be p = 848.4 lbm/fis. Analysis The atmospheric (or barometric) pressure can be expressed as Pm = pgh 4 lbm/fl3 32 2 fi/ 2 )(29 1/12 a) “bf lfiz Pa“ 50 W = 848. . s . —— ( X 32.21bm-fl/sz 144 m2 = 14.29 psia Then the absolute pressure in the tank is P \ abs = Pgagc +Patm = 50+14'29 = 64'3 pSia 1-72 The air pressure in a duct is measured by a mercurylmanontileéer. For a given mercury-level difference between the two co umns, absolute pressure in the duct is to be determined. T 3 Properties The density of mercury giventotbe p. = |13_,600 kg/m . Analysis (a) The pressure in the duct is above atmospheric pressure AIR 5 since the fluid column on the duct side is at a lower level. (b) The absolute pressure in the duct is determined fi'om P = Patm +pgh 2 1 N l kPa =(100 kPa)+(13,600 kg/m3)(9-81m/S X00451“) #1“ W52 —100‘0'N,m2 —106 kPa ' l-40C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure. A l-25C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric. ' l-23C For a system to be in thermodynamic equilibrium, the temperature has to be the same throughout but the pressure does not. HOWCver, there should be no unbalanced pressure forces present. The increasing pressure with depth in a fluid, for example, should be balanced by increasing weight. 1-34E A temperature is given in °C. It is to be expressed in °F, K, and R., Analysis Using the conversion relations between the various temperature scales, T(K] = T(°C) + 273 = 18°C + 273 = 291 K T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F RR] = JI°F) + 460 = 64.4 + 460 = 524.4 R l-37E The temperature of oil given in °F unit is to be converted to °C unit. Analysis Using the conversion relation between the temperature scales, T(°C): T( F)—32 =150—32 1.8 1.8 = 65.6°C 1-38E The temperature of air given in °C unit is to be converted to °F unit. Analysis Using the conversion relation between the temperature scales, T(°F) = 1.8T(°C) +32 = (1.8)(150) +32 = 302°F 1—2C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle. J 1-8E The weight of a man on earth is given. His weight on the moon is to be determined. Analysis Applying Newton's second law to the weight force gives 2 W 1801bf [32.1741bm-fi/s]=180.41bm W= ———>m=——= mg g 32.10%2 llbf l have the same mass on the moon. Then, his weight on the moon will be “bf J: 30.71131 32.174 lbm-fi/sz Mass is invariant and the man wil W = mg = (180.41bm)(5.47 n/s2)[ mass and weight of the air in the room are to be 1-9 The interior dimensions of a room are given. The determined. nsity of air is constant throughout the room. Assumptions The de 1.16 kg/ms. Properties The density of air is given to be p = Analysis The mass of the air in the room is m = pV =(1.16 kg/m3)(6x6><8 m3)=334.1 kg Thus, IN = =334.1k 9.31m/s2 . =3277N W mg ( g)( {lam/5,] 1-12 The acceleration of an aircraft is given in g’s. The net upward force acting on a man in the aircraft is to be determined. Analysis From the Newton‘s second law, the force applied is IN 2 =5297 N 1kg-m/s F = ma = m(6 g) = (90 kg)(6><9.81rn/sz)[ "V 1-15 Gravitational. acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reductlon 1n the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is iven to be 9.807 m/s2 at l 2 altitude “13,000 m. g I sea evel and 9.767 m/s at an Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduct1on 1n weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from ‘ %Reduction in weight = %Reduction in g = Agx 100 = 100 = 0 41% . g 9.807 ' ‘ Tiltierfore, the airplane and the people in it will weight 0.41% less at 13,000 m a 1tu e. Discussion Note that the weight loss at cruising altitudes is negligible. ...
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This note was uploaded on 04/07/2008 for the course ME 104 taught by Professor Gomatam during the Spring '08 term at Lehigh University .

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me104hw123 - 1-53 The absolute pressure in water at a...

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