KNOWN: Mass, pressure, temperature, and specific volume of water vapor.

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Teresa Green
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1 .0 The specific volume of 5 kg of water vapor at.5 MPa, 440 o C is 0.60 m /kg. Determine (a) the volume, in m, occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules. KNOWN: Mass, pressure, temperature, and specific volume of water vapor. FIND: Determine (a) the volume, in m, occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules. m = 5 kg p =.5 MPa T = 440 o C v = 0.60 m /kg. The water vapor is a closed system. ANALYSIS: (a) The specific volume is volume per unit mass. Thus, the volume occupied by the water vapor can be determined by multiplying its mass by its specific volume. m V mv (5 kg) 0.60 =.08 m kg (b) Using molecular weight of water from Table A- and applying the appropriate relation to convert the water vapor mass to gram moles gives m n M 5 kg 000 moles kg kmol 8.0 kmol = 77.5 moles (c) Using Avogadro s number to determine the number of molecules yields # Molecules Avogadro' s molecules Number # moles 6.00 (77.5 moles) mole # Molecules = molecules

2 .5 A gas contained within a piston-cylinder assembly undergoes four processes in series: Process -: Constant-pressure expansion at bar from V = 0.5 m to V = m Process -: Constant volume to bar Process -4: Constant-pressure compression to m Process 4-: Compression with pv = constant Sketch the process in series on a p-v diagram labeled with pressure and volume values at each numbered state. p (bar) V (m )

3 .0 Figure P.0 shows a tank within a tank, each containing air. Pressure gage A, which indicates pressure inside tank A, is located inside tank B and reads 5 psig (vacuum). The U-tube manometer connected to tank B contains water with a column length of 0 in. Using data on the diagram, determine the absolute pressure of the air inside tank B and inside tank A, both in psia. The atmospheric pressure surrounding tank B is 4.7 psia. The acceleration of gravity is g =. ft/s. KNOWN: A tank within a tank, each containing air. FIND: Absolute pressure of air in tank B and in tank A, both in psia. p atm = 4.7 psia Tank B L = 0 in. Tank A Gage A Water ( = 6.4 lb/ft ) g =. ft/s p gage, A = 5 psig (vacuum). The gas is a closed system.. Atmospheric pressure is exerted at the open end of the manometer.. The manometer fluid is water with a density of 6.4 lb/ft. ANALYSIS: (a) Applying Eq.. p gas,b = p atm + gl where p atm is the local atmospheric pressure to tank B, is the density of the manometer fluid (water), g is the acceleration due to gravity, and L is the column length of the manometer fluid. Substituting values lbf lb ft lbf ft p gas,b (0 in.) = 5. lbf/in. lbm ft in. ft s. 78 in. s

4 Since the gage pressure of the air in tank A is a vacuum, Eq..5 applies. p(vacuum) = p atm (absolute) p(absolute) The pressure of the gas in tank B is the local atmospheric pressure to tank A. Solving for p (absolute) and substituting values yield p(absolute) = p atm (absolute) p(vacuum) = 5. psia 5 psig = 0. psia

5 . Show that a standard atmospheric pressure of 760 mmhg is equivalent to 0. kpa. The density of mercury is,590 kg/m and g = 9.8 m/s. KNOWN: Standard atmospheric pressure of 760 mmhg. FIND: Show that 760 mmhg is equivalent to 0. kpa. Mercury vapor Hg =,590 kg/m L = 760 mm Mercury (Hg). Local gravitational acceleration is 9.8 m/s.. Pressure of mercury vapor is much less than that of the atmosphere and can be neglected. ANALYSIS: Equation. applies. p atm = p vapor + Hg gl = Hg gl Neglecting the pressure of mercury vapor and applying appropriate conversion factors yield p atm kg m N, (760 mm) m s kg m s m 000 mm kpa N 000 m = 0. kpa

6 .7 If the water pressure at the base of the water tower shown in Fig. P.7 is 4.5 bar, determine the pressure of the air trapped above the water level, in bar. The density of the water is 0 kg/m. And g = 9.8 m/s. KNOWN: Air is trapped above a column of water in a water tower. FIND: the pressure of the air trapped above the water level. Air Water L = 0 m p = 4.5 bar. Water density is 0 kg/m.. Local gravitational acceleration is 9.8 m/s. ANALYSIS: Ignoring the vertical variation in pressure of the air trapped above the water level, p = p air + gl pressure at the base p air = p gl p air = (4.5 bar) 0 kg m N 9.8 (0 m) m s kg m s bar 5 N 0 m =. bar

THERMODYNAMICS, HEAT AND MASS TRANSFER TUTORIAL NO: 1 (SPECIFIC VOLUME, PRESSURE AND TEMPERATURE)

THERMODYNAMICS, HEAT AND MASS TRANSFER TUTORIAL NO: 1 (SPECIFIC VOLUME, PRESSURE AND TEMPERATURE) 1. A vacuum gauge mounted on a condenser reads 66 cm Hg. What is the absolute pressure in the condenser