# tUT 7 - Problems 1 2 3 4 Solutions 1 The air in an...

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Unformatted text preview: Problems 1) 2) 3) 4) Solutions 1) The air in an insulated, rigid compressed-air tank is released until the pressure in the tank reduces to a specified value. The final temperature of the air in the tank is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The tank is well-insulated, and thus there is no heat transfer. Properties The gas constant of air is 0.287 kPa·m 3 /kg·K (Table A-1). The specific heats of air at room temperature are c p = 1.005 kJ/kg·K and c v = 0.718 kJ/kg·K (Table A-2 a ). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u , respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance : 2 1 1 2 system out in m m m m m m m m m e e − = − = − Δ = − Energy balance : e p e e e e e T c m T c m T c m h m u m u m u m u m h m E E E + − = + − = − = − Δ = − 1 1 2 2 1 1 2 2 1 1 2 2 energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in v v...
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## This note was uploaded on 03/09/2011 for the course ENGR 2680 taught by Professor Ahmadbarari during the Spring '11 term at UOIT.

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tUT 7 - Problems 1 2 3 4 Solutions 1 The air in an...

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