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2300_hw_9_sol

# 2300_hw_9_sol - Homework 9 Solutions#1 5-60 Air is expanded...

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Homework # 9 Solutions #1 5-60 Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats. Properties The constant pressure specific heat of air at the average temperature of (500+150)/2=325°C=598 K is c p = 1.051 kJ/kg·K (Table A-2 b ). The gas constant of air is R = 0.287 kPa m 3 /kg K (Table A-1). Analysis ( a ) There is only one inlet and one exit, and thus m m m & & & = = 2 1 . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as out in energies etc. potential, kinetic, internal, in change of Rate (steady) 0 system mass and work, heat, by nsfer energy tra net of Rate out in 0 E E E E E & & 4 4 4 3 4 4 2 1 & 43 42 1 & & = = D = - - + - = - + - = + + = + 2 ) ( 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 out out 2 2 2 2 1 1 V V T T c m V V h h m W W V h m V h m p & & & & & & The specific volume of air at the inlet and the mass flow rate are /kg m 2219 . 0 kPa 1000 K) 273 K)(500 /kg m kPa 287 . 0 ( 3 3 1 1 1 = + = = P RT v kg/s 36.06 = = = /kg m 0.2219 m/s) )(40 m 2 . 0 ( 3 2 1 1 1 v V A m & Similarly at the outlet, /kg m 214 . 1 kPa 100 K) 273 K)(150 /kg m kPa 287 . 0 ( 3 3 2 2 2 = + = = P RT v m/s 78 . 43 m 1 /kg) m 4 kg/s)(1.21 06 . 36 ( 2 3 2 2 2 = = = A m V v & ( b ) Substituting into the energy balance equation gives kW 13,260 = - + - = - + - = 2 2 2 2 2 2 2 1 2 1 out /s m 1000 kJ/kg 1 2 m/s) 78 . 43 ( m/s) 40 ( )K 150 K)(500 kJ/kg 051 . 1 ( kg/s) 06 . 36 ( 2 ) ( V V T T c m W p & & Turbin

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