HW7S - 5-30 Air is accelerated in a nozzle from 45 m/s to 180 m/s The mass flow rate the exit temperature and the exit area of the nozzle are to be

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5-30 Air is accelerated in a nozzle from 45 m/s to 180 m/s. The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is 0.287 kPa.m 3 /kg.K (Table A-1). The specific heat of air at the anticipated average temperature of 450 K is c p = 1.02 kJ/kg. ° C (Table A-2). Analysis ( a ) There is only one inlet and one exit, and thus ±±± mmm 12 == . Using the ideal gas relation, the specific volume and the mass flow rate of air are determined to be /kg m 0.4525 kPa 300 ) K 473 )( K /kg m kPa 0.287 ( 3 3 1 1 1 = = = P RT v kg/s 1.094 = = = ) m/s 45 )( m 0.0110 ( /kg m 0.4525 1 1 2 3 1 1 1 V A m ± ( b ) We take nozzle 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 ± ± ² ²³ ² ²´ µ ± µ ± ± = = Δ = Ê () 2 0 2 0 0) pe W (since /2) + ( ) 2 / ( 2 1 2 2 1 2 , 2 1 2 2 1 2 2 2 2 2 1 1 V V T T c V V h h Q V h m V h m ave p + = ⎯→ + = Δ = + ± ± ± ± Substituting, + = 2 2 2 2 2 /s m 1000 kJ/kg 1 2 ) m/s 45 ( ) m/s 180 ( ) C 200 )( K kJ/kg 1.02 ( 0 D T It yields T 2 = 185.2 ° C ( c ) The specific volume of air at the nozzle exit is /kg m 1.315 kPa 100 ) K 273 185.2 )( K /kg m kPa 0.287 ( 3 3 2 2 2 = + = = P RT m/s 180 /kg m 1.315 1 kg/s .094 1 1 2 3 2 2 2 A V A m = = ± A 2 = 0.00799 m 2 = 79.9 cm 2 AIR P 2 = 100 kPa P 1 = 300 kPa T 1 = 200 C
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5-34 Heat is lost from the steam flowing in a nozzle. The velocity and the volume flow rate at the nozzle exit are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3 There are no work interactions. Analysis We take the steam 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 Energy balance: 0) pe since 2 2 0 out 2 2 2 2 1 1 out in energies etc. potential, kinetic, internal, in change of Rate (steady) 0 system mass and work, heat,
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This note was uploaded on 01/03/2011 for the course MEEN 315 taught by Professor Ramussen during the Fall '07 term at Texas A&M.

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HW7S - 5-30 Air is accelerated in a nozzle from 45 m/s to 180 m/s The mass flow rate the exit temperature and the exit area of the nozzle are to be

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