Hw 5 solutions

Hw 5 solutions - 5-35 Steam is accelerated in a nozzle from...

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5-35 Steam is accelerated in a nozzle from a velocity of 80 m/s. The mass flow rate, the exit velocity, 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 Potential energy changes are negligible. 3 There are no work interactions. Properties From the steam tables (Table A-6) kJ/kg 3196.7 /kg m 0.057838 C 00 4 MPa 5 1 3 1 1 1 = = ° = = h T P v and kJ/kg 3024.2 /kg m 0.12551 C 00 3 MPa 2 2 3 2 2 2 = = ° = = h T P Analysis ( a ) There is only one inlet and one exit, and thus &&& mmm 12 = = . The mass flow rate of steam is kg/s 6.92 = × = = ) m 10 50 )( m/s 80 ( /kg m 0.057838 1 1 2 4 3 1 1 1 A V 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 & & 4 43 4 42 1 & 1 & & = = Δ = Ê + = Δ + = + 2 0) pe W (since /2) V + ( ) 2 / ( 2 1 2 2 1 2 out 2 2 2 out 2 1 1 V V h h m Q h m Q V h m & & & & & & Substituting, the exit velocity of the steam is determined to be () + = 2 2 2 2 2 /s m 1000 kJ/kg 1 2 m/s) (80 3196.7 3024.2 kg/s 6.916 kJ/s 120 V It yields V 2 = 562.7 m/s ( c ) The exit area of the nozzle is determined from ( )( ) 2 4 m 10 15.42 × = = = ⎯→ = m/s 562.7 /kg m 0.12551 kg/s 6.916 1 3 2 2 2 2 2 2 V m A A V m & & Steam 1 2 120 kJ/s

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5-51 Air is compressed at a rate of 10 L/s by a compressor. The work required per unit mass and the power required are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.
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Hw 5 solutions - 5-35 Steam is accelerated in a nozzle from...

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