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thermodynamics

P5.31: - Steam at 3 MPa and 400°C enters an adiabatic nozle steadily with a velocity of 40 m/s and leaves at 2.5 MPa and 300 m/s. Determine (a) the exit temperature (b) the ratio of the inlet to exit area A1/A2.
Thermodynamics
P5.64:- Refrigerant-134a at 800 kPa and 25°C is throttled to a temperature of 220°C. Determine the pressure and the internal energy of the refrigerant at the final state P1-0.8 MPa
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5-31 inlet-to-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. 4 The device is adiabatic and thus heat transfer is negligible Properties From the steam tables (Table A-6). Steam is accelerated in a nozzle from a velocity of 40 m/s to 300 m/s. The exit temperature and the ratio of the Д = 3 MPa I v1 = 0.09938 m 3 /kg 11=400°C] h,-3231.7 kJ/kg Analysis (a) There is only one inlet and one exit, and thus 깨 = 7 =771 . 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 70 (steady) Éin - Éout =0 system Rate of net energy transfer byheat, work, andmass Rate of changein intemal,kinetic, potential etc. energies P1 = 3 MPa Ti 400°C = 40 m/s Steam P2 2.5 MPa -->- у, 300 m/s out or V3231.7kJ/kg (300 m/s)2 -(40 m/s) 1 kJkg 3187.5 kJ/kg 1000 m2/s2 us T2 376.6C P - 2.5 MPa h 3187.5 kJ/kg 0.11533 m /kg (b) The ratio of the inlet to exit area is determined from the conservation of mass relation, A 242 (0.09938 m/kg) (300 m/s) (0.11533 m3/kg)(40 m/s) -6.46 V2

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