HW10S - HW 10 Solution 1. 7-108 Saturated water vapor is...

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1 HW 10 Solution 1. 7-108 Saturated water vapor is compressed in a reversible steady-flow device. The work required is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 There is no heat transfer associated with the process. 3 Kinetic and potential energy changes are negligible. Analysis The properties of water at the inlet state are 4) - A (Table /kg m 39248 . 0 kPa 16 . 476 1 C 150 3 1 1 1 1 = = = ° = v P x T Noting that the specific volume remains constant, the reversible steady-flow work expression gives kJ/kg 205.6 = = = = 3 3 1 2 1 2 1 in m kPa 1 kJ 1 476.16)kPa - /kg)(1000 m (0.39248 ) ( P P dP w Compressor 1 MPa Water 150°C sat. vap. 2. 7-112 A steam power plant operates between the pressure limits of 5 MPa and 10 kPa. The ratio of the turbine work to the pump work is to be determined. Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic. Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic, s 1 = s 2 and s 3 = s 4 . Then the properties of the steam are kJ/kg 4608.1 MPa 5 K kJ/kg .1488 8 kJ/kg 2583.9 . kPa 10 3 4 3 3 kPa 10 @ 4 kPa 10 @ 4 4 = = = = = = = = h s s P s s h h vapor sat P g g Also, 1 = f @ 10 kPa = 0.00101 m 3 /kg. The work output to this isentropic turbine is determined from the steady-flow energy balance to be ) ( 0 4 3 out out 4 3 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 h h m W W h m h m E E E E E = + = = = Δ = ± ± ± ± ± ± ± ² ²³ ² ²´ µ ± µ ± ± © H 2 O 3 4 H 2 O 2 1
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2 Substituting, kJ/kg .2 2024 9 . 2583 1 . 4608 4 3 out turb, = = = h h w The pump work input is determined from the steady-flow work relation to be () kJ/kg .041 5 m kPa 1 kJ 1 kPa ) 0 1 5000 )( /kg m 0.00101 ( 3 3 1 2 1 0 0 2 1 in pump, = = = Δ + Δ + = P P pe ke dP w v © © Thus, 402 = = 5.041 2024.2 in pump, out turb, w w 3. 7-122E Steam is compressed in an adiabatic closed system with an isentropic efficiency of 80%. The work produced and the final temperature are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The device is adiabatic and thus heat transfer is negligible. Analysis We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as 2 1 out 1 2 out energies etc. potential, kinetic, internal, in Change system mass and work, heat, by nsfer energy tra Net out in ) ( u u w u u m U W E E E = = Δ = Δ = ±² ±³ ´ ´ From the steam tables (Tables A-5 and A-6),
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HW10S - HW 10 Solution 1. 7-108 Saturated water vapor is...

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