1) Steam enters a turbine steadily at 50.0 MPa and 400°C. Neglect any change in

Question

1) Steam enters a turbine steadily at 50.0 MPa and 400°C. Neglect any change in KE and PE.
a) The steam turbine is adiabatic, ie. there is no heat exchange with the surroundings. The water leaves at 100.0 kPa with the vapour quality 60%.
i) Determine the mass flow rate if the power output is 1000 kW. (8 points)
ii) Determine the rate of entropy generation. (4 points)
b) The steam turbine is actually not well insulated. It is losing 100 kW of heat to the surroundings which is at temperature 20.0 °C. The mass flow rate of the steam is the same as above. The steam again leaves at 100.0 kPa with the vapour quality 60%.
i) Determine the power output of this steam turbine. (2 points)
ii) Determine the rate of entropy generation. (Note: a proper choice of the control volume is needed to determine the total entropy generated.) (6 points)
c) The steady-flow process in the turbine is now reversible (ie. no entropy generation) and adiabatic. The mass flow rate is the same as parts (a) and (b). The outgoing pressure of the fluid is 100.0 kPa.
i) Determine the phase of the exit fluid and its vapour quality if applicable. (6 points)
ii) Determine the power output,

Explanation / Answer

1)( there has to be 50 bar not 50 MPa)

enthaplpy of steam entering turbine from steam table:-

at 50.0 bar and 400°C :- h1=3196.59 kj / kg

steam leaving turbine at 100kpa with 60% quality

h2=hf+0.6hfg

h2=417.31+0.6(2257.96)

h2=1772.08 kj/kg

power developed= m(h1-h2)

1000=m(3196.59- 1772.08)

m= 0.701 kg/s mass flow rate

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