School of Engineering, FETBE EM203/208 (January 2013 Semester) - Assignment 31. A saturated liquid–vapor mixture of water at 150 kPa is contained in a well-insulated rigidtank. Initially, the quality of the mixture is 30%. An electric resistance heater placed in thetank is now turned on and kept on until only saturated vapor exists in the tank. If the initialmass of the vapor in the tank is 0.6 kg, determine (a) the final mass of the saturated vapor inthe tank, and (b) entropy change of the steam during this process.2. Superheated steam enters an adiabatic turbine at a rate of 5 kg/s. The pressure andtemperature of the steam at the turbine inlet is 8 MPa and550°C. The steam leaves the turbineat 20 kPa. If the isentropic efficiency of the turbine is 90%, determine (a) thetemperature atthe turbine exit and (b) the power output ofthe turbine. 8 MPa, 550°C 20kPa3. The power output of the turbine is 5 MW. Steam enters the turbine at 10MPa, 550°C, and50 m/s and leaves at 20 kPa and 120 m/s with a moisture content of 5 percent. The turbine isnot adequately insulated and it is estimated that heat is lost from the turbine at a rate of 300kW.Assuming the surroundings to be at 25°C, determine (a) the reversible power output ofthe turbine, (b) the exergy destroyed within the turbine, and (c) the second-law efficiency ofthe turbine.4.The pressure ratio of a simple Brayton cycle using air as the working fluid is 10. Theisentropic efficiencies of the compressor and turbine are 80%and 85% respectively. Assumingaminimum and maximum temperatures in the cycle of 320 and 1180 K respectively,determine(a) the air temperature at the turbine exit,(b) the net work output, and(c)the thermalefficiency.Note: Use Table A-17 to solve this problem..
.Solution1. H2O x1 = 0.3 P1 = 150 kPaFrom the steam tables (Tables A-4 through A-6)Interpolation for s2(a) The mass remains the same, so the final mass of the saturated vapor equals the total initialmass. NowThus(b) The entropy change is
2. From the steam tables:Above data are obtained from Table A-6Above data are obtained from Table A-5 (see below)From the isentropic efficiency relation,Thussince at 20 kPa, hf = 251.42 and hg = 2608.9 so that hf>h2a>hg.(b) There is only one inlet and one exit, and thus . We take the actual turbine asthe system,which is a control volume since mass crosses the boundary. The energy balancefor this steady-flow systemcan be expressed in the rate form as 0 (steady flow)ThusSubstituting
10 MPa3. (a) From the steam tables: 550°C 50 m/sAbove data are obtained from Table A-6 20 kPa x = 0.95 120 m/sAbove data are obtained from Table A-5 (see below)The enthalpy at the dead state is,The mass flow rate of steam may be determined from anenergy balance on the turbineSolving,The reversible power may be determined fromThus
(b) The exergy destroyed in the turbine is(c) The second-law efficiency is
4. The T-sdiagram of thesimple Brayton cycle is shown below. 1180 K 320 K(a) The properties of air are given in Table A-17.