Question

Water vapor at 6 MPa and 500 °C enters a turbine operating at steady state and expands to 1 bar. Mass flow rate is 2kg/s. Neglect heat transfer, kinetic energy and potential energy changes. For the actual process (1-2), water leaves the turbine with a specific entropy S2 = 7.1176 kJ / kg / k

Find: a) Plot isentropic process in the turbine (1-2s) and the actual process in the turbine (1-2) on a T-s diagram. Justify the location of each point.

b) The isentropic turbine efficiency and keep three digits after the decimal point.

Answer 1

P = 6mpa = 500°c Te mi= 2kgls P2 = 1 bar S2= 7:1176 kollege from steam Tables @ ompa & 500°C 5,= 6.8795kJ/kgk, h,=3421 Ko/kg T as S Isentropic process.

I bar Sf 1-3020 Kylkyler Sq= 7.358IKF Kyk Sfes, <Sg 2' in Saturated liquid vapour region Sy = Softy Sifog X2= 0.98 @ Ibar, Tsat T2 = 99-639 TM - 2. S, S2 S Actual process. hz = hiftz heg hifa 417.3 kJ hfg= Kg = 2257 kJ Kg hi =2584 kJ /kg Actual Work done hq -ha Za =3421-2584 =837KJ Kg

for Ideal case. S2= = 6.8795 S2 = St+XqSfg x2 = 0.92 has 24.93 kJ Kg Ideal work done =n,thal - 3421-2493 = 928 KJ Ti Isentropic efficiency 837 928 = 0.9019 90.1901)