Question

The wet steam at pressure 5 MPa and quality x = 90 % is adiabatically and reversibly decompressed to 0.5 MPa in a screw expander. Mass flow rate of steam equals to 5 kg/s. Find the final quality of the steam and the power generated in the expander. Neglect changes in kinetic and potential energies of inlet and outlet streams. Assume specific heat of water equal to 4.19 kJ/kg/K. Draw the process on the T-s diagram. The data required for the solution are given in the accompanying table. For calculation specific enthalpy and entropy of boiling water applies simplified formulae for pressure lower than 1 MPa. Ts(°C) 1 s" [MPa] 0.5 (m?/kg) (m/kg) 0,00109 0,3748 h" (kJ/kg) 2748,11 (kJ/kg/K) 6,82 151,84 5.0 263,94 0,00129 0,0394 2794,23 5,97

Answer 1

problem solved

given State- P: 5 MPa X = 90%:0.9 m: 5 kg is Reversible adiabatic CISentropic) Process State-2 P2 = 0.5 MPa 2=? Powerะ? () T-S diagram P, = 5 MPa P22 0.5 MPa T S,:52 Process 1-2 1-2 represents the decompression Process. (10) entropy at State 1 (5) S: Sf. + 3,549, at 5 MPE it canbe approximated as le S, X,88, ; From table; Sg,= S"at 5 MPa .: S = 0.9*5.97 = 5.373 kJ/kgK le. Sg, : 5.97 K3/KGK During the process 1-2: S, = S2 (isentropic Process) i S2 = S, = 5.373 KJKSK also, S2 : Sfat %25792 at 0-5 MPa. in approximated Condition S₂ = x₂992 From table Sq2 = 5,682 KJ/KGK 5.373 = x2*6.82

where : final qualilý, 2 = 0.79 (ans) (10power generated = m [hi-h2] hi= hf, tx, h fg, in approximated form hi = x, hg, = x, h at 5 MPa hi = 0.9* 2794.23 = 2514.807 k5|kg ha = xq hg2 =%₂h' at 0.5 MPa le 52 = 0.79274.8.11 = 2171.0069 kJ/kg Power generated : 5x [2514.807 - 2171.0069] P = 1719 kW (ans) le