Thermodynamics problem Water vapor at 5 bar, 400Celsius degree enters a turbine operating at steady-state with avolumetric flow rate of 1m/s and expands adiabatically to an exit state of 3 bar, 360 Celsiusdegree. Kinetic and potential energy effects are negligible. Determine for the turbine (a)the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) theisentropic turbine efficiency.

given initial pressure of water vapor, Pi=5bar initial temperature, Ti=400 degree Celsius volume flow rate, V=1 m3s when expand adiabatiacally final pressure,Pf=3bar final temperature, Tf=360 degree Celsius kinetic energy and potential energy are negligiblefrom steam table At pressure Pi=5bar , saturation temperature ,Ts1=158.919 because Ti>Ts1400>158.919 therefore steam is supersaturated steam from supersaturated steam table find corresponding values of specific Enthalpy, specific Entropy and specific volume at 5bar and 400degree Celsius h1=3270.7 KJkgs1=7.7935 KJkgv1=0.51259 m3Kg mass flow, m1=volume flow rate specific volumem1=10.51259m1=1.95087Kg from steam table At pressure Pf=3bar , saturation temperature ,Ts1=143.732 because Ti>Ts1360>143.732 therefore steam is supersaturated steam from supersaturated steam table find corresponding values of specific Enthalpy, specific Entropy and specific volume at 3bar and 360degree Celsius h2=3190.67 KJkgs2=7.906 KJkgv2=0.723297 m3Kg mass flow, m2=volume flow rate specific volumem2=10.723297m2=1.3825Kg a) power developed by turbine w=m1h1-m2h2w=1.95087×3270.7-1.3825×3190.67w=1969.42KW b) entropy generation s=m1s1-m2s2s=1.95087×7.7935-1.3825×7.906s=4.274 KWK