Given Work done by both turbines together: Wturbines=(h1−h2)+(h3−h4) From steam tables: h1=3373.6kJ/kg h2=2778.1kJ/kg Since S3=S4 S3=7.7621kJ/kg−k 7.7621=0.8319+x(7.9085−0.8319) x=0.9793 Thus, at exit of 2nd turbine will be in wet region with dryness fraction of 0.9793. Thus, h4=hf+xhfg =251.38+0.9793(2609.7–251.38) =2560.8827kJ/kg