Water is the working fluid in a Carnot vapor power cycle. Saturate liquid enters the boiler at a pressure of 10MPa, and saturated vapor enters the turbine. The condenser pressure is 6 kPa. Determine (i) the thermalefficiency, (ii) the back work ratio, (iii) the heat transfer to the working fluid in the boiler, and (iv) the heattransfer from the working fluid in the condenser. In part B of the problem, irreversibilities in the adiabaticexpansion and compression processes cannot be neglected. If the states at the turbine and pump inlet remainunchanged, repeat parts (i-iv) for the modified Carnot cycle with 77T = 0.80 and ŋp = 0.70. Finally, calculate therate of entropy generation in modified Carnot cycle having irreversibilities.

Since you have posted a question with multiple sub-parts, we will solve first four subparts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved Given: working fluid =waterCycle= Carnot vapor power cycleSaturated liquid pressure at the inlet of boiler, P1=10 MPaPressure at the outlet of condenser, P4=6 kPaIn Carnot vapor power cycle,Process 1-2: The fluid is heated reversible isothermal in a boilerProcess 2-3: Isentropic expansion in a turbineProcess 3-4: condensed reversible isothermal in a condenserProcess 4-1: Isentropic compression in a compressor to the initial state (i): The thermal efficiency of cycleQin=TH(S2-S1)Qout=TL(S3-S4)S1=S4 , S3=S2Wnet=Qin-QoutWnet=(TH-TL)(S2-S1) , assume this is equation 1ηth=WnetQin=(TH-TL)(S2-S1)TH(S2-S1)ηth=(TH-TL)THTH=T1=T2TL=T3=T4Temperature value is calculated with the help of saturated steam table by pressureP1=10 MPa , T1=TH=584.149 KP4= 6 kPa , T3= TL=309.31 KNow efficiency is ,ηth=(584.149-309.31)584.149ηth=0.47049 or 47.049%(ii): Back work ratio=Pump workTurbine workPump work=h1-h4For P=10 MPa , from steam tableh1=1408 kJ/kg , for h4, h4=hf+xhfgTo calculate x(dryness fraction)Sf1=S4=Sf+xSfg=3.36again from steam table for 6 kPa,0.521+x7.81=3.36x=0.3635h4=hf+xhfghf=151.5 , hfg=2415.5h4=151.5+0.3635×2415.5h4=1029.53 kJ/kgPump work=h1-h4=1408-1029.53Pump work=378.46 kJ/kgTurbine work=h2-h3For P=10MPah2=2725 kJ/kgfor h3, h3=hf+xhfgTo calculate x(dryness fraction)Sg2=S3=Sf+xSfg=5.614again from steam table for 6 kPa,0.521+x7.81=5.614x=0.6521h3=hf+xhfghf=151.5 , hfg=2415.5h3=151.5+0.6521×2415.5h3=1726.64 kJ/kgTurbine work=h2-h3=2725-1726.64Turbine work=998.36 kJ/kgBack work ratio=Pump workTurbine work=378.46998.36Back work ratio=0.379(iii):Heat transfer to the working fluid in the boilerQin=TH(S2-S1)From Steam table For P1=10 MPa , S1=Sf=3.36 kJ/kg-KS2=Sg=5.614 kJ/kg-KQin=584.149×(5.614 -3.36)Heat transfer to the working fluid in the boiler, Qin=1316.67 kJ/kg(iv): Heat transfer from the working fluid in the condenserQout=TL(S3-S4)S3=S2=5.614 kJ/kg-KS4=S1=3.36 kJ/kg-KQout=309.31(5.614-3.36) Heat transfer from the working fluid in the condenser, Qout=697.184 kJ/kg