Model the cycle in the new section of the power plant as a Rankine
cycle. The schematic and the Ts Diagram of this cycle is shown on the left.
(1) Determine the work output of the new system
The net work output of the cycle equals the difference between the turbine work output and the pump work input.
w_{net,out} = w_{turb,out}  w_{pump,in}
The enthalpy at each state needs to be determined first, which can be obtained from the water table.
State 1: Compressed liquid water
P_{1} = 9 kPa T_{1} =
38^{o}C (given)
v_{1} = 0.001008 m^{3}/kg
State 3: Compressed liquid water
P_{3} = 15.9 MPa T_{1} =
45^{o}C (given)
h_{3 }= 187.8 kJ/kg
State 4: Superheated vapor
P_{4} = 15.2 MPa T_{4} =
625^{o}C (given)
h_{4 }= 3641.8 kJ/kg
s_{4} = 6.7961 kJ/(kgK)
State 5: Superheated vapor
P_{5} = 15 MPa T_{5} =
600^{o}C (given)
h_{5} = 3582.3 kJ/kg
s_{5 }= 6.6776 kJ/(kgK)
State 6: Saturated mixture
P_{6} = 10 kPa (given)
s_{6s} = s_{5} = 6.6776 kJ/(kgK)
h_{6s}= 2114.9 kJ/kg
where s_{6s} and h_{6s} are the entropy and enthalpy
at state 6 if process 56 is isentropic.
Since the isentropic efficiencies of the pump and turbine are given, the
turbine work and the pump work can be determined as following:
Therefore, the net work output per mass is
w_{net,out} = w_{turb,out}  w_{pump,in}
=
1,173.9  19.0 = 1,154.9 kJ/kg
The power produced by the new system is determined from
(2) Determine the thermal efficiency
Heat input to the cycle from the boiler is determined from
q_{in} = h_{4} 
h_{3} = 3,641.8  187.8 = 3,454 kJ/kg
The thermal efficiency of the cycle is
η_{th} =
w_{net,out}/q_{in} = 1,154.9/3,454 = 33%
