Answer:
ηa=0.349
ηb=0.345
Explanation:
The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:
[tex]h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg[/tex]
The quality at state 4 is determined from the condition [tex]s_{4} =s_{3}[/tex] and the entropies of the components at the condenser pressure taken from table:
[tex]q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935[/tex]
The enthalpy at state 4 then is:
[tex]h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\[/tex]
Part A
In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:
[tex]h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg[/tex]
The enthalpy at state 2 is determined from an energy balance on the pump:
[tex]h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )[/tex]
=346.67 kJ/kg
The thermal efficiency is then determined from the heat input and output in the cycle:
[tex]=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349[/tex]
Part B
In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:
[tex]h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg[/tex]
The enthalpy at state 2 is then determined from an energy balance on the pump:
[tex]h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )[/tex]
=299.79 kJ/kg
The thermal efficiency in this case then is:
[tex]=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345[/tex]
