Answer:
The thermal efficiency of the power plant cycle is 40.95 percent.
The COP as a refrigerator is 1.442.
Explanation:
From Thermodynamics we get that thermal efficiency for power cycles is represented by the following definition:
[tex]\eta_{th} = \frac{\dot W_{net}}{\dot W_{net}+\dot Q_{L}}[/tex] (Eq. 1)
Where:
[tex]\dot W_{net}[/tex] - Net power of the power cycle, measured in kilowatts.
[tex]\dot Q_{L}[/tex] - Heat rate released from condenser, measured in kilowatts.
[tex]\eta_{th}[/tex] - Thermal efficiency of the power plant cycle, dimensionless.
The net power cycle is determined by the following expression:
[tex]\dot W_{net} = \dot W_{t}-\dot W_{p}[/tex] (Eq. 2)
Where:
[tex]\dot W_{t}[/tex] - Power generated by the turbine, measured in kilowatts.
[tex]\dot W_{p}[/tex] - Power consumed by the pump, measured in kilowatts.
If we know that [tex]\dot W_{p} = 200\,kW[/tex], [tex]\dot W_{t} = 21\times 10^{3}\,kW[/tex] and [tex]\dot Q_{L} = 30\times 10^{3}\,kW[/tex], then the thermal efficiency of the power plant cycle is:
[tex]\dot W_{net} = 21\times 10^{3}\,kW-200\,kW[/tex]
[tex]\dot W_{net} = 20.8\times 10^{3}\,kW[/tex]
[tex]\eta_{th} = \frac{20.8\times 10^{3}\,kW}{20.8\times 10^{3}\,kW + 30\times 10^{3}\,kW}[/tex]
[tex]\eta_{th} = 0.409[/tex] ([tex]40.95\,\%[/tex])
The thermal efficiency of the power plant cycle is 40.95 percent.
For refrigeration cycles we remember that the Coefficient of Performance ([tex]COP_{R}[/tex]), dimensionless, is represented by the following model:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W_{net}}[/tex] (Eq. 3)
If we know that [tex]\dot W_{net} = 20.8\times 10^{3}\,kW[/tex] and [tex]\dot Q_{L} = 30\times 10^{3}\,kW[/tex], then the Coefficient of Performance is:
[tex]COP_{R} = \frac{30\times 10^{3}\,kW}{20.8\times 10^{3}\,kW}[/tex]
[tex]COP_{R} = 1.442[/tex]
The COP as a refrigerator is 1.442.
