Answer:
a) [tex]\eta_{real} = 0.36\,(36\,\%)[/tex], [tex]Q_{out} = 640\,BTU[/tex], b) [tex]\eta_{rev} = 0.480\, (48\,\%)[/tex], [tex]Q_{out} = 520\,BTU[/tex]
Explanation:
a) The efficiency of the reversible power cycle is modelled by the Carnot's cycle:
[tex]\eta_{rev} = 1 - \frac{759.67\,R}{1459.67\,R}[/tex]
[tex]\eta_{rev} = 0.480\, (48\,\%)[/tex]
The real efficiency is:
[tex]\eta_{real} = 0.75\cdot \eta_{rev}[/tex]
[tex]\eta_{real} = 0.36\,(36\,\%)[/tex]
The energy discharged to the cold reservoir is determined herein:
[tex]Q_{out} = Q_{in} - W[/tex]
[tex]Q_{out} = (1-\eta_{real})\cdot Q_{in}[/tex]
[tex]Q_{out} = (1 - 0.36)\cdot (1000\,BTU)[/tex]
[tex]Q_{out} = 640\,BTU[/tex]
b) The efficiency of the reversible cycle is:
[tex]\eta_{rev} = 0.480\, (48\,\%)[/tex]
The energy discharge to the cold reservoir is found:
[tex]Q_{out} = Q_{in} - W[/tex]
[tex]Q_{out} = (1-\eta_{rev})\cdot Q_{in}[/tex]
[tex]Q_{out} = (1 - 0.48)\cdot (1000\,BTU)[/tex]
[tex]Q_{out} = 520\,BTU[/tex]
