Question

A power cycle receives 1000 Btu by heat transfer from a reservoir at 10008F and discharges energy by heat transfer to a reservoir at 3008F. The thermal efficiency of the cycle is 75% of that for a reversible power cycle operating between the same reservoirs, (a) For the actual cycle, determine the thermal efficiency and the energy discharged to the cold reservoir, in Btu. (b) Repeat for the reversible power cycle.

Answer 1

a)
`\eta_(real) = 0.36\,(36\,\%)`,
`Q_(out) = 640\,BTU`, b)
`\eta_(rev) = 0.480\, (48\,\%)`,
`Q_(out) = 520\,BTU`

Step-by-step explanation:

a) The efficiency of the reversible power cycle is modelled by the Carnot's cycle:

`\eta_(rev) = 1 - (759.67\,R)/(1459.67\,R)`

`\eta_(rev) = 0.480\, (48\,\%)`

The real efficiency is:

`\eta_(real) = 0.75\cdot \eta_(rev)`

`\eta_(real) = 0.36\,(36\,\%)`

The energy discharged to the cold reservoir is determined herein:

`Q_(out) = Q_(in) - W`

`Q_(out) = (1-\eta_(real))\cdot Q_(in)`

`Q_(out) = (1 - 0.36)\cdot (1000\,BTU)`

`Q_(out) = 640\,BTU`

b) The efficiency of the reversible cycle is:

`\eta_(rev) = 0.480\, (48\,\%)`

The energy discharge to the cold reservoir is found:

`Q_(out) = Q_(in) - W`

`Q_(out) = (1-\eta_(rev))\cdot Q_(in)`

`Q_(out) = (1 - 0.48)\cdot (1000\,BTU)`

`Q_(out) = 520\,BTU`