Question

A power cycle operates between a lake’s surface water at a temperature of 25 °C and water at a depth whose temperature is 6 °C. At steady state the cycle develops a power output of 14 kW, while rejecting energy by heat transfer to the lower temperature water at the rate 14,400 kJ/min. Determine the thermal efficiency of the power cycle, the maximum thermal efficiency for any such power cycle, and whether the power cycle is possible.

Answer 1

`\eta_(real) = 5.512\,\%`,
`\eta_(max) = 6.373\,\%`. Possible.

Step-by-step explanation:

The thermal efficiency of the real power cycle is determined by the following expression:

`\eta_(real) = (\dot W)/(\dot Q_(H))* 100\,\%`

Where:

`\dot W` - Power output of the power plant, in kW.

`\dot Q_(H)` - Heat transfer rate from lake surface level, in kW.

The heat transfer rate from lake surface level is the sum of the heat rejection rate and the power output. That is:

`\dot Q_(H) = \left(14,400\,(kJ)/(min) \right)\cdot \left((1)/(60)\,(min)/(s) \right) + 14\,kW`

`\dot Q_(H) = 254\,kW`

Now, the real efficiency of the power cycle is:

`\eta_(real) = (14\,kW)/(254\,kW) * 100\,\%`

`\eta_(real) = 5.512\,\%`

Besides, the maximum possible efficiency for any power cycle is based on Carnot's cycle efficiency, whose formula is:

`\eta_(max) = \left(1-(T_(L))/(T_(H)) \right)* 100\,\%`

Where:

`T_(L)` - Temperature of the cold reservoir, in K.

`T_(H)` - Temperature of the hot reservoir, in K.

The maximum efficiency of the cycle is:

`\eta_(max) = \left(1-(279.15\,K)/(298.15\,K) \right)* 100\,\%`

`\eta_(max) = 6.373\,\%`

The cycle is possible as
`\eta_(real) \leq \eta_(max)`, observing the Second Law of Thermodynamics.