Question

A power cycle operates between a lake’s surface water at a temperature of 300 K and water at a depth whose temperature is 285 K. At steady state the cycle develops a power output of 10 kW, while rejecting energy by heat transfer to the lower-temperature water at the rate 14,400 kJ/min. Determine (a) the thermal efficiency of the power cycle and (b) the maximum thermal efficiency for any such power cycle

Answer 1

Answer: a) 0.04kW = 40W

b) 0.05

Step-by-step explanation:

A)

Thermal efficiency of the power cycle = Input / output

Input = 10 kW + 14,400 kJ/min = 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output = 10 kW

Thermal Efficiency = Output / Input = 10kW / 250kW = 0.04KW = 40W

B)

Maximum Thermal Efficiency of the power cycle = 1 - T1/T2

Where T1 = 285kelvin

And T2 = 300kelvin

Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

Answer 2

a) Efficiency: 4%

b) Maximum thermal efficiency: 17%

Step-by-step explanation:

First we calculate the total heat that come from the hot reservoir. This heat is the work output plus the rejected energy to the cold reservoir.

`Q_h=W+Q_c=10kW+14400(kJ)/(min)((1min)/(60s))((kW)/(kJ/s) ) \\\\Q_h=10kW+240kW=250kW`

The efficienccy of the cycle can be calculated as the ratio between the work output and the heat input.

`\eta=(W)/(Q_h)=(10kW)/(250kW)=0.04=4\%`

The maximum thermal efficiency of this cycle can be determined in function of the temperatures of the reservoirs by this equation:

`\eta_(max)=1-(T_c)/(T_h)=1-(250)/(300)=1-0.83=0.17=17\%`