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

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

Answer:

a) Efficiency: 4%

b) Maximum thermal efficiency: 17%

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.

$Q_h=W+Q_c=10kW+14400\frac{kJ}{min}(\frac{1min}{60s})(\frac{kW}{kJ/s} ) \\\\Q_h=10kW+240kW=250kW$

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

$\eta=\frac{W}{Q_h}=\frac{10kW}{250kW}=0.04=4\%$

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

$\eta_{max}=1-\frac{T_c}{T_h}=1-\frac{250}{300}=1-0.83=0.17=17\%$