6.4 Carnot cycles

Consider a proposal to build a 500-MW steam power plant operating on a Rankine cycle by a lake. The steam generated by the boiler is at 250°C. The condenser is to be cooled by the lake water at a flow rate of 20 m³/s. As the cooling water will be discharged to the lake, the design team needs to evaluate its impact on the local aquatic ecosystem. Assume the cooling water supplied from the lake to the condenser is at a constant temperature of 18°C.

1. If the cycle were reversible (Carnot cycle), what is the thermal efficiency of the power plant? Estimate the temperature rise of the discharge water to the lake.
2. If the cycle were reversible (Carnot cycle), and the steam temperature increases to 300^oC, how would the thermal efficiency of the power plant change?
3. The above two scenarios are ideal Carnot cycles. In fact, the projected thermal efficiency of the actual cycle is about 35%. Estimate the temperature rise of the discharge water to the lake in this case.

Solution:

1. The power plant is assumed to operate on a Carnot cycle between a heat source at the steam temperature of T_H = 250°C and a heat sink at the cooling water temperature of T_L = 18°C. The thermal efficiency of the Carnot cycle is

[latex]\eta_{th,rev} = 1 - \dfrac{T_L}{T_H} = 1 - \dfrac{273.15 + 18}{273.15 + 250} = 0.443[/latex]

[latex]\eta_{th,rev} = \dfrac{\dot{W}}{\dot{Q}_H}[/latex]   and  [latex]\dot{Q}_H = \dot{Q}_L + \dot{W}[/latex]

Therefore, the heat rejected to the discharge water to the lake is

[latex]\begin{align*} \dot{Q}_L &= \dot{W}\left(\dfrac{1}{\eta_{th,rev}} - 1 \right) \\&= 500\left(\dfrac{1}{0.443} - 1 \right) = 627.48 \ \rm{MW} \end{align*}[/latex]

From Table G2, C_p = 4.181 kJ/kgK for water, and the density of water is [latex]\rho = 1000 \ \rm{kg/m^3}[/latex]; therefore,

[latex]\because \dot{Q}_{L} = \dot{m}C_p \Delta T = \rho \dot{\mathbb{V}}C_p \Delta T[/latex]

[latex]\therefore \Delta T = \dfrac{\dot{Q}_{L}}{\rho \dot{\mathbb{V}}C_p} = \dfrac{627.48 \times 10^3}{1000 \times 20 \times 4.181} = 7.5 \ {}^ \rm{o} \rm{C}[/latex]

2. For a Carnot cycle with T_H = 300°C and T_L = 18°C,

[latex]\eta_{th,rev} = 1 - \dfrac{T_L}{T_H} = 1 - \dfrac{273.15 + 18}{273.15 + 300} = 0.492[/latex]

Comment:

Compare the two Carnot cycles, increasing the heat source temperature T_H can increase the thermal efficiency of the cycle. Similarly, decreasing the heat sink temperature T_L can also increase the thermal efficiency of the cycle.

3. The thermal efficiency of an actual cycle is always less than that of the Carnot cycle operating between the same heat source and heat sink. If the projected thermal efficiency of the actual cycle is 35%, then the heat absorbed by the discharge water will be

[latex]\begin{align*} \dot{Q}_{L} &= \dot{W}\left(\dfrac{1}{\eta_{th}} - 1\right) \\& = 500 \times \left(\dfrac{1}{0.35} - 1\right) = 928.57 \ \rm{MW} \end{align*}[/latex]

The temperature rise of the discharge water will be

[latex]\Delta T = \dfrac{\dot{Q}_{L}}{\rho \dot{\mathbb{V}}C_p} = \dfrac{928.57}{1000 \times 20 \times \times 4.181} = 11.1 \ {}^ \rm{o} \rm{C}[/latex]

Comment:

The actual cycle will cause a much higher temperature rise in the discharge water than the Carnot cycle due to the existence of the irreversibilities. A large amount of heat rejection to the lake could have a negative impact on the aquatic life and should be monitored closely.

A heat pump provides 20 kW of heating to a house in winter. The house must be maintained at 24^oC. If the COP of the heat pump is 5.5, and the outdoor temperature is 0^oC, answer the following questions.

1. How much power is required to drive the heat pump? Compare the power consumption of this heat pump and an electric resistance heater providing the same amount of heating to the house.
2. If the cycle were a Carnot cycle, what would the power consumption be?
3. If the outdoor temperature decreases, will the COP of the heat pump increase, remain the same, or decrease?

Solution:
1. The heat pump provides 20 kW of heating to the house; therefore, [latex]\dot{Q}_{H} = 20 \ \rm{kW}[/latex]

[latex]\because \; COP_{HP} = \dfrac{\dot{Q}_{H}}{\dot{W}}= 5.5[/latex]

[latex]\therefore \; \dot{W} = \dfrac{\dot{Q}_{H}}{{COP}_{HP}} = \dfrac{20}{5.5} = 3.636 \ \rm{kW}[/latex]

If an electric resistance heater is used to provide 20 kW of heating to the house, the electric resistance heater will consume 20 kW of electric power, assuming the efficiency of the heater is 100%. Therefore,

[latex]\Delta \dot{W} = 20 - 3.636 = 16.364 \ \rm{kW}[/latex]

Compare the power consumption of the heat pump and electric heater, using heat pump will save 16.354 kW of power.

2. If the cycle were a Carnot cycle,

[latex]\begin{align*} COP_{HP,rev} &= \dfrac{T_{H}}{T_{H} - T_{L}} \\&= \dfrac{273.15 + 24}{(273.15 + 24) - (273.15 + 0)}= 12.38 \end{align*}[/latex]

[latex]\because \; COP_{HP,rev} = \dfrac{\dot{Q}_H}{\dot{W}_{rev}}[/latex]

[latex]\therefore \; \dot{W}_{rev} = \dfrac{\dot{Q}_{H}}{COP_{HP,rev}} = \dfrac{20}{12.38} = 1.615 \ \rm{kW}[/latex]

The actual cycle achieves a much smaller COP and consumes a much greater power than the Carnot cycle due to the irreversibilities.

3. Since [latex]COP_{HP,rev} = \dfrac{T_{H}}{T_{H} - T_{L}} = \dfrac{1}{1 - T_L/T_H}[/latex]

As [latex]T_H[/latex] is a constant room temperature, [latex]COP_{HP, rev}[/latex] will decrease if the outdoor temperature [latex]T_L[/latex] decreases. Because the Carnot cycle always has the maximum COP compared to any real cycle operating between the same heat source and heat sink, the COP of a real heat pump will also decrease if the outdoor temperature decreases; therefore, heat pumps are preferably used in mild winter conditions.