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Chapter 5: Problem 28

A power cycle operating at steady state receives energy by heat transfer from the combustion of fuel at an average temperature of \(1000 \mathrm{~K}\). Owing to environmental considerations, the cycle discharges energy by heat transfer to the atmosphere at \(300 \mathrm{~K}\) at a rate no greater than \(60 \mathrm{MW}\). Based on the cost of fuel, the cost to supply the heat transfer is \(\$ 4.50\) per GJ. The power developed by the cycle is valued at \(\$ 0.08\) per \(\mathrm{kW}\) - h. For 8000 hours of operation annually, determine for any such cycle, in $\$$ per year, (a) the maximum value of the power generated and (b) the minimum fuel cost.

Short Answer

Expert verified
The maximum power generated is 140 MW, with an annual revenue of 89.6 million USD, and the minimum fuel cost is 25.92 million USD.

Step by step solution

01

- Determine the maximum theoretical efficiency using Carnot efficiency

The Carnot efficiency is calculated using the temperatures of the heat source and sink. The formula is \[ \text{Efficiency} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \] Plugging in the values: \[ \text{Efficiency}_{\text{Carnot}} = 1 - \frac{300}{1000} = 0.70 \] Thus, the maximum theoretical efficiency is 70%.
02

- Calculate the maximum power output

The maximum heat rejection rate is given as 60 MW. The actual power developed is the product of the heat input and the efficiency. Using the Carnot efficiency from Step 1, the power developed is: \[ W_{\text{out, max}} = Q_{\text{in}} \times \text{Efficiency}_{\text{Carnot}} - Q_{\text{out}} \] Since \[ Q_{\text{in}} = \frac{Q_{\text{out}}}{1 - \text{Efficiency}_{\text{Carnot}}} \], we find \[ Q_{\text{in}} = \frac{60 \text{ MW}}{1 - 0.70} = 200 \text{ MW}. \] Then the maximum power generated is: \[ W_{\text{out, max}} = 200 \text{ MW} \times 0.70 = 140 \text{ MW}. \]
03

- Calculate the annual power generated in kWh

Convert the maximum power output to kWh: \[ 140 \text{ MW} = 140,000 \text{ kW} \] Given 8000 hours of operation annually, the total energy produced in a year is: \[ \text{Energy}_{\text{annual}} = 140,000 \text{ kW} \times 8000 \text{ hours} = 1,120,000,000 \text{ kWh}. \]
04

- Calculate the annual revenue from power generation

The value of the power developed is \$0.08 per kWh. Thus, the annual revenue is: \[ \text{Revenue}_{\text{annual}} = 1,120,000,000 \text{ kWh} \times 0.08 \text{ USD/kWh} = 89,600,000 \text{ USD}. \]
05

- Calculate the annual fuel cost

First, convert the annual heat input to GJ: \[ Q_{\text{in, annual}} = 200 \text{ MW} \times 8000 \text{ hours} = 1,600,000 \text{ MWh} \times 3.6 \text{ GJ/MWh} = 5,760,000 \text{ GJ}. \] The fuel cost is \$4.50 per GJ, thus: \[ \text{Cost}_{\text{fuel}} = 5,760,000 \text{ GJ} \times 4.50 \text{ USD/GJ} = 25,920,000 \text{ USD}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamic Efficiency
Thermodynamic efficiency measures how well a power cycle converts heat from fuel into useful work. It's calculated using temperatures at which heat is added and rejected. The formula is the Carnot efficiency: \ \ \ \ \ \ \ \ \ \[ \text{Efficiency} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \] For our power cycle: \ \[ \text{Efficiency}_{\text{Carnot}} = 1 - \frac{300}{1000} = 0.70 \] This 70% represents the highest theoretical efficiency. Actual efficiencies are lower because of real-world factors like friction and heat losses. Understanding thermodynamic efficiency helps in optimizing power cycles for better performance and minimal fuel consumption.
Carnot Cycle
The Carnot Cycle is a theoretical model representing an idealized thermodynamic cycle. It comprises four stages: two isothermal (constant temperature) and two adiabatic (no heat exchange) processes. It sets an upper limit on efficiency for any power cycle: \ \[ \text{Efficiency} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \] This efficiency depends solely on the temperature difference between the heat source and sink. For our example, the maximum efficiency is calculated using 1000 K for the source and 300 K for the sink. Although real cycles can't operate precisely like the Carnot Cycle, they aim to approach its efficiency by reducing losses.
Heat Transfer
Heat transfer plays a crucial role in power cycles. It involves moving thermal energy from one place to another. In a power cycle, heat is absorbed from a high-temperature source like fuel combustion and rejected to a low-temperature sink, such as the atmosphere. \ \ \
    Heat input (\( Q_{\text{in}} \)): This is the thermal energy absorbed from the heat source. In our problem, we calculated it as 200 MW. \ \ Heat output (\( Q_{\text{out}} \)): This is the thermal energy discharged to the heat sink. Our cycle discharges no more than 60 MW to the atmosphere. \ \ Power developed (\( W_{\text{out}} \)): This is the useful work derived from the heat input. For maximum power, it鈥檚 the product of heat input and Carnot efficiency: \
      \( W_{\text{out}} = Q_{\text{in}} \times \text{Efficiency} - Q_{\text{out}} \) \ \
        For our case: \ \[ W_{\text{out, max}} = 200 \text{ MW} \times 0.70 - 60 \text{ MW} = 80 \text{ MW} \]
Power Generation
Converting thermal energy into electrical power is the goal of power cycles. Key metrics include power output, annual energy production, and the associated costs. From our cycle, the steps involved are: \ \ \
    Calculating power output: For maximum power, we have: \
      \( W_{\text{out, max}} = 140 \text{ MW} \) \ \ \ Converting to annual energy production: Given 8000 hours of operation per year: \
        \( \text{Energy}_{\text{annual}} = 140,000 \text{ kW} \times 8000 \text{ hours} = 1,120,000,000 \text{ kWh} \) \ \ \ Calculating revenue: At \( \text{\text{USD 0.08 per kWh}} \) : \
          \( \text{Revenue}_{\text{annual}} = 1,120,000,000 \text{ kWh} \times 0.08 \text{ USD/kWh} = 89,600,000 \text{ USD} \) \ \ Power generation is optimized by maximizing efficiency and minimizing costs such as fuel. For the fuel cost: Given the cost of supplying heat \
            \( \text{\text{USD 4.50 per GJ}} \) : \
              \( \text{Cost}_{\text{fuel}} = 5,760,000 \text{ GJ} \times 4.50 \text{ USD/GJ} = 25,920,000 \text{ USD} \) \ \ Managing these factors ensures sustainable and cost-effective power generation.

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Most popular questions from this chapter

An inventor claims to have developed a refrigerator that at steady state requires a net power input of \(0.54 \mathrm{~kW}\) to remove \(12,800 \mathrm{~kJ} / \mathrm{h}\) of energy by heat transfer from the freezer compartment at \(-20^{\circ} \mathrm{C}\) and discharge energy by heat transfer to a kitchen at \(27^{\circ} \mathrm{C}\). Evaluate this claim.

Two reversible heat pump cycles operate in series. The first cycle receives energy by heat transfer from a cold reservoir at \(260 \mathrm{~K}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\) greater than 260 K. The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher- temperature reservoir at \(1200 \mathrm{~K}\). If the heat pump cycles have the same coefficient of performance, determine (a) \(T\), in \(\mathrm{K}\), and (b) the value of each coefficient of performance.

Two reversible power cycles are arranged in series. The first cycle receives energy by heat transfer from a hot reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T

Ocean temperature energy conversion (OTEC) power plants generate power by utilizing the naturally occurring decrease with depth of the temperature of ocean water. Near Florida, the ocean surface temperature is \(27^{\circ} \mathrm{C}\), while at a depth of \(700 \mathrm{~m}\) the temperature is \(7^{\circ} \mathrm{C}\). (a) Determine the maximum thermal efficiency for any power cycle operating between these temperatures. (b) The thermal efficiency of existing OTEC plants is approximately \(2 \%\). Compare this with the result of part (a) and comment.

Two reversible refrigeration cycles are arranged in series. The first cycle receives energy by heat transfer from a cold reservoir at temperature \(T_{C}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\), greater than \(T_{C .}\) The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher-temperature reservoir at \(T_{\mathrm{H}}\). Obtain an expression for the coefficient of performance of a single reversible refrigeration cycle operating directly between cold and hot reservoirs at \(T_{\mathrm{C}}\) and \(T_{\mathrm{H}}\), respectively, in terms of the coefficients of performance of the two cycles.
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