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20.5. A certain nuclear-power plant has a mechanical-power output (used to drive an electric generator) of 330 \(\mathrm{MW}\) . Its rate of heat input from the nuclear reactor is 1300 \(\mathrm{MW}\) . (a) What is the thermal efficiency of the system? (b) At what rate is heat discarded by the system?

Short Answer

Expert verified
(a) 25.38%; (b) 970 MW.

Step by step solution

01

Understand the Power and Heat Flow

The nuclear plant's mechanical power output is 330 MW (the power used to generate electricity), and its heat input from the reactor is 1300 MW. We want to find the thermal efficiency and heat discarded.
02

Calculate Thermal Efficiency

Thermal efficiency (\( \eta \)) is calculated using the formula:\[ \eta = \frac{\text{Output Power}}{\text{Input Heat Power}} \]Substituting the given values:\[ \eta = \frac{330 \text{ MW}}{1300 \text{ MW}} = 0.2538 \]This means the thermal efficiency is 25.38% when multiplied by 100.
03

Calculate Heat Discarded

The rate at which heat is discarded is found using:\[ Q_{\text{discarded}} = Q_{\text{input}} - Q_{\text{output}} \]Where \(Q_{\text{input}} = 1300 \text{ MW}\) and \(Q_{\text{output}} = 330 \text{ MW}\).\[ Q_{\text{discarded}} = 1300 \text{ MW} - 330 \text{ MW} = 970 \text{ MW} \]Thus, the rate of heat discarded by the system is 970 MW.

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Key Concepts

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

Thermal Efficiency Calculation
In a nuclear power plant, one of the most essential measures is its thermal efficiency. Thermal efficiency shows how well the energy produced from nuclear reactions is converted into mechanical power to generate electricity. To understand thermal efficiency, consider this formula:\[ \eta = \frac{\text{Output Power}}{\text{Input Heat Power}} \]This ratio, often expressed as a percentage, indicates the fraction of heat input converted into useful work. When we substitute the given values for the nuclear power plant exercise, we use:\[ \eta = \frac{330 \text{ MW}}{1300 \text{ MW}} \approx 0.2538 \] Converting it into a percentage gives approximately 25.38%. This means for every unit of heat energy the plant gets, around 25% is turned into useful electricity. The rest, unfortunately, is wasted heat.
Mechanical Power Output
Mechanical power output is the useful energy generated by the plant that is supplied to the electric generators. In our example, the plant's mechanical power output is 330 MW. This measure is crucial as it defines the capacity of the plant to convert nuclear energy into electricity. A high mechanical output indicates efficient transformation of the reactor's thermal energy into electricity. Here are some key points:
  • Mechanical power output is typically less than the total energy produced, due to energy losses in conversion processes.
  • It is a core component of a power plant's effectiveness.
Once the mechanical energy is harnessed, it can be converted into electrical energy, powering homes and industries.
Heat Input and Output
The process of converting nuclear energy into electricity involves significant heat input and output flows. Understanding these helps evaluate the efficiency and environmental impacts of the plant. **Heat Input** Heat input refers to the total thermal energy produced by the nuclear reactor. In this scenario, the reactor supplies 1300 MW of heat input. **Heat Output** The useful portion of this heat energy is transformed into mechanical energy, which is 330 MW here. The difference between heat input and mechanical power output helps determine how much energy remains unused. Heat output that is not converted into mechanical energy must be managed efficiently to reduce thermal pollution and use resources prudently.
Energy Conversion Processes
Energy conversion processes in nuclear power plants are critical for turning nuclear reactions into electricity. These processes involve several stages that impact overall efficiency. 1. **Nuclear Reaction:** - Heavy nuclei, like uranium or plutonium, go through fission. - This reaction releases a substantial amount of heat. 2. **Heat Transport:** - The generated heat is transferred to a coolant, usually a fluid. - In thermal reactors, the fluid often water, converts to steam. 3. **Mechanical Work:** - The steam drives a turbine, converting heat energy to mechanical energy. 4. **Electricity Generation:** - The turbine's mechanical energy powers a generator to produce electricity. Each step involves energy transformations with some losses. The more effective these conversions are, the higher the plant's thermal efficiency. Understanding these intricate processes is key to optimizing plant performance and sustainability.

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Most popular questions from this chapter

A Gasoline Engine. Agasoline engine takes in \(1.61 \times 10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g}\) . (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

20.10. A room air conditioner has a coefficient of performance of 29 on a hot day, and uses 850 \(\mathrm{W}\) of electrical power. (a) How many joules of heat does the air conditioner remove from the room in one minute? (b) How many joules of heat does the air conditioner deliver to the hot outside air in one minute? (c) Explain why your answers to parts (a) and (b) are not the same.

20.42. Heat Pump. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In summer it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is \(-5.0^{\circ} \mathrm{C}\) and the inside temperature is \(17.0^{\circ} \mathrm{C}\) , how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? ( b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at \(68^{\circ} \mathrm{F}\) . Show that the beat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

20.8. The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8 . (a) What is the ideal efficiency of the engine? Use \(\gamma=1.40 .\) (b) The engine in a Dodge Viper G.2 has a slightly higher compression ratio of 9.6 . How much increase in the ideal efficiency results from this increase in the compression ratio?

20.22. A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 \(\mathrm{kg}\) of ice. During this time, how much work \(W\) is performed by the engine?

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