Problem 5
Question
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?
Step-by-Step Solution
Verified Answer
(a) 25.38%; (b) 970 MW.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
Key Concepts
Thermal Efficiency CalculationMechanical Power OutputHeat Input and OutputEnergy Conversion Processes
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.
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.
Other exercises in this chapter
Problem 3
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
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20.6. (a) Calculate the theoretical efficiency for an Otto cycle engine with \(\gamma=1.40\) and \(r=9.50 .\) (b) If this engine takes in \(10,000\) J of heat f
View solution Problem 7
20\. 07. What compression ratio \(r\) must an Otto cycle have to achieve an ideal efficiency of 65.0\(\%\) if \(\gamma=1.40 ?\)
View solution