Problem 76 - Pearson Physics

Question

A nuclear power plant has a reactor that produces heat at the rate of \(835 \mathrm{MW}\). This heat is used to produce \(250 \mathrm{MW}\) of mechanical power to drive an electrical generator. (a) At what rate is heat discarded to the environment by this power plant? (b) What is the thermal efficiency of the plant?

Step-by-Step Solution

Answer

Heat discarded is 585 MW, and thermal efficiency is 29.94%.

Step 1: Understand the Problem

We know the plant generates 835 MW of heat and uses it to produce 250 MW of mechanical power. We need to find out how much heat is discarded (lost) and the efficiency of the plant.

Step 2: Use Energy Balance Equation

The heat input to the reactor is 835 MW and the mechanical output power is 250 MW. The heat lost, or discarded to the environment, can be found using the formula for energy balance: \[ Q_{discarded} = Q_{in} - W_{out} \] where \( Q_{in} = 835 \) MW and \( W_{out} = 250 \) MW.

Step 3: Calculate Heat Discarded

Substitute the given values into the energy balance equation: \[ Q_{discarded} = 835 - 250 = 585 \text{ MW} \] Therefore, the rate of heat discarded to the environment is 585 MW.

Step 4: Determine the Thermal Efficiency

The thermal efficiency \( \eta \) of a power plant is given by the formula: \[ \eta = \frac{W_{out}}{Q_{in}} \times 100\% \] This tells us the percentage of input energy converted into useful work.

Step 5: Calculate the Thermal Efficiency

Substitute the known values into the efficiency equation: \[ \eta = \frac{250}{835} \times 100\% \approx 29.94\% \] Thus, the thermal efficiency of the plant is approximately 29.94%.

Key Concepts

Heat Energy

Heat energy is produced by the nuclear reactor and is fundamental to the operation of a nuclear power plant. In this context, it is crucial as it is the primary source of energy that drives the entire process. The reactor generates heat through nuclear fission, where nuclei of atoms are split into smaller parts, releasing a significant amount of heat. This heat is measured in megawatts (MW), and for this particular exercise, the reactor supplies 835 MW of heat.

• This heat is necessary to convert water into steam.
• The steam generated is then used to drive turbines, which in turn produce mechanical power.
• Understanding the heat production is the first step towards calculating efficiency and heat loss in power plants.

The concept of heat energy is central to thermodynamics and is essential to comprehending how energy flows within a system such as a power plant.

Mechanical Power

Mechanical power is the useful energy output from the nuclear power plant provided to drive an electrical generator. This is what we aim to maximize in any power conversion system. In our case, the plant produces 250 MW of mechanical power, which is significantly lower than the 835 MW of heat generated.

• Mechanical power is generated from the conversion of steam energy through turbines.
• Turbines extract the thermal energy from steam, converting it into rotational energy.
• This rotational energy is then transformed into electrical energy via generators.

Mechanical power is a direct indicator of the plant's performance and is key to evaluating its efficiency and effectiveness in energy conversion.

Energy Balance Equation

The energy balance equation is a fundamental principle used to calculate how much energy is used, stored, or lost in a system. For a nuclear power plant, it helps determine how much heat energy is discarded. The equation applied here is:\[ Q_{discarded} = Q_{in} - W_{out} \]where:- \( Q_{in} \) is the heat energy input, which is 835 MW,- \( W_{out} \) is the mechanical power output, 250 MW.By applying this equation:

• We substitute the values to find: \( Q_{discarded} = 835 - 250 = 585 ext{ MW} \).

This equation ensures the conservation of energy principle is upheld, showing that the energy not converted into mechanical power is lost or discarded.

Thermal Efficiency

Thermal efficiency is a measure of how well a power plant converts the heat energy it receives into usable mechanical power. It is expressed as a percentage and calculated using the formula:\[ \eta = \frac{W_{out}}{Q_{in}} \times 100\%\]Where:

• \( \eta \) is the thermal efficiency,
• \( W_{out} = 250 ext{ MW} \)
• \( Q_{in} = 835 ext{ MW} \)

By substituting the given values, we find this plant's efficiency as:

• \( \eta = \frac{250}{835} \times 100\% \approx 29.94\% \)

This percentage indicates how much of the heat energy is converted into mechanical power. A higher efficiency means more energy is converted to useful work, reflecting a more effective power plant.

Heat Dissipation

Heat dissipation refers to the process of losing the energy that is not converted into mechanical power. It needs to be managed efficiently in a nuclear power plant to maintain safety and efficiency. In our exercise, the plant discards 585 MW of waste heat to the environment.

• This loss can occur through cooling towers, water bodies, or atmospheric release.
• Proper dissipation ensures the reactor does not overheat, which could be dangerous.
• Effective heat dissipation systems are essential to conserve energy.

Understanding heat dissipation helps in improving the thermal efficiency and safety measures of the nuclear power plant, indicating areas for improvement.