Problem 52 - University Physics with Modern Physics

Question

20.52. A typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40\(\%\) (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of \(265 \times 10^{7} \mathrm{J} / \mathrm{kg}\) . How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river?(d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

Step-by-Step Solution

Answer

(a) 2500 MW, (b) 8147 tonnes/day, (c) 1500 MW, (d) 717 m³/s, (e) 5.15 MW/K.

Step 1: Determine the Rate of Heat Input

To find the rate of heat input to the plant, we use the formula for efficiency: \( \eta = \frac{W}{Q_{in}} \), where \( \eta \) is the efficiency (40% or 0.4), \( W \) is the usable power (1000 MW or \( 1000 \times 10^6 \) W), and \( Q_{in} \) is the rate of heat input. Rearranging for \( Q_{in} \), we get \( Q_{in} = \frac{W}{\eta} = \frac{1000 \times 10^6}{0.4} \). Calculating this gives \( Q_{in} = 2500 \times 10^6 \) W.

Step 2: Calculate Coal Consumption Per Day

The total energy required per day is the heat input rate multiplied by the number of seconds in a day (86400 seconds). Therefore, \( Q_{in, \text{daily}} = 2500 \times 10^6 \times 86400 \). The energy per kilogram of anthracite coal is given as \( 265 \times 10^7 \) J/kg. Thus, the mass of coal required per day is \( m = \frac{Q_{in, \text{daily}}}{265 \times 10^7} \). Substituting the values and performing the calculation gives \( m \approx 8147 \) tonnes/day.

Step 3: Calculate the Rate of Heat Ejection

The rate of heat ejected to the river, \( Q_{out} \), is found by subtracting the usable power from the total heat input. So, \( Q_{out} = Q_{in} - W = 2500 \times 10^6 - 1000 \times 10^6 = 1500 \times 10^6 \) W.

Step 4: Determine the River's Flow Rate

The temperature change in the river is \( \Delta T = 18.5 - 18.0 = 0.5^{\circ} \)C. The specific heat capacity of water is \( c = 4184 \) J/(kg°C). Let \( m \) be the mass flow rate (kg/s) of the river water. The heat added to the river is \( Q_{out} \). Therefore, \( Q_{out} = m \cdot c \cdot \Delta T \). Rearranging, we find \( m = \frac{1500 \times 10^6}{4184 \times 0.5} \). Solving gives \( m \approx 716739 \) kg/s. The flow rate in terms of volume is \( \frac{m}{\rho} \), where \( \rho = 1000 \) kg/m³ (density of water), so the flow rate is approximately \( 717 \) m³/s.

Step 5: Calculate the River's Entropy Increase

The change in entropy, \( \Delta S \), is given by \( \Delta S = \frac{Q_{out}}{T_\text{river}} \). The average temperature of the river is \( T_\text{river} = 18.25^{\circ} \text{C} = 291.25 \) K. Thus, \( \Delta S = \frac{1500 \times 10^6}{291.25} \). Calculating, \( \Delta S \approx 5.15 \times 10^6 \) J/K per second.

Key Concepts

Coal-Fired Power Plant

Coal-fired power plants are a significant source of electricity generation worldwide. They work by burning coal to produce steam, which turns turbines connected to generators. These generators convert mechanical energy into electrical energy that we use daily. The process begins with coal combustion, where the chemical energy stored in coal is released as heat. This heat transforms water into steam, which powers the steam turbines to produce electricity.

Despite their efficiency in energy production, coal-fired power plants have certain drawbacks. Burning coal releases a significant amount of carbon dioxide (CO2), contributing to greenhouse gas effects.

• They also emit other pollutants like sulfur dioxide and nitrogen oxides, impacting air quality.
• Despite these emissions, advancements improve efficiency and decrease environmental impact through cleaner technologies and better emission controls.

However, understanding the efficiency and thermodynamic properties of these plants is vital for minimizing environmental impact while maximizing energy output.

Thermal Efficiency

Thermal efficiency is a measure of how well a power plant converts heat energy into usable electrical energy. For coal-fired power plants, thermal efficiency represents the fraction of heat input effectively transformed into electricity.

Calculated using the formula \( \eta = \frac{W}{Q_{in}} \), where \( \eta \) is efficiency, \( W \) is usable power, and \( Q_{in} \) is the rate of heat input. If a power plant boasts a thermal efficiency of 40%, it means only 40% of the input heat converts into electricity, while the remaining 60% becomes waste heat, often released into nearby bodies of water.

• Improving thermal efficiency involves reducing energy loss through advanced materials, better insulation, and innovative engineering designs.
• This efficiency is also crucial for evaluating the plant's environmental and economic sustainability.

Increasing thermal efficiency reduces fuel consumption and lowers greenhouse gas and pollutant emissions.

Entropy

Entropy is a thermodynamic property that measures the disorder or randomness in a system. In the context of a coal-fired power plant, it critically assesses the irreversibility of energy transformations during power generation.

During heat transfer processes in the plant, entropy often increases, indicating energy dissipation mostly as waste heat, contributing to inefficiency. The concept of entropy ties closely with the second law of thermodynamics, which states:

• In any energy exchange, if no energy enters or leaves the system, the potential energy of the state will be less than that of the initial state, reflecting increased entropy.
• Systems naturally progress towards thermodynamic equilibrium, exhibiting higher entropy.

Understanding and minimizing entropy changes can enhance plant operation efficiency by reducing energy losses and optimizing power output.

Heat Input Calculation

Calculating the heat input is integral to understanding a power plant's operational efficiency and energy balance. Heat input refers to the total heat energy introduced into the system for power generation.

In a coal-fired power plant, the heat input determines how much coal is burned to generate necessary energy. For example, if a plant operates at a 40% thermal efficiency and delivers 1000 MW of usable power, you can calculate the heat input through the formula \( Q_{in} = \frac{W}{\eta} \). This results in \( 2500 \times 10^6 \) watts, signifying significant energy throughput from coal combustion.

• Knowing the heat input helps balance fuel costs against power output.
• It also aids regulatory compliance by ensuring emissions align with environmental standards.

Therefore, accurately managing heat input is crucial for both operational and environmental perspectives.

River Temperature Change

In coal-fired power plants, a significant portion of the waste heat is released into nearby rivers. The transfer of waste heat results in a temperature change in the water body, potentially impacting local ecosystems.

Typically, rivers receiving waste heat experience a temperature rise. In the exercise, the river's temperature increases from \( 18.0^{\circ} \mathrm{C} \) to \( 18.5^{\circ} \mathrm{C} \). Such temperature shifts can affect aquatic life, altering oxygen levels and habitat conditions.

• To mitigate these effects, the design includes technologies like cooling towers that reduce thermal pollution.
• Regulating discharge temperatures ensures biodiversity protection while facilitating power plant operations.

Ensuring minimal environmental impact from temperature changes is quintessential to sustainable plant operation and preserving local wildlife.