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In \(2001,\) the United States produced 51 percent of its electricity in the amount of \(1.878 \times 10^{12} \mathrm{kWh}\) from coalfired power plants. Taking the average thermal efficiency to be 34 percent, determine the amount of thermal energy rejected by the coal-fired power plants in the United States that year.

Short Answer

Expert verified
Answer: The thermal energy rejected by the coal-fired power plants in the United States in 2001 was approximately \(3.645 \times 10^{12} \mathrm{kWh}\).

Step by step solution

01

Identify the given values

The exercise provides us with the following information: - Total electricity generated in 2001: \(1.878 \times 10^{12} \mathrm{kWh}\) - Average thermal efficiency of coal-fired power plants: 34%
02

Calculate the energy input

To calculate the energy input, we will use the formula for thermal efficiency: Thermal Efficiency = (Output Energy) / (Input Energy) We can rearrange the formula to solve for Input Energy: Input Energy = (Output Energy) / (Thermal Efficiency) Plugging in the given values: Input Energy = \((1.878 \times 10^{12} \mathrm{kWh}) / 0.34\)
03

Evaluate the expression to find the energy input

Calculating the energy input: Input Energy = \((1.878 \times 10^{12} \mathrm{kWh}) / 0.34 \approx 5.523 \times 10^{12} \mathrm{kWh}\)
04

Calculate the energy rejected

To find the energy rejected, we subtract the output energy (electricity generated) from the input energy: Energy Rejected = Input Energy - Output Energy Then, plugging in the values: Energy Rejected = \(5.523 \times 10^{12} \mathrm{kWh} - 1.878 \times 10^{12} \mathrm{kWh}\)
05

Evaluate the expression to find the energy rejected

Calculating the energy rejected: Energy Rejected = \(5.523 \times 10^{12} \mathrm{kWh} - 1.878 \times 10^{12} \mathrm{kWh} \approx 3.645 \times 10^{12} \mathrm{kWh}\) Therefore, the amount of thermal energy rejected by the coal-fired power plants in the United States in 2001 was approximately \(3.645 \times 10^{12} \mathrm{kWh}\).

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Key Concepts

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

Energy Input Calculation
When dealing with the efficiency of a power plant, one key aspect to consider is the energy input calculation. The energy input is the total amount of energy needed to produce a certain output of electricity. To find this critical value, you start with the output – the amount of electricity generated by the plant – and consider the plant's thermal efficiency.

The thermal efficiency of a power plant tells us what fraction of the input energy gets converted to electrical energy. It's calculated by taking the electrical energy output and dividing it by the total energy input. To find the energy input when we know the plant’s output and its thermal efficiency, we simply rearrange the formula:
Input Energy = Output Energy / Thermal Efficiency

For instance, if a power plant has an efficiency of 34% and produces a given output, the energy input is the output divided by 0.34. Remember, to express thermal efficiency as a decimal in calculations, you convert the percentage into a decimal by dividing by 100. Thus, 34% becomes 0.34. It's crucial to convert your units consistently to avoid any miscalculations, typically energy is measured in kilowatt-hours (kWh) for such large-scale engineering problems.
Energy Rejected by Power Plants
A significant amount of the energy generated by power plants is not converted into electricity, but is instead rejected or wasted. This is an inevitable consequence of the Second Law of Thermodynamics, which dictates that not all energy input can be transformed into useful work.

Understanding the Rejected Energy

The energy that is not used in electricity production typically takes the form of heat and is discharged into the environment. This is a loss for the plant since it represents energy that was produced (often through the burning of fuel) but did not convert to electricity. It is calculated as the difference between the total energy input and the energy that was actually converted to electricity:
Energy Rejected = Input Energy - Output Energy

To optimize the performance of power plants, engineers work on minimizing this energy rejection, enhancing the plant's thermal efficiency, and reducing the environmental impact. For example, improvements in materials and the integration of systems like combined cycle power plants, where waste heat is used to drive another cycle to produce more electricity are ways to achieve this.
Thermal Energy Conversion
Thermal energy conversion refers to the process of converting heat energy into electricity, which is a central function of thermal power plants. Coal-fired power plants, for instance, burn coal to generate heat. This heat is then used to produce steam, which drives a turbine connected to an electric generator, ultimately producing electricity.

The Efficiency of Conversion

The efficiency of this conversion process is a vital metric and is quantitatively expressed by the thermal efficiency. This implies that higher thermal efficiency equates to more electricity per unit of fuel consumed.

The efficiency is not just a measure of performance, but also of cost-effectiveness and environmental impact. A plant with higher thermal efficiency utilizes less fuel to produce the same amount of electricity, which also means it emits lower quantities of greenhouse gases if the fuel is a fossil fuel. Students and engineers alike are interested in finding ways to improve thermal efficiency, such as using higher temperature materials, waste heat recovery systems, and advanced turbine designs, making the conversion process more economical and environmentally friendly.

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Most popular questions from this chapter

Consider an office room that is being cooled adequately by a 12,000 Btu/h window air conditioner. Now it is decided to convert this room into a computer room by installing several computers, terminals, and printers with a total rated power of \(8.4 \mathrm{kW}\). The facility has several \(7000 \mathrm{Btu} / \mathrm{h}\) air conditioners in storage that can be installed to meet the additional cooling requirements. Assuming a usage factor of 0.4 (i.e., only 40 percent of the rated power will be consumed at any given time) and additional occupancy of seven people, each generating heat at a rate of \(100 \mathrm{W}\), determine how many of these air conditioners need to be installed to the room.

A refrigeration cycle is executed with \(\mathrm{R}-134 \mathrm{a}\) under the saturation dome between the pressure limits of 1.6 and \(0.2 \mathrm{MPa}\). If the power consumption of the refrigerator is \(3 \mathrm{kW},\) the maximum rate of heat removal from the cooled space of this refrigerator is \((a) 0.45 \mathrm{kJ} / \mathrm{s}\) (b) \(0.78 \mathrm{kJ} / \mathrm{s}\) \((c) 3.0 \mathrm{kJ} / \mathrm{s}\) \((d) 11.6 \mathrm{kJ} / \mathrm{s}\) \((e) 14.6 \mathrm{kJ} / \mathrm{s}\)

An air-conditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of \(32 \mathrm{kJ} / \mathrm{s}\) to maintain its temperature constant at \(20^{\circ} \mathrm{C}\) If the temperature of the outdoors is \(35^{\circ} \mathrm{C},\) the power required to operate this air-conditioning system is \((a) 0.58 \mathrm{kW}\) (b) \(3.20 \mathrm{kW}\) \((c) 1.56 \mathrm{kW}\) \((d) 2.26 \mathrm{kW}\) \((e) 1.64 \mathrm{kW}\)

A heat pump supplies heat energy to a house at the rate of \(140,000 \mathrm{kJ} / \mathrm{h}\) when the house is maintained at \(25^{\circ} \mathrm{C} .\) Over a period of one month, the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house. Consider a heat pump receiving heat from two different outside energy sources. In one application the heat pump receives heat from the outside air at \(0^{\circ} \mathrm{C} .\) In a second application the heat pump receives heat from a lake having a water temperature of \(10^{\circ} \mathrm{C}\). If electricity costs \(\$ 0.105 / \mathrm{kWh}\), determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

Using EES (or other) software, determine the maximum work that can be extracted from a pond containing \(10^{5} \mathrm{kg}\) of water at \(350 \mathrm{K}\) when the temperature of the surroundings is \(300 \mathrm{K}\). Notice that the temperature of water in the pond will be gradually decreasing as energy is extracted from it; therefore, the efficiency of the engine will be decreasing. Use temperature intervals of \((a) 5 \mathrm{K},(b) 2 \mathrm{K}\) and \((c) 1 \mathrm{K}\) until the pond temperature drops to \(300 \mathrm{K}\). Also solve this problem exactly by integration and compare the results.

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