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Based on thermal efficiency, approximately two-thirds of the energy input by heat transfer in the steam generator of a power plant is ultimately rejected to cooling water flowing through the condenser. Is the heat rejected an indicator of the inefficiency of the power plant?

Short Answer

Expert verified
The heat rejected is not an indicator of the inefficiency but a result of the second law of thermodynamics inherent in thermal power plants.

Step by step solution

01

Understanding Thermal Efficiency

Thermal efficiency is the ratio of the work output of the power plant to the heat input. It measures how effectively a power plant converts heat energy into useful work.
02

Calculating Energy Distribution

Given that two-thirds of the energy input is rejected to cooling water, it indicates that only one-third is converted into useful work. This can be expressed as: \[ \text{Useful Work} = \frac{1}{3} \times \text{Heat Input} \] and \[ \text{Heat Rejected} = \frac{2}{3} \times \text{Heat Input} \]
03

Evaluating Heat Rejected

The rejection of heat is part of the thermodynamic process. According to the second law of thermodynamics, not all the heat energy can be converted into work.
04

Assessing Power Plant Inefficiency

While the heat rejected might seem to be a waste, it is an inherent feature of thermal power plants. High rejection does not inherently indicate inefficiency but reflects the limitations imposed by thermodynamic laws.

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Key Concepts

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

thermodynamics
Thermodynamics is the study of energy, heat, and their interactions.
It involves understanding how energy transforms from one form to another and how it affects matter.
The main laws of thermodynamics help us analyze power plants and engines.
The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed.
The second law of thermodynamics introduces the concept of entropy and states that energy transformations are not 100% efficient.
This means that some energy is always lost as unusable heat.
These laws are crucial in determining how efficiently a power plant can convert heat energy into useful work. Understanding these principles can help us improve the efficiency of energy systems and develop new technologies.
heat transfer
Heat transfer is the process by which thermal energy moves from one object or substance to another.
This occurs through conduction, convection, and radiation.
In a power plant, heat transfer plays a key role in the energy conversion process.
The steam generator heats water to produce steam, which drives turbines to generate electricity.
Heat is transferred from the steam to the surrounding environment, usually through condensers and cooling systems.
Understanding heat transfer helps engineers design more efficient systems.
For example, improving heat exchangers can reduce energy losses and enhance overall thermal efficiency.
power plant efficiency
Power plant efficiency measures how well a power plant converts fuel into useful electrical energy.
It is expressed as a percentage and is calculated by comparing the energy input to the energy output.
High efficiency means more of the fuel's energy is converted into electricity, while lower efficiency means more energy is wasted.
There are many factors affecting power plant efficiency, including the type of fuel, the technology used, and how well the plant is maintained.
Improving efficiency involves optimizing the thermodynamic processes and minimizing energy losses.
This can result in lower operating costs and reduced environmental impact.
second law of thermodynamics
The second law of thermodynamics states that energy transformations are never completely efficient.
It introduces the concept of entropy, which measures the disorder or randomness of a system.
According to this law, in any energy conversion process, some energy is always lost as heat.
This is why no power plant can be 100% efficient.
In the context of a thermal power plant, a significant portion of the heat input is inevitably rejected to cooling water.
This heat rejection is not a sign of inefficiency but a natural consequence of the second law.
Recognizing this helps us set realistic expectations and work towards improving other aspects of power plant performance.

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Most popular questions from this chapter

A binary vapor power cycle consists of two ideal Rankine cycles with steam and ammonia as the working fluids. In the steam cycle, superheated vapor enters the turbine at \(6 \mathrm{MPa}\), \(640^{\circ} \mathrm{C}\), and saturated liquid exits the condenser at \(60^{\circ} \mathrm{C}\). The heat rejected from the steam cycle is provided to the ammonia cycle, producing saturated vapor at \(50^{\circ} \mathrm{C}\), which enters the ammonia turbine. Saturated liquid leaves the ammonia condenser at \(1 \mathrm{MPa}\). For a net power output of \(20 \mathrm{MW}\) from the binary cycle, determine (a) the power output of the steam and ammonia turbines, respectively, in MW. (b) the rate of heat addition to the binary cycle, in MW. (c) the thermal efficiency.

Water is the working fluid in an ideal Rankine cycle. The condenser pressure is \(8 \mathrm{kPa}\), and saturated vapor enters the turbine at (a) \(18 \mathrm{MPa}\) and (b) \(4 \mathrm{MPa}\). The net power output of the cycle is \(100 \mathrm{MW}\). Determine for each case the mass flow rate of steam, in \(\mathrm{kg} / \mathrm{h}\), the heat transfer rates for the working fluid passing through the boiler and condenser, each in \(\mathrm{kW}\), and the thermal efficiency.

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one. Steam enters the first turbine stage at \(8 \mathrm{MPa}, 480^{\circ} \mathrm{C}\), and expands to \(2 \mathrm{MPa}\). Some steam is extracted at \(2 \mathrm{MPa}\) and fed to the closed feedwater heater. The remainder expands through the second-stage turbine to \(0.3 \mathrm{MPa}\), where an additional amount is extracted and fed into the open feedwater heater, which operates at \(0.3 \mathrm{MPa}\). The steam expanding through the third-stage turbine exits at the condenser pressure of \(8 \mathrm{kPa}\). Feedwater leaves the closed heater at \(205^{\circ} \mathrm{C}, 8 \mathrm{MPa}\), and condensate exiting as saturated liquid at \(2 \mathrm{MPa}\) is trapped into the open heater. Saturated liquid at \(0.3 \mathrm{MPa}\) leaves the open feedwater heater. The net power output of the cycle is \(100 \mathrm{MW}\). If the turbine stages and pumps are isentropic, determine (a) the thermal efficiency. (b) the mass flow rate of steam entering the first turbine, in \(\mathrm{kg} / \mathrm{h}\).

An ideal Rankine cycle with reheat uses water as the working fluid. The conditions at the inlet to the first-stage turbine are \(14 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) and the steam is reheated between the turbine stages to \(600^{\circ} \mathrm{C}\). For a condenser pressure of \(6 \mathrm{kPa}\), plot the cycle thermal efficiency versus reheat pressure for pressures ranging from 2 to \(12 \mathrm{MPa}\).

Water is the working fluid in a cogeneration cycle that generates electricity and provides heat for campus buildings. Steam at \(2 \mathrm{MPa}, 320^{\circ} \mathrm{C}\), enters a two-stage turbine with a mass flow rate of \(0.82 \mathrm{~kg} / \mathrm{s}\). A fraction of the total flow, 0.141, is extracted between the two stages at \(0.15 \mathrm{MPa}\) to provide for building heating, and the remainder expands through the second stage to the condenser pressure of \(0.06\) bar. Condensate returns from the campus buildings at \(0.1 \mathrm{MPa}, 60^{\circ} \mathrm{C}\) and passes through a trap into the condenser, where it is reunited with the main feedwater flow. Saturated liquid leaves the condenser at \(0.06\) bar. Each turbine stage has an isentropic efficiency of \(80 \%\), and the pumping process can be considered isentropic. Determine (a) the rate of heat transfer to the working fluid passing through the steam generator, in \(\mathrm{kJ} / \mathrm{h}\). (b) the net power developed, in \(\mathrm{kJ} / \mathrm{h}\). (c) the rate of heat transfer for building heating, in \(\mathrm{kJ} / \mathrm{h}\). (d) the rate of heat transfer to the cooling water passing through the condenser, in \(\mathrm{kJ} / \mathrm{h}\).

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