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Chapter 8: Problem 7

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

Steam enters the turbine of a vapor power plant at 100 bar, \(520^{\circ} \mathrm{C}\) and expands adiabatically, exiting at \(0.08\) bar with a quality of \(90 \%\). Condensate leaves the condenser as saturated liquid at \(0.08\) bar. Liquid exits the pump at 100 bar, \(43^{\circ} \mathrm{C}\). The specific exergy of the fuel entering the combustor unit of the steam generator is estimated to be \(14,700 \mathrm{~kJ} / \mathrm{kg}\). No exergy is carried in by the combustion air. The exergy of the stack gases leaving the steam generator is estimated to be \(150 \mathrm{~kJ}\) per \(\mathrm{kg}\) of fuel. The mass flow rate of the steam is \(3.92 \mathrm{~kg}\) per \(\mathrm{kg}\) of fuel. Cooling water enters the condenser at \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=\) \(1 \mathrm{~atm}\) and exits at \(35^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Develop a full accounting of the exergy entering the plant with the fuel.

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage at \(12 \mathrm{MPa}, 520^{\circ} \mathrm{C}\) and expands to \(1 \mathrm{MPa}\), where some of the steam is extracted and diverted to the open feedwater heater operating at \(1 \mathrm{MPa}\). The remaining steam expands through the second turbine stage to the condenser pressure of \(6 \mathrm{kPa}\). Saturated liquid exits the open feedwater heater at \(1 \mathrm{MPa}\). For isentropic processes in the turbines and pumps, determine for the cycle (a) the thermal efficiency and (b) the mass flow rate into the first turbine stage, in \(\mathrm{kg} / \mathrm{h}\), for a net power output of \(330 \mathrm{MW}\).

Superheated steam at \(18 \mathrm{MPa}, 560^{\circ} \mathrm{C}\), enters the turbine of a vapor power plant. The pressure at the exit of the turbine is \(0.06\) bar, and liquid leaves the condenser at \(0.045\) bar, \(26^{\circ} \mathrm{C}\). The pressure is increased to \(18.2 \mathrm{MPa}\) across the pump. The turbine and pump have isentropic efficiencies of 82 and \(77 \%\), respectively. For the cycle, determine (a) the net work per unit mass of steam flow, in \(\mathrm{kJ} / \mathrm{kg}\). (b) the heat transfer to steam passing through the boiler, in kJ per \(\mathrm{kg}\) of steam flowing. (c) the thermal efficiency. (d) the heat transfer to cooling water passing through the condenser, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam condensed.

Steam enters the first turbine stage of a vapor power cycle with reheat and regeneration at \(32 \mathrm{MPa}, 600^{\circ} \mathrm{C}\), and expands to \(8 \mathrm{MPa}\). A portion of the flow is diverted to a closed feedwater heater at \(8 \mathrm{MPa}\), and the remainder is reheated to \(560^{\circ} \mathrm{C}\) before entering the second turbine stage. Expansion through the second turbine stage occurs to \(1 \mathrm{MPa}\), where another portion of the flow is diverted to a second closed feedwater heater at \(1 \mathrm{MPa}\). The remainder of the flow expands through the third turbine stage to \(0.15 \mathrm{MPa}\), where a portion of the flow is diverted to an open feedwater heater operating at \(0.15 \mathrm{MPa}\), and the rest expands through the fourth turbine stage to the condenser pressure of \(6 \mathrm{kPa}\). Condensate leaves each closed feedwater heater as saturated liquid at the respective extraction pressure. The feedwater streams leave each closed feedwater heater at a temperature equal to the saturation temperature at the respective extraction pressure. The condensate streams from the closed heaters each pass through traps into the next lower-pressure feedwater heater. Saturated liquid exiting the open heater is pumped to the steam generator pressure. If each turbine stage has an isentropic efficiency of \(85 \%\) and the pumps operate isentropically (a) sketch the layout of the cycle and number the principal state points. (b) determine the thermal efficiency of the cycle. (c) calculate the mass flow rate into the first turbine stage, in \(\mathrm{kg} / \mathrm{h}\), for a net power output of \(500 \mathrm{MW}\).

Steam at \(10 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) enters the first-stage turbine of an ideal Rankine cycle with reheat. The steam leaving the reheat section of the steam generator is at \(500^{\circ} \mathrm{C}\), and the condenser pressure is \(6 \mathrm{kPa}\). If the quality at the exit of the secondstage turbine is \(90 \%\), determine the cycle thermal efficiency.
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