Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Starting with Eq. \(10-20\), show that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as \(x_{\text {dest }}=q_{\text {in }}\left(\eta_{\text {th,Camot }}-\eta_{\text {th }}\right),\) where \(\eta_{\text {th }}\) is efficiency of the Rankine cycle and \(\eta_{\mathrm{th}, \mathrm{Camot}}\) is the efficiency of the Carnot cycle operating between the same temperature limits.

Short Answer

Expert verified
In summary, exergy destruction is directly related to the difference in efficiency between the Rankine cycle and the Carnot cycle operating between the same temperature limits. The expression for exergy destruction in terms of the efficiency of both cycles is given by: \(x_{dest} = Q_{in} (\eta_{th,Carnot} - \eta_{th})\)

Step by step solution

01

Define Efficiency of Rankine Cycle and Carnot Cycle

The efficiency of the Rankine cycle (\(\eta_{th}\)) is the ratio of net work output (\(W_{net}\)) to the heat input (\(Q_{in}\)): \(\eta_{th} = \frac{W_{net}}{Q_{in}}\) The efficiency of the Carnot cycle operating between the same temperature limits (\(\eta_{th,Carnot}\)), with the maximum temperature being \(T_H\) and minimum temperature \(T_L\), is given as: \(\eta_{th,Carnot} = 1 - \frac{T_L}{T_H}\)
02

Define Exergy Destruction

Exergy destruction (\(x_{dest}\)) is proportional to efficiency and can be defined as an energy loss that cannot be recovered due to irreversibility in a process. The exergy destruction in a Rankine cycle can be expressed as: \(x_{dest} = Q_{in} - W_{net}\)
03

Combine Efficiency and Exergy Destruction Expressions

Let's rewrite the exergy destruction expression (\(x_{dest}\)) in terms of the efficiency of the Rankine cycle and the Carnot cycle. \(x_{dest} = Q_{in} - W_{net}\) From \(\eta_{th} = \frac{W_{net}}{Q_{in}}\), we can write \(W_{net}\) as: \(W_{net} = \eta_{th} Q_{in}\) Substitute this expression into the exergy destruction equation: \(x_{dest} = Q_{in} - \eta_{th} Q_{in}\) Factor out \(Q_{in}\): \(x_{dest} = Q_{in}(1 - \eta_{th})\) Now, let's find an expression for \(\eta_{th,Carnot} - \eta_{th}\): \(\eta_{th,Carnot} - \eta_{th} = \left(1 - \frac{T_L}{T_H}\right) - \eta_{th}\) If we multiply both sides of this equation by \(Q_{in}\), we get: \(Q_{in}(\eta_{th,Carnot} - \eta_{th}) = Q_{in}\left(1 - \frac{T_L}{T_H} - \eta_{th}\right)\) Now, compare this with the equation derived for \(x_{dest}\), we can see that the two expressions are equal, therefore: \(x_{dest} = Q_{in} (\eta_{th,Carnot} - \eta_{th})\) This is the final expression we were looking for, which relates exergy destruction to the efficiency of the Rankine cycle and the efficiency of the Carnot cycle operating between the same temperature limits.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Carnot Cycle Efficiency
The Carnot cycle, named after French engineer Sadi Carnot, represents an idealized heat engine that operates on the Carnot principle. This theoretical engine exhibits the maximum possible efficiency that a heat engine can achieve between two given temperatures.

The efficiency of a Carnot cycle, denoted by \(\eta_{th,Carnot}\), depends on the temperatures of the heat source, \(T_H\), and the heat sink, \(T_L\). It is defined as the difference in heat absorbed from the hot reservoir and the work done, divided by the total heat absorbed.

\
Rankine Cycle Efficiency
The Rankine cycle is a practical model that approximates the operation of steam heat engines, commonly found in power plants. Unlike the Carnot cycle, which is a purely theoretical model and assumes reversible processes, the Rankine cycle includes the realistic irreversibility of engines.

The efficiency of a Rankine cycle, expressed as \(\eta_{th}\), is the ratio of the net work output of the cycle to the heat input. In other words, it represents how effectively the heat engine converts heat from a boiler into usable work, driving factors such as turbines. It's crucial to recognize that the Rankine cycle's efficiency is always less than that of the corresponding Carnot cycle because real processes incur losses due to factors like friction and heat loss to the environment.

Measures to improve the efficiency of the Rankine cycle include superheating the steam to higher temperatures and increasing the boiler pressure, both of which help to reduce the relative amount of energy losses compared to the energy input.

Thermodynamic Efficiency
In general, thermodynamic efficiency is a metric used to evaluate the performance of any heat engine or thermal system. It is a ratio of the useful work output to the energy input, usually expressed as a percentage. Ideally, a higher thermodynamic efficiency means that less energy is wasted and more is converted to useful work.

It's important to highlight that no real-world engine can be 100% efficient due to irreversible processes and the second law of thermodynamics. This law states that entropy, or the measure of disorder within a system, must increase over time, which gives rise to energy dissipation in heat engines such as those following the Rankine or Carnot cycles.

Efficiency improvements in thermal systems are vital for reducing energy consumption and can be achieved through various techniques such as improving insulation, using more efficient components, and designing systems to minimize resistive losses.

Entropy and Exergy
Entropy is a fundamental concept in thermodynamics, often interpreted as a measure of disorder or randomness in a system. The second law of thermodynamics establishes that the total entropy of an isolated system can never decrease over time. In practical terms, entropy quantifies the amount of energy that is no longer capable of doing work due to being dispersed or 'spread out' across a system.

Exergy, on the other hand, measures the quality or 'usefulness' of energy. It reflects the maximum work potential of a system or process. Exergy takes into account both the system and its environment to determine how much work could theoretically be extracted. When a system undergoes a change, the reduction in exergy represents the lost opportunity to do work, also known as exergy destruction. This destruction is primarily due to irreversibilities within the system, such as friction, inelastic deformation, and heat transfer to lower temperatures.

Understanding entropy and exergy is crucial for designing more efficient systems and processes. Reducing exergy destruction can significantly enhance the operational efficiency and sustainability of thermodynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the boiler at \(5000 \mathrm{kPa},\) the reheat section at \(1200 \mathrm{kPa}\), and the condenser at 20 kPa. The mixture quality at the exit of both turbines is 96 percent. Determine the temperature at the inlet of each turbine and the cycle's thermal efficiency.

Consider a cogeneration plant for which the utilization factor is \(0.5 .\) Can the exergy destruction associated with this plant be zero? If yes, under what conditions?

A steam power plant operates on an ideal reheat regenerative Rankine cycle and has a net power output of \(80 \mathrm{MW}\). Steam enters the high-pressure turbine at \(10 \mathrm{MPa}\) and \(550^{\circ} \mathrm{C}\) and leaves at \(0.8 \mathrm{MPa}\). Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater. The rest of the steam is reheated to \(500^{\circ} \mathrm{C}\) and is expanded in the low-pressure turbine to the condenser pressure of \(10 \mathrm{kPa}\). Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine \((a)\) the mass flow rate of steam through the boiler and ( \(b\) ) the thermal efficiency of the cycle.

Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet state the same, \((a)\) the turbine work output will decrease. \((b)\) the amount of heat rejected will decrease. \((c)\) the cycle efficiency will decrease. \((d)\) the moisture content at turbine exit will decrease. \((e)\) the pump work input will decrease.

Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the cycle is modified with regeneration that involves one open feedwater heater (select the correct statement per unit mass of steam flowing through the boiler), \((a)\) the turbine work output will decrease. \((b)\) the amount of heat rejected will increase. \((c)\) the cycle thermal efficiency will decrease. \((d)\) the quality of steam at turbine exit will decrease. \((e)\) the amount of heat input will increase.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free