/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Use the exergy balance equation ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the exergy balance equation to find the efficiency of a steady-state Carnot heat engine operating between two fixed temperature reservoirs.

Short Answer

Expert verified
The efficiency of a steady-state Carnot heat engine is \( \eta = 1 - \frac{T_C}{T_H} \).

Step by step solution

01

Understand the Carnot Heat Engine

A Carnot heat engine is an idealized engine that operates on the Carnot cycle. It is considered the most efficient possible heat engine that can operate between two thermal reservoirs. Its efficiency depends solely on the temperatures of the two reservoirs.
02

State the Temperatures

Identify the temperatures of the two reservoirs. Let the hot reservoir have a temperature of \( T_H \) and the cold reservoir have a temperature of \( T_C \), both measured in Kelvin.
03

Express Efficiency with Exergy Balance

The efficiency \( \eta \) of a Carnot engine can be defined using the exergy balance, emphasizing that the efficiency is a function of the temperature of the reservoirs: \[ \eta = 1 - \frac{T_C}{T_H} \]. This shows that efficiency increases as the temperature of the hot reservoir increases or as the temperature of the cold reservoir decreases.
04

Interpretation of the Formula

This formula, derived from the exergy balance principle for a Carnot engine, indicates that some of the input energy (exergy) is unavoidably lost due to the necessity of transferring heat to the cold reservoir. Thus, the Carnot efficiency represents the maximum conceivable efficiency.

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 91Ó°ÊÓ!

Key Concepts

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

Carnot heat engine
The Carnot heat engine is a theoretical concept designed to understand the limits of efficiency for heat engines. It operates on what is known as the Carnot cycle, which is an idealized cycle because it assumes no energy is lost due to friction or other inefficiencies.
The main insight behind a Carnot engine is that it achieves maximum efficiency by extracting work from heat transferred between two thermal reservoirs.
In reality, no engine can achieve this idealized efficiency due to real-world constraints, but it provides a benchmark for measuring how efficient a real engine can be. The cycle consists of two isothermal processes (constant temperature) and two adiabatic processes (no heat exchange), making the Carnot engine a reversible cycle.
Efficiency
Efficiency in the context of thermal engines refers to the ratio of useful work output to the heat input. For the Carnot heat engine, this efficiency is maximized and is expressed through the equation: \[ \eta = 1 - \frac{T_C}{T_H} \]where \( \eta \) is the efficiency, \( T_H \) is the temperature of the hot reservoir, and \( T_C \) is the temperature of the cold reservoir, all measured in Kelvin.
This formula shows that
  • Efficiency increases as \( T_H \) increases.
  • Efficiency increases as \( T_C \) decreases.
In essence, to improve the efficiency of a Carnot engine, one should increase the temperature of the hot reservoir or decrease the temperature of the cold reservoir.
However, in practical applications, there are limitations in how much we can actually increase or decrease these temperatures.
Thermal reservoirs
Thermal reservoirs are a crucial component in the working of a Carnot heat engine. These reservoirs act as infinite sinks or sources of heat, meaning they can absorb or supply heat without changing their own temperature significantly.
  • The hot reservoir provides heat energy to the engine.
  • The cold reservoir absorbs the waste heat from the engine.
The temperatures of these reservoirs are denoted as \( T_H \) for the hot reservoir and \( T_C \) for the cold reservoir.
These temperatures are fundamental in calculating the Carnot efficiency and understanding the flow of energy within the system. Essentially functioning as the starting and ending points for the heat cycle, the reservoirs are essential for the theoretical framework of the Carnot engine.
Carnot cycle
The Carnot cycle is the idealized cycle on which the Carnot heat engine operates. It comprises four stages:
  • Isothermal expansion: The system absorbs heat from the hot reservoir at \( T_H \) while doing work on the surroundings.
  • Adiabatic expansion: The system continues to expand without heat exchange, causing the temperature to drop.
  • Isothermal compression: The system releases heat to the cold reservoir at \( T_C \) while the surroundings do work on it.
  • Adiabatic compression: The system is compressed further, increasing its temperature back to \( T_H \) without heat exchange.
This cycle demonstrates the maximum efficiency of a heat engine and provides insights into the fundamental thermodynamic principles of energy conversion.
While real engines cannot practically achieve the Carnot cycle's efficiency due to irreversible processes present in actual systems, it sets a valuable standard for evaluating engine performance.

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

A 200-L insulated tank contains nitrogen gas at \(200 \mathrm{kPa}, 300 \mathrm{~K}\). A line with nitrogen at \(500 \mathrm{~K}\) \(500 \mathrm{kPa}\) adds \(40 \%\) more mass to the tank with a flow through a valve. Use constant specific heats to find the final temperature and the exergy destruction.

A steady-flow device receives \(\mathrm{R}-410 \mathrm{a}\) at \(40^{\circ} \mathrm{C}\) \(800 \mathrm{kPa}\) and it exits at \(40^{\circ} \mathrm{C}, 100 \mathrm{kPa}\). Assume a reversible isothermal process. Find the change in specific exergy.

An adiabatic and reversible air compressor takes air in at 15 psia, \(560 \mathrm{R}\). The air exits at 90 psia at the rate of \(0.8 \mathrm{lbm} / \mathrm{s}\). Determine the minimum compressor work input and repeat for an inlet at \(530 \mathrm{R}\) instead. Why is the work less for a lower inlet \(T ?\)

Water as saturated liquid at 200 kPa goes through a constant-pressure heat exchanger, as shown in Fig. \(\mathrm{P} 8.79 .\) The heat input is supplied from a reversible heat pump extracting heat from the surroundings at \(17^{\circ} \mathrm{C}\). The water flow rate is \(2 \mathrm{~kg} / \mathrm{min}\) and the whole process is reversible; that is, there is no overall net entropy change. If the heat pump receives \(40 \mathrm{~kW}\) of work, find the water exit state and the increase in specific exergy of the water.

An air conditioner on a hot summer day removes \(8 \mathrm{~kW}\) of energy from a house at \(21^{\circ} \mathrm{C}\) and pushes energy to the outside, which is at \(31^{\circ} \mathrm{C}\). The house has a 15000 -kg mass with an average specific heat of \(0.95 \mathrm{~kJ} / \mathrm{kgK}\). In order to do this, the cold side of the air conditioner is at \(5^{\circ} \mathrm{C}\) and the hot side is at \(40^{\circ} \mathrm{C}\). The air conditioner (refrigerator) has a \(\mathrm{COP}\) that is \(60 \%\) that of a corresponding Carnot refrigerator. Find the actual air conditioner COP, the power required to run the air conditioner, the rate of exergy destruction inside the air conditioner, and the total rate of exergy destruction due to the air conditioner and the house.

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.