/*! 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 56 A cyclic machine, shown in Fig. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A cyclic machine, shown in Fig. P5.56, receives \(300 \mathrm{~kJ}\) from a \(1000-\mathrm{K}\) energy reservoir. It rejects \(120 \mathrm{~kJ}\) to a \(400-\mathrm{K}\) energy reservoir, and the cycle produces \(180 \mathrm{~kJ}\) of work as output. Is this cycle reversible, irreversible, or impossible?

Short Answer

Expert verified
The cycle is reversible.

Step by step solution

01

Identify Known Values

We are given the following values: the energy taken from the high-temperature reservoir, \(Q_H = 300 \, \text{kJ}\), the energy rejected to the low-temperature reservoir, \(Q_C = 120 \, \text{kJ}\), the work output, \(W = 180 \, \text{kJ}\), the high reservoir temperature, \(T_H = 1000 \, \text{K}\), and the low reservoir temperature, \(T_C = 400 \, \text{K}\).
02

Verify Energy Balance

Verify the energy balance in the cycle using the first law of thermodynamics, which states that \(Q_H = W + Q_C\). Substitute the known values: \(300 \, \text{kJ} = 180 \, \text{kJ} + 120 \, \text{kJ}\). This equation holds true, so the energy balance is confirmed.
03

Calculate Carnot Efficiency

The maximum efficiency for a heat engine operating between two reservoirs is given by the Carnot efficiency formula: \(\eta_c = 1 - \frac{T_C}{T_H}\). For this problem, \(\eta_c = 1 - \frac{400}{1000} = 0.6\, (60\%\)).
04

Calculate Actual Efficiency

The actual efficiency of the cycle is given by \(\eta = \frac{W}{Q_H}\). With \(W = 180 \, \text{kJ}\) and \(Q_H = 300 \, \text{kJ}\), the efficiency is \(\eta = \frac{180}{300} = 0.6\, (60\%\)).
05

Compare Efficiencies

Compare the actual efficiency to the Carnot efficiency. Since the actual efficiency (60%) is equal to the Carnot efficiency (60%), the cycle is operating at maximum theoretical efficiency, which means it is reversible.

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.

First Law of Thermodynamics
The first law of thermodynamics, often called the law of energy conservation, is a fundamental principle in physics. It states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of thermodynamic cycles, this means the energy going into a system must equal the energy coming out of it.
This can be understood through the equation:
  • \( Q_H = W + Q_C \)
Here, \( Q_H \) is the heat energy taken from a high-temperature reservoir, \( W \) is the work done by the system, and \( Q_C \) is the heat energy released to a low-temperature reservoir.
In the given exercise, by assigning the values provided, we have \( 300 \, \text{kJ} = 180 \, \text{kJ} + 120 \, \text{kJ} \). This confirms the equation balances correctly, adhering to the first law of thermodynamics. Understanding this principle is crucial as it connects directly to energy conservation in a closed system.
Carnot Efficiency
Carnot efficiency represents the theoretical maximum efficiency a heat engine can achieve operating between two reservoirs at different temperatures. It is an ideal standard and serves as a benchmark to compare real-world engines.
The efficiency is given by:
  • \( \eta_c = 1 - \frac{T_C}{T_H} \)
where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir, both measured in Kelvin.
For the exercise, this results in a Carnot efficiency of \( 60\% \), calculated as \( \eta_c = 1 - \frac{400}{1000} \). This calculation shows the potential for heat engine's performance between the specific temperature limits. Generally, no real engine can exceed this efficiency, as it would require a perfectly reversible process where no entropy is produced.
Reversible Cycles
Reversible cycles are ideal processes where the system returns to its initial state without any net change in the surroundings. These cycles are theoretical because real processes always involve some form of irreversibility, such as friction or unbalanced forces.
The hallmark of a reversible process is that it can achieve maximum efficiency without entropy increase.
In our exercise, the cycle is considered reversible because it achieves the Carnot efficiency of \( 60\% \), the theoretical limit for engines between given reservoir temperatures.
Recognizing reversible cycles helps in understanding the best possible scenario for thermodynamic systems, giving insights into how far real systems can come to ideal behavior.

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

Hydrogen gas is used in a Carnot cycle having an efficiency of \(60 \%\) with a low temperature of \(300 \mathrm{~K}\). During heat rejection, the pressure changes from \(90 \mathrm{kPa}\) to \(120 \mathrm{kPa}\). Find the high- and lowtemperature heat transfers and the net cycle work per unit mass of hydrogen.

An air conditioner on a hot summer day removes \(8 \mathrm{Btu} / \mathrm{s}\) of energy from a house at \(70 \mathrm{~F}\) and pushes energy to the outside, which is at \(88 \mathrm{~F}\). The house has \(50000 \mathrm{lbm}\) mass with an average specific heat of \(0.23 \mathrm{Btu} / \mathrm{lbm}-\mathrm{R}\). In order to do this, the cold side of the air conditioner is at \(40 \mathrm{~F}\) and the hot side is at \(100 \mathrm{~F}\). The air conditioner (refrigerator) has a COP that is \(50 \%\) that of a corresponding Carnot refrigerator. Find the actual COP of the air conditioner and the power required to run it.

A heat pump receives energy from a source at \(80^{\circ} \mathrm{C}\) and delivers energy to a boiler that operates at \(350 \mathrm{kPa}\). The boiler input is saturated liquid water and the exit is saturated vapor, both at \(350 \mathrm{kPa}\). The heat pump is driven by a \(2.5-\mathrm{MW}\) motor and has a COP that is \(60 \%\) that of a Carnot heat pump. What is the maximum mass-flow rate of water the system can deliver?

Consider the setup with two stacked (temperature-wise) heat engines, as in Fig. P5.2. Let \(T_{H}=1500 \mathrm{R}, T_{M}=1000 \mathrm{R}\), and \(T_{L}=650\) R. Find the two heat engine efficiencies and the combined overall efficiency assuming Carnot cycles.

A heat pump is used to heat a house during the winter. The house is to be maintained at \(20^{\circ} \mathrm{C}\) at all times. When the ambient temperature outside drops to \(-10^{\circ} \mathrm{C}\), the rate at which heat is lost from the house is estimated to be \(25 \mathrm{~kW}\). What is the minimum electrical power required to drive the heat pump?

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.