91Ó°ÊÓ

A gas flows through a one-inlet, one-exit control volume operating at steady state. Heat transfer at the rate \(\dot{Q}_{\mathrm{cv}}\) takes place only at a location on the boundary where the temperature is \(T_{\mathrm{b}}\). For each of the following cases, determine whether the specific entropy of the gas at the exit is greater than, equal to, or less than the specific entropy of the gas at the inlet: (a) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}=0\). (b) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (c) no internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}>0\). (d) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}}<0\). (e) internal irreversibilities, \(\dot{Q}_{\mathrm{cv}} \geq 0\)

Step by step solution

01

- Understand the entropy balance equation

For a control volume operating at steady state, the entropy balance can be expressed as: \[ \frac{dS_{cv}}{dt} = \frac{\text{d}\text{mi}_{in}}{T_{in}} - \frac{\text{d}\text{mi}_{out}}{T_{out}} + \frac{Q_{cv}}{T_b} + \text{entropy generation} \] Since the system is at steady state, there is no accumulation of entropy within the control volume, so \[ \frac{dS_{cv}}{dt} = 0 \].

02

- Simplify the entropy balance equation

Given that \( \frac{dS_{cv}}{dt} = 0 \), the entropy balance equation becomes: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} - \frac{\text{d}\text{mi}_{out}}{T_{out}} + \frac{Q_{cv}}{T_b} + \text{entropy generation} = 0 \]

03

- Analyze the case \( \text{(a)} \)

For case \( \text{(a)} \), no internal irreversibilities and \( \text{Q}_{\text{cv}} = 0 \): Since there are no internal irreversibilities, entropy generation is zero. The entropy balance reduces to: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} = \frac{\text{d}\text{mi}_{out}}{T_{out}} \] Therefore, the specific entropy at the exit is equal to that at the inlet.

04

- Analyze the case \( \text{(b)} \)

For case \( \text{(b)} \), no internal irreversibilities and \( \text{Q}_{\text{cv}} < 0 \): Negative heat transfer means heat is leaving the system. Since there are no internal irreversibilities, entropy generation is zero. The entropy balance reduces to: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} = \frac{\text{d}\text{mi}_{out}}{T_{out}} - \frac{|Q_{cv}|}{T_b} \] Therefore, specific entropy at the exit is less than that at the inlet.

05

- Analyze the case \( \text{(c)} \)

For case \( \text{(c)} \), no internal irreversibilities and \( \text{Q}_{\text{cv}} > 0 \): Positive heat transfer means heat is entering the system. Since there are no internal irreversibilities, entropy generation is zero. The entropy balance reduces to: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} = \frac{\text{d}\text{mi}_{out}}{T_{out}} + \frac{Q_{cv}}{T_b} \] Therefore, specific entropy at the exit is greater than that at the inlet.

06

- Analyze the case \( \text{(d)} \)

For case \( \text{(d)} \), internal irreversibilities and \( \text{Q}_{\text{cv}} < 0 \): Negative heat transfer means heat is leaving the system. Since there are internal irreversibilities, entropy generation is positive. The entropy balance equation becomes: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} = \frac{\text{d}\text{mi}_{out}}{T_{out}} - \frac{|Q_{cv}|}{T_b} + \text{entropy generation} \] Therefore, specific entropy at the exit is not necessarily less than the inlet because the entropy generation might counteract heat loss.

07

- Analyze the case \( \text{(e)} \)

For case \( \text{(e)} \), internal irreversibilities and \( \text{Q}_{\text{cv}} \text{is greater than or equal to} 0 \): Positive or zero heat transfer means heat is entering or there is no heat transfer, respectively. Since there are internal irreversibilities, entropy generation is positive. The entropy balance equation becomes: \[ \frac{\text{d}\text{mi}_{in}}{T_{in}} = \frac{\text{d}\text{mi}_{out}}{T_{out}} + \frac{Q_{cv}}{T_b} + \text{entropy generation} \] Therefore, specific entropy at the exit is greater than the inlet due to positive entropy generation and possible heat transfer.

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.

steady-state analysis

In thermodynamics, a steady-state analysis is crucial for understanding systems where conditions remain constant over time. This means that properties such as temperature, pressure, and specific entropy do not change as time passes.
Analysis of a system at steady state simplifies our calculations because we can ignore time-dependent changes. We assume that all inflows and outflows are constant.
In our exercise, the gas flows through a control volume in steady state. This implies a steady flow of mass and energy, which is helpful for applying the entropy balance equation to determine specific entropy changes.

heat transfer

Heat transfer is the movement of thermal energy from one place to another due to temperature differences. In the context of our exercise, heat transfer at a given rate \(\theta{Q}_{cv}\) takes place at the system's boundary where the temperature is \(T_{b}\).
Heat transfer can affect the entropy of a system. When heat is transferred into the system (positive \(Q_{cv}\)), it increases the system's entropy. Conversely, when heat is transferred out of the system (negative \(Q_{cv}\)), it decreases the system's entropy.
Understanding how heat transfer interacts with entropy helps us determine whether the specific entropy at the exit is greater than, equal to, or less than the specific entropy at the inlet.

internal irreversibilities

Internal irreversibilities refer to inefficiencies within a thermodynamic system that generate additional entropy. These can be caused by friction, turbulence, or other non-ideal processes.
In the exercise, internal irreversibilities influence whether entropy generation is zero or positive. For scenarios without internal irreversibilities, it's assumed there is no additional entropy generation.
Understanding the presence of internal irreversibilities helps us analyze the changes in specific entropy. For example, if internal irreversibilities are present, it means positive entropy generation, leading to potentially higher specific entropy at the exit compared to the inlet.

specific entropy

Specific entropy is a measure of the disorder or randomness per unit mass in a system. It's denoted as \(s\) and is typically measured in units of J/(kg·K).
In the entropy balance equation, specific entropy helps us understand how the entropy values at the inlet and outlet of the control volume compare.
For this exercise, we want to determine the effect of different scenarios on specific entropy. Factors like heat transfer, presence of internal irreversibilities, and steady-state analysis provide the context for determining whether specific entropy at the exit is greater than, equal to, or less than at the inlet.

thermodynamic systems

A thermodynamic system is a defined space or quantity of matter where we study energy and mass interactions with its surroundings.
In this exercise, the gas flowing through a one-inlet, one-exit control volume is our thermodynamic system. To analyze it, we apply principles from thermodynamics, such as the entropy balance equation, under the assumption of steady state.
Understanding the behavior of thermodynamic systems enables us to predict how variables like heat transfer and internal irreversibilities impact specific entropy, helping us solve for the conditions at the inlet and exit.