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Chapter 6: Problem 60

The temperature of an incompressible substance of mass \(m\) and specific heat \(c\) is reduced from \(T_{0}\) to \(T\left(

Short Answer

Expert verified
Plot \ \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \ versus \ x = \frac{T}{T_0} \ for \ 0.8 \leq x \leq 1.0.

Step by step solution

01

- Understand the Problem

The goal is to find the minimum theoretical work input required by a refrigeration cycle and then plot a specific ratio involving that work over a certain range of temperatures.
02

- Define the Minimum Work Input

Using the second law of thermodynamics for an ideal refrigeration process, the minimum work input, \(W_{\min}\), can be derived via the coefficient of performance (COP). For refrigeration, \(COP = \frac{Q_{\text{out}}}{W_{\text{min}}}\), where \(Q_{\text{out}} = m c (T_0 - T)\).
03

- Apply the Coefficient of Performance

For an ideal refrigeration cycle, \(COP = \frac{T}{T_0 - T}\). Rearranging this formula to find \(W_{\min}\) yields: \[ W_{\min} = m c (T_0 - T) \frac{T_0}{T} \].
04

- Simplify the Expression

Next, simplify the work expression into the requested form. The equation becomes \[ \frac{W_{\min}}{m c T_0} = \frac{T_0 - T}{T} \].
05

- Define the Ratio

Let \(x = \frac{T}{T_0}\), then \( \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \).
06

- Plot the Function

Plot the function \[ \frac{W_{\min}}{m c T_0} = \frac{1 - x}{x} \] versus \(x\) for \(x \) ranging from 0.8 to 1.0. This can be done using graphing software or by hand on graph paper.

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Key Concepts

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

incompressible substance
An incompressible substance is a material whose volume or density does not change significantly with changes in pressure. This property is typical of most liquids.

In our exercise, we assume the substance is incompressible to simplify the analysis. This assumption is crucial because it means the substance's volume and specific heat (denoted as c) remain constant throughout the process. These constants simplify our calculations for determining the minimum theoretical work input.
specific heat
The specific heat of a substance, represented by the symbol c, is the amount of heat required to change the temperature of a unit mass of the substance by one degree. It is defined by the equation:

\[ \text{specific heat} = \frac{Q}{m \times \triangle T} \ \ \] where \(Q\) is the heat added or removed, \(m\) is the mass, and \( \triangle T\) is the change in temperature.

In our context, knowing the specific heat helps us relate the heat transfer needed to achieve a particular temperature change (from \(T_0\) to \(T\)) in the refrigeration cycle. This relationship is critical for calculating the required work input.
second law of thermodynamics
The Second Law of Thermodynamics is fundamental in refrigeration cycle analysis. It states that heat transfer occurs spontaneously from higher to lower temperatures. For refrigeration, we need to input work to transfer heat in the opposite direction.

This principle is key to understand minimum work input using the ideal coefficient of performance (COP). By leveraging the second law, we derive the relationship:

\[ \text{COP} = \frac{Q_{\text{out}}}{W_{\text{min}}} \ \]

Where:
  • COP is the coefficient of performance
  • \(Q_{\text{out}}\) is the heat removed
  • \(W_{\text{min}}\) is the minimum work input
coefficient of performance
The coefficient of performance (COP) is a measure of a refrigeration cycle's efficiency. It is defined as the ratio of useful heating or cooling provided to the work required:

\[ \text{COP} = \frac{T}{T_0 - T} \ \]

For our problem, the equation helps us determine the work required for ideal scenarios. Replacing the COP value in our equations and rearranging gives us:

\[ W_{\text{min}} = m c (T_0 - T) \frac{T_0}{T} \ \ \]
This equation shows how the work required depends on temperature changes and the specific heat capacity of the substance. Simplifying further allows plotting the desired function:

\[ \frac{W_{\text{min}}}{m c T_0} = \frac{T_0 - T}{T} \ \] Letting \( x = \frac{T}{T_0} \), we get: \[ \frac{W_{\text{min}}}{m c T_0} = \frac{1 - x}{x} \ \]

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Most popular questions from this chapter

What can be deduced from energy and entropy balances about a system undergoing a thermodynamic cycle while receiving energy by heat transfer at temperature \(T_{\mathrm{C}}\) and discharging energy by heat transfer at a higher temperature \(T_{\mathrm{H}}\), if these are the only energy transfers the system experiences?

A system executes a power cycle while receiving \(1000 \mathrm{~kJ}\) by heat transfer at a temperature of \(500 \mathrm{~K}\) and discharging energy by heat transfer at a temperature of \(300 \mathrm{~K}\). There are no other heat transfers. Applying Eq. 6.2, determine \(\sigma_{\text {cycle }}\) if the thermal efficiency is (a) \(60 \%\), (b) \(40 \%\), (c) \(20 \%\). Identify the cases (if any) that are internally reversible or impossible.

A system consists of \(2 \mathrm{~m}^{3}\) of hydrogen gas \(\left(\mathrm{H}_{2}\right)\), initially at \(35^{\circ} \mathrm{C}, 215 \mathrm{kPa}\), contained in a closed rigid tank. Energy is transferred to the system from a reservoir at \(300^{\circ} \mathrm{C}\) until the temperature of the hydrogen is \(160^{\circ} \mathrm{C}\). The temperature at the system boundary where heat transfer occurs is \(300^{\circ} \mathrm{C}\). Modeling the hydrogen as an ideal gas, determine the heat transfer, in \(\mathrm{kJ}\), the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\), and the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\). For the reservoir, determine the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Why do these two entropy changes differ?

A patent application describes a device that at steady state receives a heat transfer at the rate \(1 \mathrm{~kW}\) at a temperature of \(167^{\circ} \mathrm{C}\) and generates electricity. There are no other energy transfers. Does the claimed performance violate any principles of thermodynamics? Explain.

One-tenth kilogram of water executes a Carnot power cycle. At the beginning of the isothermal expansion, the water is a saturated liquid at \(160^{\circ} \mathrm{C}\). The isothermal expansion continues until the quality is \(98 \%\). The temperature at the conclusion of the adiabatic expansion is \(20^{\circ} \mathrm{C}\). (a) Sketch the cycle on \(T-s\) and \(p-v\) coordinates. (b) Determine the heat added and net work, each in kJ. (c) Evaluate the thermal efficiency.
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