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A reversible power cycle receives energy \(Q_{\mathrm{H}}\) from a reservoir at temperature \(T_{\mathrm{H}}\) and rejects \(Q_{\mathrm{C}}\) to a reservoir at temperature \(T_{\mathrm{C}}\). The work developed by the power cycle is used to drive a reversible heat pump that removes energy \(Q_{\mathrm{C}}^{\prime}\) from a reservoir at temperature \(T_{\mathrm{C}}^{\top}\) and rejects energy \(Q_{\mathrm{H}}^{\prime}\) to a reservoir at temperature \(T_{\mathrm{H}}^{\prime}\) (a) Develop an expression for the ratio \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) in terms of the temperatures of the four reservoirs. (b) What must be the relationship of the temperatures \(T_{\mathrm{H}}, T_{\mathrm{C}}, T_{\mathrm{C}}^{\prime}\), and \(T^{\prime}{ }_{\mathrm{H}}\) for \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) to exceed a value of unity?

Step by step solution

01

Understand the problem

A reversible power cycle and a reversible heat pump are coupled. The goal is to find the ratio \(\frac{Q_{\mathrm{H}}^{'}}{Q_{\mathrm{H}}}\) and the temperature conditions for this ratio to be greater than 1.

02

Apply the First Law of Thermodynamics

For the power cycle, the First Law gives us \(W = Q_{\mathrm{H}} - Q_{\mathrm{C}}\). Since the work produced by the power cycle is used to drive the heat pump, we get \(W = Q_{\mathrm{H}}^{'} - Q_{\mathrm{C}}^{'}\).

03

Express work in terms of cycle performance

For a reversible power cycle, its efficiency \(\eta\) is given by \(\eta = 1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\). So, \(W = Q_{\mathrm{H}} \left(1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}\right)\).

04

Express work in terms of heat pump performance

For a reversible heat pump, the coefficient of performance (COP) is given by \(\mathrm{COP} = \frac{Q_{\mathrm{H}}^{'}}{W} = \frac{T_{\mathrm{H}}^{'}}{T_{\mathrm{H}}^{'} - T_{\mathrm{C}}^{\prime}}\). Therefore, \(W = \frac{Q_{\mathrm{H}}^{'}}{\mathrm{COP}} = Q_{\mathrm{H}}^{'} \left(\frac{T_{\mathrm{H}}^{'} - T_{\mathrm{C}}^{\prime}}{T_{\mathrm{H}}^{'}}\right)\).

05

Equate the two expressions for work

Setting \(Q_{\mathrm{H}} (1 - \frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}) =Q_{\mathrm{H}}^{'} \left(\frac{T_{\mathrm{H}}^{'} - T_{\mathrm{C}}^{\prime}}{T_{\mathrm{H}}^{'}}\right)\) and solving for \(\frac{Q_{\mathrm{H}}^{'}}{Q_{\mathrm{H}}}\), we get \(\frac{Q_{\mathrm{H}}^{'}}{Q_{\mathrm{H}}} = \frac{T_{\mathrm{H}} (T_{\mathrm{H}}^{'} - T_{\mathrm{C}}^{\prime})}{T_{\mathrm{H}}^{'} (T_{\mathrm{H}} - T_{\mathrm{C}})}\).

06

Condition for \(\frac{Q_{\mathrm{H}}^{'}}{Q_{\mathrm{H}}}\) to exceed unity

For \(\frac{Q_{\mathrm{H}}^{'}}{Q_{\mathrm{H}}} > 1\), the numerator must be greater than the denominator, giving us the condition \( T_{\mathrm{H}} (T_{\mathrm{H}}^{'} - T_{\mathrm{C}}^{\prime}) > T_{\mathrm{H}}^{'} (T_{\mathrm{H}} - T_{\mathrm{C}}) \).

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Key Concepts

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

First Law of Thermodynamics

In any thermodynamic process, the First Law of Thermodynamics is fundamental. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another.
For the given exercise, we apply this principle to both the power cycle and the heat pump.
The energy balance for the power cycle is expressed as:
\(W = Q_{\text{H}} - Q_{\text{C}}\), where:

• \(W\) is the work output
• \(Q_{\text{H}}\) is the heat input from the high-temperature reservoir
• \(Q_{\text{C}}\) is the heat rejected to the low-temperature reservoir

This work produced by the power cycle is then used to drive the heat pump:
\(W = Q_{\text{H}}' - Q_{\text{C}}'\), where:
• \(Q_{\text{H}}'\) is the heat output to the high-temperature reservoir
• \(Q_{\text{C}}'\) is the heat extracted from the low-temperature reservoir
These equations reflect the First Law of Thermodynamics by showing the conservation of energy in both cycles.

Efficiency

Efficiency is a measure of how well a system converts energy from one form to another. For a reversible power cycle, efficiency (\(\eta\)) is an important factor.
The efficiency of a reversible power cycle is given by:
\(\eta = 1 - \frac{T_{\text{C}}}{T_{\text{H}}}\), where:

• \(T_{\text{C}}\) is the temperature of the cold reservoir
• \(T_{\text{H}}\) is the temperature of the hot reservoir

Using this formula, we can express the work output in terms of the heat input and temperature difference:
\(W = Q_{\text{H}} \left(1 - \frac{T_{\text{C}}}{T_{\text{H}}}\right)\)
This is crucial for understanding how much work can be extracted from a given amount of heat and how efficiently the power cycle operates.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is used to assess the efficiency of heat pumps and refrigerators. It is defined as the ratio of useful heating or cooling provided to the work required.
For the reversible heat pump in the exercise:
\(\text{COP} = \frac{Q_{\text{H}}'}{W} = \frac{T_{\text{H}}'}{T_{\text{H}}' - T_{\text{C}}'}\), where:

• \(Q_{\text{H}}'\) is the heat output to the high-temperature reservoir
• \(T_{\text{H}}'\) is the temperature of the heat pump's high-temperature reservoir
• \(T_{\text{C}}'\) is the temperature of the heat pump's low-temperature reservoir

By rearranging the COP equation, we can express the work input in terms of the temperatures:
\(W = \frac{Q_{\text{H}}'}{\text{COP}} = Q_{\text{H}}' \left(\frac{T_{\text{H}}' - T_{\text{C}}'}{T_{\text{H}}'}\right)\)
This helps in understanding how the heat pump operates and how the temperatures influence its efficiency.
Combining this with the power cycle's equations allows us to derive the ratio \(\frac{Q_{\text{H}}'}{Q_{\text{H}}}\) and the temperature conditions for this ratio to exceed unity.