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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 \(T_{\mathrm{C}}^{\prime}\) and rejects energy \(Q_{\mathrm{H}}^{\prime}\) to a reservoir at temperature \(T_{\mathrm{H}^{\prime}}^{\prime}\) (a) Develop an expression for the ratio \(Q_{H}^{\prime} / Q_{H}\) 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 \(T_{\mathrm{H}}^{\prime}\) for \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) to exceed a value of unity? (c) Letting \(T_{\mathrm{H}}^{\prime}=T_{\mathrm{C}}=T_{0}\), plot \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) versus \(T_{\mathrm{H}} / T_{0}\) for \(T_{\mathrm{C}}^{\prime} / T_{0}=0.85,0.9\), and \(0.95\), and versus \(T_{\mathrm{C}}^{\prime} / T_{0}\) for \(T_{\mathrm{H}} / T_{0}\) \(=2,3\), and 4 .

Step by step solution

01

- Understanding the System

A power cycle takes in energy \(Q_\text{H}\) and rejects energy \(Q_\text{C}\). The work done by the cycle, \(W\), is used to power a heat pump. The heat pump removes energy \(Q_\text{C}'\) and rejects energy \(Q_\text{H}'\).

02

- Define Efficiency for Power Cycle

The thermal efficiency of the power cycle is \[ \text{Efficiency} = \frac{W}{Q_\text{H}} = 1 - \frac{Q_\text{C}}{Q_\text{H}} = 1 - \frac{T_\text{C}}{T_\text{H}} \]

03

- Work Relation

The work done by the power cycle is therefore: \[ W = Q_\text{H} \times \text{Efficiency} = Q_\text{H} \times \frac{T_\text{H} - T_\text{C}}{T_\text{H}} \]

04

- Heat Pump Coefficient of Performance (COP)

The COP of the heat pump is given by: \[ \text{COP} = \frac{Q_\text{H}'}{W} = \frac{Q_\text{H}'}{Q_\text{C}'} = \frac{T_\text{H}'}{T_\text{H}' - T_\text{C}'} \]

05

- Setting Up the Relation

The work done by the power cycle is the input for the heat pump. Rearranging terms, we get: \[ \frac{Q_\text{H}'}{Q_\text{H}} = \text{COP} \times \frac{W}{Q_\text{H}} \]

06

- Substituting Expressions

Using the expressions for \text{Efficiency\ and COP, we substitute them in: \[ \frac{Q_\text{H}'}{Q_\text{H}} = \frac{T_\text{H}'}{T_\text{H}' - T_\text{C}'} \times \frac{T_\text{H} - T_\text{C}}{T_\text{H}} \]

07

- Relation of Temperatures for Q_\text{H}'/Q_\text{H}>1

For \(Q_\text{H}'/Q_\text{H}\) to be greater than 1, the ratio \[ \frac{T_\text{H}'}{T_\text{H}' - T_\text{C}'} \times \frac{T_\text{H} - T_\text{C}}{T_\text{H}} \] must be greater than 1.

08

- Assign Special Conditions for Plotting

Under the condition \(T_\text{H}'=T_\text{C}=T_0\), the expression simplifies, so \[ \frac{Q_\text{H}'}{Q_\text{H}} = \frac{T_0}{T_0 - T_\text{C}'} \times \frac{T_\text{H} - T_0}{T_\text{H}} \]. Now plot \(Q_\text{H}'/Q_\text{H}\) versus \(T_\text{H}/T_0\) with \(T_\text{C}'/T_0 = 0.85, 0.9, 0.95\) and versus \(T_\text{C}'/T_0\) for \(T_\text{H}/T_0 = 2, 3, 4\).

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Key Concepts

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

Reversible Power Cycle

A reversible power cycle is a theoretical model in thermodynamics. It represents the ideal case where there are no losses due to friction, unrestrained expansion, or heat transfer. These cycles can be thought of as perfectly efficient. They absorb heat (energy) from a high-temperature reservoir and convert part of this energy into work, rejecting the remainder to a low-temperature reservoir. In a reversible power cycle, everything happens in such a way that the system and environment can be returned to the original state without any net energy change or entropy increase.

Thermal Efficiency

Thermal efficiency is a measure of a power cycle’s effectiveness in converting heat into work. The higher the efficiency, the more work is generated from a given amount of heat. For a power cycle, the thermal efficiency \text{Efficiency} is given by: \[ \text{Efficiency} = \frac{W}{Q_\text{H}} = 1 - \frac{Q_\text{C}}{Q_\text{H}} = 1 - \frac{T_\text{C}}{T_\text{H}} \]
Where:

• Q_\text{H} is the heat extracted from the high-temperature reservoir.
• Q_\text{C} is the heat rejected to the low-temperature reservoir.

The efficiency formula shows that the higher the temperature difference between the high (T_\text{H}) and low (T_\text{C}) temperature reservoirs, the higher the efficiency.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of the effectiveness of a heat pump or refrigerator. It is defined as the ratio of the heat output to the work input. For a heat pump extracting heat from a cold reservoir and rejecting it to a hot reservoir, the COP is given by: \[ \text{COP} = \frac{Q_\text{H}'}{W} = \frac{Q_\text{H}'}{Q_\text{C}'} = \frac{T_\text{H}'}{T_\text{H}' - T_\text{C}'} \]
Where:

• Q_\text{H}' is the heat rejected to the high-temperature reservoir.
• Q_\text{C}' is the heat absorbed from the cold reservoir.

The COP indicates how effectively the heat pump transfers heat compared to its energy input. A higher COP value means greater efficiency.

Reservoir Temperatures

Reservoir temperatures are crucial in thermodynamic cycles because they directly affect both thermal efficiency and COP. Key temperatures are:

• T_\text{H}: The temperature of the high-temperature reservoir providing heat to the power cycle.
• T_\text{C}: The temperature of the low-temperature reservoir receiving rejected heat from the power cycle.
• T_\text{C}': The temperature of the cold reservoir where the heat pump extracts heat.
• T_\text{H}': The temperature of the hot reservoir where the heat pump rejects heat.

To maximize the performance of a thermodynamic system with a power cycle and heat pump operating together, the relationships between these temperatures play a significant role. For instance, ensuring that \[ \frac{T_\text{H}'}{T_\text{H}' - T_\text{C}'} \times \frac{T_\text{H} - T_\text{C}}{T_\text{H}} \] is greater than 1 can optimize the effectiveness of the system, resulting in more energy transferred and utilized efficiently.