Question

Complete the following involving reversible and irreversible cycles: (a) Reversible and irreversible power cycles each discharge energy \(Q_{\mathrm{C}}\) to a cold reservoir at temperature \(T_{\mathrm{C}}\) and receive energy \(Q_{\mathrm{H}}\) from hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}>T_{\mathrm{H}}\). (b) Reversible and irreversible refrigeration cycles each discharge energy \(Q_{\mathrm{H}}\) to a hot reservoir at temperature \(T_{\mathrm{H}}\) and receive energy \(Q_{C}\) from cold reservoirs at temperatures \(T_{C}\). and \(T_{C}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{C}}^{\prime}>T_{\mathrm{C}}\). (c) Reversible and irreversible heat pump cycles each receive energy \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{\mathrm{C}}\) and discharge energy \(Q_{\mathrm{H}}\) to hot reservoirs at temperatures \(T_{\mathrm{H}}\) and \(T_{\mathrm{H}}^{\prime}\), respectively. There are no other heat transfers. Show that \(T_{\mathrm{H}}^{\prime}

Short answer

(a) \(T_{H}' > T_{H}\)(b) \(T_{C}' > T_{C}\)(c) \(T_{H}' < T_{H}\)

Step-by-step solution

- Understand Power Cycles Definitions

In a power cycle, energy is transferred from a hot reservoir to a cold reservoir, doing work in the process. For reversible cycles, the efficiency is maximized, while irreversible cycles have lower efficiency due to entropy generation.

- Efficiency Equations for Power Cycles

For a reversible power cycle, the efficiency is given by \[ \eta_{\text{rev}} = 1 - \frac{T_{C}}{T_{H}} \]For an irreversible power cycle, the efficiency is lower:\[ \eta_{\text{irr}} < 1 - \frac{T_{C}}{T_{H}'} \]

- Compare Efficiencies to Show Temperature Relations

Given that irreversible cycles are less efficient, we have:\[ 1 - \frac{T_{C}}{T_{H}'} < 1 - \frac{T_{C}}{T_{H}} \]which simplifies to:\[ T_{H}' > T_{H} \]

- Understand Refrigeration Cycles Definitions

In a refrigeration cycle, energy is transferred from a cold reservoir to a hot reservoir, requiring work. For reversible cycles, the efficiency (coefficient of performance) is maximized, while irreversible cycles have a lower coefficient of performance.

- Coefficient of Performance Equations for Refrigeration Cycles

For a reversible refrigeration cycle, the coefficient of performance (COP) is:\[ \text{COP}_{\text{rev}} = \frac{T_{C}}{T_{H} - T_{C}} \]For an irreversible refrigeration cycle, the COP is lower: \[ \text{COP}_{\text{irr}} < \frac{T_{C}'}{T_{H} - T_{C}'} \]

- Compare COPs to Show Temperature Relations

Given that irreversible cycles have a lower COP, we get:\[ \frac{T_{C}}{T_{H} - T_{C}} > \frac{T_{C}'}{T_{H} - T_{C}'} \]which simplifies to:\[ T_{C}' > T_{C} \]

- Understand Heat Pump Cycles Definitions

In a heat pump cycle, energy is transferred from a cold reservoir to a hot reservoir, doing work in the process. For reversible cycles, the efficiency (coefficient of performance) is maximized, while irreversible cycles have a lower coefficient of performance.

- Coefficient of Performance Equations for Heat Pump Cycles

For a reversible heat pump cycle, the coefficient of performance (COP) is:\[ \text{COP}_{\text{rev}} = \frac{T_{H}}{T_{H} - T_{C}} \]For an irreversible heat pump cycle, the COP is lower:\[ \text{COP}_{\text{irr}} < \frac{T_{H}'}{T_{H}' - T_{C}} \]

- Compare COPs to Show Temperature Relations

Given that irreversible cycles have a lower COP, we get:\[ \frac{T_{H}}{T_{H} - T_{C}} > \frac{T_{H}'}{T_{H}' - T_{C}} \]which simplifies to:\[ T_{H}' < T_{H} \]

Key concepts

Reversible Cycles

Reversible cycles are ideal thermodynamic processes. They operate with maximum efficiency and no entropy generation. In these cycles, every step can be reversed without leaving any change in the system or surroundings. Examples include the Carnot cycle. For power cycles, the efficiency \( \eta_{rev} \) is given by \[ \eta_{rev} = 1 - \frac{T_{C}}{T_{H}} \]. In refrigeration and heat pump cycles, the coefficient of performance (COP) is optimal in reversible operations.

Irreversible Cycles

Irreversible cycles are real-world processes. They incur losses due to factors like friction, unrestrained expansion, and thermal resistance, leading to entropy generation. These cycles are less efficient compared to reversible cycles. In power cycles, the efficiency \( \eta_{irr} \) is always less than \[ 1 - \frac{T_{C}}{T_{H}'} \]. For refrigeration, the COP is less than ideal: \[ COP_{irr} < \frac{T_{C}'}{T_{H} - T_{C}'} \]. Similarly, for heat pumps, the COP is also lower than that of a reversible heat pump: \[ COP_{irr} < \frac{T_{H}'}{T_{H}' - T_{C}} \].

Efficiency

Efficiency in thermodynamic cycles measures how well the cycle converts heat into work. For power cycles, efficiency (\( \eta \)) is the ratio of work output to heat input. For reversible cycles, the equation is \[ \eta_{rev} = 1 - \frac{T_{C}}{T_{H}} \]. However, for irreversible cycles, the efficiency can never reach the ideal values and is always lower: \[ \eta_{irr} < 1 - \frac{T_{C}}{T_{H}'} \]. This is due to the additional entropy generated in irreversible processes.

Coefficient of Performance

The coefficient of performance (COP) measures the effectiveness of refrigeration and heat pump cycles. For refrigeration cycles, COP is defined as \[ COP_{rev} = \frac{T_{C}}{T_{H} - T_{C}} \] for reversible cycles. For irreversible cycles, this ratio is lower: \[ COP_{irr} < \frac{T_{C}'}{T_{H} - T_{C}'} \]. In heat pumps, COP represents the heat delivered to the hot reservoir relative to the work input, given by \[ COP_{rev} = \frac{T_{H}}{T_{H} - T_{C}} \] in reversible cycles. For irreversible heat pumps, it's less efficient: \[ COP_{irr} < \frac{T_{H}'}{T_{H}' - T_{C}} \].

Power Cycles

Power cycles generate work by transferring energy from a hot reservoir to a cold reservoir. In an ideal, reversible power cycle, such as the Carnot cycle, efficiency is maximized: \[ \eta_{rev} = 1 - \frac{T_{C}}{T_{H}} \]. In irreversible cycles, efficiency is reduced because not all the heat energy is converted to work, resulting in the relation \[ T_{H}' > T_{H} \]. This shows that irreversible cycles are less efficient, requiring a higher temperature difference to produce the same work output.

Refrigeration Cycles

Refrigeration cycles transfer energy from a cold reservoir to a hot reservoir, requiring work input. The reversed Carnot cycle represents an ideal refrigeration cycle, where efficiency is high, defined by \[ COP_{rev} = \frac{T_{C}}{T_{H} - T_{C}} \]. For real, irreversible cycles, the efficiency is always less, showing that the temperature of the cold reservoir must be higher in an irreversible cycle: \[ T_{C}' > T_{C} \]. This is because extra work must account for the inefficiencies of real-world processes.

Heat Pump Cycles

Heat pump cycles transfer energy from a cold reservoir to a hot reservoir, essentially heating spaces. In ideal heat pump cycles, the reversible COP is given by \[ COP_{rev} = \frac{T_{H}}{T_{H} - T_{C}} \]. In practice, reversible operation is unattainable, leading to lower performance. For irreversible heat pumps, COP is reduced, and consequently, the temperature of the hot reservoir in an irreversible cycle must be less than that in a reversible cycle: \[ T_{H}' < T_{H} \]. The extra work lost due to inefficiencies impacts the achievable temperatures.