Question

A reversible power cycle \(\mathrm{R}\) and an irreversible power cycle I operate between the same two reservoirs. Each receives \(Q_{\mathrm{H}}\) from the hot reservoir. The reversible cycle develops work \(W_{R}\), while the irreversible cycle develops work \(W_{\mathrm{l}}\). The reversible cycle discharges \(Q_{\mathrm{C}}\) to the cold reservoir, while the irreversible cycle discharges \(Q_{c}^{\prime}\). (a) Evaluate \(\sigma_{\text {cycle }}\) for cycle \(\mathrm{I}\) in terms of \(W_{\mathrm{1}}, W_{\mathrm{R}}\), and temperature \(T_{\mathrm{C}}\) of the cold reservoir only. (b) Demonstrate that \(W_{1}Q_{\mathrm{C}}\).

Short answer

For cycle I, \(\sigma_{\text{cycle}} \) is \( \frac{W_R - W_I}{T_C} \). Since \(W_R > W_I\), \(Q_{C}^{\text{'}} > Q_C\).

Step-by-step solution

Understanding the Given Information

Identify the terms and understand the context: A reversible power cycle (R) and an irreversible power cycle (I) operate between the same hot and cold reservoirs. They both receive heat, \(Q_{H}\), from the hot reservoir. The reversible cycle develops \(W_{R}\) work, while the irreversible cycle develops \(W_{I}\) work. A reversible cycle discharges \(Q_{C}\) to the cold reservoir, and the irreversible cycle discharges \(Q_{C}^{\text{'}}\).

Formulate the Second Law of Thermodynamics for Irreversible Cycle

By the second law of thermodynamics, for an irreversible cycle, the entropy generation \(\sigma_{\text{cycle}}\) is given by: \[ \sigma_{\text{cycle}} = \frac{Q_H}{T_H} - \frac{Q_C^{\text{'}}}{T_C} > 0 \]

Express the Heat Transfer in Irreversible Cycle

In the irreversible cycle, the total heat added from the hot reservoir must equal the sum of the work done and the heat rejected to the cold reservoir: \[ Q_H = W_I + Q_C^{\text{'}} \]

Express the Heat Transfer in Reversible Cycle

For the reversible cycle, the heat from the hot reservoir equals the work done plus the heat rejected to the cold reservoir: \[ Q_H = W_R + Q_C \]

Derive Entropy Generation for Cycle I

Using the second law of thermodynamics, for the irreversible cycle: \[ \sigma_{\text{cycle}} = \frac{Q_H}{T_H} - \frac{Q_C^{\text{'}}}{T_C} \] Rearrange to solve for \(Q_C^{\text{'}}\): \[ \sigma_{\text{cycle}} \times T_C = \frac{Q_H \times T_C}{T_H} - Q_C^{\text{'}} \] \[ Q_C^{\text{'}} = \frac{Q_H \times T_C}{T_H} - \sigma_{\text{cycle}} \times T_C \]

Comparing Work Done by Cycles

Since \(\sigma_{\text{cycle}} > 0\), the entropy generation causes a total reduction in useful work: \[ W_{R} - W_{I} = \sigma_{\text{cycle}} \times T_C \] Therefore, it can be stated that: \[ W_{R} > W_{I} \]

Comparing Heat Rejection

Since \( Q_H = W_R + Q_C \) for the reversible and \( Q_H = W_I + Q_C^{\text{'}} \) for the irreversible, and we have established that \( W_R > W_I \), it follows that: \[ W_R - W_I = Q_C^{\text{'}} - Q_C \] Since \(W_R - W_I > 0\), \( Q_C^{\text{'}} > Q_C \)

Key concepts

Second Law of Thermodynamics

The Second Law of Thermodynamics fundamentally describes the direction in which processes occur. This law states that entropy, or disorder, within an isolated system will always increase or remain the same over time. For power cycles, this means that no process is perfectly efficient. In practical terms, when we talk about reversible and irreversible cycles, the reversible cycle is the ideal scenario with no entropy generation. This cycle follows a theoretical path where all processes can be reversed without any loss of energy. On the other hand, an irreversible cycle is more realistic and includes unavoidable inefficiencies, resulting in entropy generation. For any irreversible cycle, the Second Law can be expressed in terms of entropy change: \text {For an irreversible cycle,} \frac {Q_H}{T_H} - \frac{Q_C^{'}}{T_C} > 0 where \(\sigma_{\text{cycle}} \) is the net increase in entropy for the cycle, \(Q_H \) is the heat received from the hot reservoir, \(Q_C^{\{'}}\) is the heat discharged to the cold reservoir, and \(T_H \) and \(T_C \) are the temperatures of the hot and cold reservoirs, respectively.

Entropy Generation

Entropy generation is a key concept that helps quantify the irreversibilities in any cycle or process. When a power cycle is irreversible, it generates more entropy compared to a reversible cycle operating between the same two reservoirs. Entropy generation can be calculated using the heat transfer and temperature differences in the cycle: \text {\(\sigma_{\text{cycle}} \mathrel{\mathop:}= \frac {Q_H}{T_H} - \frac{Q_C^{'}}{T_C} \)} The greater the value of \(\sigma_{\text{cycle}}\), the less efficient the cycle is, as more energy is wasted. This wasted energy translates into lesser work output from the power cycle as compared to a reversible cycle.

Heat Transfer Equations

Heat transfer equations illustrate how energy is exchanged between the reservoirs and work generation in power cycles. For a reversible cycle, the heat received from the hot reservoir \(Q_H\) is divided into work \(W_R\) and heat rejected \(Q_C\) to the cold reservoir: \[ Q_H = W_R + Q_C \] For an irreversible cycle, the heat received from the hot reservoir results in work \(W_I \) and heat rejection \(Q_C^{\{'}} \): \[ Q_H = W_I + Q_C^{'} \] Comparing these, it’s clear from the entropy generation relationship that irreversible cycles will always reject more heat to the cold reservoir compared to reversible cycles.

Comparison of Work Done

In power cycles, one of the key performance metrics is the amount of work done. Reversible cycles are ideal and produce the maximum possible work ( \(W_R \)) for the same amount of heat input \(Q_H\). For an irreversible cycle, due to entropy generation and system inefficiencies, less work ( \(W_I \)) is generated: \[ W_R - W_I = T_C \sigma_{\text{cycle}} \] This difference ( \(T_C \sigma_{\text{cycle}} \)) represents the lost work potential due to irreversibilities in the system. Thus, it is evident that \(W_R > W_I \) and correspondingly, more heat is rejected to the cold reservoir: \[ Q_C^{'} > Q_C \] These comparisons succinctly illustrate the intrinsic advantage of reversible cycles in terms of efficiency and work output.