Question

A refrigeration cycle operating between two reservoirs receives energy \(Q_{\mathrm{C}}\) from a cold reservoir at \(T_{\mathrm{C}}=280 \mathrm{~K}\) and rejects energy \(Q_{\mathrm{H}}\) to a hot reservoir at \(T_{\mathrm{H}}=320 \mathrm{~K}\). For each of the following cases determine whether the cycle operates reversibly, irreversibly, or is impossible: (a) \(Q_{\mathrm{C}}=1500 \mathrm{~kJ}, W_{\text {cycle }}=150 \mathrm{~kJ}\). (b) \(Q_{\mathrm{C}}=1400 \mathrm{~kJ}, Q_{\mathrm{H}}=1600 \mathrm{~kJ}\). (c) \(Q_{\mathrm{H}}=1600 \mathrm{~kJ}, W_{\text {cycle }}=400 \mathrm{~kJ}\). (d) \(\beta=5\).

Short answer

Case (a) is impossible. Case (b) is reversible. Case (c) is irreversible. Case (d) is irreversible.

Step-by-step solution

Understand the Problem

Identify the temperatures of the cold and hot reservoirs: At the cold reservoir, \( T_{\text{C}} = 280 \text{ K} \) At the hot reservoir, \( T_{\text{H}} = 320 \text{ K} \). For each case, check if the cycle operates reversibly, irreversibly, or is impossible. Use the coefficient of performance (COP) of the ideal refrigeration cycle as a reference: \( \beta_{\text{ideal}} = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}} \).

Calculate the Ideal Coefficient of Performance (COP)

Calculate the ideal COP using the given temperatures: \( \beta_{\text{ideal}} = \frac{T_{\text{C}}}{T_{\text{H}} - T_{\text{C}}} = \frac{280}{320 - 280} = \frac{280}{40} = 7 \).

(a): Check for Reversibility, Irreversibility, or Impossibility for Case (a)

For case (a): Given: \( Q_{\text{C}} = 1500 \text{ kJ}, \) \( W_{\text{cycle}} = 150 \text{ kJ} \) We use the relationship: \( Q_{\text{H}} = Q_{\text{C}} + W_{\text{cycle}} \): \( Q_{\text{H}} = 1500 + 150 = 1650 \text{ kJ} \). Calculate the actual COP: \( \beta = \frac{Q_{\text{C}}}{W_{\text{cycle}}} = \frac{1500}{150} = 10 \). Since \( \beta > \beta_{\text{ideal}} \), this cycle is impossible.

(b): Check for Reversibility, Irreversibility, or Impossibility for Case (b)

For case (b): Given: \( Q_{\text{C}} = 1400 \text{ kJ}, \) \( Q_{\text{H}} = 1600 \text{ kJ} \) Use the relationship: \( W_{\text{cycle}} = Q_{\text{H}} - Q_{\text{C}} \): \( W_{\text{cycle}} = 1600 - 1400 = 200 \text{ kJ} \). Calculate the actual COP: \( \beta = \frac{Q_{\text{C}}}{W_{\text{cycle}}} = \frac{1400}{200} = 7 \). Since \( \beta = \beta_{\text{ideal}} \), this cycle operates reversibly.

(c): Check for Reversibility, Irreversibility, or Impossibility for Case (c)

For case (c): Given: \( Q_{\text{H}} = 1600 \text{ kJ}, \) \( W_{\text{cycle}} = 400 \text{ kJ} \) Use the relationship: \( Q_{\text{C}} = Q_{\text{H}} - W_{\text{cycle}} \): \( Q_{\text{C}} = 1600 - 400 = 1200 \text{ kJ} \). Calculate the actual COP: \( \beta = \frac{Q_{\text{C}}}{W_{\text{cycle}}} = \frac{1200}{400} = 3 \). Since \( \beta < \beta_{\text{ideal}} \), this cycle operates irreversibly.

(d): Determine the Type of Operation for Case (d)

For case (d): Given: Coefficient of performance, \( \beta = 5 \). Compare this COP with \( \beta_{\text{ideal}} \): Since \( \beta = 5 < 7 = \beta_{\text{ideal}} \), this cycle operates irreversibly.

Key concepts

Coefficient of Performance

The coefficient of performance (COP) is a key concept in refrigeration cycles. It measures the efficiency of the refrigeration cycle. The COP is given by the formula \(\beta = \frac{Q_C}{W_{cycle}}\), where \Q_C\ is the energy received from the cold reservoir, and \W_{cycle}\ is the work input to the cycle. In an ideal refrigeration cycle, the COP can also be determined using temperatures of the cold and hot reservoirs: \(\beta_{\text{ideal}} = \frac{T_C}{T_H - T_C}\). A higher COP indicates a more efficient refrigeration cycle. Ideal COP represents the maximum achievable efficiency, often used as a benchmark to assess if a cycle is operating efficiently, irreversibly, or is impossible.

Reversible Cycle

A reversible cycle is an idealized process that operates with the maximum possible efficiency. In thermodynamics, this means the process can be reversed without any net change to the system or the surroundings. For refrigeration cycles, if the actual COP is equal to the ideal COP, the cycle is considered reversible. This condition signifies that there are no losses in the form of heat, friction, or any other form of entropy production. For example, in the step-by-step solution provided, case (b) shows a situation where the actual COP matches the ideal COP of 7, indicating a reversible cycle.

Irreversible Cycle

An irreversible cycle is a practical, real-world situation where the efficiency is lower than the theoretical maximum. This occurs due to factors like friction, unrestrained expansion, and other types of inefficiencies. In the context of the refrigeration cycle, if the actual COP is less than the ideal COP, the cycle is deemed irreversible. Steps 5 (c) and 6 (d) of the solution highlight scenarios where the calculated COP values of 3 and 5 are below the ideal COP of 7, indicating that these processes are irreversible due to inherent inefficiencies.

Thermodynamic Reservoirs

Thermodynamic reservoirs refer to large systems with a constant temperature which can absorb or supply finite amounts of heat without undergoing any significant change in temperature. In a refrigeration cycle, there are typically two reservoirs - a cold reservoir and a hot reservoir. The cold reservoir at temperature \( T_C \) provides the energy \(Q_C \) to the cycle, and the hot reservoir at higher temperature \( T_H \) absorbs the energy \(Q_H \) rejected by the cycle. In the given exercise, the cold reservoir is at 280 K and the hot reservoir is at 320 K. The relationship between these temperatures helps determine the ideal COP and play a critical role in assessing the performance of the refrigeration cycle.