Question

A reversible refrigeration cycle \(\mathrm{R}\) and an irreversible refrigeration cycle I operate between the same two reservoirs and each removes \(Q_{\mathrm{C}}\) from the cold reservoir. The net work input required by \(\mathrm{R}\) is \(W_{\mathrm{R}}\), while the net work input for \(\mathrm{I}\) is \(W_{\mathrm{I}}\). The reversible cycle discharges \(Q_{\mathrm{H}}\) to the hot reservoir, while the irreversible cycle discharges \(Q_{\mathrm{H}}^{\prime}\). Show that \(W_{1}>W_{\mathrm{R}}\) and \(Q_{\mathrm{H}}^{\prime}>Q_{\mathrm{H}}\).

Short answer

Since the irreversible cycle has a lower COP, it requires more work input (\(W_{I} > W_{R}\)) and discharges more heat to the hot reservoir (\(Q_{H}^{\text{'}} > Q_{H}\)).

Step-by-step solution

Understand the Concept of Reversible and Irreversible Cycles

A reversible cycle operates in an ideal manner without any losses, thus requiring the minimum possible work input, while an irreversible cycle has inefficiencies and requires more work input. Both cycles remove the same amount of heat, \(Q_{\text{C}}\), from the cold reservoir.

Apply the First Law of Thermodynamics

For both cycles, the first law of thermodynamics gives the relationship between heat and work. For the reversible cycle, the energy balance equation is: \[Q_{\text{H}} = Q_{\text{C}} + W_{\text{R}}.\]For the irreversible cycle, the equation is: \[Q_{\text{H}}^{\text{'}} = Q_{\text{C}} + W_{\text{I}}.\]

Determine the Coefficient of Performance (COP)

The coefficient of performance for a refrigeration cycle is defined as the ratio of the heat removed from the cold reservoir to the work input: \[\text{COP} = \frac{Q_{\text{C}}}{W}.\]For the reversible cycle, the COP is maximized: \[\text{COP}_{R} = \frac{Q_{\text{C}}}{W_{\text{R}}} = \frac{Q_{\text{C}}}{Q_{\text{H}} - Q_{\text{C}}};\] for the irreversible cycle, the COP is lower: \[\text{COP}_{I} = \frac{Q_{\text{C}}}{W_{\text{I}}} < \text{COP}_{R}.\]

Use the Relationship between COP and Work Input

Since \[\text{COP}_{I} < \text{COP}_{R},\]we can infer that \[\frac{Q_{\text{C}}}{W_{\text{I}}} < \frac{Q_{\text{C}}}{W_{\text{R}}},\]which simplifies to \[W_{\text{I}} > W_{\text{R}}.\]

Analyze the Heat Discharged to the Hot Reservoir

From the first law of thermodynamics equations, we can see that since \[W_{\text{I}} > W_{\text{R}},\] it must follow that \[Q_{\text{C}} + W_{\text{I}} > Q_{\text{C}} + W_{\text{R}},\] which implies \[Q_{\text{H}}^{\text{'}} > Q_{\text{H}}.\]

Key concepts

reversible cycle

A reversible cycle is an ideal process in thermodynamics. Think of it as a perfect machine with no energy losses. This kind of cycle maximizes efficiency and requires the least amount of work input.

When we talk about refrigeration cycles, the reversible cycle has several remarkable properties:

• It operates in such a way that every stage of the cycle can be reversed without leaving any change in the surroundings.
• It is an idealized model that doesn't exist in real-world applications because all real processes involve some losses or inefficiencies.

Understanding the reversible cycle is crucial to grasping the fundamental limits of performance for refrigeration systems. It sets the benchmark for how efficient any refrigeration process can be.

Mathematically, for a reversible refrigeration cycle, the relationship between the heat transferred to the hot reservoir \(Q_H\), the heat taken from the cold reservoir \(Q_C\), and the work done \(W_R\) is given by the equation: \[ Q_{\text{H}} = Q_{\text{C}} + W_{\text{R}} \] Substituting into the coefficient of performance (COP), we get: \[ \text{COP}_R = \frac{Q_{\text{C}}}{W_{\text{R}}} = \frac{Q_{\text{C}}}{Q_{\text{H}} - Q_{\text{C}}} \]
This represents the highest possible efficiency achievable by any refrigeration cycle. It shows that the reverse cycle does the same amount of cooling with the minimum work input, thus highlighting its ideal nature.

irreversible cycle

An irreversible cycle is a real-world version of a refrigeration cycle. Unlike its reversible counterpart, an irreversible cycle is subject to inefficiencies such as friction, unrestrained expansion, or heat loss. This makes it less efficient, as it requires more work input than a reversible cycle to achieve the same amount of cooling.

Some key points to remember about irreversible cycles are:

• Inefficiencies increase the amount of work needed to achieve the same cooling effect.
• It’s representative of actual refrigeration systems and includes real-world constraints.

The relationship between the heat transferred to the hot reservoir \(Q_{\text{H}}'\), the heat absorbed from the cold reservoir \(Q_{\text{C}}\), and the work done \(W_{\text{I}}\) is described by: \[ Q_{\text{H}}' = Q_{\text{C}} + W_{\text{I}} \] From the comparison with the reversible cycle, we find: \[ W_{\text{I}} > W_{\text{R}} \] since more work is needed due to inefficiencies. This difference in work input results in a lower coefficient of performance (COP) for the irreversible cycle as shown below: \[ \text{COP}_{I} = \frac{Q_{\text{C}}}{W_{\text{I}}} \] Because \[ \text{COP}_I < \text{COP}_R, \] it indicates that an irreversible cycle has a reduced efficiency compared to a reversible cycle. This means that the energy cost of achieving the same cooling effect is higher, which has practical implications for designing and operating refrigeration systems.

thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and energy. It's fundamental to understanding refrigeration cycles as it governs how energy transformations occur.

Some core principles in thermodynamics relevant to refrigeration cycles include:

• First Law of Thermodynamics: Energy can neither be created nor destroyed, only transferred or converted from one form to another. This law forms the basis for the energy balance equations used in refrigeration cycles.
• Second Law of Thermodynamics: Heat naturally flows from hot to cold bodies, and systems tend to increase in entropy or disorder. It explains why work is needed to transfer heat from a colder place to a hotter one, as in refrigeration.

In refrigeration cycles, the First Law gives us energy balance equations:
For a reversible cycle: \[ Q_{\text{H}} = Q_{\text{C}} + W_{\text{R}} \]
For an irreversible cycle: \[ Q_{\text{H}}' = Q_{\text{C}} + W_{\text{I}} \]
The Second Law of Thermodynamics underlines why reversible cycles have maximum efficiency and why COP is higher for reversible cycles compared to irreversible cycles.

By applying these laws to refrigeration cycles, we can work out important performance metrics, such as work input and heat transfer, for both ideal and real-world situations. This helps in designing more efficient and practical refrigeration systems.
Understanding these thermodynamic principles is essential for anyone who wants to deeply understand refrigeration cycles and improve their efficiency.