Question

A reversible power cycle receives \(Q_{H}\) from a hot reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy by heat transfer to the surroundings at temperature \(T_{0}\). The work developed by the power cycle is used to drive a refrigeration cycle that removes \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{\mathrm{C}}\) and discharges energy by heat transfer to the same surroundings at \(T_{0}\). (a) Develop an expression for the ratio \(Q_{\mathrm{C}} / Q_{\mathrm{H}}\) in terms of the temperature ratios \(T_{\mathrm{H}} / T_{0}\) and \(T_{\mathrm{C}} / T_{0}\). (b) Plot \(Q_{\mathrm{C}} / Q_{\mathrm{H}}\) versus \(T_{\mathrm{H}} / T_{0}\) for \(T_{\mathrm{C}} / T_{0}=0.85,0.9\), and \(0.95\), and versus \(T_{C} / T_{0}\) for \(T_{H} / T_{0}=2,3\), and 4.

Short answer

The ratio \( \frac{Q_C}{Q_H} \) is \[ \frac{T_C \( 1 - \frac{T_0}{T_H} \)}{T_0 - T_C} \].

Step-by-step solution

- Understand the Carnot Efficiency of the Power Cycle

Since the power cycle is reversible, it operates as a Carnot engine. The efficiency \( \eta_{carnot} \) of a Carnot engine is given by: \[ \eta_{carnot} = 1 - \frac{T_0}{T_H} \] The work developed by the power cycle is given by: \[ W = Q_H \eta_{carnot} = Q_H \( 1 - \frac{T_0}{T_H} \) \]

- Determine the Coefficient of Performance (COP) of the Refrigeration Cycle

For a reversible refrigeration cycle, the coefficient of performance (COP) is given by: \[ \text{COP} = \frac{Q_C}{W} = \frac{T_C}{T_0 - T_C} \]

- Relate Work Developed to Heat Rejected in Refrigeration Cycle

The work developed by the power cycle is used to drive the refrigeration cycle. Therefore, we have: \[ W = Q_H \( 1 - \frac{T_0}{T_H} \) \] Substituting this into the COP equation, we get: \[ \frac{Q_C}{Q_H \( 1 - \frac{T_0}{T_H} \)} = \frac{T_C}{T_0 - T_C} \]

- Isolate and Simplify the Expression for \( \frac{Q_C}{Q_H} \)

Multiply both sides of the equation by \( Q_H \( 1 - \frac{T_0}{T_H} \) \) to get: \[ Q_C = Q_H \(1 - \frac{T_0}{T_H} \) \frac{T_C}{T_0 - T_C} \] Dividing both sides by \( Q_H \), we get an expression for \( \frac{Q_C}{Q_H} \): \[ \frac{Q_C}{Q_H} = \frac{T_C \( 1 - \frac{T_0}{T_H} \)}{T_0 - T_C} \]

- Plotting the Expression

Plot the expression for \( \frac{Q_C}{Q_H} \) versus \( \frac{T_H}{T_0} \) for \( \frac{T_C}{T_0} = 0.85, 0.9 \, \text{and} \ 0.95 \). Then plot \( \frac{Q_C}{Q_H} \) versus \( \frac{T_C}{T_0} \) for \( \frac{T_H}{T_0} = 2, 3 \, \text{and} \ 4 \).

Key concepts

Carnot efficiency

To understand Carnot efficiency, we need to focus on the most efficient heat engine possible, called the Carnot engine. The efficiency of this engine, \( \eta_{carnot} \), is a measure of how well it converts heat from a hot reservoir into work. Importantly, it only depends on the temperatures of the hot and cold reservoirs. The formula is: \ \ \[ \eta_{carnot} = 1 - \frac{T_0}{T_H} \]
Where:

• \( T_H \) is the absolute temperature of the hot reservoir.
• \( T_0 \) is the absolute temperature of the cold reservoir/surroundings.

This relation shows that as \( T_H \) increases or as \( T_0 \) decreases, efficiency goes up. No real engine can achieve this efficiency because all practical engines have irreversibilities.

coefficient of performance

The coefficient of performance (COP) measures the efficiency of a refrigeration cycle. Unlike engines, refrigerators aim to transfer heat from a cool area to a warmer one. The COP is a ratio expressing how effective this heat transfer is:
\[ \text{COP} = \frac{Q_C}{W} = \frac{T_C}{T_0 - T_C} \]
Where:

• \( Q_C \) is the heat removed from the cold reservoir.
• \( T_C \) is the absolute temperature of the cold reservoir.
• \( W \) is the work input to the refrigerator.

This formula shows that the COP increases as \( T_C \) gets closer to \( T_0 \), meaning less work is needed for a smaller temperature spread.

heat transfer ratios

The ratio of heat transferred in different stages of thermodynamic cycles is key to understanding their efficiency. For the problem at hand, the ratio \( \frac{Q_C}{Q_H} \) emerges from combining expressions for Carnot efficiency and COP:
\[ \frac{Q_C}{Q_H} = \frac{T_C \( 1 - \frac{T_0}{T_H} \)}{T_0 - T_C} \]
This ratio gives a comparative measure of the amounts of heat taken in and released under ideal conditions, governed by the temperatures of the respective reservoirs. In practical terms, this helps in designing efficient cycles both for engines and refrigerators.

refrigeration cycle

A refrigeration cycle aims to move heat from a cold space to a warmer one. By doing so, it makes the cold space cooler. Reversible cycles like the one in the problem use work to drive this process, effectively transferring heat against the natural direction of flow.
Refrigeration cycles are evaluated largely by their COP and the efficiency of engines driving them. In the problem, the refrigerating cycle is driven by a Carnot engine, and the ideal scenarios help in simplifying and understanding the principles involved.

power cycle analysis

Analyzing power cycles helps us to understand how engines and refrigerators work together in thermodynamic systems. In this problem, a power cycle drives a refrigeration cycle, showing how the work produced can be reutilized to achieve different ends.
In a broader context, energy analyses of power cycles involve:

• Calculating work and heat transfers.
• Determining efficiencies and performance coefficients.
• Using temperature ratios to predict and enhance performance.

By plotting results as described in the exercise, we can visually grasp the impacts of changing temperatures on the ratio \( Q_C/Q_H \) and thus the overall efficiency of the combined cycles.