Question

A reversible power cycle \(\mathrm{R}\) and an irreversible power cycle I operate between the same two reservoirs. (a) If each cycle receives the same amount of energy \(Q_{\mathrm{H}}\) from the hot reservoir, show that cycle I necessarily discharges more energy \(Q_{C}\) to the cold reservoir than cycle \(R\). Discuss the implications of this for actual power cycles. (b) If each cycle develops the same net work, show that cycle I necessarily receives more energy \(Q_{\mathrm{H}}\) from the hot reservoir than cycle R. Discuss the implications of this for actual power cycles.

Short answer

The irreversible cycle discharges more energy to the cold reservoir and requires more energy from the hot reservoir compared to the reversible cycle for the same work. This highlights the inefficiency introduced by irreversibilities.

Step-by-step solution

Understand the Second Law of Thermodynamics

According to the Second Law of Thermodynamics, for a reversible cycle, the efficiency \( \eta _{R} = \frac{W}{Q_{H}} \) is higher compared to an irreversible cycle. Let's denote the efficiencies as \( \eta_{R} > \eta_{I} \) where \( \eta_{R} \) is the efficiency of the reversible cycle and \( \eta_{I} \) is that of the irreversible cycle. This applies because irreversibilities (like friction, unrestrained expansion, etc.) reduce the efficiency.

Analyze Energy Discharge to Cold Reservoir (Part a)

The efficiency of a power cycle is given by \( \eta = \frac{W}{Q_{H}} = 1 - \frac{Q_{C}}{Q_{H}} \) which can be rearranged to \( Q_{C} = Q_{H}(1 - \eta) \). For the reversible cycle, \( Q_{C_{R}} = Q_{H}(1 - \eta_{R}) \) and for the irreversible cycle, \( Q_{C_{I}} = Q_{H}(1 - \eta_{I}) \). Given that \( \eta_{R} > \eta_{I} \), it follows that \( (1 - \eta_{R}) < (1 - \eta_{I}) \), thus \( Q_{C_{R}} < Q_{C_{I}} \). This means cycle I discharges more energy to the cold reservoir than cycle R.

Discussion on Implications (Part a)

Since cycle I discharges more energy to the cold reservoir than cycle R, it means that the irreversible cycle is less efficient. This highlights the importance of minimizing irreversibilities in actual power cycles to ensure they operate closer to the reversible cycle efficiency, thereby reducing waste heat and improving performance.

Understand Net Work for Both Cycles (Part b)

Let's denote the net work output as \( W \). By definition, if both cycles produce the same net work, \( W = Q_{H} - Q_{C} \). We already established that \( Q_{C_{R}} < Q_{C_{I}} \). Given this relationship and the fact that \( Q_{H_{R}} - Q_{C_{R}} = Q_{H_{I}} - Q_{C_{I}} = W \), it follows that \( Q_{H_{I}} > Q_{H_{R}} \). This means that cycle I needs more energy input from the hot reservoir to generate the same amount of net work as cycle R.

Discussion on Implications (Part b)

The fact that cycle I requires more energy input from the hot reservoir to produce the same work as cycle R reiterates the inefficiency caused by irreversible processes. In actual power cycles, this means that reducing irreversibilities can lead to a reduction in fuel consumption and higher overall efficiency.

Key concepts

reversible cycle efficiency

The efficiency of a cycle is a measure of how well it converts heat from the hot reservoir into work. For a reversible cycle, the efficiency, denoted as \( \eta_R \), is determined by the Second Law of Thermodynamics. This law states that the efficiency of a reversible cycle is always the highest possible for given reservoirs. It can be expressed as follows: \( \eta_R = \frac{W}{Q_H} \), where \( W \) is the work done and \( Q_H \) is the heat taken from the hot reservoir.
A key takeaway is that no real cycle can surpass the efficiency of a reversible cycle. This is because all real cycles involve some form of irreversibility, such as friction or unrestrained expansion, which reduce efficiency.

irreversible processes

Irreversibilities are factors that cause real cycles to fall short of the ideal, reversible cycle. These can include friction, rapid expansion, and heat losses. Because of these irreversibilities, the efficiency \( \eta_I \) of an actual cycle is always lower than that of a reversible cycle.
This lower efficiency from irreversibilities is mathematically represented as:
\( \eta_I = \frac{W}{Q_H} \)
Comparing this to the reversible cycle, we can see:
\( \eta_R > \eta_I \)
Meaning actual cycles discharge more energy as waste heat to the cold reservoir than reversible cycles. Consequently, actual power cycles are less efficient, making it essential to minimize irreversibilities in engineering designs.

power cycles efficiency

Power cycles are systems designed to convert heat energy into mechanical work. They typically operate between two reservoirs: a hot one and a cold one. The efficiency of a power cycle depends on how much useful work it can extract from the heat energy supplied.
The efficiency of a power cycle, \( \eta \), is given by:
\( \eta = \frac{W}{Q_H} \)
or equivalently,
\( \eta = 1 - \frac{Q_C}{Q_H} \)
Where \( Q_C \) is the heat discharged to the cold reservoir, and \( Q_H \) is the heat absorbed from the hot reservoir.
For a given amount of work, if the cycle is irreversible, it needs more heat from the hot reservoir (\( Q_H \)) than a reversible cycle, because inefficient processes waste more energy. Thus:
- Reversible cycles: Less heat input needed to produce the same amount of work.
- Irreversible cycles: More heat input needed to produce the same amount of work.
Understanding and optimizing power cycle efficiency is crucial for enhancing energy conversion and reducing fuel consumption in real-world applications.