Question

A system executes a power cycle while receiving \(1000 \mathrm{~kJ}\) by heat transfer at a temperature of \(500 \mathrm{~K}\) and discharging energy by heat transfer at a temperature of \(300 \mathrm{~K}\). There are no other heat transfers. Applying Eq. 6.2, determine \(\sigma_{\text {cycle }}\) if the thermal efficiency is (a) \(60 \%\), (b) \(40 \%\), (c) \(20 \%\). Identify the cases (if any) that are internally reversible or impossible.

Short answer

(a) Impossible, \(\sigma_{\text{cycle}} = -\frac{2}{3}\). (b) Internally reversible, \(\sigma_{\text{cycle}} = 0\). (c) Irreversible, \(\sigma_{\text{cycle}} = \frac{2}{3}\).

Step-by-step solution

Identify Given Information

The system executes a power cycle with the following details: - Heat input, \(Q_H = 1000 \mathrm{~kJ}\) at \(T_H = 500 \mathrm{~K}\).- Heat rejection at \(T_C = 300 \mathrm{~K}\).- No other heat transfers.We need to determine \(\sigma_{\text {cycle }}\) using the given thermal efficiencies of 60%, 40%, and 20% and then identify whether each case is internally reversible or impossible.

Calculate Useful Work Output

Thermal efficiency \(\eta\) is given by \[ \eta = \frac{W_{net}}{Q_H} \]Rearrange to find \(W_{net}\): \[ W_{net} = \eta \cdot Q_H \]

Determine the Heat Rejected, \(Q_L\)

Using the first law of thermodynamics, the heat rejected \(Q_L\) can be found using the equation:\[ Q_H = W_{net} + Q_L \]Rearranging, we get:\[ Q_L = Q_H - W_{net} \]

Calculate Entropy Generation, \(\sigma_{\text{cycle}}\)

Using Eq. 6.2, the entropy generation \(\sigma_{\text{cycle}}\) is given by: \[ \sigma_{\text{cycle}} = \frac{Q_L}{T_C} - \frac{Q_H}{T_H} \]

Case (a) - Calculate for 60% Efficiency

Given \(\eta = 0.60\):\[ W_{net} = 0.60 \cdot 1000 \mathrm{~kJ} = 600 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 600 \mathrm{~kJ} = 400 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{400}{300} - \frac{1000}{500} = \frac{4}{3} - 2 = -\frac{2}{3} \]Since \(\sigma_{\text{cycle}} < 0\), this process is impossible.

Case (b) - Calculate for 40% Efficiency

Given \(\eta = 0.40\):\[ W_{net} = 0.40 \cdot 1000 \mathrm{~kJ} = 400 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 400 \mathrm{~kJ} = 600 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{600}{300} - \frac{1000}{500} = 2 - 2 = 0 \]Since \(\sigma_{\text{cycle}} = 0\), this process is internally reversible.

Case (c) - Calculate for 20% Efficiency

Given \(\eta = 0.20\):\[ W_{net} = 0.20 \cdot 1000 \mathrm{~kJ} = 200 \mathrm{~kJ} \]\[ Q_L = 1000 \mathrm{~kJ} - 200 \mathrm{~kJ} = 800 \mathrm{~kJ} \]Now, calculate entropy generation:\[ \sigma_{\text{cycle}} = \frac{800}{300} - \frac{1000}{500} = \frac{8}{3} - 2 = \frac{2}{3} \]Since \(\sigma_{\text{cycle}} > 0\), this process is irreversible.

Key concepts

entropy generation

Entropy generation \( \sigma_{\text {cycle}} \) is a key concept in thermodynamics, helping us understand the irreversibility of processes. In simple terms, entropy generation measures how much disorder or randomness increases due to a process. For a reversible process, the entropy generation is zero, meaning no irreversibility.
As entropy generation increases, the process becomes more irreversible. It's important to remember that a negative entropy generation is impossible as it would violate the second law of thermodynamics.
When calculating \( \sigma_{\text {cycle}} \) for a power cycle, we need the heat rejected ( \( \ Q_L \ \) ) and the heat input ( \( \ Q_H \ \) ) along with the respective temperatures at which these heat transfers occur:
\( \sigma_{\text{cycle}} = \frac{Q_L}{T_C} - \frac{Q_H}{T_H} \ \textrm{ .} \)
From the exercise, we can see the calculation for each thermal efficiency case:

• For 60% efficiency, entropy generation is negative, making the process impossible.
• For 40% efficiency, entropy generation is zero, indicating an internally reversible process.
• For 20% efficiency, entropy generation is positive, showing an irreversible process.

thermal efficiency

Thermal efficiency \( \eta \ \) is a measure of how well a power cycle converts heat into work. It is a ratio defined by:
\( \eta = \frac{W_{net}}{Q_H} \)
\