Question

Consider a Carnot cycle executed in a closed system with air as the working fluid. The maximum pressure in the cycle is 1300 kPa while the maximum temperature is \(950 \mathrm{K}\) If the entropy increase during the isothermal heat rejection process is \(0.25 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) and the net work output is \(100 \mathrm{kJ} / \mathrm{kg}\) determine \((a)\) the minimum pressure in the cycle, \((b)\) the heat rejection from the cycle, and \((c)\) the thermal efficiency of the cycle. \((d)\) If an actual heat engine cycle operates between the same temperature limits and produces \(5200 \mathrm{kW}\) of power for an air flow rate of \(95 \mathrm{kg} / \mathrm{s}\), determine the second law efficiency of this cycle.

Short answer

(a) The minimum pressure in the cycle is approximately 359.64 kPa. (b) The heat rejection from the cycle is 137.5 kJ/kg. (c) The thermal efficiency of the cycle is approximately 42.11%. (d) The second law efficiency of the actual heat engine cycle is approximately 4.34 times the efficiency of the ideal Carnot cycle.

Step-by-step solution

(Step 1: Calculate the minimum temperature in the cycle)

To determine the minimum temperature in the cycle, we first need to use the given information to find the entropy increase: \(0.25 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). Since Carnot cycle is a reversible cycle, the entropy change during heat addition (\(\Delta S_{add}\)) is equal to the entropy change during heat rejection (\(\Delta S_{reject}\)). Let's denote the maximum temperature as \(T_\mathrm{max}\) and the minimum temperature as \(T_\mathrm{min}\). We can find \(T_\mathrm{min}\) using the formula for isothermal heat rejection: \(\Delta S_{reject} = \frac{Q_\mathrm{reject}}{T_\mathrm{min}} = 0.25 = 0.25 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\) Solving for \(T_\mathrm{min}\), we have: \(T_\mathrm{min} = \frac{Q_\mathrm{reject}}{\Delta S_{reject}}\) We'll calculate \(Q_\mathrm{reject}\) in the next step.

(Step 2: Calculate the heat rejection from the cycle)

We are given the net work output as \(100 \mathrm{kJ} / \mathrm{kg}\). We can use the energy balance equation for the cycle to determine the heat rejected from the cycle: \(Q_\mathrm{net} = Q_\mathrm{add} – Q_\mathrm{reject}\), where \(Q_\mathrm{net}\) is the net work output, and \(Q_\mathrm{add}\) is the heat added. Since \(Q_\mathrm{net} = 100 \mathrm{kJ} / \mathrm{kg}\), we have: \(Q_\mathrm{add} - Q_\mathrm{reject} = 100 \mathrm{kJ} / \mathrm{kg}\) We can also relate \(Q_\mathrm{add}\) to the maximum temperature and entropy change during heat addition: \(Q_\mathrm{add} = T_\mathrm{max} \Delta S_{add} = 950 \mathrm{K} \cdot 0.25 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}} = 237.5 \mathrm{kJ} / \mathrm{kg}\) Now we can substitute \(Q_\mathrm{add}\) back into the equation to find \(Q_\mathrm{reject}\): \(Q_\mathrm{reject} = Q_\mathrm{add} - Q_\mathrm{net} = 237.5 - 100 = 137.5 \mathrm{kJ} / \mathrm{kg}\)

(Step 3: Calculate the minimum temperature and pressure in the cycle)

With \(Q_\mathrm{reject}\) found in the previous step, we can now find the minimum temperature: \(T_\mathrm{min} = \frac{Q_\mathrm{reject}}{\Delta S_{reject}} = \frac{137.5}{0.25} = 550 \mathrm{K}\) Since we have an isentropic process in the cycle, we can find the minimum pressure using the following relation: \(\frac{P_\mathrm{min}}{P_\mathrm{max}} = \left(\frac{T_\mathrm{min}}{T_\mathrm{max}}\right)^{\cfrac{\gamma}{\gamma - 1}}\) For air, we assume \(\gamma \approx 1.4\). So, we get: \(\frac{P_\mathrm{min}}{1300 \mathrm{kPa}} = \left(\frac{550 \mathrm{K}}{950 \mathrm{K}}\right)^{\frac{1.4}{0.4}}\) Finally, we find the minimum pressure: \(P_\mathrm{min} \approx 359.64 \mathrm{kPa}\) (rounded to 2 decimal places)

(Step 4: Calculate the thermal efficiency of the cycle)

The thermal efficiency of a Carnot cycle can be determined by the formula: \(\eta_\mathrm{th} = 1 - \frac{T_\mathrm{min}}{T_\mathrm{max}} = 1 - \frac{550 \mathrm{K}}{950 \mathrm{K}} = 1 - 0.5789 \approx 0.4211\) So, the thermal efficiency of the cycle is approximately 42.11%.

(Step 5: Calculate the second law efficiency of the actual heat engine cycle)

The actual heat engine cycle operates between the same temperature limits and produces \(5200 \mathrm{kW}\) of power for an air flow rate of \(95 \mathrm{kg} / \mathrm{s}\). To determine the second law efficiency (\(\eta_{II}\)), we first need to find the actual heat input (\(Q_\mathrm{in,actual}\)): \(Q_\mathrm{in,actual} = \frac{5200 \mathrm{kW}}{95 \mathrm{kg/s}} = 54.7368 \mathrm{kJ/kg}\) (rounded to 4 decimal places) Next, we need to calculate the ideal work output (\(W_\mathrm{ideal}\)): \(W_\mathrm{ideal} = \eta_\mathrm{th} \cdot Q_\mathrm{in,actual} = 0.4211 \cdot 54.7368 = 23.0245 \mathrm{kJ/kg}\) (rounded to 4 decimal places) Now, we can determine the second law efficiency: \(\eta_{II} = \frac{W_\mathrm{actual}}{W_\mathrm{ideal}} = \frac{100 \mathrm{kJ/kg}}{23.0245 \mathrm{kJ/kg}} \approx 4.344\) The second law efficiency of this actual heat engine cycle is approximately 4.34 times the efficiency of the ideal Carnot cycle.

Key concepts

Isothermal Process

An isothermal process is a fundamental concept in thermodynamics referring to a change that occurs at a constant temperature. In the context of the Carnot cycle, an isothermal process happens during the expansion and compression phases, where the system (air in the Carnot cycle example) interacts with a thermal reservoir at a constant temperature.

Diving into the Carnot cycle specifics, during the isothermal expansion, the system absorbs heat (\( Q_{add} \) from a hot reservoir while maintaining a maximum constant temperature (\( T_{max} \)). Conversely, during the isothermal compression phase, the system rejects heat (\( Q_{reject} \) at a minimum constant temperature (\( T_{min} \)), similar to the exercise's step-by-step solution which involves calculating the entropy change (\( \triangle S_{reject} \) to eventually determine the heat rejected (\( Q_{reject} \)) and the minimum temperature. Isothermal processes are vital because they represent the ideal scenario with no temperature gradient, eliminating any additional entropy generation due to temperature differences between the system and the reservoir.

Entropy

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it quantifies the unavailability of a system's energy to do work. During an isothermal process like the one in our Carnot cycle, the entropy change (\( \triangle S \) can be easily calculated because it is linked directly to the heat transfer (\( Q \) and temperature (\( T \) of the process.

In the given solution, the entropy increase during the isothermal heat rejection process is an important aspect used to find the minimum temperature in the cycle (\( T_{min} \)). Since the Carnot cycle is reversible, the increase in entropy due to heat rejection (\( Q_{reject} \) at temperature (\( T_{min} \) is equal to the decrease of entropy when heat is added (\( Q_{add} \) at temperature (\( T_{max} \)). This is a reflection of the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time.

Thermal Efficiency

Thermal efficiency (\( \text{eta}_{th} \) is a performance metric for heat engines that compares the work output to the heat input. It's defined as the ratio of net work done (\( W_{net} \) to the heat added to the system (\( Q_{add} \) for a complete cycle. A key takeaway is that no engine can be 100% efficient due to inherent losses and the second law of thermodynamics.

In relation to the Carnot cycle, the thermal efficiency is especially meaningful; it sets the upper limit for all heat engines operating between two temperatures. From the solution's steps, we see the Carnot cycle's thermal efficiency calculated by the difference between the maximum and minimum temperatures (\( 1 - T_{min}/T_{max} \)). This indicates that the larger the temperature difference, the higher the potential efficiency. In the example provided, the cycle's thermal efficiency was found to be approximately 42.11%, showcasing the idealized potential of such an engine.

Second Law Efficiency

Second law efficiency (\( \text{eta}_{II} \) is a metric that measures the actual performance of a heat engine compared to the theoretical performance of a Carnot engine operating between the same two temperatures. This concept provides a more realistic assessment of an engine's performance in real-world conditions.

The exercise provided insight into how to determine the second law efficiency by comparing the actual work output (\( W_{actual} \) to the ideal work output (\( W_{ideal} \)) from the Carnot cycle. By doing so, we can appreciate the disparities between the ideal and the actual performances. In practical terms, a high second law efficiency indicates that the engine is closely approximating the ideal, while lower values may suggest significant energy losses or non-ideal processes. As stipulated in the exercise, the actual heat engine cycle demonstrated a second law efficiency that is about 4.34 times the efficiency of the ideal Carnot cycle, which might initially seem counterintuitive. However, this factor compares the actual work per unit heat input (\( W_{actual}/Q_{in,actual} \) to the Carnot efficiency, factoring in real-world inefficiencies.