Question

In an ideal Brayton cycle, air is compressed from \(95 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) to \(1100 \mathrm{kPa} .\) Under cold-air-standard conditions, the thermal efficiency of this cycle is \((a) 45\) percent (b) 50 percent \((c) 62\) percent \((d) 73\) percent \((e) 86\) percent

Short answer

Choose from the following options: A) 1.24% B) 2.45% C) 1.54% D) 2.25% Answer: None of the provided options are correct. The thermal efficiency of this ideal Brayton cycle under cold-air-standard conditions is approximately 0.658%.

Step-by-step solution

Find the properties of air at the initial state

To begin, we need to determine the specific enthalpy (\(h_{1}\)) and specific entropy (\(s_{1}\)) at the initial state (state 1), which is at a pressure of 95 kPa and temperature of 25°C. Using air property tables or software, we find: \(h_{1} = 298.8 \thinspace \mathrm{kJ/kg}\) and \(s_{1} = 6.85 \thinspace \mathrm{kJ/(kg\cdot K)}\)

Find the properties of air at the end of the compression process

During the isentropic compression process (1-2), entropy remains constant. Therefore, we have \(s_{2} = s_{1}\). The final pressure after compression is given as 1100 kPa. Using air property tables or software, again and knowing that the compression process is isentropic, we find: \(h_{2} = 686.5 \thinspace \mathrm{kJ/kg}\)

Calculate the work done during the compression process

The work done during the compression process (W\(_{c}\)) can be calculated as: \(W_{c} = h_{1} - h_{2} = 298.8 - 686.5 = -387.7 \thinspace \mathrm{kJ/kg}\)

Find the properties of air at the end of the isobaric heat addition process

During the isobaric heat addition process (2-3), pressure remains constant at 1100 kPa. The heat added (Q\(_{in}\)) during this process can be calculated as: \(Q_{in} = W_{c} + |W_{c}|= -387.7 + 387.7 = 774.4 \thinspace \mathrm{kJ/kg}\) With this value, we can find the specific enthalpy at the end of the heat addition process (state 3): \(h_{3} = h_{2} + Q_{in} = 686.5 + 774.4 = 1460.9 \thinspace \mathrm{kJ/kg}\) Using air property tables or software, we have: \(s_{3} = 7.534 \thinspace \mathrm{kJ/(kg\cdot K)}\)

Find the properties of air at the end of the expansion process

During the isentropic expansion process (3-4), entropy remains constant. Therefore, we have \(s_{4} = s_{3}\). The final pressure after expansion is 95 kPa (same as the initial pressure). Using air property tables or software for an isentropic process, we find: \(h_{4} = 1068.1 \thinspace \mathrm{kJ/kg}\)

Calculate the work done during the expansion process

The work done during the expansion process (W\(_{t}\)) can be calculated as: \(W_{t} = h_{3} - h_{4} = 1460.9 - 1068.1 = 392.8 \thinspace \mathrm{kJ/kg}\)

Calculate the net work done and the thermal efficiency

The net work done (W\(_{net}\)) can be calculated as: \(W_{net} = W_{t} - |W_{c}| = 392.8 - 387.7 = 5.1 \thinspace \mathrm{kJ/kg}\) The thermal efficiency (\(\eta\)) can now be calculated as: \(\eta = \frac{W_{net}}{Q_{in}} = \frac{5.1}{774.4} = 0.00658\) Converting to percent, we have: \(\eta = 0.658 \%\) Comparing our result with the given options, we can conclude that none of the provided options are correct. The thermal efficiency of this ideal Brayton cycle under cold-air-standard conditions is approximately 0.658%.

Key concepts

Isentropic Process

An isentropic process is a theoretical process in which entropy remains constant. This is significant in thermodynamics and can often be idealized in many real-world applications, like compressors and turbines in a Brayton cycle, where the process is assumed to be reversible and adiabatic, meaning no heat is transferred to or from the fluid during the process.

In the context of the Brayton cycle, during the compression (from state 1 to 2) and expansion (from state 3 to 4) stages, the assumption of an isentropic process allows for the simplification of calculations. It ensures that the specific entropy at the beginning and end of these stages remains the same. Remember, in reality, all processes incur some inefficiency, leading to entropy production, but assuming an isentropic process provides a useful benchmark for efficiency.

Isobaric Process

An isobaric process is characterized by constant pressure throughout the process. In the Brayton cycle, the heat addition step (from state 2 to 3) is an example of an isobaric process, as it occurs at constant pressure.

The significance of an isobaric process in thermodynamics is the direct relationship between heat added or removed and the change in specific enthalpy. This relationship simplifies the calculation of heat transfer, which is essential for evaluating the energy efficiency of a cycle. During this step in the Brayton cycle, knowing that the process is isobaric allows for the straightforward calculation of heat added to the system by simply adjusting the specific enthalpy.

Specific Enthalpy

Specific enthalpy is the total energy content per unit mass of a system and is a crucial property in thermodynamics, particularly for the evaluation of heat engines like the Brayton cycle. It combines both the internal energy and the product of pressure and volume of a substance.

In the Brayton cycle, the specific enthalpy at different states of the air (the working fluid) determines the work done during compression and expansion, as well as the heat added during combustion. It is central to calculating the work and heat interactions within the cycle. Changes in specific enthalpy, as shown in the step-by-step solution, are used to determine the net work output and, subsequently, the cycle's thermal efficiency.

Specific Entropy

Specific entropy quantifies the disorder or randomness within a system at the molecular level per unit mass. It's a measure of energy dispersion in thermodynamics. During processes within a thermodynamic cycle such as the Brayton cycle, knowing the specific entropy at various states allows engineers to understand how close the processes are to ideality.

In the Brayton cycle solution provided, specific entropy is crucial for identifying the isentropic compression and expansion processes by ensuring that the initial and final entropy values for these processes are equal. Recognizing the importance of specific entropy in these calculations helps in determining the actual work done by the system, and any deviation from the isentropic assumption indicates losses and inefficiencies within the cycle.