91Ó°ÊÓ

Thermal efficiency in a Brayton cycle measures how effectively the cycle converts heat into work. It's a ratio of the net work output and the heat added to the cycle. To find the thermal efficiency (\u03b7_{thermal}), you use the formula: \[ \u03b7_{thermal} = \frac{W_{net}}{Q_{in}} \times 100 \] where \( W_{net} \) is the net work output, and \( Q_{in} \) is the heat added. A higher thermal efficiency means a more efficient engine. In our exercise, it comes out to be approximately 30.03%. This value signifies that only 30.03% of the input energy is converted into useful work, while the rest is wasted.

An isentropic process is one that is both adiabatic (no heat transfer) and reversible. In thermodynamics, it's a key idealization used to analyze cycles. During an isentropic process in the Brayton cycle, entropy remains constant, which aids in calculating temperatures and pressures at various points.
To calculate the temperature after adiabatic compression (T2s) or expansion (T4s), you can use the isentropic relation: \[ \frac{T2s}{T1} = \left( \frac{P2}{P1} \right)^{\frac{k-1}{k}} \] For the turbine, this becomes: \[ \frac{T4s}{T3} = \left( \frac{P4}{P3} \right)^{\frac{k-1}{k}} \] where \( T \) is temperature, \( P \) is pressure, and \( k \) is the specific heat ratio (1.4 for air in this case).

Compressor work is the energy required to compress the air from a lower pressure to a higher pressure in the Brayton cycle. It can be calculated using the mass flow rate, specific heat at constant pressure \( c_p \), and the temperature difference.
Using the relation: \[ W_{c} = m \times c_p \times (T2 - T1) \] where \( W_{c} \) is the compressor work in kW, \( m \) is the mass flow rate in kg/s, \( c_p \) is the specific heat capacity at constant pressure, and \( \Delta T \) (\text{T2 - T1}) is the temperature rise through the compressor. Given our values: \[ W_{c} \approx 895.44 \text{ kW} \]

Turbine work is the energy produced when high-temperature, high-pressure air expands through the turbine. This expansion drives the turbine blades, producing work. Similar to compressor work, it can be calculated with mass flow rate, specific heat capacity, and temperature drop.
Using: \[ W_{t} = m \times c_p \times (T3 - T4) \] where \( W_{t} \) is the turbine work output in kW, \( T3 \) is the temperature before expansion, and \( T4 \) is the temperature after expansion. With the given values, it results in about 1533.50 kW of turbine work.

The back work ratio (BWR) is an indicator of how much work generated by the turbine is used to drive the compressor. It is the ratio of the compressor work \( W_{c} \) to the turbine work \( W_{t} \).
Mathematically, it is expressed as: \[ BWR = \frac{W_{c}}{W_{t}} \] For the Brayton cycle in our problem, it is approximately 0.58. This implies that 58% of the work produced by the turbine goes back into driving the compressor. A lower BWR is generally desired as it means more net work is available for useful purposes.

Specific heat capacity (\( c_p \)) is a measure of the heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. In the Brayton cycle, the specific heat at constant pressure for air is used to calculate the heat added and removed.
It's vital in formulas like: \[ Q = m \times c_p \times \text{ \Delta T} \] for calculating the heat transfer in and out of the working fluid. For air, \( c_p \) is given as 1.005 kJ/kg·K. This value helps in determining how much energy is needed to heat or cool the air in different parts of the cycle.