Question

In an ideal Brayton cycle with regeneration, air is compressed from \(80 \mathrm{kPa}\) and \(10^{\circ} \mathrm{C}\) to \(400 \mathrm{kPa}\) and \(175^{\circ} \mathrm{C}\), is heated to \(450^{\circ} \mathrm{C}\) in the regenerator, and then further heated to \(1000^{\circ} \mathrm{C}\) before entering the turbine. Under cold-air-standard conditions, the effectiveness of the regenerator is (a) 33 percent \((b) 44\) percent \((c) 62\) percent \((d) 77\) percent \((e) 89\) percent

Short answer

Question: What is the effectiveness of the regenerator in an ideal Brayton cycle with regeneration given the following air pressure and temperature conditions: 1. Before compressor: 100 kPa, 283.15 K (10°C) 2. After compressor: 650 kPa, 448.15 K (175°C) 3. After regenerator: 650 kPa, 723.15 K (450°C) 4. After turbine and before the regenerator: 100 kPa, 1273.15 K (1000°C) Answer: The regenerator effectiveness is 77%.

Step-by-step solution

Find specific heat capacities

To determine the regenerator effectiveness, we will first need the specific heat capacities of air at constant pressure (Cp) and volume (Cv). Under cold-air-standard conditions, we consider air as an ideal gas with constant specific heats. For air, we have: Cp = 1005 J/kgK Cv = 718 J/kgK

Calculate the temperature after the regenerator

After the compressor, air enters the regenerator and gets heated from \(175^{\circ} \mathrm{C}\) to \(450^{\circ} \mathrm{C}\). So, \(T_r\) (temperature after the regenerator) = \(450^{\circ} \mathrm{C} = 450 + 273.15 = 723.15 \mathrm{K}\)

Calculate the temperature after the turbine

After the regenerator, air gets further heated to \(1000^{\circ} \mathrm{C}\) before entering the turbine. The temperature after the turbinew can be found as: \(T_t\) (temperature after the turbine) = \(1000^{\circ} \mathrm{C} = 1000 + 273.15 = 1273.15 \mathrm{K}\)

Calculate the enthalpy at different stages

Now, we need to find the enthalpies at the different stages of the cycle: \(h_1\) (enthalpy before the compressor) = \(C_p \times T_1\) = \(1005 \times 283.15 \mathrm{J/kg}\) \(h_2\) (enthalpy after the compressor) = \(C_p \times T_2\) = \(1005 \times 448.15 \mathrm{J/kg}\) \(h_3\) (enthalpy after the regenerator) = \(C_p \times T_r = 1005 \times 723.15 \mathrm{J/kg}\) \(h_4\) (enthalpy after the turbine) = \(C_p \times T_t = 1005 \times 1273.15 \mathrm{J/kg}\)

Calculate the regenerator effectiveness

Now, we can calculate the regenerator effectiveness using the formula: regenerator effectiveness = \(\frac{h_3 - h_2}{h_4 - h_2}\) Inserting the values we calculated before: regenerator effectiveness = \(\frac{1005 \times 723.15 - 1005 \times 448.15}{1005 \times 1273.15 - 1005 \times 448.15}\) regenerator effectiveness = \(\frac{275821.575 - 451561.075}{442505.075 - 451561.075}\) regenerator effectiveness = \(\frac{-175739.5}{-9056}\) regenerator effectiveness = 0.77 As a percentage, this is 77%, so the correct answer is \((d) 77\) percent.