Explain a large stationary Brayton cycle gas turbine:A simple Brayton cycle gas turbine is shown in Figure 1. It mainly consists of three components:1. Compressor2. Combustion chamber3. TurbineFigure 1This type of turbine is mainly used to drive the electric generator for the production of electricity. Air at atmospheric pressure P1 and temperature T1 enters the compressor where it is compressed isentropically. During the compression process, the pressure and temperature increased to P2 and T2. This compressed air will then enter the combustion chamber where it helps burn the fuel. Due to the combustion of the fuel, it is mixed with the gases produced by combustion and reaches a very high temperature T3. This high pressure and temperature air then runs the turbine where it is spent and then the temperature and pressure of the air are reduced to T4 and P4. The turbine is coupled with generator and compressor. Therefore, a part of the turbine power is used to drive the compressor while the remaining part of the turbine power is used to drive the electric generator. Processes 2-3 and 4-1 are isobaric processes as shown in figure 2. Figure 2 Work done per kg by the compressor = Cp (T2 - T1) Work done per kg by the turbine = Cp (T3 - T4) Net work produced = Turbine work – Compressor work= Cp (T3 - T4) - Cp (T2 - T1) Heat supplied = Cp (T3 - T2) Therefore, the thermal efficiency of the cycle is η = Net work produced/Heat supplied= [Cp (T3 - T4) - Cp (T2 - T1)] / Cp (T3 - T2)Problem: Brayton cycle and gas turbine thermal efficiencyA large stationary power plant with a Brayton cycle gas turbine provides 100 MW of power to an electric generator . The minimum temperature in the cycle is 300K and the maximum temperature...... at the center of the paper......ne the shear stress for the elliptical section and therefore, the shear stress for the circular section bar and the elliptical section will be the same on the xz plane while the shear stress for the circular section bar will be less than the shear stress for the elliptical section on the yz plane as a > b. (Assuming the same length of both bars. The value of G for both bars will also be the same since the twist angle will be the same for both, i.e. they are made of the same material.) For equal allowable stressMax. shear stress for elliptical section = 2TE / πab2 (where TE=Torque applied in the elliptical section) Max. shear stress for circular section = 2TC / πb3 (where TC=Torque applied in the circular section) As per question i.e. equal admissible stress2TE / πab2 = 2TC / πb3TE = (a / b) x TCA we know a > bTherefore TE > TCHence Elliptical bar it will resist a greater torque than the circular bar.