Explain a large stationary Brayton cycle gas-turbine:
A simple Brayton Cycle Gas Turbine is shown in figure 1. It consists of mainly three components :
1. Compressor
2. Combustion Chamber
3. Turbine
Figure 1
This type of turbine is mainly used to run the electric generator for the electric power production. Air at atmospheric pressure P1 and temperature T1 enters the compressor where it is compressed isentropically. During compression process, pressure and temp increased to P2 and T2. This compressed air will then enter the combustion chamber where it helps in burning the fuel. Due to burning of fuel, it is mixed with gases produced due to burning and attains a very high temp T3. This high pressure and high temperature air then runs the turbine where it is expended and hence the temp and pressure of air is reduced to T4 and P4. The turbine is coupled with generator as well as compressor. Therefore, some part of the turbine output is used to run the compressor while rest part of the turbine output is used to run the electric generator. Process 2-3 and 4-1 are isobaric process as shown in the figure 2.
Figure 2
Work done per Kg by the Compressor = Cp (T2 - T1)
Work output per Kg by the Turbine = Cp (T3 - T4)
Net Work output = Turbine Work – Compressor Work
= Cp (T3 - T4) - Cp (T2 - T1)
Heat Supplied = Cp (T3 - T2)
Therefore, Thermal efficiency of the cycle is η =Net Work output/Heat supplied
= [Cp (T3 - T4) - Cp (T2 - T1)] / Cp (T3 - T2)
Problem : Brayton Cycle and Gas Turbine Thermal Efficiency
A large stationary Brayton cycle gas-turbine power plant delivers a power output of 100MW to an electric generator. The minimum temperature in the cycle is 300K, and the maximum temperatur...
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...n and Shear stress for elliptical section and Therefore, Shear Stress for circular section bar and elliptical section will be same on xz plane while the Shear Stress for circular section bar will be less than the shear stress for elliptical section on yz plane as a > b. (Assuming equal length of both bar. Also the value of G for both bars will also be same as angle of twist will be same for both i.e. they are made of same material).
For equal allowable stress
Max. shear stress for elliptical section = 2TE / πab2 (where TE=Torque applied in elliptical section)
Max. shear stress for circular section = 2TC / πb3 (where TC=Torque applied in circular section)
As per Question i.e. equal allowable stress
2TE / πab2 = 2TC / πb3
TE = (a / b) x TC
As we know a > b
Therefore TE > TC
Hence Elliptical bar will resist larger twisting moment than circular bar.
For the other material ASTM A216 Gr WCB same pressure of 16 Mpa is applied and the stress developed is approximately as similar to the connecting rod made with material of cast iron. Figure no. 9 indicates the maximum and minimum stress developed in connecting rod at small & big end. The equivalent stress maximum and minimum values are 71.347 MPa and 4.4955e-5 MPa respectively.
Circular thickness: The distance of the arc along the pitch circle from one side of a gear tooth to the other
Torque= Force x radius , since the radius is half the diameter , T=F x D/2
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