Answer Question A) B) C) ## Brayton Cycle in a Turbine Engine### The ProcessThe process of burning fuel in a turbine engine can be modeled by a heat engine that utilizes a Brayton Cycle. The Brayton Cycle consists of the following steps:1. **Adiabatic Compression**2. **Isobaric Expansion due to Heating**3. **Adiabatic Expansion**4. **Isobaric Compression**### Given ConditionsConsider a turbine that starts with:- **Initial Pressure**: \( P_1 = 101,325 \, \text{Pa} \)- **Initial Volume**: \( V = 2.0 \, \text{m}^3 \)- **Initial Temperature**: \( T_1 = 300.0 \, \text{K} \)**Process:**- The gas is compressed adiabatically to a pressure \( P_2 = 1,013,250 \, \text{Pa} \).- The gas then expands isobarically via heating.- Finally, the gas expands adiabatically to a pressure \( P_4 = 101,325 \, \text{Pa} \).### Tasks#### a) Draw a PV Diagram- Create a PV diagram for one cycle of this heat engine.- Label the step with \( Q_h \) (heat addition).- Label the step with \( Q_c \) (heat rejection).- Mark the initial point, second point, third point, and fourth point.#### b) Determine the QuantitiesFind the values of:- \( V_2, V_3, V_4 \)- \( T_2, T_3, T_4 \)#### c) Work Done from Points 1 to 2Calculate the work done during the adiabatic compression from point 1 to point 2.#### d) Work Done from Points 3 to 4Calculate the work done during the adiabatic expansion from point 3 to point 4.#### e) Power OutputIf the engine operates at \( 465 \, \text{rev/min} \), determine its power output.---This exercise guides students through analyzing the thermodynamic cycle of a turbine engine, honing their understanding of adiabatic and isobaric processes.

a) The PV diagram is drawn as follows. b) In the above diagram process 1→2 and 3→4 are adiabatic processes. So we can write first for the process 1→2 P1 V1γ=P2 V2γ →(1)Now P1=101,325 Pa; P2=1,013,250 Pa V1=2.0 m3; V2=?Now in gas turbine, normally natural gas is used such as "Nitrogen""Carbon dioxide" etc.So we consider the gas to be "diatomic" in nature.Hence the value of γ=1.4 . Where γ is "adiabatic exponent".Substituting the values in equation (1) we get(V2V1)γ=P1P2V2=(P1P2)1γ V1 =(1013251013250)11.4×2 =0.4Hence V2 is 0.4 m3 .Again for process 1→2 we can write,T1 V1γ-1=T2 V2γ-1 →(2)T2=T1×(V1V2)γ-1 =(300)×(20.4)1.4-1 =(300)×1.9 =570Hence T2 is 570 K.NOTE: VALUE OF T3 OR V3 SHOULD BE MENTIONED OTHERWISETHE PROBLEM CANNOT BE SOLVED FURTHER.c) Work done during the adiabatic process 1→2 is calculated as follows W=Rγ-1[T1-T2] →(1) =8.3141.4-1[300-570] =-5611.95 JNegative sign shows work is done on the system.Hence, work done on the system is 5611.95 J.

Answered: Answer Question A) B) C)

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Answer Question A) B) C)