The last 3 parts B,D,E a(a)Airin(b)(c)DiffuserCompressor1StateState 1State 2State 3State 4State 5TUMCombustors2400803TurbineFigure 1: Figure for Problem 2.Productgases outIn a modern jet engine, air passes through the following states from the inlet to the outlet,as shown in Figure 1:42607801500900640Nozzle5→Pressure (kPa) Temperature (K)8033003200justify them) to find the compressor specific work.For this probler, you may neglect any heat transfer, as well as neglect kinetic energy exceptat the outlet (state 5). Use the tables for obtaining properties (do not assume constantspecific heat). Assume air is an ideal gas with ideal gas constant of R = 0.287 kJ/kg-K.Use the appropriate conservation equations and make approximations (andIn a similar manner, find the turbine specific work.And finally, using similar arguments, find the nozzle exit velocity.(d)Suppose the jet engine has a mass flow rate of m = 250 kg/s and a circularinlet with an inlet diameter of 2 meters. Is the assumption to neglect kinetic energyvalid at state 1? You can assume cross section at state 1 is the same as the inlet.

As per Bartleby code of conduct, it is recommended only to solve the first three subparts of a question with multiple subpart. Kindly post the last subpart to get answered To find: The specific work of compressor and turbine and the nozzle exit velocity. Given: The characteristic gas constant value, R = 0.287 kJ.kg.k. Formula used: Specific work of compressor: wc = h2-h1 Specific work of turbine: wt = h3 - h4 Exit velocity of nozzle: h4 = h5 + v522 where, v5 is the exit velocity of nozzle. The compressor, turbine and nozzle are all steady state single flow devices and they are adiabatic. Using air properties the value of enthalpies can be calculated: h1 = 260.32 kJ/kg, h2 = 800.28 kJ/kg, h3 = 1635.80 kJ.kg, h4 = 933.15 kJ/kg, h5 = 649.53 kJ/kg.a) The specific work of compressor can be calculated as follows: wc = 800.28 - 260.32 = 539.36 kJ/kg. b) The specific work of turbine can be calculated as follows: wT = 1635.80 - 260.32 = 702.65 kJ/kg.c) Since, it is mentioned in the question that we can only used the energy equation at the outlet of the nozzle. Therefore: 933.15 = 649.53 + v522000v5 = 2000(933.15-649.53)v5= 753.15 m/s.