An industrial process discharges 5,700 m³/min of gaseous products at 200°C, 100 kPa. The figure belowshows a proposed system for utilizing the combustion products and its steady state conditions. Heattransfer from the outer surface of the steam generator (heat exchanger) and turbine can be ignored, ascan the changes in kinetic and potential energies of the streams. There is no pressure drop through theheat exchanger. The combustion product can be modeled as air as an ideal gas.Determine:a) The turbine inlet temperature, T4;b) the power developed by the turbine; andc) the rates of entropy generation in the heat exchanger and turbine, each in kW/K.Combustion products inP1 = 100 kPa1T1 = 480 K(AV), = 5700 m³/minTurbine4PoweroutP4 = 300 kPa%3DCombustion products outP2 = 100 kPaT2 = 400 KSteamgenerator3 Water inP3 = 300 kPaT3 = 40°Cm3 = 120 kg/minWater outPs = 5 kPaX5 = 97%%3D%3Dwwwwwwww

state 1 by ideal gas equation P1V1 = ma RT1100×570060 = ma ×0.287×480ma = 68.9605 kg/sstate 2…

Answered: the streams. There is no odeled as air…

Engineering

Mechanical Engineering

the streams. There is no odeled as air as an ideal

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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