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Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor, and (b) the exit temperature of the air if the process were reversible.

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Answer:isentropic efficiency = 0.818Explanation:given data pressure P1 = 95 kPatemperature = 27°Cpressure P2 = 600 kPatemperature = 277°Cto find outisentropic efficiency of the compressor and exit temperature of the airsolutionwe know from ideal gas of properties of air is Pr1 at 27°C = 1.3860and h1 at 300 K = 300.19 kJ/kgand h2 at 550 K = 555.74 kJ/kgand we know equation for isentropic process that is Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor, and (b) the exit temperature of the air if the process were reversible.    .........................1put here value we get Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor, and (b) the exit temperature of the air if the process were reversible. solve we get Pr2 Pr2 = 8.75 by  ideal gas of properties of air will be at Pr2 h2s = 508.66 T2s = 505.5 K 'so isentropic efficiency will be here as isentropic efficiency = Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor, and (b) the exit temperature of the air if the process were reversible. isentropic efficiency = Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor, and (b) the exit temperature of the air if the process were reversible. isentropic efficiency = 0.818

Answerd by jsmehra
11 months ago 10 5.0