A large air separation plant takes ambient air (79% N2 , 21% O2 bymole) at 100 k

Question

A large air separation plant takes ambient air (79% N2 , 21% O2 bymole) at 100 kPa and 20 C at a rate of 25 kg/s. It discharges astream of pure O2 gas at 200 kPa and 100 C and a stream of pure N2gas at 100 kPa and 20 C. the plant operates on an electrical powerinput of 2000 kW. Calculate the net rate of entropy change for theprocess

How would you do this problem if the air was givenas 79% N2 and 21% O2 by VOLUME instead? My instructor gave us thisproblem for hmwk and I cannot figure it out! PLEASEHELP!!!!!!

Explanation / Answer

Main use of giving these % composition by mole is to know the massfraction with the help of molar mass. mass fraction =yiMi /yiMi where, yi is molar fraction and Mi isthe molar mass for CO2 and N2 now mass flow rate of co2 and N2 can be calculatedas: mass fraction*mass flow rate ofair Thus, we know mass in rate and mass out rates then using energy balance , we can calculate the net heat flowrate. and then using entropy rate eq. for control volume, entropychange can be calculate. NOW, Suppose your instructor gives the fraction as a volume % calculate the mass fraction using density, mass fraction =xii/xiiwherexi is the volume fraction and i is the density forCO2 and N2.rest of the steps will besame.

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