Question

Mixed stream Ti = 300 K Q3.25 pts An ideal gas mixture consisting of 20% -300K O2 and 80% N2 (by mol) is separated into pure 02 P 1 atm and pure N2 streams. The gas mixture enters at 300 K, 1 atm, and pure gases leave at 300 K, 1 atm. Assuming steady-state, calculate the minimum work that is required to generate 1 kmol of pure 02. Assume that the heat transfer between the separator device and the surroundings occurs at 300 K. Hint: Entropy generation cannot be negative. R = 8.314 kJ/(kmol.) Tds = dh - ūdp 20% Os. 80% Ne (by vol.) T-300 K Plat 1 012 0.). min 1 MP Steady state energy balance: 0 – w + Şniñu - niche = 0 Steady state entropy balance: 0 = _ + > --Ştes + der S-A, Game Os tria [Saya se) toy Sess] {niki - {sołe = 1 khol a honi + 4 kun i hans het house and thus) = 0

Answer 1

Given Ř = 8.314 KJ/Kmalik T = 3ook 204.02, 80%N2 Molar maxs of O2 = 32 kg/kmol moler mass of H2 = 28 kg/kmol dš= EAS; = Ř Eni In (ni) = 8.3146.2 116. 2.) +0.8 Inco.pl] . for On ni = 0. 2 7 T for Nz ni= 0.8 l = ds = 4.16 kJ/kmolok Total molecular mass Mm = En Mi => Mm = (0,2)(82) 760.8)x(28) . = 28.8 kg /kmol. > 15 = ds = 4:16 KJ/K molek Man 28.8 kg / kg. K => A5 = 0.144 kJ /kg.k so Minimum work required W = TXA5 = 600k) (0.144kJ/kg. =) Tw= 43.33 kJ/kg