Question

Problem 1 A mixture having mole fractions of 37% N,, 50% CO,,10% 0, and 3% CO flows through a turbine with T, - 100°C and P, = 200 kPa. Analyze the gas mixture to find: (a) the mass fractions of the components (b) the equivalent molar mass (c) the gas constant for the mixture (d) the specific heat at constant pressure (C, ) and the specific heat ratio (k) for the mixture (e) the power output by a turbine with exit pressure of 100 kPa if the mass flow rate of the gas mixture is 10 kg/s. Assume the process is reversible and adiabatic

Answer 1

Mole fractions: 374. Ne Yu, 0.37 T: 100Pc 50 % CO2 You 0.5D Prz 200 kPa 11 10 % O2 you P₂ = 100kPa 010 1 Y co 0:03 31. CO. la) Mass fractions (X): assuming 100 kmot of mixtore kg 10 36 kg Ny 37* (28) mcor 50 X (14) kes 700 kg 11 14 320 mor ks 10 X (32) les 84 " ks mco 3x (28) Mtotala mNet mort moat mlo 2140 kg Ang MNL 0.48 1036 2440 metotal 110) - 0.33 100 2140 320 2140 do2 = 0.15 Xco 84 0.04 2140

(Ь) Equivalent Molar mass, Mom Antolal we assumed, Notodal = 100 kmol Ntotal 214 0 100 21.4 kg/leme t Ans. Mmix (C) Gas constant for mixture 15 kkmol-te) Ru 8.314 Rmix 21.4 kg Ikimos Rmix 0.389 kJ kg. k 10) Specific heat at constant poe asure, I cplmiy (2) Constant Yolume, ((), mix E di (Cp,i) G individual Specifre heat at Const pressure Cp, mix to! bly. Cvonly = di (Cu, i)

ki/kg.k raha ko legok 0.743 (CP 1.039 (Cu) ce (0) 0.657 (CP.COM 0.846 (Gup) or (Cp) or 0.918 0.658 (Cp) Co 1.ouo (UP) LO 0.744 (Spaniya (0.48)1.039)+ (0.33) 10.846) + 10.1510.918) of lououll 1.040) KI Pimix= 0.9572 kg.k (Cr) mix 0-65'8 E10.48)10.743) + (0.33 / 09657) +10.15)1 BAN) + (0.04) 0.744) 0.702 p.71111 Cumiy th kgok cp, min 0.9572 k = Specific heat ratio : 0-11410:702 kil36

le) Is entropic proces: Ti: 100°C = 373 k K- k To 7) le 1.361 1.36 T, - (313) (2000 318/ * 310 k Tz = 37°c 2 & F 4514 Power out but by a turbine, 1 AT=(1,-92) Pa mi (cpm,AT 37 10 x (0.9572)/100-48) 1 + 4/4819 Ky 47 anys. P= 603 KW error Not! Had to Strike off for calculation instead of re-doing it again