Question

Problem 1 A mixture having mole fractions of 37% N2, 50% CO2, 10% O2 and 3% CO flows through a turbine with Ti = 100° C and P1 = 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 (Cp) and the specific heat ratio (k) for the mixture 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 (e) the

Answer 1

37% Ne 50 % 10 O2 Co2 Talooc P,= 200 kPa P2 = 100 kPa na 10kals 3 % co for convenience, consider 100 kg of mixture. Maz 3 i7 kg mar so kg moz lo kg mo 3 kg > Number of moles of each component MN2 37 28 1.321 kmoles Nost mor Mcoz 1.136 kmoles 74 Noz - 0.3125 kmoles. 32 moz Moz mco Mco Nco 3 28 = 0.107 kmoles Total moles Nm = 2.87 65 kmoles

is Some mole fraction, so volume fraction Y Not NNz Nm 1.321 2.8765 = 0.4592 = 45.92% You 0.3125 2.8765 d.1086 10.86 %. Nm Yeaza Ncor 1-136 2.8765 = 0.3949 = 39.49% Nm Yco Nco Orlof 2.8765 = 0.0372 = 3.72.1. b) molar mass! Mina MM Nm loo 2.8765 34.76 kg/kmo) ) Gas constant! Rm Ru 8-314 34.76 = 0.2392 kJ / kg K d) Cpm =(MIN₂ CPN 2) + (Mfo CP02) + (mf.com Cpco) + (mtco2. CP CO2) (0.37% 1.039) + (0-1X0.918 )+(C-03X 1.04) +10.5X0.846)

0.93043 K+ lkg k --Ans Also, R=C-CV => = 0.69123 45/16gx k - e 0.93043 0.69123 1-346 Ans e) for reversible adiabatic expansion process 182 let ta T Pi 1.346-) 1.346 T2=373x (loo 20 T2 = 312-12K Power output :- Parch, - ha) rice (Ti-T2) 10x 0.93043x(373-312.12) P = 566.41 kW Ans.

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