Question

n-butane burns with 15% excess air, both streams enter at 25°C. The gas burns completely producing 90 mol of CO2 / 10mol of CO. The combustion gas comes out at 400ºC and 1 atm. Taking as a basis of calculation 1000 moles/h of the gas that feeds the oven:
Calculate the air supply speed and the composition of the combustion gases.

Answer 1

Solution Basis 1000 moles/h 1000 moles y Holhe 25°0 Reactor n (mole coln n (mole coln N₂ (mole Holly ny (mote O2 ln) nel mole Nylws 15% excess air 25°c 400°C, latm

२. 1000 mole Catholh 17.5 mole O, (now) theoretical I mole lithio 2 7500 mole. Oslu 415 % excess air olo 0.21 no z 1115 (7500 mole 0,1W) mole air in no - 41071.43 mit in 2) Oq fed (nor). 0.21 no , 8625 mole 621h 32446.43 mole Nz/h inn) 0.79 no .- we have , ny 90 mole co, 2 10 mole co 2 ni an

Atounc c balance; 1000 mole Cuttoli u mole " = an, mole coulhli molec Imoleco I mole (utio the motel molecolli I molec i moleco lon, 4000 is. mole colu nr 400 mole 그 na an 23600 mole con la

Atomie H balance: N mole Hoolh 2 mol H - 1000 mole in Ho Chillo molett linole Cullio 의 그 I mol H₂O 2 nz loooo C 2 mole H₂Olh n 5.000 2 zo

Atomic balance! oxygen 360o mole oalh 2 mole o 8625 mol Olhi 2 mol o I mol Or otally Imole e + 400 molecolhli mole o I'moleco + 5000 mole H20/h i moletop I mole o I moleto + nu mole Orlh 2 mote o l.mole O₂ 8625x2 = 360082 +400 15000+ inu 2 th nu 23:25 mole Olh คป น is ineet so / ng 324 46.43 mole Nz/h

composition of product gus; Total molar flow rate of product gus n. th, и; thy the no 3600+ 400 +5000+2325+92 446.43 - 43771.43 mole ih 9.22 mole % Co cor 2600 x 100% 43771-43 0.91 mole % co 2 co чоо co; x 160 % 43771.43 11:42 mole % H2O C H2O 500 x 100% 43771143 5:31 mole % ₂ 23 25 x 100 % 43771.43 32n46:43 74.13 mol % N₂ z Nr's X 100 % 4377.1.43