Question

4. a. (18 points) Ethane (CH) is burned with dry air, giving the products having a molar analysis on a dry basis: CO2, 7.8%; CO, 0.2%, 02, 7%; N2, 85% (actual combustion). Determine: (3 points each) (a) the stoichiometric air-fuel ratio. (b) the actual air fuel ratio on molar and mass bases based on the given products of combustion. (c) the gravimetric wet analysis (mass fractions) of the products of the actual combustion. (d) the gas constant of the products of combustion and the excess air factor. (e) the percentage of excess air used in the actual combustion. (f) the partial pressure of each product of the products of combustion if the actual product mixture pressure is 100 kPa (atmospheric pressure).

Answer 1

of call we write Sol" First stochionsabrir eq" for combustion of ethone Calto 0-4C, Ho +20.90₂ + BSN → 78C02+0.260 +70, +85 N₂ + 12 H₂O This air lean burning of love is fuel mister is eness ongen m that why the right sech of egth. Otherwise the ea" would be (a) (H + a₂ + a3.76N2 - 260 +3th ta3.76N₂ 3.5 stochometre equ a So becom H + 3.50₂ + 13.16 N₂ - 200₂ + 3 H₂O +13.16N2

(a Mass of full 12x 2 + 6x1=30 Mass of cu 3.58 32 + 13.16828 = 480.80 sloctis metrui aur to fuel ralio 480.80 30 (6) nchua cur For this we fuel ratio will consider consider au li Mola bass = 4 No. of moles of fuel No. of moles of our = 20.9+85 = 105.9 ner fuel ratio 105.9 = 26.475 4

(1) Mass bca Mass of fuel - 4x (24+6)= 96+24 wa - 120 Mass ge cur 20.92x32 +85x28 3049.44 Aur feed ratio = 3049.44 120 u 25.412 Total moles of prodent & 7.8+0.2 +7 +85+ 12 = (12 Tolal mass of products = 7.8 (12+32) +0.2 (12+16) +7(32) +85028) +12 (2+16) Total = 7.8844+ 0.2 x 28 +7+32+85628 + 12x 18 = 3168.8

Mass fraction of co, 7.8 (12 +32)_0.108 3168.8 fracher of co-0.2 (12 +26) = 0.00176 3168.8 -0.0706 Mass fraction of ₂ 7x32 3168.8 0.75lo Mass fraction of N = 85x28 3168.8 Mass feachia of H₂O = 12 (2+1) = 0.0681 3168.8 d) Average molecular mass of products Emi m.fi Em. A mi > molar mass of in gass fracha of a gas mit; nass E mif

0. 108 x 4 4+.001768 28 +0.0706x32 + 7510 x 28 + .0681 14 now 29.04 This alerc molecular mass Gaso constant 8 3 14.5_ 286.23 29.04 -t H fo erst یا will calculate stiochioneatrii ratio Al prew a Neta (1/a) (FIA) 25 16 (012) achun = 1.5675

Now enuss aur factor n we can Calculat h = - (2-1)x100 (1.56-1) xlo 0% h = h 56% 56% of excess aur used (F) Sunce we know partial pressure by gas enerted is proportional to its mole fraction Mole fraction of co, 7.8 112 CO = 0.2 112 O=] 112 =) N₂=85 N₂=85 H₂O = 12 112 112

so partial pressure of co₂7.8 x 100 6.96 KP a for co 0.2 xloo 112 0.178 K P a Oza 7 112 x loo z 6.25 kPa For N2 85 112 10o a 85 kPa t For H₂O = 12 (12. 10o = 10.7 kPa +