Question

During the test, (C10H22) is burnt using an air-fuel ratio of 13:1 by weight. Determine the

complete volumetric analysis of the products of combustion, if the whole amount of

hydrogen burns to form water vapour and there is neither any free oxygen nor any free

carbon. The carbon burns to CO2 and CO.

Answer 1

Solution [Actual combustion equation] Ciotz2+ a[02+3.56N2] b Co₂ tcco + d H₂O + enz. [to find a, b, c, d & e] a [ 32+ 3.76X28] (AF) = 13 I 2X10 +22 a 13. 447 - Ву H' balance on both sides: 2 0 -- 22 d = 1) By 'c' balance: b+c=10 By 'о balance 2 btctd - 2a

2b + 6 + 11 = 2 x 13.447 i, 2b+c= 15.834 from 0 & 2 b=5.894 4.106 By 'n' balance ee= 2x ax 3.76 :. e - 3.76X 13.447 e=50-56 Volumetric analysis on dry & wet basis: component moles (ni) % by dry analysis (nikenal xiso I. by wet analysis. (nilenw)x100, Co2 b -- 5.894 9.732 8.236 со c=4.106 6.78 5.738 H2O d IS - 372 Nz e=50.56 83.488 70.654 Total dry= btcte źna - 60.56 Total Wet=b+c+dte Enw- 71.56

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