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One kilogram of butane (C4H10) is burned with 25 kg of air that is at 308C and 90 kPa. Assuming that the combustion is complete and the pressure of the products is 90 kPa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products.

Respuesta :

Manetho

Explanation:

Combustion of butane gives following reaction

C4H10 + 13/2 O2 → 5H20 + 4CO2

Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.

17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.

Weight % of O2 in Air = 20 %

Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %

The dew point temperature of H2O is :

D = (237.3×B)/(1-B)

B = ln(E/6.108)/17.27

E is vapor pressure at given temp.

At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars

B = 0.1114

D = 29.74°C

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