The products of combustion of an unknown hydrocarbon CxHy have the following composition as measured by an Orsat apparatus: - CO2= 8.0%, CO = 0.9% O2 = 8.8% and N2 = 82.3% - Mumbai university > MECH > SEM 3 > THERMO

Marks: 8M

Year: May 2014

Solution:

1. Composition of the fuel.

   From the given Orsat analysis the combustion equation is written as follows:

   $aC +bH +cO_2+ (\frac{79}{21})cN_2 → 8CO_2 +0.9CO + 8.8O_2+ x H_2O +82.3N_2$

   Now, Carbon balance: $a = 8 + 0.9; \ \ \ a = 8.9$

   Nitrogen balance:

   $(79/21)c = 82.3 ; \ \ c = 21.877$

   Oxygen balance:

   c = 8 + 0.45 + 8.8 + 0.5 x

   21.877 = 17.25 + 0.5 x

   x =9.254

   Hydrogen balance:

   b = 2x; b = 18.5

2. Air fuel ratio.

   The air supplied per 100 moles of dry products is = 21.877 × 32 + (79/21) × 21.877 × 28

   = 3004.441 kg

   Air-fuel ratio = $\frac{3004.441}{12 × 8.9 + 18.5 × 1} $= 23.978kg/kg of fuel

3. The percentage excess air used.

   Mass fraction of carbon = $\frac{12 × 8.9}{12 × 8.9 + 18.5 × 1} = 0.8523$

   Mass fraction of Hydrogen = 1 – 0.8523 = 0.1477

   Considering 1 kg of fuel,

   The air required for complete combustion is = $0.8523 × \frac{8}{3} × \frac{100} {23.3} + 0.1477 × 8 × \frac{100}{23.3}$

   = 14.8257 kg

   Per cent excess air required for combustion = (23.978-14.8257)/14.8257 x 100 = 61.732 %