Question

Calculate the heat of combustion and the stoichiometric fuel to air ratio (mass and molar) for 1) Methane and 2) Octane. Comment on whether the assumption of f<<1 is valid for these conditions.

Answer 1

COMBUSTION or Oxidation is the process of addition of Oxygen.

chemical reaction for Methane:

CH₄ + bO₂ cCO₂ + dH₂O

on balancing we get c=1; d=2; b=1

heat of combustion is also know by the name as heat of reaction =

{heat of formation of CO₂ + 2 x heat of formation of H₂O} -
{heat of formation of CH₄ + 2 x heat of formation of O₂ }

For the calculation of these you require to learn some values or they me be provided in the question paper.

Note: we have taken heat of formation of O₂ as reference so its value is taken as 0kJ/kg

= { -393.5 kJ + 2x(-286 kJ)} - {-74.8 kJ + 2x(0 kJ)}
= - 890.7 kJ/mol

Similarly heat of combustion for Octane is

C₈H₁₈ + bO₂ cCO₂ + dH₂O

On balancing no. of moles of C, H, O, we will be getting 3 equations and we have 3 unknowns in the form b, d &c

so on solving we get, b=12.5; c=8;d=9;

heat of combustion of Octane :

{8 x heat of formation of CO₂ + 9 x heat of formation of H₂O} -
{heat of formation of C₈H₁₈ + 12.5 x heat of formation of O₂ }

Note: You need enthalpy of formation of Octane which may be different in different temperature condition Taking heat of reaction of C₈H₁₈ as -269kJ/mol.

On putting values you can get the final answer as -5722+269=-5453 kJ/mol.

STOCHIOMETRIC RATIO:

1. CH₄:

CH₄ + b(O₂+3.76N₂) cCO₂ + dH₂O + (3.76 x b)N₂

Note: I am assuming that air contains 21% O₂ 79% N₂ So, if the number of moles O₂ are 1. So the number of moles of N₂: 79/21=3.76

On balancing we get b=2; c=1; d=2;

so stochiometric ratio(A/F) ON MOLAR BASIS = = 2:1

ON MASS BASIS(A/F) == 17.16:1

STOCHIOMETRIC RATIO FOR OCTANE:

C₈H₁₈ + b* (O₂ +3.76N₂) cCO₂ + dH₂O + (b*3.76) N₂

On solving, we get c=8; d=9; b=12.5;

STOCHIOMETRIC RATIO (A/F) =

MASS BASIS = = 15.05:1

MOLAR BASIS = =12.5:1

Kindly comment about the symbol 'f'