Question

At a coal-fired power plant, assume that lignite coal is being burned as shown in the following stoichiometric combustion reaction equation, which assumes that the nitrogen in the lignite and in the air is all converted to NO and that combustion is complete:

CH0.96O0.18N0.016S0.012 + x (O2 + 3.78 N2) → CO2 + 0.48 H2O + 0.016 NO + 0.012 SO2 + 3.78x N2

(a) What is the value of the coefficient, x?

(b) To ensure complete combustion, air must be provided in an amount greater than that required by the stoichiometry shown in the above reaction equation. In this case, assume that the air is provided at an equivalence ratio of 0.95, meaning that the air is provided at a ratio of (1/0.95) times that required by chemical stoichiometry (meaning that about 5.3% excess air must be provided). Write a revised combustion reaction equation that takes into account the excess air requirement.

(c) Based on your revised reaction equation, how many moles of combustion gas are produced per mole of carbon combusted?

(d) Again based on your revised reaction equation, what is the mole fraction (moles/mole) of each gas (CO2, H2O, etc.) in the combustion gas?

(e) Again based on your reaction equation, what is the concentration (ppm) of each gas in the combustion gas?

Answer 1

(a): The balanced chemical reaction is

CH0.96O0.18N0.016S0.012 + x (O2 + 3.78 N2) → CO2 + 0.48 H2O + 0.016 NO + 0.012 SO2 + 3.78x N2

Equating the coefficients of O2 in the above reaction

0.18 O + 2x O = 2 O + 0.48 O + 0.016 O + 2*0.012 O

=> 2x O = 2.34 O

=> x = 1.17 (answer)

(b): Stoichiometric amount of air required = (O2 + 3.78 N2) * x = (O2 + 3.78 N2) * 1.17 mol

We need an excess of 5.3 % excess air.

Hence moles of air given =  (O2 + 3.78 N2) * 1.17 mol* (105.3 / 100) = (O2 + 3.78 N2) *1.232 mol

Hence the  revised combustion reaction equation is

CH0.96O0.18N0.016S0.012 + 1.232 (O2 + 3.78 N2)

--- > CO2 + 0.48 H2O + 0.016 NO + 0.012 SO2 + 4.657 N2 + 0.062 O2 (answer)

(c): Here the combustion gases are CO2, H2O, NO, SO2. N2 and O2

Hence moles of combustion gases produced per mole of C combusted

= 1 mol CO2 + 0.48 mol H2O + 0.016 mol NO + 0.012 mol SO2 + 4.657 N2 + 0.062 O2

= 6.227 mol combustion gas / mol C (answer)

(d): Mole fraction of CO2 = 1 mol CO2 / 6.227 mol = 0.1606

Mole fraction of H2O = 0.48 mol H2O / 6.227 mol = 0.0771

Mole fraction of NO = 0.016 mol NO / 6.227 mol = 0.00257

Mole fraction of SO2 = 0.012 mol SO2 / 6.227 mol = 0.00193

Mole fraction of N2 = 4.657 mol N2 / 6.227 mol = 0.7479

Mole fraction of O2 = 0.062 mol O2 / 6.227 mol = 0.00996