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A coal from Texas which has an ultimate analysis (by mass) of 39.25 percent C, 6.93 percent H2,H_{2}, 41.11 percent O2,O_{2}, 0.72 percent N2,N_{2}, 0.79 percent S, and 11.20 percent ash (non-combustibles) is burned steadily with 40 percent excess air in a power plant boiler. The coal and air enter this boiler at standard conditions, and the products of combustion in the smokestack are at 127C.127^{\circ} \mathrm{C}. Calculate the heat transfer, in kJ/kg fuel, in this boiler. Include the effect of the sulfur in the energy analysis by noting that sulfur dioxide has an enthalpy of formation of -297,100 kJ/kmol and an average specific heat at constant pressure of cp=41.7 kJ/kmolK.c_{p}=41.7\ \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K}.

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First the mole numbers of the components of the coal are determined from their masses:

NC=(mM)C=39.2512kmol=3.2708kmol\begin{align*} N_{\text{C}}&=\bigg(\dfrac{m}{M}\bigg)_{\text{C}}\\ &=\dfrac{39.25}{12}\:\text{kmol}\\ &=3.2708\:\text{kmol} \end{align*}

NH2=(mM)H2=6.932kmol=3.4650kmol\begin{align*} N_{\text{H}_{2}}&=\bigg(\dfrac{m}{M}\bigg)_{\text{H}_{2}}\\ &=\dfrac{6.93}{2}\:\text{kmol}\\ &=3.4650\:\text{kmol} \end{align*}

NO2=(mM)O2=41.1132kmol=1.2847kmol\begin{align*} N_{\text{O}_{2}}&=\bigg(\dfrac{m}{M}\bigg)_{\text{O}_{2}}\\ &=\dfrac{41.11}{32}\:\text{kmol}\\ &=1.2847\:\text{kmol} \end{align*}

NN2=(mM)N2=0.7228kmol=0.0257kmol\begin{align*} N_{\text{N}_{2}}&=\bigg(\dfrac{m}{M}\bigg)_{\text{N}_{2}}\\ &=\dfrac{0.72}{28}\:\text{kmol}\\ &=0.0257\:\text{kmol} \end{align*}

NS=(mM)S=0.7932kmol=0.0247kmol\begin{align*} N_{\text{S}}&=\bigg(\dfrac{m}{M}\bigg)_{\text{S}}\\ &=\dfrac{0.79}{32}\:\text{kmol}\\ &=0.0247\:\text{kmol} \end{align*}

The total number of moles is Nm=8.0709kmolN_{m}=8.0709\:\text{kmol}. The mole fractions are then:

yC=NCNm=3.27088.0709=0.4053\begin{align*} y_{\text{C}}&=\dfrac{N_{\text{C}}}{N_{m}}\\ &=\dfrac{3.2708}{8.0709}\\ &=0.4053 \end{align*}

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