## Related questions with answers

A coal from Texas which has an ultimate analysis (by mass) of 39.25 percent C, 6.93 percent $H_{2},$ 41.11 percent $O_{2},$ 0.72 percent $N_{2},$ 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 $127^{\circ} \mathrm{C}.$ 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 $c_{p}=41.7\ \mathrm{kJ} / \mathrm{kmol} \cdot \mathrm{K}.$

Solution

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

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

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

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

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

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

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

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

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