Chapter 15.7, Problem 45P

To determine

The higher and lower heating values of a coal.

Answer to Problem 45P

The higher and lower heating values of a coal is $\text{30,000}\text{kJ}/\text{kg}\text{and}\text{28,700}\text{kJ}/\text{kg}$ respectively.

Explanation of Solution

Express the total mass of the coal when the ash is substituted.

$m_{\text{total}}=100-m_{\text{ash}}$ (I)

Here, mass of ash is $m_{\text{ash}}$.

Express the mass fraction of carbon.

$m{f}_{\text{C}}=\frac{m_{\text{C}}}{m_{\text{total}}}$ (II)

Here, mass of carbon is $m_{\text{C}}$.

Express the mass fraction of hydrogen.

$m{f}_{{\text{H}}_2}=\frac{m_{{\text{H}}_2}}{m_{\text{total}}}$ (III)

Here, mass of hydrogen is $m_{{\text{H}}_2}$.

Express the mass fraction of oxygen.

$m{f}_{{\text{O}}_2}=\frac{m_{{\text{O}}_2}}{m_{\text{total}}}$ (IV)

Here, mass of oxygen is $m_{{\text{O}}_2}$.

Express the mass fraction of nitrogen.

$m{f}_{{\text{N}}_2}=\frac{m_{{\text{N}}_2}}{m_{\text{total}}}$ (V)

Here, mass of nitrogen is $m_{{\text{N}}_2}$.

Express the mass fraction of sulphur.

$m{f}_{\text{S}}=\frac{m_{\text{S}}}{m_{\text{total}}}$ (VI)

Here, mass of sulphur is $m_{\text{S}}$.

Express the number of moles of carbon.

$N_{\text{C}}=\frac{m{f}_{\text{C}}}{M_{\text{C}}}$ (VII)

Here, molar mass of carbon is $M_{\text{C}}$.

Express the number of moles of hydrogen.

$N_{{\text{H}}_{\text{2}}}=\frac{m{f}_{{\text{H}}_{\text{2}}}}{M_{{\text{H}}_{\text{2}}}}$ (VIII)

Here, molar mass of hydrogen is $M_{{\text{H}}_{\text{2}}}$.

Express the number of moles of oxygen.

$N_{{\text{O}}_{\text{2}}}=\frac{m{f}_{{\text{O}}_{\text{2}}}}{M_{{\text{O}}_{\text{2}}}}$ (IX)

Here, molar mass of oxygen is $M_{{\text{O}}_{\text{2}}}$.

Express the number of moles of nitrogen.

$N_{{\text{N}}_{\text{2}}}=\frac{m{f}_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (X)

Here, molar mass of nitrogen is $M_{{\text{N}}_{\text{2}}}$.

Express the number of moles of sulphur.

$N_{\text{S}}=\frac{m{f}_{\text{S}}}{M_{\text{S}}}$ (XI)

Here, molar mass of sulphur is $M_{\text{S}}$.

Express the total number of moles.

$N_m=N_{\text{C}}+N_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{\text{S}}$ (XII)

Express the mole fraction of carbon.

$y_{\text{C}}=\frac{N_{\text{C}}}{N_m}$ (XIII)

Express the mole fraction of hydrogen.

$y_{{\text{H}}_{\text{2}}}=\frac{N_{{\text{H}}_{\text{2}}}}{N_m}$ (XIV)

Express the mole fraction of oxygen.

$y_{{\text{O}}_{\text{2}}}=\frac{N_{{\text{O}}_{\text{2}}}}{N_m}$ (XV)

Express the mole fraction of nitrogen.

$y_{{\text{N}}_{\text{2}}}=\frac{N_{{\text{N}}_{\text{2}}}}{N_m}$ (XVI)

Express the mole fraction of sulphur.

$y_{\text{S}}=\frac{N_{\text{S}}}{N_m}$ (XVII)

Express the heat transfer from the combustion chamber.

$\begin{array}{c}q=h_C\\ =H_P-H_R\\ ={\displaystyle \sum N_P}{\overline{h}}_{f,P}^{\text{o}}-{\displaystyle \sum N_R}{\overline{h}}_{f,R}^{\text{o}}\\ ={\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{CO}}_{\text{2}}}+{\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{H}}_{\text{2}}\text{O}}+{\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{SO}}_2}\end{array}$ (XVIII)

Here, number of moles of products is $N_P$, number of moles of reactants is $N_R$, enthalpy of formation is ${\overline{h}}_f$, enthalpy of combustion is $h_C$, enthalpy of products and reactants is $H_P\text{and}{H}_R$ respectively.

Express apparent molecular weight of the coal.

$\begin{array}{c}{M}_m=\frac{m_m}{N_m}\\ =\frac{N_{\text{C}}{M}_{\text{C}}+N_{{\text{H}}_{\text{2}}}{M}_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}+N_{\text{S}}{M}_{\text{S}}}{N_{\text{C}}+N_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{\text{S}}}\end{array}$ (XIX)

Here, number of moles of carbon, hydrogen, oxygen, nitrogen and sulfur is $N_{\text{C}},N_{{\text{H}}_{\text{2}}},N_{{\text{O}}_{\text{2}}},N_{{\text{N}}_{\text{2}}}\text{and}{N}_{\text{S}}$ respectively.

Express higher value of coal.

$\begin{array}{c}\text{HHV}=\frac{-h_C}{m_m}\\ =\frac{-h_{\text{C}}}{N_{\text{C}}{M}_{\text{C}}+N_{{\text{H}}_{\text{2}}}{M}_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}+N_{\text{S}}{M}_{\text{S}}}\end{array}$ (XX)

Express lower value of coal.

$\begin{array}{c}\text{LHV}=\frac{-h_C}{m_m}\\ =\frac{-h_{\text{C}}}{N_{\text{C}}{M}_{\text{C}}+N_{{\text{H}}_{\text{2}}}{M}_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}+N_{\text{S}}{M}_{\text{S}}}\end{array}$ (XXI)

Conclusion:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

$\begin{array}{l}{M}_{\text{C}}=12\text{kg}/\text{kmol}\\ M_{{\text{H}}_2}=2\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=32\text{kg}/\text{kmol}\\ M_{\text{S}}=32\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{\text{air}}=29\text{kg}/\text{kmol}\\ M_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}\end{array}$

Here, molar mass of air is $M_{\text{air}}$.

Substitute $5$ for $m_{\text{ash}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{total}}=100-5\\ =95\text{kg}\end{array}$

Substitute $\text{61}\text{.40}\text{kg}$ for $m_{\text{C}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (II).

$\begin{array}{c}m{f}_{\text{C}}=\frac{\text{61}\text{.40}\text{kg}}{\text{95}\text{kg}}\\ =0.6463\\ =64.63\text{kg}\end{array}$

Substitute $\text{5}\text{.79}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (III).

$\begin{array}{c}m{f}_{{\text{H}}_{\text{2}}}=\frac{5.79\text{kg}}{\text{95}\text{kg}}\\ =0.06095\\ =6.095\text{kg}\end{array}$

Substitute $\text{25}\text{.31}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (IV).

$\begin{array}{c}m{f}_{{\text{O}}_{\text{2}}}=\frac{\text{25}\text{.31}\text{kg}}{\text{95}\text{kg}}\\ =0.2664\\ =26.64\text{kg}\end{array}$

Substitute $\text{1}\text{.09}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (V).

$\begin{array}{c}m{f}_{{\text{N}}_{\text{2}}}=\frac{\text{1}\text{.09}\text{kg}}{\text{95}\text{kg}}\\ =0.01147\\ =1.147\text{kg}\end{array}$

Substitute $\text{1}\text{.41}\text{kg}$ for $m_{\text{S}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (VI).

$\begin{array}{c}m{f}_{\text{S}}=\frac{\text{1}\text{.41}\text{kg}}{\text{95}\text{kg}}\\ =0.01484\\ =1.484\text{kg}\end{array}$

Substitute $\text{64}\text{.63}\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (VII).

$\begin{array}{c}{N}_{\text{C}}=\frac{\text{64}\text{.63}\text{kg}}{12\text{kg}/\text{kmol}}\\ =5.386\text{kmol}\end{array}$

Substitute $\text{6}\text{.095}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (VIII).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{6}\text{.095}\text{kg}}{2\text{kg}/\text{kmol}}\\ =3.048\text{kmol}\end{array}$

Substitute $\text{26}\text{.64}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{26.64\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.8325\text{kmol}\end{array}$

Substitute $\text{1}\text{.147}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (X).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{1.147\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.04096\text{kmol}\end{array}$

Substitute $\text{1}\text{.484}\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (XI).

$\begin{array}{c}{N}_{\text{S}}=\frac{1.484\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.04638\text{kmol}\end{array}$

Substitute $5.386\text{kmol}$ for $N_{\text{C}}$, $3.048\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.8325\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.04096\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.04638\text{kmol}$ for $N_{\text{S}}$ in Equation (XII).

$\begin{array}{c}{N}_m=5.386\text{kmol}+3.048\text{kmol}+0.8325\text{kmol}+0.04096\text{kmol}+0.04638\text{kmol}\\ =9.354\text{kmol}\end{array}$

Substitute $5.386\text{kmol}$ for $N_{\text{C}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XIII).

$\begin{array}{c}{y}_{\text{C}}=\frac{5.386\text{kmol}}{9.354\text{kmol}}\\ =0.5758\end{array}$

Substitute $3.048\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XIV).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}}=\frac{3.048\text{kmol}}{9.354\text{kmol}}\\ =0.3258\end{array}$

Substitute $0.8325\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XV).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{0.8325\text{kmol}}{9.354\text{kmol}}\\ =0.0890\end{array}$

Substitute $0.04096\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XVI).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.04096\text{kmol}}{9.354\text{kmol}}\\ =0.00438\end{array}$

Substitute $0.04638\text{kmol}$ for $N_{\text{S}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XVII).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.04638\text{kmol}}{9.354\text{kmol}}\\ =0.00496\end{array}$

Express the combustion equation.

$\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +a_{\text{th}}\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.5758{\text{CO}}_{\text{2}}+0.3258{\text{H}}_{\text{2}}\text{O}+0.00496{\text{SO}}_2+k{\text{N}}_{\text{2}}\end{array}\right]$ (XXII)

Perform the species balancing:

Oxygen balance:

$\begin{array}{l}0.0890+a_{\text{th}}=0.5758+0.5\left(0.3258\right)+0.00496\\ a_{\text{th}}=0.6547\end{array}$

Nitrogen balance:

$\begin{array}{c}k=0.00438+3.76\left(0.6547\right)\\ =2.466\end{array}$

Substitute $0.6547$ for $a_{\text{th}}$ and $2.466$ for $k$ in Equation (XXII).

$\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +0.6547\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to \left\{\begin{array}{l}0.5758{\text{CO}}_{\text{2}}+0.3258{\text{H}}_{\text{2}}\text{O}\\ +0.00496{\text{SO}}_2+2.466{\text{N}}_{\text{2}}\end{array}\right\}\end{array}\right]$ (XXIII)

Refer Equation (XIII) and write the number of moles of carbon dioxide, water and sulfur oxide.

$\begin{array}{l}{N}_{{\text{CO}}_{\text{2}}}=0.5758\text{kmol}\\ N_{{\text{H}}_{\text{2}}\text{O}}=0.3258\text{kmol}\\ N_{{\text{SO}}_{\text{2}}}=0.00496\text{kmol}\end{array}$

Refer Table A-26, “enthalpy of formation, Gibbs function of formation and entropy at $\text{25°C,1}\text{atm}$”, and write the enthalpy of formation of carbon dioxide, water and ethane.

$\begin{array}{l}{\overline{h}}_{f,{\text{CO}}_{\text{2}}}^{\text{o}}=-393,520\text{kJ}/\text{kmol}\\ {\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^{\text{o}}\left(l\right)=-285,830\text{kJ}/\text{kmol}\\ {\overline{h}}_{f,{\text{SO}}_{\text{2}}}^{\text{o}}=-297,100\text{kJ}/\text{kmol}\end{array}$

Substitute $0.5758\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $-393,520\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{CO}}_{\text{2}}}^{\text{o}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $-285,830\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^{\text{o}}$, $0.00496\text{kmol}$ for $N_{{\text{SO}}_{\text{2}}}$ and $-297,100\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{SO}}_{\text{2}}}^{\text{o}}$ in Equation (XVIII).

$\begin{array}{c}{h}_C=\left[\begin{array}{c}\left(0.5758\text{kmol}\right)\left(-393,520\text{kJ}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(-285,830\text{kJ}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(-297,100\text{kJ}/\text{kmol}\right)\end{array}\right]\\ =-321,200\text{kJ}/\text{kmol}\end{array}$

Refer Equation (XXIII), and write the number of moles.

$\begin{array}{l}{N}_{\text{C}}=0.5758\text{kmol}\\ N_{{\text{H}}_{\text{2}}}=0.3258\text{kmol}\\ N_{{\text{O}}_{\text{2}}}=0.0890\text{kmol}\\ N_{{\text{N}}_{\text{2}}}=0.00438\text{kmol}\end{array}$

$N_{\text{S}}=0.00496\text{kmol}$

Substitute $0.5758\text{kmol}$ for $N_{\text{C}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.0890\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.00438\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.00496\text{kmol}$ for $N_{\text{S}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (XIX).

$\begin{array}{c}{M}_m=\frac{\left[\begin{array}{l}\left(0.5758\text{kmol}\right)\left(12\text{kg}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(2\text{kg}/\text{kmol}\right)\\ +\left(0.0890\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)+\left(0.00438\text{kmol}\right)\left(28\text{kg}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)\end{array}\right]}{0.5758\text{kmol}+0.3258\text{kmol}+0.0890\text{kmol}+0.00438\text{kmol}+0.00496\text{kmol}}\\ =\frac{10.69\text{kg}}{1\text{kmol}}\\ =10.69\text{kg}/\text{kmol}\end{array}$

Substitute $-321,200\text{kJ}/\text{kmol}$ for $h_C$, $0.5758\text{kmol}$ for $N_{\text{C}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.0890\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.00438\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.00496\text{kmol}$ for $N_{\text{S}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (XX).

$\begin{array}{c}\text{HHV}=\frac{-\left(-321,200\text{kJ}/\text{kmol}\right)}{\left[\begin{array}{l}\left(0.5758\text{kmol}\right)\left(12\text{kg}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(2\text{kg}/\text{kmol}\right)\\ +\left(0.0890\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)+\left(0.00438\text{kmol}\right)\left(28\text{kg}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)\end{array}\right]}\\ =\frac{321,200\text{kJ}/\text{kmol}}{10.69\text{kg}/\text{kmol}}\\ =\text{30,000}\text{kJ}/\text{kg}\end{array}$

Hence, the higher heating value of a coal is $\text{30,000}\text{kJ}/\text{kg}$.

For the LHV (lower heating value), the water in the products is taken to be vapor.

Refer Table A-26, “enthalpy of formation, Gibbs function of formation and entropy at $\text{25°C,1}\text{atm}$”, and write the enthalpy of gaseous water.

${\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^{\text{o}}\left(g\right)=-241,820\text{kJ}/\text{kmol}$

Substitute $0.5758\text{kmol}$ for $N_{{\text{CO}}_{\text{2}}}$, $-393,520\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{CO}}_{\text{2}}}^{\text{o}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $-241,820\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{H}}_{\text{2}}\text{O}}^{\text{o}}$, $0.00496\text{kmol}$ for $N_{{\text{SO}}_{\text{2}}}$ and $-297,100\text{kJ}/\text{kmol}$ for ${\overline{h}}_{f,{\text{SO}}_{\text{2}}}^{\text{o}}$ in Equation (XVIII).

$\begin{array}{c}{h}_C=\left[\begin{array}{c}\left(0.5758\text{kmol}\right)\left(-393,520\text{kJ}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(-241,820\text{kJ}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(-297,100\text{kJ}/\text{kmol}\right)\end{array}\right]\\ =-306,850\text{kJ}/\text{kmol}\end{array}$

Substitute $-306,850\text{kJ}/\text{kmol}$ for $h_C$, $0.5758\text{kmol}$ for $N_{\text{C}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.0890\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.00438\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $0.00496\text{kmol}$ for $N_{\text{S}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (XXI).

$\begin{array}{c}\text{LHV}=\frac{-\left(-306,850\text{kJ}/\text{kmol}\right)}{\left[\begin{array}{l}\left(0.5758\text{kmol}\right)\left(12\text{kg}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(2\text{kg}/\text{kmol}\right)\\ +\left(0.0890\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)+\left(0.00438\text{kmol}\right)\left(28\text{kg}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)\end{array}\right]}\\ =\frac{306,850\text{kJ}/\text{kmol}}{10.69\text{kg}/\text{kmol}}\\ =\text{28,700}\text{kJ}/\text{kg}\end{array}$

Hence, the lower heating value of a coal is $\text{28,700}\text{kJ}/\text{kg}$.