Chapter 15.7, Problem 49P

To determine

The higher heating value of the coal.

The lower heating value of the coal.

Answer to Problem 49P

The higher heating value of the coal is $33,650\text{kJ}/\text{kg coal}$.

The lower heating value of the coal is $31,370\text{kJ}/\text{kg coal}$.

Explanation of Solution

Write the expression to calculate the number of moles of constituent $\left(N\right)$ in a mixture.

$N=\frac{m}{M}$ (I)

Here, mass of the gas constituent is $m$ and the molecular mass of the constitute is $M$.

Write the expression to calculate the total number of moles of gases $\left(N_m\right)$ in products.

$N_m=N_A+N_B+\mathrm{..............}+N_Z$ (II)

Here, number of moles of compound A is $N_A$, number of moles of compound B is $N_B$, and the number of moles of compound Z is $N_Z$.

Write the expression to calculate the mole fraction of a constituent $\left(y\right)$ in a mixture.

$y=\frac{N}{N_m}$ (III)

Write the expression to calculate the mass of the mixture of gases $\left(m_{\text{total}}\right)$.

$m_{\text{total}}=N_A{M}_A+N_B{M}_B+\mathrm{.......}+N_Z{M}_Z$ (IV)

Here, number of moles of the compound A in the product is $N_A$, the molecular weight of the compound A is $M_A$, number of moles of compound B is $N_B$, molecular weight of compound B is $M_B$

Write the expression to calculate the apparent molecular weight of the product gas $\left(M_m\right)$.

$M_m=\frac{m_m}{N_m}$ (V)

Here, mass of the product gas is $m_m$.

Write the expression to calculate the heat transfer $\left(q\right)$ for the combustion.

$q=h_C=H_P-H_R$

$q={\displaystyle \sum N_P{\overline{h}}_{f,P}^o}-{\displaystyle \sum N_R{\overline{h}}_{f,R}^o}$ (VI)

Here, enthalpy of combustion is $h_C$, enthalpy of the products is $H_P$, enthalpy of reactants is $H_R$, number of moles of product gas is $N_P$, enthalpy of formation of product gas is ${\overline{h}}_{f,P}^o$, number of moles of reactants is $N_R$, and the enthalpy of formation of reactants is ${\overline{h}}_{f,R}^o$.

Write the expression to calculate the heating value of coal.

$\text{HV}=\frac{-h_C}{M_m}$ (VII)

Conclusion:

From the Table A-1 of “Molar mass, gas constants, and critical-point properties”, select the molar masses of carbon, hydrogen, oxygen, Sulphur, and air as,

$\begin{array}{l}{M}_{\text{C}}=12\text{kg}/\text{kmol}\\ M_{{\text{H}}_2}=2\text{kg}/\text{kmol}\\ M_{{\text{O}}_2}=32\text{kg}/\text{kmol}\\ M_{\text{S}}=32\text{kg}/\text{kmol}\\ M_{\text{air}}=29\text{kg}/\text{kmol}\end{array}$

Consider for $100\text{kg}$ of coal. Substitute $67.40\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{C}}=\frac{67.40\text{kg}}{12\text{kg}/\text{kmol}}\\ =5.617\text{kmol}\end{array}$

Substitute $5.31\text{kg}$ for $m_{{\text{H}}_2}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{H}}_2}=\frac{5.31\text{kg}}{2\text{kg}/\text{kmol}}\\ =2.655\text{kmol}\end{array}$

Substitute $15.11\text{kg}$ for $m_{{\text{O}}_2}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_2}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{O}}_2}=\frac{15.11\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.4722\text{kmol}\end{array}$

Substitute $1.44\text{kg}$ for $m_{{\text{N}}_2}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$ in Equation (I).

$\begin{array}{c}{N}_{{\text{N}}_2}=\frac{1.44\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.05143\text{kmol}\end{array}$

Substitute $2.36\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{S}}=\frac{2.36\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.07375\text{kmol}\end{array}$

Substitute $5.617\text{kmol}$ for $N_{\text{C}}$, $2.655\text{kmol}$ for $N_{{\text{H}}_2}$, $0.4722\text{kmol}$ for $N_{{\text{O}}_2}$, $0.05143\text{kmol}$ for $N_{{\text{N}}_2}$, and $0.07375\text{kmol}$ for $N_{\text{S}}$ in Equation (II).

$\begin{array}{c}{\left(N_m\right)}_{\text{reactants}}=\left(\begin{array}{c}5.617\text{kmol}+2.655\text{kmol}+0.4722\text{kmol}\\ +0.05143\text{kmol}+0.07375\text{kmol}\end{array}\right)\\ =8.869\text{kmol}\end{array}$

Substitute $5.617\text{kmol}$ for $N_{\text{C}}$ and $8.869\text{kmol}$ for $N_m$ in Equation (III).

$\begin{array}{c}{y}_{\text{C}}=\frac{5.617\text{kmol}}{8.869\text{kmol}}\\ =0.6333\end{array}$

Substitute $2.655\text{kmol}$ for $N_{{\text{H}}_2}$ and $8.869\text{kmol}$ for $N_m$ in Equation (III).

$\begin{array}{c}{y}_{{\text{H}}_2}=\frac{2.655\text{kmol}}{8.869\text{kmol}}\\ =0.2994\end{array}$

Substitute $0.4722\text{kmol}$ for $N_{{\text{O}}_2}$ and $8.869\text{kmol}$ for $N_m$ in Equation (III).

$\begin{array}{c}{y}_{{\text{O}}_2}=\frac{0.4722\text{kmol}}{8.869\text{kmol}}\\ =0.05323\end{array}$

Substitute $0.05143\text{kmol}$ for $N_{{\text{N}}_2}$ and $8.869\text{kmol}$ for $N_m$ in Equation (III).

$\begin{array}{c}{y}_{{\text{N}}_2}=\frac{0.05143\text{kmol}}{8.869\text{kmol}}\\ =0.00580\end{array}$

Substitute $0.07375\text{kmol}$ for $N_{\text{S}}$ and $8.869\text{kmol}$ for $N_m$ in Equation (III).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.07375\text{kmol}}{8.869\text{kmol}}\\ =0.00832\end{array}$

Write the chemical reaction equation for complete combustion.

$\left\{\begin{array}{l}0.6333\text{C}+0.2994{\text{H}}_2\\ +0.05323{\text{O}}_2+0.00580{\text{N}}_2\\ +0.00832\text{S}+a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_2\right)\end{array}\right\}\to \left(\begin{array}{l}x{\text{CO}}_2+y{\text{H}}_2\text{O}\\ +z{\text{SO}}_2+k{\text{N}}_2\end{array}\right)$ (VIII)

Balance for the Carbon from Combustion reaction Equation (VIII).

$x=0.6333$

Balance for the Hydrogen from Combustion reaction Equation (VIII).

$y=0.2994$

Balance for the Sulphur from Combustion reaction Equation (VIII).

$z=0.00832$

Balance for the Oxygen from Combustion reaction Equation (VIII).

$0.05323+a_{\text{th}}=x+0.5y+z$

Substitute $0.6333$ for $x$, $0.2994$ for $y$, and $0.00832$ for $z$.

$\begin{array}{l}0.05323+a_{\text{th}}=0.6333+0.5\left(0.2994\right)+0.00832\\ a_{\text{th}}=0.7381\end{array}$

Balance for the Nitrogen from Combustion reaction Equation (VIII).

$k=0.00580+3.76a_{\text{th}}$

Substitute $0.7381$ for $a_{\text{th}}$.

$\begin{array}{l}k=0.00580+3.76\left(0.7381\right)\\ k=2.781\end{array}$

Rewrite the complete balanced chemical reaction for combustion as follows:

$\left\{\begin{array}{l}0.6333\text{C}+0.2994{\text{H}}_2\\ +0.05323{\text{O}}_2+0.00580{\text{N}}_2\\ +0.00832\text{S}+0.7381\left({\text{O}}_2+3.76{\text{N}}_2\right)\end{array}\right\}\to \left(\begin{array}{l}0.6333{\text{CO}}_2+0.2994{\text{H}}_2\text{O}\\ +0.00832{\text{SO}}_2+2.781{\text{N}}_2\end{array}\right)$ (IX)

Substitute $0.6333\text{kmol}$ for $N_{\text{C}}$, $0.2994\text{kmol}$ for $N_{{\text{H}}_2}$, $0.00832\text{kmol}$ for $N_{\text{S}}$, $0.05323\text{kmol}$ for $N_{{\text{O}}_2}$, and $0.00580\text{kmol}$ for $N_{{\text{N}}_2}$ in Equation (II).

$\begin{array}{c}{\left(N_m\right)}_{\text{reactants}}=\left(\begin{array}{c}0.6333\text{kmol}+0.2994\text{kmol}+0.00832\text{kmol}\\ +0.05323\text{kmol}+0.000580\text{kmol}\end{array}\right)\\ \simeq 1.0\text{kmol}\end{array}$

Substitute $0.6333\text{kmol}$ for $N_{\text{C}}$, $0.2994\text{kmol}$ for $N_{{\text{H}}_2}$, $0.00832\text{kmol}$ for $N_{\text{S}}$, $0.05323\text{kmol}$ for $N_{{\text{O}}_2}$, $0.00580\text{kmol}$ for $N_{{\text{N}}_2}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_2}$, $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$, and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (IV).

$\begin{array}{c}{m}_{\text{total}}=\left\{\begin{array}{l}\left(0.6333\text{kmol}\right)12\text{kg}/\text{kmol}+\left(0.2994\text{kmol}\right)2\text{kg}/\text{kmol}\\ +\left(0.05323\text{kmol}\right)32\text{kg}/\text{kmol}+\left(0.00580\text{kmol}\right)28\text{kg}/\text{kmol}\\ +\left(0.00832\text{kmol}\right)32\text{kg}/\text{kmol}\end{array}\right\}\\ =10.3304\text{kg}\end{array}$

Substitute $10.3304\text{kg}$ for $m_m$ and $1\text{kmol}$ for $N_m$ in Equation (V).

$\begin{array}{c}{M}_m=\frac{10.3304\text{kg}}{1\text{kmol}}\\ =10.33\text{kg}/\text{kmol}\end{array}$

From the Table A-26 of “Enthalpy of formation, Gibbs function of formation, and the absolute entropy at $25°\text{C, 1 atm}$”, obtain the enthalpy of formations as,

$\begin{array}{l}{\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}=0\\ {\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}=-393,520\text{kJ}/\text{kmol}\\ {\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}l}=-285,830\text{kJ}/\text{kmol}\\ {\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}g}=-241,820\text{kJ}/\text{kmol}\\ {\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}=-297,100\text{kJ}/\text{kmol}\end{array}$

Here, the enthalpy of formation of ${\text{N}}_2$ is ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$, the enthalpy of formation of ${\text{CO}}_2$ is ${\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}$, the enthalpy of formation for ${\text{H}}_2\text{O}$ in vapor is ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}g}$, the enthalpy of formation for ${\text{H}}_2\text{O}$ in liquid is ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}l}$, and the enthalpy of formation of ${\text{SO}}_2$ is ${\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}$.

Rewrite the Equation (VI) for the higher heating value.

$h_C=N_{{\text{CO}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}+N_{{\text{H}}_2\text{O}}{\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}l}+N_{{\text{SO}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}+N_{{\text{N}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$ (X)

Substitute $0.6333\text{kmol}$ for $N_{{\text{CO}}_2}$, $-393,520\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}$, $-285,830\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}l}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$, $0.2994\text{kmol}$ for $N_{{\text{H}}_2\text{O}}$, $-297,100\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}$, and $0.00832\text{kmol}$ for $N_{{\text{SO}}_2}$ in Equation (X).

$\begin{array}{l}{h}_C=\left\{\begin{array}{l}0.6333\text{kmol}\left(-393,520\text{kJ}/\text{kmol}\right)\\ +0.2994\text{kmol}\left(-285,830\text{kJ}/\text{kmol}\right)\\ +0.00832\text{kmol}\left(-297,100\text{kJ}/\text{kmol}\right)+N_{{\text{N}}_2}\left(0\right)\end{array}\right\}\\ h_C=-337,270\text{kJ}/\text{kmol coal}\end{array}$

Substitute $10.33\text{kg}/\text{kmol}$ for $M_m$ and $-337,270\text{kJ}/\text{kmol coal}$ for $h_C$ in Equation (VII).

$\begin{array}{c}\text{HHV}=\frac{-\left(-337,270\text{kJ}/\text{kmol coal}\right)}{10.33\text{kg}/\text{kmol}}\\ \simeq 33,650\text{kJ}/\text{kg coal}\end{array}$

Thus, the higher heating value of the coal is $33,650\text{kJ}/\text{kg coal}$.

Rewrite the Equation (VI) for the higher heating value.

$h_C=N_{{\text{CO}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}+N_{{\text{H}}_2\text{O}}{\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}g}+N_{{\text{SO}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}+N_{{\text{N}}_2}{\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$ (XI)

Substitute $0.6333\text{kmol}$ for $N_{{\text{CO}}_2}$, $-393,520\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}$, $-241,820\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O,}g}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$, $0.2994\text{kmol}$ for $N_{{\text{H}}_2\text{O}}$, $-297,100\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{SO}}_2}$, and $0.00832\text{kmol}$ for $N_{{\text{SO}}_2}$ in Equation (XI).

$\begin{array}{l}{h}_C=\left\{\begin{array}{l}0.6333\text{kmol}\left(-393,520\text{kJ}/\text{kmol}\right)\\ +0.2994\text{kmol}\left(-241,820\text{kJ}/\text{kmol}\right)\\ +0.00832\text{kmol}\left(-297,100\text{kJ}/\text{kmol}\right)+N_{{\text{N}}_2}\left(0\right)\end{array}\right\}\\ h_C=-324,090\text{kJ}/\text{kmol coal}\end{array}$

Substitute $10.33\text{kg}/\text{kmol}$ for $M_m$ and $-324,090\text{kJ}/\text{kmol coal}$ for $h_C$ in Equation (VII).

$\begin{array}{c}\text{LHV}=\frac{-\left(-324,090\text{kJ}/\text{kmol coal}\right)}{10.33\text{kg}/\text{kmol}}\\ \simeq 31,370\text{kJ}/\text{kg coal}\end{array}$

Thus, the lower heating value of the coal is $31,370\text{kJ}/\text{kg coal}$.