Chapter 15.7, Problem 82P

To determine

The estimated temperature of the combustion products.

Answer to Problem 82P

The estimated temperature of the combustion products is $\text{1956}\text{K}$.

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)

Apply energy balance under steady flow conditions on the combustion chamber.

$\begin{array}{c}{\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_R\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum N_R}{\overline{h}}_{f,R}^{\text{o}}\end{array}$ (XVIII)

Here, number of moles of products is $N_P$, number of moles of reactants is $N_R$, enthalpy of formation is ${\overline{h}}_f$.

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 $8.62$ for $m_{\text{ash}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{total}}=100-8.62\\ =91.38\text{kg}\end{array}$

Substitute $\text{79}\text{.61}\text{kg}$ for $m_{\text{C}}$ and $\text{91}\text{.38}\text{kg}$ for $m_{\text{total}}$ in Equation (II).

$\begin{array}{c}m{f}_{\text{C}}=\frac{\text{79}\text{.61}\text{kg}}{\text{91}\text{.38}\text{kg}}\\ =0.8712\\ =87.12\text{kg}\end{array}$

Substitute $\text{4}\text{.66}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $\text{91}\text{.38}\text{kg}$ for $m_{\text{total}}$ in Equation (III).

$\begin{array}{c}m{f}_{{\text{H}}_{\text{2}}}=\frac{\text{4}\text{.66}\text{kg}}{\text{91}\text{.38}\text{kg}}\\ =0.05100\\ =5.10\text{kg}\end{array}$

Substitute $\text{4}\text{.76}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $\text{91}\text{.38}\text{kg}$ for $m_{\text{total}}$ in Equation (IV).

$\begin{array}{c}m{f}_{{\text{O}}_{\text{2}}}=\frac{\text{4}\text{.76}\text{kg}}{\text{91}\text{.38}\text{kg}}\\ =0.05209\\ =5.209\text{kg}\end{array}$

Substitute $\text{1}\text{.83}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $\text{91}\text{.38}\text{kg}$ for $m_{\text{total}}$ in Equation (V).

$\begin{array}{c}m{f}_{{\text{N}}_{\text{2}}}=\frac{\text{1}\text{.83}\text{kg}}{\text{91}\text{.38}\text{kg}}\\ =0.02003\\ =2.003\text{kg}\end{array}$

Substitute $\text{0}\text{.52}\text{kg}$ for $m_{\text{S}}$ and $\text{91}\text{.38}\text{kg}$ for $m_{\text{total}}$ in Equation (VI).

$\begin{array}{c}m{f}_{\text{S}}=\frac{\text{0}\text{.52}\text{kg}}{\text{91}\text{.38}\text{kg}}\\ =0.00569\\ =0.569\text{kg}\end{array}$

Substitute $\text{87}\text{.12}\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (VII).

$\begin{array}{c}{N}_{\text{C}}=\frac{\text{87}\text{.12}\text{kg}}{12\text{kg}/\text{kmol}}\\ =7.26\text{kmol}\end{array}$

Substitute $\text{5}\text{.10}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (VIII).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{5}\text{.10}\text{kg}}{2\text{kg}/\text{kmol}}\\ =2.55\text{kmol}\end{array}$

Substitute $\text{5}\text{.209}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\text{5}\text{.209}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.1628\text{kmol}\end{array}$

Substitute $\text{2}\text{.003}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (X).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{\text{2}\text{.003}\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.07154\text{kmol}\end{array}$

Substitute $\text{0}\text{.569}\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (XI).

$\begin{array}{c}{N}_{\text{S}}=\frac{\text{0}\text{.569}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.01778\text{kmol}\end{array}$

Substitute $7.26\text{kmol}$ for $N_{\text{C}}$, $2.55\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.1628\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.07154\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.01778\text{kmol}$ for $N_{\text{S}}$ in Equation (XII).

$\begin{array}{c}{N}_m=7.26\text{kmol}+2.55\text{kmol}+0.1628\text{kmol}+0.07154\text{kmol}+0.01778\text{kmol}\\ =10.06\text{kmol}\end{array}$

Substitute $7.26\text{kmol}$ for $N_{\text{C}}$ and $10.06\text{kmol}$ for $N_m$ in Equation (XIII).

$\begin{array}{c}{y}_{\text{C}}=\frac{7.26\text{kmol}}{10.06\text{kmol}}\\ =0.7125\end{array}$

Substitute $2.55\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$ and $10.06\text{kmol}$ for $N_m$ in Equation (XIV).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}}=\frac{2.55\text{kmol}}{10.06\text{kmol}}\\ =0.2535\end{array}$

Substitute $0.1628\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $10.06\text{kmol}$ for $N_m$ in Equation (XV).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{0.1628\text{kmol}}{10.06\text{kmol}}\\ =0.01618\end{array}$

Substitute $0.07154\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $10.06\text{kmol}$ for $N_m$ in Equation (XVI).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.07154\text{kmol}}{10.06\text{kmol}}\\ =0.00711\end{array}$

Substitute $0.01778\text{kmol}$ for $N_{\text{S}}$ and $10.06\text{kmol}$ for $N_m$ in Equation (XVII).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.01778\text{kmol}}{10.06\text{kmol}}\\ =0.00177\end{array}$

Express the combustion equation.

$\left[\begin{array}{l}0.7215\text{C}+0.2535{\text{H}}_{\text{2}}+0.01618{\text{O}}_{\text{2}}+0.00711{\text{N}}_{\text{2}}+0.00177\text{S}\\ +1.5a_{\text{th}}\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to x{\text{CO}}_{\text{2}}+y{\text{H}}_{\text{2}}\text{O}+z{\text{SO}}_2+k{\text{N}}_{\text{2}}+0.5a_{\text{th}}{\text{O}}_{\text{2}}\end{array}\right]$ (XIX)

Perform the species balancing:

Carbon balance:

$x=0.7215$

Hydrogen balance:

$y=0.2535$

Sulphur balance:

$z=0.00177$

Oxygen balance:

$\begin{array}{l}0.01618+1.5a_{\text{th}}=x+0.5y+z+0.5a_{\text{th}}\\ 0.01618+1.5a_{\text{th}}=0.7215+0.5\left(0.2535\right)+0.00177+0.5a_{\text{th}}\\ a_{\text{th}}=0.7215+0.12675+0.00177-0.01618\\ a_{\text{th}}=0.8339\end{array}$

Nitrogen balance:

$\begin{array}{l}0.00711+1.5\left(3.76a_{\text{th}}\right)=k\\ k=0.00711+1.5\left(3.76\times 0.8339\right)\\ k=4.710\end{array}$

Substitute $0.7215$ for $x$, $0.2535$ for $y$, $0.00177$ for $z$, $0.8339$ for $a_{\text{th}}$ and $4.710$ for $k$ in Equation (XIX).

$\begin{array}{l}\left[\begin{array}{l}0.7215\text{C}+0.2535{\text{H}}_{\text{2}}+0.01618{\text{O}}_{\text{2}}+0.00711{\text{N}}_{\text{2}}+0.00177\text{S}\\ +1.5\left(0.8339\right)\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.7215{\text{CO}}_{\text{2}}+0.2535{\text{H}}_{\text{2}}\text{O}+0.00177{\text{SO}}_2\\ +4.710{\text{N}}_{\text{2}}+0.5\left(0.8339\right){\text{O}}_{\text{2}}\end{array}\right]\\ \left[\begin{array}{l}0.7215\text{C}+0.2535{\text{H}}_{\text{2}}+0.01618{\text{O}}_{\text{2}}+0.00711{\text{N}}_{\text{2}}+0.00177\text{S}\\ +1.2509\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.7215{\text{CO}}_{\text{2}}+0.2535{\text{H}}_{\text{2}}\text{O}+0.00177{\text{SO}}_2\\ +4.710{\text{N}}_{\text{2}}+0.4170{\text{O}}_{\text{2}}\end{array}\right]\end{array}$ (XX)

Refer Equation (XX), and write the number of moles of products and reactants.

$\begin{array}{l}{N}_{\text{C}}=0.7215\text{kmol}\\ N_{{\text{H}}_{\text{2}}}=0.2535\text{kmol}\\ N_{{\text{O}}_2}=0.01617\text{kmol}\end{array}$

$\begin{array}{l}{N}_{{\text{N}}_2}=\text{0}\text{.00711}\text{kmol}\\ N_{\text{S}}=0.00177\text{kmol}\end{array}$

Refer Appendix Table A-18, A-19, A-20 and A-23 and write the property table for products and reactants as in Table (1).

Substance	$\overline{h_f^{\text{o}}}$ $\left(\text{kJ}/\text{kmol}\right)$	${\overline{h}}_{298\text{K}}$ $\left(\text{kJ}/\text{kmol}\right)$	${\overline{h}}_{400\text{K}}$ $\left(\text{kJ}/\text{kmol}\right)$
${\text{O}}_{\text{2}}$	0	8682	11,171
${\text{N}}_{\text{2}}$	0	8669	11,640
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	$-241,820$	9904
${\text{CO}}_{\text{2}}$	$-393,520$	9364

Substitute the values form Table (I) into Equation (XVIII) to get,

$\begin{array}{l}\left[\begin{array}{l}0.7215\left(-393,520+{\overline{h}}_{{\text{CO}}_{\text{2}}}-9364\right)\\ +\left(0.2535\right)\left(-241,820+{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}-9904\right)\\ +\left(0.4170\right)\left(0+{\overline{h}}_{{\text{O}}_2}-8682\right)+\left(4.71\right)\left(0+{\overline{h}}_{{\text{N}}_2}-8669\right)\\ =\left(1.2509\right)\left(0+11,711-8682\right)\\ +\left(4.703\right)\left(11,640-8669\right)\end{array}\right]\\ 0.7215{\overline{h}}_{{\text{CO}}_{\text{2}}}+0.2535{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+0.4170{\overline{h}}_{{\text{O}}_2}+4.71{\overline{h}}_{{\text{N}}_2}=416,706\text{kJ}\end{array}$ (XXI)

Perform trial and error method to balance the Equation (XXI).

Iteration I:

Take $T_P=2000\text{K}$

$\begin{array}{c}\left[\begin{array}{c}0.7215{\overline{h}}_{{\text{CO}}_{\text{2}}}+0.2535{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}\\ +0.4170{\overline{h}}_{{\text{O}}_2}+4.71{\overline{h}}_{{\text{N}}_2}\end{array}\right]=\left[\begin{array}{l}\left(0.7125\right)\left(100,804\right)+\left(0.2535\right)\left(82,593\right)\\ +\left(0.4170\right)\left(67,881\right)+\left(4.71\right)\left(64,810\right)\end{array}\right]\\ =427,229\text{kJ}\left(\text{higher}\text{than}\text{416,706}\text{kJ}\right)\end{array}$

Iteration II:

Take $T_{\text{prod}}=1980\text{K}$

$\begin{array}{c}\left[\begin{array}{c}0.7215{\overline{h}}_{{\text{CO}}_{\text{2}}}+0.2535{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}\\ +0.4170{\overline{h}}_{{\text{O}}_2}+4.71{\overline{h}}_{{\text{N}}_2}\end{array}\right]=\left[\begin{array}{l}\left(0.7125\right)\left(99,606\right)+\left(0.2535\right)\left(81,573\right)\\ +\left(0.4170\right)\left(67,127\right)+\left(4.71\right)\left(64,090\right)\end{array}\right]\\ =422,400\text{kJ}\left(\text{higher}\text{than}\text{416,706}\text{kJ}\right)\end{array}$

Perform the interpolation method to obtain the adiabatic flame temperature of the product gases.

Write the formula of interpolation method of two variables.

$y_1=\frac{\left(x_2-x_1\right)\left(y_2-y_3\right)}{\left(x_3-x_1\right)}+y_2$ (XXII)

Here, the variables denote by x and y is enthalpy and estimated temperature respectively

Show the adiabatic flame temperature corresponding to enthalpy as in Table (1).

Enthalpy $\overline{h}\left(\text{kJ}\right)$	Estimated temperature $T_P\left(\text{K}\right)$
$416,706$$\left(x_1\right)$	$\left(y_1=?\right)$
$422,400$$\left(x_2\right)$	1980 $\left(y_2\right)$
$427,229$$\left(x_3\right)$	2000$\left(y_3\right)$

Substitute $416,706\text{kJ}$ for $x_1$, $422,400\text{kJ}$ for $x_2$, $427,229\text{kJ}$ for $x_3$, $\text{1980}\text{K}$ for $y_2$ and $\text{2000}\text{K}$ for $y_3$ in Equation (XXII).

$\begin{array}{c}{y}_1=\frac{\left(422,400\text{kJ}-416,706\text{kJ}\right)\left(\text{1980}\text{K}-\text{2000}\text{K}\right)}{\left(427,229\text{kJ}-416,706\text{kJ}\right)}+\text{1980}\text{K}\\ =\text{1956}\text{K}\end{array}$

Thus, the estimated temperature of the combustion product gases is,

$T_P=1956\text{K}$

Here, estimated temperature of the combustion product gases is $T_P$.

Hence, the estimated temperature of the combustion products is $\text{1956}\text{K}$.