Chapter 15.7, Problem 58P

To determine

The heat transfer of fuel in the boiler.

Answer to Problem 58P

The heat transfer of fuel in the boiler is $23,022\text{kJ}/\text{kg}\text{fuel}$.

Explanation of Solution

Express the number of moles of carbon.

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

Here, molar mass of carbon is $M_{\text{C}}$ and mole fraction of carbon is $m{f}_{\text{C}}$.

Express the number of moles of hydrogen.

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

Here, molar mass of hydrogen is $M_{{\text{H}}_{\text{2}}}$ and mole fraction of hydrogen is $m{f}_{{\text{H}}_{\text{2}}}$.

Express the number of moles of oxygen.

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

Here, molar mass of oxygen is $M_{{\text{O}}_{\text{2}}}$ and mole fraction of oxygen is $m{f}_{{\text{O}}_{\text{2}}}$.

Express the number of moles of nitrogen.

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

Here, molar mass of nitrogen is $M_{{\text{N}}_{\text{2}}}$ and mole fraction of nitrogen is $m{f}_{{\text{N}}_{\text{2}}}$.

Express the number of moles of sulfur.

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

Here, molar mass of sulphur is $M_{\text{S}}$ and mole fraction of sulfur is $m{f}_{\text{S}}$.

Express the total number of moles.

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

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

Express the mole fraction of carbon.

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

Express the mole fraction of hydrogen.

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

Express the mole fraction of oxygen.

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

Express the mole fraction of nitrogen.

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

Express the mole fraction of sulfur.

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

Express the heat transfer from the combustion chamber.

$\begin{array}{c}-{\overline{Q}}_{\text{out}}={\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P-{\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_R\\ ={\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+{\overline{h}}_{400\text{K}}-{\overline{h}}_{298\text{K}}^{\text{o}}\right)}_P-{\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+{\overline{h}}_{400\text{K}}-{\overline{h}}_{298\text{K}}^{\text{o}}\right)}_R\end{array}$ (XII)

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

Express the enthalpy change of sulfur dioxide between the standard temperature and the product temperature.

$\begin{array}{c}\Delta {\overline{h}}_{{\text{SO}}_{\text{2}}}=c_p\Delta T\\ =c_p\left(T_2-T_{\text{stand}}\right)\end{array}$ (XIII)

Here, specific heat constant pressure is $c_p$ net temperature is $\Delta T$, temperature at state 2 is $T_2$ and temperature at standard condition is $T_{\text{stand}}$.

Express the heat transfer of fuel in the boiler.

$Q_{\text{out}}=\frac{{\overline{Q}}_{\text{out}}}{N_{\text{C}}{M}_{\text{C}}+N_{{\text{H}}_{\text{2}}}{M}_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}{M}_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_2}+N_{\text{S}}{M}_{\text{S}}}$ (XIV)

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

Conclusion:

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

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

$M_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}$

Substitute $\text{39}\text{.25}\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (I).

$\begin{array}{c}{N}_{\text{C}}=\frac{39.25\text{kg}}{12\text{kg}/\text{kmol}}\\ =3.271\text{kmol}\end{array}$

Substitute $\text{6}\text{.93}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (II).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{6}\text{.93}\text{kg}}{2\text{kg}/\text{kmol}}\\ =3.465\text{kmol}\end{array}$

Substitute $\text{41}\text{.11}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (III).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\text{41}\text{.11}\text{kg}}{32\text{kg}/\text{kmol}}\\ =1.285\text{kmol}\end{array}$

Substitute $\text{0}\text{.72}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (IV).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{\text{0}\text{.72}\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.0257\text{kmol}\end{array}$

Substitute $\text{0}\text{.79}\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (V).

$\begin{array}{c}{N}_{\text{S}}=\frac{\text{0}\text{.79}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.0247\text{kmol}\end{array}$

Substitute $3.271\text{kmol}$ for $N_{\text{C}}$, $3.465\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $1.285\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.0257\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.00247\text{kmol}$ for $N_{\text{S}}$ in Equation (VI).

$\begin{array}{c}{N}_m=3.271\text{kmol}+3.465\text{kmol}+1.285\text{kmol}+0.0257\text{kmol}+0.0247\text{kmol}\\ =8.071\text{kmol}\end{array}$

Substitute $3.271\text{kmol}$ for $N_{\text{C}}$ and $8.071\text{kmol}$ for $N_m$ in Equation (VII).

$\begin{array}{c}{y}_{\text{C}}=\frac{3.271\text{kmol}}{8.071\text{kmol}}\\ =0.4052\end{array}$

Substitute $3.465\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$ and $8.071\text{kmol}$ for $N_m$ in Equation (VIII).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}}=\frac{3.465\text{kmol}}{8.071\text{kmol}}\\ =0.4293\end{array}$

Substitute $1.285\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $8.071\text{kmol}$ for $N_m$ in Equation (IX).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{1.285\text{kmol}}{8.071\text{kmol}}\\ =0.1592\end{array}$

Substitute $0.0257\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $8.071\text{kmol}$ for $N_m$ in Equation (X).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.0257\text{kmol}}{8.071\text{kmol}}\\ =0.00319\end{array}$

Substitute $0.0247\text{kmol}$ for $N_{\text{S}}$ and $8.071\text{kmol}$ for $N_m$ in Equation (XI).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.0247\text{kmol}}{8.071\text{kmol}}\\ =0.00306\end{array}$

Express the combustion equation.

$\left[\left\{\begin{array}{l}0.4052\text{C}+0.4293{\text{H}}_{\text{2}}+0.1592{\text{O}}_{\text{2}}\\ +0.00319{\text{N}}_{\text{2}}+0.00306\text{S}\\ +1.4a_{\text{th}}\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\end{array}\right\}\to \left\{\begin{array}{l}0.4052{\text{CO}}_{\text{2}}+0.4293{\text{H}}_{\text{2}}\text{O}\\ +0.4a_{\text{th}}{\text{O}}_{\text{2}}+0.00306{\text{SO}}_{\text{2}}\\ +1.4a_{\text{th}}\times 3.76{\text{N}}_{\text{2}}\end{array}\right\}\right]$ (XV)

Here, stoichiometric coefficient of air is $a_{\text{th}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, carbon dioxide is ${\text{CO}}_{\text{2}}$, water is ${\text{H}}_{\text{2}}\text{O}$ and hydrogen is ${\text{H}}_{\text{2}}$.

Express the stoichiometric coefficient of air by ${\text{O}}_{\text{2}}$ balancing.

$\begin{array}{l}0.1592+1.4a_{\text{th}}=0.4052+0.5\times 0.4293+0.4a_{\text{th}}+0.00306\\ a_{\text{th}}=0.4637\end{array}$

Substitute $0.4637$ for $a_{\text{th}}$ in Equation (XV).

$\begin{array}{l}\left[\left\{\begin{array}{l}0.4052\text{C}+0.4293{\text{H}}_{\text{2}}+0.1592{\text{O}}_{\text{2}}\\ +0.00319{\text{N}}_{\text{2}}+0.00306\text{S}\\ +1.4\times 0.4637\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\end{array}\right\}\to \left\{\begin{array}{l}0.4052{\text{CO}}_{\text{2}}+0.4293{\text{H}}_{\text{2}}\text{O}\\ +0.4\times 0.4637{\text{O}}_{\text{2}}\\ +0.00306{\text{SO}}_{\text{2}}\\ +1.4\times 0.4637\times 3.76{\text{N}}_{\text{2}}\end{array}\right\}\right]\\ \left[\left\{\begin{array}{l}0.4052\text{C}+0.4293{\text{H}}_{\text{2}}+0.1592{\text{O}}_{\text{2}}\\ +0.00319{\text{N}}_{\text{2}}+0.00306\text{S}\\ +0.6492\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\end{array}\right\}\to \left\{\begin{array}{l}0.4052{\text{CO}}_{\text{2}}+0.4293{\text{H}}_{\text{2}}\text{O}\\ +0.1855{\text{O}}_{\text{2}}+0.00306{\text{SO}}_{\text{2}}\\ +2.441{\text{N}}_{\text{2}}\end{array}\right\}\right]\end{array}$ (XVI)

Refer Equation (XVI) and write the number of moles of reactants.

$\begin{array}{l}{M}_{\text{C}}=0.4052\text{kmol}\\ M_{{\text{H}}_{\text{2}}}=0.4293\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=0.15920\text{kmol}\end{array}$

$\begin{array}{l}{M}_{{\text{N}}_{\text{2}}}=0.00319\text{kmol}\\ M_{\text{S}}=0.00306\text{kmol}\end{array}$

Perform unit conversion of temperature at state 2 from degree Celsius to Kelvin.

$\begin{array}{c}{T}_2=127\text{°C}\\ =\left(127+273\right)\text{K}\\ =\text{400}\text{K}\end{array}$

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

$\begin{array}{l}{N}_{P,{\text{CO}}_2}=0.4052\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=0.4293\text{kmol}\\ N_{P,{\text{O}}_2}=0.1855\text{kmol}\\ N_{P,{\text{SO}}_2}=0.00306\text{kmol}\end{array}$

$N_{P,{\text{N}}_{\text{2}}}=2.441\text{kmol}$

Here, number of moles of product carbon dioxide, water, oxygen, sulfur dioxide and nitrogen is $N_{P,{\text{CO}}_2},N_{P,{\text{H}}_{\text{2}}\text{O}},N_{P,{\text{O}}_2},N_{P,{\text{SO}}_{\text{2}}}\text{and}{N}_{P,{\text{N}}_{\text{2}}}$ respectively.

Take the temperature at standard condition as,

$T_{\text{stand}}=25\text{°C}$

Substitute $\text{41}\text{.7}\text{kJ}/\text{kmol}\cdot \text{K}$ for $c_p$ and $\text{127°C}$ for $T_2$ and $25\text{°C}$ for $T_{\text{stand}}$ in Equation (XIII).

$\begin{array}{c}\Delta {\overline{h}}_{{\text{SO}}_{\text{2}}}=\left(\text{41}\text{.7}\text{kJ}/\text{kmol}\cdot \text{K}\right)\left(\text{127°C}-25\text{°C}\right)\\ =\left(\text{41}\text{.7}\text{kJ}/\text{kmol}\cdot \text{K}\right)\left[\left(\text{127}-25\right)\text{K}\right]\\ =4253\text{kJ}/\text{kmol}\end{array}$

Refer Appendix Table A-18, A-19, A-20, A-23 and A-26 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	11,711
${\text{N}}_{\text{2}}$	0	8669	11,640
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	$-241,820$	9904	13,356
${\text{CO}}_{\text{2}}$	$-393,520$	9364	13,372
${\text{SO}}_2$	$-297,100$

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

$\begin{array}{c}-{\overline{Q}}_{\text{out}}=\left\{\begin{array}{l}\left[\left(0.4052\right)\left(-393,520+13,372-9364\right)\right]+\left[\begin{array}{l}\left(04293\right)\\ \left(-241,820+13,356-9904\right)\end{array}\right]\\ +\left[\left(0.1855\right)\left(0+11,711-8682\right)\right]+\left[\left(2.441\right)\left(0+11,640-8669\right)\right]\\ +\left[\left(0.00306\right)\left(-297,100+4253-0\right)\right]\end{array}\right\}\\ =-253,244\text{kJ}/\text{kmol}\text{fuel}\\ {\overline{Q}}_{\text{out}}=253,244\text{kJ}/\text{kmol}\text{fuel}\end{array}$

Substitute $253,244\text{kJ}/\text{kmol}\text{fuel}$ for ${\overline{Q}}_{\text{out}}$, $0.4052\text{kmol}$ for $M_{\text{C}}$, $0.4293\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$, $0.15920\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $0.00319\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$, $0.00306\text{kmol}$ for $M_{\text{S}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (XIV).

$\begin{array}{c}{Q}_{\text{out}}=\frac{253,244\text{kJ}/\text{kmol}\text{fuel}}{\left[\begin{array}{l}\left(0.4052\times 12\right)+\left(0.4293\times 2\right)+\left(0.1592\times 32\right)\\ +\left(0.00319\times 28\right)+\left(0.00306\times 32\right)\end{array}\right]\text{kg}/\text{kmol}}\\ =\frac{253,244\text{kJ}/\text{kmol}\text{fuel}}{11\text{kg}/\text{kmol}}\\ =23,022\text{kJ}/\text{kg}\text{fuel}\end{array}$

Hence, the heat transfer of fuel in the boiler is $23,022\text{kJ}/\text{kg}\text{fuel}$.