Chapter 16.6, Problem 35P

(a)

To determine

The equilibrium composition of the product gases from combustion of liquid propane.

Answer to Problem 35P

The equilibrium composition of the product gases from combustion of liquid propane is

$3{\text{CO}}_2+7.5{\text{O}}_2+4{\text{H}}_2\text{O}+47{\text{N}}_2$.

Explanation of Solution

Write the stoichiometric equation for combustion of 1 kmol of liquid propane $\left({\text{C}}_3{\text{H}}_8\right)$.

$\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(l\right)+a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to 3{\text{CO}}_2+4{\text{H}}_2\text{O}+3.76a_{\text{th}}{\text{N}}_2\end{array}$ (I)

Write the chemical reaction equation for dissociation of carbon dioxide.

${\text{CO}}_2\iff \text{CO}+\frac{1}{2}{\text{O}}_2$ (II)

Write the formula to calculate the equilibrium constant $\left(K_P\right)$ for the reaction in Equation (II).

$K_P=\frac{N_{\text{CO}}^{v_{\text{CO}}}{N}_{{\text{O}}_2}^{v_{{\text{O}}_2}}}{N_{{\text{CO}}_2}^{v_{{\text{CO}}_2}}}{\left(\frac{P}{N_{\text{total}}}\right)}^{\left(v_{\text{CO}}+v_{{\text{O}}_2}-v_{{\text{CO}}_2}\right)}$ (III)

Here, number of moles of ${\text{CO}}_2$ is $N_{{\text{CO}}_2}$, number of moles of $\text{CO}$ is $N_{\text{CO}}$, number of moles of ${\text{O}}_2$ is $N_{{\text{O}}_2}$, pressure is $P$, total number of moles of the mixture is $N_{\text{total}}$, stoichiometric coefficient of ${\text{CO}}_2$ is $v_{{\text{CO}}_2}$, stoichiometric coefficient of $\text{CO}$ is $v_{\text{CO}}$, and stoichiometric coefficient of ${\text{O}}_2$ is $v_{{\text{O}}_2}$.

Conclusion:

The stoichiometric coefficients from the Equation (II),

$\begin{array}{l}{v}_{\text{CO}}=1\\ v_{{\text{O}}_2}=0.5\\ v_{{\text{CO}}_2}=1\end{array}$

Balance for ${\text{O}}_2$ from the Equation (I).

$\begin{array}{l}2.5a_{\text{th}}=3+2+1.5a_{\text{th}}\\ a_{\text{th}}=5\end{array}$

Re-write the actual combustion equation for the combustion of 1 kmol of liquid propane $\left({\text{C}}_3{\text{H}}_8\right)$ with 150% excess air.

$\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(l\right)+2.5a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to x{\text{CO}}_2+\left(3-x\right)\text{CO}+\left(9-0.5x\right){\text{O}}_2+4{\text{H}}_2\text{O}+3.76\left(2.5a_{\text{th}}\right){\text{N}}_2\end{array}$

$\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(l\right)+12.5\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to x{\text{CO}}_2+\left(3-x\right)\text{CO}+\left(9-0.5x\right){\text{O}}_2+4{\text{H}}_2\text{O}+47{\text{N}}_2\end{array}$ (IV)

Calculate the total number of moles of products $\left(N_{\text{total}}\right)$.

$\begin{array}{c}{N}_{\text{total}}=x+\left(3-x\right)+\left(9-0.5x\right)+4+47\\ =63-0.5x\end{array}$

From the Table A-28 of “Natural logarithms of the equilibrium constant $K_P$”, the value of equilibrium constant $K_P$ at the temperature of $1200\text{K}$ is

$\begin{array}{l}\ln K_P=-17.871\\ K_P=1.73\times {10}^{-8}\end{array}$

Substitute $1.73\times {10}^{-8}$ for $K_P$, 1 for $v_{\text{CO}}$, 0.5 for $v_{{\text{O}}_2}$, 1 for $v_{{\text{CO}}_2}$, $2\text{ atm}$ for $P$, and $\left(63-0.5x\right)$ for $N_{\text{total}}$ in Equation (III), and rewrite the Equation (III) using Equation (IV).

$\begin{array}{l}1.73\times {10}^{-8}=\frac{\left(3-x\right){\left(9-0.5x\right)}^{0.5}}{x}{\left(\frac{2}{63-0.5x}\right)}^{\left(1+0.5-1\right)}\\ x\simeq 3\end{array}$

Substitute 3 for $x$ in Equation (IV).

$\begin{array}{l}{\text{C}}_3{\text{H}}_8\left(l\right)+12.5\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to 3{\text{CO}}_2+\left(3-3\right)\text{CO}+\left(9-0.5\times 3\right){\text{O}}_2+4{\text{H}}_2\text{O}+47{\text{N}}_2\\ {\text{C}}_3{\text{H}}_8\left(l\right)+12.5\left({\text{O}}_2+3.76{\text{N}}_2\right)\\ \to 3{\text{CO}}_2+7.5{\text{O}}_2+4{\text{H}}_2\text{O}+47{\text{N}}_2\end{array}$

Thus, the equilibrium composition of the product gases from combustion of liquid propane is

$3{\text{CO}}_2+7.5{\text{O}}_2+4{\text{H}}_2\text{O}+47{\text{N}}_2$.

(b)

To determine

The amount of heat transfer for the combustion of liquid propane.

Answer to Problem 35P

The amount of heat transfer for the combustion of liquid propane is $5066\text{kJ}/\min$.

Explanation of Solution

Write the energy balance equation to determine the heat transfer $\left(Q_{\text{out}}\right)$ for the combustion process.

$-Q_{\text{out}}={\displaystyle \sum N_P{\left({\overline{h}}_f^o+\overline{h}-{\overline{h}}^o\right)}_P-N_R{\left({\overline{h}}_f^o+\overline{h}-{\overline{h}}^o\right)}_R}$ (V)

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

Write the expression to calculate the molar flow rate of liquid propane $\left(\dot{N}\right)$.

$\dot{N}=\frac{\dot{m}}{M}$ (VI)

Here, mass flowrate of propane is $\dot{m}$ and the molar mass of propane is $M$.

Write the expression to calculate the rate of heat transfer $\left({\dot{Q}}_{\text{out}}\right)$.

${\dot{Q}}_{\text{out}}=\dot{N}{Q}_{\text{out}}$ (VII)

Conclusion:

Rewrite the Equation (V) for the combustion of liquid propane.

$-Q_{\text{out}}=\left\{\begin{array}{l}{\left[\begin{array}{l}{N}_{{\text{CO}}_2}{\left({\overline{h}}_f^o+{\overline{h}}_{1200\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{CO}}_2}\\ +N_{{\text{H}}_2\text{O}}{\left({\overline{h}}_f^o+{\overline{h}}_{1200\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{H}}_2\text{O}}\\ +N_{{\text{O}}_2}{\left({\overline{h}}_f^o+{\overline{h}}_{1200\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{O}}_2}\\ +N_{{\text{N}}_2}{\left({\overline{h}}_f^o+{\overline{h}}_{1200\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{N}}_2}\end{array}\right]}_P\\ -{\left[\begin{array}{l}{N}_{{\text{C}}_3{\text{H}}_8}{\left({\overline{h}}_f^o\right)}_{{\text{C}}_3{\text{H}}_8}+N_{{\text{O}}_2}{\left({\overline{h}}_f^o+{\overline{h}}_{285\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{O}}_2}\\ +N_{{\text{N}}_2}{\left({\overline{h}}_f^o+{\overline{h}}_{285\text{K}}-{\overline{h}}_{298\text{K}}\right)}_{{\text{N}}_2}\end{array}\right]}_R\end{array}\right\}$ (VIII)

Here, enthalpy of formation at $1200\text{K}$ is ${\overline{h}}_{1200\text{K}}$, enthalpy of formation at $298\text{K}$ is ${\overline{h}}_{\text{298 K}}$, number of moles of water is $N_{{\text{H}}_2\text{O}}$, number of moles of ${\text{N}}_2$ is $N_{{\text{N}}_2}$, and number of moles of liquid propane is $N_{{\text{C}}_3{\text{H}}_8}$.

From the Table A-18 through Table A-26, obtain the enthalpies of vaporization and enthalpies of formation for different substances as in Table 1.

Substance	Enthalpy of vaporization ${\overline{h}}_f^o\left(\text{kJ}/\text{kmol}\right)$	Enthalpy of formation at 285 K ${\overline{h}}_{285\text{K}}\left(\text{kJ}/\text{kmol}\right)$	Enthalpy of formation at 298 K ${\overline{h}}_{298\text{K}}\left(\text{kJ}/\text{kmol}\right)$	Enthalpy of formation at 1200 K ${\overline{h}}_{1200\text{K}}\left(\text{kJ}/\text{kmol}\right)$
${\text{C}}_3{\text{H}}_8$	$-118,910$	-	-	-
${\text{O}}_2$	0	8696.5	8682	38,447
${\text{N}}_2$	0	8286.5	8669	36,777
${\text{H}}_2\text{O}\left(g\right)$	$-241,820$	-	9904	44,380
${\text{CO}}_2$	$-393,520$	-	9364	53,848

Substitute $3\text{kmol}$ for $N_{{\text{CO}}_2}$, $-393,520\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{CO}}_2}$, $53,848\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{1200\text{K}}\right)}_{{\text{CO}}_2}$, $9364\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{298\text{K}}\right)}_{{\text{CO}}_2}$, $4\text{kmol}$ for $N_{{\text{H}}_2\text{O}}$, $-241,820\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{H}}_2\text{O}}$, $44,380\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{1200\text{K}}\right)}_{{\text{H}}_2\text{O}}$, $9904\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{298\text{K}}\right)}_{{\text{H}}_2\text{O}}$, $7.5\text{kmol}$ for $N_{{\text{O}}_2,P}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{O}}_2}$, $38,447\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{1200\text{K}}\right)}_{{\text{O}}_2}$, $8682\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{298\text{K}}\right)}_{{\text{O}}_2}$, $47\text{kmol}$ for $N_{{\text{N}}_2}$, 0 for ${\left({\overline{h}}_f^o\right)}_{{\text{N}}_2}$, $36,777\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{1200\text{K}}\right)}_{{\text{N}}_2}$, $8669\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{298\text{K}}\right)}_{{\text{N}}_2}$, $1\text{kmol}$ for $N_{{\text{C}}_3{\text{H}}_8}$, $-118,910\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_f^o\right)}_{{\text{C}}_3{\text{H}}_8}$, $12.5\text{kmol}$ for $N_{{\text{O}}_2,R}$, $8696.5\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{285\text{K}}\right)}_{{\text{O}}_2}$, and $8286.5\text{kJ}/\text{kmol}$ for ${\left({\overline{h}}_{285\text{K}}\right)}_{{\text{N}}_2}$ in Equation (VI).

$\begin{array}{c}-Q_{\text{out}}=\left\{\begin{array}{l}\left[\begin{array}{l}3\left(-393,520+53,848-9364\right)+4\left(-241,820+44,380-9904\right)\\ +7.5\left(0+38,447-8682\right)+47\left(0+36,777-8669\right)\end{array}\right]\\ -\left[\begin{array}{l}1\left(-118,910\right)+12.5\left(0+8696.5-8682\right)\\ +47\left(0+8286.5-8669\right)\end{array}\right]\end{array}\right\}\\ =-185,764\text{kJ}/{\text{kmol of C}}_3{\text{H}}_8\end{array}$

$Q_{\text{out}}=185,764\text{kJ}/{\text{kmol of C}}_3{\text{H}}_8$

Substitute $1.2\text{kg}/\min$ for $\dot{m}$ and $44\text{kg}/\text{kmol}$ for $M$ in Equation (VI).

$\begin{array}{c}\dot{N}=\frac{1.2\text{kg}/\min }{44\text{kg}/\text{kmol}}\\ =0.02727\text{kmol}/\min \end{array}$

Substitute $0.02727\text{kmol}/\min$ for $\dot{N}$ and $185,764\text{kJ}/{\text{kmol of C}}_3{\text{H}}_8$ for $Q_{\text{out}}$.

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(0.02727\text{kmol}/\min \right)\left(185,764\text{kJ}/{\text{kmol of C}}_3{\text{H}}_8\right)\\ \simeq 5066\text{kJ}/\min \end{array}$

Thus, the amount of heat transfer for the combustion of liquid propane is $5066\text{kJ}/\min$.