Chapter 15.7, Problem 106RP

Propane gas (C₃H₈) enters a steady-flow combustion chamber at 1 atm and 25°C and is burned with air that enters the combustion chamber at the same state. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 200 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 90 percent theoretical air.

To determine

The adiabatic flame temperature for complete combustion with 100 percent theoretical air.

Answer to Problem 106RP

The adiabatic flame temperature for complete combustion with 100 percent theoretical air is $\text{2394}\text{K}$.

Explanation of Solution

Express the theoretical combustion equation of propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right)$ with stoichiometric amount of air.

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right)+5\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\to 3{\text{CO}}_{\text{2}}+4{\text{H}}_{\text{2}}\text{O}+18.8{\text{N}}_{\text{2}}$ (I)

Here, propane is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, carbon dioxide is ${\text{CO}}_{\text{2}}$ and water is ${\text{H}}_{\text{2}}\text{O}$.

As the combustion process is a steady flow process, thus heat lost is equal to the heat gained.

$\begin{array}{l}{H}_{\text{prod}}=H_{\text{react}}\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_R\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h_T}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum {\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}}\end{array}$ (II)

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 is $\overline{h}$, enthalpy of products and reactants is $H_{\text{prod}}\text{and}{H}_{\text{react}}$

Conclusion:

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

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=3\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=4\text{kmol}\\ N_{P,{\text{N}}_{\text{2}}}=18.8\text{kmol}\\ N_{R,{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}=1\text{kmol}\end{array}$

Here, number of moles of products carbon dioxide, water and nitrogen is $N_{P,{\text{CO}}_{\text{2}}},N_{P,{\text{H}}_{\text{2}}\text{O}}\text{and}{N}_{P,{\text{N}}_{\text{2}}}$ respectively and number of mole of reactant propane is $N_{R,{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}$.

Refer Appendix Table A-18, A-19, A-20, A-21, A-23 and A-26 and write the property table for products and reactants as in Table (1).

Substance	$\overline{h_f^{\text{o}}}$ $\left(\text{kJ}/\text{kmol}\right)$	${\overline{h}}_{298\text{K}}$ $\left(\text{kJ}/\text{kmol}\right)$
${\text{C}}_3{\text{H}}_8\left(g\right)$	$-103,850$
${\text{O}}_{\text{2}}$	0	8682
${\text{N}}_{\text{2}}$	0	8669
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	$-241,820$	9904
$\text{CO}$	$-110,530$	8669
${\text{CO}}_{\text{2}}$	$-393,520$	9364

Substitute the values from Table (I) into Equation (II) to get,

$\begin{array}{l}\left[\begin{array}{l}\left(3\right)\left(-393,520+{\overline{h}}_{{\text{CO}}_{\text{2}}}-9364\right)+\left(4\right)\left(-241,820+{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}-9904\right)\\ +\left(18.8\right)\left(0+{\overline{h}}_{{\text{N}}_{\text{2}}}-8669\right)=\left(1\right)\left(-103,850\right)\end{array}\right]\\ 3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+18.8{\overline{h}}_{{\text{N}}_{\text{2}}}=2,274,675\text{kJ}\end{array}$ (III)

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

Iteration I:

Take $T_P=2400\text{K}$

$\begin{array}{c}3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+18.8{\overline{h}}_{{\text{N}}_{\text{2}}}=3\left(125,152\right)+4\left(103,508\right)+18.8\left(79,320\right)\\ =2,280,704\text{kJ}\left(\text{higher}\text{than}\text{2,274,675}\text{kJ}\right)\end{array}$

Iteration II:

Take $T_P=2350\text{K}$

$\begin{array}{c}3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+18.8{\overline{h}}_{{\text{N}}_{\text{2}}}=3\left(122,091\right)+4\left(100,846\right)+18.8\left(77,496\right)\\ =2,226,582\text{kJ}\left(\text{lower}\text{than}\text{2,274,675}\text{kJ}\right)\end{array}$

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

Write the formula of interpolation method of two variables.

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

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

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

Enthalpy $\overline{h}\left(\text{kJ}\right)$	Adiabatic flame temperature $T_P\left(\text{K}\right)$
$2,226,582$$\left(x_1\right)$	2350 $\left(y_1\right)$
$2,274,675$$\left(x_2\right)$	$\left(y_2=?\right)$
$2,280,704$$\left(x_3\right)$	2400 $\left(y_3\right)$

Substitute $2,226,582\text{kJ}$ for $x_1$, $2,274,675\text{kJ}$ for $x_2$, $2,280,704\text{kJ}$ for $x_3$, $\text{2350}\text{K}$ for $y_1$ and $\text{2400}\text{K}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(2,274,675\text{kJ}-2,226,582\text{kJ}\right)\left(\text{2400}\text{K}-\text{2350}\text{K}\right)}{\left(2,280,704\text{kJ}-2,226,582\text{kJ}\right)}+\text{2350}\text{K}\\ =\text{2394}\text{K}\end{array}$

Thus, the adiabatic flame temperature when burning with 100% theoretical air is,

$T_P=2394\text{K}$

Hence, the adiabatic flame temperature for complete combustion with 100 percent theoretical air is $\text{2394}\text{K}$.

To determine

The adiabatic flame temperature for complete combustion with 200 percent theoretical air.

Answer to Problem 106RP

The adiabatic flame temperature for complete combustion with 200 percent theoretical air is $\text{1510}\text{K}$.

Explanation of Solution

Express the balanced combustion equation of propane $\left({\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\right)$ with 200% theoretical air.

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right)+10\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\to 3{\text{CO}}_{\text{2}}+4{\text{H}}_{\text{2}}\text{O}+5{\text{O}}_{\text{2}}+37.6{\text{N}}_{\text{2}}$ (V)

As the combustion process is a steady flow process, thus heat lost is equal to the heat gained.

$\begin{array}{l}{H}_{\text{prod}}=H_{\text{react}}\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_R\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h_T}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum {\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}}\end{array}$ (VI)

Conclusion:

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

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=3\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=4\text{kmol}\\ N_{P,{\text{N}}_{\text{2}}}=37.6\text{kmol}\\ N_{P,{\text{O}}_{\text{2}}}=5\text{kmol}\end{array}$

$N_{R,{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}=1\text{kmol}$

Here, number of moles of products carbon dioxide, water nitrogen and oxygen is $N_{P,{\text{CO}}_{\text{2}}},N_{P,{\text{H}}_{\text{2}}\text{O}}\text{,}{N}_{P,{\text{N}}_{\text{2}}}\text{and}{N}_{P,{\text{O}}_2}$ respectively and number of mole of reactant propane is $N_{R,{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}$.

Substitute the values from Table (I) into Equation (VI) to get,

$\begin{array}{l}\left[\begin{array}{l}\left(3\right)\left(-393,520+{\overline{h}}_{{\text{CO}}_{\text{2}}}-9364\right)+\left(4\right)\left(-241,820+{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}-9904\right)\\ +\left(5\right)\left(0+{\overline{h}}_{{\text{O}}_{\text{2}}}-8682\right)+\left(37.6\right)\left(0+{\overline{h}}_{{\text{N}}_{\text{2}}}-8669\right)=\left(1\right)\left(-103,850\right)\end{array}\right]\\ 3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+5{\overline{h}}_{{\text{O}}_{\text{2}}}+37.6{\overline{h}}_{{\text{N}}_{\text{2}}}=2,481,060\text{kJ}\end{array}$ (VII)

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

Iteration I:

Take $T_P=1540\text{K}$

$\begin{array}{c}3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+5{\overline{h}}_{{\text{O}}_{\text{2}}}+37.6{\overline{h}}_{{\text{N}}_{\text{2}}}=\left[\begin{array}{l}\left(3\right)\left(73,417\right)+\left(4\right)\left(59,888\right)\\ +\left(5\right)\left(50,756\right)+\left(37.6\right)\left(48,470\right)\end{array}\right]\\ =2,536,055\text{kJ}\left(\text{higher}\text{than}2,481,060\text{kJ}\right)\end{array}$

Iteration II:

Take $T_P=1500\text{K}$

$\begin{array}{c}3{\overline{h}}_{{\text{CO}}_{\text{2}}}+4{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}+5{\overline{h}}_{{\text{O}}_{\text{2}}}+37.6{\overline{h}}_{{\text{N}}_{\text{2}}}=\left[\begin{array}{l}\left(3\right)\left(71,078\right)+\left(4\right)\left(57,999\right)\\ +\left(5\right)\left(49,292\right)+\left(37.6\right)\left(47,073\right)\end{array}\right]\\ =2,461,630\text{kJ}\left(\text{lower}\text{than}2,481,060\text{kJ}\right)\end{array}$

Show the adiabatic flame temperature corresponding to enthalpy as in Table (2).

Enthalpy $\overline{h}\left(\text{kJ}\right)$	Adiabatic flame temperature $T_P\left(\text{K}\right)$
$2,461,630$$\left(x_1\right)$	1500 $\left(y_1\right)$
$2,481,060$$\left(x_2\right)$	$\left(y_2=?\right)$
$2,536,055$$\left(x_3\right)$	1540 $\left(y_3\right)$

Substitute $2,461,630\text{kJ}$ for $x_1$, $2,481,060\text{kJ}$ for $x_2$, $2,536,055\text{kJ}$ for $x_3$, $\text{1500}\text{K}$ for $y_1$ and $\text{1540}\text{K}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(2,481,060\text{kJ}-2,461,630\text{kJ}\right)\left(\text{1540}\text{K}-\text{1500}\text{K}\right)}{\left(2,536,055\text{kJ}-2,461,630\text{kJ}\right)}+\text{1500}\text{K}\\ =\text{1510}\text{K}\end{array}$

Thus, the adiabatic flame temperature when burning with 200% theoretical air is,

$T_P=1510\text{K}$

Hence, the adiabatic flame temperature for complete combustion with 200 percent theoretical air is $\text{1510}\text{K}$.

To determine

The adiabatic flame temperature for incomplete combustion with 90 percent theoretical air.

Answer to Problem 106RP

The adiabatic flame temperature for incomplete combustion with 90 percent theoretical air is $\text{2287}\text{K}$.

Explanation of Solution

Express the balanced combustion equation for incomplete combustion with 90% theoretical air.

${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\left(g\right)+4.5\left({\text{O}}_{\text{2}}+3.76{\text{N}}_{\text{2}}\right)\to 2{\text{CO}}_{\text{2}}+1\text{CO}+4{\text{H}}_{\text{2}}\text{O}+16.92{\text{N}}_{\text{2}}$ (VIII)

As the combustion process is a steady flow process, thus heat lost is equal to the heat gained.

$\begin{array}{l}{H}_{\text{prod}}=H_{\text{react}}\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum N_R}{\left({\overline{h}}_f^{\text{o}}+\overline{h}-{\overline{h}}^{\text{o}}\right)}_R\\ {\displaystyle \sum N_P}{\left({\overline{h}}_f^{\text{o}}+\overline{h_T}-{\overline{h}}^{\text{o}}\right)}_P={\displaystyle \sum {\left(N{\overline{h}}_f^{\text{o}}\right)}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}}\end{array}$ (IX)

Conclusion:

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

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=2\text{kmol}\\ N_{P,\text{CO}}=1\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=4\text{kmol}\\ N_{P,{\text{N}}_{\text{2}}}=16.92\text{kmol}\end{array}$

$N_{R,{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}=1\text{kmol}$

Here, number of moles of products carbon monoxide is $N_{P,\text{CO}}$.

Substitute the values from Table (I) into Equation (IX) to get,

$\begin{array}{l}\left[\begin{array}{l}\left(2\right)\left(-393,520+{\overline{h}}_{{\text{CO}}_{\text{2}}}-9364\right)+\left(1\right)\left(-110,530+{\overline{h}}_{\text{CO}}-8669\right)\\ +\left(4\right)\left(-241,820+{\overline{h}}_{{\text{H}}_{\text{2}}\text{O}}-9904\right)+\left(16.92\right)\left(0+{\overline{h}}_{{\text{N}}_{\text{2}}}-8669\right)\\ =\left(1\right)\left(-103,850\right)\end{array}\right]\\ 2{\overline{h}}_{{\text{CO}}_{\text{2}}}+1{\overline{h}}_{\text{CO}}+4{\overline{h}}_{{\text{H}}_2\text{O}}+16.92{\overline{h}}_{{\text{N}}_{\text{2}}}=1,974,692\text{kJ}\end{array}$ (X)

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

Iteration I:

Take $T_P=2250\text{K}$

$\begin{array}{c}2{\overline{h}}_{{\text{CO}}_{\text{2}}}+1{\overline{h}}_{\text{CO}}+4{\overline{h}}_{{\text{H}}_2\text{O}}+16.92{\overline{h}}_{{\text{N}}_{\text{2}}}=\left[\begin{array}{l}2\left(115,984\right)+\left(1\right)\left(74,516\right)\\ +\left(4\right)\left(95,562\right)+\left(16.92\right)\left(73,856\right)\end{array}\right]\\ =1,938,376\text{kJ}\left(\text{lower}\text{than}1,974,692\text{kJ}\right)\end{array}$

Iteration II:

Take $T_P=2300\text{K}$

$\begin{array}{c}2{\overline{h}}_{{\text{CO}}_{\text{2}}}+1{\overline{h}}_{\text{CO}}+4{\overline{h}}_{{\text{H}}_2\text{O}}+16.92{\overline{h}}_{{\text{N}}_{\text{2}}}=\left[\begin{array}{l}2\left(119,035\right)+\left(1\right)\left(76,345\right)\\ +\left(4\right)\left(98,199\right)+\left(16.92\right)\left(75,676\right)\end{array}\right]\\ =1,987,649\text{kJ}\left(\text{higher}\text{than}1,974,692\text{kJ}\right)\end{array}$

Show the adiabatic flame temperature corresponding to enthalpy as in Table (3).

Enthalpy $\overline{h}\left(\text{kJ}\right)$	Adiabatic flame temperature $T_P\left(\text{K}\right)$
$1,938,376$$\left(x_1\right)$	2250 $\left(y_1\right)$
$1,974,692$$\left(x_2\right)$	$\left(y_2=?\right)$
$1,987,649$$\left(x_3\right)$	2300 $\left(y_3\right)$

Substitute $1,938,376\text{kJ}$ for $x_1$, $1,974,692\text{kJ}$ for $x_2$, $1,987,649\text{kJ}$ for $x_3$, $\text{2250}\text{K}$ for $y_1$ and $\text{2300}\text{K}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(1,974,692\text{kJ}-1,938,376\text{kJ}\right)\left(\text{2300}\text{K}-\text{2250}\text{K}\right)}{\left(1,987,649\text{kJ}-1,938,376\text{kJ}\right)}+\text{2250}\text{K}\\ =\text{2287}\text{K}\end{array}$

Thus, the adiabatic flame temperature for incomplete combustion with 90 percent theoretical air,

$T_P=2287\text{K}$

Hence, the adiabatic flame temperature for incomplete combustion with 90 percent theoretical air is $\text{2287}\text{K}$.