Chapter 15.7, Problem 83P

Liquid octane (C₈H₁₈) enters a steady-flow combustion chamber at 25°C and 1 atm at a rate of 0.25 kg/min. It is burned with 50 percent excess air that also enters at 25°C and 1 atm. After combustion, the products are allowed to cool to 25°C. Assuming complete combustion and that all the H₂O in the products is in liquid form, determine (a) the heat transfer rate from the combustion chamber, (b) the entropy generation rate, and (c) the exergy destruction rate. Assume that T₀ = 298 K and the products leave the combustion chamber at 1 atm pressure.

FIGURE P15–87

To determine

The rate of heat transfer from the combustion chamber.

Answer to Problem 83P

The rate of heat transfer from the combustion chamber is $\underset{\_}{11,997\text{kJ}/\text{min}}$.

Explanation of Solution

Write the energy balance equation using steady-flow equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $0$ for $E_{\text{in}}$, $-Q_{\text{out}}$ for $E_{\text{out}}$, and $\Delta U$ for $\Delta E_{\text{system}}$ in Equation (I)

$\begin{array}{l}\left(0\right)-Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-\overline{h^{°}}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-\overline{h^{°}}\right)}}_R\\ -Q_{\text{out}}={\displaystyle \sum N_P{\overline{h}}_{f,P}^{°}}-{\displaystyle \sum N_R{\overline{h}}_{f,R}^{°}}\end{array}$ (II)

Here, the enthalpy of formation for product is ${\overline{h}}_{f,P}^{°}$, the enthalpy of formation for reactant is ${\overline{h}}_{f,R}^{°}$, the mole number of the product is $N_P$, and the mole number of the reactant is $N_R$.

Calculate the molar mass of the ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$.

$M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}=\left[\left({\text{N}}_{\text{C}}\right)\left(M_{\text{C}}\right)+\left({\text{N}}_{\text{H}}\right)\left(M_{\text{H}}\right)\right]$ (III)

Here, the number of carbon atoms is ${\text{N}}_{\text{C}}$, the molar mass of the carbon is $M_{\text{C}}$, the number of hydrogen atoms is ${\text{N}}_{\text{H}}$, the molar mass of the hydrogen is $M_{\text{H}}$.

Determine the rate of mole flow rates of the product.

$\dot{N}=\frac{\dot{m}}{M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}}$ (IV)

Here, the mass flow rate is $\dot{m}$ and the molar mass of the ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$.

Determine the heat transfer rate from the combustion chamber.

${\dot{Q}}_{\text{out}}=\dot{N}\cdot Q_{\text{out}}$ (V)

Conclusion:

Write the theoretical combustion equation of for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$.

$\left\{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+1.5a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}\\ +0.5a_{\text{th}}{\text{O}}_2+\left(1.5\right)\left(3.76\right)a_{\text{th}}{\text{N}}_{\text{2}}\end{array}\right\}$ (VI)

Here, liquid octane is ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$, stoichiometric coefficient of air is $a_{\text{th}}$, oxygen is ${\text{O}}_{\text{2}}$, nitrogen is ${\text{N}}_{\text{2}}$, carbon dioxide is ${\text{CO}}_{\text{2}}$ and water is ${\text{H}}_{\text{2}}\text{O}$.

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

$\begin{array}{l}1.5a_{\text{th}}=8+4.5+0.5a_{\text{th}}\\ a_{\text{th}}=12.5\end{array}$

Substitute $12.5$ for $a_{\text{th}}$ in Equation (VI)

$\begin{array}{l}\left\{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+1.5\left(12.5\right)\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}+0.5\left(12.5\right){\text{O}}_2\\ +\left(1.5\right)\left(3.76\right)\left(12.5\right){\text{N}}_{\text{2}}\end{array}\right\}\\ \left\{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)+18.75\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}\\ +6.25{\text{O}}_2+\left(70.5\right){\text{N}}_{\text{2}}\end{array}\right\}\end{array}$ (VII)

From the Table-26, “Enthalpy of formation, Gibbs function of formation, and absolute entropy at $25\text{°C}$, 1 atm”, obtain the enthalpy of formation for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)$, ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}\left(l\right)$, and ${\text{CO}}_{\text{2}}$ is given in a table (I) as:

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$
${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)$	-249,950
${\text{O}}_{\text{2}}$	0
${\text{N}}_{\text{2}}$	0
${\text{H}}_{\text{2}}\text{O}\left(l\right)$	-285,830
${\text{CO}}_{\text{2}}$	-393,520

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

$\begin{array}{l}{N}_{R,{\text{C}}_8{\text{H}}_{18}}=1\text{kmol}\\ N_{R,{\text{O}}_{\text{2}}}=18.75\text{kmol}\\ N_{R,{\text{N}}_{\text{2}}}=70.5\text{kmol}\end{array}$

Here, number of moles of reactant octane, oxygen and nitrogen is $N_{R,{\text{C}}_8{\text{H}}_{18}},N_{R,{\text{O}}_{\text{2}}}\text{and}{N}_{R,{\text{N}}_{\text{2}}}$ respectively.

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

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=8\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=9\text{kmol}\\ N_{P,{\text{O}}_2}=6.25\text{kmol}\\ N_{P,{\text{N}}_2}=70.5\text{kmol}\end{array}$

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

Substitute the value table (I) of substance in Equation (II).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(8\right)\left(-393,520\text{kJ}/\text{kmol}\right)+\left(9\right)\left(-285,830\text{kJ}/\text{kmol}\right)\\ +\left(6.25\right)\left(0\right)+70.5\left(0\right)-\left(1\right)\left(-249,950\text{kJ}/\text{kmol}\right)\\ -18.75\left(\left(0\right)+3.76\left(0\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(8\right)\left(-393,520\text{kJ}/\text{kmol}\right)+\left(9\right)\left(-285,830\text{kJ}/\text{kmol}\right)\\ +\left(0\right)+\left(0\right)-\left(1\right)\left(-249,950\text{kJ}/\text{kmol}\right)-\left(0\right)-\left(0\right)\end{array}\right]\\ =-5,470,680\text{kJ}/\text{kmol}\end{array}$

Therefore the heat transfer for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $5,470680\text{kJ}/\text{kmol}$.

Substitute 8 for ${\text{N}}_{\text{C}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, 18 for ${\text{N}}_{\text{H}}$, and $1\text{kg}/\text{kmol}$ for $M_{\text{H}}$ in Equation (III).

$\begin{array}{c}{M}_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}=\left[\left(8\right)\left(12\right)+\left(18\right)\left(1\right)\right]\text{kg}/\text{kmol}\\ =\left[\left(96\right)+\left(18\right)\right]\text{kg}/\text{kmol}\\ =114\text{kg}/\text{kmol}\end{array}$

Substitute $0.25\text{kg}/\text{min}$ for $\dot{m}$ and $114\text{kg}/\text{kmol}$ for $M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}$ in Equation (IV).

$\begin{array}{c}\dot{N}=\frac{\left(0.25\text{kg}/\text{min}\right)}{\left(114\text{kg}/\text{kmol}\right)}\\ =0.002193\text{kmol}/\text{min}\end{array}$

Substitute $0.002193\text{kmol}/\text{min}$ for $\dot{N}$ and $5,470680\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(0.002193\text{kmol}/\text{min}\right)\left(5,470680\text{kJ}/\text{kmol}\right)\\ =11,997\text{kJ}/\text{min}\end{array}$

Thus, the rate of heat transfer from the combustion chamber is $\underset{\_}{11,997\text{kJ}/\text{min}}$.

To determine

The entropy generation rate from the combustion chamber.

Answer to Problem 83P

The entropy generation rate from the combustion chamber is $\underset{\_}{39.02\text{kJ}/\text{min}\cdot \text{K}}$.

Explanation of Solution

Write the expression for entropy generation during this process.

$S_{\text{gen}}=S_P-S_R+\frac{Q_{\text{out}}}{T_{\text{surr}}}$ (VIII)

Write the combustion equation of Equation (VI)

$S_{\text{gen}}=S_P-S_R+\frac{Q_{\text{out}}}{T_{\text{surr}}}\to S_{\text{gen}}={\displaystyle \sum N_P{\overline{s}}_P}-{\displaystyle \sum N_R{\overline{s}}_R}+\frac{Q_{\text{out}}}{T_{\text{surr}}}$ (IX)

Here, the entropy of the product is ${\overline{s}}_P$, the entropy of the reactant is ${\overline{s}}_R$, the heat transfer for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $Q_{\text{out}}$, and the surrounding temperature is $T_{\text{surr}}$.

Determine the entropy at the partial pressure of the components.

$\begin{array}{c}{S}_i=N_i{\overline{s}}_i\left(T,P_i\right)\\ =N_i{\overline{s}}_i^{°}\left(T,P_0\right)-R_u\ln \left(y_i{P}_m\right)\end{array}$ (X).

Here, the partial pressure is $P_i$, the mole fraction of the component is $y_i$, the total pressure of the mixture is $P_m$, and the universal gas constant is $R_u$.

Determine the entropy generation rate from the combustion chamber.

${\dot{S}}_{\text{gen}}=\dot{N}{S}_{\text{gen}}$ (XI)

Conclusion:

Refer Equation (X) for reactant and product to calculation the entropy in tabular form as:

For reactant entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$	1	1.00	360.79	---	360.79
${\text{O}}_2$	18.75	0.21	205.14	-12.98	4089.75
${\text{N}}_{\text{2}}$	70.50	0.79	191.61	-1.96	13646.69
$S_R=18,097.23\text{kJ}/\text{K}$

For product entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$	8	0.0944	213.80	-19.62	1867.3
${\text{H}}_{\text{2}}\text{O}\left(l\right)$	9	---	69.92	---	629.3
${\text{O}}_2$	6.25	0.0737	205.04	-21.68	1417.6
${\text{N}}_{\text{2}}$	70.50	0.8319	191.61	-1.53	13616.3
$S_P=17,531\text{kJ}/\text{K}$

Substitute $17,531\text{kJ}/\text{K}$ for $S_P$, $18,097.23\text{kJ}/\text{K}$ for $S_R$, $298\text{K}$ for $T_{\text{surr}}$, and $5,470680\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ in Equation (VIII).

$\begin{array}{c}{S}_{\text{gen}}=\left(17531-18097\right)\text{kJ}/\text{K}+\frac{5,470,523\text{kJ}}{298\text{K}}\\ =\left(-566\right)\text{kJ}/\text{K}+\left(18357.46\right)\text{kJ}/\text{K}\\ =17791.46\text{kJ}/\text{K}\end{array}$

Substitute $0.002193\text{kmol}/\text{min}$ for $\dot{N}$ and $17,791.46\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (XI).

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left(0.002193\text{kmol}/\text{min}\right)\times \left(17,791.46\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =39.016\text{kJ}/\text{min}\cdot \text{K}\\ \cong 39.02\text{kJ}/\text{min}\cdot \text{K}\end{array}$

Thus, the entropy generation rate from the combustion chamber is $\underset{\_}{39.02\text{kJ}/\text{min}\cdot \text{K}}$.

To determine

The exergy destruction rate from the combustion chamber.

Answer to Problem 83P

The exergy destruction rate from the combustion chamber is $\underset{\_}{193.8\text{kW}}$.

Explanation of Solution

Write the expression for exergy destruction during this process.

${\dot{X}}_{\text{destroyed}}=T_0{\dot{S}}_{\text{gen}}$ (XII)

Here, the thermodynamic temperature of the surrounding is $T_0$

Conclusion:

Substitute $298\text{K}$ for $T_0$ and $39.02\text{kJ}/\text{min}\cdot \text{K}$ for ${\dot{S}}_{\text{gen}}$ in Equation (XII).

$\begin{array}{c}{\dot{X}}_{\text{destroyed}}=\left(298\text{K}\right)\left(39.02\text{kJ}/\text{min}\cdot \text{K}\right)\\ =11627.96\text{kJ}/\text{min}\\ =11627.96\text{kJ}/\text{min}\times \left(\frac{0.016667\text{kW}}{1\text{kJ}/\text{min}}\right)\\ =193.8\text{kW}\end{array}$

Thus, the exergy destruction rate from the combustion chamber is $\underset{\_}{193.8\text{kW}}$.