Chapter 15.7, Problem 92P

n-Octane [C₈H₁₈(l)] is burned in the constant-pressure combustor of an aircraft engine with 70 percent excess air. Air enters this combustor at 600 kPa and 327°C, liquid fuel is injected at 25°C, and the products of combustion leave at 600 kPa and 1227°C. Determine the entropy generation and exergy destruction per unit mass of fuel during this combustion process. Take T0 = 25°C.

To determine

The entropy generation from the combustion chamber per unit mass of fuel.

Answer to Problem 92P

The entropy generation from the combustion chamber per unit mass of fuel is $\underset{\_}{84.7\text{kJ}/\text{kg}\cdot \text{K}}$.

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)

$\begin{array}{l}\left(0\right)-Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_R\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+{\overline{h}}_{1500\text{K}}-{\overline{h}}_{298\text{K}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+{\overline{h}}_{600\text{K}}-{\overline{h}}_{298\text{K}}^{°}\right)}}_R\end{array}$ (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 ${\text{C}}_8{\text{H}}_{\text{18}}$.

$M_{{\text{C}}_8{\text{H}}_{\text{18}}}=\left[\left({\text{N}}_{\text{C}}\right)\left(M_{\text{C}}\right)+\left({\text{N}}_{\text{H}}\right)\left(M_{\text{H}}\right)\right]$ (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 $M_{\text{H}}$.

Write the expression for entropy generation during this process.

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

Write the combustion equation of Equation (IV)

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

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.

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

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 $R_u$.

Write the expression for exergy destruction during this process.

$X_{\text{destroyed}}=T_0{S}_{\text{gen}}$ (VII)

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

Determine the entropy generation per unit mass of the fuel.

$\begin{array}{l}{S}_{\text{gen}}=\frac{{\overline{S}}_{\text{gen}}}{M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}}\\ {\overline{S}}_{\text{gen}}=S_{\text{gen}}\times M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\end{array}$ (VIII)

Conclusion:

Perform unit conversion of temperature at state 1 from degree Celsius to Kelvin.

For air temperature enter in the combustion chamber,

$\begin{array}{c}{T}_{\text{enter}}=327\text{°C}\\ =\left(327+273\right)\text{K}\\ =\text{600}\text{K}\end{array}$

For then liquid injected temperature in the combustion chamber,

$\begin{array}{c}{T}_{\text{injected}}=25\text{°C}\\ =\left(25+273\right)\text{K}\\ =\text{298}\text{K}\end{array}$

For air temperature exit in the combustion chamber,

$\begin{array}{c}{T}_{\text{exit}}=1227\text{°C}\\ =\left(1227+273\right)\text{K}\\ =\text{1500}\text{K}\end{array}$

Write the combustion equation of 1 kmol for ${\text{C}}_8{\text{H}}_{\text{18}}$.

$\left\{{\text{C}}_8{\text{H}}_{\text{18}}\left(l\right)+1.7a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}+0.7a_{\text{th}}{\text{O}}_{\text{2}}\\ +\left(1.7\right)\left(3.76\right)a_{\text{th}}{\text{N}}_{\text{2}}\end{array}\right\}$ (IX)

Here, liquid octane is ${\text{C}}_8{\text{H}}_{\text{18}}$, 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 ${\text{H}}_{\text{2}}\text{O}$.

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

$\begin{array}{l}1.7a_{\text{th}}=8+4.5+0.7a_{\text{th}}\\ a_{\text{th}}=12.5\end{array}$

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

$\left\{{\text{C}}_8{\text{H}}_{\text{18}}\left(l\right)+\left(21.25\right)\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}\\ +8.75{\text{O}}_{\text{2}}+\left(79.9\right){\text{N}}_{\text{2}}\end{array}\right\}$ (X).

Refer Appendix Table A-18, A-19, A-20, and A-23, obtain the enthalpy of formation, at 298 K, 600 K, and 1500 K for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$, ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, Co, ${\text{H}}_{\text{2}}\text{O}\left(g\right)$, and ${\text{CO}}_{\text{2}}$ is given in a table (I) as:

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{298\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{600\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{1500\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$
${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$	-249,950	---	---	---
${\text{O}}_{\text{2}}$	0	8682	17,929	49,292
${\text{N}}_{\text{2}}$	0	8669	17,563	47,073
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	-241,820	9904	---	57,999
${\text{CO}}_{\text{2}}$	-393,520	9364	---	71,078

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

$\begin{array}{l}{N}_{R,{\text{C}}_8{\text{H}}_{18}}=1\text{kmol}\\ N_{R,{\text{O}}_{\text{2}}}=21.25\text{kmol}\\ N_{R,{\text{N}}_{\text{2}}}=79.9\text{kmol}\end{array}$

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 (X), and write the number of moles of products.

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=8\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=9\text{kmol}\\ N_{P,{\text{O}}_2}=8.75\text{kmol}\\ N_{P,{\text{N}}_2}=79.9\text{kmol}\end{array}$

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

Substitute the value of substance in Equation (II).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(8\right)\left(-393,520\text{kJ}/\text{kmol}\cdot \text{K}+71,078\text{kJ}/\text{kmol}\cdot \text{K}-9364\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(9\right)\left(-241,820\text{kJ}/\text{kmol}\cdot \text{K}+57,999\text{kJ}/\text{kmol}\cdot \text{K}-9904\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(8.75\right)\left(0+49,292\text{kJ}/\text{kmol}\cdot \text{K}-8682\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(79.9\right)\left(0+47,073\text{kJ}/\text{kmol}\cdot \text{K}+8669\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(1\right)\left(-249,950\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ -\left(21.25\right)\left(0+17,929\text{kJ}/\text{kmol}\cdot \text{K}-8682\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ -\left(79.9\right)\left(0+17,563\text{kJ}/\text{kmol}\cdot \text{K}-8669\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\\ =-1,631,335\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Therefore the heat transfer for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $1,631,335\text{kJ}/\text{kmol}\cdot \text{K}$.

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

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

Refer Equation (VI) 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)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{C}}_8{\text{H}}_{\text{18}}$	1	---	466.73	14.79	451.94
${\text{O}}_2$	21.25	0.21	226.35	1.81	4771.48
${\text{N}}_{\text{2}}$	79.9	0.79	212.07	12.83	15,919.28
$S_R=21,142.70\text{kJ}/\text{K}$

For product entropy,

Substance	$N_i$	$y_i$	${\overline{s}}_i^{°}$ (T, 1 atm)	$R_u\ln \left(y_i{P}_m\right)$	$N_i{\overline{s}}_i$
${\text{CO}}_2$	8	0.0757	292.11	-6.673	2390.26
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	9	0.0852	250.45	-5.690	2305.26
${\text{O}}_{\text{2}}$	8.75	0.0828	25.97	-5.928	2309.11
${\text{N}}_{\text{2}}$	79.9	0.7563	241.77	12.46	18,321.87
$S_P=25,326.50\text{kJ}/\text{K}$

Substitute $25,326.50\text{kJ}/\text{K}$ for $S_P$, $21,142.70\text{kJ}/\text{K}$ for $S_R$, $25°\text{C}$ for $T_{\text{surr}}$, and $1,631,335\text{kJ}/\text{kmol}\cdot \text{K}$ for $Q_{\text{out}}$ in Equation (IV).

$\begin{array}{c}{S}_{\text{gen}}=\left(25,326.50-21,142.70\right)\text{kJ}/\text{K}+\frac{1,631,335\text{kJ}/\text{kmol}\cdot \text{K}}{25°\text{C}}\\ =\left(4183.8\text{kJ}/\text{K}\right)+\frac{1,631,335\text{kJ}/\text{kmol}\cdot \text{K}}{25°\text{C+273}}\\ =9658.1\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $25°\text{C}$ for $T_0$ and $9658.1\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (VII).

$\begin{array}{c}{X}_{\text{dest}}=\left(25°\text{C}\right)\left(9658.1\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =\left(25°\text{C+273}\right)\left(9658.1\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =2,878,108\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $114\text{kJ}/\text{kmol}\cdot \text{K}$ for $M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}$ and $9658.1\text{kJ}/\text{kmol}\cdot \text{K}$ for $S_{\text{gen}}$ in Equation (VIII).

$\begin{array}{c}{\overline{S}}_{\text{gen}}=\frac{\left(9658.1\text{kJ}/\text{kmol}\cdot \text{K}\right)}{\left(114\text{kg}/\text{kmol}\right)}\\ =84.7\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Thus, the entropy generation from the combustion chamber per unit mass of fuel is $\underset{\_}{84.7\text{kJ}/\text{kg}\cdot \text{K}}$.

To determine

The exergy destruction from the combustion chamber per unit mass of fuel.

Answer to Problem 92P

The exergy destruction from the combustion chamber per unit mass of fuel is $\underset{\_}{25,247\text{kJ}/\text{kg}}$.

Explanation of Solution

Determine the exergy destruction from the combustion chamber per unit mass of the fuel.

$\begin{array}{l}{X}_{\text{destory}}=\frac{{\overline{X}}_{\text{destrory}}}{M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}}\\ {\overline{X}}_{\text{destrory}}=X_{\text{destrory}}\times M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\end{array}$ . (XI)

Conclusion:

Substitute $114\text{kJ}/\text{kmol}\cdot \text{K}$ for $M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}$ and $2,878,108\text{kJ}/\text{kmol}\cdot \text{K}$ for $X_{\text{dest}}$ in Equation (XI).

$\begin{array}{c}{\overline{X}}_{\text{dest}}=\frac{\left(2,878,108\text{kJ}/\text{kmol}\cdot \text{K}\right)}{\left(114\text{kg}/\text{kmol}\right)}\\ =25,247\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Thus, the exergy destruction from the combustion chamber per unit mass of fuel is $\underset{\_}{390,800\text{kJ}}$.