Chapter 15.7, Problem 110RP

Liquid octane (C₈H₁₈) enters a steady-flow combustion chamber at 25°C and 8 atm at a rate of 0.8 kg/min. It is burned with 200 percent excess air that is compressed and preheated to 500 K and 8 atm before entering the combustion chamber. After combustion, the products enter an adiabatic turbine at 1300 K and 8 atm and leave at 950 K and 2 atm. Assuming complete combustion and T₀ = 25°C, determine (a) the heat transfer rate from the combustion chamber, (b) the power output of the turbine, and (c) the reversible work and exergy destruction for the entire process.

To determine

The rate of heat transfer from the combustion chamber.

Answer to Problem 110RP

The rate of heat transfer from the combustion chamber is $\underset{\_}{770\text{kJ}/\min }$.

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)

$\begin{array}{l}\left(0\right)-Q_{\text{out}}=\Delta U\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+\overline{h}-{\overline{h}}^{°}\right)}}_R\\ -Q_{\text{out}}={{\displaystyle \sum N_P\left({\overline{h}}_f^{°}+{\overline{h}}_{1300\text{K}}-{\overline{h}}_{298\text{K}}^{°}\right)}}_P-{{\displaystyle \sum N_R\left({\overline{h}}_f^{°}+{\overline{h}}_{500\text{K}}-{\overline{h}}_{298\text{K}}^{°}\right)}}_R\end{array}$ (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 $M_{\text{H}}$.

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:

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

For air temperature enter in the combustion chamber,

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

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

$\left\{{\text{C}}_8{\text{H}}_{\text{18}}\left(l\right)+3a_{\text{th}}\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}+2a_{\text{th}}{\text{O}}_{\text{2}}\\ +\left(3\right)\left(3.76\right)a_{\text{th}}{\text{N}}_{\text{2}}\end{array}\right\}$ (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.

$\begin{array}{l}3a_{\text{th}}=8+4.5+2a_{\text{th}}\\ a_{\text{th}}=12.5\end{array}$

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

$\left\{{\text{C}}_8{\text{H}}_{\text{18}}\left(l\right)+\left(37.5\right)\left({\text{O}}_2+3.76{\text{N}}_{\text{2}}\right)\right\}\to \left\{\begin{array}{l}8{\text{CO}}_{\text{2}}+9{\text{H}}_{\text{2}}\text{O}\\ +25{\text{O}}_{\text{2}}+\left(141\right){\text{N}}_{\text{2}}\end{array}\right\}$ (X).

Refer Appendix Table A-18, A-19, A-20, and A-23, obtain the enthalpy of formation, at 298 K, 500 K, 950 K, and 1500 K for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$, ${\text{O}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}\left(g\right)$, and ${\text{CO}}_{\text{2}}$ is given in a table (I) as:

Substance	$\begin{array}{l}{\overline{h}}_f^{°}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{500\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{298\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{1300\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$	$\begin{array}{l}{\overline{h}}_{950\text{K}}\\ \text{kJ}/\text{kmol}\end{array}$
${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\left(g\right)$	-249,950	---	---	---	---
${\text{O}}_{\text{2}}$	0	14,770	8682	42,033	26,652
${\text{N}}_{\text{2}}$	0	14,581	8669	40,170	28,501
${\text{H}}_{\text{2}}\text{O}\left(g\right)$	-241,820	---	9904	48,807	33,841
${\text{CO}}_{\text{2}}$	-393,520	---	9364	59,552	40,070

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

$\begin{array}{l}{N}_{R,{\text{C}}_8{\text{H}}_{18}}=1\text{kmol}\\ N_{R,{\text{O}}_{\text{2}}}=37.5\text{kmol}\\ N_{R,{\text{N}}_{\text{2}}}=141\text{kmol}\end{array}$

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.

$\begin{array}{l}{N}_{P,{\text{CO}}_{\text{2}}}=8\text{kmol}\\ N_{P,{\text{H}}_{\text{2}}\text{O}}=9\text{kmol}\\ N_{P,{\text{O}}_2}=25\text{kmol}\\ N_{P,{\text{N}}_2}=141\text{kmol}\end{array}$

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 from table (I) of substance in Equation (II).

$\begin{array}{c}-Q_{\text{out}}=\left[\begin{array}{l}\left(8\right)\left(-393,520\text{kJ}/\text{kmol}\cdot \text{K}+59,522\text{kJ}/\text{kmol}\cdot \text{K}-9364\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(9\right)\left(-241,820\text{kJ}/\text{kmol}\cdot \text{K}+48,807\text{kJ}/\text{kmol}\cdot \text{K}-9904\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(25\right)\left(0+42,033\text{kJ}/\text{kmol}\cdot \text{K}-8682\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(141\right)\left(0+40,170\text{kJ}/\text{kmol}\cdot \text{K}+8669\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(1\right)\left(-249,950\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ -\left(37.5\right)\left(0+14,770\text{kJ}/\text{kmol}\cdot \text{K}-8682\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ -\left(141\right)\left(0+14,581\text{kJ}/\text{kmol}\cdot \text{K}-8669\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\\ =-109,675\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Therefore the heat transfer for ${\text{C}}_{\text{8}}{\text{H}}_{\text{18}}$ is $109,675\text{kJ}/\text{kmol}\cdot \text{K}$.

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).

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

Substitute $0.8\text{kg}/\text{min}$ for $\dot{m}$ and $114\text{kg}/\text{kmol}$ for $M_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}$ in Equation (IV).

$\begin{array}{c}\dot{N}=\frac{\left(0.8\text{kg}/\text{min}\right)}{\left(114\text{kg}/\text{kmol}\right)}\\ =0.007018\text{kmol}/\text{min}\end{array}$

Substitute $0.007018\text{kmol}/\text{min}$ for $\dot{N}$ and $109,675\text{kJ}/\text{kmol}\cdot \text{K}$ for $Q_{\text{out}}$ in Equation (V).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(0.007018\text{kmol}/\text{min}\right)\left(109,675\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =770\text{kJ}/\text{min}\end{array}$

Thus, the rate of heat transfer from the combustion chamber is $\underset{\_}{770\text{kJ}/\min }$.

To determine

The rate of power output of the turbine.

Answer to Problem 110RP

The rate of power output of the turbine is $\underset{\_}{262.6\text{kW}}$.

Explanation of Solution

Determine the power output of the adiabatic turbine from the steady-flow energy balance equation for non-reacting gas mixture.

$\begin{array}{l}-W_{\text{out}}={\displaystyle \sum N_P\left({\overline{h}}_e-{\overline{h}}_i\right)}\\ W_{\text{out}}={\displaystyle \sum N_P\left({\overline{h}}_{1300\text{K}}-{\overline{h}}_{950\text{K}}\right)}\end{array}$ . (XI)

Determine the rate of work done of the adiabatic turbine.

${\dot{W}}_{\text{out}}=\dot{N}\times W_{\text{out}}$ (XII)

Conclusion:

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

$\begin{array}{c}{W}_{\text{out}}=\left[\begin{array}{l}\left(8\right)\left(59,522\text{kJ}/\text{kmol}\cdot \text{K}-40,070\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(9\right)\left(48,807\text{kJ}/\text{kmol}\cdot \text{K}-33,841\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(25\right)\left(42,033\text{kJ}/\text{kmol}\cdot \text{K}-29,652\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ +\left(141\right)\left(40,170\text{kJ}/\text{kmol}\cdot \text{K}-28,501\text{kJ}/\text{kmol}\cdot \text{K}\right)\end{array}\right]\\ =2,245,164\text{kJ}/\text{kmol}\cdot \text{K}\end{array}$

Substitute $0.007018\text{kmol}/\text{min}$ for $\dot{N}$ and $2,245,164\text{kJ}/\text{kmol}\cdot \text{K}$ for $W_{\text{out}}$ inn Equation (XII).

$\begin{array}{c}{\dot{W}}_{\text{out}}=\left(0.007018\text{kmol}/\text{min}\right)\times \left(2,245,164\text{kJ}/\text{kmol}\cdot \text{K}\right)\\ =15,756\text{kJ}/\text{min}\\ =15,756\text{kJ}/\text{min}\times \left(\frac{0.016667\text{kW}}{1\text{kJ}/\text{min}}\right)\\ =262.6\text{kW}\end{array}$

Thus, the rate of power output of the turbine is $\underset{\_}{262.6\text{kW}}$.

To determine

The exergy destruction rate from the combustion chamber.

The rate of reversible work done in the combustion chamber.

Answer to Problem 110RP

The exergy destruction rate from the combustion chamber is $\underset{\_}{251.3\text{kW}}$.

The rate of reversible work done in the combustion chamber is $\underset{\_}{513.9\text{kW}}$.

Explanation of Solution

Write the expression for entropy generation during this process.

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

Write the combustion equation of Equation (VI)

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

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.

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

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.

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

Write the expression for exergy destruction during this process.

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

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

Determine the rate of the reversible work done of the combustion chamber.

${\dot{W}}_{\text{rev}}=\dot{W}+{\dot{X}}_{\text{dest}}$ (XVIII)

Conclusion:

Refer Equation (XV) 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	17.288	360.79
${\text{O}}_2$	37.5	0.21	220.589	4.313	8,110.34
${\text{N}}_{\text{2}}$	141	0.79	206.630	15.329	26,973.44
$S_R=35,427.28\text{kJ}/\text{K}$

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	0.0437	266.444	-20.260	2,293.63
${\text{H}}_{\text{2}}\text{O}\left(l\right)$	9	0.0490	230.499	-19.281	2,248.02
${\text{O}}_2$	25	0.1366	241.689	-10.787	6,311.90
${\text{N}}_{\text{2}}$	141	0.7705	226.389	3.595	31,413.93
$S_P=42,267.48\text{kJ}/\text{K}$

Substitute $42,267.48\text{kJ}/\text{K}$ for $S_P$, $35,427.28\text{kJ}/\text{K}$ for $S_R$, $298\text{K}$ for $T_{\text{surr}}$, and $109,675\text{kJ}/\text{kmol}$ for $Q_{\text{out}}$ in Equation (XIII).

$\begin{array}{c}{S}_{\text{gen}}=\left(42,267.48+35,427.28\right)\text{kJ}/\text{K}+\frac{109,675\text{kJ}/\text{kmol}}{298\text{K}}\\ =\left(262.6053\right)\text{kJ}/\text{K}+\left(368.0369\right)\text{kJ}/\text{kmol}\\ =7208.237\text{kJ}/\text{kmol}\end{array}$

Substitute $0.007018\text{kmol}/\text{min}$ for $\dot{N}$ and $7208.237\text{kJ}/\text{kmol}$ for $S_{\text{gen}}$ in Equation (XVI).

$\begin{array}{c}{\dot{S}}_{\text{gen}}=\left(0.007018\text{kmol}/\text{min}\right)\times \left(7208.237\text{kJ}/\text{kmol}\right)\\ =50.587\text{kJ}/\text{min}\cdot \text{K}\\ \cong 50.59\text{kJ}/\text{min}\cdot \text{K}\end{array}$

Substitute $298\text{K}$ for $T_0$ and $50.59\text{kJ}/\text{min}\cdot \text{K}$ for ${\dot{S}}_{\text{gen}}$ in Equation (XVII).

$\begin{array}{c}{\dot{X}}_{\text{destroyed}}=\left(298\text{K}\right)\left(50.59\text{kJ}/\text{min}\cdot \text{K}\right)\\ =15075.82\text{kJ}/\text{min}\\ =15075.82\text{kJ}/\text{min}\times \left(\frac{0.016667\text{kW}}{1\text{kJ}/\text{min}}\right)\\ =251.26\text{kW}\end{array}$

             $\cong 251.3\text{kW}$

Thus, the exergy destruction rate from the combustion chamber is $\underset{\_}{251.3\text{kW}}$.

Substitute $262.6\text{kW}$ for $\dot{W}$ and $251.3\text{kW}$ for ${\dot{X}}_{\text{dest}}$ in Equation (XVIII).

$\begin{array}{c}{\dot{W}}_{\text{rev}}=\left(262.6+251.3\right)\text{kW}\\ =\text{513}\text{.9}\text{kW}\end{array}$

Thus, the rate of reversible work done in the combustion chamber is $\underset{\_}{513.9\text{kW}}$.