Evaluating Entropy Production for a Reactor Fueled by Liquid Octane
Liquid octane at 25°C, 1 atm enters a well-insulated reactor and reacts with air entering at the same temperature and pressure. The products of combustion exit at 1 atm pressure. For steady-state operation and negligible effects of kinetic and potential energy, determine the rate of entropy production, in kJ/K per kmol of fuel, for complete combustion with (a) the theoretical amount of air, (b) 400% theoretical air.
Known Liquid octane and air, each at 25°C and 1 atm, burn completely within a well-insulated reactor operating at steady state. The products of combustion exit at 1 atm pressure.
Find Determine the rate of entropy production, in kJ/K per kmol of fuel, for combustion with (a) the theoretical amount of air, (b) 400% theoretical air.
Schematic and Given Data:
Engineering Model
1. The control volume shown on the accompanying figure by a dashed line operates at steady state and without heat transfer with its surroundings.
2. Combustion is complete. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.
3. The combustion air can be modeled as an ideal gas mixture, as can the products of combustion.
4. The reactants enter at 25°C, 1 atm. The products exit at a pressure of 1 atm.
Analysis The temperature of the exiting products of combustion T_{ P } was evaluated in Example 13.8 for each of the two cases. For combustion with the theoretical amount of air, T_{ P }=2395 K. For complete combustion with 400% theoretical air, T_{ P }=962 K.
a. For combustion of liquid octane with the theoretical amount of air, the chemical equation is
C _{8} H _{18}( l )+12.5 O _{2}+47 N _{2} \rightarrow 8 CO _{2}+9 H _{2} O ( g )+47 N _{2}
With assumptions 1 and 3, the entropy rate balance on a per mole of fuel basis, Eq. 13.24, takes the form
\begin{aligned}0=&\sum_{i} \frac{\dot{Q}_{j} / T_{j}}{\dot{n}_{ F }}+\bar{s}_{ F }+\underline{\left( a +\frac{ b }{4}\right) \bar{s}_{ O _{2}}+\left( a +\frac{ b }{4}\right) 3.76 \bar{s}_{ N _{2}}}] \\&-\left[ \underline{a \bar{s}_{ CO _{2}}+\frac{ b }{2} \bar{s}_{ H _{2} O }+\left( a +\frac{ b }{4}\right) 3.76 \bar{s}_{ N _{2}}}\right]+\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}\end{aligned} (13.24)
0=\sum_{j} \frac{\dot{Q}_{j} /T_{j}^{\nearrow0}}{\dot{n}_{ F }}+\bar{s}_{ F }+\left(12.5 \bar{s}_{ O _{2}}+47 \bar{s}_{ N _{2}}\right)
-\left(8 \bar{s}_{ CO _{2}}+9 \bar{s}_{ H _{2} O ( g )}+47 \bar{s}_{ N _{2}}\right)+\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}
or on rearrangement
\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}=\left(8 \bar{s}_{ CO _{2}}+9 \bar{s}_{ H _{2} O ( g )}+47 \bar{s}_{ N _{2}}\right)-\bar{s}_{ F }-\left(12.5 \bar{s}_{ O _{2}}+47 \bar{s}_{ N _{2}}\right) (1)
Each coefficient of this equation is the same as for the corresponding term of the balanced chemical equation.
The fuel enters the reactor separately at T_{\text {ref }}, p_{\text {ref }}. The absolute entropy of liquid octane required by the entropy balance is obtained from Table A-25 as 360.79 kJ/kmol ⋅ K.
The oxygen and nitrogen in the combustion air enter the reactor as components of an ideal gas mixture at T_{\text {ref }}, p_{\text {ref }}. With Eq. 13.23 and absolute entropy data from Table A-23
\bar{s}_{i}\left(T, p_{i}\right)=\bar{s}_{i}^{\circ}(T)-\bar{R} \ln \frac{y_{i} p}{p_{\text {ref }}} (component of an ideal gas mixture) (13.23)
\bar{s}_{ O _{2}}=\bar{s}_{ O _{2}}^{\circ}\left(T_{ ref }\right)-\bar{R} \ln \frac{y_{ O _{2}} p_{ ref }}{p_{ ref }}
=205.03-8.314 \ln 0.21=218.01 kJ / kmol \cdot K
\bar{s}_{ N _{2}}=\bar{s}_{ N _{2}}^{\circ}\left(T_{ ref }\right)-\bar{R} \ln \frac{y_{ N _{2}} p_{ ref }}{p_{ ref }}
=191.5-8.314 \text { ln } 0.79=193.46 kJ / kmol \cdot K
The product gas exits as an ideal gas mixture at 1 atm, 2395 K with the following composition: y_{ CO _{2}}=8 / 64=0.125, \quad y_{ H _{2} O ( g )}=9 / 64=0.1406, \quad y_{ N _{2}}=47 / 64=0.7344. With Eq. 13.23 and absolute entropy data at 2395 K from Tables A-23
\begin{aligned}\bar{s}_{ CO _{2}} &=\bar{s}_{ CO _{2}}^{\circ}-\bar{R} \ln y_{ CO _{2}} \\&=320.173-8.314 \ln 0.125=337.46 kJ / kmol \cdot K \\\bar{s}_{ H _{2} O } &=273.986-8.314 \ln 0.1406=290.30 kJ / kmol \cdot K \\\bar{s}_{ N _{2}} &=258.503-8.314 \ln 0.7344=261.07 kJ / kmol \cdot K\end{aligned}
Inserting values into Eq. (1), the expression for the rate of entropy production, we get
\begin{aligned}\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}=& 8(337.46)+9(290.30)+47(261.07) \\&-360.79-12.5(218.01)-47(193.46)\end{aligned}
=5404 kJ / kmol \text { (octane) } \cdot K
Alternative Solution
As an alternative, the following IT code can be used to determine the entropy production per mole of fuel entering, where sigma denotes \dot{\sigma}_{ cv } / \dot{n}_{ F }, and sN2_R and sN2_P denote the entropy of N _{2} in the reactants and products, respectively, and so on. In the Units menu, select temperature in K, pressure in bar, and amount of substance in moles.
Using the Solve button, the result is sigma = 5404 kJ/kmol (octane) ⋅ K, which agrees with the result obtained above.
b. The complete combustion of liquid octane with 400% theoretical air is described by the following chemical equation:
\begin{aligned}&C _{8} H _{18}(1)+50 O _{2}+188 N _{2} \rightarrow \\& 8 CO _{2}+9 H _{2} O ( g )+37.5 O _{2}+188 N _{2}\end{aligned}
The entropy rate balance on a per mole of fuel basis takes the form
\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}=\left(8 \bar{s}_{ CO _{2}}+9 \bar{s}_{ H _{2} O ( g )}+37.5 \bar{s}_{ O _{2}}+188 \bar{s}_{ N _{2}}\right)
-\bar{s}_{ F }-\left(50 \bar{s}_{ O _{2}}+188 \bar{s}_{ N _{2}}\right)
The specific entropies of the reactants have the same values as in part (a). The product gas exits as an ideal gas mixture at 1 atm, 962 K with the following composition: y_{ CO _{2}}=8 / 242.5=0.033, y_{ H _{2} O ( g )}=9 / 242.5=0.0371, y_{ O _{2}}=37.5 / 242.5=0.1546, y_{ N _{2}}=0.7753. With the same approach as in part (a)
\begin{aligned}\bar{s}_{ CO _{2}} &=267.12-8.314 \ln 0.033=295.481 kJ / kmol \cdot K \\\bar{s}_{ H _{2} O } &=231.01-8.314 \ln 0.0371=258.397 kJ / kmol \cdot K \\\bar{s}_{ O _{2}} &=242.12-8.314 \ln 0.1546=257.642 kJ / kmol \cdot K \\\bar{s}_{ N _{2}} &=226.795-8.314 \ln 0.7753=228.911 kJ / kmol \cdot K\end{aligned}
Inserting values into the expression for the rate of entropy production
\begin{aligned}\frac{\dot{\sigma}_{ cv }}{\dot{n}_{ F }}=& 8(295.481)+9(258.397)+37.5(257.642) \\&+188(228.911)-360.79-50(218.01)-188(193.46)\end{aligned}
2 = 9754 kJ/kmol (octane) ⋅ K
The use of IT to solve part (b) is left as an exercise.
1 For several gases modeled as ideal gases, IT directly returns the absolute entropies required by entropy balances for reacting systems. The entropy data obtained from IT agree with values calculated from Eq. 13.23 using table data.
2 Although the rates of entropy production calculated in this example are positive, as required by the second law, this does not mean that the proposed reactions necessarily would occur, for the results are based on the assumption of complete combustion. The possibility of achieving complete combustion with specified reactants at a given temperature and pressure can be investigated with the methods dealing with chemical equilibrium. For further discussion, see Sec. 14.4.1.
Skills Developed
Ability to…
• apply the control volume entropy balance to a reacting system.
• evaluate entropy values appropriately based on absolute entropies.
Quick Quiz
How do combustion product temperature and rate of entropy production vary, respectively, as percent excess air increases? Assume complete combustion. Ans. Decrease, increase.
TR = 25 + 273.15 // K
p = 1.01325 // bar
TP = 2394 // K (Value from the IT alternative
solution of Example 13.8)
// Determine the partial pressures
pO2_R = 0.21 * p
pN2_R = 0.79 * p
pCO2_P = (8/64) * p
pH2O_P = (9/64) * p
pN2_P = (47/64) * p
1 // Evaluate the absolute entropies
sC8H18 = 360.79 // kJ/kmol ⋅ K (from Table A-25)
sO2_R = s_TP(“O2”, TR, pO2_R)
sN2_R = s_TP(“N2”, TR, pN2_R)
sCO2_P = s_TP(“CO2”, TP, pCO2_P)
sH2O_P = s_TP(“H2O”, TP, pH2O_P)
sN2_P = s_TP(“N2”, TP, pN2_P)
// Evaluate the reactant and product
entropies, sR and sP, respectively
sR = sC8H18 + 12.5 * sO2_R + 47 * sN2_R
sR = 8 * sCO2_P + 9 * sH2O_P + 47 * sN2_P
// Entropy balance, Eq. (1)
sigma = sP − sR