Question 14.13: Emission Calculation Methane is burned with 125% theoretical...

Set up the overall reaction in terms of six unknown coefficients. Use atom balance equations to reduce the number of unknowns to one. Use a single stoichiometric reaction to provide closure.

Assumptions
The IG mixture model is applicable.

Analysis
For 125% theoretical air (Chapter 13), the overall reaction can be expressed in terms of three unknown coefficients—a, b, and c—as:

CH _{4}+2.5\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +a O _{2}+b N _{2}+c NO

Using the atom balance equations, b and c can be expressed in terms of a:

b = a + 8.9; c = 1 – 2a;

With only one unknown, just one elementary step involving some of the components in the equilibrium mixture is needed. We select 0.5 N _{2}+0.5 O _{2} \leftrightarrows NO , for which ln K at 2000 K is listed in Table G-3 as -3.931. Alternatively, if you use the IGE state TESTcalc and set the reactants and products, evaluate State-1 for reactants and State-2 for products at 100 kPa, 2000 K, ln(K) is displayed as -3.916 in the composition panel. The mole fractions of the components that appear in this reaction can be expressed as in Example 14-10 as:

y_{ N _{2}}=\frac{a+8.9}{12.9} ; \quad y_{ O _{2}}=\frac{a}{12.9} ; \quad \text { and } \quad y_{ NO }=\frac{1-2 a}{12.9}

Substituting these expressions in the equilibrium equation, Eq. (14.50), we obtain:

K \equiv \frac{\prod_{p} y_{p}^{\nu_{p}}}{\prod_{r} y_{r}^{\nu_{r}}}\left(\frac{p}{p_{0}}\right)^{\sum_{p} v_{p}-\sum_{r} v_{r}}                                   (14.50)

\begin{aligned}&K=\frac{y_{ NO }}{y_{ N _{2}}^{0.5} y_{ O _{2}}^{0.5}}\left(\frac{p}{p_{0}}\right)^{1-0.5-0.5}=\left(\frac{1-2 a}{12.9}\right)\left(\frac{a+8.9}{12.9}\right)^{-0.5}\left(\frac{a}{12.9}\right)^{-0.5} \\&\Rightarrow \quad e^{-3.931}=\frac{1-2 a}{\sqrt{a(a+8.9)}} ; \quad \text { Solving, } \quad a=0.4792\end{aligned}

The overall reaction, therefore, is evaluated as:

CH _{4}+2.5\left( O _{2}+3.76 N _{2}\right) \rightarrow CO _{2}+2 H _{2} O +0.4792 O _{2}+9.3792 N _{2}+0.0416 NO

From the products composition, the mole fraction of NO can be found:

y_{ NO }=\frac{\nu_{ NO }}{\sum_{p} \nu_{p}}=\frac{0.0416}{12.9}=0.003225 \text { or } 3225 ppm

The heat transfer can be evaluated from the energy equation, Eq. (13.27), for the overall reaction:

0=\bar{h}_{R}-\bar{h}_{P}+\frac{\dot{Q}}{\dot{n}_{F}}-\frac{\dot{W}_{\text {ext }}}{\dot{n}_{F}} ;\left[\frac{ kJ }{ kmol Fuel }\right]                                 (13.27)

\frac{\dot{Q}}{\dot{n}_{F}}=\sum_{p} \nu_{p} \bar{h}_{p}-\sum_{r} \nu_{r} \bar{h}_{r}=\bar{h}_{P}-\bar{h}_{R}

where \bar{h}_{P} \text { and } \bar{h}_{R} can be evaluated using the procedure described in Example 13-5 (Fig. 14.33):

\begin{aligned}\bar{h}_{P} &=\bar{h}_{ CO _{2}}+2 \bar{h}_{ H _{2} O }+0.479 \bar{h}_{ O _{2}}+9.379 \bar{h}_{ N _{2}}+0.0416 \bar{h}_{ NO } \\&=-302,082+2(-169,137)+0.479(59,199)+9.379(56,141)+0.0416(148,150) \\&=-79,290 kJ / kmol \\\bar{h}_{R} &=\bar{h}_{ CH _{4}}+2.5 \bar{h}_{ O _{2}}+9.4 \bar{h}_{ N _{2}}=\bar{h}_{ CH _{4}} \\&=-74,876 kJ / kmol\end{aligned}

Therefore, \dot{Q} / \dot{n}_{F}=\bar{h}_{P}-\bar{h}_{R}=-79,290-(-74,876)=-4,414 kJ / kmol

TEST Analysis
Launch the open-steady chemical equilibrium TESTcalc from the Specific, Combustion and Equilibrium branch. In the reaction panel, choose CH4 in the fuel block, enter lambda = 1.25, and select Excess/Deficient Air from the action menu. The reaction is balanced for a complete reaction. In the composition panel, the reactants—1 kmol of CH4, 9.4 kmol of N2, and 2.5 kmol of O2—are automatically populated. Add NO to the default list of products (scroll through the list to find NO and select the checkbox). Evaluate the reactants state, State-1, with p1=1 MPa and T1=298 K, and the products state, State-2, with p2=p1 and T2=2000 K (select the Products radio-button). The mole fraction of NO is displayed in the composition (or I/O) panel as 0.00328. To determine the heat transfer, enter mdot1=361 kg/s and mdot2=mdot1 (total amount of products for burning 1 kmol of fuel). In the device panel, load the inlet and exit states and enter Wdot_ext=0 to obtain Qdot as -3617 kW, which is somewhat different from the manual solution due to slight difference between the products composition evaluated by the TESTcalc.

What-if scenario
Include CO in the products mixture and click Super-Calculate. A very small amount of CO (y = 0.00017) is reported in the products mixture. There seems to be no appreciable effect on the amount of NO.

Discussion
If the overall reaction was assumed to be complete (no NO in the products), the exit enthalpy at the exit can be shown (by using TEST) to be \bar{h}_{E}=\bar{h}_{ CO _{2}}+2 \bar{h}_{ H _{2} O }+0.5 \bar{h}_{ O _{2}}+9.4 \bar{h}_{ N _{2}}= -84,432 kJ / kmol . Therefore, as shown in Figure 14.33, heat transfer calculated on the assumption of complete combustion is larger in magnitude. If all likely species are included in the equilibrium calculations, the enthalpy of the products (shown by the dotted lines in Fig. 14.33) may significantly deviate from the corresponding complete combustion results. For adiabatic combustion, the exit temperature based on equilibrium composition at the exit is called the equilibrium flame temperature. As shown in Figure 14.34, the equilibrium flame temperature is always smaller than the adiabatic flame temperature for complete combustion (Sec. 13.3.2). The actual flame temperature is likely to be slightly less than the equilibrium temperature due to radiative losses from the flame.