Emission Calculation Methane is burned with 125% theoretical air in an open-steady combustor. The products leave at 1 MPa, 2000 K at equilibrium with the following species in the products: CO2, H2O, O2, N2, and NO. Determine (a) the mole fraction of NO in the mixture and (b) heat transfer, per unit

Question

Emission Calculation

Methane is burned with 125% theoretical air in an open-steady combustor. The products leave at 1 MPa, 2000 K at equilibrium with the following species in the products: [latex] CO _{2}, H _{2} O , O _{2}, N _{2} [/latex], and NO. Determine (a) the mole fraction of NO in the mixture and (b) heat transfer, per unit kmol of the fuel, if the fuel and air enter the chamber at 1 MPa and 298 K. What-if scenario: (c) What would the answer in part (a) be if CO was included in the products?

Accepted Answer

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:

[latex] CH _{4}+2.5\left( O _{2}+3.76 N _{2} ight) ightarrow CO _{2}+2 H _{2} O +a O _{2}+b N _{2}+c NO [/latex]

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 [latex] 0.5 N _{2}+0.5 O _{2} \leftrightarrows NO [/latex], 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:

[latex] y_{ N _{2}}=\frac{a+8.9}{12.9} ; \quad y_{ O _{2}}=\frac{a}{12.9} ; \quad ext { and } \quad y_{ NO }=\frac{1-2 a}{12.9} [/latex]

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

[latex] K \equiv \frac{\prod_{p} y_{p}^{ u_{p}}}{\prod_{r} y_{r}^{ u_{r}}}\left(\frac{p}{p_{0}} ight)^{\sum_{p} v_{p}-\sum_{r} v_{r}} [/latex] (14.50)

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

The overall reaction, therefore, is evaluated as:

[latex] CH _{4}+2.5\left( O _{2}+3.76 N _{2} ight) ightarrow CO _{2}+2 H _{2} O +0.4792 O _{2}+9.3792 N _{2}+0.0416 NO [/latex]

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

[latex] y_{ NO }=\frac{ u_{ NO }}{\sum_{p} u_{p}}=\frac{0.0416}{12.9}=0.003225 ext { or } 3225 ppm [/latex]

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

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

[latex] \frac{\dot{Q}}{\dot{n}_{F}}=\sum_{p} u_{p} \bar{h}_{p}-\sum_{r} u_{r} \bar{h}_{r}=\bar{h}_{P}-\bar{h}_{R} [/latex]

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

[latex]\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}[/latex]

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

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 [latex] \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 [/latex]. 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.

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

Related Answered Questions

Use of van’t Hoff Equation (a) Use van’t Hoff equation to estimate the heat of combustion for the reaction CO + 0.5O2 → CO2 at 2000 K. (b) Use TEST to verify the result. ...

Verified Answer:

Equilibrium Composition One kmol of hydrogen and one kmol of oxygen react to produce an equilibrium mixture of water, hydrogen, and oxygen at 2500 K. (a) Determine the equilibrium molar composition if the mixture pressure is 1 atm. What-if scenario: What would the mole fraction of water in ...

Verified Answer:

Equilibrium of an Ideal Gas Mixture A natural gas reservoir contains a mixture of methane and nitrogen at 298 K. At the top of the well, the mixture pressure is 100 kPa and the molar composition is found to be 50% methane and 50% nitrogen. Determine (a) the pressure and (b) the molar composition ...

Verified Answer:

Verified Answer:

Ionization and Neutral Component A mixture containing 95, helium and 5, cesium on molar basis is heated to 2000 K. The equilibrium constants for ionization of helium and cesium are 5000 and 15.63, respectively. Determine the pressure at which the mole fraction of electron’s is 0.01. ...

Verified Answer:

Verified Answer:

Equilibrium Constant Evaluate ln K at 298 K for the following elementary steps: (a) H2 + 0.5O2 S H2O and (b) 2H2O S 2H2 + O2. Assume water to be in the vapor phase. What-if scenario: What would the answer in part (a) be if the temperature was (c) 1000 K or (d) 2500 K? ...

Verified Answer:

Verified Answer:

Desalination through Reverse Osmosis (a) Determine the minimum useful work required for desalination per unit mass of ocean water at 10°C and 1 atm through reverse osmosis. (b) Compare it with the energy required for desalination by distillation. (c) Also calculate the osmotic head. Use ...

Verified Answer:

Verified Answer: