CO2 emission rate calculation Natural gas is fed into a burner at a rate of 1,000 m³/h at 1 atm and 25 °C. Assuming that the air is premixed with Φ = 1, and the final products at equilibrium under 1,000 K and 1 atm contain O2, N2, CO2, CO, H2O, and H2, determine the emission rate of CO2 generation.

Question

[latex]CO_{2}[/latex] emission rate calculation
Natural gas is fed into a burner at a rate of 1,000 m³/h at 1 atm and 25 °C. Assuming that the air is premixed with [latex]\phi[/latex] = 1, and the final products at equilibrium under 1,000 K and 1 atm contain [latex]O_{2}, N_{2}, CO_{2}, CO, H_{2}O[/latex], and [latex]H_{2}[/latex], determine the emission rate of [latex]CO_{2}[/latex] generation.

Accepted Answer

Assuming the natural gas is pure methane, with [latex]\phi[/latex] = 1, α = 1 and β = 4 Eq. (12.5) becomes
[latex]C_{\alpha }H_{\beta }+\frac{1}{\phi}\left(\alpha +\frac{\beta }{4} ight)(O_{2}+3.76 N_{2})[/latex]
[latex] ightarrow [xCO_{2}+(1 - x)CO]+[yH_{2}O + (1 - y)H_{2}]+\frac{3.76}{\phi}\left(\alpha +\frac{\beta}{4} ight)N_{2}[/latex]
[latex]+\left[\frac{1}{\phi}\left(\alpha +\frac{\beta }{4} ight)-\frac{1 + x + y}{2} ight]O_{2}[/latex] (12.5)

[latex]CH_{4} + 2(O_{2} + 3.76N_{2})[/latex]
→ [latex][xCO_{2} + (1 - x)CO] + [yH_{2}O + (1 - y)H_{2}] + 7.52N_{2} + \left(\frac{3 + x + y}{2} ight)O_{2}[/latex]

For each mole of [latex]CH_{4}[/latex], the total mole amount of products is
[latex]n=11.02+\frac{x + y}{2}[/latex]

Consider two equilibrium reactions as follows:
[latex]CO_{2} + H_{2} ↔ CO + H_{2}O[/latex] (1)
[latex]CO_{2} ↔ CO + ½O_{2}[/latex] (2)

The chemical equilibrium constants at 1,000 K can be found in Table 12.2.
[latex]K_{P1}=e^{-0.366}=0.6935=\frac{(1 - x)y}{x(1 - y)}[/latex]

[latex]K_{P2}=e^{-23.529}=6.046 imes 10^{-11}=\frac{1 - x}{x}\frac{\left(\frac{3 + x + y}{2} ight)^{1/2}}{\left(11.02 + \frac{x + y}{2} ight)^{1/2}}[/latex]

Reorganizing the equations leads to
[latex]\begin{cases}x=\frac{1}{1 + 0.6935\left(\frac{1 - y}{y} ight) }\\frac{1 - x}{x}\sqrt{\frac{3 + x + y}{22.04 + x + y} }=6.046 imes 10^{-11}\end{cases}[/latex]

Solving these equations with the assistance of software we can get
[latex]x ≈ y ≈ 1; 1 - x = 1.32 × 10^{-10}; 1 - y = 1.91 × 10^{-10}[/latex]

The extremely low CO and [latex]H_{2}[/latex] concentration is a result of stoichiometric combustion at high temperature. Anyway, for each mole of [latex]CH_{4}[/latex], there is about 1 mol of [latex]CO_{2}[/latex] produced. At 1 atm and 25 °C, one mole of gas corresponds to 0.248 m³/mol. Thus, the [latex]CH_{4}[/latex] feeding rate is 4032.26 mol/h. As a result, the [latex]CO_{2}[/latex] production rate is the same as 4032.26 mol/h, or
[latex]\dot{m}_{CO_{2}}=44\frac{kg}{kmole} imes 4.032\frac{kmole}{h}=177.42 kg/h[/latex]

Table 12.2 Equilibrium constants based on partial pressure for chemical reactions with [latex]CO_{2}[/latex] and CO

T (K)	[latex]ln(K_{p})[/latex]
T (K)	[latex]CO_{2} + H_{2} ↔ CO + H_{2}O[/latex]	[latex]CO_{2} ↔ CO + ½O_{2}[/latex]
298	−11.554	−103.762
500	−4.9252	−57.616
1,000	−0.366	−23.529
1,200	0.3108	−17.871
1,400	0.767	−13.842
1,600	1.091	−10.83
1,800	1.328	−8.497
2,000	1.51	−6.635
2,200	1.648	−5.12
2,400	1.759	−3.86
2,600	1.847	−2.801
2,800	1.918	−1.894
3,000	1.976	−1.111
3,200	2.022	−0.429
3,400	2.061	0.169
3,600		0.701
3,800		1.176
4,000		1.599

Question 12.1: CO2 emission rate calculation Natural gas is fed into a burn...

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