A sewage (bio) gas has the following composition (volume basis)

CH_4 – 65%

CO_2 – 33%

N_2 – 1%

H_2S – 1%

Now determine:

1. The heating value

2. Air required for combustion

3. Amount of flue gases with no excess air

4. Composition of flue gases

5. Theoretical flame temperature

Step-by-Step

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The given composition of the gas shows that methane is the only major combustible component. H_2S is minor.

The methane reaction is

At 65% CH_4 by volume the heat evolved by combustion of methane from 1 m³ gas is

35,800 × 0.65 = 23,270 kJ

The H_2S reaction is

\mathrm{H}_2 \mathrm{~S}+1.5 \mathrm{O}_2 \rightarrow \mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}+23,170 \mathrm{~kJ} / \mathrm{m}^3At 1% H_2S in the gas, the heat contribution from H_2S is

23,170 × 0.01 = 231 kJ

The total heat available from 1 m³ gas is

\begin{aligned}23,270+231 & =23,501 \\Q^h & =23,500 \ \mathrm{~kJ} / \mathrm{m}^3\end{aligned}The methane reaction shows that 1 m³ methane requires 2 m³ oxygen.

Hence, 0.65 m³ methane will require

2 × 0.65 = 1.30 m³ O_2

The H_2S reaction shows that 1 m³ H_2S requires 1.5 m³ oxygen.

Hence, 0.01 m³ will require

1.5 × 0.01 = 0.015 m³ O_2

Total oxygen required is

1.30 + 0.015 = 1.315 m³

At N_2:O_2 volume ratio 3.762 nitrogen corresponding to the above oxygen is

1.315 × 3.762 = 4.947 m³

The air required for combustion is

1.315 + 4.947 = 6.262 m³/m³ gas

On combustion, 1 m³ methane produces 1 m³ each of CO_2 and H_2O, hence 0.65 m³ will produce

0.65 m³ CO_2

0.65 m³ H_2O

Similarly, 1 m³ H_2S produces 1 m³ each of SO_2 and CO_2 Hence, 0.01 m³ H_2S will produce

0.01 m³ SO_2

0.01 m³ H_2O

Total gases produced by combustion are

\left.\begin{array}{l}0.65 \ \mathrm{~m}^3 \mathrm{CO}_2 \\0.65+0.01=0.66 \ \mathrm{~m}^3 \mathrm{H}_2 \mathrm{O} \\0.01 \ \mathrm{~m}^3 \mathrm{SO}_2\end{array}\right\}As 1 m³ of gas already contains 0.33 m³ CO_2 and 0.01 m³ N_2, the combustion products will contain

0.65 + 0.33 = 0.98 m³ CO_2

0.65 m³ H_2O

0.01 m³ SO_2

4.947 + 0.01 = 4.957 m³ N_2

Total amount of flue gases Vcp is 6.597 m³, say 6.6 m³

The % composition of combustion products is

\left.\begin{array}{l}\mathrm{CO}_2-14.8 \% \\\mathrm{H}_2 \mathrm{O}-9.8 \% \\\mathrm{SO}_2-0.16 \% \\\mathrm{~N}_2-76.0 \%\end{array}\right\}Let us assume a flame temperature of 1800°C. At this temperature the specific heats of constituent gases are (kJ/m³°C)

CO_2 – 2.4226

H_2O – 1.9055

N_2 – 1.4705

Since the percentage of SO_2 is very small, it can be neglected. The specific heat contribution of these gases to the combustion gas mixture in Equation (5.26) is in proportion to their amount. Hence,

\begin{aligned}& \mathrm{CO}_2-2.4226 \times 0.148=0.3585 \\& \mathrm{H}_2 \mathrm{O}-1.9055 \times 0.098=0.1867 \\& \mathrm{~N}_2-1.4705 \times 0.76=\underline{1.1176} \\& \text { Total } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.6628 \ \mathrm{~kJ} / \mathrm{m}^3{ }^{\circ} \mathrm{C} \\& C_{c p}=1.6628 \ \mathrm{~kJ} / \mathrm{m}^3{ }^{\circ} \mathrm{C} \\& Q^h=V_{c p} \times C_{c p} \times t_c \\& 23,500 \times 10^3=6.6 \times 1.6628 \times 10^3 \times t_c \\& \therefore t_c=2141^{\circ} \mathrm{C} \\&\end{aligned}Due to excess air and incomplete burning, this temperature will be about 1900°C.

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