# Question 5.3: A sewage (bio) gas has the following composition (volume bas......

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

C$H_4$ – 65%

C$O_2$ – 33%

$N_2$ – 1%

$H_2$S – 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. $H_2$S is minor.
The methane reaction is

$\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}+35,800 \ \mathrm{~kJ} / \mathrm{m}^3$

At 65% C$H_4$ by volume the heat evolved by combustion of methane from 1 m³ gas is

35,800 × 0.65 = 23,270 kJ

The $H_2$S reaction is

$\mathrm{H}_2 \mathrm{~S}+1.5 \mathrm{O}_2 \rightarrow \mathrm{SO}_2+\mathrm{H}_2 \mathrm{O}+23,170 \mathrm{~kJ} / \mathrm{m}^3$

At 1% $H_2$S in the gas, the heat contribution from $H_2$S 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 $H_2$S reaction shows that 1 m³ $H_2$S 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 C$O_2$ and $H_2$O, hence 0.65 m³ will produce

0.65 m³ C$O_2$
0.65 m³ $H_2$O

Similarly, 1 m³ $H_2$S produces 1 m³ each of S$O_2$ and C$O_2$ Hence, 0.01 m³ $H_2$S will produce

0.01 m³ S$O_2$
0.01 m³ $H_2$O

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 C$O_2$ and 0.01 m³ $N_2$, the combustion products will contain

0.65 + 0.33 = 0.98 m³ C$O_2$

0.65 m³ $H_2$O

0.01 m³ S$O_2$

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)

C$O_2$ – 2.4226
$H_2$O – 1.9055
$N_2$ – 1.4705

Since the percentage of S$O_2$ 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.

Question: 5.1

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The combustion reaction for oil can be written as ...
Question: 5.4

## A fuel oil has the following composition (%) Carbon – 86.5 Hydrogen – 11.5 Oxygen – 1.0 N, S and ash – 1.0 Now determine: 1. The heating (calorific) value per kg 2. Air required for combustion 3. Volume and composition of combustion products 4. Theoretical combustion temperature 5. Composition and ...

Calorific value as per Mendeleev formula Q^...