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 amount of combustion product at

20% excess air

6. Theoretical combustion temperature at 20% excess air

Step-by-Step

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Calorific value as per Mendeleev formula

Q^h=4.187(81 \times \mathrm{C}+300 \times \mathrm{H}-26(\mathrm{O}-\mathrm{S})) \mathrm{kJ} / \mathrm{kg}Neglecting the last term, as % O and S are small, and substituting

\begin{aligned}Q^h & =4.187(81 \times 86.5+300 \times 11.5) \\& =43.78 \ \mathrm{MJ} / \mathrm{kg} \\& \cong 44 \ \mathrm{MJ} / \mathrm{kg}\end{aligned}We can determine Q^h from heat given by the C and H reactions.

1 \mathrm{~kg} \mathrm{C} \rightarrow 34,070 \ \mathrm{~kJ} / \mathrm{kg} \\ Hence, \ 0.865 \mathrm{C} \rightarrow 34,070 \times 0.865 \\ = 29,470.55 \ kJSimilarly, 1 kg H_2 → 143,112 kJ/kg

Hence, 0.115 H_2 → 143,112 × 0.115

= 16,457.88 kJ

Adding Equation (5.28) and Equation (5.29) and neglecting O, S, and N

Q^h = 45.928 ~ 46.0 MJ/kg

This value of Q^h is higher than that obtained in Equation (5.27). This is expected because the second method assumes that oil contains elemental C and H, which is not true, hence, we will take Q^h = 44.0 MJ/kg for further calculations.

To calculate air required (theoretical) for combustion of 1 kg oil consider the reactions for C and O

Hence, total O_2 required for 1 kg of oil

= 2.306 + 0.92

= 3.226 kg

On weight basis N_2 : O_2 for air is 3.31

∴ N_2 corresponding to 3.226 kg O_2 is

3.31 × 3.226 = 10.678 kg

Air required for combustion of 1 kg oil is

3.226 + 10.678 = 13.904 kg

On volume basis taking density of air as 1.293 kg/m³

Air required = 10.75 m³

From above combustion equations

1 kg C will produce 3.666 kg CO_2

∴ 0.865 kg → 3.17 kg CO_2

Similarly,

1 kg will produce 9 kg H_2O

∴ 0.115 → 0.115 × 9 → 1.035 kg H_2O

Nitrogen coming in from air = 10.678 kg

Total gases 3.17 + 1.035 + 10.678 = 13.883 kg

Moles of combustion gases

CO_2 = 3.17/44 = 0.0720

H_2O = 1.035/18 = 0.0575

N_2 = 10.68/28 = 0.3813

Total moles = 0.5108

As 1 k mole occupies 22.4 m³ (STP),

thus, volume of combustion gases

= 11.442 m³/kg oil

As 1 kg oil gives 44 MJ

combustion gases per MJ

Volume of individual gases can be obtained from moles of gases on multiplying by 22.4

CO_2 = 0.0720 × 22.4 = 1.6128 m³

H_2O = 0.0575 × 22.4 = 1.288 m³

N_2 = 0.3813 × 22.4 = 8.541 m³

The percentage composition of combustion gases by volume

\left.\begin{array}{c}\mathrm{CO}_2-14.11 \% \\\mathrm{H}_2 \mathrm{O}-11.25 \% \\\mathrm{~N}_2-74.64 \%\end{array}\right\}Assuming that all the heat produced by combustion is taken up by the gases, and the specific heat of composite flue gas as 1.68 kJ/m³°C, then

Q^h=V_{cp} \times C_{cp} \times t_cwhere V_{cp} = Volume of combustion products (m³/kg)

C_{cp} = Specific heat of combustion gases (at constant pressure) (kJ/m³°C)

t_c = Theoretical combustion temperature (°C)

Substituting

44 × 10³ = 11.4 × 1.68 × t_c

t_c = 2289°C

Air required for combustion is 10.75 m³/kg oil.

N_2 : O_2 volume ratio for air is 3.76 : 1

For 20% excess air the quantity is

10.75 × 0.20 = 2.150 m³ which will contain

N_2 – 1.698 m³

O_2 – 0.452 m³

Combustion gases at theoretical air are

CO_2 – 1.6128 m³

H_2O – 1.288 m³

N_2 – 8.541 m³

Composition of combustion products will be

\left.\begin{array}{l}\mathrm{CO}_2-1.6128 \ \mathrm{~m}^3 \\\mathrm{H}_2 \mathrm{O}-1.288 \ \mathrm{~m}^3 \\\mathrm{~N}_2-8.541+1.698=10.239 \ \mathrm{~m}^3 \\\mathrm{O}_2-0.452 \ \mathrm{~m}^3 \\\text { Total }-13.592 \ \mathrm{~m}^3\end{array}\right\}On percentage basis

\left.\begin{array}{l}\mathrm{CO}_2-11.84 \% \\\mathrm{H}_2 \mathrm{O}-9.47 \% \\\mathrm{~N}_2-75.73 \% \\\mathrm{O}_2-3.32 \%\end{array}\right\}Assuming same specific heat, i.e., 1.68 kJ/m³°C the theoretical combustion temperature for 20% excess air is

\begin{aligned}44 \times 10^3 & =13.592 \times 1.68 \times t_c \\t_c & =1927^{\circ} \mathrm{C}\end{aligned}Question: 5.1

The combustion reaction for oil can be written as
...

Question: 5.3

The given composition of the gas shows that methan...

Question: 5.2

The sample contains 25% moisture. In incineration ...