# Question 5.4: A fuel oil has the following composition (%) Carbon – 86.5 H......

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.

$1 \mathrm{~kg} \mathrm{C} \rightarrow 34,070 \ \mathrm{~kJ} / \mathrm{kg} \\ Hence, \ 0.865 \mathrm{C} \rightarrow 34,070 \times 0.865 \\ = 29,470.55 \ kJ$

Similarly, 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

\begin{aligned}& \mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \\& 1 \mathrm{~kg} \mathrm{C} \text { will require } 2.666 \ \mathrm{~kg} \mathrm{O}_2 \\& \therefore \quad 0.865 \mathrm{~kg} \rightarrow 2.66 \times 0.865 \\& =2.306 \ \mathrm{~kg} \mathrm{O}_2 \\& \text { Similarly, } \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \\& 1 \mathrm{~kg} \mathrm{H}_2 \text { will require } 8 \mathrm{~kg} \mathrm{O}_2 \\& \therefore 0.115 \ \mathrm{~kg} \rightarrow 0.115 \times 8 \\& =0.92 \ \mathrm{~kg} \mathrm{O}_2 \\&\end{aligned}

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 C$O_2$
∴ 0.865 kg → 3.17 kg C$O_2$
Similarly,
1 kg will produce 9 kg $H_2$O
∴ 0.115 → 0.115 × 9 → 1.035 kg $H_2$O
Nitrogen coming in from air = 10.678 kg
Total gases 3.17 + 1.035 + 10.678 = 13.883 kg
Moles of combustion gases
C$O_2$ = 3.17/44 = 0.0720
$H_2$O = 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

$=\frac{1142}{44}=0.260 \ \mathrm{~m}^3 / \mathrm{MJ}$

Volume of individual gases can be obtained from moles of gases on multiplying by 22.4
C$O_2$ = 0.0720 × 22.4 = 1.6128 m³
$H_2$O = 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

$Q^h=V_{cp} \times C_{cp} \times t_c$

where $V_{cp}$ = 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
C$O_2$ – 1.6128 m³
$H_2$O – 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

## A fuel oil has following properties: Analysis W +% Carbon – 86.20 Hydrogen – 12.30 Sulfur – 1.50 Calorific value – 43.40 MJ/kg Theoretical air for combustion – 10.71 m³/kg oil Volume of combustion products – 11.40 m³/kg Composition of combustion products, % volume CO2 – 13.5% H2O – 12.3% N2 – 74.2% ...

The combustion reaction for oil can be written as ...
Question: 5.3