# Question 5.1: A fuel oil has following properties: Analysis W +% Carbon – ......

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

C$O_2$     – 13.5%

$H_2$O      – 12.3%

$N_2$      – 74.2%
Theoretical flame temperature is 2240°C. If the oil is supplied with 10% deficit air, now determine:
1. Volume and composition of combustion products
2. Amount of heat and fuel lost
3. Theoretical flame temperature

Step-by-Step
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The combustion reaction for oil can be written as

Only 90% air is made available

∴ 0.9 × 10.71 = 9.64 m³  available air

This air will contain

$O_2$ – 2.025 m³

$N_2$ – 7.615 m³

1 kg oil requires 2.25 m³$O_2$ for complete combustion.

Hence, 2.025 m³$O_2$ will combust.

$\frac{2.025}{2.25}=0.9 \ \mathrm{~kg} \text { oil }$

Hence, unburned oil is 0.1 kg.
It will contain 0.0862 C

$\begin{gathered}\text { (in } \mathrm{kg} \text { ) } 0.0123 \mathrm{H}_2 \\0.015 \mathrm{~S}\end{gathered}$

Let us assume that the carbon in unburned fuel reacts with water vapor.

$\mathrm{C}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CO}+\mathrm{H}_2 \\ Mole \ 1+1 \rightarrow 1+1$

Unburned oil contains 0.0862 kg C, i.e.,

$\frac{0.0862}{12}=0.00718 \ \text { mole } \mathrm{C}$

This requires 0.00718 mole of $H_2$O and gives 0.00718 mole each of CO and $H_2$.

Volume of $H_2$O required = 0.00718 × 22.4

= 0.1608 m³

Volumes of CO and $H_2$ formed are the same, i.e., 0.160 m³ each.

Volume of combustion products is 11.40 m³/kg. For 0.9 kg oil, the products of combustion will be 10.26 m³. It will contain (vol %)

\begin{aligned}& \mathrm{CO}_2-13.5=1.385 \ \mathrm{~m}^3 \\& \mathrm{H}_2 \mathrm{O}-12.3=1.262 \ \mathrm{~m}^3 \\& \mathrm{~N}_2-74.2=7.613 \ \mathrm{~m}^3\end{aligned}

To this we have to add CO and $N_2$ and subtract $H_2$O from incomplete combustion.

\begin{aligned} \mathrm{CO}_2 & =1.385 \mathrm{~m}^3 \\ \mathrm{H}_2 \mathrm{O} & =1.262-0.1608=1.1012 \mathrm{~m}^3 \\ \mathrm{~N}_2 & =7.613 \mathrm{~m}^3 \\ \mathrm{CO} & =0.1608 \mathrm{~m}^3 \\ \mathrm{H}_2 & =0.1608+0.0123 \text { (from fuel) }=0.1731 \end{aligned}

Total volume of incomplete combustion products is 10.433. The composition (vol %) is

\begin{aligned}& \mathrm{CO}_2-13.28 \\& \mathrm{H}_2 \mathrm{O}-10.55 \\& \mathrm{~N}_2-72.97 \\& \mathrm{CO}-1.54 \\& \mathrm{H}_2-1.66\end{aligned}

The calorific value of oil is 43.4 MJ/kg.
Out of 1 kg only 0.9 kg is burned. Hence,

43.4 × 0.9 = 39.06 MJ evoloved.

For the reaction heat absorbed is 9.902 MJ/kg C.
Unburned fuel is 0.1 kg and contains 0.0862 kg C, hence,
heat absorbed is

9.902 × 0.0862 = 0.853 MJ

Net heat available

= 39.06 – 0.853

= 38.207 MJ

hence,

$\frac{38.2}{43.4}=8.8 \% \text { heat is lost }$

Theoretical flame temperature is

$t=\frac{Q^h(\text { available })}{C_c \times V_c}$

Assuming $C_c$ = 1.65 kJ/m³°C
$t=\frac{38,207}{1.65 \times 10.433}=2,219^{\circ} \mathrm{C}$

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^...
Question: 5.3