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
CO_2 – 13.5%
H_2O – 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
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 m³O_2 for complete combustion.
Hence, 2.025 m³O_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
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+1Unburned 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_2O and gives 0.00718 mole each of CO and H_2.
Volume of H_2O 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_2O 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}
