---
title: "Calorific value and air from ultimate analysis"
subject: "Energy Resources"
chapter: 1
key: "semester_6/energy_resources:dulong-calorific-value-fuel-oil"
source: https://www.bctengineer.com/q/semester_6/energy_resources/dulong-calorific-value-fuel-oil
site: BCT Engineering
---

# Calorific value and air from ultimate analysis

Energy Resources · Chapter 1: Fossil fuels and their characteristics

- Asked in: year not given, 6 marks
- Video: [Solving HCV and LCV Numericals Using Dulong's Formula | Engineering Chemistry 2](https://www.youtube.com/watch?v=Euz3ZIkv6P8) — Saarang Maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/dulong-calorific-value-fuel-oil
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-1

## Question

A fuel oil has the following ultimate analysis by mass: C = 82 %, H = 12 %, O = 4 %, S = 1 %, N = 1 %. Using Dulong's formula, calculate (a) the higher calorific value, (b) the lower calorific value (latent heat of steam 2442 kJ/kg), and (c) the minimum mass of air required for complete combustion per kg of fuel.

## Short answer

Using Dulong's formula: HCV = 33800C + 144000(H - O/8) + 9270S. For C=0.82, H=0.12, O=0.04, S=0.01, HCV = 44,370 kJ/kg. LCV = HCV - 9H×2442 = 41,730 kJ/kg. Minimum air = (100/23)[(8/3)C + 8(H-O/8) + S] = 13.55 kg air per kg fuel.

## Answer

#### Formulas
Dulong's formula (mass fractions):

$$
HCV = 33\,800\,C + 144\,000\left(H - \frac{O}{8}\right) + 9\,270\,S\ \ \text{kJ/kg}
$$

Here $C = 0.82$, $H = 0.12$, $O = 0.04$, $S = 0.01$.

#### (a) Higher calorific value
$$
\begin{aligned}
H - \frac{O}{8} &= 0.12 - 0.005 = 0.115 \\
HCV &= 33\,800(0.82) + 144\,000(0.115) + 9\,270(0.01) \\
&= 27\,716 + 16\,560 + 92.7 \\
&= 44\,369\ \text{kJ/kg}
\end{aligned}
$$

#### (b) Lower calorific value
Each kg of hydrogen forms 9 kg of steam, whose latent heat is not recovered.

$$
\begin{aligned}
LCV &= HCV - 9H\,(2442) \\
&= 44\,369 - 9(0.12)(2442) \\
&= 44\,369 - 2\,637 \\
&= 41\,731\ \text{kJ/kg}
\end{aligned}
$$

#### (c) Minimum air
The oxygen needed per kg of fuel is $\frac{8}{3}C + 8\left(H - \frac{O}{8}\right) + S$. Air contains 23 % O₂ by mass, so:

$$
\begin{aligned}
A_{min} &= \frac{100}{23}\left[\frac{8}{3}C + 8\left(H-\frac{O}{8}\right) + S\right] \\
&= 4.348\,[2.187 + 0.920 + 0.010] \\
&= 4.348 \times 3.117 = 13.55\ \text{kg air/kg fuel}
\end{aligned}
$$

Here $\tfrac{8}{3}(0.82) = 2.187$ kg of O₂ is for carbon (C + O₂ → CO₂ needs 32/12 kg O₂ per kg C), $8(0.115) = 0.920$ kg is for hydrogen, and $1 \times 0.01$ kg is for sulphur.

**Answer:** HCV = 44 370 kJ/kg (about 44.4 MJ/kg); LCV = 41 730 kJ/kg (about 41.7 MJ/kg); minimum air = 13.55 kg per kg of fuel.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In Dulong's formula why is (H - O/8) used?
   A. Oxygen adds extra heat to hydrogen
   B. Part of the hydrogen is already combined with oxygen in the fuel as water and gives no heat
   C. It converts mass to volume fraction
   D. It accounts for the sulphur content
   Answer: B. O/8 is the hydrogen bound to the fuel oxygen; only the remainder, 0.115 here, is available to burn. (Key point: Effective hydrogen term)

2. For C=82%, H=12%, O=4%, S=1%, what is the HCV by Dulong's formula?
   A. About 44 370 kJ/kg
   B. About 41 730 kJ/kg
   C. About 27 716 kJ/kg
   D. About 13 560 kJ/kg
   Answer: A. 27 716 + 16 560 + 92.7 = 44 369 kJ/kg. (Key point: HCV value)

3. How is LCV obtained from HCV?
   A. Add 9H times 2442 for steam heat
   B. Subtract 2442 times the carbon fraction
   C. Multiply HCV by 0.23
   D. Subtract 9H times 2442, since each kg of H forms 9 kg of steam whose latent heat is lost
   Answer: D. 9(0.12)(2442) = 2637 kJ/kg, so LCV = 44 369 - 2637 = 41 731 kJ/kg. (Key point: LCV from HCV)

4. Why is the oxygen requirement divided by 0.23 to get minimum air?
   A. Air contains 23% nitrogen by volume
   B. Only 23% of the fuel burns
   C. Air contains 23% oxygen by mass
   D. 23% excess air is always supplied
   Answer: C. Minimum air = (100/23) x O2 needed = 4.35 x 3.117 = 13.56 kg air/kg fuel. (Key point: Minimum air)

## Related in Energy Resources

- [Fuel cell voltage, efficiency and hydrogen use](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Afuel-cell-stack-numerical) — Energy Resources, chapter 6
- [Flat plate collector useful heat and outlet temperature](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Aflat-plate-collector-efficiency-numerical) — Energy Resources, chapter 2

## Related in other subjects

- [Calorific value of fuel by Dulong's formula](https://www.bctengineer.com/question.md?key=semester_6%2Finternal_combustion_engines%3Acalorific-value-dulong-formula-numerical) — Internal Combustion Engines, chapter 3