Fuels & Combustion Notes FE Sem 1 Applied Chemistry | MU

This module introduces fuels, the properties that make a fuel good, and how the heat value of a fuel is measured and calculated. It then covers proximate and ultimate analysis, the two standard laboratory tests that decide the quality and rank of coal. Finally it shows how to calculate the minimum oxygen and air needed to burn a coal completely. These numericals appear in almost every MU paper and are directly useful for boiler, furnace and power plant work.

Summary

A fuel is a combustible substance containing carbon as the main constituent which, on proper burning, gives a large amount of heat that can be used economically for domestic and industrial purposes. Fuels are classified as primary (natural: wood, coal, crude oil, natural gas) or secondary (derived: coke, petrol, kerosene, coal gas), and each type may be solid, liquid or gaseous.

A good fuel should have a high calorific value, a moderate ignition temperature, low moisture and low non-combustible (ash) content, a moderate and controllable rate of combustion, and should give harmless combustion products. It should also be cheap, easily available, and easy to store and transport, and should contain little sulphur.

Calorific value is the total quantity of heat liberated when a unit mass (or unit volume for gases) of a fuel is burnt completely in air or oxygen. The higher or gross calorific value (HCV or GCV) is the heat liberated when the products of combustion are cooled to room temperature, so the latent heat of the condensed steam is included. The lower or net calorific value (LCV or NCV) is the heat liberated when the products, including steam, escape without condensing, so LCV = HCV minus the latent heat of the steam formed from the hydrogen of the fuel (0.09 H x 587). Common units are cal/g, kcal/kg, J/kg and kcal/m3 for gases.

The calorific value of solid and non-volatile liquid fuels is determined with the bomb calorimeter: a weighed sample is burnt in oxygen at about 25 atm in a steel bomb immersed in a known mass of water, and the heat released is found from the rise in water temperature and the water equivalent of the calorimeter. Corrections are applied for the fuse wire, acid formation (H2SO4 and HNO3), cotton thread and cooling. For gaseous fuels the Boys gas calorimeter is used. When no calorimeter is available, the HCV can be estimated from the ultimate analysis using Dulong's formula, which assumes that the heat comes from carbon, hydrogen and sulphur, and that the oxygen of the fuel is already combined with one-eighth of its weight of hydrogen.

Proximate analysis gives the percentages of moisture, volatile matter, ash and fixed carbon. Moisture is the loss in weight on heating at 105 to 110 degree C for one hour; volatile matter is the loss on heating the dried sample in a covered crucible at 925 plus or minus 5 degree C for exactly 7 minutes; ash is the residue left after burning the coal in an open crucible at 700 plus or minus 50 degree C; and fixed carbon is 100 minus the sum of the other three. High moisture, volatile matter and ash all lower the heating value, while high fixed carbon indicates good quality coal.

Ultimate analysis gives the elemental composition: carbon, hydrogen, nitrogen, sulphur, ash and oxygen. Carbon and hydrogen are found by burning the coal in oxygen and absorbing CO2 in KOH tubes and H2O in anhydrous CaCl2 tubes; nitrogen by the Kjeldahl method; sulphur by precipitating sulphate from the bomb calorimeter washings as BaSO4; ash as in proximate analysis; and oxygen by difference. High carbon and hydrogen give high calorific value, whereas oxygen, nitrogen and sulphur are undesirable.

Combustion is the exothermic oxidation of the fuel: C + O2 gives CO2, H2 + 1/2 O2 gives H2O and S + O2 gives SO2. From these equations, 1 kg of carbon needs 32/12 kg of oxygen, 1 kg of hydrogen needs 8 kg, and 1 kg of sulphur needs 1 kg. The oxygen present in the fuel is subtracted, nitrogen and ash need none. Air contains 23 percent oxygen by mass and 21 percent by volume, so these factors convert oxygen to air. One mole of any gas occupies 22.4 L at NTP, which converts mass of oxygen to volume. In practice excess air is supplied, and the actual air is the minimum air multiplied by (1 + percent excess/100).

Mind map

Determination of calorific value with a bomb calorimeter

Important definitions

Formulas

Key points to remember

• HCV is always greater than LCV because it includes the latent heat of the steam formed from hydrogen.
• A fuel without hydrogen (pure carbon, CO) has HCV equal to LCV.
• The bomb calorimeter gives the GCV, since the steam condenses inside the bomb.
• In Dulong formula, only (H - O/8) hydrogen is available because the oxygen in the fuel is assumed to be combined with hydrogen as water.
• Volatile matter is determined in a covered crucible to avoid burning; ash in an open crucible to allow burning.
• In the C and H estimation, the CaCl2 tube is placed before the KOH tube because KOH absorbs both water and CO2.
• In the Kjeldahl method K2SO4 raises the boiling point of the acid and CuSO4 acts as catalyst.
• Nitrogen in coal has no calorific value; sulphur adds a little heat but forms corrosive and polluting SO2.
• For combustion: 12 kg C needs 32 kg O2; 4 kg H2 needs 32 kg O2 (1 kg H needs 8 kg O2); 32 kg S needs 32 kg O2.
• Always subtract the oxygen already present in the fuel from the oxygen required.
• Air: 23% O2 by weight, 21% O2 by volume. Do not mix the two factors.
• Molar volume at NTP is 22.4 L, so 32 g O2 occupies 22.4 L.

Solved examples

Example 1. In a bomb calorimeter, 0.95 g of a coal sample was burnt. Mass of water in the calorimeter = 2200 g, water equivalent of the calorimeter = 570 g, rise in temperature = 2.42 C. The coal contains 6% hydrogen. Calculate GCV and NCV.

Step 1: GCV = (W + w)(t2 - t1)/x = (2200 + 570) x 2.42 / 0.95. Step 2: (2770 x 2.42) = 6703.4; 6703.4/0.95 = 7056.2 cal/g. Step 3: NCV = GCV - 0.09 H x 587 = 7056.2 - 0.09 x 6 x 587 = 7056.2 - 317.0 = 6739.2 cal/g. Answer: GCV = 7056.2 cal/g (kcal/kg), NCV = 6739.2 cal/g.

Example 2. A coal has the following composition by weight: C = 84%, H = 5%, O = 6%, S = 1.5%, N = 1.5%, ash = 2%. Calculate HCV and LCV using Dulong formula.

Step 1: HCV = 1/100 [8080 C + 34500 (H - O/8) + 2240 S]. Step 2: 8080 x 84 = 678720. Available H = 5 - 6/8 = 4.25; 34500 x 4.25 = 146625. 2240 x 1.5 = 3360. Step 3: Sum = 828705; HCV = 828705/100 = 8287.05 kcal/kg. Step 4: LCV = HCV - 0.09 H x 587 = 8287.05 - 0.09 x 5 x 587 = 8287.05 - 264.15 = 8022.90 kcal/kg. N and ash do not contribute.

Example 3. 2.5 g of an air-dried coal sample was heated at 110 C for 1 hour and the residue weighed 2.415 g. The residue was then heated in a covered crucible at 925 C for 7 minutes and weighed 1.528 g. On complete burning, the ash left weighed 0.245 g. Calculate the proximate analysis.

Moisture: loss = 2.5 - 2.415 = 0.085 g; % moisture = 0.085/2.5 x 100 = 3.40%. Volatile matter: loss = 2.415 - 1.528 = 0.887 g; % VM = 0.887/2.5 x 100 = 35.48%. Ash: % ash = 0.245/2.5 x 100 = 9.80%. Fixed carbon = 100 - (3.40 + 35.48 + 9.80) = 100 - 48.68 = 51.32%.

Example 4. 0.25 g of coal was burnt in a combustion apparatus. The increases in weight of the KOH tube and the CaCl2 tube were 0.72 g and 0.12 g respectively. Calculate % C and % H.

Step 1: % C = (increase in KOH tube x 12 x 100)/(44 x wt of coal) = 0.72 x 12 x 100/(44 x 0.25) = 864/11 = 78.55%. Step 2: % H = (increase in CaCl2 tube x 2 x 100)/(18 x wt of coal) = 0.12 x 2 x 100/(18 x 0.25) = 24/4.5 = 5.33%.

Example 5. 1.2 g of coal was Kjeldahlised and the ammonia evolved was absorbed in standard acid; it neutralised 9.0 mL of 0.1 N H2SO4. In a separate experiment 1.5 g of coal gave 0.12 g of BaSO4 in the bomb washings. Calculate % N and % S.

Step 1: % N = (V x N x 1.4)/wt of coal = 9.0 x 0.1 x 1.4/1.2 = 1.26/1.2 = 1.05%. Step 2: % S = (wt BaSO4 x 32 x 100)/(233 x wt of coal) = 0.12 x 32 x 100/(233 x 1.5) = 384/349.5 = 1.10%.

Example 6. A coal contains C = 80%, H = 5%, O = 6%, S = 1%, N = 2% and ash = 6%. Calculate the minimum weight of air required for complete combustion of 1 kg of the coal. Also find the weight of air if 40% excess air is supplied.

Step 1: O2 for C: C + O2 gives CO2; 12 kg C needs 32 kg O2, so 0.80 kg C needs 0.80 x 32/12 = 2.133 kg. Step 2: O2 for H: H2 + 1/2 O2 gives H2O; 2 kg H needs 16 kg O2, so 0.05 kg needs 0.05 x 8 = 0.400 kg. Step 3: O2 for S: 32 kg S needs 32 kg O2, so 0.01 kg needs 0.010 kg. Step 4: Total = 2.543 kg; subtract O2 in fuel 0.060 kg; net minimum O2 = 2.483 kg. Step 5: Minimum air = 2.483 x 100/23 = 10.80 kg. Step 6: With 40% excess, air = 10.80 x 1.40 = 15.12 kg.

Example 7. Calculate the volume of air at NTP required for complete combustion of 1 kg of coal containing C = 90%, H = 4%, S = 1%, O = 3%, rest ash.

Step 1: O2 needed (kg) = 0.90 x 32/12 + 0.04 x 8 + 0.01 x 1 - 0.03 = 2.400 + 0.320 + 0.010 - 0.030 = 2.700 kg = 2700 g. Step 2: Moles O2 = 2700/32 = 84.375 mol. Volume at NTP = 84.375 x 22.4 = 1890 L = 1.89 m3. Step 3: Air has 21% O2 by volume, so volume of air = 1.89 x 100/21 = 9.0 m3.

Important questions and answers

Q1. Define calorific value. Distinguish between HCV and LCV. (3 marks)

Calorific value is the total heat liberated when unit mass or volume of a fuel is burnt completely. HCV is the heat liberated when the combustion products are cooled to room temperature, so latent heat of condensed steam is included. LCV is the heat liberated when the products escape without condensing, so it excludes that latent heat. LCV = HCV - 0.09 H x 587; HCV is always greater for hydrogen-containing fuels.

Q2. State the characteristics of a good fuel. (4 marks)

A good fuel should have: (1) high calorific value; (2) moderate ignition temperature, neither too low (fire hazard) nor too high (hard to start); (3) low moisture content; (4) low non-combustible matter and ash; (5) moderate, controllable rate of combustion; (6) harmless, non-polluting combustion products and low sulphur; (7) low cost and ready availability; (8) easy storage and transport.

Q3. Describe the determination of calorific value of a solid fuel by bomb calorimeter, with a neat labelled diagram. (6 marks)

Diagram labels: stainless steel bomb, oxygen valve, electrodes, crucible with fuel, fuse wire, copper calorimeter with water, stirrer, Beckmann thermometer, air jacket and water jacket. Working: x g of fuel in the crucible is connected by a fuse wire, the bomb is filled with O2 at 25 atm and placed in W g of water; the water equivalent of the apparatus is w g. Initial temperature t1 is noted, the fuel is ignited electrically, and the maximum temperature t2 is noted. Heat liberated = heat absorbed, so GCV = (W + w)(t2 - t1)/x cal/g. LCV = GCV - 0.09 H x 587. Corrections for fuse wire, acid, cotton thread and cooling may be applied.

Q4. Explain the proximate analysis of coal and its significance. (6 marks)

Moisture: 1 g powdered coal heated at 105-110 C for 1 hour; % moisture = loss/wt x 100. Significance: moisture lowers CV and wastes heat in evaporation. Volatile matter: dried coal heated in a covered crucible at 925 plus or minus 5 C for 7 min; % VM = loss/wt x 100. Significance: high VM gives a long smoky flame and low CV; such coal suits gas making, low VM coal suits coke making. Ash: coal burnt in open crucible at 700 plus or minus 50 C; % ash = residue/wt x 100. Significance: ash lowers CV, forms clinkers, adds disposal cost. Fixed carbon = 100 - (M + VM + A); higher FC means higher CV and better coal.

Q5. Explain ultimate analysis of coal and its significance. (6 marks)

C and H: a known weight of coal is burnt in O2; H2O is absorbed in a weighed anhydrous CaCl2 tube and CO2 in a weighed KOH tube; % C = increase in KOH x 12 x 100/(44 x wt), % H = increase in CaCl2 x 2 x 100/(18 x wt). N (Kjeldahl): coal digested with conc. H2SO4, K2SO4 and CuSO4 to ammonium sulphate, made alkaline with NaOH, NH3 distilled into standard acid and the unused acid back-titrated; % N = V N x 1.4/wt. S: sulphate in bomb washings precipitated as BaSO4; % S = wt BaSO4 x 32 x 100/(233 x wt). Ash as in proximate analysis. O = 100 - (C + H + N + S + ash). Significance: high C and H raise CV; N has no heating value; S causes corrosion and SO2 pollution; high O lowers CV and indicates low-rank, moisture-holding coal.

Q6. State Dulong formula and explain the term (H - O/8). (3 marks)

HCV = 1/100 [8080 C + 34500 (H - O/8) + 2240 S] kcal/kg, where C, H, O, S are percentages. The oxygen in the fuel is assumed to be already combined with hydrogen as water in the ratio 8:1 by weight (H2O: 2 g H with 16 g O). So O/8 parts of hydrogen are not available for burning, and only (H - O/8) contributes heat.

Q7. Calculate GCV and NCV: mass of coal 0.95 g, water 2200 g, water equivalent 570 g, rise in temperature 2.42 C, H = 6%. (4 marks)

GCV = (2200 + 570) x 2.42/0.95 = 7056.2 cal/g. NCV = 7056.2 - 0.09 x 6 x 587 = 6739.2 cal/g.

Q8. Calculate the weight and volume of air required for complete combustion of 1 kg coal with C 90%, H 4%, S 1%, O 3%, ash 2%. (5 marks)

O2 = 0.9 x 32/12 + 0.04 x 8 + 0.01 - 0.03 = 2.70 kg. Weight of air = 2.70 x 100/23 = 11.74 kg. Volume of O2 at NTP = 2700/32 x 22.4 = 1890 L = 1.89 m3; volume of air = 1.89 x 100/21 = 9.0 m3.

Q9. Why is the oxygen content of coal subtracted in combustion calculations, and why does nitrogen not appear? (3 marks)

Oxygen present in the coal is available for combustion, so it reduces the amount of oxygen that must be supplied from air. Nitrogen in coal is not oxidised during normal combustion (it escapes mainly as N2), so it needs no oxygen. Ash is non-combustible and also needs no oxygen.

Q10. Calculate the proximate analysis: 2.5 g coal; after heating at 110 C 2.415 g; after heating at 925 C 1.528 g; ash 0.245 g. (4 marks)

Moisture = 0.085/2.5 x 100 = 3.40%; VM = 0.887/2.5 x 100 = 35.48%; ash = 0.245/2.5 x 100 = 9.80%; FC = 100 - 48.68 = 51.32%.

Exam tips

• Always write the formula first, then substitute; marks are given for the formula even if arithmetic slips.
• Write units in every answer: cal/g or kcal/kg for solids, kcal/m3 for gases.
• In combustion problems, write the three balanced equations and the weight ratios before calculating.
• Check whether the question asks for weight or volume of air and use 23 or 21 accordingly.
• In proximate analysis numericals, see whether volatile matter was found on the same (dried) sample or a separate sample; the loss is always divided by the original sample weight.
• Draw the bomb calorimeter diagram neatly with at least eight labels for the 6-mark question.

FAQs

Why is HCV greater than LCV?

Because HCV includes the latent heat released when the steam formed from the hydrogen of the fuel condenses, while LCV assumes the steam escapes uncondensed.

Which calorific value does the bomb calorimeter measure?

The gross (higher) calorific value, since the products cool and the water condenses inside the bomb. LCV is then calculated.

Why is 587 used in the LCV formula?

587 kcal/kg (cal/g) is the latent heat of condensation of steam used in standard textbooks; 9 parts water are formed per part hydrogen, hence 0.09 H x 587.

Is volatile matter determined in an open or covered crucible?

Covered, so that the coal is heated without burning; ash is determined in an open crucible.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Applied Chemistry) and were prepared with AI help and checked.