The energy released by a unit quantity of fuel when it is burnt is known as its calorific value and is usually quoted in kJ/kg. The calorific value of a fuel depends on its make up. The constituents of fuel which release energy when they are burnt are Hydrogen, carbon and sulphur Hydrogen releases the most energy; 144000kJ/kg. Carbon releases 33700kJ/kg. Sulphur is the lowest, only releasing 9300kJ/kg.
It should be evident therefore, that the less sulphur, and the more hydrogen a fuel contains, the greater its calorific value will be.
For example, a typical marine heavy fuel oil may contain 85% carbon, 12% hydrogen and 3% sulphur by mass.
. The higher calorific value (HCV) of the fuel will be (0.85 × 33700) + (0.12 × 144000) + (0.03 × 9300) = 46204kJ/kg.
Not all the energy released can be utilised; when hydrogen burns, the water produced is as steam, and this takes energy from the burning fuel. This energy lost is known as the enthalpy of evaporation, and is 2442kJ/kg (of water produced). The useful energy in the fuel is known as it's Lower Calorific Value (LCV) and is calculated by subtracting the energy required to turn the water produced from burning the hydrogen into steam from the Higher Calorific Value (HCV)
When a mass of Hydrogen is burnt, 9 times that mass is produced as water.
So LCV = HCV - (9 × mass of hydrogen/kg fuel × 2442)
For the example above LCV = 46204 - (9 × 0.12 × 2442) = 43566.64 kJ/kg
Download the spreadsheet here and calculate the HCV & LCV of a fuel
Engine Manufacturers quote their fuel consumption figures in g/kWh. Typical figures for a modern engine are between 165 and 170g/kWh. This gives a guide to the efficiency of the engine.
Because the specific fuel consumption would vary depending on its calorific value (the lower the value, the more fuel must be burnt to produce a certain amount of power), engine manufacturers base their figures on a calorific value of 42700kJ/kg
for example if a fuel with a lower calorific value of 42700kJ/kg is burnt then 1g will release 427kJ of useful energy and 170grams (0.17kg) will release 7259kJ of useful energy. If this is the energy released in one hour to provide 1kW of power for one hour then the energy released per second by the fuel is 7259 ÷ 3600 = 2.02kJ
This means that for every kW of power produced at the output shaft, the engine must produce 2.02kJ energy from the fuel, making it just under 50% efficient.
The calorific value of a fuel can be determined without knowing its precise ratio of carbon hydrogen and sulphur. To do this, a piece of equipment called a bomb calorimeter is used.
Heat released by burning fuel (Q) = mass of water (m) × specific heat capacity of water (C) × change in temperature (DT)
The specific heat capacity is defined as the quantity of energy required to raise 1kg of the substance through 1°C, and for water is 4.2kJ/kg°C
The heat energy absorbed by the bomb and the water container must also be taken into account. To simplify the calculation the manufacturer of the equipment gives a water equivalent of the bomb and container which is added to the mass of water to allow for the heat energy absorbed by them.
The equation therefore becomes Q = (mass of water + water equivalent) × 4.2 × DT
For example: A fuel sample of 0.7g raises the temperature of 2kg water + bomb and container with a water equivalent of 0.5kg by 3°C. Calculate the calorific value of the fuel.
Q = 2.5 × 4.2 × 3 = 31.5kJ.
This is the heat energy released by 0.7 grams of fuel. Therefore the heat released by 1kg fuel = 31.5 ÷ 0.0007 = 45000kJ/kg
The higher calorific value (HCV) of the fuel is therefore 45000kJ/kg
This is the higher calorific value because the hydrogen burns to water (in the form of steam). This steam then condenses on the inside surfaces of the bomb giving up its heat energy.
To calculate the lower calorific value (LCV) the amount of water produced by burning the fuel sample must be measured. The water produced by burning 1kg of fuel can then be calculated. and if this figure is multiplied by 2442 (the enthalpy of evaporation) and subtracted from the HCV. The LCV will be found.
For example: After the combustion of the sample above, the bomb is opened and found to contain 0.75g of water.
LCV = HCV - {(0.75 ÷ 0.7) × 2442} = 45000 - 2616.4 = 42383.6kJ/kg
The spreadsheet allows you to input data and view the results of the experiment. |

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