N L N G : C O M B U S T I O N G R A N T I N G R A M 6 T H D E C E M B E R 2 0 1 7 This short lecture provides an introduction into the basic principles of combustion and some of the background to the use of fossil fuels. Combustion This lecture provides a very brief overview of the major chemical processes that take place during combustion. Combustion is a rapid reaction usually with a flame which occurs between a fuel and an oxygen carrier such as air. Molecules of fuel and air have energy stored in the bonds between atoms. In the new molecules formed by combustion the energy stored is reduced. The difference is the energy released during combustion. Combustion is said to be exothermic as it releases heat to the surroundings. Figure 1: Fossil fuels are a popular choice for aircraft engines as they have a high energy density Combustion is important in engineering as it provides the heat for a lot of thermal power cycles. Most common fuels consist of hydrogen and carbon whether solid (e.g. coal), liquid (e.g. diesel) or gaseous (e.g North Sea gas) see Table 1. Fuel Composition Table 1: Common Fuel Compositions Standard LNG 89.63% CH 4, 6.32% C 2 H 6, 2.16% n-c 3 H 8, 1.20% n-c 4 H 10, 0.69% N 2 Bitumous Coal 74% C, 5.98% H, 13.01% O, 2.26% N + S, 4.75% Ash Petrol 85.5% C, 14.4% H, 0.1% S North Sea Gas 93% CH 4, 4.8% C 2 H 6, 2% N 2, 0.2% C0 2 To sum up: Combustion is a rapid reaction (usually with a flame) that occurs between a fuel and an oxygen carrier (normally air). Chemical energy is turned into heat. Combustion processes are described in a chemical equation which shows how the atoms of the reactants are rearranged to from the products of combustion, for example Carbon and Oxygen can be combusted to produce Carbon Dioxide: C + O 2 CO 2 Chemical equations are the conservation of mass in terms of the conservation of atoms. Stoichiometric combustion occurs when the quantity of air supplied is just sufficient to completely burn a given quantity of fuel. The mole is defined as the amount of substance in a system that contains as many elementary entities as there are in exactly 0.012kg of carbon 12 ( 12 C). Those elementary entities may be atoms, molecules, electrons etc. Figure 2: Combustion Process
nlng: combustion 2 The SI abbreviation for the mole is mol although kmol is often a more convenient unit. The number of moles in a system is related to the system s mass by the atomic or molecular weight. The number of particles (atoms, molecules etc.) in a mole is defined by the Avagadro Constant (6.022 14 10 23 particles/mol) One of the key advantages of hydrocarbon fuels is that they have a high energy density - see Table 2). Fuel Energy Density Methane 50 000 kj/kg LNG 46 100 kj/kg JP8 Jet Fuel 43 180 kj/kg Lithium-Ion Battery 721 kj/kg Hydro scheme at 100 m head 981 kj/kg Wind at 10 m/s 50 kj/kg Average power in wave-front North Atlantic 109 kw/m Peak solar irradiance in Mingulay, Scotland 0.6 kw/m 2 Table 2: Energy Densities of Various Power Sources Example 1.1. Methane is burnt in Oxygen to produce Carbon Dioxide and Water. Determine the molal composition after combustion, given that CH 4 + ao 2 xco 2 + yh 2 O where a,x and y are unknown. Determine also the mass composition after combustion. If burning in air, Nitrogen is present, N 2 (Molecular mass of 1 atom, N = 14 so then N 2 = 28): O 2 N 2 Table 3: Composition of Air Volumetric 21.0% 79.0% Gravimetric 23.2% 76.8% So this has to be taken into account, for example burning CH4 in air: CH 4 + 2O 2 + 2 (79/21)N 2 CO 2 + 2H 2 O + 2 (79/21)N 2 by mass (kg): 16 + 64 + 210.67 44 + 36 + 210.67 Stoichiometric air: fuel ratio (by mass) is: Air:Fuel = 64+210.67:16 = 17.2:1. This is with all fuel burnt, no excess oxygen.
nlng: combustion 3 Example 1.2. What is the stoichiometric air-fuel ratio, by mass, for burning gasoline (petrol) containing 85.5% C and 14.5% H in atmospheric air? The molal composition of the exhaust in the example was: 7.12 CO 2, 7.25 H 2 0 and 40.42 N 2, to convert to a mass fraction you need to use the molecular mass. Moles (= volume ) % Molecular Mass Mass / [kg] % CO 2 7.12 13.0 44 44 7.12 = 313.28 19.9 H 2 O 7.25 13.2 18 18 7.25 = 103.50 8.3 N 2 79 21 10.745 = 40.42 73.8 28 28 40.42 = 1131.8 71.8 54.79 100 1575.9 100 Here we have assumed that H 2 O has remained as a vapour and none of it has condensed. These equations are for complete reactions; in practice the reaction may be incomplete, with both CO and O 2 in the exhaust and even some CH (unburnt hydrocarbons) or C (particulates). The high combustion temperature may also cause N 2 + O 2 NO, NO 2 and so on (NO X ) Table 4: Exhaust Composition If more air than stoichiometric - mixture is fuel lean. If less air than stoichiometric - mixture is fuel rich. Disadvantages of Fossil Fuels C0 2 is produced by burning Hydrocarbons. This is a fundamental part of the process! C0 2 is a greenhouse gas which is causing climate change. Climate change is generally regarded as a bad thing and a reduction in C0 2 production is a key driver of energy policies. To reduce C0 2 production when using fossil fuels the following options are available: A Green house gas is simply put a gas that traps heat in the atmosphere. The main greenhouse gases are: Carbon Dioxide (C0 2 ), Methane (CH 4 ), Nitrous Oxide (N 2 O) and Flourinated gases improve cycle efficiency (in a car use a higher pressure ratio diesel cycle) use less energy for the application (use a lighter car) don t use hydrocarbon fuel (use an electric car) NO X is a byproduct of high temperature combustion. This can be manipulated by advanced combustion techniques. The supply of fossil fuels is also finite. Many of the current conflicts around the world are strongly influenced by energy security concerns. This and other concerns led to the idea of sustainability coined by a UN commission in 1987 1 it is a straightforward concept: Figure 3: ECS Electric Car. This requires no combustion in the car itself. Source: http://experiencedurhamev. blogspot.co.uk/ 1 Our Common Future. Also known as the Brundtland report after the Norweigan Prime Minister who chaired the committee. http://www.un-documents.net/ wced-ocf.htm
nlng: combustion 4 Humanity has the ability to make development sustainable to ensure that it meets the needs of the present without compromising the ability of future generations to meet their own needs. Conservation of Energy Applied to Combustion Consider a steady flow calorimeter which is has a continuous flow through the system. This has three steps: Reactants (fuel+o 2 ) at pressure P 0, temperature T 0 are introduced to the combustion chamber Combustion takes place and heat is extracted The products of combustion leave at pressure P 0 and temperature T 0. So the heat extracted is quite large as it has to taken away most of the energy in combustion. Apply the SFEE to this system where we have products denoted by subscript P and reactants with subscript R: Q + Ẇ + ṁ i (h i + U2 i 2 + gz i ) = 0 W = 0 and z P = z R and we can also arranged for U P = U R which means that: Figure 4: Steady Flow Calorimeter Q = m P h P m R h R = H P H R = H The datum for these tests is by convention 1 bar and 25 C. A superscript or subscript 0 is often used to denote this. So H = H 0. The relation between H 0 and U 0 is determined in the same way any relationship between internal energy and enthalpy is obtained: H 0 = H P0 H R0 = U P0 + (p O V) P U R0 (p O V) R = U 0 + (p 0 V) P (p 0 V) R But we can write p 0 V = nr 0 T 0 where of course n is the number of moles for the gaseous component. So (p 0 V) P (p 0 V) R = R 0 T 0 (n P n R ) and therefore: H 0 = U 0 + R 0 T 0 (n P n R ) If n P = n R clearly H 0 = U 0 which is the case for some common cases. Even if n P = n R, the second term is small i.e. H 0 U 0. H 0 and U 0 are both known as the Calorific Value and tables of values for different fuels are widely available. Calorific Value (also known as heat of reaction, or enthalpy of combustion) is usually expressed in J/kg but sometimes as J/mol. For fuel containing hydrogen, two values are possible depending on whether the water in the combustion products is condensed or not. These are:
nlng: combustion 5 HCV - Higher Calorific Value (water condensed) LCV - Lower Calorific Value (water as vapour) The difference is latent heat of condensation. HCV kj/kg LCV kj/kg Hydrogen H 2 142 000 120 000 Methane CH 4 55 500 50 010 Natural Gas 50 200 46 100 n-octane C 8 H 12 petrol (vapour) 48 270 44 800 JP8 Jet Fuel C 12 H 23.8 46 087 43 180 Table 5: Typical Values of H 0 This section on combustion is intended to provide you with an introduction to the topic. Much more information on the topic is found in Rogers and Mayhew 2 or Potter 3 detailing much more advanced topics such as enthalpy of formation and dissociation. Combustion Problem 2 GFC Rogers amd YR Mayhew. Engineering Thermodynamics, Work and Heat Transfer. Longman Group, 1992 3 Merle C Potter. Thermodynamics Demystified. McGraw Hill, 2009 Butane (C 4 H 10 ) is to be burned in dry air. (a) Calculate the stoichiometric air-fuel ratio. (b) If the actual air-fuel ratio is 25 calculate the volumetric percentage of C0 2 in the exhaust. Answers: a) 15.39:1 b) 7.5%
nlng: combustion 6 References GFC Rogers amd YR Mayhew. Engineering Thermodynamics, Work and Heat Transfer. Longman Group, 1992. Merle C Potter. Thermodynamics Demystified. McGraw Hill, 2009.