COMBUSTION
COMBUSTION
In order to operate a heat engine we need a hot source In order to operate a heat engine we need a hot source together with a cold sink
together with a cold sink
Occasionally these occur together in nature Occasionally these occur together in nature
eg:-geothermal sites or solar powered engines, but usually the geothermal sites or solar powered engines, but usually the heat source has to be 'artificially' provided.
heat source has to be 'artificially' provided.
The most common way of doing this is by the combustion of The most common way of doing this is by the combustion of a fuel. (Nuclear fission/fusion & solar are alternatives )
a fuel. (Nuclear fission/fusion & solar are alternatives ) Engines
Engines
Where the heat from combustion is transferred from the Where the heat from combustion is transferred from the products (or process ) of combustion to a separate working products (or process ) of combustion to a separate working fluid , the engine
fluid , the engine is referred to as an EXTERNAL combustionis referred to as an EXTERNAL combustion engine
engine
eg:-Steam engines / turbines Steam engines / turbines
Stirling engines Stirling engines
These constitute true 'heat engines' as we have studied to These constitute true 'heat engines' as we have studied to date.
date.
Where the combustion occurs within the working fluid ( usually Where the combustion occurs within the working fluid ( usually air ) the engine is referred to as an INTERNAL combustion air ) the engine is referred to as an INTERNAL combustion engine ie:
engine ie:
Petrol & Diesel engines Petrol & Diesel engines
Gas turbines Gas turbines
These are not true heat engines although we often analyse These are not true heat engines although we often analyse
THE COMBUSTION PROCESS THE COMBUSTION PROCESS
FUEL:-Solid:- Coal, wood - consists mainly of C, H,
Solid:- Coal, wood - consists mainly of C, H, & O + impurities& O + impurities Liquid:- Large hydrocarbon molecules of varying boiling points Liquid:- Large hydrocarbon molecules of varying boiling points mainly C & H : Petrol, Diesel, Fuel oil etc;
mainly C & H : Petrol, Diesel, Fuel oil etc;
Gas:- Small hydrocarbon molecules - methane, ethane, Gas:- Small hydrocarbon molecules - methane, ethane, propane, butane, etc; plus a range of manufactured gases ; propane, butane, etc; plus a range of manufactured gases ; eg
eg HH22 , Acetylene etc.., Acetylene etc..
OXIDANT:-Usually air ( the oxygen in the air ) Usually air ( the oxygen in the air )
Where air is unavailable the oxidant has
Where air is unavailable the oxidant has to be carried as wellto be carried as well as the fuel - space vehicles, rockets etc;
as the fuel - space vehicles, rockets etc; PRODUCTS
PRODUCTS ( ( OF OF COMBUSTION) COMBUSTION) HEATHEAT
FUEL OXIDANT FUEL OXIDANT MIXING MIXING IGNITION IGNITION COMBUSTION HEAT COMBUSTION HEAT
MIXING MIXING
SOLIDS :- pulverising to powder or small lumps SOLIDS :- pulverising to powder or small lumps
LIQUIDS :- spray nozzles, atomisers, vaporisers, LIQUIDS :- spray nozzles, atomisers, vaporisers, carburettors, 'burners'.
carburettors, 'burners'. GASES :- mixing valves,
GASES :- mixing valves, chambers, burners;chambers, burners;
usually need precautions to avoid explosion and flash back . usually need precautions to avoid explosion and flash back .
IGNITION IGNITION
Simply mixing methane and air will not cause it to burn. Simply mixing methane and air will not cause it to burn.
The molecules need to reach a certain threshold energy level The molecules need to reach a certain threshold energy level before the combustion process will proceed.
before the combustion process will proceed.
This may be provided initially by another flame, a spark or This may be provided initially by another flame, a spark or hot surface. The combustion process itself ( if sustained ) hot surface. The combustion process itself ( if sustained ) then continues the ignition process.
then continues the ignition process.
The chemical dissociation of the fuel and it's
The chemical dissociation of the fuel and it's recombinationrecombination with oxy
with oxygen. gen. Energy Energy and Maand Mass arss are conse conserved.erved. COMBUSTION
COMBUSTION
PRODUCTS PRODUCTS C
C COCO22 completecomplete combustioncombustion partialpartial combustioncombustion
H
H22 HH22OO
Other products may also be formed Other products may also be formed :-N
N22 NOx etc; often harmful pollutants ( SONOx etc; often harmful pollutants ( SO22, SO, SO33 ))
C C CO CO (soot) (soot) (carbon monoxide) (carbon monoxide)
COMBUSTION OF FUELS IN AIR COMBUSTION OF FUELS IN AIR Use an example
Use an example
:-Say we have a fuel that is 80% CH
Say we have a fuel that is 80% CH44 ( methane ) and( methane ) and
20% C
20% C22HH66 ( ethane ) by volume (UK natural gas) .( ethane ) by volume (UK natural gas) .
We could choose any quantity of fuel but for convenience we We could choose any quantity of fuel but for convenience we shall use 1 kmole, which can be written as
shall use 1 kmole, which can be written as :-( 0.8 CH
( 0.8 CH44 + 0.2 C+ 0.2 C22HH66 ))
Let us say that this mixes with Y kmoles of air and that the Let us say that this mixes with Y kmoles of air and that the combustion processes go to completion.
combustion processes go to completion. Now air is - 79% N
Now air is - 79% N22 & 21% O& 21% O22
Therefore
Therefore 1 kmole 1 kmole of air of air may be may be written as written as :- :-( 0.79 N
( 0.79 N22 + 0.21 O+ 0.21 O22 ))
We may therefore write a combustion equation We may therefore write a combustion equation :-( 0.8 CH ( 0.8 CH44 + 0.2 C+ 0.2 C22HH66 ) + Y( 0.79 N) + Y( 0.79 N22 + 0.21 O+ 0.21 O22 )) (the reactants) (the reactants) a CO a CO22 + b H+ b H22O + c OO + c O22 + d N+ d N22
(the products of complete combustion) (the products of complete combustion)
Because the combustion process does not create or
Because the combustion process does not create or destroydestroy atoms, the number of atoms on the L.H.S. of the equation atoms, the number of atoms on the L.H.S. of the equation must exactly balance the number on the R.H.S.
must exactly balance the number on the R.H.S.
No. of atoms in kmoles No. of atoms in kmoles
L.H.S. R.H.S.
L.H.S. R.H.S.
CARBON
CARBON C C 0.8 0.8 + + 2 2 x x 0.2 0.2 a a (i)(i) OXYGE
OXYGEN N O O 2 2 x x 0.21 0.21 x x Y Y 2a 2a + + b b + + 2c 2c (ii)(ii) HYDROGE
HYDROGEN N H 4 H 4 x x 0.8 0.8 + + 6 6 x x 0.2 0.2 2b 2b (iii)(iii) NITROGEN
NITROGEN N N 2 2 x x 0.79 0.79 x x Y Y 2d 2d (iv)(iv)
Solving equations (i) to (iv) simultaneously Solving equations (i) to (iv) simultaneously :-from
from (i) (i) a a = = 1.21.2 from
from (iii) (iii) b b = = 2.22.2 from
from (ii) (ii) c c = = 0.21 0.21 Y Y - - 2.32.3 from
EXHAUST GAS ANALYSES EXHAUST GAS ANALYSES
The R.H.S. of a combustion equation indicates the products of The R.H.S. of a combustion equation indicates the products of combustion ie: the exhaust gases.
combustion ie: the exhaust gases. Stoichiometric Air-Fuel ratio
Stoichiometric Air-Fuel ratio This is the air -
This is the air - fuel ratio (AFR) necessary to achieve completefuel ratio (AFR) necessary to achieve complete combustion of the fuel
combustion of the fuel and no more.and no more.
We can find this value from the combustion equation by
We can find this value from the combustion equation by settingsetting the number of kmoles of Oxygen on the R.H.S. equal to zero. the number of kmoles of Oxygen on the R.H.S. equal to zero. ie In the above equation c = 0
ie In the above equation c = 0 Therefore
Therefore Y Y = = = = 10.9510.95
Stoichiometric AFR = 10.95 : 1 Stoichiometric AFR = 10.95 : 1 N.B. this ratio is a ratio of
N.B. this ratio is a ratio of kmoles kmoles (volume for gases)(volume for gases)
How would we convert it to a mass ratio ? How would we convert it to a mass ratio ?
10.95 kmol air = 10.95 x 29 kg/kmol = 317.62 kg 10.95 kmol air = 10.95 x 29 kg/kmol = 317.62 kg 1 kmol fuel = 0.8 x 16 + 0.2 x 30 = 18.80 kg
1 kmol fuel = 0.8 x 16 + 0.2 x 30 = 18.80 kg
Therefore the AFR (
Therefore the AFR (by mass)by mass) is 317.62:18.80 or 16.89:1is 317.62:18.80 or 16.89:1 2 x 1.2 + 2.2 2 x 1.2 + 2.2 2 x 0.21 2 x 0.21 (CH (CH44)) (C(C22HH66))
(i) If the AFR is LESS than stoichiometric complete (i) If the AFR is LESS than stoichiometric complete combustion cannot occur ie: we have an excess of fuel, or a combustion cannot occur ie: we have an excess of fuel, or a rich mixture, leading to unburnt or incompletely burnt fuel. CO rich mixture, leading to unburnt or incompletely burnt fuel. CO will be pr
will be present. esent. This clearly This clearly is undesirable is undesirable as we as we will not bewill not be releasing the maximum available heat energy from the fuel. releasing the maximum available heat energy from the fuel. (ii) If the AFR is GREATER than stoichiometric we can obtain (ii) If the AFR is GREATER than stoichiometric we can obtain complete combustion but we may be supplying too much air complete combustion but we may be supplying too much air thus reducing the temperature of the combustion products* thus reducing the temperature of the combustion products* and therefore reducing the heat transfer from them.
and therefore reducing the heat transfer from them.
* N.B. we sometimes do this deliberately eg: in diesel & gas * N.B. we sometimes do this deliberately eg: in diesel & gas turbine engines.
turbine engines.
Excess air is usually expressed as a percentage of the Excess air is usually expressed as a percentage of the stoichiometric AFR
stoichiometric AFR
ie:-% excess air = Y - Y
% excess air = Y - Y stoicstoic x 100%x 100%
If, in the above equation, Y = 12, If, in the above equation, Y = 12, % excess air = 12 - 10.95 % excess air = 12 - 10.95 Y Y stoicstoic = 9.6 % = 9.6 % 10.95 10.95 x 100%x 100%
FINDING THE AFR FROM AN EXHAUST GAS ANALYSIS FINDING THE AFR FROM AN EXHAUST GAS ANALYSIS If we know the composition of a fuel and we measure the If we know the composition of a fuel and we measure the composition of the exhaust gas we can find the AFR.
composition of the exhaust gas we can find the AFR. Use an example
Use an example
:-Same fuel as above ( 0.8 CH
Same fuel as above ( 0.8 CH44 + 0.2 C+ 0.2 C22HH66 ))
( 0.8 CH
( 0.8 CH44 + 0.2 C+ 0.2 C22HH66 ) + Y ( 0.21 O) + Y ( 0.21 O22 + 0.79 N+ 0.79 N22 ))
a CO
a CO22 + b H+ b H22O + c OO + c O22 + d N+ d N22
Normally, when we analyse a gas sample we dry it first: ie Normally, when we analyse a gas sample we dry it first: ie we condense
we condense out ( out ( or absorb or absorb ) all t) all the water. he water. The remainingThe remaining gases are then expressed as a percentage ( by volume) of gases are then expressed as a percentage ( by volume) of the total dry gas constituents - in this case CO
the total dry gas constituents - in this case CO22, O, O22, & N, & N2.2.
This is known as a DRY analysis. This is known as a DRY analysis. Say we measure the %CO
Say we measure the %CO22 = 10.5% ( by vol.)= 10.5% ( by vol.)
Therefore
Therefore aa = 0.105= 0.105 a
but from the combustion equation but from the combustion equation :-a = 1.2 a = 1.2 c = 0.21 Y - 2.3 c = 0.21 Y - 2.3 d = 0.79 Y d = 0.79 Y Therefore Therefore = = 0.1050.105 or
or AFR AFR = = 12.53 12.53 : : 1 1 ( ( by by vol vol ))
%
% excess excess air air = = x x 100% 100% = = 14.4 14.4 %%
We could also have done the analysis using a figure for the We could also have done the analysis using a figure for the percentage O
percentage O22 in the exhaust, but because this is normallyin the exhaust, but because this is normally
small and is therefore more difficult to measure accurately, small and is therefore more difficult to measure accurately, the calculation is less reliable.
the calculation is less reliable.
A CO detector could be used to indicate a rich mixture . A CO detector could be used to indicate a rich mixture . The calculation assumes complete combustion.
The calculation assumes complete combustion.
If combustion is incomplete, further information is needed If combustion is incomplete, further information is needed otherwise the combustion equation cannot be solved.
otherwise the combustion equation cannot be solved. 1.2 1.2 1.2 + 0.21 Y - 2.3 + 0.79 Y 1.2 + 0.21 Y - 2.3 + 0.79 Y = 0.105 = 0.105 Y - 1.1 Y - 1.1 1.2 1.2 Y = 12.53 Y = 12.53 12.53 - 10.95 12.53 - 10.95 10.95 10.95
CALORIFIC VALUE OF FUELS CALORIFIC VALUE OF FUELS
The calorific value of a fuel is the standard heat of reaction The calorific value of a fuel is the standard heat of reaction at constant pressure where the fuel burns completely with at constant pressure where the fuel burns completely with oxygen.
oxygen.
It is measured using a gas calorimeter ( for gases ) It is measured using a gas calorimeter ( for gases ) bomb calorimeter ( for liquids & solids )
bomb calorimeter ( for liquids & solids ) ( See S & C p.396 ff for description ) ( See S & C p.396 ff for description )
Note :- minor corrections are required when using bomb Note :- minor corrections are required when using bomb calorimeters because it is constant volume combustion.
calorimeters because it is constant volume combustion.
HIGHER AND LOWER CALORIFIC VALUES HIGHER AND LOWER CALORIFIC VALUES
For fuels containing Hydrogen, water is a product of For fuels containing Hydrogen, water is a product of combustion and
combustion and is always is always formed as formed as steam. steam. If we If we 'allow' the'allow' the steam to condense to water then we will obtain
steam to condense to water then we will obtain a higher heata higher heat output/unit mass of fuel than if we did not allow i
output/unit mass of fuel than if we did not allow it to condense.t to condense. This measure of the calorific value is called the HIGHER This measure of the calorific value is called the HIGHER CALORIFIC VALUE ( HCV or gross ).
CALORIFIC VALUE ( HCV or gross ).
If the steam does not condense ( and this is more usually the If the steam does not condense ( and this is more usually the case ) we obtain the LOWER CALORIFIC VALUE ( LCV or case ) we obtain the LOWER CALORIFIC VALUE ( LCV or net ). ie the 'latent' heat of steam is unavailable.
The two values are related by The two values are related by :-HCV = LCV + m h
HCV = LCV + m hfgfg
where m = mass H
where m = mass H22O produced / kg of fuelO produced / kg of fuel
h
hfgfg = enthalpy of evaporation of water at the standard= enthalpy of evaporation of water at the standard
temperature of 25 degrees Centigrade ( 2442.5 kJ/kg ) temperature of 25 degrees Centigrade ( 2442.5 kJ/kg )
Some
Some typical typical values values :- :- H.C.V. H.C.V. L.C.V.L.C.V. Anthracite
Anthracite coal coal 34.583 34.583 33.913 33.913 MJ/kgMJ/kg Wood Wood 15.826 15.826 14.31914.319 Petrol Petrol 46.892 46.892 43.71043.710 Diesel Diesel 45.971 45.971 43.16643.166 U.K.Natural
U.K.Natural gas gas 36.38 36.38 32.75 MJ/m³ 32.75 MJ/m³ **** Hydrogen
Hydrogen 11.92 11.92 10.0510.05
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