CALCULATION OF AIR QUANTITY
1. Law of definite proportion: Substances always combine in definite proportions and these proportions are determined by the molecular masses of reactants consumed and the products formed.
C (g) + O2(g) == CO2 (g)
12 32 44
2. The law of gaseous volumes ( Gay Lussac 1808): according to this law When gas combine they do so in volumes which bear a simple ratio to each other and also to the product formed, provided all gases are measured under similar conditions
H2(g) + ½ O2(g) = H2O(v)
1 vol 0.5 vol 1 vol
3. Avogadros law ; Equal volumes of gases under similar conditions of pressure and temperature posses equal no of moles or molecules
At 1 atm pressure and 0 deg C I mol of gas contains 6.023 X 10 23 molecules of gas
One mole of an ideal gas at 0 deg C and 1 Atm pressure occupies 22.4 litre volume.
Significance of these laws
A combustion equation
CH4 (g) + 2 O2 (g) = CO2 (g) + 2 H2O (L)
1) Weight ratio . 16 gm 64 gm 44 gm 36 gm
2) Volume ratio 1 vol 2 vol 1 vol 2 vol
3) Mol ratio 1 mol 2 mol 1 mol 2 mol
4)Molecule ratio 6.023x 10 23 2x6.023x 10 23 6.023x 10 23 2x6.023x 10 23
In actual practice the combustion is carried out in presence of air. Therefore amount of air is calculated
1) Air contains 21 % of oxygen by volume, and 23% of Oxygen by mass. Hence from the amount of Oxygen required by the fuel the amount of air can be calculated
I kg of oxygen is supplied by 1x 100/23= 4.35 kg of air
1 M3 of oxygen is supplied by 1x100/21= 4.76 M3 of air
Mean molecular mass of air is taken as 28.94 g/mol
Minimum oxygen required for combustion = Theoritical oxygen reqd- O2 present in the fuel
Minimum O2 required should be calculated on the basis of complete combustion. If the combustion product contain CO and O2 then excess O2 is found by subtracting the amount of O2 required to burn CO to CO2
The mass of any gas can be converted to its volume at certain temp. and pressure by assuming that the gas behaves ideally and using the gas equation
PV= nRT
The volume of a gas at a given temp. and pressure can be redused to the corresponding volume at any other specified conditions of temp. and pressure with the help of
P1V1/T1 = P2V2/T2
CONVERSION OF VOLUME OF AIR INTO WEIGHT OR VICEVERSA
Density of Air ( d air) = Molar mass of air(M)/ Molar volume of air (V)
28.94 g/mol
= ---------------
22.4 L/mol
28.94
= --------------
22.4 g/L
Let us assume that for the combustion of a particular fuel m gm of air is reqd
The corresponding volume of air ( let v litre) requirement can be calculated by assumimg that the density of air remains constant.
Thus m/v = M/V
m/v = 28.94/22.4
v= 22.4/28.94 m
Guide line for combustion calculation
1) Determine the amounts of various combustibles present in a given fuel
2) Write the balanced combustion equation for all the combustibles
3) Calculate the requirement of oxygen for complete combustion of fuel constituents either by weight or by volume or by moles.
4) Find the total oxygen required by adding up the oxygen requirement for individual combustibles substance of the given fuel .
5) Calculate net oxygen required by subtracting from the total oxygen required the amount of oxygen already present in the fuel
6) Convert net oxygen requirement into total air requirement.
Combustion calculation by weight
S no Fuel constituent Amount of constituent in fuel (gm) Combustion equation Weight of oxygen
1
2
3
4
5 C
CO
H
S
O a
b
c
d
e C + O2 = CO2
12 32
CO + ½ O2= CO2
28 16
H2 + ½ O2
2 16
S + O2 = SO2
32 32
-----------------------
32/12 * a = A
16/28 * b = B
16/2 * c = C
32/32 * d = D
------------------
Total weight of oxygen needed for combustion A+ B+ C+D = T ( LET)
Weight of oxygen already present in fuel = e gm
Thus net weight of Oxygen required for combustion T-e gm = N gm (LET)
Theoretical weight of air required for combustion = N * 100/23 gms
COMBUSTION CALCULATION BY VOLUME
S No Fuel constituent Amount of constituent in fuel (M3) Combustion equation Volume of oxygen needed for combustion (M 3)
1
2 CH4
C2H6 A
B CH4 + 2O2 = CO2 + 2 H2O
1 M3 2 M3
C2H6 + 3.5 O2 = 2 CO2 +3 H2O 2/1* A
3.5/1 * B
Total volume of oxygen needed for combustion A+B+C+……… = T (M3)
Volume of Oxygen already present = H (M3)
Thus net volume of Oxygen required = T- H ( M3)
Theoretical volume of air required = ( T- H) * 100/21 M3
COMBUSTION CALCULATION BY MOLES
S No Fuel constituent Amount of constituent in fuel Combustion equation Moles of oxygen needed for combustion
1
2
3 C
H
S A gm = A/12 mol
B gm = B/2 mol
C gm= C/32 mol C + O2 = CO2
1 Mole 1 Mole
H2 + 0.5 O2 = H2O
1 mol 0.5 mol
S + O2 = SO2
1 mole 1 mole A/12 *1= A1
B/2* 0.5 = B1
C/32 *1=C1
Total moles of oxygen needed for combustion A1+B1+C1+ ……. + = T Moles (LET)
As moles of oxygen already present in fuel = G Moles
Net moles of Oxygen needed for combustion = T-G moles= N Moles
Weight of oxygen needed for combustion = N * 32 Gm = N1 gm
Thus weight of Air required for complete combustion of fuel = N1 *100/23
Now volume of air required for complete combustion of fuel = N1 * (22.4 L/mol/ 28.94 g/mol)
A hydro carbon fuel on combustion gave a flue gas of the following volume % composition CO2= 13.53%, N2 = 82.91% and O2 = 3.56 . Determine composition of fuel by weight b) the % of excess air and c) volume of air supplied per kg of the fuel
Solution:
a) As the flue gas contains CO2 so one of the constituents in fuel must be carbon
C + O2 = CO2
That indicates that moles of CO2=Moles of O2= Moles of C
Assuming total amount of dry flue gas = 100 Kmol
So mols of CO2= 13.53 K mol
Thus amount of O2 in flue gas = 13.53 Kmol + 3.56 Kmol = 17.09 Kmol
Given amount of N2 in flue gas = 82.91 kmol
As in 100 Kmol of air 79 Kmol N2 is present
Thus, Amount of air supplied for combustion = 100/79* 82.91 = 104.95 Kmol
Therefore amount of oxygen supplied = 104.95 *21/100= 22.039 kmol
Amount of O2 combined with H2= 22.04- 17.09= 4.95 Kmol
Combustion equation of H2= H2 + ½ O2 = H20
So amount of H2 burnt= 2X4.95= 9.90 Kmol
And weight of H2 burnt= 9.9x2= 19.8 Kg
As 100 Kmol of flue gas containg 13.53 Kmol of CO2 = 13.53 Kmol of C
Therefore weight of C in fuel 13.53 X12= 162.36 Kg
Now % of C= [162.36/( 162.36+ 19.80)] *100=89.130
And % of H= [19.80/( 162.36+ 19.80) ]* 100= 10.87
b) Theoretically amount of O2 required for producing 13.53 kmol of Co2= 13.53 kmol of O2
And amount of O2 required for burning H2 = 4.95 Kmol
So total O2 theoretically required = 13.53+4.95= 18.48 Kmol
Oxygen supplied 22.04 kmol
So % of excess air = [(22.04-18.48)/18.48]*100 = 19.264
c) Total air supplied = 104.95 Kmol
Volume of air supplied= 104.95x 22.4= 2350.93 M3
Thus volume of air supplied per kg of fuel= 12.91 M3/kg
in doubt ask at kaacconsultant@gmail.com
1. Law of definite proportion: Substances always combine in definite proportions and these proportions are determined by the molecular masses of reactants consumed and the products formed.
C (g) + O2(g) == CO2 (g)
12 32 44
2. The law of gaseous volumes ( Gay Lussac 1808): according to this law When gas combine they do so in volumes which bear a simple ratio to each other and also to the product formed, provided all gases are measured under similar conditions
H2(g) + ½ O2(g) = H2O(v)
1 vol 0.5 vol 1 vol
3. Avogadros law ; Equal volumes of gases under similar conditions of pressure and temperature posses equal no of moles or molecules
At 1 atm pressure and 0 deg C I mol of gas contains 6.023 X 10 23 molecules of gas
One mole of an ideal gas at 0 deg C and 1 Atm pressure occupies 22.4 litre volume.
Significance of these laws
A combustion equation
CH4 (g) + 2 O2 (g) = CO2 (g) + 2 H2O (L)
1) Weight ratio . 16 gm 64 gm 44 gm 36 gm
2) Volume ratio 1 vol 2 vol 1 vol 2 vol
3) Mol ratio 1 mol 2 mol 1 mol 2 mol
4)Molecule ratio 6.023x 10 23 2x6.023x 10 23 6.023x 10 23 2x6.023x 10 23
In actual practice the combustion is carried out in presence of air. Therefore amount of air is calculated
1) Air contains 21 % of oxygen by volume, and 23% of Oxygen by mass. Hence from the amount of Oxygen required by the fuel the amount of air can be calculated
I kg of oxygen is supplied by 1x 100/23= 4.35 kg of air
1 M3 of oxygen is supplied by 1x100/21= 4.76 M3 of air
Mean molecular mass of air is taken as 28.94 g/mol
Minimum oxygen required for combustion = Theoritical oxygen reqd- O2 present in the fuel
Minimum O2 required should be calculated on the basis of complete combustion. If the combustion product contain CO and O2 then excess O2 is found by subtracting the amount of O2 required to burn CO to CO2
The mass of any gas can be converted to its volume at certain temp. and pressure by assuming that the gas behaves ideally and using the gas equation
PV= nRT
The volume of a gas at a given temp. and pressure can be redused to the corresponding volume at any other specified conditions of temp. and pressure with the help of
P1V1/T1 = P2V2/T2
CONVERSION OF VOLUME OF AIR INTO WEIGHT OR VICEVERSA
Density of Air ( d air) = Molar mass of air(M)/ Molar volume of air (V)
28.94 g/mol
= ---------------
22.4 L/mol
28.94
= --------------
22.4 g/L
Let us assume that for the combustion of a particular fuel m gm of air is reqd
The corresponding volume of air ( let v litre) requirement can be calculated by assumimg that the density of air remains constant.
Thus m/v = M/V
m/v = 28.94/22.4
v= 22.4/28.94 m
Guide line for combustion calculation
1) Determine the amounts of various combustibles present in a given fuel
2) Write the balanced combustion equation for all the combustibles
3) Calculate the requirement of oxygen for complete combustion of fuel constituents either by weight or by volume or by moles.
4) Find the total oxygen required by adding up the oxygen requirement for individual combustibles substance of the given fuel .
5) Calculate net oxygen required by subtracting from the total oxygen required the amount of oxygen already present in the fuel
6) Convert net oxygen requirement into total air requirement.
Combustion calculation by weight
S no Fuel constituent Amount of constituent in fuel (gm) Combustion equation Weight of oxygen
1
2
3
4
5 C
CO
H
S
O a
b
c
d
e C + O2 = CO2
12 32
CO + ½ O2= CO2
28 16
H2 + ½ O2
2 16
S + O2 = SO2
32 32
-----------------------
32/12 * a = A
16/28 * b = B
16/2 * c = C
32/32 * d = D
------------------
Total weight of oxygen needed for combustion A+ B+ C+D = T ( LET)
Weight of oxygen already present in fuel = e gm
Thus net weight of Oxygen required for combustion T-e gm = N gm (LET)
Theoretical weight of air required for combustion = N * 100/23 gms
COMBUSTION CALCULATION BY VOLUME
S No Fuel constituent Amount of constituent in fuel (M3) Combustion equation Volume of oxygen needed for combustion (M 3)
1
2 CH4
C2H6 A
B CH4 + 2O2 = CO2 + 2 H2O
1 M3 2 M3
C2H6 + 3.5 O2 = 2 CO2 +3 H2O 2/1* A
3.5/1 * B
Total volume of oxygen needed for combustion A+B+C+……… = T (M3)
Volume of Oxygen already present = H (M3)
Thus net volume of Oxygen required = T- H ( M3)
Theoretical volume of air required = ( T- H) * 100/21 M3
COMBUSTION CALCULATION BY MOLES
S No Fuel constituent Amount of constituent in fuel Combustion equation Moles of oxygen needed for combustion
1
2
3 C
H
S A gm = A/12 mol
B gm = B/2 mol
C gm= C/32 mol C + O2 = CO2
1 Mole 1 Mole
H2 + 0.5 O2 = H2O
1 mol 0.5 mol
S + O2 = SO2
1 mole 1 mole A/12 *1= A1
B/2* 0.5 = B1
C/32 *1=C1
Total moles of oxygen needed for combustion A1+B1+C1+ ……. + = T Moles (LET)
As moles of oxygen already present in fuel = G Moles
Net moles of Oxygen needed for combustion = T-G moles= N Moles
Weight of oxygen needed for combustion = N * 32 Gm = N1 gm
Thus weight of Air required for complete combustion of fuel = N1 *100/23
Now volume of air required for complete combustion of fuel = N1 * (22.4 L/mol/ 28.94 g/mol)
A hydro carbon fuel on combustion gave a flue gas of the following volume % composition CO2= 13.53%, N2 = 82.91% and O2 = 3.56 . Determine composition of fuel by weight b) the % of excess air and c) volume of air supplied per kg of the fuel
Solution:
a) As the flue gas contains CO2 so one of the constituents in fuel must be carbon
C + O2 = CO2
That indicates that moles of CO2=Moles of O2= Moles of C
Assuming total amount of dry flue gas = 100 Kmol
So mols of CO2= 13.53 K mol
Thus amount of O2 in flue gas = 13.53 Kmol + 3.56 Kmol = 17.09 Kmol
Given amount of N2 in flue gas = 82.91 kmol
As in 100 Kmol of air 79 Kmol N2 is present
Thus, Amount of air supplied for combustion = 100/79* 82.91 = 104.95 Kmol
Therefore amount of oxygen supplied = 104.95 *21/100= 22.039 kmol
Amount of O2 combined with H2= 22.04- 17.09= 4.95 Kmol
Combustion equation of H2= H2 + ½ O2 = H20
So amount of H2 burnt= 2X4.95= 9.90 Kmol
And weight of H2 burnt= 9.9x2= 19.8 Kg
As 100 Kmol of flue gas containg 13.53 Kmol of CO2 = 13.53 Kmol of C
Therefore weight of C in fuel 13.53 X12= 162.36 Kg
Now % of C= [162.36/( 162.36+ 19.80)] *100=89.130
And % of H= [19.80/( 162.36+ 19.80) ]* 100= 10.87
b) Theoretically amount of O2 required for producing 13.53 kmol of Co2= 13.53 kmol of O2
And amount of O2 required for burning H2 = 4.95 Kmol
So total O2 theoretically required = 13.53+4.95= 18.48 Kmol
Oxygen supplied 22.04 kmol
So % of excess air = [(22.04-18.48)/18.48]*100 = 19.264
c) Total air supplied = 104.95 Kmol
Volume of air supplied= 104.95x 22.4= 2350.93 M3
Thus volume of air supplied per kg of fuel= 12.91 M3/kg
in doubt ask at kaacconsultant@gmail.com
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