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Combustion Equation For Hydrogen

1. The combustion equation for hydrogen shows that two molecules of hydrogen react with one molecule of oxygen to form two molecules of water. 2. The stoichiometric air-fuel ratio for anthracite coal is calculated to be 11.245 kg air per kg of coal based on the mass fractions of carbon, hydrogen, sulfur, and oxygen in the coal. 3. For actual combustion with 20% excess air, the actual air-fuel ratio is calculated to be 13.494, resulting in a mixture strength of 83.3% which falls within the normal working range.

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133 views4 pages

Combustion Equation For Hydrogen

1. The combustion equation for hydrogen shows that two molecules of hydrogen react with one molecule of oxygen to form two molecules of water. 2. The stoichiometric air-fuel ratio for anthracite coal is calculated to be 11.245 kg air per kg of coal based on the mass fractions of carbon, hydrogen, sulfur, and oxygen in the coal. 3. For actual combustion with 20% excess air, the actual air-fuel ratio is calculated to be 13.494, resulting in a mixture strength of 83.3% which falls within the normal working range.

Uploaded by

nabeel
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Combustion Equation

Combustion equation for hydrogen:

2 H 2+O 2→ 2 H 2 O

1. Hydrogen reacts with oxygen to form steam or water.


2. Two molecules of H2 reacts with one molecule of O2.

In Volume,
2 vol . of H 2+1 vol . of O 2 →2 vol . of H 2 o

By Mass,

2 × ( 2 ×1 ) + ( 2× 16 ) →2 × ( (2 ×1 ) +16 )
4 Kg H 2+32 Kg O 2→ 36 Kg H 2O

3 volume in reactant + 2 volume in product shows volumetric contraction in combustion

O2 accompanied by nitrogen for atmospheric air,


Hence equation can be written as, N2 as Inert Gas
79 79
2 H 2+O 2+ N 2→ 2 H 2 O+ N 2
21 21
By mass,
79
4Kg H2 + 32Kg O2 + 28 N2 → 36Kg O2 + 105.3 Kg N2
21
For,
1
H2 + O2 → H2O
2
By mass,
2KG H2 + 16Kg O2 → 18 Kg H20
By volume,
1
1 vol of H2 + vol of O2 → 1 vol of H2O
2
With Nitrogen,
1 79 79
H2 + O2 + N 2 → 2H2O + N2
2 21 21

Combustion Equations

1. To complete combustion of Carbon to Carbon dioxide:

C +O2→CO 2

And including
79 79
C +O2+ N 2→CO 2+ N 2
21 21
79
As with 1kmol of oxygen there are kmol of nitrogen
21
Considering volumes of reactants and products
79 79
O volume + 1 volume O 2 + Volume N 2 → 1 volume CO 2 + Volume N 2
21 21

By mass,
79
12kg C + 32 kg O2 + 28 kg N 2(reactant) → 44kgCO 2 + 105.3 kg N 2 (product)
21

2. For incomplete combustion:


2 C+O 2→ 2 CO
With N2
79 79
2 C+O 2+ N 2 → 2CO + N 2
21 21
By mass,
79
24 Kg C+32 KgO 2+ N 2(reactants)→ 56 KgCO +105.3 Kg N 2( products)
21
If further supply of O2 is available combustion can come to completion
79 79
2 CO+O 2+ N 2 → 2CO 2+ N 2
21 21
By mass ,
56 kg CO +32 Kg O 2+ 105.3 Kg N 2 ( reactants ) → 88 Kg CO 2+105.3 Kg N 2 ( products )

Calculations for Stiochiometric A/F Ratio:

We take a sample of dry anthracite has the following composition by mass:


Anthracite(type of Coal):
 C=90%, H=3%, O=2.5%, N=1%, S=0.5%, Ash=3%
Firstly, each constituent is taken separately and the amount of oxygen is found from relevant
chemical equation.
Carbon:
C+ O2 →CO 2
By mass:
12 kg C+32 kg O 2 → 44 kg CO 2
44
Oxygen Required=0.9 × ( mass of 1 carbon ) =2.4 kg of coal
12

Hydrogen :

H2 + 1/2 O2 → H2O (1)

General equation :
2kgH2 + 16O2 → 18kgH2O
For our equation no. (1)
1kgH2 + 8kgO2 → 9kgH2O
Oxygen required = 0.03 x 16/2
= 0.24kg/kg coal
Sulphur :

S + O2 → SO2

Our mass equation ;


1kgS + 1kgO2 → 2kgSO2
General equation ;
32kgS + 32kgO2 → 64kgSO2
Oxygen required = 0.005 x 1/1
=0.0005kg/kgcoal

Constituent Mass Fraction Oxygen Product mass


required(kj/kg)coal (kj/kg)coal
Carbon(C) 0.900 2.40 3.30(Co2)
Hydrogen(H) 0.030 0.240 0.27(H2O)
Sulphur(S) 0.005 0.005 0.01(SO2)
Oxygen(O) 0.025 0.025
Nitrogen(N) 0.010 0.01(N2)
Ash 0.030
Total 2.620

O2 required per kg at coal = 2.62kg


2.62
Stoichiometric air-fuel ratio = =11.245 kg
0.233
Stoichiometric A/F ratio = 11.245 kg

For Actual A/F ratio:


We take O2 20% excess air:

Using percentage excess air formula:


A
acutal −StoichiometricA / f
F
%excess air=
StoichiometricA / f

% excess of air x stoichiometric air/fuel ratio + stoichiometric air/fuel = Actual air/fuel ratio

11.245 + (20/100 x 11.245) = Actual air/fuel ratio

Actual A/F ratio = 13.494

To find mixture strength:-

Mixture strength = stoichiometric A/F/ Actual A/F ratio

Mixture strength = 11.245 / 13.494


Mixture strength = 0.833

Mixture strength = 0.833 x 100

Mixture strength = 83.3%

Working valve ranges between 80% (weak) and 120% (rich)

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