Fuel / Air Ratio

Relative Mass - the mass of an object or substance in comparison to the mass of another object or a standard reference mass

EXAMPLE

Relative Density “specific gravity”

RD = density of a substance / density of freshwater

Pseawater = 1025 kg/m³
Pfreshwater = 1000 kg/m³

RDseawater = 1025 / 1000
= 1.025

ATOM

Mass of Carbon = 1.99 × 10-26

Mass number = number of protons + number of neutrons

A - atomic
M - mass
U - unit

Symbol

H2

H2O

CO2

O2

SO2

Name

Hydrogen

Water

Carbon Dioxide

Oxygen

Sulphur Dioxide

relative mass

2

18

44

32

64

Calculate the Relative Mass of CH4
C = 12
H x 4 = 4
CH4 = 16

Calculate the mass of oxygen required to burn 1kg of carbon completely to carbon dioxide.

C + O2 ——→ CO2
12 32 44

for 1kg → 32/12 = 2.67
1 + 2.67 = 3.67 kg

Calculate the mass of oxygen required to burn 1kg of hydrogen completely to water.

2H2 + O2 —> H2O
4 32 36

for 1kg → 32/4 = 8
4/4 = 1
1 + 8 = 9 kg

Calculate the mass of oxygen required to burn 1kg of sulphur completely to sulphur dioxide.

S + O2 —> SO2
32 32 64

for 1kg → 32/32 = 1
32/32 = 1
1 + 1 = 2 kg

Air consists of 23% oxygen and 77% nitrogen by mass

Oxygen is an active element
Nitrogen is an inert gas and takes to active part.

An excess supply of air is always necessary for a complete combustion

To obtain 1kg of oxygen, what would be the supply of air?

23 % x Air = oxygen
Air = Oxygen / 23 %
= 1 / 23
= 4.348 kg

For complete combustion, 4kg of air supply is required. What would be the supply of air in KG for the following conditions?

Excess Air supply

KG

40%

4 + (4 × 40%) = 5.6

100%

4 + (4 × 100%) = 8

120%

4 + (4 ×120%) = 8.8

220%

4 + (4 × 220%) = 12.8