Combustion 2

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• Combustion
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combustion 2

At the end of this lesson

• You know what happens during combustion

• You can calculate the amount of air needed for an complete combustion, theoratical air flow.

• You can calculate the amount emmision of CO2 after combustion

What do you already know about emmisions

Propane C3H8

How many oxiden molecules are needed for a complete combustion of propane

A

3

B

5

C

8

D

11

Weight of the Atomes

Atomic mass

• Hydrogen = 1

• Carbon = 12

• Oxygen = 16

• Sulphur = 32

• 1 mol = atomic mass in grams 1kmol = atomic mass in kilograms

write down the reaction equation of a clean combustion of methane CH4

What is the weight of one methane molecule?

CH4

What is the weight of 2 oxyden molecules?

Atomic mass

Hydrogen = 1 Carbon = 12 Oxygen = 16 Sulphur = 32

1 CH4 + 2 O2 = 1 CO2 + 2 H20

• CH4 = 12 + 4 = 16

• 2 O2 = (16+16) x 2 = 64

• CO2 = 12 + (16+16) = 44

• 2 H20 = 2 x (1+1+16) = 36

Law of Lavoisier

1 CH4 + 2 O2 = 1 CO2 + 2 H20

16 + 64 = 44 + 36

So when you want to burn 16 kg Methane you need

64 kg O2 and your product will be 44 kg CO2 and 36 kg water (evaporated)

How much oxiden is needed to burn 1 kg of methane

1 CH4 + 2 O2 = 1 CO2 + 2 H20

16 kg + 64 kg = 44 kg + 36 kg

1 CH4 + 2 O2 = 1 CO2 + 2 H20

16 kg + 64 kg = 44 kg + 36 kg

1 kg + 64/16 kg = 44/16 kg + 36/16 kg

1 kg fuel + 4 kg O2 = 2,75 kg CO2 + 2,25 kg H20

Theoretical air flow rate

1 kg CH4 + 4 kg O2 = 2,75 kg CO2 + 2,25 kg H20

But have much air is this?

Theoretical air flow rate

4 kg O2

Air in the atmosphere contains about 23% oxygen

4 kg O2

23% oxygen in air

How much air do you need to burn 1 kg of methane

Theoretical air flow rate

4 kg O2

23% oxygen in air

So, theoretical you need for burning 1 kg of Methane:

100/23% x 4 kg O2 = 17,392 kg Air

how much air is needed for 1kg Ethane?

C2H6

how much air is needed for 1 kg Pentane?

C5H12

how much air is needed for Propane?

C3H8

That's the theory...

• In practice, it is not possible to know exactly which molecules are in the fuel. That is why it is calculated for each element how much air is needed and how many emissions are generated.

Example:

A fuel has the following composition: C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO S = 1% =>0,01 kg/kg FO With these elements, we will calculate how much air we need.

C + O2 = CO2

C + O2 = CO2

12 + 32 = 44

1 + 32/12 = 44/12 

1 + 8/3 = 11/3

C + 8/3C = 11/3 C

Hydrogen = 1

Carbon = 12

Oxygen = 16

Sulfur = 32

Combustion of Carbon

Atomic Weight

C=87% => 0,87 kg C/ kg fuel

2H2 + O2 = 2H20

2H2 + O2 = 2H20

4 + 32 = 36

1 + 8 = 9

H + 8H = 9H

Hydrogen = 1

Carbon = 12

Oxygen = 16

Sulfur = 32

Combustion of hydrogen

Atomic Weight

H= 12% => 0,12 kg H / kg fuel

S + O2 = SO2

S + O2 = SO2

32 + 32 = 64

1 + 1 = 2

S + S = 2S

Hydrogen = 1

Carbon = 12

Oxygen = 16

Suphur = 32

Combustion of sulfur

Atomic Weight

sulphur = 1%

Example:

A fuel has the following composition: C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

weight of oxiden needed: 8/3C + 8H + S

weight of oxiden needed: 8/3C + 8H + S

Air consists of 23% oxygen

Theoretical air flow is:

100/23 x (8/3C + 8H + S)

theoretical air needed

C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

fuel composition:

100/23 x (8/3C + 8H + S)

100/23 x (8/3x0,87 + 8x0,12 + 0,01)

=>14.3 kg air / kg fuel

emission CO2

oO

C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

fuel composition:

C + O2 = CO2

C + 8/3C = 11/3 C

CO2 = 11/3 C = 11/3 x 0.87

=> 3,19 kg/kg FO

emission SO2

C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

fuel composition:

S + O2 = SO2

S + S = 2S

= 2 x 0,01

=> 0,02 kg/kg FO

Emissions exhaust gas

• Now that we know the theoretical amount of air required for this kg fuel, we are now going to look at the engine and its consumption of fuel and air needed for the process of combustion.

Task:

Data:

• 4 cilinder, 4 stroke engine

• Rpm: 540 per minute

• Fuel: 2 gram per injection

• Air th 14,3 kg/kg Fuel

• λ cleaning: 1,25

• λ combustion: 2

W hat is the mass flow of fuel for this engine in kg per second?

• 4 cilinder, 4 stroke engine

• Rpm: 540 per minute

• Fuel: 2 gram per injection

• Air th 14,3 kg/kg Fuel

• λ cleaning: 1,25

• λ combustion: 2

A. Mf: mass flow fuel (kg/per sec.)

The rotation of 540 per minute must be converted to per second:

=> 540/60 => n= 9 per second.

Number of processes per second => rpm x cylinders/ process constant 2 n x z/a => 9 x 4 / 2 => 18 per sec.

Mass flow per injection = 2 grams => 2 x 18 = 36 grams of fuel per sec => 0.036 kg per sec (=> 3.11 tons/day)

What is the Mass Flow Air (kg/sec.)

• 4 cilinder, 4 stroke engine

• Rpm: 540 per minute

• Fuel: 2 gram per injection

• Air th 14,3 kg/kg Fuel

• λ cleaning: 1,25

• λ combustion: 2

B. Mass Flow Air (kg/sec.)

• Ma: mf x Air th x λ combustion x λ cleaning/cooling

=> 0,036 x 14,3 x 2 x 1,25

=> 1,287 kg air per second

(density air≈1,29 kg/m²)

=> ≈ 1 m² per second air.

total mass flow in

• Fuel 0,036 kg/sec

• Air 1,287 kg/sec

mf + ma => 0.036 +1,287 => 1,323 kg/sec

C. Mass Flow exhaust Gasses in kg/sec

• 4 cilinder, 4 stroke engine

• Rpm: 540 per minute

• Fuel: 2 gram per injection

• Air th 14,3 kg/kg Fuel

• λ cleaning: 1,25

• λ combustion: 2

Law of Lavoisier

C. Mass flow exhaust gasses

• Law of lavoisier : Mass in is mass out

• Fuel 0,036 kg/sec

• Air 1,287 kg/sec

mf + ma = mg => 0.036 +1,287 => 1,323 kg/sec Exhaust Gas

Mass Flow of CO2 en SO2 in kg/sec

• This fuel had the following composition:

• C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO S = 1% =>0,01 kg/kg FO

What is the mass flow of SO2?

C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

Mass flow of SO2?

• S + S = 2S

• 2 x S = 2 x 0,01 = 0,02 kg SO2/Kg FO

• Mf = 0,036 kg/sec

• 0.02 x 0,036 => 0,00072 kg SO2/ sec

What is the mass flow of CO2?

C = 87% =>0,87 kg/ kg FO H = 12% =>0,12 kg/kg FO

S = 1% =>0,01 kg/kg FO

Emission CO2?

• C + 8/3C = 11/3 C

• CO2 = 11/3 C

= 11/3 x 0,87 = 3.19 kg/kg FO

• Mf = 0,036 kg/ sec

• => 0,115 CO2 kg/sec

Nitric oxide

In the previous picture you can also see the NOx which is also an important GHG. Nitric oxide has 3 formation mechanisms:

1. nitrogen as an element in the fuel; Fuel Nox

2. nitrogen from the excess air present reacts with O2 during combustion; Thermal NOx (the higher the flash point, the more NOx is formed)

3. Reaction of nitrogen (compounds) with free oxygen radians; Prompt NOx

PIC: products of incomplete combustion

• CO Carbon Monoxide

• HC Hydrocarbon

• C Carbon

• PM Particle Matter

What is theoretical airflow in a diesel engine?

A

Airflow measured under real conditions

B

Actual airflow during engine operation

C

Airflow affected by engine speed

D

Specific airflow based on fuel

What causes practical airflow to differ?

A

Consistent atmospheric pressure

B

Perfect fuel combustion

C

Engine load and speed variations

D

Fixed engine temperature

What did you learn today?