CombustionCombustion

UnderUnderControlControl

OutsideOutsideControlControl

• Fuel Type• Fuel handling• Heat Transfer• Excess Air• Burner momentum• Emission

• Secondary air momentum• Tertiary air momentum• Kiln aerodynamics• Calciner aerodynamic

Combustion since is the since of exothermic chemical reactions in flow of heat and mass transferCombustion since is the since of exothermic chemical reactions in flow of heat and mass transfer

Fuel :Fuel :Fuel :Fuel :

Gas H2 or natural gasGas H2 or natural gas

Liquid Alcohol or oilLiquid Alcohol or oil

Solid Na, pure carbon or coalSolid Na, pure carbon or coal

Hydro Carbons (fuels)

Hydro Carbons (fuels)

Carbon CO2

Hydrogen H2O

Sulfur SO2

Carbon CO2

Hydrogen H2O

Sulfur SO2

C + O2 CO2 + ( 94 kCal/mole)C + O2 CO2 + ( 94 kCal/mole)

2H + O2 2H2O2H + O2 2H2O

+ 116 kCal/moleWater as steam

+ 116 kCal/moleWater as steam

+ 137 kCal/moleWater condensed

+ 137 kCal/moleWater condensed

G.C.V:G.C.V: Gross calorific valuesGross calorific values

H.H.V:H.H.V: High heating valuesHigh heating valuesWater condensed

N.C.V:N.C.V: Net calorific valuesNet calorific values

L.H.V:L.H.V: Low heating valuesLow heating valuesWater remain as steam

The incomplete Oxidation of carbon

Not all of the carbon in the fuel will be oxidized to carbon dioxide but some will be partially oxidized to carbon monoxide

Carbon monoxide: Reduce the heat release from the fuel

2C + O2 2CO + ( 53 kCal/mole)2C + O2 2CO + ( 53 kCal/mole)

Reversible chemical reaction

2CO + O2 2CO2 + ( 41 kCal/mole)2CO + O2 2CO2 + ( 41 kCal/mole)

Natural gas requires more combustion air per kCal of heat released than most other fuels, produces more exhaust gases, smallest CO2

Natural gas requires more combustion air per kCal of heat released than most other fuels, produces more exhaust gases, smallest CO2

Physics of Combustion

Combustion Stages

Mixing Ignition Chemical reaction Dispersal of product

slow Very fast

If it’s mixed, it’s burnIf it’s mixed, it’s burn

11 Kg fuel Nm3 fuel11Or

Oxygen is required

Combustion equation

Fuel analysis is known

Empirical formula

Heat Value is known

Air → 21% Oxygen

Minimum air required = 4.762 times of the minimum Oxygen

A = Amin * n

A = Air supplied to the kiln system.

Nm3/kg fuel or Nm3/Nm3 fuel

A min= minimum air required for combustion the kiln system.

n = Excess Air factor.Ratio of air supplied

Minimum air

Heat of Combustion or “heat value”

If combustion product contain water.

High heat value – low heat value

= the heat of vaporization of the water

• combustion products• excess air of combustion• false air• gas from the raw meal

The kiln gas consists of :

carbon dioxide (CO2 )

water vaporsulfur absorbed (Circulation Phenomena)

Analysis of the Kiln GasesBy ORSAT (dry)

The Orsat - analysis is used to analyze the dry kiln gases

For Water contentCombustion

Water content of the raw meal

Can be calculated from the dew point.

Orsat-analysis

•Excess air factor n•Incomplete combustion•Heat consumption•False air

Oxygen required for combustion of 1 kg fuel

(liquid + solid fuel)

Nm3 / kg fuel

C + O2 CO2C + O2 CO21 KM

12 Kg

1 KM

32 Kg

1 KM

44 Kg

22.4

12)C) = 1.864 (C)

22.4 = Volume occupied by 1 Mol

12 = Molecular wt of carbon

Nm3/Mol

Kg(C)/Mol= =

Nm3

Kg(C)

H2 + ½ O2 H2OH2 + ½ O2 H2O1 KM

2 Kg

½ KM

16 Kg

1 KM

18 Kg

22.4

4)H) = 5.553 (H)

S + O2 SO2S + O2 SO222.4

32)S) = 0.6982 (S)

1 KM 1 KM 1 KM

O2 is used for above reaction 22.4

32)O) = 0.6997 (O)

Omin. = 1.864 * C + 5.553 * H + 0.6982 * S - 0.6997 * 0

Nm3 of Oxygen / kg fuel

Oxygen Required for Combustion of 1 Nm3 Fuel

H2 + ½ O2 H2O

CxHy + ( X + Y/4 ) O2 X CO2 + (Y/2) H2O

CO + ½ O2 CO2

Omin = 0.5 CO + (X + Y / 4 ) CxHy + 0.5 H2 – O2

Nm3 of Oxygen / Nm3 of fuel

Nm3 fuel (Gas Fuel( Nm3 / Nm3 fuel

By Weight

0.232 O2

0.768 N2

By Volume

0.21 O2 0.79 N2

H4 + O2 = 2H2O C + O2 = CO2

CH4 + 2 O2 = CO2 + 2H2O

(12+4) : 2 x 32

16 : 64 1 kg fuel : 4 kg of Oxygen

weight of air required per kg

Kg CH4

= 4

0.232= 17.24

Kg air

Kg Ch4

Volume of air required Nm3

Nm3 CH4

= 2

0.21= 9.521

Nm3 air

Nm3 Ch4

Amin = Omin

0.21

Combustion Products

CO2 , H2O , some SO2

a) Complete Combustion

CO2 , H2O , N2 , & O2 excess from combustion

b) Incomplete Combustion

CO2 , H2O , N2 , & CO

Combustion products of 1 kg fuel without excess air

CO2

1.855 C SO2

0.6841 S+

0.8 N2 + 0.79 A min N2

+1.244 H2O + 11.21 H

V min =

Water Vapor

Dry Combustion+

1.855 =12

22.4CO2

0.6841 =32

22.4SO2

0.8 =28

22.4N2

1.244 =18

22.4H2O

11.21 =2

22.4H

Combustion products of Nm3 fuel without excess air

CO2 : V = CO + CO2 + X Cx Hy Nm3Co2 / Nm3 fuel

H2O : V = Y/2 CxHy + H2 Nm3 H2O / Nm3 fuel

Vmin = (CO + CO2 + XCxHy) + (Y/2 Cx Hy + H2 ) + N2 + 0.79 Amin

CO2 H2O

Excess Air in Combustion

A = n* Amin Nm3 air / kg or Nm3 fuel

V = Vmin + (n-1)* Amin Nm3 gas kg or Nm3 fuel

Heat of Combustion

Heat values are determined experimentally by calorimeters in which products of combustion are cooled to the initial temperature and the heat absorbed by the cooling medium is measured

The low heat value is evaluated assuming no water vapor condensed

Hu =

Ho =the high heat value is calculated assuming all water vapor condensed

Hu = Ho – ( H2O + 9 H2) 2.499 MJ/kg fuel

Hu = Ho – (H2O + 9 H2) 597) Kcal / kg fuel

1 Cal = 4.18 J

Empirical Rules for Air Requirement and Combustion

Minimum Air : Amin = 0.26 * Hu

Minimum Combustion Products (wet) :

Vmin = 0.28* Hu

Each MJ (Megajoul) burnt in the firing

requires 0.26 Nm3 minimum air and produces

0.28 Nm3 minimum combustion gas.

GAS FORM RAW MEAL GAS FORM RAW MEAL

CaCO3 CaO + CO2

1kMol 1 kMol 1 kMol

100.09kg 56.08 kg 22.26 Nm3

Mg CO3 MgO + CO2

Nm3Co2 Kg R.M

22.26CO2 R.M =100.09

Ca CO3

84.33

Mg Co3

+

R = Raw Mill Factor (1.55 – 1.95)

CO2 produced /kg Clinker

Carbon Dioxide from Raw Meal

Nm3Co2 Kg clinker

CO2 RMCO2 RC = * R

CO2 RC =22.26

100 *Nm3Co2 Kg clinker

Titration * R

Titration = % Weight of carbonate kg/kg RM

R =

Kg Raw Meal

Kg Clinker

Water from Raw Meal mw

1 - mw

m H2O R.M = Kg H2O / kg dry R.M

mw

1 - mw

m H2O = Kg H2O / kg clinker* R

mw

1 - mw

V H2O = Nm3 H2O / kg clinker*

R0.8038

0.8038 = 18

22.4

Kg/mol

Nm3/mol

kgNm3=

18 = Molecular weight of water 22.4 = Volume occupied by 1 mol

Orsat Analysis

Sulfur from Raw Meal SOSO22 SO SO33

CO2 KOH solution, (Pyrogllic acid) + Potas hydroxide

O2

CO

N2

KOH solution, (Pyrogllic acid) + Potas hydroxide

cuprous chloride ( Cu2Cl2)

final unabsorbed gas

Errors due to leakage and poor sampling

Excessive Air Factor n

A Amin

n =N2

N2 min

n = n =N2

N2 - N2 excess air

From the Orsat analysis the oxygen content O2

N2

O2

0.790.21

= 3.762=

N2 excess air = 3.762 Oxygen excess air

n =N2

N2 - 3.762 O2

If there is CO

n =N2

(N2 - 3.762) * (O2 - 0.5 CO)

N2

OR

n =1

(1 - 3.762) * (O2 - 0.5 CO)

N2

Incomplete Combustion

explosion in the kiln system

The specific heat consumption of

the kiln increases

If the false air increase then the secondary air decreases

Co2 = 12.64 MJ /Nm3 ( 3020 kcal/Nm3 )

H2 = 10.80 MJ / Nm3 ( 2580 kcal/Nm3 )

CH4 = 64.34 MJ / Nm3 ( 15370 kcal/Nm3 )

The low heat value

Density of the Kiln Gas

GasComponen

t

Molecular

Weight

Density at Oo C, 760 mm

Hg kg/m3

RelativeWeight

(Air = 1)

Gas constant

R kgm/ kg

k

CO2 44.00 1.9768 1.5291 19.25

O2 32.00 1.4289 1.1053 26.49

CO 28.00 1.2500 0.9669 30.28

H2 2.016 0.0899 0.0695 420.75

N2

(from air)28.15 1.2567 0.9721 30.12

H2O 18.016 -- -- --

CH4 16.031 0.7168 0.5545 52.89

Heat Consumption

The specific heat consumption qq

CO2 RC

0.79 AminHu

CO2t

1-CO2t

CO2f

1-CO2f

-q =

CO2 RC, Nm3 / kg Cli , CO2 from raw meal

Amin Nm3 / kg fuel, Nm3 / Nm3 fuel

Hu MJ / kg, low heat value of fuel

CO2t = CO2 content in the exit gas if not excess air were present

CO2 + CO

1 - O2 - 0.5 co0.21

CO2t =

CO2, CO and O2 are the volume fraction of the dry gas, as obtained

by the Orsat analysis.

CO2f = CO2 from combustion of fuel if no excess air were present

CO2f = CO2 =1

V min dry

1.855 C

C = weight fraction of carbon of the fuel

CO2f = CO2 =1

V min dry

( CO + CO2 + X CX Hy)

CO, CO2 and CXHy are volume fraction of the gaseous fuel.

False Air

A Amin

n =n Excessive Air Factor

From the Orsat analysis

The false air is the difference of the total amounts A2 of air upstream and downstream A1of the leak.

A false Air = A2 – A1 = ( n2 –n1) Amin Nm3/kg Clinker

A false Air = ( n2 –n1) Amin * K Nm3/kg Clinker

K = specific fuel consumption.Kg fuel /kg Clinker

Nm3 fuel/kg Clinker