mining Combustion and fuels tutorials finalfluid.wme.pwr.wroc.pl/.../combustion_mpe_tutorials.pdfCourse: Combustion and fuels – tutorials Project co-financed by European Union within

Project co-financed by European Union within European Social Fund

THE DEVELOPMENT OF THE POTENTIAL AND ACADEMIC PROGRAMMES OF WROCŁAW UNIVERSITY OF TECHNOLOGY

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Mining and Power Engineering

Author: dr inż. Tomasz Hardy (PhD Eng.)

Course: Combustion and fuels – tutorials



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The content of the classes

1. Exchangeability of gases

2. Flammability limits

3. Stoichiometry of combustion processes

4. Calorimetry of combustion processes

5. Temperature of combustion processes

6. Calculations of combustion chambers

Materials for theoretical preparation

1. Spalanie i paliwa, ed. by W. Kordylewski, Oficyna Wydawnicza Politechniki

Wrocławskiej, Wrocław, 2008 (ed. V) or 2005 (ed. IV) (in Polish)

2. Presentations for the lecture Combustion and fuels available online at

www.spalanie.pwr.wroc.pl (in English)

3. Handbook of combustion, Vol. 1-5, ed. by M. Lackner, F.Winter, A.K. Agarwal, Wiley-

VCH Verlag GmbH & Co. KGaA, Weinheim, 2010 (in English)



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1. Exchangeability of gases

The gases are exchangeable with each other if they can burn properly in the same burners

without the need for their change (adaptation).

The condition of exchangeability of gases is determined by Wobbe number (Wb):

v

s

d

QWb =

where dv is the relative density of gas:

air

gvd

ρρ

=

and ρg, ρair denote the density of gas and air.

The condition of exchangeability of gas

Two gases are exchangeable if at the same overpressure (∆p) the thermal power of a gas burner

(N) will be the same: N1 = N2

a) Determination:

1 – the first gas,

2 – the second gas.

ρ1, ρ2 [kg/m3] - density of gas 1 and 2.

Qs1, Qs2 [J/m3] – heat of combustion gas 1 and 2 (upper caloric value UCV).

V*

1, V*

2 [m

3/s] - flow rate of gas 1 and 2.

N1, N2 [W] – (the thermal power of a gas burner powered with gas 1 and 2.

b) The condition of exchangeability of gas 1 and 2: N1 = N2 besides: ∆p = const

c) Justification:

N1 = V*

1· Qs1

N2 = V*

2 · Qs2

Hence:

V*

1 · Qs1 = V*

2 · Qs2 --- >

2

2

2

1

V

V

Q

Q

s

s

&

&

= .

Overpressure of the gas (∆p) before the gas burner is equal to the resistance of the gas

flow:

∆p = ½ ζρgU2, where is the velocity of the gas flow [m/s].



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For both gases:

∆p1 = ½ ζρg1U12, ∆p2 = ½ ζρg2U2

2

∆p = ∆p1 = ∆p2

that is: ρg1U12 = ρg2U2

2

and next: 21

22

2

1

U

U

g

g =ρρ

because: F

VU

&

= , therefore 2

1

22

2

21

2

22

2

1

V

V

F

VF

V

g

g

&

&

&

&

==ρρ

so:

2

1

2

1

2

1

1

2

v

v

pow

g

pow

g

g

g

d

d

V

V===

ρρρρ

ρρ

&

&

.

From the condition of equal power N1 = N2 :

2

1

2

22

1

v

v

s

s

d

d

V

V

Q

Q==

&

&

.

So, there is:

vn

ns

v

s

v

s

v

s

d

Q

d

Q

d

Q

d

Q==== ...

3

3

2

2

1

1

constd

Q

v

s =

and the resulting expression is called the Wobbe number

v

s

d

QWb = .



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Density of gases

Density is the ratio of the mass (m) of a gas and its volume (V)

ρ = m/V, [kg/m3].

Specific volume (v) is the ratio of the volume (V) of a material and its mass (m) - so it is the reciprocal

of density

v = V/m, [m3/kg].

Density of a gas can be changed by modifying either the pressure or the temperature. Increasing the

pressure increases the density, whereas increasing the temperature decreases it. The change can be

calculated using the Clapeyron’s ideal gas law:

RTp

orRTpv ==ρ

where: p, N/m2 – the absolute pressure of the gas,

v, m3/kg – the specific volume or ρ, kg/m

3 – the density

R, J/(kgK) – the individual gas constant

T, K – the absolute temperature.

Due to this law, the following relationship can be obtained:

1

2

2

1

2

1

T

T

p

p ⋅=ρρ

so at a constant pressure (p)

1

2

2

1

T

T=

ρρ

and at constant temperature (T)

2

1

2

1

p

p=

ρρ

.

The task

1. Perform a sample calculation of the Wobbe number for the selected gaseous fuels.

2. Calculate the value of the Wobbe number for methane CH4.

Take for the air: ρpow(20oC) = 1,164 kg/m3

Qs,CH4 = 39,9 MJ/m3



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2. Flammability limits

Flammability limits are the minimum (lower) and the maximum (upper) contents of fuel in the

fuel/air mixture, between which flame propagation in this mixture is possible (the mixture is

flammable). Flammability limits mean the same as ignition limits or explosion limits.

Table 2.1 Flammability limits of gases

Flammability limits

% gas in air % gas in oxygen Gas

lower upper lower upper

hydrogen H2

carbon monoxide CO

methane CH4

ethane C2H6

propane C3H8

n-buthane C4H10

n-pentane C5H12

n-hexane C6H14

ethylene C4H10

acetylene C2H2

benzene C6H6

methanol CH3OH

ethanol C2H6OH

4,1

12,5

5,0

3,2

2,4

1,9

1,4

1,25

3,75

2,5

1,41

6,72

3,28

74,2

74,2

15,0

12,5

9,5

8,4

7,8

6,9

29,6

80,0

6,75

36,5

18,95

4,0

15,5

5,1

3,0

2,3

1,8

–

–

2,9

2,5

2,6

–

–

94,0

94,0

61,0

66,0

55,0

48,0

–

–

79,9

89,4

30,0

–

–

The flammability limit of the combustible mixture without inert components ( [CO2]+[N2] < 5%) can

be calculated from the Le Chatelier formula knowing the explosion limits of the components of the

mixture:

,...

100

),(),(2

2

),(1

1,

dgn

i

dgdg

dg

L

a

L

a

L

aL

+++= (2.1)

where:

Lg d – lower or upper limit of ignition, %;

ai – concentration of a single, combustible component of the mixture (∑ ai = 100%).

The flammability limit of mixtures containing the inert gases ([CO2]+[N2] > 5%), can be determined

from the formula:

,

1100

1001

1

dg,

dg,dg,

σ

σL

σ

σ

LLσ

−+

−+

= (2.2)

where: σ = CO2 + N2.



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Example 1. Calculate the lower and upper flammability limit of a gas of a given composition:

CO=28%, H2=4%, CH4=1%, CO2=8%, N2=58%, H2O=1%.

Solution

In the composition of the gas there are significant amounts of CO2 i N2 (CO2+N2>5%). Therefore,

we calculate first the lower and upper flammability limit of the gas according to the formula 2.1

and next according to the formula 2.2.

- the lower flammability limit:

%2,10

5

1

5

4

5,12

28100100

42

42

=++

=++

=

dCHdHdCO

d

L

CH

L

H

L

COL

σ = CO2 + N2 = 0,58+0,08 = 0,66

%0,25

66,0166,0

2,10100

10066,01

66,01

2,10

1100

1001

1

d

dd =

−⋅+

−+

=

−+

−+

=

σ

σL

σ

σ

LLσ

- the upper flammability limit:

%3,66

15

1

2,74

4

2,74

28100100

42

42

=++

=++

=

gCHgHgCO

g

L

CH

L

H

L

COL

%3,85

66,0166,0

3,66100

10066,01

66,01

3,66

1100

1001

1

g

gg =

−⋅+

−+

=

−+

−+

=

σ

σL

σ

σ

LLσ

Example 2. Calculate the range of the stoichiometric ratio (λ) in which it is possible to burn the gas

with the composition shown in Example 1

Solution

Flammability limits were calculated previously: Ld= 25% i Lg=85,3%.

Knowing the flammability limit given by the concentration of gas (Vg ) in the gas/air mixture (Vm):



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%100%100 ⋅+

=⋅=airg

g

m

g

VV

V

V

VL

and the formula for the excess air ratio (λ)

gt

air

air

VV

V

⋅=λ , where V

tair – unit demand for air given in m

3air/1m

3fuel

we can calculate:

gt

airair VVV ⋅⋅= λ , m3/s

tair

tairg

g

VVV

VL

⋅+=

⋅+⋅

=λλ 1

100

)1(

100, %

LV

Lt

air ⋅−= 100λ

Vairt = (100/21)·[0,5(0,28+0,04)+2·0,01] = 0,857 m

3/m

3

5,325857,0

25100 =⋅

−=λ

2,03,85857,0

3,85100 =⋅

−=λ

The solution - the gas is flammable within the range 0,2 < λ < 3,5.



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3. Stoichiometry of combustion processes

Fossil fuels are mainly compounds of carbon and hydrogen (hydrocarbons CmHn). The reaction of

their oxidation can be expressed by an equation of stoichiometry:

CmHn + (m+n/4)O2 → mCO2 + n/2H2O

1 mole + (m+n/4)mole → (m+n/2)mole.

Therefore, for one mole of fuel there is necessary exactly: (m+n/4) moles of oxygen.

When the fuel and oxidizer content in the fuel mixture results from the equation of stoichiometry,

we say that the mixture is stoichiometric.

If combustion of a stoichiometric mixture is complete, there cannot be neither fuel nor oxygen

present in the flue gas.

Example:

Burning of a stoichiometric methane mixture in air (there is no fuel or oxygen in flue gas):

CH4 + 2O2 + 7.52N2 →→→→ CO2 + 2H2O + 7.52N2 (where air: 79% N2 + 21% O2).

The number of moles of flue gas N equals:

N = 1 mole CO2 + 2 moles H2O + 7.52 mole N2 = 10.52 moles.

Flue gas composition:

[CO2] = 1mol CO2/10,52 = 9,5% CO2 vol.

[H2O] = 2 moles H2O/10,52 = 19% H2O vol.

[N2] = 7.52 mole N2/10,52 = 71,5% N2 vol.

Burning of a lean methane mixture in air (the content of air in the mixture is 5% larger than in the

stoichiometric one):

CH4 + 2.1O2 + 7.9N2 →→→→ CO2 + 2H2O + 0.1O2 + 7.9N2 .

The number of moles of flue gas N equals:

N = 1 mole CO2 + 2 moles H2O + 7.9 mole N2 + 0.1 mole O2 = 11 moles.

According to the wet (water is liquid) analysis of flue gas, the concentration of the components is as

follows:

[CO2] = 1 mole CO2/11 = 9.09% CO2 vol.

[H2O] = 2 moles H2O/11 = 18.2% H2O vol.

[O2] = 0.1 mole O2/11 = 0.91% O2 vol.

[N2] = 7.52 moles N2/11 = 71.82 N2 vol.



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Stoichiometric air

The theoretical air required to complete combustion of fuel results from the equation of

stoichiometry of oxygen/fuel reaction.

Stoichiometric air means the minimum amount of air in a stoichiometric mixture.

The real combustion air depends on the assumed air excess n (the equivalence ratio or the

stoichiometric ratio λ).

air tricStoichiome

air Actual=λ

[ ]29.20

9.20

O−=λ

The type of flame/ type of mixture:

λ<1 → Rich (too much fuel)

λ=1 → Stoichiometric (stoichiometric content of fuel and oxygen)

λ>1 → Lean (too much oxidizer)

The air excess (n):

air tricstoichiome

air tricstoichiome - actual=n .

The relationship between the stoichiometric ratio λ and the air excess n:

n = (λ -1) ·100%.

Example:

CH4 + 2O2 →→→→ CO2 + 2H2O

1 kmol CH4 + 2 kmoles O2 → 1 kmol CO2 + 2 kmoles H2O

1 nm3 CH4 + 2 nm

3 O2 → 1 nm

3 CO2 + 2 nm

3 H2O (In normal conditions!)

LO2t = 2 m3O2/m3CH4

Lairt = 2/0,21 = 9,52 m3air/m3CH4

So, the stoichiometric air for methane burning (CH4) is 9.52 m3

air/m3

CH4 .

Assuming for example:

- the rate feeding of methane V = 10 m3

CH4/h

- the rate feeding of air V = 110 m3

air/h (Actual air)

we have: λ = 110 / (9,52*10) = 1.16

n = (1.16 – 1.0)·100% = 16%.

Theoretical air requirement based on elementary analysis of fuel:

+−+= sohcLo 83

8

232,0

1, kg/kgfuel,

+−+= sohcLo 83

833,3 , m

3/kgfuel.

where: c, h, o, s – carbon, hydrogen, oxygen and sulphur content in fuel.



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Theoretical oxygen and air requirement for the solid and liquid fuels

The overall composition of the solid fuels is: c + h + s + o + n + w + a = 100%, where c, h, s, o, n, w, a

are the weight fractions of compounds (respectively: carbon, hydrogen, sulphur, oxygen, nitrogen,

moisture and ash) based on elementary analysis.

To calculate the theoretical and actual oxygen (and air) requirement (in different units!) we can write

the reaction of the main components oxidation (c, h, s):

C + O2 = CO2

1 kmol C + 1 kmol O2 → 1 kmol CO2

12 kg C + 1 kmol O2 → 1 kmol CO2

1 kg C + 1/12 kmol O2 → 1/12 kmol CO2

H2 + ½ O2 = H2O

1 kmol H2 + 1/2 kmol O2 → 1 kmol H2O

2 kg H2 + 1/2 kmol O2 → 1 kmol H2O

1 kg H2 + 1/4 kmol O2 → 1/2 kmol H2O

S + O2 = SO2

1 kmol S + 1 kmol O2 → 1 kmol SO2

32 kg S + 1 kmol O2 → 1 kmol SO2

1 kg S + 1/32 kmol O2 → 1/32 kmol SO2

1 kg of fuel contains (from elem. analysis): c kmol C/kg, h kmol H2/kg, s kmol S/kg and o kmol O2/kg,,

so theoretical air requirement is:

−++=32324122

oshcL t

O kmol O2/kgfuel (we always must take into account the oxygen present in the fuel)

C + O2 = CO2


12 kg C + 32 kg O2 → 1 kmol CO2

1 kg C + 8/3 kg O2 → 44 kg CO2

H2 + ½ O2 = H2O


2 kg H2 + 16 kg O2 → 18 kg H2O

1 kg H2 + 8 kg O2 → 9 kg H2O

S + O2 = SO2


32 kg S + 32 kg O2 → 64 kg SO2

1 kg S + 1 kg O2 → 1/2 kg SO2



−++= oshcL tO 8

3

82

kg O2/kgfuel

+−+= sohcL tair 8

3

8

232,0

1 kgair/kgfuel

where 0,232 is the mass fraction of oxygen in the air (it is not volume fraction!)



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C + O2 = CO2


12 kg C + 22,4 m3 O2 → 22,4 m

3 CO2

1 kg C + 22,4/12 m3 O2 → 1/12 m

3 CO2 1 kg C + 1,86 kmol O2 → 1/12 kmol CO2

H2 + ½ O2 = H2O


2 kg H2 + 22,4/2 m3 O2 → 22,4 m

3 H2O

1 kg H2 + 22,4/4 m3 O2 → 1/2 m

3 H2O 1 kg H2 + 5,6 kmol O2 → 1/2 kmol H2O

S + O2 = SO2


32 kg S + 22,4 m3 O2 → 22,4 m

3 SO2

1 kg S + 0,7 km3 O2 → 0,7 m

3 SO2



LO2t = 1,86*c + 5,6*h + 0,7*s – 0,7*o m

3 O2/kgfuel

( )oshcL tair 7,07,06,586,1

21,0

1 −++= , m3

air/kgfuel,

We can write:

⋅−⋅+⋅+⋅⋅⋅=7,0

7,07,06,586,17,0

21,0

1 oshcL t

air m

3air/kgfuel

Then:

+−+= sohcL tair 8

3

833,3 m

3air/kgfuel

Theoretical oxygen and air requirement for the mixtures of gaseous fuels

The average composition of the gas fuel is as follows (the volume fraction):

CO + H2 + CH4 + CnHm + O2 + SO2 + CO2 + N2 = 100%

The main combustible compounds are CO, H2, CH4 and CnHm.(for example ethane C2H6)

The stoichiometric equations for combustion of gaseous fuels:

CO + 1/2 O2 = CO2

1 kmol CO + 1/2 kmol O2 → 1 kmol CO2

1 m3 CO + 1/2 m

3 O2 → 1 m

3 CO2

H2 + 1/2 O2 = H2O

1 kmol H2 + ½ kmol O2 → 1 kmol H2O

1 m3 H2 + 1/2 m

3 O2 → 1 m

3 H2O

CH4 + 2O2 = CO2 + 2H2O

1 kmol CH4 + 2kmol O2 → 1 kmol CO2 + 2 kmol H2O



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C2H6 + 3 ½ O2 = 2CO2 + 3H2O

1 kmol C2H6 + 3 ½ kmol O2 → 2 kmol CO2 + 3 kmol H2O

So we can write:

Lo2t = CO/2 + H2/2 + 2CH4 + 3 ½ C2H6 – O2 m

3O2/m

3fuel (In normal conditions)

Lairt = Lo2

t /0,21 m

3air/m

3fuel

*********************************************************************************

The helpful information about the molar mass and density

The definition of density is as follows:

DENSITY is defined as a ratio of mass to occupied volume:

ρ = m/V, [kg/m3]

but we also know that mass of substance can be express by multiplication:

m = n*M, [kg]

where:

n – the amount of substance expressed in [mole] or [kmol] units M – the molar mass (it is the mass of 1 mol in normal conditions expressed in [g/mol] or [kg/kmol])

The molar mass is defined for each compounds – the values of molar mass can be found in periodic

table (Mendeleev’s Table);

– for example (rounded values): MH=1, MC=12, MO=16 MN=14 MS=32 g/mol

MH2= 2 g/mol, MO2= 2*16=32 g/mol, MN2= 2*14=28 g/mol, MH2O= 2*1+16=18 g/mol,

MCO2= 12+2*16=44 g/mol, MCH4= 12+4*1=16 g/mol, etc.

Volume of gases (in normal conditions):

V = n*22,4, [m3],

where 22,4 m3 is the volume of 1 kmol of each compounds (1 mol occupies 22,4 dm

3 !)

So we can write down that:

ρ = m/V = (n·M)/(n·22,4) = M/22.4

For example we can calculate the density of methane (in normal conditions):

MCH4 = 12 + 4 = 16 kg/kmol (or g/mol)

so: ρ CH4 = MCH4/22,4 = 16/22.4 = 0.714 kg/m3,

The gas density can be also calculated using Clapeyron’s formula: pV = mRT

(The possible difference between results is the consequence of rounding.)



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4. Calorimetry of combustion processes

Every fuel is described by two caloric parameters:

- upper caloric value: HCV [J/kg]

- lower caloric value: LCV [J/kg].

The difference between the upper and the lower caloric value results from the fact that water in flue

gas is in liquid form during the determination of UCV and in gaseous form (steam) when LCV is

determined .

Upper caloric value is higher than lower caloric value (UCV > LCV).

Standard lower caloric value: LCVa

st = UCVa

st – r· (Wa+8.94H

a)

where:

r – heat of evaporation of water at the temperature of 25oC corresponding to 1% of water in

fuel [J/kg], r = 24.42 [J/kg],

Wa – moisture content in analytical sample of fuel, %

Ha – hydrogen content in analytical sample of fuel, %

8.94 – coefficient for water.

The caloric effect of combustion results from the oxidation reaction of a fuel due to the change of its

chemical structure in the combustion process. It is calculated by subtracting the formation enthalpy

of subtracts (air and fuel) from the formation enthalpy of products (flue gas).

Due to the Hess law, the caloric effect of a chemical reaction depends only on the initial and final

state of the reacting system and is independent of the route of conversion.

Standard upper caloric value:

∆H = ∆hst

(prod) – ∆hst

(sub) (T=298K, p= 101.3kPa)

where:

sub- air and fuel,

prod – flue gas

∑∑ ∆⋅−∆⋅=∆=i

subiii

prodii hnhnHQ )()(

ni – stoichiometric coefficients.



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Table 4.1 The values of the enthalpies of formation for selected compounds

Compound State Δh, kJ/mol

O2 Gas 0

N2 Gas 0

H2 Gas 0

C Solid 0

O Gas 249,19

N Gas 472,66

H Gas 217,94

CO2 Gas -393,52

H2O Gas -241,83

H2O Liquid -285,84

CO Gas -110,54

CH4 Gas -74,87

C2H2 Gas 227

C2H4 Gas 52

C2H6 Gas -85

HCl Gas -92

H2SO4 Liquid -908

SO3 Gas -395

Example:

Evaluate UCV and LCV for hydrogen.

- hydrogen combustion reaction:

)(2)(2)(2 22 lgg OHOH ⋅→+⋅

- calculation of upper caloric value UCV, when water is liquid:

)(298)(298)(298 22222 gO

ogH

olOH

o HHHH ∆−∆⋅−∆⋅=∆

68,571−=∆H kJ/mol

assuming the molar mass of hydrogen M = 2H2 = 4,04 g/mol, we obtain:

5,14104,4

68,571 −=−=∆=M

HUCV , MJ/kg

- calculation of lower caloric value UCV, when water remains in gaseous form (steam):

)(2)(2)(2 22 ggg OHOH ⋅→+⋅

)(298)(298)(298 22222 gO

ogH

ogOH

o HHHH ∆−∆⋅−∆⋅=∆

66,483−=∆H kJ/mol,

7,11904,4

66,483 −=−=∆=M

HLCV , MJ/kg.



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Approximate formulas for the lower caloric value of solid and liquid fuels (Mendeleev formula)

( ) rrrrrri WSOHCQ ⋅−−⋅−⋅+⋅= 1,259,108103015,339

, kJ/kg

where Cr, H

r, O

r, S

r are weight contents of elements and W

r - water in a fuel (%) in raw conditions.

Approximate formulas for the lower caloric values for liquid fuels:

15

3106,1265,27

ρ⋅+=r

iQ, MJ/kg

and

1000

2935,1354230 15r

ri

SQ

⋅−⋅−=

ρ, MJ/kg

where ρ15 is the density of the fuel in the temperature of 15 oC (kg/m

3), and S

r is the sulphur content

in fuel (%) in raw conditions.

The lower caloric values for gaseous fuels can be determined from the formula:

( )[ ]∑=

⋅=n

iii

ri

ri vLCVLCV

1

, MJ/m3.



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5. Temperature of combustion processes

Temperature of combustion Ts – can be different depending on the heat loss from flame.

Calorimetric temperature of combustion Ts,c – the highest temperature of non-dissociated flue gas as

a result of adiabatic and isobaric combustion of fuel and stoichiometric air.

Theoretical temperature of combustion Ts,t – the highest temperature of dissociated flue gas as a

result of adiabatic and isobaric combustion of fuel and not necessary stoichiometric air.

Theoretical temperature of combustion

The theoretical temperature of combustion is calculated under constant pressure for the equation of

the enthalpy balance of subtracts and products.

(5.1)

where: mpp – rate of fuel supply, mpow – rate of air supply, msp – stream of flue gas.

Simplified method of the temperature of combustion calculation

Assuming equal mean specific heat and equal initial temperature:

cp(fuel) = cp(air) = cp(flue gas) (5.2)

Tf = Ta = To (5.3)

we obtain:

(Tc – To) · cp,flue gas · mflue gas = QHCV · mfuel (5.4)

Tc = To + (QHCV·mfuel)/(mflue gas·cp,flue gas) . (5.5)

The real temperature of combustion is lower than the calorimetric temperature of combustion due

to:

- heat losses from the flame,

- incomplete combustion (dissociation),

- non-total combustion.

∫

∫ ∫

+∆

=++∆

Ts

Tst

psttw

sp

Tpal

Tst

Tpow

Tst

ppowpsttw

dTspcsphm

dTpowcmdTpalcpalh

))()((

)())()((m

,

,pp



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Table 5.1. Examples of the measured temperature of combustion of selected fuels

Fuel Oxidizer Fuel in mixture % Temperature, oC

Hydrogen

Hydrogen

CO

Methane

Butane

Acetylene

Acetylene

Air

Oxygen

Air

Air

Air

Air

Oxygen

31,6

78

20

10

3,2

9

33

2045

2660

1650

1875

1895

2325

3007

Example:

Calculate the calorimetric temperature (adiabatic, without dissociation) of combustion:

Assumption: Vm = 100 m3/h

Composition of the flammable mixture: 10% CH4 + 90% air

Content of components:

[CH4] = 0.1

[O2] = 0.21·0.9 = 0.189

[N2] = 0.79*0.9 = 0.711

Volumetric flow of components :

Vm = 100 m3/h

VCH4 = 0.1·100 m3/h = 10 m

3/h

VO2 = 0.189·100 m3/h = 18.9 m

3/h

VN2 = 0.711·100 m3/h = 71.1 m

3/h

The demand of oxygen for the combustion of methane

VO2,teoret = 2·VCH4 =20 m3/h

Combustion stoichiometry

CH4 + 2O2 = CO2 + 2H2O

3 moles = 3 moles

Conclusions:

1. The numbers of moles of fuel mixture and flue gas are equal.

2. The volumetric flows of fuel mixture Vm and of flue gas determined in the standard

conditions (T = 0oC, patm) are equal:

Vm = Vs.

The excess air ratio (λ)

λ = (Vair, real/Vair, teoret) = VO2, real/VO2, teoret = (18.9 m3/h)/(20 m

3/h) = 0.945 < 1 (rich mixture)

Vm Vs



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Conclusion: CH4 remains in the flu gas.

Composition of flue gas

The real (non-stoichiometric) reaction of oxidation:

1CH4 + 1.89O2 = 0.945CO2 + 1.89H2O + 0.055CH4.

The number of moles of the gases N: left = 1+1.89 = 2.89, right = 0.945+1.89+0.055 = 2.89.

It can be calculated how much of each component (m3) of flue gas is obtained during the combustion

of 10m3 of CH4.

+ 9.45 m

3 CO2

+ 18.9 m3 H2O

+ 0.55 m3 CH4

Consumption of the combustible mixture:

- 18.9 m3 O2

- 10 m3 CH4

Composition of the flue gas (CH4, CO2, H2O i N2):

Vs = 100 m3/h the concentration of the components:

VCH4 = 0.55 m3/h, [CH4] = 0.0055

VCO2 = 9.45 m3/h, [CO2] = 0.0945

VH2O = 18.9 m3/h, [H2O] = 0.189

VN2 = 71.1 m3/h, [N2] = 0.711

Specific heat of gas (cp) [kJ/(m3K)] – determined in normal conditions ((T = 0

oC, patm):

cp,av [kJ/(m3 o

C)] = cpch4·[CH4] + cpCO2·[CO2] + cpH2O·[H2O] + cpN2·[N2] =

2.826·0.0055 + 2.45·0.0945 + 2.026·0.189 + 1.5·0.711 = 1.695 kJ/(m3 o

C )

cp,av – average specific heat of gas (T from 0 to Ts,oC)

To calculate the cp,śr , values were taken for the temperature of 2500 oC.

The combustion temperature (flue gas)

Ts = To + LCV·VCH4/(Vs cp,av), To = 0 oC

Ts = LCV·Vm/(Vs·cp,av) = (35 330 kJ/m3·9.45 m

3/h)/(100 m

3/h·1.695 kJ/(m

3 oC ))

The lower calorific value (LCV = 35 330 kJ/m3)

was assumed for methane, because the water in the

flue gas is gaseous (steam).

The flow rate of burnt methane VCH4 = VCH4pal - VCH4spal = 10 - 0,55 = 9.45 m3/h

Ts = 1969.7 oC.

The measured temperature: (see the table 5.1): Ts = 1875 oC.

The relative error is ∆T/Ts·100% = (1969.7 – 1875)/(1875 + 273.15) = 94.7/2148.15 = 4.41%.



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Note 1: To calculate the relative error, the temperature difference is divided by the absolute

temperature.

Note 2: The calculations are iterative. This is the first approximation – the second approximation is

obtained when the average specific heat of flue gas components will be taken for the calculated

temperature 1970oC, and so on.

Note 3: The dissociation of H2O i CO2 was not included in the calculations.



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6. Calculations of combustion chambers

The initial dimensions of the combustion chamber are determined on the basis of the unit volumetric

heat load of the combustion chamber:

qv=B·LCV/Vch, kW/m3 (6.1)

where: B – fuel flow rate, kg/s or m3/s

LCV – lower caloric value, kJ/kg or kJ/m3

Vk – combustion chamber volume, m3 .

The following equation describes the unit heat load of cross-section area:

qF=B·LCV/Fch, kW/m2 (6.2)

where: Fch – cross-section area of the combustion chamber, m2.

The approximate height of the chamber: h = Vk/Fk, m.

For stoker-fired boilers the combustion chamber size depends on the grate size and it is calculated

from this formula:

FG = B·LCV/qG, m2 (6.3)

where: FG – the grate area, m2

qG – acceptable unit heat load of the grate, kW/m2.

In the case of the initial design of steam boilers combustion chambers, usually the following values of

the unit heat load are taken:

- boilers with a mechanical stoker qV = 250 – 300

kW/m3

qF = 1000 – 2000 kW/m2

- pulverized boilers: qv = 150 – 250 kW/m3

qF = 2000 – 4000 kW/m2

- heavy oil or natural gas boilers: qv = 300 – 350 kW/m3 .

According to other sources, for the dry-slag pulverized coal furnace the following values of the

unit volumetric heat load can be assumed:

- for coals qv= 200 – 230 kW/m3,

- for lignites qv= 150 – 180 kW/m3,

- for oil and natural gas qv= 280 – 350 kW/m3.



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The unit heat load of cross-sectional area of the combustion chamber is assumed within the range:

- for the dry-slag pulverized coal boilers depending on the heat capacity:

- up to 50 kg/s qF = 2300 – 2900 kW/m2

- up to 100 kg/s qF = 2900 – 3500 kW/m2

- up to 180 kg/s qF = 3500 – 4100 kW/m2

- above 180 kg/s qF = 4100 – 4700 kW/m2

- for the light oil and natural gas boilers qF= 3500 – 6000 kW/m2

Note: The loss of incomplete combustion can be considered:

Bcal = B·(1-Sn/100) (6.4)

where: Sn – the loss of incomplete combustion, % (the loss associated with the participation of the

combustible particles in ash and slag).

Example 1:

Determine the size of the combustion chamber of a pulverized coal fired boiler for combustion 8 t/h

of hard coal with LCV = 22500 kJ/kg.

Solution

- The volume of combustion chamber (qv= 250 kW/m3):

Vch = B·LCV/qv = 8000·22500/(3600·250) = 200 m3

- Cross-section area of combustion chamber (qF= 2000 kW/m2):

Fch = B·LCV/qF = 8000·22500/(3600·2000) = 25 m2

- Height of the chamber: h=Vch/Fch= 200/25 = 8 m

Example 2:

The steam boiler is fired with coal (Cr=65%, H

r=6%, S

r=5%, N

r=4%, O

r=10%, W

r=10%). The efficiency of

the boiler is η=0.78. Steam parameters: ms=1150Mg/h, ps=13Mpa, ts=560oC, the temperature of

supply water tw=80oC. Calculate the mass flow of burnt coal.

Solution:

The lower caloric value of coal: LCV= 339.15·C+1030·H-108.9(O-S)-25.1·W = 21311 kJ/kg

From the i-s graph (for p=13MPa and T=833K) → is=3510 kJ/kg

From the tables (for tw=80oC) → iw=258.8 kJ/kg

LCVB

iim wpp

⋅−⋅

=)(

η , LCV

iimB wpp

⋅−⋅

=η

)(

h

t

kg

kJkg

kJ

h

t

B 2252131178,0

)8,2583510(1150

=⋅

−⋅=



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For the pulverized boiler we assume: qv= 200 kW/m3, qF= 4500 kW/m

3 .

The volume of the combustion chamber:

Vch = B·LCV/qv = 225000·21311/(3600·200) = 6660 m3.

The cross-section area of the combustion chamber:

Fch = B·LCV/qF = 225000·21311/(3600·4500) = 296m2

The height of the combustion chamber:

h=Vk/Fk= 6660/296 = 22,5 m

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