les écoulements de fluides dans l'industrie

Production de la chaleur - Chaudière

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Combustion and Heat production equipment

Jean-Marie SEYNHAEVE

• PART 1 : Theoretical background

1.1 Combustion = Chemical reaction1.2 Heating values of fuels

• PART 2 : Heat production : equipment

2.1 Energy balance and equipment efficiency2.2 Combustion control2.3 Illustration of burners2.4 Practical problems


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Part 1 : Theoretical background - Table of contents

I - COMBUSTION : CHEMICAL REACTION

1.1 Examples of Chemical Reaction of Combustion, Fundamental Laws

1.2 Air Required for Combustion - Flue Gases contents1.2.1 Solid or Liquid Fuel1.2.2 Gas Fuel 1.2.3 Practical Problems

II - HEATING VALUES OF FUELS

2.1 Heat of Reaction - Some Examples

2.2 Definition of the Heating Value of a Fuel

2.3 How to evaluate the Heating Value of Fuels2.3.1 Solid or Liquid Fuel2.3.2 Gas Fuel

2.4 Practical Examples of Evaluation


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Combustible Elements in Fuels :

C S H2 CO CH4 C2H4 C2H6 C3H8 ......Alcans : CnH2n+2

Alcens : CnH2n

Aromatics ...

I. Combustion – Chemical reaction : Fundamental laws

Combustion :

C + O2 CO2

S + O2 SO2

H2 + ½ O2 H2OCO + ½ O2 CO2

CH4 + 2 O2 CO2 + 2 H2OC2 H4 + 3 O2 2 CO2 + 2 H2O........CnH2n+2 + ½ ( 3n + 1 ) O2 n CO2 + ( n + 1 ) H2O

Chemical Reaction of Oxidation + HEAT


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Chemical BALANCED reactions in terms of number of MOLES CnH2n+2 + ½ ( 3n + 1 ) O2 n CO2 + ( n + 1 ) H2O

n C n C + n O2

( n + 1 ) H2 ( n + 1 ) H2 + ½ ( n + 1 ) O2

½ ( 3n + 1) O2 ½ ( 3n + 1 ) O2Molecular weight of molecules :

TYPE OF MOLECULE MOLECULAR WEIGHT Carbon C 12 kg/kmole Suffer S 32 kg/kmole

Hydrogen H2 2 kg/kmole Oxygen O2 32 kg/kmole Nitrogen N2 28 kg/kmole

Example : C2H6 : 2x12 kg/kmole + 3x2 kg/kmole = 30 kg/kmole


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Some important laws :

AVOGADRO

At same pressure and temperature, 1 kmole of every gas occupies the same volume.

For Normal Conditions ( t= 0 °C, p = 760 mm Hg) :322.414kmoleV m

LAVOISIER The total MASS in reaction remains constant

Application to air composition (combustive) : O2 N2 ( Ar Ne Xe … ) - Molar Fractions = Volume Fractions for air ( Avogadro Law ) :

[O2] = 0.210 (kmole of O2 / kmole of Air or m3 of O2 / m3 of Air)[N2] = 0.790 " N2 " " N2 "

- For Air : 1 kmole of O2 + 0.79/0.21 ( 3.76 ) kmole of N2

[i]

- Mass fraction for air : (i)0.210 kmole of O2 = 6.72 kg of O20.790 kmole of N2 = 22.12 kg of N2 1 kmole of Air = 28.84 kg of Air

2

2

6.72 28.84 0.233

22.12 28.84 0.767

O

N


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Air required for Combustion – Production of flue gases Pouvoir comburivore – Pouvoir fumigène

Fuel composition• For solid and liquid fuel : elementary mass composition

• Combustible Elements : C (H) (S) ...• Oxygen in Fuel : (O)• Humidity : (H2O)• Inert Elements : (N) (Ash)

1

1n

i

i

• For gas fuel : molar (volume) composition• Combustible components : [H2 ] [CO ] [CH4 ] [C2H4 ] [C2H6] ...• (Oxygen in Fuel : [O2 ])• Humidity : [H2O ]• Inert Elements : [N2 ] [CO2 ] …Definition : Excess of air

1

[ ] 1n

i

i

Practical Air RequirementTheoretical need of Air

1 Stoechiometry


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SOLID and LIQUID FUEL - Combustion accounting

Air requirement for combustion in m 3N /kg of fuel

22.414/0.21O2

Flue gases produced by combustion in m 3N /kg of fuel

22.414 [(C)/12+(S)/32+(H)/2+(H2O)/18+(N)/28+3.76O2+(-1) O2]

kmole CO 2 kmole SO 2 kmole H 2O kmole N 2 kmole O 2

(C)/12 - - - -- - (H)/2 - -- (S)/32 - - -- - - - -- - - (N)/28 -- - (H2O)/18 - -- - - - -

… … … … …(C)/12 (S)/32 (H)/2+(H2O)/18 (N)/28+3.76 O2 -

3.76( -1) O2 ( -1) O2

(C)/12 (S)/32 (H)/2+(H2O)/18 (N)/28+3.76O2 ( -1) O2

Elements after combustion - Flue gases1 kg of fuel kmole of fuel kmole of O 2

(C) (C)/12 of C (C)/12(H) (H)/2 of H2 (H)/4(S) (S)/32 of S (S)/32 (O) (O)/32 of O2 -(O)/32(N) (N)/28 of N2 -

(H2O) (H2O)/18 of H2O -(Ash) - -

…. … …Stoechiometry O2

Excess of Air (-1) O2

TOTAL O2

Elements before combustion


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SOLID and LIQUID FUEL – Flue gases composition

The flue gases composition is usually given for a DRY SAMPLE

2' 4.76 112 32 28C S N

Den O

2

2

22

2

22

12[ ]''

32[ ]''

3.76 28[ ]''

1[ ]'

'

CCO

DenS

SODen

NON

DenO

ODen

Possible measurements …

Dry sample


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GAS FUEL - Combustion accounting

1 m 3N of fuel m 3

N of O 2

[H2] 1/2 [H2][CO] 1/2 [CO]

[CH4] 2 [CH4] [C2H4] 3 [C2H4] [C2H6] 3.5 [C2H6]

… … [H2O] - [CO2] - [N2] -… …

Stoechiometry O2

Excess of Air ( -1) O2

TOTAL O2

Elements before combustion m 3

N of CO 2 m 3N of H 2O m 3

N of N 2 m 3N of O 2

- [H2] - -1/2 [CO] - - -

[CH4] 2 [CH4] - -2 [C2H4] 2 [C2H4] - -2 [C2H6] 3 [C2H6] - -

… … … …- [H2O] - -

[CO2] - - -- [N2]

… … … …CO2,fg

2Ofg [N2]+3.76 O2 -

3.76( -1) O2 ( -1) O2CO2,fg

2Ofg [N2]+3.76O2 ( -1) O2

Elements after combustion - Flue gases

Air requirement for combustion in m 3N / m 3

N of fuel/0.21O2

Flue gases produced by combustion in m 3N / m 3

N of fuelCO2,fg+H2Ofg+[N2]+3.76O2+(-1) O2


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GAS FUEL – Flue gases composition

The flue gases composition is usually given for a DRY SAMPLE

2, 2 2' [ ] 4.76 1 ...fgDen CO N O

2,2

2 22

22

[ ]''

[ ] 3.76[ ]'

'1

[ ]' ...'

fgCOCO

DenN O

NDen

OO

Den

Possible measurements …

Dry sample!


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Objectives of those accounting tables

Practical point of view Theoretical point of view

Stack

Flue Gases Analysis Device

Purging

Volume fraction onDry Flue Gases

• Excess of air evaluation• Combustion control• Heat production equipment efficiency


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II. Heating values of fuels (Pouvoir calorifique)Combustible Elements in Fuels :

C S H2 CO CH4 C2H4 C2H6 C3H8 ......Alcans : CnH2n+2

Alcens : CnH2n

Aromatics ... Combustion :

N° Chemical Reaction Ref. : Standard Conditions

Heat kJ/kmole

Heat kJ/kg

Heat kJ/m3

N 1 C + O2 CO2 404400 33700 - 2 C + ½ O2 CO 121400 10117 - 3 S + O2 SO2 296930 9279 - 4 CO + ½ O2 CO2 283000 10107 12626 5 C + 2 H2 CH4 85800 7140 6 H2 + ½ O2 H2O VAP 241800 120900 10790 7 H2 + ½ O2 H2O LIQ 285800 142900 12750 8 CH4 + 2 O2 CO2 + 2 H2OVAP 802200 50040 35790 9 CH4 + 2 O2 CO2 + 2 H2OLIQ 890200 55637 39710

/ / 22.414


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Principle of Energy balance :

Reaction 1 = Reaction 2 + Reaction 4 404400 = 121400 + 283000 Reaction 8 = Reaction 1 + 2 x Reaction 6 - Reaction 5 802200 = 404400 + 2 x 241800 - 85800

Heating (calorific) value – Pouvoir calorifique : Definition

Calorimeter of MALHERFuel + AirPatm , t = 25 °C

Adiabatic combustion

Hot flue gas

Cooling

Cold flue gas At t = 25°C

Heat which is extracted = Heating value

Calorimeter of YOUNGER

Experiment at constant VOLUME

Experiment at constant PRESSURE


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Type of Heating value 4 types

• 1st distinction : Constant Volume or Constant Pressure ?

In practice CVv never used …• 2nd distinction :

Flue gas at 25 °C

H2O Vapour Lower (Net, inférieur) CVH2O Liquid Higher (Gross, supérieur) CV

In practice: HCV USA, UK, Natural Gas …LCV G,F,I,H,B, …

Conclusion : HCVv LCVv HCVp LCVp


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Evaluation of Heating Values of fuels

Heating Values of Solid or Liquid Fuels

• Fuel composition : reference 1 kg of fuel Elementary composition : (C) (H) (S) … (O) (H2O) (N) …

• Heating for solid or liquid fuel :

/

/

ii combustible

ii combustible

HCV i HCV in kJ kg

LCV i LCV in kJ kg


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• Heating for gas fuel :

3

3

/

/

i Ni combustible

i Ni combustible

HCV i HCV in kJ m

LCV i LCV in kJ m

2 4

32 4 2 6

2 4

32 4 2 6

12750 12620 39710

58950 69210 ... /

10790 12620 35790

56920 63200 ... /

N

N

HCV H CO CH

C H C H in kJ m

LCV H CO CH

C H C H in kJ m

Heating Values of Gas Fuels• Composition of gas fuel : reference 1 m3

N Combustible components

Inert components

2 4 2 4 ...H CO CH C H

2 2 2 2... ...O N CO H O


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Combustibles solides

Combustibles liquides

PCI = 44000 … 42000 kJ/kg

fractions légères lourdes

CHARBON

BOIS

PCIBRUT = [1 - (cen) - (H2O)] PCIPUR & SEC - 2501 (H2O)

PCIPUR & SEC = 35000 … 35500 kJ/kg

PCIPUR & SEC 18500 kJ/kg

si (H2O) = 0.20

PCIBRUT 14300 kJ/kg


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Combustibles gazeux

PCI = 802 400 kJ/kmole = 50 150 kJ/kg

1 kmole CH4 22.414 m3N

PCI = 35 800 kJ/m3N

1 kmole CH4 2 kmole H2O = 36 kg H2O

PCS - PCI = 36 x 2501 = 90 036 kJ/kg

PCS = 892 436 kJ/kmole = 39 816 kJ/m3N

CH4

PCS 1.11PCI


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Examples of Evaluation

1. Liquid fuel (fuel-oil, gas-oil, gasoline, kerosene …) :

1.5 ... 1.9H

C

nan

Where a is the number of atoms of hydrogen per atom of carbon

Calculate the HCV and the LCV

a (H) (C) HCV LCV1.50 0.111 0.889 45833 433781.55 0.114 0.886 46192 436631.60 0.118 0.882 46547 439471.65 0.121 0.879 46900 442291.70 0.124 0.876 47250 445081.75 0.127 0.873 47598 447851.80 0.130 0.870 47943 450611.85 0.134 0.866 48286 453341.90 0.137 0.863 48627 45606

Calorific Values of Liquid Fuels

42000

44000

46000

48000

50000

1.50 1.60 1.70 1.80 1.90 2.00a (number of hydrogen per carbon)

Cal

orifi

c va

lue

(kJ/

kg)

HCVLCV


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2. Natural gas :

4

2

2

0.820

0.030 ( ...3... ...8...)

0.140

0.010

n m

CH

C H with n and m

N

CO

Calculate the HCV and the LCV

3

3

39710*0.820 69210*0.030 34639 /

35790*0.820 63200*0.030 31244 /N

N

HCV kJ m

LCV kJ m


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Part 2 : Heat Production Equipment

I - ENERGY BALANCE AND EQUIPMENT EFFICIENCY

1.1 Scheme of heat production equipment

1.2 Input and output of energy

1.3 Efficiency and Description of Losses

II - COMBUSTION CONTROL

III - ILLUSTRATION

3.1 Gas Burners, Oil Burners, Coal-Fired Stokers

3.2 Pulverised Fuel System

3.3 Multiple Fuel Burners

2.4 Boilers (fire tubes, water tubes)

IV - PRACTICAL PROBLEM


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I. Energy balance and efficiency

1. Scheme of heat production equipment

Air

Fuel

Flue GasBurner

Heat exchanger

Load to be heated


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Fuel

Air

Flue Gas

Load to be heated

- Mass Flow Rate : qC ( kg/s ) - Calorific Value : HCV or LCV ( kJ/kg )

- Air required for combustion : VA, ( kg of air/kg of fuel ) - Specific : hA ( kJ/kg )

- Flue gas produced : VF, ( kg of flue gas/kg of fuel ) - Specific enthalpy of flue gas : hF=cF.tF ( kJ/kg )

- Useful Thermal Power transferred to load : QUSE ( kW )


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2. Input and Output of Energy - INPUT OF ENERGY QC ( LCV + VA, hA ) ( kJ/s or kW ) kg/s kJ/kg kg/kg kJ/kg

- OUTPUT OF ENERGY (in kW) . Useful Thermal Power ( Energy transferred to load ) : QUSE . Energy Loss in Flue Gases : qC VF, hF

. Energy Loss from Walls (convection, radiation): QWALL = r qC LCV . Energy Loss from Unburn Fuel (proportion i) : i qC LCW


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3. Efficiency of heat production equipment

Input of Energy

Output of Energy

,C A AQ LCV V h ,USE C F F C CQ Q V h iQ LCV rQ LCV

Definition of

Efficiency

USE USE

Available C

Q QQ Q LCV

After calculation

, ,

1 1 ( )

F F A A

Losses e i r

V h V hwith e

LCV

- i : unburn fuel loss - e : loss in flue gases ( loss in stack ) - r : radiation ( convection ) heat loss


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II. Combustion control 1. Optimisation of combustion - Criteria 1st point : The e loss is a function of . VF, and VA, : both depending on . hF or temperature of flue gases tF : role of economiser !!! Conclusion : e if or tF 2nd point : Unburn loss i : i if with a limit ( See diagram )

Criteria of combustion optimisation 1 - minimum , but no unburn fuel in flue gases 2 - Temperature of flue gases tF minimum ( Attention !!! )


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Combustion control -Diagram


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Combustion control -Diagram


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2. Parameters controlling the combustion . Volume fractions of CO2 , O2 on dry flue gas :

[CO2]' or / and [O2]' in flue gas ( excess of air ) e loss Dry Sample of Flue Gas !

. Control of unburn fuel : - [CO]' in flue gases : some rules 100 ppm for combustion of gas or fuel 400 ppm for combustion of coal - Unburn solid fuel : . Carbon particles in flue gases : Checking particles by "smoke pump" device . Unburn fuel in ash : Checking by chemical analysis . Flue gas temperature : e loss evaluation ( comparison between burners ... ) . Vacuum in fire place : control of dirtiness of flue gas circuits


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Corrosion acideCorrosion acide

Gasoil ; gaz naturelFraction molaire dans gaz brûlés

[H2O] = 0.12 ; 0.19 à = 1

[H2O] = 0.114 ; 0.181 à = 1.05

psat 0.114 ; 0.181

tsat 49°C 58°C

3 2 2 4

2 2 2 2 4

SO H O H SO1SO O H O H SO2

Seuil de condensation : tp = tsat tfs = 2 tsat - ta0

si ta0 = 0°C tfs = 98°C ; 116°C

( )fs a0p

t tt

2

+=

fst

a0t


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Charge partielleCharge partielle

Pu*

10.5

Pu

R

P0*

P0

k P0*

• Charge / Régime : PuRPu*

=

• Pertes fixes (parois) : 0kP *@

• Pertes proportionnelles : 0πRP *@

• Puissance utile :

u 0 0 0P P kP πRP *= - -• Rendement nominal :

u 0 0 0

0 0

P * P * kP * πRP *η* 1 k πP * P *

- -= = = - -


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Charge partielleCharge partielle• Rendement à charge partielle :

u

u 0 0

PηP kP * πRP *

= + +

or u u 0P RP * Rη* P *= =

On obtient finalement

( )Rη* η*η

1R k 1 R 1 k 1R

= = æ ö+ - ÷ç+ - ÷ç ÷çè øFORMULE DE DITTRICH


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Charge partielleCharge partielleRendement d'une chaudière à charge partielle

= 0.12

0.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Coefficient de charge (R)

Rend

emen

t k = 0.02k = 0.04k = 0.06

0.86

0.840.82


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III. Illustrations

Gas burners

GAZ - AIRMIXTURE

CONNECTION TO BURNER


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Liquid fuel burners


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Cas d’étude : Économiseur d’une chaudière

Brûleur

Air de combustion

T = 20 °C

Gaz naturel

Eau d’alimentation

Vapeur

3 bars, 135 °C

Cheminée

[O2] = 0.03

Tf = 180 °C

Economiseur

Chaudière

• Caractéristiques chaudièrePuissance thermique : 10 MWUtilisation : 4000 h/anPrix du combustible : 0.25 €/m3

N

• Caractéristiques économiseur(AU) : 0.7 kW/K Courant croisé « fluides non mélangés »Durée de vie : 10 ansCoût de l’installation : 30000 €Entretien : 2000 €


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Évaluation des conditions de fonctionnement de la chaudière

Combustible

• Pouvoir calorifique :

PCI

• Composition : [CH4] = 0.81 [C2H6] = 0.05 [N2] = 0.11 [CO2] = 0.03 ...

2 4

2 4 2 6

12750 12620 39710

58950 69210 ...

PCS H CO CH

C H C H

PCS = 35626 kJ/m3N

• Débit nominal : thcomb

PQPCS

Qcomb = 0.287 m3N /s

• Consommation annuelle : C = 0.287 x 3600 x 4000 = 4132800 m3N /an

• Coût annuel combustible : combustible fF C Fcombustible= 1033202 €/an


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Combustion

• Excès d ’air :

PCI

'2

1.795 10.03

6.75 1.05O

• Pouvoir comburivore:,

1.7950.21aV

= 1.147

• Pouvoir fumigène :

Va, = 9.804 m3N d ’air / m3

N de gaz

• Débit d’air :

, 0.940 1.770 6.75 0.11 1.795 1fV

Combustible 1 Nm3 Nm3 O2 Nm3 CO2 Nm3 H2O Nm3 N2 Nm3 O2[CH4] 0.81 1.620 0.810 1.620 - -

[C2H6] 0.05 0.175 0.100 0.150 - -[N2] 0.11 - - - 0.110 -

[CO2] 0.03 - 0.030 - - -Air - - - - 6.749 -

Stoechiométrie 1 1.795 0.940 1.770 6.859 -Pratique 1.795 0.940 1.770 6.75+0.11 1.795(

Après combustionAvant combustion

Vf, = 10.826 m3N d ’air / m3

N de gaz

• Débit de fumée :

,air comb aQ Q V

,fumée comb fQ Q V

Qair = 2.814 m3N /s = 3.638 kg/s

Qfumée = 3.107 m3N /s = 4.017 kg/s

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