boiler total air optimization

7/31/2019 Boiler Total Air Optimization

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Efficient operation of boiler depends onoptimisation of CO2 and O2

This involved elimination of source of air ingress.

As different losses of boiler is affected by variationof air, the optimum value is determined by plottingthe total loss and thereby determining CO2 and O2.


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Oxygen % at various locations in boiler

0

2

4

6

8

10

Furn Outlet AH Inlet AH Outlet ID outlet

O2%

210 MW 210 MW 500 MW 210 MW


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4. Boiler Air Ingress

Cold air leaks into the boiler from openings in the furnaceand convective pass and through open observation doors.

Some of the boiler leakage air aids the combustionprocess; some air that leaks into the boiler in the low

temperature zones causes only a dilution of the flue gas. This portion of air appears as a difference in O2 level

between the furnace exit and oxygen analysers ateconomizer exit. Actual oxygen in the furnace could be

much less. Also, boiler casing and ducting air ingress affects ID fans

power consumption and margins in a major way.


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FurnaceOutlet

Air-in-

leakage

Zirconia

O2 Probe

AH

Seal

Lkg

ESP

Expansion Joints

Air Ingress Points Furnace Roof , Expansionjoints, Air heaters, Ducts, ESP Hoppers, PeepHoles, Manholes, Furnace Bottom


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Air Ingress Calculations

Air ingress quantification is done with the same formulae as those

used for calculation of AH leakage

Air ingress = O2out - O2in * 0.9 * 100(21- O2out)

The basis of O2 or CO2 calculation should be the same eitherwet or dry.


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Find Dry gas, unburnt gas, combustible in ash andauxiliary power (fans) to find the total loss.

Repeat the same for different test air (% of CO2 at

A/H inlet) settings.

Find the minimum total loss.

Find air setting corresponding to minimum total

loss.


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Fuel firing rate = 5599.17 kg/hr

Steam generation rate = 21937.5 kg/hr

Steam pressure = 43 kg/cm2(g)

Steam temperature = 377oC

Feed water temperature = 96oC

%CO2 in Flue gas = 14%CO in flue gas = 0.55

Average flue gas temperature = 190oC

Ambient temperature = 31oC

Humidity in ambient air = 0.0204 kg / kg dry air

Surface temperature of boiler = 70oC

Wind velocity around the boiler = 3.5 m/s

Total surface area of boiler = 90 m2

GCV of Bottom ash = 800 kCal/kg

GCV of fly ash = 452.5 kCal/kg

Ratio of bottom ash to fly ash = 90:10

Fuel Analysis (in %)

Ash content in fuel = 8.63

Moisture in coal = 31.6

Carbon content = 41.65

Hydrogen content = 2.0413

Nitrogen content = 1.6

Oxygen content = 14.48

GCV of Coal = 3501 kCal/kg

The data collected arefor a boiler using coal

as the fuel.

Find out the boilerefficiency by indirectmethod.


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Boiler efficiency by indirect method

Step1 Find theoretical air

requirement

Theoretical air required for

complete combustion

= [(11.43 x C) + {34.5 x (H2O2/8)} + (4.32 x S)] /

100 kg/kg of coal

= [(11.43 x 41.65) + {34.5 x (2.041314.48/8)} +

(4.32 x 0)] / 100

= 4.84 kg / kg of coal


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Step2 Find theoretical CO2 %

% CO2 at theoretical condition( CO2 )t

=

Moles of C

Moles of N2 + Moles of C

Where,

Moles of N2 =

4.84 x 77/100 0.016

+ = 0.133228 28

Where moles of C = 0.4165/12 = 0.0347

( CO2 )t =

0.0347

0.1332 + 0.0347

= 20.67


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Step3 To find Excess airsupplied

Actual CO2 measured in flue gas = 14.0%

% Excess air supplied (EA) = 7900 x [ ( CO2)t(CO2)a]

(CO2)a x [100 (CO2)t ]

= 7900 x [20.67 14 ]

14a x [100

20.67]

= 47.44 %


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Step 4 to find actual mass of air supplied

Actual mass of air supplied = {1 + EA/100} x theoretical air

= {1 + 47.44/100} x 4.84

= 7.13 kg/kg of coal

Step5 to find actual mass of dry flue gasMass of dry flue gas consists of Mass of CO2 +Mass of N2 content in the fuel+ Mass

of N2 in the combustion air supplied + Mass of

oxygen in combustion air supplied

Mass of dry flue gas =0.4165 x 44 7.13 x 77 (7.13-4.84) x 23

+ 0.016 + +

12 100 100

= 7.562 kg / kg of coal


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Step

6 to find all losses

1. % Heat loss in dry flue gas (L1) = m x cp x (TfTa )

x 100GCV of fuel

=7.562 x 0.23 x (190

31)

x 1003501

L1 = 7.89 %


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2. % Heat loss due to formation

of water from H2 in fuel (L2)

= 9 x H2 x {584 + Cp (TfTa )}

x 100

GCV of fuel

=

9 x .02041 x {584 + 0.45(190-31)}

x 100

3501L2 = 3.44 %

3. % Heat loss due to moisture in

fuel (L3)

=

M x {584 + Cp ( TfTa )}

X 100

GCV of fuel

=0.316 x {584 + 0.45 ( 190 31) }

x 100

3501

L3 = 5.91 %


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4. % Heat loss due to moisture in

air (L4)

=AAS x humidity x Cp x (TfTa ) x 100

GCV of fuel

=7.13 x 0.0204 x 0.45 x (190

31) x 100

3501

L4 = 0.29 %

5. % Heat loss due to partial

onversion of C to CO (L5)

=%CO x %C 5744

x x 100

% CO + (% CO2)a GCV of fuel

=

0.55 x 0.4165 5744

x x 1000.55 + 14 3501

L5 = 2.58 %


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6. Heat loss due to radiation andonvection (L6)

= 0.548 x [ (343/55.55)4 (304/55.55)

4] + 1.957 x

(343 - 304)1.25

x sq.rt of [(196.85 x 3.5 + 68.9) /

68.9]

= 633.3 w/m2= 633.3 x 0.86

= 544.64 kCal / m2

otal radiation and convection

oss per hour

= 544.64 x 90

= 49017.6 kCal

% radiation and convection loss = 49017.6 x 100

3501 x 5591.17L6 = 0.25 %


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7. % Heat loss due to unburnt in fly ash

% Ash in coal = 8.63

Ratio of bottom ash to fly ash = 90:10

GCV of fly ash = 452.5 kCal/kg

Amount of fly ash in 1 kg of coal = 0.1 x 0.0863

= 0.00863 kg

Heat loss in fly ash = 0.00863 x 452.5

= 3.905 kCal / kg of coal

% heat loss in fly ash = 3.905 x 100 / 3501

L7 = 0.11 %

8. % Heat loss due to unburnt in fly ash

GCV of bottom ash = 800 kCal/kg

Amount of bottom ash in 1 kg of

oal

= 0.9 x 0.0863

= 0.077 kg

Heat loss in bottom ash = 0.077 x 800

= 62.136 kCal/kg of coal

% Heat loss in bottom ash = 62.136 x 100 / 3501

L8 = 1.77 %


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Boiler efficiency by indirectmethod

= 100 (L1+ L2+ L3+ L4+ L5+ L6+ L7+ L8)

= 100-(7.89 + 3.44+ 5.91+ 0.29+ 2.58+ 0.25+0.11+1.77)

= 100-22.24

= 77.76 %

Summary of Heat Balance for Coal Fired Boiler

Input/Output Parameter kCal / kg of

coal

% loss

Heat Input = 3501 100Losses in boiler

1. Dry flue gas, L1 = 276.23 7.89

2. Loss due to hydrogen in fuel, L2 = 120.43 3.44

3. Loss due to moisture in fuel, L3 = 206.91 5.91

4. Loss due to moisture in air, L4 = 10.15 0.29

5. Partial combustion of C to CO, L5 = 90.32 2.586. Surface heat losses, L6 = 8.75 0.25

7. Loss due to Unburnt in fly ash, L7 = 3.85 0.11

8. Loss due to Unburnt in bottom ash,L8

= 61.97 1.77

Boiler Efficiency = 100 (L1 + L2+ L3+ L4+ L5+ L6+ L7+ L8) = 77.76 %


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boiler total air optimization

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