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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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