B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
Emissions from LCP – Project #3
1
B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
 Basic combustion parameters
Enthalpy of the return water hw
ℎ𝑤 = 𝑐𝑤 ∙ 𝑡𝑤 [
𝑘𝐽
]
𝑘𝑔
Where,
tw – feedwater temperature; [K]
cw – specific heat for water temperature tw = 240°C; [kJ/(kg*K)]
Thus,
ℎ𝑤 = 4.756 ∙ (240 + 273.15) = 2440.54 [
𝑘𝐽
]
𝑘𝑔
Boiler efficiency W
𝑘𝐽
𝑊 = 𝐷 ∙ (ℎ𝑝 − ℎ𝑤 ) [ ]
ℎ
Where,
D – maximum continuous rating; [t/h]
hp – enthalpy of live steam; [kJ/kg]
hw – enthalpy of the return water; [kJ/kg]
The value for hp is determined by reading the H-S plot for the given live steam temperature tp and live
steam pressure pp (13.8 MPa ≈ 138 bar) and hp = 3450 [kJ/kg]
Thus,
𝑘𝐽
𝑊 = 650 ∙ 103 ∙ (3450 − 2440.54) = 656 149 000 [ ]
ℎ
Thermal power of the source M
𝑀=
𝑊
[𝑀𝑊𝑡 ]
3600 ∙ 103
Thus,
𝑀=
656 149 000
= 182.2636 [𝑀𝑊𝑡 ]
3600 ∙ 103
Fuel consumption B
𝐵=
𝑊 𝑘𝑔
[ ]
𝜂∙𝑄 ℎ
Where,
W – Thermal power of the source; [kJ/h]
η – Gross efficiency of the boiler;
Q – Calorific value of the fuel; [kJ/kg]
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B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
Thus,
𝐵=
656 149 000
𝑘𝑔
= 37 281.2 [ ]
0.88 ∙ 20 000
ℎ
Theoretical volume of air to burn 1 kg of fuel Vpow,t
𝑉𝑝𝑜𝑤,𝑡 =
0.241 ∙ 𝑄
𝑁𝑚3
+ 0.5 [
]
1000
𝑘𝑔
Where,
Q – Calorific value of the fuel; [kJ/kg]
Thus,
0.241 ∙ 20000
𝑁𝑚3
𝑉𝑝𝑜𝑤,𝑡 =
+ 0.5 = 5.32 [
]
1000
𝑘𝑔
Unit volume of the wet flue gas for λ = 1 Vsp,w
𝑉𝑠𝑝,𝑤 =
0.212 ∙ 𝑄
𝑁𝑚3
+ 1.65 [
]
1000
𝑘𝑔
Thus,
𝑉𝑠𝑝,𝑤 =
0.212 ∙ 20000
𝑁𝑚3
+ 1.65 = 5.89 [
]
1000
𝑘𝑔
Wet flue fas volume for the real λ Vspw
𝑁𝑚3
𝑤
𝑉𝑠𝑝
= 𝑉𝑠𝑝,𝑤 + (𝜆 − 1) ∙ 𝑉𝑝𝑜𝑤,𝑡 + 1.602 ∙ 𝜆 ∙ 𝑉𝑝𝑜𝑤,𝑡 ∙ 𝑌𝑤 [
]
𝑘𝑔
Where,
λ – Excess air ratio
Yw – Moisture content in the air; [kg/kg of air]
Thus,
𝑁𝑚3
𝑤
𝑉𝑠𝑝
= 5.89 + (1.6 − 1) ∙ 5.32 + 1.602 ∙ 1.6 ∙ 5.32 ∙ 15 ∙ 10−3 = 9.29 [
]
𝑘𝑔
Volume of dry flue gas for the real λ Vsp
𝑤
𝑉𝑠𝑝 = 𝑉𝑠𝑝
∙ (1 − 𝑋𝑤 )
Where,
Xw – Moisture content in exhaust gases
Thus,
𝑁𝑚3
𝑉𝑠𝑝 = 9.29 ∙ (1 − 0.15) = 7.90 [
]
𝑘𝑔
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B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
Basic combustion parameters: summary table
Parameter
hw
W
M
B
Vpow,t
Vsp,w
Vspw
Vsp
Value
2 440.54 [kJ/kg]
656 149 000 [kJ/h]
182.2636 [MWt]
37 281.2 [kg/h]
5.32 [Nm3/kg]
5.89 [Nm3/kg]
9.29 [Nm3/kg]
7.90 [Nm3/kg]
 Pollutants’ (SO2, NOx, ash) concentration in flue gases
Concentration of SO2
𝐶𝑆𝑂2 =
2 ∙ 𝑠 𝑘𝑔
[
]
𝑉𝑠𝑝 𝑁𝑚3
Where,
s – Sulphur content in the fuel;
Vsp – Volume of dry flue gas for the real λ; [Nm3/kg]
Thus,
𝐶𝑆𝑂2 =
2 ∙ 0.008
𝑘𝑔
𝑚𝑔
= 0.0020253 [
] ∙ 106 = 2 025.3 [
]
3
7.90
𝑁𝑚
𝑁𝑚3
Concentration of NO2
𝐶𝑁𝑂2 =
𝐶𝑁𝑂2 ,𝑝𝑝𝑚 ∙ 𝑀𝑁𝑂2 𝑘𝑔
[
]
22.41
𝑁𝑚3
Where,
CNO2,ppm – NO2 concentration in flue gases; [ppm]
MNO2 – Molar mass of NO2 [mg/mmol]
22.41 – volume occupied by one mole of an ideal gas at standard temperature (0°C) and pressure (1
atm)
Thus,
𝐶𝑁𝑂2 =
270 ∙ 48
𝑚𝑔
𝑘𝑔
= 578.3 [
] ÷ 106 = 0.0005783 [
]
3
22.41
𝑁𝑚
𝑁𝑚3
Concentration of ash
𝐶𝑎𝑠ℎ =
𝑎 ∙ 𝑢 ∙ (1 − 𝜂𝑑𝑒𝑑𝑢𝑠𝑡 ) 𝑘𝑔
[
]
𝑉𝑠𝑝
𝑁𝑚3
Where,
4
B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
a – ash content;
u – ash uplift from the combustion bed;
ηdedust – primary dedusting efficiency;
Thus,
𝐶𝑎𝑠ℎ =
0.15 ∙ 0.40 ∙ (1 − 0.81)
𝑘𝑔
𝑚𝑔
= 0.0014430 [
] ∙ 106 = 1 443.0 [
]
3
7.90
𝑁𝑚
𝑁𝑚3
Pollutants’ concentration in exhaust gases: summary table
Parameter
Value [kg/Nm3]
0.0020253
Concentration of SO2 (CSO2)
0.0014430
Concentration of NO2
Value [mg/Nm3]
2 025.3
1 443.0
(CNO2)
Concentration of ash (Cash)
578.3
0.0005783
 Pollutants’ emissions
Emission of pollutant X (X = SO2, NO2, ash)
𝐸𝑋 = 𝑉𝑠𝑝 ∙ 𝐶𝑋 ∙ 𝐵 ∙ 𝑛 [
𝑘𝑔
]
ℎ
Where,
CX – pollutant X’s concentration in flue gases;
B – fuel consumption of one boiler unit
n – number of boilers
Emission of pollutant SO2
𝐸𝑆𝑂2 = 7.90 ∙ 0.0020253 ∙ 37 281.2 ∙ 4 = 2 385.98 [
𝑘𝑔
]
ℎ
Emission of pollutant NO2
𝐸𝑁𝑂2 = 7.90 ∙ 0.0014430 ∙ 37 281.2 ∙ 4 = 1 699.98 [
𝑘𝑔
]
ℎ
Emission of pollutant ash
𝐸𝑎𝑠ℎ = 7.90 ∙ 0.0005783 ∙ 37 281.2 ∙ 4 = 681.29 [
Pollutants’ emissions: summary table
Parameter
ESO2
ENO2
Eash
𝑘𝑔
]
ℎ
Value [kg/h]
2 385.98
1 699.98
681.29
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B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
 Required efficiency of air pollutants’ removal
Oxygen content in exhaust gases
𝑥1 =
0.21 ∙ (𝜆 − 1) ∙ 𝑉𝑝𝑜𝑤,𝑡
[– ]
𝑤
𝑉𝑠𝑝
∙ (1 − 𝑋𝑤 )
Thus,
𝑥1 =
0.21 ∙ (1.6 − 1) ∙ 5.32
= 0.0849[– ]
9.29 ∙ (1 − 0.15)
𝑥2 = 0.06
Concentration of SO2 calculated for 6% of oxygen
𝑥
𝐶𝑆𝑂22 = 𝐶𝑆𝑂2 ∙
0.21 − 𝑥2 𝑚𝑔
[
]
0.21 − 𝑥1 𝑁𝑚3
Thus,
𝑥
𝐶𝑆𝑂22 = 2025.3 ∙
0.21 − 0.06
𝑚𝑔
= 2 428 [
]
0.21 − 0.0849
𝑁𝑚3
Concentration of NO2 calculated for 6% of oxygen
0.21 − 𝑥2 𝑚𝑔
[
]
0.21 − 𝑥1 𝑁𝑚3
𝑥
𝐶𝑁𝑂2 2 = 𝐶𝑁𝑂2 ∙
Thus,
𝑥
𝐶𝑁𝑂2 2 = 1443.0 ∙
0.21 − 0.06
𝑚𝑔
= 1 730 [
]
0.21 − 0.0849
𝑁𝑚3
Concentration of ash calculated for 6% of oxygen
𝑥
2
𝐶𝑎𝑠ℎ
= 𝐶𝑎𝑠ℎ ∙
0.21 − 𝑥2 𝑚𝑔
[
]
0.21 − 𝑥1 𝑁𝑚3
Thus,
𝑥
2
𝐶𝑎𝑠ℎ
= 578.3 ∙
0.21 − 0.06
𝑚𝑔
= 693 [
]
0.21 − 0.0849
𝑁𝑚3
Pollutants’ concentrations calculated for 6% of oxygen: summary table
Parameter
Value
0.0849
x1
0.06
x2
2 428 [mg/Nm3]
CSO2
1 730 [mg/Nm3]
CNO2
693 [mg/Nm3]
Cash
The studied installation is treated as a “new new installation” with thermal power equal to 182.2
MWt.
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B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
Limit values established by the LCP directive
For “new” installations fuelled by coal and with a power in the range 100-300 MWt:
SO2
200 [mg/Nm3]
NOx
200 [mg/Nm3]
PM
30 [mg/Nm3]
Limit values established by the IED directive
For “new” installations fuelled by coal and with a power in the range 100-300 MWt:
SO2
250 [mg/Nm3]
NOx
200 [mg/Nm3]
PM
25 [mg/Nm3]
Minimum rates of reduction calculated from LCP standards
𝜂𝑖 =
𝐶𝑖6% − 𝐷𝑖
𝐶𝑖6%
∙ 100%
Where,
Di – limit value established by the LCP directive
Thus,
𝜂𝑆𝑂2 =
2 428 − 200
∙ 100% = 92%
2 428
𝜂𝑁𝑂2 =
1 730 − 200
∙ 100% = 89%
1 730
𝜂𝑎𝑠ℎ =
693 − 30
∙ 100% = 96%
693
Minimum rates of reduction calculated from IED standards
𝜂𝑆𝑂2 =
2 428 − 250
∙ 100% = 90%
2 428
𝜂𝑁𝑂2 =
1 730 − 200
∙ 100% = 89%
1 730
𝜂𝑎𝑠ℎ =
693 − 25
∙ 100% = 97%
693
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B.Sc. Environmental Engineering – Year 3 – Semester 5
Subject : Air Pollution Control
Minimum rates of reduction calculated from LCP and IED standards: summary
Minimum rate of reduction
Value from LCP standards
Value from IED standards
ηSO2
92%
90%
ηNO2
89%
89%
ηash
96%
97%
A quick analysis of the minimum rates of reduction issued from LCP and IED directives for
those three pollutants reveals that they are very similar. However, it is noticeable that for SO2 the
minimum rate of reduction decreases with the new IED directive in comparison with the former LCP
directive. In contrast, the minimum rate of reduction for ash increases with the new IED directive in
comparison with the former LCP directive. And as for NO2, the minimum rate of reduction remains
the same in the IED directive as in the LCP directive.
 Technologies necessary for meeting emission standards
In this case, only LCP standards are taken into account, therefore the minimum rates of
reduction for SO2, NO2 and ash are, respectively 92%, 89% and 96%.
Desulphurisation by 92% can only be achieved either by wet absorption (up to 95% removal
efficiency) or by adsorption (95-99% removal efficiency). However, the latter option being more
expensive, it is more profitable to choose the first option by adding a wet absorber to our facility.
NOx removal by 89% is achievable only by means of Selective Catalytic Reduction (SCR) which yields a
80-90% removal efficiency. Thus, this facility will have to be implemented to our installation.
Finally, dust removal by 96% can be achieved either by Electrostatic precipitator (ESP) yielding a
removal efficiency of 96-99% or by a Baghouse with fabric filters yielding a removal efficiency of up
to 99.9%. Similarly to desulphurisation, the first option will be chosen over the latter one, merely for
economic reasons as an Electrostatic precipitator will be cheaper to maintain than a Baghouse with
fabric filters that easily and quickly get clogged with particulate matter.
 Scheme of the studied installation with flue gas cleaning devices
8