Transcript Example Calculations 2012 Emissions Inventory Workshop 1

Example Calculations
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For most emission sources the following equation is used:
E = (Q*EF*(1-ER/100))
Where
E = Calculated emissions in tons per year (tpy)
Q = Activity rate (process rate)
EF = Emission factor-Determined by the Methods of Calculation (e.g.,
AP42, Mfg Data). For combustion emissions, the sulfur percent and/or the
ash content of the fuel will affect the emission factor
ER = Overall Control Efficiency (Overall Emission Reduction efficiency), %.
This is the combination of the capture efficiency and the
control/destruction/removal efficiency. To calculate, multiply the capture
efficiency by the control/destruction/removal efficiency and divide by 100
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Process:
Stone Quarrying - Primary Crushing
SCC:
30502001 or 30532001
Annual Rate:
717600 tons of Limestone
Permit Factor:
PM10 – 0.00059 lb/ton of rock
http://cfpub.epa.gov/oarweb/index.cfm?action=fire.main
http://www.epa.gov/ttn/chief/ap42/ch11/final/c11s1902.pdf
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Stone Quarrying - Primary Crushing
Process Rate
Emission Factor
717,600 tons 0.00059 lbs of PM
year
tons of rock
= 423.384 lbs
1 ton
year 2000 lbs
= 0.212 PM10 TPY
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Process:
4-cycle rich burn engine
SCC:
20200253
Factor:
12 grams NOx/hp-hr from 500 hp engine
Hours/Year: 8760
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4-cycle rich burn engine
Emission Factor
Conversion
12 grams NOx
1 lb
hp-hour 454 grams
Rated Horsepower
500 hp
= 13.216 lbs
hour
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4-cycle rich burn engine
Emissions Amount
Actual Hours
13.216 lbs 8760 hours
hour
year
Conversion
1 ton
2000 lbs
= 57.886 tpy of NOx
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Process:
4-cycle rich burn engine
SCC:
20200253
Factor:
PM2.5 – 9.500E-3 lb/MMBtu Fuel Input
Annual Rate:
45 mmscf
Fuel Heat Content: missing from inventory
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Factor:
PM2.5 – 9.500E-3 lb/mmBtu Fuel Input
Annual Rate:
45 mmscf
To convert from (lb/mmBtu) to (lb/mmscf), multiply by the heat
content of the fuel. If the heat content is not available, use
1020 Btu/scf.
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4-cycle rich burn engine
Heat Content-Conversion
1020 mmBtu
1 mmscf
Emission Factor
0.0095 lb PM 2.5
1 mmBtu
= 9.69 lb PM 2.5
1 mmscf
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4-cycle rich burn engine
Actual Emissions Amount
9.69 lb PM 2.5
mmscf
Process Rate
45 mmscf
year
= 436.05 lbs PM 2.5 1 ton
year
2000 lbs
= 0.218 tpy PM 2.5
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Process:
Industrial Boiler
SCC:
10200502
Fuel:
No. 2 Fuel Oil: 140,000 Btu per gallon
Annual Rate:
5000 gallons per year
AP42 Factor:
SO2 142 (S) lb per 1000 gallons burned
Sulfur:
0.4 % sulfur
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Industrial Boiler
Emission Factor
142 lb SO2
1000 gallons
=
Conversion-fuel contaminant
0.4 % Sulfur in fuel
56.8 lb SO2 per 1000 gallons
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Industrial Boiler
Process Rate
5000 gallons
1 year
Emission Factor
56.8 lb SO2
1000 gallons
= 284 lbs
year
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Industrial Boiler
Actual Emissions Amount Conversion
284 lbs SO2
year
1 ton
2000 lbs
= 0.142 tpy SO2
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Process:
Grain Handling
SCC:
30200752
Annual Rate:
5,000 tons
Factor:
0.87 lb PM/ton grain
Particle distributions: 15% PM-10 & 1% PM-2.5
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Grain Handling
Emission Factor
0.87 lb PM
1 ton of grain
Conversion-Particle distributions
15 PM10
100
0.1305 lb PM 10
1 ton of grain
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Grain Handling
Process Rate
Emission Factor
5,000 tons of grain 0.1305 lbs PM10
year 1 tons of grain
= 652.5 pounds per year of PM10
652.5 pounds of PM10 1 ton
year 2000 pounds
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Grain Handling
Emission Factor
0.87 lb PM
1 ton of grain
Conversion-Particle distributions
1 PM 2.5
100
Equals
0.0087 lb PM
1 ton of grain
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Grain Handling
Process Rate
Emission Factor
5,000 tons of grain 0.0087 lbs PM 2.5
year 1 tons of grain
= 43.5 pounds per year of PM 2.5
43.5 pounds of PM 2.5 1 ton
year
2000 pounds
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Grain Handling
Actual Emissions Amount
0.33 tons of PM-10
&
0.02 tons of PM-2.5
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Process:
SCC:
Surface Coating-Spray Painting
40200101
Annual Rate:
1600 gallons per year
Density:
7.5 lbs/gal as applied
VOC Content
6.2 lbs/gal
Transfer Efficiency
60.00%
Overall Control Efficiency
99.00%
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VOCs =
VOC content x Annual usage
6.2 lbs/gal x 1600 gal/yr = 9920 lbs/yr
9920 lbs/yr x 1 ton/2000 lbs = 4.96 tpy
Solid Content
Coating Density –VOC content
7.5 lbs/gal – 6.2 lbs/gal = 1.3 lbs of solids /gallon
Uncontrolled PM
Solid Content x annual usage x (1 – transfer efficiency)
1.3 lbs /gallon x 1600 gal/yr x (1 – 0.6) =
832.0 lbs/yr or 0.416 tpy
Controlled PM
Uncontrolled PM x (1 – overall control efficiency) =
0.416 tpy x (1 – 99/100) =
0.00416 tpy
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Surface Coating- VOC calculations
Example:
• VOC content: 6.2 lbs/gal
• Annual usage= 1600 gal/yr
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Surface Coating- Solid Content
Example:
• Coating density = 7.5 lbs/gal
• VOC content = 6.2 lbs/gal
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Surface Coating- Uncontrolled PM
Example:
• Solid content: 1.3 lbs/gal
• Annual usage= 1600 gal/yr
• Transfer efficiency = 60 %
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Surface Coating- Controlled PM
Example:
• Uncontrolled PM=0.416 tons/yr
• Control efficiency = 99%
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Surface Coating- MSDS Xylene
Example:
• VOC content: 8.75 lbs/gal
• Annual usage= 800 gal/yr
• Xylene content= 8%
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Capture Efficiency- the percentage of air emission that is
collected and routed to the control equipment.
Control Efficiency- the percentage of air pollutant that is
removed from the air stream. (control/destruction/removal
efficiency)
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Example:
• Capture efficiency is 80%
• Device has a control efficiency of 95%
Overall Control Efficiency=
% Captured * % Control Efficiency
(0.80*0.95)= 0.76 or 76 %
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Multiple emission control devices, affecting a common
air stream.
Common occurrences:
Dual Catalytic convertors
Combination of bag house(s) and cyclone(s)
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Primary control device A
• The capture efficiency is 90%.
• The control equipment removes 80% of the air
pollutant from the emission stream
Secondary control device B
• The capture efficiency is 100%.
• The control equipment removes 98% of the air
pollutant from the emission stream
Note: secondary controls most always have 100% capture efficiency.
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Primary control device A
• The capture efficiency is 90%.
• The control equipment removes 80% of the air pollutant from the emission
stream
Emissions reduction = 72%
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Secondary control device B
• The capture efficiency is 100%.
• The control equipment removes 98% of the air pollutant from the emission
stream
Emission reduction = 98%
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Step 1: Total un-controlled emissions = 100 TPY
Primary emission reduction= 72%
Step 2: Remaining emissions = 28 TPY
Secondary emission reduction= 98%
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1
100.0 lbs/hr of
PM generated
4
2
10.0 lbs/hr of PM
emitted “not captured”
5
3
Amount to Hopper
90.0 lbs/hr * 0.80 = 72.0 lbs/hr
90.0 lbs/hr – 72 lbs/hr =
18.0 lbs/hr to the Baghouse
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Amount to Hopper
18.0 lbs/hr * 0.98 = 17.64 lbs/hr
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Amount emitted to the atmosphere
18.0 lbs/hr – 17.64 lbs/hr = 0.36 lbs/hr
0.36 lbs/hr (stack emissions + 10.0
lbs/hr (“not captured, emitted as a
fugitive) =
90.0 lbs/hr of
PM “captured” ,
sent to cyclone
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100 pennies
Only 90 pennies go to control device A
Control device A removes 80% of the 90 pennies leaving 18 pennies
Control device B sees only 18 pennies, and removes 98% leaving 0.36 pennies.
10 lbs/hr was directly emitted as fugitives (not captured by Control device A).
10.0 + 0.36 = 10.36 lbs/hr
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Program Manager:
Mark Gibbs
[email protected]
Emission Inventory Staff:
Michelle Horn
[email protected]
Jenafer Icona
[email protected]
Justin Milton
[email protected]
Carrie Schroeder
[email protected]
Matt Weis
[email protected]
http://www.deq.state.ok.us/aqdnew/emissions/index.htm
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Given:
• 1200-hp Natural gas compressor engine
• Actual annual hours = 6500
• Emission factor for CO = 0.557 lbs/mmBtu (AP-42 table 3.2-2)
• Process rate= 10,000 mmBtu/yr
Find the total CO emission amount in tons.
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Annual CO emissions
• Emission factor for CO = 0.557 lbs/mmBtu (AP-42 table 3.2-2)
• Process rate= 10,000 mmBtu/yr
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Given:
• 1200-hp Natural gas compressor engine
• PM10 Emission Factor = 0.009987 lbs/mmBtu (AP-42 table 3.2-2)
• Annual Process Rate= 20 mmscf of natural gas
• Fuel heat content= 1020 mmBtu/mmscf
Find the total PM10 emission amount in tons.
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Step 2- Find annual emission amount for PM10
• Emission Factor = 0.009987 lbs/MMBtu (AP-42 table 3.2-2)
• Process Rate = 20,400 mmbtu
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