example calculations 2012 emissions inventory workshop 1
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Example Calculations
2012 Emissions Inventory Workshop
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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
717,600 tons 0.00059 lbs of PM
year tons of rock
= 423.384 lbs 1 ton 1 ton year 2000 lbs 2000 lbs
Stone Quarrying - Primary Crushing
Process Rate Emission Factor
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= 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
12 grams NOx 1 lb 500 hp
hp-hour 454 grams
= 13.216 lbs hour
4-cycle rich burn engine
Emission Factor Conversion Rated Horsepower
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13.216 lbs 8760 hours 1 ton
hour year 2000 lbs
= 57.886 tpy of NOx
4-cycle rich burn engine
Emissions Amount Actual Hours Conversion
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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.
1020 mmBtu 0.0095 lb PM 2.5
1 mmscf 1 mmBtu
= 9.69 lb PM 2.5 1 mmscf
4-cycle rich burn engine
Heat Content-Conversion
Emission Factor
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9.69 lb PM 2.5 45 mmscf
mmscf year
= 436.05 lbs PM 2.5 1 ton year 2000
lbs
= 0.218 tpy PM 2.5
4-cycle rich burn engine
Actual Emissions Amount
Process Rate
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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
142 lb SO2 0.4 % Sulfur in fuel
1000 gallons
= 56.8 lb SO2 per 1000 gallons
Industrial Boiler
Emission Factor Conversion-fuel contaminant
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5000 gallons 56.8 lb SO2
1 year 1000 gallons
= 284 lbs
year
Industrial Boiler
Process Rate Emission Factor
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284 lbs SO2 1 ton year 2000 lbs
= 0.142 tpy SO2
Industrial Boiler
Actual Emissions Amount
Conversion
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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
Grain Handling
Emission Factor Conversion-Particle distributions
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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
Grain Handling
Process Rate Emission Factor
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0.87 lb PM 1 PM 2.5
1 ton of grain 100
Equals
0.0087 lb PM 1 ton of grain
Grain Handling
Emission Factor Conversion-Particle distributions
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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
Grain Handling
Process Rate Emission Factor
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0.33 tons of PM-10 &
0.02 tons of PM-2.5
Grain Handling
Actual Emissions Amount
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Process: Surface Coating-Spray Painting
SCC: 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
Surface Coating- VOC calculationsSurface Coating- VOC calculations
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Example:• VOC content: 6.2 lbs/gal• Annual usage= 1600 gal/yr
Surface Coating- Solid ContentSurface Coating- Solid Content
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Example:• Coating density = 7.5 lbs/gal• VOC content = 6.2 lbs/gal
Surface Coating- Uncontrolled PMSurface Coating- Uncontrolled PM
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Example:• Solid content: 1.3 lbs/gal• Annual usage= 1600 gal/yr• Transfer efficiency = 60 %
Surface Coating- Controlled PMSurface Coating- Controlled PM
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Example:• Uncontrolled PM=0.416 tons/yr• Control efficiency = 99%
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Surface Coating- MSDS XyleneSurface Coating- MSDS Xylene
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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 convertorsCombination 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%
100.0 lbs/hr of PM generated
1
10.0 lbs/hr of PMemitted “not captured”
2
90.0 lbs/hr of PM “captured” , sent to cyclone
3
4 Amount to Hopper 90.0 lbs/hr * 0.80 = 72.0 lbs/hr
5 90.0 lbs/hr – 72 lbs/hr = 18.0 lbs/hr to the Baghouse 7
6 Amount to Hopper18.0 lbs/hr * 0.98 = 17.64 lbs/hr
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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