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Example Calculations 2012 Emissions Inventory Workshop 1

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2 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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2012 Emissions Inventory Workshop 3

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4 Process: Stone Quarrying - Primary Crushing SCC: or Annual Rate: tons of Limestone Permit Factor:PM10 – lb/ton of rock

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717,600 tons lbs of PM year tons of rock 1 ton= lbs 1 ton 2000 lbs year 2000 lbs Stone Quarrying - Primary Crushing Process RateEmission Factor 2012 Emissions Inventory Workshop 5 = PM10 TPY

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2012 Emissions Inventory Workshop 6 Process:4-cycle rich burn engine SCC: Factor:12 grams NOx/hp-hr from 500 hp engine Hours/Year:8760

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12 grams NOx 1 lb 500 hp hp-hour 454 grams = lbs hour 4-cycle rich burn engine Emission FactorConversionRated Horsepower 2012 Emissions Inventory Workshop 7

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lbs 8760 hours 1 ton hour year 2000 lbs = tpy of NOx 4-cycle rich burn engine Emissions AmountActual HoursConversion 2012 Emissions Inventory Workshop 8

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9 Process:4-cycle rich burn engine SCC: Factor:PM2.5 – 9.500E-3 lb/MMBtu Fuel Input Annual Rate:45 mmscf Fuel Heat Content: missing from inventory

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2012 Emissions Inventory Workshop 10 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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1020 mmBtu lb PM mmscf 1 mmBtu = 9.69 lb PM mmscf 4-cycle rich burn engine Heat Content-ConversionEmission Factor 2012 Emissions Inventory Workshop 11

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9.69 lb PM mmscf mmscf year = lbs PM ton year 2000 lbs = tpy PM cycle rich burn engine Actual Emissions AmountProcess Rate 2012 Emissions Inventory Workshop 12

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2012 Emissions Inventory Workshop 13 Process:Industrial Boiler SCC: Fuel:No. 2 Fuel Oil: 140,000 Btu per gallon Annual Rate:5000 gallons per year AP42 Factor:SO (S) lb per 1000 gallons burned Sulfur:0.4 % sulfur

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142 lb SO % Sulfur in fuel 1000 gallons = 56.8 lb SO 2 per 1000 gallons Industrial Boiler Emission FactorConversion-fuel contaminant 2012 Emissions Inventory Workshop 14

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5000 gallons 56.8 lb SO 2 1 year 1000 gallons = 284 lbs year Industrial Boiler Process RateEmission Factor 2012 Emissions Inventory Workshop 15

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284 lbs SO 2 1 ton year 2000 lbs = tpy SO 2 Industrial Boiler Actual Emissions AmountConversion 2012 Emissions Inventory Workshop 16

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2012 Emissions Inventory Workshop 17 Process:Grain Handling SCC: 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 FactorConversion-Particle distributions 2012 Emissions Inventory Workshop 18

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5,000 tons of grain lbs PM10 year 1 tons of grain = pounds per year of PM pounds of PM10 1 ton year 2000 pounds Grain Handling Process RateEmission Factor 2012 Emissions Inventory Workshop 19

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0.87 lb PM 1 PM ton of grain 100 Equals lb PM 1 ton of grain Grain Handling Emission FactorConversion-Particle distributions 2012 Emissions Inventory Workshop 20

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5,000 tons of grain lbs PM 2.5 year 1 tons of grain = 43.5 pounds per year of PM pounds of PM ton year 2000 pounds Grain Handling Process RateEmission Factor 2012 Emissions Inventory Workshop 21

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0.33 tons of PM-10 & 0.02 tons of PM-2.5 Grain Handling Actual Emissions Amount 2012 Emissions Inventory Workshop 22

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2012 Emissions Inventory Workshop 23 Process:Surface Coating-Spray Painting SCC: Annual Rate:1600 gallons per year Density:7.5 lbs/gal as applied VOC Content6.2 lbs/gal Transfer Efficiency60.00% Overall Control Efficiency99.00%

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2012 Emissions Inventory Workshop 24 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 ContentCoating Density –VOC content 7.5 lbs/gal – 6.2 lbs/gal = 1.3 lbs of solids /gallon Uncontrolled PMSolid Content x annual usage x (1 – transfer efficiency) 1.3 lbs /gallon x 1600 gal/yr x (1 – 0.6) = lbs/yr or tpy Controlled PMUncontrolled PM x (1 – overall control efficiency) = tpy x (1 – 99/100) = tpy

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2012 Emissions Inventory Workshop 25 Example: VOC content: 6.2 lbs/gal Annual usage= 1600 gal/yr

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2012 Emissions Inventory Workshop 26 Example: Coating density = 7.5 lbs/gal VOC content = 6.2 lbs/gal

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2012 Emissions Inventory Workshop 27 Example: Solid content: 1.3 lbs/gal Annual usage= 1600 gal/yr Transfer efficiency = 60 %

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2012 Emissions Inventory Workshop 28 Example: Uncontrolled PM=0.416 tons/yr Control efficiency = 99%

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January

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2012 Emissions Inventory Workshop 30 Example: VOC content: 8.75 lbs/gal Annual usage= 800 gal/yr Xylene content= 8%

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2012 Emissions Inventory Workshop 31

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Emissions Inventory Workshop 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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2012 Emissions Inventory Workshop 33 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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Emissions Inventory Workshop 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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Emissions Inventory Workshop 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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Emissions Inventory Workshop 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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Emissions Inventory Workshop 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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Emissions Inventory Workshop 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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100.0 lbs/hr of PM generated lbs/hr of PM emitted not captured lbs/hr of PM captured, sent to cyclone 3 4 Amount to Hopper 90.0 lbs/hr * 0.80 = 72.0 lbs/hr lbs/hr – 72 lbs/hr = 18.0 lbs/hr to the Baghouse 7 6 Amount to Hopper 18.0 lbs/hr * 0.98 = lbs/hr January

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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) = lbs/hr 2012 Emissions Inventory Workshop

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41 Program Manager: Mark Gibbs Emission Inventory Staff: Michelle Horn Jenafer Icona Justin Milton Carrie Schroeder Matt Weis

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2012 Emissions Inventory Workshop 42

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January Given: 1200-hp Natural gas compressor engine Actual annual hours = 6500 Emission factor for CO = 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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January Annual CO emissions Emission factor for CO = lbs/mmBtu (AP-42 table 3.2-2) Process rate= 10,000 mmBtu/yr

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January Given: 1200-hp Natural gas compressor engine PM10 Emission Factor = 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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January

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January Step 2- Find annual emission amount for PM10 Emission Factor = lbs/MMBtu (AP-42 table 3.2-2) Process Rate = 20,400 mmbtu

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