# Example Calculations 2012 Emissions Inventory Workshop 1.

## Presentation on theme: "Example Calculations 2012 Emissions Inventory Workshop 1."— Presentation transcript:

Example Calculations 2012 Emissions Inventory Workshop 1

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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4 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 1 ton= 423.384 lbs 1 ton 2000 lbs year 2000 lbs Stone Quarrying - Primary Crushing Process RateEmission Factor 2012 Emissions Inventory Workshop 5 = 0.212 PM10 TPY

2012 Emissions Inventory Workshop 6 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 FactorConversionRated Horsepower 2012 Emissions Inventory Workshop 7

13.216 lbs 8760 hours 1 ton hour year 2000 lbs = 57.886 tpy of NOx 4-cycle rich burn engine Emissions AmountActual HoursConversion 2012 Emissions Inventory Workshop 8

9 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

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.

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-ConversionEmission Factor 2012 Emissions Inventory Workshop 11

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 AmountProcess Rate 2012 Emissions Inventory Workshop 12

2012 Emissions Inventory Workshop 13 Process:Industrial Boiler SCC:10200502 Fuel:No. 2 Fuel Oil: 140,000 Btu per gallon Annual Rate:5000 gallons per year AP42 Factor:SO 2 142 (S) lb per 1000 gallons burned Sulfur:0.4 % sulfur

142 lb SO 2 0.4 % Sulfur in fuel 1000 gallons = 56.8 lb SO 2 per 1000 gallons Industrial Boiler Emission FactorConversion-fuel contaminant 2012 Emissions Inventory Workshop 14

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

284 lbs SO 2 1 ton year 2000 lbs = 0.142 tpy SO 2 Industrial Boiler Actual Emissions AmountConversion 2012 Emissions Inventory Workshop 16

2012 Emissions Inventory Workshop 17 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 FactorConversion-Particle distributions 2012 Emissions Inventory Workshop 18

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 RateEmission Factor 2012 Emissions Inventory Workshop 19

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 FactorConversion-Particle distributions 2012 Emissions Inventory Workshop 20

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 RateEmission Factor 2012 Emissions Inventory Workshop 21

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

2012 Emissions Inventory Workshop 23 Process:Surface Coating-Spray Painting SCC:40200101 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%

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) = 832.0 lbs/yr or 0.416 tpy Controlled PMUncontrolled PM x (1 – overall control efficiency) = 0.416 tpy x (1 – 99/100) = 0.00416 tpy

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

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

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

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

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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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32 2012 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)

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 %

34 2012 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)

35 2012 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.

36 2012 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%

37 2012 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%

38 2012 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%

100.0 lbs/hr of PM generated 1 10.0 lbs/hr of PM emitted 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 Hopper 18.0 lbs/hr * 0.98 = 17.64 lbs/hr January 2012 39

40 100 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 2012 Emissions Inventory Workshop

41 Program Manager: Mark Gibbs Mark.Gibbs@deq.ok.gov Emission Inventory Staff: Michelle Horn Michelle.Horn@deq.ok.gov Jenafer Icona Jenafer.Icona@deq.ok.gov Justin Milton Justin.Milton@deq.ok.gov Carrie Schroeder Carrie.Schroeder@deq.ok.gov Matt Weis Matt.Weis@deq.ok.gov http://www.deq.state.ok.us/aqdnew/emissions/index.htm

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January 2012 43 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.

January 2012 44 Annual CO emissions Emission factor for CO = 0.557 lbs/mmBtu (AP-42 table 3.2-2) Process rate= 10,000 mmBtu/yr

January 2012 45 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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January 2012 47 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