Example Calculations 2012 Emissions Inventory Workshop 1.

Example Calculations

2012 Emissions Inventory Workshop

1

2012 Emissions Inventory Workshop 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


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

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


= 0.212 PM10 TPY


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


13.216 lbs 8760 hours 1 ton

hour year 2000 lbs

= 57.886 tpy of NOx


Emissions Amount Actual Hours Conversion



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


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


Heat Content-Conversion

Emission Factor


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


Actual Emissions Amount

Process Rate



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


5000 gallons 56.8 lb SO2

1 year 1000 gallons

= 284 lbs

year

Industrial Boiler



284 lbs SO2 1 ton year 2000 lbs

= 0.142 tpy SO2

Industrial Boiler


Conversion



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


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



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


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



0.33 tons of PM-10 &

0.02 tons of PM-2.5

Grain Handling




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%


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


Example:• VOC content: 6.2 lbs/gal• Annual usage= 1600 gal/yr

Surface Coating- Solid ContentSurface Coating- Solid Content


Example:• Coating density = 7.5 lbs/gal• VOC content = 6.2 lbs/gal

Surface Coating- Uncontrolled PMSurface Coating- Uncontrolled PM


Example:• Solid content: 1.3 lbs/gal• Annual usage= 1600 gal/yr• Transfer efficiency = 60 %

Surface Coating- Controlled PMSurface Coating- Controlled PM


Example:• Uncontrolled PM=0.416 tons/yr• Control efficiency = 99%

January 2012 29

Surface Coating- MSDS XyleneSurface Coating- MSDS Xylene


Example:• VOC content: 8.75 lbs/gal• Annual usage= 800 gal/yr• Xylene content= 8%


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)


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 %


Multiple emission control devices, affecting a common air stream.

Common occurrences:Dual Catalytic convertorsCombination of bag house(s) and cyclone(s)


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.


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%


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%


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

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



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

http://www.deq.state.ok.us/aqdnew/emissions/index.htm

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.

January 2012 46

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

Example Calculations 2012 Emissions Inventory Workshop 1.

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