how to calculate your emissions
DESCRIPTION
How to Calculate Your Emissions. E = A × EF . CO. PM. 2000 lb /ton. VOC. NOx. 2014 Emissions Inventory Workshop. General Equation. For most emission sources the following equation is used: E = A * EF Where E = Calculated emissions A = Activity or annual process rate - PowerPoint PPT PresentationTRANSCRIPT
2014 Emissions Inventory Workshop
How to Calculate Your Emissions
CO
VOC
E = A × EF
PMNOx
2000 lb/ton
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General EquationFor most emission sources the following equation is used:
E = A * EFWhere
E = Calculated emissions
A = Activity or annual process rate
EF = Emission factor - determined by the method of calculation
Example sources:
• AP-42
• WebFire
• Permit
Process: 10-100 mmbtu/hr boiler
SCC: 10200602
Fuel: Natural Gas
Capacity: 84 mmbtu/hr
Control: Uncontrolled
Industrial Boiler – CO Calculation
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US EPA CHIEF Website: http://www.epa.gov/ttn/chief/index.html
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Industrial Boiler – CO Calculation
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E = A * EF = ??? tons
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A EFECO=¿673mmscf ∗84 lbsmmscf ¿56,532 lbs
CO=¿56,532 lbs∗ ton2,000 lbs ¿28.266 tons
Convert to tons:
Industrial Boiler – CO Calculation
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Process: 2 - Cycle Lean Burn Natural Gas Engine
SCC: 20200252
Fuel: Natural Gas
Capacity: 1100 hp
Control: Uncontrolled
Engine – Formaldehyde Calculation
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US EPA CHIEF Website: http://www.epa.gov/ttn/chief/index.html
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E = A * EF = ??? tons
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A EFEF ormaldehyde=¿47000mmBTU ∗0.0552 lbsmmBTU¿2,594 lbs
Formaldehyde=¿2,594 lbs∗ton2,000 lbs¿1.297 tons
Convert to tons:
Engine – Formaldehyde Calculation
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Process: Reciprocating emergency generator
SCC: 20200102
Fuel: Diesel
Capacity: 400 hp
Pollutant: NOx
Factor: 604 lbs/1000 gal
Control: Uncontrolled
Operation: 50 hours
Activity: 500 gal
Heat Content: 137 mmbtu/1000 gal
Emergency Generator – NOx Calculation
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E = A * EF = ??? tons
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A EFE
NOx=¿500 gal∗604 lbs1000gal¿302 lbs
NOx=¿302 lbs∗ton2,000 lbs ¿0.151 tons
Convert to tons:
Process: Surface coating – spray painting
SCC: 40200101
Pollutant: Xylene, VOC (non-HAP)
Paint: Acrylic Red
Surface Coating – Xylene/VOC Calculation
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E = A * VOC Content * (Xylene:VOC Ratio) = ??? tons
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TotalVOC=¿ ¿7,000 lbs=3.5 tons∗ 8.75 lbsgal800 gal
= ¿2.38 tons
Now, for non-HAP VOC:
Xylene=¿ ¿1.12 tons ∗ 0.080.25
VOC (non-HAP)
VOC (non-HAP) = Total VOC Xylene ̶
Process: Pneumatic Controllers: High Bleed
SCC: 31000242
Pollutant: VOC
Bleed Rate: 32.1 scf/hr whole gas
Natural Gas Production – Pneumatic Devices
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E = Gas Vented * VOC Content = ??? tons
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Gas vented=¿ ∗Devices∗HoursBleed rate
19.25 0.13% VOC
VOC =
Gas vented=¿ ∗8760hr5devices∗ 32.1𝑠𝑐𝑓1hr
1406 MscfVOC =
= 1406 Mscf
Gas MW wt. % VOC Conversion
Gas Vented
= 9,284 lb
VOC Emissions = 9,284 lb = 4.642 tons VOC
Control Equipment Calculations
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Capture Efficiency:
The percentage of air emission that is collected and routed to the control equipment. Most control devices have a 100% capture efficiency.
Control Efficiency:
The percentage of air pollutant that is removed from the air stream by the control device.
Control Efficiency Calculations
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Process: Primary Crushing
SCC: 30532001
Pollutant: PM10
Control: Water Spraying
Stone Quarry – Primary Crushing Calculation
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Stone Quarry – Primary Crushing Calculation
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E = A * EF * (1 – Control Efficiency) = ??? tons
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A EFEPM 10=¿1,361,120 tons∗0.0024 lbston ¿3,267 lbs
∗ (1−0.777 )
Account for control:
PM 10=¿729 lbs∗ ton2,000 lbs ¿0.36 tons
PM 10=¿3,267 lbs ¿729 lbs
Convert to tons:
Process: Grain handling for feed manufacture
SCC: 30200801
Activity: 10,000 tons
Pollutant: PM 10
Factor: 3 lbs/ton
Primary Control: Cyclone
Secondary Control: Baghouse
Two Controls – Grain Handling Calculation
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A Complex Situation: Capture Efficiency, Primary Efficiency & Secondary Efficiency
Capture Device: Hood
Therefore, the other 10% is emitted directly to the atmosphere
Secondary control device: Baghouse• The control equipment removes 98% of PM 10 from the emission stream.
Two Controls – Grain Handling Calculation
Primary Efficiency (PE)=80%=0.80
Secondary Efficiency (SE)=98%=0.98
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Primary control device: Cyclone• The control equipment removes 80% of PM 10 from the emission stream.
C apture Efficiency (CE )=90% =0.90
Total E = Captured E + Uncaptured E = ??? tons
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Captured E = A * EF * CE * (1 – PE) * (1 – SE)
Uncaptured E = A * EF * (1 – CE)
Captured E = A * EF * CE * (1 – PE) * (1 – SE)
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A EF
PM 10=¿10,000 tons∗3 lbston¿30,000 lbs
∗ (1−0.80 )PM 10=¿27,000 lbs ¿108 lbs∗ (1−0.98 )
∗ (0.90 )
Account for capture and controls:
Captured PM 10=¿30,000 lbs ¿27,000 lbsCE
PE SE
Uncaptured E = E * (1 – CE)
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Uncaptured PM 10¿30,000 lbs∗(1−0.90)
Account for all PM 10 emissions:
Total PM 10=¿3,108 lbs∗ton2,000 lbs¿1.554 tons
Total PM 10=¿C aptured PM 10+Uncaptured PM 10
Convert to tons:
¿3,000 lbs
Total PM 10=¿108 lbs+3,000 lbs=3,108 lbs
30,000 lbs of PM generated
1
3,000 lbs of PMemitted “not captured”
2
27,000 lbs of PM “captured” , sent to cyclone
3
4 Amount to Hopper 27,000 lbs* 0.80 = 21,600 lbs
5 27,000 lbs– 21,600 lbs = 5,400 lbs to the Baghouse
Amount emitted to the atmosphere5,400 lbs – 5,292 lbs = 108 lbs
108 lbs stack emissions + 3,000 lbs “not captured,” emitted as a fugitive = 3,108 lbs = 1.554 tons
7
6 Amount to Hopper5,400 lbs * 0.98 = 5,292 lbs
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Other Examples
If you would like to see an example calculation for another industry, process, or control equipment not presented today:
Please email [email protected] and we will create an example for you.
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ODEQ Emissions Inventory Section Program Manager:
Mark Gibbs [email protected]
Emissions Inventory Staff:
Cooper Garbe [email protected]
Cody Lathrop [email protected]
Michelle Horn [email protected]
Justin Milton [email protected]
Carrie Schroeder [email protected]
http://www.deq.state.ok.us/aqdnew/emissions/index.htm
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