2013 Emissions Inventory Workshop EXAMPLE CALCULATIONS.

2013 Emissions Inventory WorkshopEXAMPLE CALCULATIONS

General Equation

For 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

Examples sources:

• AP-42

• WebFire

• Permit

2013 Emissions Inventory Workshop 2

Process: 10-100 mmbtu/hr boiler

SCC: 10200602

Fuel: Natural Gas

Capacity: 84 mmbtu/hr

Control: Uncontrolled

Industrial Boiler – CO & Benzene Calculation


2013 Emissions Inventory Workshop

US EPA CHIEF Website: http://www.epa.gov/ttn/chief/index.html

4

Industrial Boiler – CO Calculation


E = A * EF = ??? tons

A EFE

CO=¿673mmscf ∗84 lbsmmscf

¿56,532 lbs

CO=¿56,532 lbs∗ ton2,000 lbs

¿28.266 tons

Convert to tons:



US EPA CHIEF Website: http://www.epa.gov/ttn/chief/index.html

12

E = A * EF = ??? tons

A EFE

Benzene=¿673mmscf ∗0.0021lbsmmscf ¿1.41 lbs

Benzene=¿1.41 lbs∗ton2,000 lbs ¿0.001 tons

Convert to tons:


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


E = A * EF = ??? tons

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


E = A * VOC Content * (Xylene:VOC Ratio) = ??? tons

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 ̶�

Control Equipment Calculations


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


Process: Primary Crushing

SCC: 30532001

Pollutant: PM10

Control: Water Spraying

Stone Quarry – Primary Crushing Calculation


E = A * EF * (1 – Control Efficiency) = ??? tons

A EFE

PM 10=¿136,112 tons∗0.0024 lbston ¿326.7 lbs

∗ (1−0.777 )

Account for control:

PM 10=¿72.9 lbs∗ ton2,000 lbs ¿0.036 tons

PM 10=¿326.7 lbs ¿72.9 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



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


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


Captured E = A * EF * CE * (1 – PE) * (1 – SE)

Uncaptured E = A * EF * (1 – CE)

Captured E = A * EF * CE * (1 – PE) * (1 – SE)

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 )2013 Emissions Inventory Workshop 34

∗ (0.90 )

Account for capture and controls:

Captured PM 10=¿30,000 lbs ¿27,000 lbsCE

PE SE

Uncaptured E = E * (1 – CE)

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

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6 Amount to Hopper5,400 lbs * 0.98 = 5,292 lbs


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.


ODEQ Emissions Inventory Section

Program Manager:

Mark Gibbs [email protected]

Emissions Inventory Staff:

Michelle Horn [email protected]

Justin Milton [email protected]

Carrie Schroeder [email protected]

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


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

2013 Emissions Inventory Workshop EXAMPLE CALCULATIONS.

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