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e-CFR Data is current as of February 12, 2009


Title 40: Protection of Environment
PART 1065—ENGINE-TESTING PROCEDURES
Subpart G—Calculations and Data Requirements

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§ 1065.667   Dilution air background emission correction.

(a) To determine the mass of background emissions to subtract from a diluted exhaust sample, first determine the total flow of dilution air, n dil, over the test interval. This may be a measured quantity or a quantity calculated from the diluted exhaust flow and the flow-weighted mean fraction of dilution air in diluted exhaust, x dil/exh. Multiply the total flow of dilution air by the mean concentration of a background emission. This may be a time-weighted mean or a flow-weighted mean (e.g., a proportionally sampled background). The product of n diland the mean concentration of a background emission is the total amount of a background emission. If this is a molar quantity, convert it to a mass by multiplying it by its molar mass, M . The result is the mass of the background emission, m . In the case of PM, where the mean PM concentration is already in units of mass per mole of sample, M PM, multiply it by the total amount of dilution air, and the result is the total background mass of PM, m PM. Subtract total background masses from total mass to correct for background emissions.

(b) You may determine the total flow of dilution air by a direct flow measurement. In this case, calculate the total mass of background as described in §1065.650(c), using the dilution air flow, n dil. Subtract the background mass from the total mass. Use the result in brake-specific emission calculations.

(c) You may determine the total flow of dilution air from the total flow of diluted exhaust and a chemical balance of the fuel, intake air, and exhaust as described in §1065.655. In this case, calculate the total mass of background as described in §1065.650(c), using the total flow of diluted exhaust, n dexh, then multiply this result by the flow-weighted mean fraction of dilution air in diluted exhaust, x dil/exh. Calculate x dil/exhusing flow-weighted mean concentrations of emissions in the chemical balance, as described in §1065.655. You may assume that your engine operates stoichiometrically, even if it is a lean-burn engine, such as a compression-ignition engine. Note that for lean-burn engines this assumption could result in an error in emission calculations. This error could occur because the chemical balances in §1065.655 correct excess air passing through a lean-burn engine as if it was dilution air. If an emission concentration expected at the standard is about 100 times its dilution air background concentration, this error is negligible. However, if an emission concentration expected at the standard is similar to its background concentration, this error could be significant. If this error might affect your ability to show that your engines comply with applicable standards, we recommend that you remove background emissions from dilution air by HEPA filtration, chemical adsorption, or catalytic scrubbing. You might also consider using a partial-flow dilution technique such as a bag mini-diluter, which uses purified air as the dilution air.

(d) The following is an example of using the flow-weighted mean fraction of dilution air in diluted exhaust, x dil/exh, and the total mass of background emissions calculated using the total flow of diluted exhaust, n dexh, as described in §1065.650(c):





Example:

M NOx= 46.0055 g/mol

x bkgnd= 0.05 µmol/mol = 0.05·10−6mol/mol

n dexh= 23280.5 mol

x dil/exh= 0.843

mbkgndNOxdexh= 46.0055·0.05·10−6· 23280.5

m bkgndNOxdexh= 0.0536 g

m bkgndNOx= 0.843·0.0536

m bkgndNOx= 0.0452 g

(e) The following is an example of using the fraction of dilution air in diluted exhaust, x dil/exh, and the mass rate of background emissions calculated using the flow rate of diluted exhaust, n dexh, as described in §1065.650(c):





Example:

M NOX= 46.0055 g/mol

x bkgnd= 0.05 µmol/mol = 0.05 · 10−6mol/mol

n dexh= 23280.5 mol/s

x dil/exh= 0.843

m bkgndNOxdexh= 46.0055 · 0.05 · 10−6· 23280.5

m bkgndNOxdexh= 0.0536 g/hr

m bkgndNOx= 0.843 · 0.0536

m bkgndNOx= 0.0452 g/hr

[73 FR 59339, Oct. 8, 2008]

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