40 CFR Ch. I (7–1–17 Edition) § 1065.665

§ 1065.665 40 CFR Ch. I (7–1–17 Edition)

Where: the nonmethane cutter, according to § 1065.365(f). xCH4 = concentration of CH4. xTHC[NMC–FID]cor = concentration of THC, initial PFCH4[NMC–FID] = nonmethane cutter CH4 pene- THC contamination (optional) and dry- tration fraction, according to § 1065.365(f). to-wet corrected, as measured by the RFCH4[THC–FID] = response factor of THC FID to NMC FID during sampling through the CH4, according to § 1065.360(d). NMC. Example: x = concentration of THC, initial THC[THC–FID]cor x = 10.4 μmol/mol THC contamination and dry-to-wet cor- THC[NMC–FID]cor μ rected, as measured by the THC FID dur- xTHC[THC–FID]cor = 150.3 mol/mol ing sampling while bypassing the NMC. RFPFC2H6[NMC–FID] = 0.019 RFPFC2H6[NMC–FID] = the combined ethane re- PFCH4[NMC–FID] = 0.990 sponse factor and penetration fraction of RFCH4[THC–FID] = 1.05

xCH4 = 7.78 μmol/mol first calculate its molar concentration in the exhaust sample stream from (2) For a GC–FID or FTIR, xCH4 is the which the sample was taken (raw or di- actual dry-to-wet corrected CH4 concentration as measured by the ana- luted exhaust), and convert this into a lyzer. C1-equivalent molar concentration. (e) C2H6 determination. For a GC–FID Add these C1-equivalent molar con- or FTIR, xC2H6 is the C1-equivalent, dry- centrations to the molar concentration to-wet corrected C2H6 concentration as of non-oxygenated total hydrocarbon measured by the analyzer. (NOTHC). The result is the molar con- [76 FR 57462, Sept. 15, 2011, as amended at 81 centration of total hydrocarbon equiv- FR 74184, Oct. 25, 2016] alent (THCE). Calculate THCE concentration using the following equa- § 1065.665 THCE and NMHCE deter- tions, noting that Eq. 1065.665–3 is re- mination. quired only if you need to convert your (a) If you measured an oxygenated oxygenated hydrocarbon (OHC) con- hydrocarbon’s mass concentration, centration from mass to moles:

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Where: RFOHCi[THC–FID] = the response factor of the FID to species i relative to propane on a xTHCE = the sum of the C1-equivalent concentrations of non-oxygenated hydro- C1-equivalent basis. C# = the mean number of carbon atoms in the carbon, alcohols, and aldehydes. particular compound. xNOTHC = the sum of the C1-equivalent con- Mdexh = the molar mass of diluted exhaust as centrations of NOTHC. determine in § 1065.340. xOHCi = the C1-equivalent concentration of mdexhOHCi = the mass of oxygenated species i in oxygenated species i in diluted exhaust, dilute exhaust. not corrected for initial contamination. MOHCi = the C1-equivalent molecular weight xOHCi-init = the C1-equivalent concentration of of oxygenated species i. the initial system contamination (op- mdexh = the mass of diluted exhaust tional) of oxygenated species i, dry-to- ndexhOHCi = the number of moles of oxygenated wet corrected. species i in total diluted exhaust flow.

xTHC[THC–FID]cor = the C1-equivalent response to ndexh = the total diluted exhaust flow. NOTHC and all OHC in diluted exhaust, (b) If we require you to determine HC contamination and dry-to-wet cor- nonmethane hydrocarbon equivalent rected, as measured by the THC–FID. (NMHCE), use the following equation:

Where: (CH3OH), acetaldehyde (C2H4O), and xNMHCE = the sum of the C1-equivalent con- formaldehyde (CH2O) as C1-equivalent centrations of nonoxygenated non- molar concentrations: methane hydrocarbon (NONMHC), alco- μ hols, and aldehydes. xTHC[THC-FID]cor = 145.6 mol/mol μ RF = the response factor of THC– xCH4 = 18.9 mol/mol CH4[THC–FID] μ FID to CH . xC2H5OH = 100.8 mol/mol 4 μ x = concentration of CH HC contamina- xCH3OH = 1.1 mol/mol CH4 4, μ tion (optional) and dry-to-wet corrected, xC2H4O = 19.1 mol/mol as measured by the gas chromatograph xCH2O = 1.3 μmol/mol FID. RFCH4[THC-FID] = 1.07 RFC2H5OH[THC-FID] = 0.76 (c) The following example shows how RFCH3OH[THC-FID] = 0.74 to determine NMHCE emissions based RFH2H4O[THC-FID] = 0.50 on ethanol (C2H5OH), methanol RFCH2O[THC-FID] = 0.0 225

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xNMHCE = xTHC[THC-FID]cor ¥ (xC2H5OH · this option, the molar flow of dilution RFC2H5OH[THC-FID] + xCH3OH · RFCH3OH[THC-FID] air is calculated by multiplying the di- + xC2H4O · RFC2H4O[THC-FID] + xCH2O · lute exhaust flow by the mole fraction RFCH2O[THC-FID]) + xC2H5OH + xCH3OH + xC2H4O of dilution gas to dilute exhaust, x , + x ¥ (RF · x ) dil/exh CH2O CH4[THC-FID] CH4 from the dilute chemical balance. This xNMHCE = 145.6 ¥ (100.8 · 0.76 + 1.1 · 0.74 + 19.1 · 0.50 + 1.3 · 0) + 100.8 + 1.1 + 19.1 + 1.3 ¥ may be done by totaling continuous (1.07 · 18.9) calculations or by using batch results. xNMHCE = 160.71 μmol/mol For example, to use batch results, the [79 FR 23805, Apr. 28, 2014, as amended at 81 total flow of dilution air is calculated FR 74187, Oct. 25, 2016] by multiplying the total flow of diluted exhaust, ndexh, by the flow-weighted § 1065.667 Dilution air background mean mole fraction of dilution air in emission correction. diluted exhaust, x¯ dil/exh. Calculate x¯ dil/exh (a) To determine the mass of back- using flow-weighted mean concentra- ground emissions to subtract from a di- tions of emissions in the chemical bal- luted exhaust sample, first determine ance, as described in § 1065.655. The the total flow of dilution air, ndil, over chemical balance in § 1065.655 assumes the test interval. This may be a meas- that your engine operates ured quantity or a calculated quantity. stoichiometrically, even if it is a lean- Multiply the total flow of dilution air burn engine, such as a compression-ig- by the mean mole fraction (i.e., con- nition engine. Note that for lean-burn centration) of a background emission. engines this assumption could result in This may be a time-weighted mean or a an error in emission calculations. This flow-weighted mean (e.g., a proportion- error could occur because the chemical ally sampled background). Finally, balance in § 1065.655 treats excess air multiply by the molar mass, M, of the passing through a lean-burn engine as associated gaseous emission con- if it was dilution air. If an emission stituent. The product of ndil and the concentration expected at the standard mean molar concentration of a back- is about 100 times its dilution air background emission and its molar mass, ground concentration, this error is neg- M, is the total background emission ligible. However, if an emission con- mass, m. In the case of PM, where the centration expected at the standard is mean PM concentration is already in similar to its background concentra- ¯ units of mass per mole of sample, MPM, tion, this error could be significant. If multiply it by the total amount of di- this error might affect your ability to lution air flow, and the result is the show that your engines comply with total background mass of PM, mPM. applicable standards, we recommend Subtract total background mass from that you either determine the total total mass to correct for background flow of dilution air using one of the emissions. more accurate methods in paragraph (b) You may determine the total flow (b) or (c) of this section, or remove of dilution air by a direct flow meas- background emissions from dilution air urement. by HEPA filtration, chemical adsorp- (c) You may determine the total flow tion, or catalytic scrubbing. You might of dilution air by subtracting the cal- also consider using a partial-flow dilu- culated raw exhaust molar flow as de- tion technique such as a bag mini-di- scribed in § 1065.655(g) from the meas- luter, which uses purified air as the di- ured dilute exhaust flow. This may be lution air. done by totaling continuous calcula- (e) The following is an example of tions or by using batch results. using the flow-weighted mean fraction (d) You may determine the total flow of dilution air in diluted exhaust, of dilution air from the measured di- x¯ dil/exh, and the total mass of back- lute exhaust flow and a chemical bal- ground emissions calculated using the ance of the fuel, intake air, and dilute total flow of diluted exhaust, ndexh, as exhaust as described in § 1065.655. For described in § 1065.650(c):

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