# 40 CFR § 1066.610 - Dilution air background correction.

§ 1066.610 Dilution air background correction.

(a) Correct the emissions in a gaseous sample for background using the following equation:

$\begin{array}{c}{x}_{\left[\mathrm{emission}\right]}={x}_{\text{[emission]dexh}}-{x}_{\text{[emission]bkgnd}}·\left(1-\left(\frac{1}{\mathrm{DF}}\right)\right)\\ \text{Eq. 1066.610-1}\end{array}$

Where:
x[emission]dexh = measured emission concentration in dilute exhaust (after dry-to-wet correction, if applicable).
x[emission]bkgnd = measured emission concentration in the dilution air (after dry-to-wet correction, if applicable).
DF = dilution factor, as determined in paragraph (b) of this section.

$\begin{array}{c}\text{Example:}\\ {x}_{\mathrm{NOxdexh}}=1.08305\phantom{\rule{0ex}{0ex}}\mathrm{ppm}\\ {x}_{\mathrm{NOxbkgnd}}=0.12456\phantom{\rule{0ex}{0ex}}\mathrm{ppm}\\ DF=9.14506\\ {x}_{\mathrm{NOx}}=1.08305-0.12456·\left(1-\left(\frac{1}{9.14506}\right)\right)=0.97211\phantom{\rule{0ex}{0ex}}\mathrm{ppm}\\ \phantom{\rule{0ex}{0ex}}\\ \phantom{\rule{0ex}{0ex}}\text{(b) Except as specified in paragraph (c) of this section, determine the dilution factor,}DF,\\ \text{over the test interval using the following equation:}\\ \phantom{\rule{0ex}{0ex}}\\ DF=\frac{1}{\left(1+\frac{\alpha }{2}+3.76·\left(1+\frac{\alpha }{4}-\frac{\beta }{2}\right)\right)·\left({x}_{\mathrm{CO}2}+{x}_{\mathrm{NMHC}}+{x}_{\mathrm{CH}4}+{x}_{\mathrm{CO}}\right)}\\ \phantom{\rule{0ex}{0ex}}\\ \text{Eq. 1066.610-2}\end{array}$

Where:
xCO2 = amount of CO2 measured in the sample over the test interval.
xNMHC = amount of C1-equivalent NMHC measured in the sample over the test interval.
xCH4 = amount of CH4 measured in the sample over the test interval.
xCO = amount of CO measured in the sample over the test interval.
a = atomic hydrogen-to-carbon ratio of the test fuel. You may measure a or use default values from Table 1 of 40 CFR 1065.655.
b = atomic oxygen-to-carbon ratio of the test fuel. You may measure b or use default values from Table 1 of 40 CFR 1065.655.

$\begin{array}{c}\text{Example:}\\ {x}_{\mathrm{CO}2}=1.456%=0.01456\\ {x}_{\mathrm{NMHC}}=0.84\phantom{\rule{0ex}{0ex}}\mathrm{ppm}=0.00000084\\ {x}_{\mathrm{CH}4}=0.26\phantom{\rule{0ex}{0ex}}\mathrm{ppm}=0.00000026\\ {x}_{\mathrm{CO}}=80.4\phantom{\rule{0ex}{0ex}}\mathrm{ppm}=0.0000804\\ \alpha =1.92\\ \beta =0.03\\ DF=\frac{1}{\left(1+\frac{1.92}{2}+3.76·\left(1+\frac{1.92}{4}-\frac{0.03}{2}\right)\right)·\left(0.01456+0.00000084+0.00000026+0.0000804\right)}=9.14506\end{array}$

(c) Determine the dilution factor, DF, over the test interval for partial-flow dilution sample systems using the following equation:

$\begin{array}{c}DF=\frac{{V}_{\mathrm{dexhstd}}}{{V}_{\mathrm{exhstd}}}\\ \text{Eq. 1066.610-3}\end{array}$

Where:

Vdexhstd = total dilute exhaust volume sampled over the test interval, corrected to standard reference conditions.

Vexhstd = total exhaust volume sampled from the vehicle, corrected to standard reference conditions.

$\begin{array}{c}\text{Example:}\\ {V}_{\mathrm{dexhstd}}=170.9{\phantom{\rule{0ex}{0ex}}m}^{3}\\ {V}_{\mathrm{exhstd}}=15.9{\phantom{\rule{0ex}{0ex}}m}^{3}\\ \mathrm{DF}=\frac{170.9}{15.4}=11.1\end{array}$

(d) Determine the time-weighted dilution factor, DFw, over the duty cycle using the following equation:

${DF}_{w}=\frac{\sum _{i=1}^{N}{t}_{i}}{\sum _{i=1}^{N}\frac{1}{{DF}_{i}}\phantom{\rule{0ex}{0ex}}·\phantom{\rule{0ex}{0ex}}{t}_{i}}\phantom{\rule{0ex}{0ex}}\text{Eq. 1066.620-4}$

Where:
N = number of test intervals.
i = test interval number
t = duration of the test interval.
DF = dilution factor over the test interval.
Example:

$\begin{array}{c}N=3\\ {\mathrm{DF}}_{1}=14.40\\ {t}_{1}=505\phantom{\rule{0ex}{0ex}}s\\ {\mathrm{DF}}_{2}=24.48\\ {t}_{2}=867\phantom{\rule{0ex}{0ex}}s\\ {\mathrm{DF}}_{3}=17.28\\ {t}_{3}=505\phantom{\rule{0ex}{0ex}}s\\ {\mathrm{DF}}_{w}=\frac{505+867+505}{\left(\frac{1}{14.40}·505\right)+\left(\frac{1}{24.48}·867\right)+\left(\frac{1}{17.28}·505\right)}=18.82\end{array}$