§ 1065.642 PDP, SSV, and CFV molar flow rate calculations.

This section describes the equations for calculating molar flow rates from various flow meters. After you calibrate a flow meter according to § 1065.640, use the calculations described in this section to calculate flow during an emission test.

(a) PDP molar flow rate.

(1) Based on the speed at which you operate the PDP for a test interval, select the corresponding slope, a1, and intercept, a0, as calculated in § 1065.640, to calculate PDP molar flow rate,, as follows:

$\begin{array}{c}\stackrel{\xb7}{n}-{f}_{\mathrm{nPDP}}\phantom{\rule{0ex}{0ex}}\xb7\phantom{\rule{0ex}{0ex}}\frac{{V}_{\mathrm{rev}}\xb7{p}_{\mathrm{in}}}{R\xb7{T}_{\mathrm{in}}}\\ \text{Eq. 1065.642-1}\end{array}$

(2) Calculate Vrev using the following equation:

$\begin{array}{c}{V}_{\mathrm{rev}}=\frac{{a}_{1}}{{f}_{\mathrm{nPDP}}}\xb7\sqrt{\frac{{p}_{\mathrm{out}}-{p}_{\mathrm{in}}}{{p}_{\mathrm{out}}}}+{a}_{0}\\ \text{Eq. 1065.642-2}\end{array}$

pout = static absolute pressure at the PDP outlet.

Example:

a1 = 0.8405 (m
3/s)

fnPDP = 12.58 r/s

Pout = 99.950 kPa

Pin = 98.575 kPa = 98575 Pa = 98575 kg/(m·s
2)

a0 = 0.056 (m
3/r)

R = 8.314472 J/(mol·K) = 8.314472 (m
2·kg)/(s
2·mol·K)

Tin = 323.5 K

${V}_{\mathrm{rev}}=\frac{0.8405}{12.58}\phantom{\rule{0ex}{0ex}}\xb7\sqrt{\frac{99.950-98.575}{99.950}}+0.056$

$\stackrel{\xb7}{n}=12.58\phantom{\rule{0ex}{0ex}}\xb7\phantom{\rule{0ex}{0ex}}\frac{98575\xb70.06383}{8.314472\xb7323.5}$

(b) SSV molar flow rate. Calculate SSV molar flow rate, n

, as follows:

Example:

At = 0.01824 m
2

pin = 99.132 kPa = 99132 Pa = 99132 kg/(m·s
2)

Z = 1

Mmix = 28.7805 g/mol = 0.0287805 kg/mol

R = 8.314472 J/(mol·K) = 8.314472 (m
2·kg)/(s
2·mol·K)

Tin = 298.15 K

Re# = 7.232·10
5

γ = 1.399

β = 0.8

Δp = 2.312 kPa

Using Eq. 1065.640-7:

rssv = 0.997

Using Eq. 1065.640-6:

Cf = 0.274

Using Eq. 1065.640-5:

Cd = 0.990

(c) CFV molar flow rate. If you use multiple venturis and you calibrate each venturi independently to determine a separate discharge coefficient, Cd (or calibration coefficient, Kv), for each venturi, calculate the individual molar flow rates through each venturi and sum all their flow rates to determine CFV flow rate, n

. If you use multiple venturis and you calibrated venturis in combination, calculate

n
using the sum of the active venturi throat areas as

At, the square root of the sum of the squares of the active venturi throat diameters as

dt, and the ratio of the venturi throat to inlet diameters as the ratio of the square root of the sum of the active venturi throat diameters (

dt) to the diameter of the common entrance to all the venturis (

D).

(1) To calculate n

through one venturi or one combination of venturis, use its respective mean C

d and other

constants you determined according to

§ 1065.640 and calculate

n
as follows:

Example:

Cd = 0.985

Cf = 0.7219

At = 0.00456 m
2

pin = 98.836 kPa = 98836 Pa = 98836 kg/(m·s
2)

Z = 1

Mmix = 28.7805 g/mol = 0.0287805 kg/mol

R = 8.314472 J/(mol·K) = 8.314472 (m
2·kg)/(s
2·mol·K)

Tin = 378.15 K

(2) To calculate the molar flow rate through one venturi or a combination of venturis, you may use its respective mean, Kv, and other constants you determined according to § 1065.640 and calculate its molar flow rate n

during an emission test. Note that if you follow the permissible ranges of dilution air

dewpoint versus

calibration air

dewpoint in Table 3 of § 1065.640, you may set

Mmix-cal and

Mmix equal to 1. Calculate

n
as follows:

$\begin{array}{c}\stackrel{\xb7}{n}=\frac{{K}_{v}\xb7{p}_{\mathrm{in}}}{\sqrt{{T}_{\mathrm{in}}}}\xb7\frac{{p}_{\mathrm{std}}}{{T}_{\mathrm{in}}\xb7R}\xb7\frac{\sqrt{{M}_{\mathrm{mix}-\mathrm{cal}}}}{\sqrt{{M}_{\mathrm{mix}}}}\\ \text{Eq. 1065.642-5}\end{array}$

Example:

Vstdref = 0.4895 m
3

Tin-cal = 302.52 K

Pin-cal = 99.654 kPa = 99654 Pa = 99654 kg/(m·s
2)

pin = 98.836 kPa = 98836 Pa = 98836 kg/(m·s
2)

pstd = 101.325 kPa = 101325 Pa = 101325 kg/(m·s
2)

Mmix-cal = 28.9656 g/mol = 0.0289656 kg/mol

Mmix = 28.7805 g/mol = 0.0287805 kg/mol

Tin = 353.15 K

Tstd = 293.15 K

R = 8.314472 J/(mol·K) = 8.314472 (m
2·kg)/(s
2·mol·K)

$\begin{array}{c}{K}_{v}=\frac{0.4895\xb7\sqrt{302.52}}{99654}=0.000074954{\phantom{\rule{0ex}{0ex}}m}^{4}\xb7s\xb7{K}^{0.5}/\mathrm{kg}\\ \stackrel{\xb7}{n}=\frac{0.000074954\xb798936}{\sqrt{353.15}}\xb7\frac{101325}{293.15\xb78.314472}\xb7\frac{\sqrt{0.0289656}}{\sqrt{0.0287805}}\end{array}$