The measurement data for the calculation of the individual mode results are shown below. In this example, CO and NO_x are measured on a dry basis, HC on a wet basis. The HC concentration is given in propane equivalent (C3) and has to be multiplied by 3 to result in the C1 equivalent. The calculation procedure is identical for the other modes.

The following example calculation is given for CO; the calculation procedure is identical for the other components.

The emission mass flow rates of the individual modes are multiplied by the respective weighting factors, as indicated in Annex III, Appendix 1, Section 2.7.1, and summed up to result in the mean emission mass flow rate over the cycle:

CO	=	$\left(\mathrm{6,7\; \times \; 0,15}\right)+\left(\mathrm{24,6\; \times \; 0,08}\right)+\left(\mathrm{20,5\; \times \; 0,10}\right)+\left(\mathrm{20,7\; \times \; 0,10}\right)+\left(\mathrm{20,6\; \times \; 0,05}\right)+\left(\mathrm{15,0\; \times \; 0,05}\right)+\left(\mathrm{19,7\; \times \; 0,05}\right)+\left(\mathrm{74,5\; \times \; 0,09}\right)+\left(\mathrm{31,5\; \times \; 0,10}\right)+\left(\mathrm{81,9\; \times \; 0,08}\right)+\left(\mathrm{34,8\; \times \; 0,05}\right)+\left(\mathrm{30,8\; \times \; 0,05}\right)+\left(\mathrm{27,3\; \times \; 0,05}\right)$
	=	30,91 g/h

The engine power of the individual modes is multiplied by the respective weighting factors, as indicated in Annex III, Appendix 1, Section 2.7.1, and summed up to result in the mean cycle power:

$P\left(n\right)$	=	$\left(\mathrm{0,1\; \times \; 0,15}\right)+\left(\mathrm{96,8\; \times \; 0,08}\right)+\left(\mathrm{55,2\; \times 0,10}\right)+\left(\mathrm{82,9\; \times \; 0,10}\right)+\left(\mathrm{46,8\; \times \; 0,05}\right)+\left(\mathrm{70,1\; \times \; 0,05}\right)+\left(\mathrm{23,0\; \times \; 0,05}\right)+\left(\mathrm{114,3\; \times \; 0,09}\right)+\left(\mathrm{27,0\; \times \; 0,10}\right)+\left(\mathrm{122,0\; \times \; 0,08}\right)+\left(\mathrm{28,6\; \times \; 0,05}\right)+\left(\mathrm{87,4\; \times \; 0,05}\right)+\left(\mathrm{57,9\; \times \; 0,05}\right)$
	=	60,006 kW

CO‾= 30,9160,006 = 0,0515 g/kWh

Assume the following values have been determined on the random point:

n_Z

1 600 min^-1

M_Z

495 Nm

NO_{x mass,Z}

487,9 g/h (calculated according to the previous formulae)

P(n)_Z

83 kW

NO_x,Z

487,9/83 = 5,878 g/kWh

Assume the values of the four enveloping modes on the ESC to be as follows:

ETU = 5,889 + 4,973-5,889 × 1 600-1 368 / 1 785-1 368 = 5,377 g/kWh

ERS = 5,943 + 5,565-5,943 × 1 600-1 368 / 1 785-1 368 = 5,732 g/kWh

MTU = 681 + 601-681 × 1 600-1 368 / 1 785-1 368 = 641,3 Nm

MRS = 515 + 460-515 × 1 600-1 368 / 1 785-1 368 = 484,3 Nm

EZ = 5,732 + 5,377-5,732 × 495-484,3 / 641,3-484,3 = 5,708 g/kWh

Particulate measurement is based on the principle of sampling the particulates over the complete cycle, but determining the sample and flow rates (M_SAM and G_EDF) during the individual modes. The calculation of G_EDF depends on the system used. In the following examples, a system with CO₂ measurement and carbon balance method and a system with flow measurement are used. When using a full flow dilution system, G_EDF is directly measured by the CVS equipment.

Assume the following measurement data of mode 4. The calculation procedure is identical for the other modes.

1. (a)

   carbon balance method

   $G_{\mathrm{EDFW}}=\frac{\mathrm{206,5\; \times \; 10,76}}{\mathrm{0,657-0,040}}=\mathrm{3\; 601,2}\mathrm{kg/h}$
2. (b)

   flow measurement method

   $\mathrm{q\; =}\frac{\mathrm{6,0}}{\mathrm{6,0-5,4435}}\mathrm{=\; 10,78}$$G_{\mathrm{EDFW}}\mathrm{=\; 334,02\; \times \; 10,78\; =}\mathrm{3\; 600,7}\mathrm{kg/h}$

The G_EDFW flow rates of the individual modes are multiplied by the respective weighting factors, as indicated in Annex III, Appendix 1, Section 2.7.1, and summed up to result in the mean G_EDF over the cycle. The total sample rate M_SAM is summed up from the sample rates of the individual modes.

${\bar{G}}_{\mathrm{EDFW}}$	=	$\left(\mathrm{3\; 567}\mathrm{\times \; 0,15}\right)+\left(\mathrm{3\; 592}\mathrm{\times \; 0,08}\right)+\left(\mathrm{3\; 611}\mathrm{\times \; 0,10}\right)+\left(\mathrm{3\; 600}\mathrm{\times \; 0,10}\right)+\left(\mathrm{3\; 618}\mathrm{\times \; 0,05}\right)+\left(\mathrm{3\; 600}\mathrm{\times \; 0,05}\right)+\left(\mathrm{3\; 640}\mathrm{\times \; 0,05}\right)+\left(\mathrm{3\; 614}\mathrm{\times \; 0,09}\right)+\left(\mathrm{3\; 620}\mathrm{\times \; 0,10}\right)+\left(\mathrm{3\; 601}\mathrm{\times \; 0,08}\right)+\left(\mathrm{3\; 639}\mathrm{\times \; 0,05}\right)+\left(\mathrm{3\; 582}\mathrm{\times \; 0,05}\right)+\left(\mathrm{3\; 635}\mathrm{\times \; 0,05}\right)$
	=	3 604,6 kg/h
$M_{\mathrm{SAM}}$	=	0,226 + 0,122 + 0,151 + 0,152 + 0,076 + 0,076 + 0,076 + 0,136 + 0,151 + 0,121 + 0,076 + 0,076 + 0,075
	=	1,515 kg

Assume the particulate mass on the filters to be 2,5 mg, then

PTmass = 2,51,515 × 360,41 000 = 5,948 g/h

Assume one background measurement with the following values. The calculation of the dilution factor DF is identical to Section 3.1 of this Annex and not shown here.

Md = 0,1 mg; MDIL = 1,5 kg

Sum of DF	=	$\left[\left(\mathrm{1-1/119,15}\right)\mathrm{\times \; 0,15}\right]+\left[\left(\mathrm{1-1/8,89}\right)\mathrm{\times \; 0,08}\right]+\left[\left(\mathrm{1-1/14,75}\right)\mathrm{\times \; 0,10}\right]+\left[\left(\mathrm{1-1/10,10}\right)\mathrm{\times \; 0,10}\right]+\left[\left(\mathrm{1-1/18,02}\right)\mathrm{\times \; 0,05}\right]+\left[\left(\mathrm{1-1/12,33}\right)\mathrm{\times \; 0,05}\right]+\left[\left(\mathrm{1-1/32,18}\right)\mathrm{\times \; 0,05}\right]+\left[\left(\mathrm{1-1/6,94}\right)\mathrm{\times \; 0,09}\right]+\left[\left(\mathrm{1-1/25,19}\right)\mathrm{\times \; 0,10}\right]+\left[\left(\mathrm{1-1/6,12}\right)\mathrm{\times \; 0,08}\right]+\left[\left(\mathrm{1-1/20,87}\right)\mathrm{\times \; 0,05}\right]+\left[\left(\mathrm{1-1/8,77}\right)\mathrm{\times \; 0,05}\right]+\left[\left(\mathrm{1-1/12,59}\right)\mathrm{\times \; 0,05}\right]$
	=	0,923

PTmass = 2,51,515-0,11,5 × 0,923 × 3 604,61 000 = 5,726 g/h

Assume the values calculated for mode 4 above, then

WFE,i = 0,152 × 3 604,6/1,515 × 3 600,7 = 0,1004

This value is within the required value of 0,10 ± 0,003.

In this example a Bessel algorithm is designed in several steps according to the above iteration procedure which is based upon Annex III, Appendix 1, Section 6.1.

For the opacimeter and the data acquisition system, the following characteristics are assumed:

• physical response time t_p 0,15 s

• electrical response time t_e 0,05 s

• overall response time t_Aver 1,00 s (by definition in this Directive)

• sampling rate 150 Hz

f_c

$\frac{\mathrm{3,1415}}{\mathrm{10\; \times \; 0,987421}}\mathrm{=\; 0,318152\; Hz}$

Δt

1/150 = 0,006667 s

Ω

$\frac{1}{\tan \left[\mathrm{3,1415\; \times \; 0,006667\; \times \; 0,318152}\right]}\mathrm{=\; 150,07664}$

E

$\frac{1}{\mathrm{1\; +\; 150,076644\; \times }\sqrt{\mathrm{3\; \times \; 0,618034}}{\mathrm{+\; 0,618034\; +\; 150,076644}}^2}{\mathrm{=\; 7,07948\; \times \; 10}}^{\mathrm{-5}}$

K

${\mathrm{2\; \times \; 7,07948\; \times \; 10}}^{\mathrm{-5}}\times \left({\mathrm{0,618034\; \times \; 150,076644}}^2\mathrm{-1}\right)\mathrm{-1\; =\; 0,970783}$

This gives the Bessel algorithm:

Yi = Yi-1 + 7,07948 E - 5 × Si + 2 × Si-1 + Si-2-4 × Yi-2 + 0,970783 × Yi-1-Yi-2

where S_i represents the values of the step input signal (either ‘0’ or ‘1’) and Y_i represents the filtered values of the output signal.

The Bessel filter response time t_F is defined as the rise time of the filtered output signal between 10 % and 90 % on a step input signal. For determining the times of 10 % (t₁₀) and 90 % (t₉₀) of the output signal, a Bessel filter has to be applied to a step input using the above values of f_c, E and K.

The index numbers, the time and the values of a step input signal and the resulting values of the filtered output signal for the first and the second iteration are shown in Table B. The points adjacent to t₁₀ and t₉₀ are marked in bold numbers.

In Table B, first iteration, the 10 % value occurs between index number 30 and 31 and the 90 % value occurs between index number 191 and 192. For the calculation of t_F,iter the exact t₁₀ and t₉₀ values are determined by linear interpolation between the adjacent measuring points, as follows:

t10 = tlower + Δt × 0,1-outlower/outupper-outlower

t90 = tlower + Δt × 0,9-outlower/outupper-outlower

where out_upper and out_lower, respectively, are the adjacent points of the Bessel filtered output signal, and t_lower is the time of the adjacent time point, as indicated in Table B.

t10 = 0,200000 + 0,006667 × 0,1-0,099208/0,104794-0,099208 = 0,200945 s

t90 = 0,273333 + 0,006667 × 0,9-0,899147/0,901168-0,899147 = 1,276147 s

|Δ| ≤ 0,01 is required. Since 0,081641 > 0,01, the iteration criteria is not met and a further iteration cycle has to be started. For this iteration cycle, a new cut-off frequency is calculated from f_c and Δ as follows:

fc,new = 0,318152 × 1 + 0,081641 = 0,344126 Hz

This new cut-off frequency is used in the second iteration cycle, starting at step 2 again. The iteration has to be repeated until the iteration criteria is met. The resulting values of the first and second iteration are summarised in Table A.

As soon as the iteration criteria has been met, the final Bessel filter constants and the final Bessel algorithm are calculated according to step 2. In this example, the iteration criteria has been met after the second iteration (Δ = 0,006657 ≤ 0,01). The final algorithm is then used for determining the averaged smoke values (see next Section 2.3).

Yi = Yi-1 + 8,272777 × 10-5 × Si + 2 × Si-1 + Si-2-4 × Yi-2 + 0,968410 × Yi-1-Yi-2

k =-1/0,430 × ln 1-16,783/100 = 0,427252 m-1

This value corresponds to S₂₇₂ in the following equation.

In the following equation, the Bessel constants of the previous Section 2.2 are used. The actual unfiltered k-value, as calculated above, corresponds to S₂₇₂ (S_i). S₂₇₁ (S_i-1) and S₂₇₀ (S_i-2) are the two preceding unfiltered k-values, Y₂₇₁ (Y_i-1) and Y₂₇₀ (Y_i-2) are the two preceding filtered k-values.

$Y_{272}$	=	${\mathrm{0,542383\; +\; 8,272777\; \times \; 10}}^{\mathrm{-5}}\times \left(\mathrm{0,427252\; +\; 2\; \times \; 0,427392\; +\; 0,427532-4\; \times \; 0,542337}\right)\mathrm{+\; 0,968410\; \times }\left(\mathrm{0,542383-0,542337}\right)$
	=	${\mathrm{0,542389\; m}}^{\mathrm{-1}}$

This value corresponds to Y_max1,A in the following equation.

From each smoke trace, the maximum filtered k-value is taken for the further calculation.

Assume the following values

SVA = 0,5424 + 0,5435 + 0,5587 / 3 = 0,5482 m- 1

SVB = 0,5596 + 0,5400 + 0,5389 / 3 = 0,5462 m- 1

SVC = 0,4912 + 0,5207 + 0,5177 / 3 = 0,5099 m- 1

SV = 0,43 × 0,5482 + 0,56 × 0,5462 + 0,01 × 0,5099 = 0,5467 m- 1

Before calculating SV, the cycle must be validated by calculating the relative standard deviations of the smoke of the three cycles for each speed.

In this example, the validation criteria of 15 % are met for each speed.

Assuming a diesel fuel of the composition C₁H_1,8

FS = 100 × 11 + 1,82 + 3,76 × 1 + 1,84 = 13,6

DF = 13,60,723 + 9,00 + 38,9 × 10- 4 = 18,69

NOx conc = 53,7-0,4 × 1-1/18,69 = 53,3 ppm

COconc = 38,9-1,0 × 1-1/18,69 = 37,9 ppm

HCconc = 9,00-3,02 × 1-1/18,69 = 6,14 ppm

1. (a)

   GC method

   ${\mathrm{NMHC}}_{\mathrm{conce}}\mathrm{=\; 27,0-18,0\; =\; 9,0\; ppm}$
2. (b)

   NMC method

   Assuming a methane efficiency of 0,04 and an ethane efficiency of 0,98 (see Annex III, Appendix 5, Section 1.8.4)

   ${\mathrm{NMHC}}_{\mathrm{conce}}=\frac{\mathrm{27,0\; \times }\left(\mathrm{1-0,04}\right)\mathrm{-18,0}}{\mathrm{0,98-0,04}}\mathrm{=\; 8,4\; ppm}$

Assuming a G₂₀ reference fuel (100 % methane) of the composition C₁H₄:

FS = 100 × 11 + 42 + 3,76 × 1 + 44= 9,5

DF = 9,50,723 + 27,0 + 44,3 × 10- 4 = 13,01