meet these RRA criteria, it is out of control . 

(i) For all three data points, the PM CEMS response 

value can be no greater than t he gre atest PM CEMS response 

value used to develop your correlation curve. 

(ii) For two of the three da t a points, the PM CEMS 

response value must lie within the PM CEMS output range 

used to develop your correlation curve . 

(iii) At least two of the three sets of PM CEMS and 

reference method measurements must fall within the same 

speci fied area on a graph of the correlation regression 

line as requ ired for t he RCA and described in paragraph 

(5) (iii) of this section . 

10 . 5 What do I do if my PM CEHS is out of con trol? 

If your PM CEMS is out of control, you must take the 

actions listed in paragraphs (1) and (2) of this section. 

(1) You must take necessary corrective action to 

eliminate the problem and perform tests, as appropriate, 

to ensure that the corrective action was successful . 

(i) Fol l01~i ng corrective action , you must repeat the 

previously failed audit to confirm t hat your ~ CE ~S is 

operating within the specifications . 

(ii) If your P~l CEMS tailed an RRA, you must take 

corrective action until your PM CEMS passes the RRA 

criteria. If the RRA criteria cannot be achieved, you 

VES009700 
must perform an RCA . 

(iii) I f your PM CEMS fai l ed an RCA, you must follow 

procedures specified in section 10 . 6 of this procedure . 

(2) You must report both the audit showi ng your PM 

CEMS to be out of control and the results of the audit 

fol l owing corrective act ion show1n9 your PM CEMS to be 

operating within specifi cations. 

10 . 6 Wha t do I do if my PM CEMS fails an RCA? After 

an RCA failure, you must take all applicable actions 

listed in paragraphs (1) through (3) of this section. 

(1) Combine RCA data l<ith data from the active P~l 

CEMS correlation and perform the mathematical evaluations 

defined in PS-11 for development of a PM CEMS correlation, 

including examination of alternate correl at1on models 

(i . e . , linear, pol ynomial , logarithmic , exponential, and 

power} . If the expanded data base and revised correlation 

meet PS- 11 stat istical criteria, use the revised 

correlation. 

(2} If the criteria specified in paragraph (ll of 

t his section are not achieved, you must develop a new PM 

CEMS correlation based on revised data. The revised data 

set must consist of the test results from only the RCA. 

The new data must meet all requirements of PS- 11 t o 

develop a revi sed PM CEI~S correlation, except that t he 

VES009701 
minimum numbe r of sets of PM CEMS and reference method 

mebsureme nts is 12 i nst ead of t he minimum o f 15 sets 

required by PS-11 . Your PM cgMs is considered to be back 

i n control l ed status when the revised correla t ion meets 

all of the performance criteria specified in s ection 13. 2 

of PS-1 1 . 

{3) If t he actions in paragraphs {1) and (2) of t his 

section do not result in an acceptable correlat i on , you 

must eva luate the cause(s) and comply with t he actions 

listed i n paragraphs {3) {i J through (iv) of this section 

withi n 90 days after the completion o f t he fa iled RCA. 

{i) Completely i nspect your PM CEMS for mechanical 

or operational problems . lf you find a mechanical or 

operational problem, repa i r your PM CEMS and repeat the 

RCA . 

(ii) You may need to relocate your PM CEMS to a more 

appropriate measurement l ocation . If you relocate your PM 

CEMS , you must pe rform a new correlation test according to 

the procedures specified in PS-11 . 

(ii i) The characteristics of t he PM or gas in your 

source's f l ue gas stream may have changed such that your 

PM CEMS measu rement t echnology i s no longer appropriate . 

I f t his is the case, you muse instal l a PM CEMS wi th 

measurement technology that is appropriate fo r your 

VES009702 
source's flue gas characteristics . You must perform a new 

corre lat i on test according to the procedures specified in 

PS-11. 

(i v) If t he corrective actions i n pa ragraphs (3) (i) 

through (iii) of this section were not successful, you 

must petit ion us, the State, or local agency for approval 

of alternati ve criteria or an alternative for continuous 

PM monitoring . 

10 . 7 When does the out-of-control period begin 

and end? The out-of-control period begins immediately 

after the last test run or c heck of an unsuccessful RCA, 

RRA, ACA, SVA, drift check, or sample vol ume chec k. The 

out - of- control period ends immediatel y after the last test 

run or check of the subsequent successful audit or drift 

check. 

10 . 8 Can I use the data recorded by my PM CEMS 

during out - of- control peri ods? During any period when 

your PM CEMS is out of control, you may not use your PM 

GEMS data to calculate emission compli ance or to meet 

mi nimum data availability requir ements described i n the 

applicable regulation . 

10 . 9 What are the QA/QC reporting requirements for 

my ~M CEMS? You must report the accuracy results for your 

~M GEMS, specified in section 10 . 4 of this procedure , at 

VE$009703 
the interval specified in the applicable regul a t ion. 

Report the drift and accuracy i nforma tion as a Da ta 

Assessment Report (DAR), and include one copy of this DAR 

for each quarterly a udit with t he report of emissions 

required under the applicable regulation . An example DAR 

is p rovided in Pr ocedure 1 , Appendix F of th~s part. 

10 . 10 What minimum i nformation must I i ncl ude in my 

DAR? As a minimum, you must include the information 

listed in paragraphs (1 ) through (5) of this sec t ion in 

the DAR : 

(1) Your name and address . 

(2) I dentification and location of monitors in your 

CEMS . 

(3) Manufacturer and model number of each monitor in 

your CEMS . 

(1) Assessment of PM CEMS data 

accuracy/acceptability, and date of assessment, as 

determi ned by an RCA , RRA, ACA, or SVA described in 

section 10, i ncluding the acceptability determination f or 

the RCA or RRA, the accuracy for the ACA or SVA, t he 

reference me thod results, t he audit standards, your PM 

CEMS r esponses, and the calculation results a s defined in 

section 12 . I f the accuracy audit res ults show your PM 

CEMS to be out of control, you must report bot h the audit 

VES009704 
results showing your PM CEMS to be out of control and the 

resu l t s of the aud it fo l lowing correc tive action showing 

your PM CEMS t o be operating within specifications. 

(5) Summary of a ll corrective actions you t ook when 

you determined your PM CEMS to be out of control, as 

described i n section 10 . 5 , or a fter failing an RCA, as 

des cribed in section 10 . 6 . 

10.7 Where and ho'• long must I r etain the QA data 

t hat this procedure requires me to record for my PM CEMS? 

You must keep the records required by t hi s procedure for 

your PM CEMS onsite and a vai l able for inspection by us, 

the State, and/or local enfor cement agency for a pe r iod of 

5 years. 

11 . 0 Wh at anal ytical p rocedures apply to this procgdt1re? 

Sample collection and analysis are concurrent for 

this procedure . You must r efer to the appropriate 

reference method for t he specific analytical procedures . 

12 .0 What calculatjons and data anal ysis must I perform 

for mv pM CE;MS? 

( 1) Hovl do I determine RCA and RRA acceptability? 

You must plot each of your PM CEMS and reference method 

data sets from an RCA or RRA on a graph based on your PM 

CEMS correlation l ine to determine i f the criteria in 

paragraphs 10 . 4(5) or (6), respectively, a re met. 

VES009705 
(2) How do r calculate ACA accuracy? You mus t use 

Equation 2- 1 t o ca lculate ACA accuracy for e ac h of t he 

three audit points: 

ACA !Rca -Rvl Accuracy= x 100 R, (Eq . 2- 1) 

where : 

ACA Accuracy • The ACA accuracy a t each audit point, 

in percent, 

Your PM CE!~S response to the reference 

standard, and 

The reference standard val ue . 

(3) How do I ca lculate daily upscale a nd zero drift? 

You must calculate the upsca le drift using to Equation 2-2 

and the zero drift according to Equation 2-3 : 

I

R - R I 

UD = 0

' " 

0

x 100 

Ro 

(Eq . 2- 2) 

"'here : 

UD - The upscale drift of your PM CEMS , in 

pe rcent , 

Your PM CEMS response to the upscale check 

value , and 

VES009706 
= The upscale check value . 

(Eq . 2-3) 

where : 

ZD = The zero (low-level) drift of your PM CEMS, 

in percent, 

R~ Your PM CEMS response of the zero check 

value , 

The zero check va lue , and 

- The upscale check value . 

(4) How do I cal culate SVA accuracy? You must use 

Equati on 2-4 to calculate the accuracy, in percent, for 

each of the three SVA tests or the daily sample volume 

check : 

~<here : 

(v- v) Accuracy= " • x 100 

FS (Eq . 2- 4) 

Sample gas volume determined/reported by 

your PM CEMS (e . g ., dscm} , 

Sample gas volume measured by the 

independent calibrated reference device 

(e.g. , dscm) for the SVA or the reference 

VES009707 
value for the daily sample volume check, 

and 

FS ~ull-sca le value . 

Note : Before calculating SVA accuracy , you must 

cor rect the sample gas vol umes measured by your PM CE~ S 

and the independent calibrated reference device to the 

same basis o f temperature , p ressure, and moisture content . 

You must document all data and calculations. 

13 . 0 Methqd Performance . (Reserved! 

14 . 0 Pqll ption preven tion . (Reserved] 

15 . 0 Wast e Management . (Reserved] 

16 . 0 Whj ch references are relevant t o this method? 

(Reserved] 

17 . 0 What tables. diagrams. flowc harts. and ya lidati qn 

data are r e l gyant to t his me jhod? {Reserved) 

VES009708 
VES009709.pdf 
Y.O EXAMPLE CALCULATIONS-INITIAL CORRELATION TEST 

This section presents several examples of initial correlation test calculations. For each of the 

examples, a brief description of the correlation test scenario is provided, followed by a summary 

and discussion of the results of the calculations. For the fmt example, all of the equations used 

to evaluate the data and some sample calculations arc presented. The calculations for all of the 

exumples were pcrfonned using the Correlation Test Calculation Spreadsheet, which is described 

in Section X. Appendix A includes the entire spreadsheet output for each of the examples. 

Y .I EXAMPLE I: HOW DO I DEVELOP AN INITIAL CORRELATION? 

Y .1.1 Correlation Test Scenario 

A correlation test is performed on a boiler at Facility A. The boiler is equipped with an in siru 

PM CBMS that responds to PM concentrations at acrual conditions. The emission limit, when 

converted to units that are consistent with the PM CEMS sampling conditions, is 20.0 mglacm. 

Twenty-four test runs are performed using a single Method 5 sampling train. The facility 

detcnnines that four of the test runs (Runs 7, 8, 9, and 18) were suspect because of problems 

with the sampling train. Table Y -I summarizes the correlation test data. Figure Y-1 shows a 

plot of the data. 

Table Y-1. Summary of Correlation Test Data for Facility A 

PM O:MS response. Pl\1 concentTation, 

Ruo no. mA mg/acm 

I 4.90 2.1 

2 7.60 2.9 

3 7.80 5. 1 

4 9.20 4.8 

s 10.10 2.9 

6 11.06 6.3 

7 13.40 0.2 

8 16.10 0.6 

9 15.80 0.3 

10 12.00 8.4 

(collfinued) 

Y-1 

VES009709 
Run no. 

20 

18 

E 16 

X 

a. 14 

E 

.: 12 

0 

~ 10 

t a t! 

c 6 8 

:Iii 4 .,_ 

2 

0 

11 

12 

13 

14 

IS 

16 

17 

18 

19 

20 

21 

22 

23 

24 

- -

0 

Table Y-1. (continued) 

PM E~fS r esponse, PM concentration, 

rnA mglacm 

18.20 8.4 

15.70 4.9 

13.90 6.3 

19.70 13.0 

21.02 9.1 

22.30 13.9 

24.10 8.7 

25.90 0.6 

24.70 12.6 

26.40 15.9 

29.08 16.2 

29.30 13.5 

30.50 15.9 

33.10 18.5 

• 

- - -· -- ---· -- - • • • 

• • • • 

• • 

• • • 

• • 

# • 

5 10 15 20 25 30 

PM CEMS Response, mA 

Figure Y -1. Plot of tbe correlation test data for Fadllty A. 

Y-2 

35 

VES009710 
Must I use all of the test data to develop the correlation? 

Section 8.6(3) ofPS-1 1 requires a minimum of 15 valid test runs for developing the correlation, 

and Section 8.6{3 )(iii) allows you to discard up to 5 t~t runs without explanation. Additional 

test runs can be discarded if the basis for rejecting the runs is specified in the manual test 

method, in PS-1 1, or in the quality assurance plan tb r the source. In this example, an additional 

nine te st runs were performed, so it is not necessary to use a ll of tl1e test da ta. Five of the runs 

can be discarded without explanation. As indicoted in Figure Y -I, the results from test Runs 7, 

8, 9, and 18 appear to be outliers. If the four outliers are discarded, there are 20 runs remaining, 

which satisfies the minimum requirement of 15 runs for developing the initial correlation. Note 

that Section 8.6(3Xii) ofPS- 11 stiU requ ires you to report all of the data. even for those nms that 

are rejected. 

How do I deyelop a linear con·eJation curve fiom the data? 

To develop a linear correlation model, follow the procedures described in Section 12.3(1) of 

PS-11. The form of the linear model is provided in Equation I 1-3 of PS-I I: 

where 

y = b0 + b,x 

y = the predicted PM concentration 

b0 = the intercept for th~ correlation curve 

b1 : the slope of the correlation curve 

x - the PM CEMS response value. 

To begin, the mean PM CEMS response (X) and mean PM concentration (y) are calculated as 

follows: 

I ~ I 

x = - ~ x1 = -

0

(4.9 + 7.6+ 7.8 + .... + 33.1) ~ 1853 

II t•l 2 

I • I .Y= -Ly, = 20

(2.1+ 2.9 + 5.1 + .... +185)~ 9.47 

n r-1 

The variables S., and S,,.. which are presented in Equation II· 7 of PS-11, nrc calculated next as 

follows: 

s M = f (x, - x)' • ( ( 4.9- 1 KS3)' + (7.6- 1 H.S3)2 + ... +(33.1 - 1 H.S3)') = 1,442.2 

••• 

• 

s.,. ~ L: (x1 - x)(y, - y) = (4.9 - 18.53){2.1 - 9.47) + (7.6 - 18.53)(2.9- 9.47)+ ... + 

I • I 

(33.1- 18.53)(18.5- 9.47)= 785.2 

Y-3 

VES009711 
The slope of the correlation curve (b,) is determined by Equation 11-6 of PS-11: 

b - s,, = 785.2 - 0.544 

~ 1,442.2 

The intet-cept of the correlution curve is detennined using Equation 11-4 ofPS- 11 : 

b. = y- b, . X. 9.47 - (0.544 X 18.53) = - 0.621 

The resulting equation for the correlation curve is as follows: 

f = - 0.621 + 0.544x 

How do I determjne the ljnear correlation coefficient? 

The linear correlation coefficient (r) is calculated using the parameters Sc and Sy. The parameter 

Sc is defined in Equation 11-9 as follows: 

SL = -'- f (;,-y,)' n - 2 ,.. 1 

To calculate Sv the linear correlation equation must be used to determine the predicted PM 

concenlnltion (j) for each test run. For example, for Run 3, the PM CEMS response is 7.8 and 

the predicted PM concentration is 

y = - 0.621 + 0.544(7.8) = 3.626 

Table Y -2 shows the results of the calculations of y. Using these values, the calculation for Sc is 

as follows: 

SL = - 1

- ((2.047 - 2.1)

2 

+ (3.517- 2.9)2 + (3.626- 5.1)

2 

+ ... +(17.401 - 18.5)' ) = 1.910 

20 - 2 

Table Y -2. Predicted Pl\1 Concentrations (j) for Linear Correlation 

PM CUtS Pl\f concutrallon, Predlcted PM 

Run no. response, mA mglacm concentration If)' 

I 4.9 2.1 2.047 

2 7.6 2.9 3.517 

3 7.8 5.1 3.626 

4 9.2 4.8 4.388 

5 10.1 2.9 I 4.878 - (continued) 

Y-4 

VES009712 
Table Y-2. (continued) 

PM CEMS PM oonccntratkm, Predicted PM 

Run no. response, rnA mg/ucm concentration (j)' 

6 I 1.06 6.3 5.401 

10 12 8.4 5.913 

II 18.2 8.4 9.289 

12 15.7 4.9 7.928 

13 13.9 6.3 6.947 

14 19.7 13 10.105 

IS 21.02 9.1 10.824 

16 22.3 13.9 11.52 1 

17 24. 1 8.7 12.501 

19 24.7 12.6 12.828 

20 26.4 15.9 13.753 

21 29.08 16.2 15.213 

22 29.3 13.5 15.332 

23 30.5 15.9 15.986 

24 33.1 18.5 17.401 

• Calculated using the correlation equation: .r~ ..0.62 I + O.S44.r. 

Tbe parameter S, is calculated using Equation 1 I -15 of PS-I J as lollows: 

f (y,- y)' 

s = 11-"'""''- - - = ·" n - l 

(2.1 - 9.47)' + (2.9- 9.47)' + (5.1 - 9.47)'+ ... +(185 - 9.47)' 

20 - I = 

5

·

095 

With the values for ~, and S" the correlation coefficient is calculated using Equation 11- 14 as 

follows: · 

The correlation coefficient (0.927) is greater than 0.&5 and, therefore, satisfies PS- 11. 

Note: For dam sets that show little to no correlation, the value for SL can be greater than S,. In 

such cases, the correlation coefficient (r) calculated using Equation I I-I 4 is undefined because it 

eot<tils determining the square root of a negative number. Even though the correlation 

coefficient cannot be determined, it can be assumed that the value for r does not meet the 

criterion of 0.85 specified by PS-11. 

Y-5 

VES009713 
How do I calculate the confidence interval half range percentage for the ljnear correlation? 

The half range for the 95 percent confidence interval (CI) for the predicted PM coocemrntion at 

tlte mean value of x is determined using Equation 11 -8 ofPS-11: 

Cl • t4f.•-·;: · S,-Jf 

The variable t .... ••• is obtained fTOm n table of !-statistic values. The degrees of freedom (dl) 

equals n-2, and the value for the t statistic for this example is 2.10 I. With this value and the 

value for S, calculated above, Cl is calculated as follows: 

a= 2.JOI(L9t)Ho = 0.897 

The emission limit io this example is 20.0 mglacm. To determine Cl as a percentage of the 

emission limit, Equation Il-l 0 is used as follows: 

Cl 0.897 

CJ% = EL x 100% = 2

0.0 x 1000/o = 4.49% 

The resulting value for CI% is 4.49 percent Because this value is less than 10 percent, the 

correlation satisfies the criterion for confidence interval half range. 

How do I calculate tbe tolerance interval half range percentage for the linear C!>rrelation? 

Equation 11-11 of PS-11 is used to calculated tbe tolerance interval half mngc (Tl) at the mean 

value ofx: 

The parameter k1 is determined using Equation 11-1 2 of PS-11 : 

The parameters u,. and v., are detennined from Table I of PS-I I. In this example, n' is the 

number of test runs used for the corre lation (20), u,. • l.l 77, and v « equals 1.384 for n-2 degrees 

of freedom. With these values. k1 and n 3rc calculated as follows: 

k T = u

11, • v df : 1.177( 1.384) = 1.629 

Tl = k, · S, = 1.629(1.91) = 3.111 

With an emission limit of20.0 mg/acm, the tolerance interval as a percentage of the emission 

limit is determined using Equation J 1-13 as follows: 

Y-6 

VES009714 
TT 3.1 11 

Tl %= -x 100% = - x 100%= 156% EL 20.0 . 

The resulting value for Tl% is 15.6 pe:rcenL Because litis value is less than 25 percent, the 

correlation satisfies the criterion for tolerance interval half range. In summary, the linear 

con-elation developed trorn the data satisfies all of the criteria specified in PS- 11 . 

How do I develop a OOIVI19rnial correlation curve from the data? 

A polynomial correlation curve can be expressed by the following: 

The procedures for developing a polynomial correlation curve are described in Section 12.3(2) or 

PS-11. The first step is to determine the values for the parameters S, through S,, as define~ in 

Equations 11-17 and 11 -18 of PS- 11. The.se calculations are summarized as follows: 

• 

S, = I;(x,)= (4.9+ 7.6 + 7.8+ ... +33.1),.. 370.7 i=l 

s l = :f;(x/ )= (4.92 + 7.67 + 7.82+ ... +33.1

2

)= 8,312 

,., 

• 

S, = I;{x/) = (4.93 + 7.63 + 7.83 +.. .+33.1')= 208,377 

,., 

" S, a L (x,') = (4.9' + 7.64 + 7 .8' + ... + 33.14 )+ 5,574,977 i.s.l 

• 

s, = I: (y,) = c2.1 + 2.9 + s.It ... + t8.5l = 189.4 

I• I 

• 

S6 = L (x,y,) = ((4.9)(2.1) + (7.6)(2.9) + (7.8)(5.l)t ... +(33.1Xl85)) = 4,295 

i=-1 

s, = :f: (x,' r,). ((4.9'X2.1) + (7.6'X2ol + (7.s' Xs.I)+ ... + (33.l')(ls.sJ) = 1os,s61 ,_, 

Next, the parameter detA must be calculated using Equation I 1-22 ofPS.-11. For this example, 

the calculation is as follows: 

dct A = n · S, · S, - S, · S, · S, + S, · S, · S, - S, · S, · n. + s, · S, · S, - S, · S, · S, = 

(2ol(s.mXs,s74,977)- (s,312Xs.312Xs.3 12) + 37o.7(20s.mXs.312)- (2os.mX2os.J77)(2o) + 

(8,312X370.7)(208.377)- (5,574,977X370.7)(370.7) = 2.120x to• 

The values for the correlation curve pnrruneters b., b1• and b, can now be detennincd using 

Equations 11-19 to 11-2 1 ofPS-1 1: 

Y-7 

VES009715 
(s,. s, . s, + s,. s,. s, + s,. s •. s,- s, . s, . s, -s,. s,. s,-s •. s •. s,) 

bo = det A 

= 2.t2 ~ 10• f(ls9J(s,312Xs,s74,977)+ (37o.7)(2os,mXJos,s67) + (zos,377X4,29s)(2os,m)-

(tos,s67Xs,mXs.3 12)- (2oS,377X2os.mXls9)- (s,s74,977)(4.295X370.7)J 

= 0.420 

(n ·S,·S, t S,·S,·S, t S2 ·S1 ·S, - S, ·S. · S,- S, · S,·n - S, ·S, ·S,) ,=~~~--~~~~~ ~~~ ~--~~~~ 

del A 

= 2 2 

1 

o• [{2o)(4,295Xs,s74,977)+ (I 89)(2os,mXs.J 12) + (&,3 t2X370.7l(l os,s67)- .I X I 

(s,3t2X4,295X8.3 12)- (tos,s61X2os,m)2oJ- (s,s74,977)37o.7Xts9ll 

= 0.406 

b, = (n · s, · s, + s, · S6 • s, + s, · s, · s,-s, · s, · s,-s, · s •. , -s,. s,. s,) 

• detA 

= I • [{20)(8,312Xl08.567)+ (370.7)(4,295X8,312)+ (189){370.7)(208,377)- 2.12 X 10 

(s,mXs.J12Xts9) - (20S.J77X4.295X2o} - (1os.s67X37o.7){37o.7} J 

= 0.00368 

The resulting correlation curve is as follows: 

j = 0.420 + 0.406.t + 0.0036&x> 

How do I determine the polynomial correlation coefficient? 

The polynomial C<lrrelation coefficient (r) is calculated using the paranteters S, and s.. The 

pardmcter s, is defined in Equation 11-26 as follows: · 

I I • ' 

sp = "n - 3 L (Ji, - yJ y ,., 

To calculates •• the predicted PM concentrations must be detennined by substituting the PM 

CEMS response values into the correlation curve. For example, the predicted PM concentration 

for Run 10 is 

Y-8 

YES009716 
y = 0.420+ 0.406(12)+ 0.00368(12) 2 

= 5.819 

Table Y ·3 shows the predicted PM concentrations for all of the 20 test runs used to develop the 

correlation. Substitute these values into Equation I 1-26 to calculate the value for s.: 

s, = 20 ~ 3 ((2.497 - 2.1)

2 

+ (3.71 7- 2.9f + (3.809 - 5.1) 2 + ... +(1 7.884- 185)') = 1.951 

Table Y -3. Predicted PM Concentrations (9) for Polynomial Correla tion 

Pllf CEMS PM concentration, Predicted PM 

Run no. response, rnA m&focm eoncentrolion (!)' 

1 4.9 2. 1 2.497 

2 7.6 2.9 3.717 

3 7.8 S.l 3.809 

4 9.2 4.8 4.465 

5 10.1 2.9 4.894 

6 11.06 6.3 5.358 

10 12 8.4 5.819 

II 18.2 8.4 9.024 - 12 15.7 4.9 7.698 

13 13.9 6.3 6.771 

14 19.7 13 9.842 

15 21.02 9.1 10.576 

16 22.3 13.9 11.299 

17 24.1 8.7 12.337 

19 24.7 12.6 12.688 

20 26.4 15.9 13.698 

21 29.08 16.2 15.333 

22 29.3 13.5 15.469 

23 30.5 15.9 16.221 

24 33.1 18.5 17.884 

• Calculated usmg the correlauon equanon: j • 0.420 + 0.406x + 0.00368..-'. 

The parameterS, was determined previously to be 5.095. As shown below, substituting S, and 

S, into Equation 11-33 results in a correlation coefficient of0.924, which satislies the minimum 

value of0.85 specified by I'S-1 I. 

I- ( 1.9Sl)' - 0.924 

(5.o95)' 

Y-9 

VES009717 
How do 1 calculate the confidence intezyal half range oorcentage for the nolynomjal correlation? 

The half range for the 95 percent CI for the predicted PM concentration must be determined at 

the mioim1m1 value of the parameter delta (Am;.). The expression for calculating A is provided in 

Equation 1 1-25 of PS-1 1: 

6 = C0 + 2C1.t + (2C, + c,)x' + 2C,x' + C.',x' 

Before calculating A, it is first necessary to calculate tbe parameters C. to Cs, whlcb are defined 

in Equation 1 1-23 ofPS-11 as follows: 

(S2 ·S, - S,') 

Co = D 

C, = (s, ·S2 - S, ·S,) 

D 

c, = s, . s, - s,> - D 

(ns, - S/ ) 

c, = D 

(s,. s,-ns,) 

C, = D 

(ns,-s,') c, = D 

The variable Din the denominator of these expressions is determined using Equation 11 -24. For 

this example, this calculation is as follows: 

D= n(s, ·S,-s,')+ s,(s, ·S, - S,·S,)+ s,(s, .s,-s,') 

= 2~ (s,312X5,574,977) - (208,377)') + 370.7( (208,377)(8,312) - (370.7)(5.574,577)) + 

8,3 t2((37o.7X208,377) - (8,312)' ) = 2.120 x 10• 

Substituting the values for D and for the variables S1 to S,, whlch were defoned previously, you 

find the following coefficients for determining A: 

((s,mXs,s74,977)- (208,377Y) _ 

Co 2.1 2 x l0• -1375 

Y-10 

VES009718 
((zos.m Xs,3r2) - (370.7)(s.s74,977)) c, • 2 2 0

• = - 0.1578 .I xl 

((310.7)(208,377)- (8,312)') 

C2 = 2.1 2 x 10• = o.003846 

((2o)(s,s74,977) - (s.mY) 

C, : 2.l

2 X J09 - 0.02001 

((370.7)(8,3 12)- (20}{208,377)) 

C,= 2.1 2 xro• = - 0.0005126 

((20)(8,3 12)- (370.7)') -s 

C, = 9 1360x 10 2J2 X 10 

When these values are substituted, Equation l l-25 becomes the following: 

A = 1.375+ 2(- 0.1578).<+ ((2)( 0.003846) + 0.02001)x' + 2(- 0.0005126)x' + 1.36 x W ' x' 

= 1.3 75- 03155.<+ 0.0277 x' - 0.00 I 025x3 + 136 x I o-s .<4 

Table Y -4 summarizes the results wben the PM CEMS response values for each test mn are 

substituted into the above equation. 

Table Y-4. Values for 6. 

Run PMCEMS PM coocentralion, 

no. respoJUe, mA mglacm r x> x' 6. 

I 4.9 2.1 24.01 118 576 0.3817 

2 7.6 2.9 57.76 439 3,336 0. 1727 

3 7.8 5.1 60.84 475 3,702 0.1634 

4 9.2 4.8 84.64 779 7,164 0.1162 

5 10.1 2.9 102.01 1,030 10,406 0.0995 

6 11.06 6.3 122.32 1,353 14,963 0.0906 

10 12.0 8.4 144.00 1,728 20,736 0.0884 

II 18.2 8.4 331.24 6,029 109.720 0.1202 

12 15.7 4.9 246.49 3,870 60,757 0.1084 

13 13.9 6.3 193.21 2,686 37,330 0.0960 

14 19.7 13.0 388.09 7,645 150,614 0.1204 

15 21.02 9.1 441.84 9,287 195,223 0.1162 

16 22.3 13.9 497.29 11,090 247,297 0.1092 

(continued) 

Y-1 1 

VES009719 
Ta ble Y-4. (continued) 

Ra.n Pl\lCEMS PM cooctntr.ltion, 

no. response, mA mglacm x' x> x' 4 

17 24.1 8.7 580.81 13,998 337,340 0.0985 

19 24.7 12.6 610,09 15,069 372,210 0.0959 - -

20 26.4 15.9 696.96 18,400 485,753 0.0960 

21 29.08 16.2 845.65 24,591 715,118 0. 1417 

22 29.3 13.5 858.49 25,154 737,005 0.1492 

23 30.5 15.9 930.25 28,373 865,365 0.2046 

24 33.1 18.5 1,095.61 36,265 I 200,361 0.4312 

The minimum value for 11 is 0.0884 and corresponds to Run I 0. Using I his value for 11 ... " the 

value lb r S, calculated previously, and the 1 statistic with n-3 degrees of freedom, you can 

calculale the confidence interval half range using Equation 11-27. 

ct = ,., . s,.Jt. ... = (2.11X 19SI)~o.o884 = 1.223 

The confidence interval half range as a percentage of the emission limi1 is calculated using 

Equation I I -28 as follows: 

CI 1 .• 224 Cl% = - x 100"/o = - -x 100"/o = 612% EL 20 . 

The rcsulling value for Cl% is 6.12 percent Because this value is less than I 0 percenl, the 

correlalion satisfies the crilcrion for confidence inlerval half range. 

How do I deteonine the tolerance interval half range percentage for the oolynomial correlation? 

To calculate the tolerance interval half range, tbe parameter n' mUSt be calculated using 

Equation I I -31 of PS-II. 

I I 

n' ; --= - -= 11.31 

"'""' 0.0884 

With n' rounded to the nearest inleger (II), the values for the variables u,.. and v« with n-3 

degrees of freedom are taken from Table I ofPS-1 1 and substituted into Equation 11-30 to 

determine kr: 

k,= .- ·v~r = (1.195)(1.400) = 1.673 

Using this value for kr and the value for S, calculated previously, the tolerance interval half 

range is calculated using Equation 11-29 as follows: 

Y- 12 

VES009720 
TI= kr ·S, • (1.673)(1.951)= 3.263 

Finally, to determine the tolerance imerval as a percentage of the emission limit (TI%), 

Equation I I -32 is used as foUows: 

Tl 3.263 

TI% = - X I 000/o = - - X I 000/o = I 6.3% EL 20.0 

The resulting value for TI% is 16.3 percent Because this value is less than 25 percent, the 

c01relation satisfies tile criterion for tolerance iJnerval half range. In summary, Ote polynomial 

correlation developed ltom the data satisfies all of the criteria specified in PS-11 . 

How do I develop a logarithmic correlation curve from the data? 

A logarithmic correlation curve is expressed as follows: 

y = b0 + b1/11(x) = b0 + b,x' 

Developing a logarithmic correlation entails following the same procedures used to develop a 

linear correlation. However, instead of using the PM CEMS response (x) values, the natural 

logarithm ofOte PM CEMS values (x' values) are used in the calculations. Table Y-5 

summarizes the test data, including the corresponding values of x'. 

Table Y-S. Data for Developing LogarithmJc Correlatlon 

Logarithm of PM 

PM CEMS COtS r'"'ponse PM ooncentrallon, 

Run no. response, rnA s' • In (x) mg/acm 

l 4 .9 1.589 2.1 

2 7.6 2.028 2.9 

3 7.8 2.054 5.1 

4 9.2 2.219 4.8 

5 10.1 2.313 2.9 

6 11.06 2.403 6.3 

lO 12 2.485 8.4 

ll 18.2 2.901 8.4 

12 15.7 2.754 4.9 

13 13.9 2.632 6.3 

14 19.7 2.981 13 

15 21.02 3.045 9.1 

(colllilrued) 

Y-13 

VES009721 
Table Y-5. (continued) 

Logaritbm of PM 

PM CEMS CEMS response PM tonccntntion, 

Run no. response, mA x' • In (x) mg/acnt 

16 22.3 3.105 13.9 

17 24.1 3.182 8.7 

19 24.7 3.207 12.6 

20 26.4 3.273 15.9 

21 29.08 3.370 16.2 

22 29.3 3.378 13.S 

23 30.S 3.418 15.9 

24 33.1 3.500 18.5 

Following the same procedures described previously for developing a linear correlation curve, 

you can calculate values for the parameters x' and y, using Equation 11-5; S,.,. and S,.Y, using 

Equation I J-7; b1

, using Equation 11-6; aod b0, using Equation 11-4: 

I x' = 

20(L589 + 2.028+ 2.054+ ... +3.500) = 2.79 

I 

y = 20 (2.1 + 29 .. 5.1+ ... + 18.5) " 9.47 

s •.•. = (L589 - 2.79)2 + (2.028 - 2.79) 2 + (2.054 - 2.79)' + ... +(3.50 - 2.79)2 • 5.74 

s ...... = (1.589- 2.79)(2.1 - 9.47) + (2.028 - 2.79)(2.9 - 9.47)+ ... + 

(3.50- 2.79)(18.5 - 9.47) = 47.8 

47.8 

b, = 5.74 = 8.323 

b. " 9.47 - (8.323)(2.79) = - 13.77 

The resulting equation for the logarithmic correlation curve is 

y • - 13.8 + 8.32· ltr(x) 

How do I determine the logarithmic correlation coefficient? 

Determining the logarithmic correlation coefficient requires calculating the predicted PM 

concentration (Y) for each test ron, the parameter St, and the parameters, following the same 

procedure used 10 calculate the linear correlation coefficient. The only difference is that the 

predicted PM concentrations are calculated using the logarithmic correlation equation. 

Table Y-6 shows the PM concentrations calculated using the logarithmic correlation coefficient. 

Y-14 

VES009722 
Table Y-6. Pr edicted PM Concentntions for tbe Logarithmic Correlation 

Logarithm or I' M 

PM CEMS CEMS response PM conceolrlltion, Predlctcd PM 

Run no. resjH)n.sc, rnA x' = In (x) mg/acm concentration (j)' 

I 4.9 1.589 2.1 -0.539 

2 7.6 2.028 2.9 3.114 

3 7.8 2.054 5.1 3.330 

4 9.2 2.219 4.8 4.704 

5 10.1 2.313 2.9 5.481 

6 11.06 2.403 6.3 6.237 

10 12 2.485 8.4 li.916 

11 18.2 2.901 8.4 10.382 

12 15.7 2.754 4.9 9.152 - 13 13.9 2.632 6.3 8.139 

14 19.7 2.981 13 11.041 

15 21.02 3.045 9.1 11.581 

16 22.3 3.105 13.9 12.073 

17 24.1 3.182 8.7 12.719 

19 24.7 3.207 12.6 12.924 

20 26.4 3.273 15.9 13.478 

21 29.08 3.370 16.2 14.283 

22 29.3 3.378 13.5 14.345 

23 30.5 3.418 15.9 14.679 

24 33. 1 3.500 18.5 15.360 

' Cnlculatcd using the logarithmic correlation equation: j= - 13.8 + 8.32 · /n(x). 

With the values for y shown in Table Y -6, the calculation for S, is as follows: 

S L : (

2

0 ~ 2

) (<- 0539 - 2.1)

2 

+ (3.114 - 2.9) 2 t.. .t(J5.J6 - 185)

1

) ; 2.305 

The value for s, is tbe same as was calculated for the linear correlation (5.095). Substituting the 

values for S, and S, into Equation ll-14, the logarithmic correlation coemcient is 

r = 1- (

2.30S)' = 0.892 

(5.095)' 

The logarithmic correlation coefficient (0.892) is greater than the minimum value specified by 

PS-1 1 (0.85) and is acceptable. 

Y-15 

VES009723 
How do I determine the confidence interval halfmnge oerceutage for the logarithmic 

correlation? 

The confidence interval half range for the logarithmic correlation is detcnnined using the same 

procedures used for the linear correlation. Substituting the value for SL calculated above (2.305) 

and the same value for the /-statistic (2.1 01) used for the linear correlation, Equation It-~ yields 

CJ = (2.l01)(2J05),{fa = 1.083 

The emission limit in this example is 20.0 mg/ncm. To determine Cl as a percentage of the 

emission limit, Equation 11-10 is used as follows: 

1.083 Tox 100"/o = 5.41% 

The resulting value for C!% is 5.41 percent. Because this value is less than 10 percent, the 

correlation satisfies the criterion for confidence interval half range. 

How do I determine the 10lergnce interval half range oecceorage for the logarithmic correlation? 

The tolerance interval half range for the logarithmic correlation is determined using the same 

procedures used for the linear correlation. Substituting the value for s .. calculated above (2.305) 

and the same value for the kT used for the linear correlation ( 1.629), Equation 11-1 1 yields 

TI = (L629)(2.305) = 3.75 

The emission limit in this exumple is 20.0 mglucm. To determine Tl as u percentage of the 

emission limit, Equation 11-13 is used as follows: 

3.75 

--x 100% • 18.8% 

20.0 

The resulting value for Tl% is 18.8 percent. Because this value is less than 25 percent, the 

correlation satisfies t11e criterion for confidence interval half range. 

How do I develop an exoonep)jal correlation curve from )he data? 

An exponential correlation curve is expressed as foUows: 

To develop a.n exponential con-elation curve, follow the same procedures used to develop a 

linear correlation curve. However, instead of using the PM CEMS concentrations (y) values, use 

Y-16 

VES009724 
the natural logarithm of the PM CEMS concentrations (y' values) to perform the calculations. 

Table Y -7 summarizes the test data, including the corresponding values ofy'. 

Table Y-7. Transformed Daw for Developing Exponentlal Corr elation 

Logarithm ofl'M 

PM CEMS PM concentration, CEMS response 

Run no. response, rnA mg/acm v' =In (y) 

I 4.9 2.1 0.742 

2 7.6 2.9 1.065 -- - 3 7.8 5.1 1.629 

4 9.2 4.8 1.569 

5 10.1 2.9 1.065 

6 11.06 6.3 1.841 

10 12 8.4 2. 128 

I I 18.2 8.4 2. 128 

-· .. 12 15.7 4.9 1.589 

13 13.9 6.3 1.841 

14 19.7 13 2.565 

15 2 1.02 9.1 2.208 

16 22.3 13.9 2.632 

17 24.1 8.7 2. 163 

19 24.7 12.6 2.534 

20 26.4 15.9 2.766 

21 29.08 16.2 ' 2.785 

22 29.3 13.5 2.603 

23 30.5 15.9 2.766 

24 33.1 18.5 2.918 

Following the same procedures described previously for developing a linear correlation curve, 

calculate values for the parameters lt and y', using Equation 11-5; S,.. and S,-, using 

Equation 11-7; b1', using Equation 11-6; nod b~', using Equation 11-4: 

x= ;

0

(4.947.6+7.8+ ... +33.1): 18.53 

I Y' = 20 (0.742 + 1.065+ 1.629+ ... +2.918) = 2.077 

su = ((4.9 - 18.53)' + (7.6- 18.53)2+ ... +{33.1- 1853)' ) = 1,4422 

s, .. = ((49- 1853)(0.742 - 2.077) + (7.6 - 18.53)(1.065- 2.077)+ ... + 

(33.1 - 185 3)(2.9 18 - 2.077)) " 96.58 

Y-17 

VES009725 
96.58 

b, = 1442.2 = 0.0670 

b0 = 2.on- <o.o67o)(Js.s3l = o.s3s7 

The resulting equation is expressed in the following form: 

For this example, the correlation equation becomes 

y' = 0.836+ 0.0670x 

Converting this to the fonoat that yields the predicted PM concentration in the correct units 

entails taking the exponential of each side of the equation, as follows: 

E.rp{y·)= &p(O.S357 + 0.0670x) 

y = 2.3le0.0670x 

In the resulting correlation equation, the cocfficient2.31 is simply the exponential ofb0. 

How do I dete nujnc the exnonential correlation coefficient? 

Determining the exponential correlation coefficient requires calculating the predicted PM 

concentration ()'') for each test run, the parameter S.., and the parameter s,.. following the same 

procedure used to calculate the linear correlation coefficient. Table Y-8 shows the PM 

concentrations calculated using the linear form (i.e., r = bo. + b,'x) of the exponential 

correlation equation. 

Table Y -8. Predicted PM ConcenCrations Using Exponential Correlation Equation 

Logarithm of PM 

PM CEMS mpo~ Pl\t rontenfnltlon, CEMS response Predicted PM 

Run no. ntA mglorm y' =In (y) <oncentration (y')" 

I 4.9 2.1 0.742 1.164 

2 7.6 2.9 1.065 1.345 

3 7.8 5. 1 1.629 1.358 

4 9.2 4.8 1.569 1.452 

5 10.1 2.9 1.065 1.512 

6 11.06 6.3 1.841 1.576 

10 12 8.4 2.128 1.639 . - 1 I 18.2 8.4 2.128 2.055 

12 15.7 4.9 1.589 1.887 

(contmued) 

Y-18 

VES009726 
Table Y -8. (continued) 

LogarUtlm ofl'M 

PM CEMS response, PM eoncentratlon, CEMS response Predicted Pl\1 

Run no. mA mwacm y• • In (y) concentration cr>· 

13 13.9 6.3 1.841 1.767 

14 19.7 13 2.565 2.155 

15 21.02 9.1 2.208 2.243 

16 22.3 13.9 2.632 2.329 

17 24.1 8.7 2.163 2.450 

19 24.7 12.6 2.534 2.490 

20 26.4 15.9 2.766 2.604 

21 29.08 16.2 2.785 2.783 

22 29.3 13.5 2.603 2.798 

23 30.5 15.9 2.766 2.R1R 

24 33.1 18.5 2.918 3.052 

• Calculated usmg the exponenhal correlatron equat10<1: j' ~ 0.836 + 0.067x. 

Using the vulucs for 9' shown in Table Y-8. calculate SL as follows: 

SL - (20 ~ 2) ((1.164 - 0.742)

2 + ( 1345- 1.065)1 

t . .. +(3.052 - 2.918)') = 0218 

The value for S,.. is detennined using Equation 11-15 with the values for y ' substituted for the 

values of y, as follows: 

(( 0.742- 2.077)2 + ( 1.065 - 2.077)2 + ... +(2.91 8- 2.077)

2

) 

sy = (20- 1) = 0.643 

Substitute the values for SL and SY. into Equation 11-14 to calculate the exponential correlation 

coefficient: 

(0278) 2 

I - ( )' = 0.902 0.643 

r = 

The exponential con·elation coefficient (0.902) is greater than the minimum value spec ified by 

PS-11 (0.85) and is acceptable. 

How do I detennine the confidence interval half range percentage for the exponeotjal 

correlation? 

The confidence interval half range for the exponential correlation is detennined using 

Equation 11-8, which is the same equation used to calculate the confidence interval balfrange 

Y-19 

·-

VE$009727 
for linear correlations. Because the confidence interval balf range is calculated on the log scale, 

it is designated asCI' for the exponential model. Using Equation 11-8, Cl' is calculated as 

follows: 

Cf': t,lf./·•/>' sL{f D (2.IOJ){0.278)Ho = O.JJ07 

The lower confidence limit (LCL') and upper confidence limit (UCL') arc calculated using 

Equations 11-40 aud 11-41, respectively' 

LCL' = 5'' - Cl' = 2.077 • 0.1307 = 1.946 

UCL' = 9' + Cl' = 2.077 + 0.1307 = 2.207 

Next, the confidence interval half range (CI) must be calculated on the original PM 

concentration scale using Equation 11-42 !Tom PS-11 

(e<ICL' _ eLCL') (e'·20' _ e'5>6) 

CI = 2 = 2 "' 1.045 

The confidence interval as a percentage of the emission limit is calculated using Equation Il-l 0 

as follows: 

1.045 

CI% =--X 1000/o = 5.23% 20 

The resulting value for Cl% is 5.23 percent. Because this value is less than 10 percent, the 

correlation satisfies the criterion for confidence interval half range. 

I low do I detennioe the tolerance interval balfrapge percentage for the exnonep(jal correlation? 

The tolerance interval balfrange for the exponential correlation also is calculated using the same 

procedure as is used for linear models. Because it is calculated on the log scale, it is denoted 

here as Tl'. Using Equation 11-11, Tl' is calculated as follows: 

Tl' = k1 • SL = (I .629)(0.278) = 0.453 

The lower tolerance limit (LTL') and upper tolerance limit (UTL') are calculated using 

Equationsl l-43 and ll-44, respectively 

L TL' • 'J' - Tl' = 2.077 • 0.453 = 1.624 

UTL'-"/' + Tl' - 2.077 + 0.453 = 2.530 

1 Tbe equations presented in thi.s section for exponential and power models correspood to lbc amended 

version of PS .. It, which wns not yet publisbed ot the tune this was prepared. 

Y-20 

VES009728 
Next, the tolerance inteJVal half range (TI) must be calculated on the original PM concentration 

scale using Equation I 1-45 from PS-I I 

The tolerance interval as a percentage of the emission limit is calculated using Equation II-I 3 as 

follows: 

3.74 Tl%= --x JOO%= 18.7% 20 

The resulting value for TJ% is 18.7 percent. Because this value is less than 25 percent, the 

correlation satisfies the criterion for confidence interval half range. In summary, the exponential 

model satisfies the PS-11 criteria for correlation coefficient, ronfidence interval half rdoge 

percentage, and tolerance inte"•al half range percentage. 

How do I develop a cower correlation curve from the data"! 

A power correlation curve has the following form: 

Developing a power correlation requires transforming both the PM CEMS response (x) values, 

as described previously for a logarithmic correlation, and the PM concentration (y) values, as 

was described above for an exponential correlation. The procedures used to develop a linear 

correlation are then followed using the transfonned values for x and y. Table Y -9 shows the test 

data and transformed values that must be used. 

Run no. 

I 

2 

3 

4 

5 I 

6 

10 

J1 

Table Y-9. Summary of Test Data and Transformed Values for 

Developing Power Correlation 

Logarilllm of PM Logarithm of PM 

PM CEMS respon¥e, CEMS response PM concentration, CEMS response 

DIA x' • In (x) milia em Y' • In cY) 

4.9 1.589 2.1 0.742 

7.6 2.028 ' 2.9 1.065 ' 

7.8 2.054 5.1 1.629 

92 2.219 4.8 l.S69 

10.1 2.313 2.9 1.065 

11.06 2.403 6.3 1.841 

12 I 2.485 8.4 2.128 

18.2 I 2.901 8.4 2.128 

(continued) 

Y-21 

VES009729 
Table Y -9. (continued) 

Logarithm or PM Logarithm or PM 

PM CEMS response, CEMS response PM concentr ation, CEMS mponse 

Run no- mA x' • In {x) ml!facm y' • Ln (y) 

12 15.7 2.754 4.9 1.589 

13 13.9 2.632 6.3 1.841 

14 19.7 2.981 13 2.565 

15 21.02 3.045 9.1 2.208 

16 22.3 3. 105 13.9 2.632 

17 24.1 3.182 8.7 2.163 

19 ' 24.7 3.207 12.6 2.534 

20 26.4 3.273 15.9 2.766 

21 29.08 3.370 16.2 2.785 

22 29.3 3.378 13.5 2.603 

23 30.5 3.418 15.9 2.766 

24 33.1 3.500 18.5 2.918 

Following the same procedures described previously for developing a linear correlation curve, 

calculate values for the parameters l<' and y', using Equation 11-5; S,., . and S,y, using 

Equation 11-7; b1 ',using Equation l l-6; and bu', using Equation ll-4: 

I 

X' = 20 ( 1589 t 2.02R t 2.054+ ... + 35) = 2.792 

I Y' = 20(0.742+ 1.065+ 1.629+ ... +2.918)= 2.077 

s, ... = ((1589 - 2.792)

2 + (2.028- 2.792)'+ ... +(350- 2.792)')= 5.74 

S,y = (1.588 - 2.792)(0.742- 2.077) + (2.028- 2.792)(1.065- 2.077)+ ... + 

(350 - 2.792)(2.918- 2.077) = 6.22 

6.22 

b, = 5.74 = 1.0839 

b~ = 2.077- (1.0839)(2.792) . - 0.9492 

The resulting equation is expressed in the following fonn: 

For this example, lhe correlation equation becomes 

y' = -0.9492 + 1.084x' 

Converting this to ihe format that yields the predicted PM concentration in the correct units 

entails taking the exponential of each side of the equation, as follows: 

Y-22 

VES009730 
Exp(f) = E.xp( -0.9492 + !.084x') 

j • 0.387x

1

·"" 

In thls equation, the coefticient 0.387 is the exponential of -0.9492, the value for b0 ' . 

How do I determine the power correlation coefficienfl 

To detem1ine the power correlation coefficient, the predicted PM concentration (f) must be 

calculated for each test run. Table Y - I 0 shows the predicted PM concentrations calculated using 

tbe linear form (i.e., y = b0 • + b1 'x') o f the power correlation equation. 

Table Y-10. Predicted PM Conce.ntmtions Using the Power Correlation Equation 

Logarithm of PM PM Loga.ritbm of PM 

PMCEMS CEMS response concentration, CE.MS response Predlct•'ll PM 

Run no. n!$poose, mA x• • In (x) mglacm y' • In (y) concentration (Y1

)• 

I 4.9 1.589 2.1 0.742 0.773 

2 7.6 2.028 2.9 \ 1.065 1.249 

3 7.8 2.054 5.1 1.629 1.277 

4 9.2 2.219 4.8 1.569 1.456 . s 10.1 2.313 2.9 1.065 1.557 

6 11.06 2.403 63 1.841 1.656 

10 12 2.485 8.4 2.128 1.744 

11 18.2 2.901 8.4 2.128 2.196 

12 15.7 2.754 4.9 1.589 2.035 

13 13.9 2.632 6.3 1.841 1.903 

14 19.7 2.981 13 2.565 2.281 

15 21.02 3.045 9.1 2.208 2.352 

16 22.3 3. 105 13.9 I 2.632 2.416 

17 24.1 3.182 8.7 2.163 2.500 

19 24.7 3.207 12.6 2.534 2.527 

20 26.4 3.273 15.9 2.766 2.599 

21 29.08 3.370 16.2 2.785 2.704 

22 29.3 3.37R 13.5 2.603 2.71 2 

23 30.5 3.418 15.9 2.766 2.755 

24 33.1 3.500 18.5 2.918 2.844 

• Calculated usmg the exponentaa1 correlation equatton: f' = - 0.9492 + 1.084x' 

Y-23 

VES009731 
The parameters. is calculated using Equation 11-9 with the predicted values (9') given in the 

above table, and the parameters,. is calculated lbllowing the same procedure used for the 

exponential correlation coefficient, as follows: 

s, = (20~ 2) ((0.773- 0.742)

2 

+ ( 1.249- 1.065)

2 

+.. .+(2.844- 2.918)2

) = 0249 

((0.742 - 2.077) 2 + ( 1.065- 2.077)' .f-... +(2.918- 2.077) 2

) 

s,.. = 20 - l 2 0.643 

Substituting the values for SL and Sy. into Equation 11-14, the power correlation coefficient is 

(0249)

2 

r = I - ( )' = 0.922 

0.643 

The correlation coefficient for the power model (0.922) is greater than the minimum value 

specified by PS-I I (0.85) and is acceptable. 

How do I dctennine lhe contjdence interval half range oercentage for the nower correlation? 

The confidence interval half range for the power correlation is detennined using the same 

procedures used for the exponential correlation. Because the confidence interval half range is 

calculated on the log scale, it is designated as CI' for the exponential modeL Using 

Equation 11-l!, CJ' is calculated as follows: 

Cl' = (2.101X0.249)~ = 0.1 17 

Tbe lower confidence limit (LCL ')and upper confidence limit (UCL') arc calculated using 

Equations 11-40 and 11-41, respectively 

LCL' = y' -Cl' - 2.077-0.117 = 1.96 

UCL' = y' + Cl' = 2.077 + 0.1 17 = 2. I 94 

Next, the confidence interval half range (CI) rnu~t be calculated on the original PM 

concentration scale using Equation 11-42 from l'S- 11 

The confidence interval as a percentage of the emission limit is calculated using Equation 11- 10 

as foUows: 

Y-24 

VES009732 
0.936 

Cl% ~ 2o X 100"/o = 4.68% 

The resulting value for CI% is 4.68 pcrcenl. Because this value is less than 10 percent, the 

correlation satisfies the criterion for confidence interval half range. 

How do I determine the tolerance jnteryal half range Percentage for the oower correlation'? 

The tolerance interval half range. for the power correlation also is calculated using the same 

procedure used for the exponential model. Because it is calculated on the Jog scale, it is denoted 

here as Tl '. Using Equation 11-11, Tl' is calculated as follows: 

TI' = kT· SL = (1.629)(0.249) = 0.406 

The lower tolerance limit (L TL') and upper tolerance limit (UTL') are calculated using 

Equationsll-43 and 11-44, respectively 

LTL' = 9'- TI' , 2.077- 0.406 = 1.671 

UTL' = y• + Tl' = 2.077 + 0.406 = 2.483 

Next, the tolerance interval half range (TI) must be calculated on the original PM concentration 

scale using Equation 11-45 from PS- I I 

(eun: - e=·) 

Tl = - 2 

The tolerance interval as a percentage of the emission limit is calculated using Equation 11-13 as 

follows: 

3327 

Tl% = 20 X 100"/o = 16.6% 

The resulting value for Tl% is 16.6 percent. Because this value is less than 25 percent, the 

correlation satisfies the criterion for conJidcuce interval half range. In summary, the power 

mode.! satisfies the PS-1 1 criteria for correlation coefficient, confidence interval half range 

pe.rcenlage, and tolerance interval half range percentage. 

Whjch model is the best fit for the data? 

Table Y -1 1 summarizes the results of the analysis for all of the correlation models. As indicated 

in the table. all five models met the acceptance criteria specified in Section 13.2 ofPS-11. The 

linear model bad the highest correlation coefficient and would be considered the best lit for the 

data. However, any of the models could be used to show compliance. The complete output for 

the spreadsheet for this example is provided in Appendix A. I. 

Y-25 

VES009733 
Table Y-U. Summary of e.~ ulls for Facility A Correlation Test 

ConOdcnce Tolerance 

Correlation lnterva.l half inten•al half 

lltodel type Corrclation equation coeffic\eat (r) range range 

Acceptance Criteria for PS.Il ~ 0.85 s IOOh s25% 

Linear y • -0.621 + 0.544x 0.927 4.49% I 15.6% . 

Polynomial y • 0.420 + 0.406x + 0.00368 x• 0.924 6.12% 16.3% 

Logarithmic y a -13.8 + 8.32Ln(x) 0.892 5.41% 18.8% 

Exponential 9- 2.31 e"·""" 0.902 5.23% 18.7% 

Power Y a 0.387x1

-"' 0.922 4.68% 16.6% 

Y.2 EXAMPLE 2: HOW DO I DEVELOP A CORRELATION WITH PAIRED SAMPLING 

TRAIN DATA? 

Y .2. I Correlation Test Scenario 

A correlation test is perfonned on an incinerator at Facility B. The incinerator is equipped with 

an in situ PM CE!MS that responds to PM concentrations at actual conditions. The emission 

limit, wl1en converted to units that are consistent with the PM CEMS sampling conditions, is 

24.3 mgfacm. Fifteen test runs are perfom1cd using Method 5i with paired sampling trains 

(Train A and Train D). No problems were reported for the test. Table Y -1 2 summarizes the 

correlation test data. Figure Y-2 shows a plot of the data. 

Table Y-1 2. Summary of Correlation Test Data for Facility B 

PM CEMS PM coa ten tration, mgfoom 

Run 11(1. rl'Sponse, mA Sampling Train A Sampling Traln B 

I 6.4 1.3 1.2 - 2 7.5 2.3 2.5 

3 11.2 2.5 2.8 

4 16.4 7.7 7.9 

5 29.9 12.9 12.5 

6 222 7.8 7.3 

7 19.5 5.& 5.9 

8 27.4 11 .7 12.0 

9 32.8 20.4 20.9 

10 26.5 20.9 21.3 

II 33.5 22.3 22.6 

12 28.1 18.4 17.58 

13 9.6 3.7 3.9 

14 5.3 4.2 4.3 

IS 18.1 6.8 7.1 

Y-26 

VES009734 
25 

• 

E 20 

" 

• • 

a. * E 

c 15 

0 

"' 1': • • 

3 10 ·- c 

8 

:E 5 

a.. 

• • * • • • • • • 0 

0 5 10 15 20 25 30 35 40 

PM CEMS Response, rnA 

Figure Y-2. Plot of co.-relation test data for Facility n. 

Should I averoge the two PM concentrations for each test run? 

You should nm average the rwo PM concentrations for any of the test runs. Instead, you should 

treat each PM concentration as a separate data point, as specified in Section 12.3 of PS-1 1. This 

means that for each run, the data set includes two PM concentrations and one PM CEMS 

response value. In other words, the correlation should be developed as if there were 30 test runs 

with one PM concentration for each run. Table Y-1 3 shows the data in the f01111 that should be 

used for developing tlte correlation. 

Before proceeding with the correlation, paired sampling train data should be checked for 

precision and bias. Guidance on precision and bias is provided separately. For this example, it 

is assumed that the data meet the criteria of precision and bias and are acceptable. 

Y-27 

VES009735 
Table V-13. Paired Sampling Train Test Data in Format 

for Developing Correlation 

PM CEMS Response, PM Concentration, 

Run No. mA mg/acru 

lA 6.4 1.3 

1B 6.4 1.2 

2A 7.5 2.3 

2B 7.5 2.5 

3A 11.2 2.5 

38 11.2 2.8 

4A 16.4 7.7 

4B 16.4 7.9 

SA 29.9 12.9 

58 29.9 12.5 

6A \ 22.2 7.8 

6B ' 22.2 7.3 

7A 19.5 5.8 

78 19.5 5.9 

SA 27.4 I 1.7 

88 27.4 12.0 

9A 32.8 20.4 

9B 32.8 20.9 

lOA i 26.5 20.9 

lOB 26.5 21.3 

IIA 33.5 22.3 . - -·- liB 33.5 22.6 

12A 28.1 18.4 

128 28.1 17.8 

13A 9.6 3.7 

138 9.6 3.9 

14A i 5.3 4.2 

148 53 4.3 

ISA 18.1 6.8 

158 18.1 7.1 

V-28 

VES009736 
What are the results of the correlation analysis? 

Table Y -14 sununarizes the results of the analysis. The complete output for the spreadsheet for 

this example is provided in Appendix A.2. 

Table Y-14. Summary of Results for Facility B Correlation Test 

I 

Confidence Tolenmcc 

Correlation intern! half interval half 

Model type Correlation equation coefficient (r) range ra nge 

Acceptance Criteria for PS-11 ~ 0.85 s JO% s 25% 

Linear j = -3.24 + 0.672x 0.895 4.98% 20.0% 

Pol}'ltomial f= 2.85 - 0.173x + 0.0220 >' ! 0.9 19 5.91% 18.0% ' ; 

Logarithmic y; -17.7 + 9.81Ln(x} 0.822 6.37% ; 25.6% 

Exponential f = 1.43e0

·

0

"'' 0.921 3.80% 15.9% 

Power f= 0.201x 7 0.892 4.41~-'o 18.7% 

Which model should I use? 

All of the models except the logarithmic model satisfy the acceptance criteria for PS-11 . The 

logarithmic model fails to meet the minimum allowable correlation coefficient of0.85. In 

addition, the tolerance interval half range percentage for the logarithmic model exceeds dte 

allowable percentage of25 percent. Based on the correlation coefficient, the exponential model 

is the best fit. However, the linear, polynomial, and power models could also be used. 

Y.3 EXAMPLE 3: HOW DO I DEVELOP THE INITIAL CORRELATION'? 

Y.3.1 Correlation Test Scenario 

A correlation test is performed on a hazardous waste incinerator at Facility C. The incinerator is 

equipped with an extractive PM CEMS that extracts and heats the samples to 121 °C (250°F). 

The incinerator is subject to 40 CFR 63, subpart BEE, and must meet an emission limit of 

34.0 mg/dscm corrected to 7 percent oxygen. Fifteen test runs are perfonued using a Method 5 

sampling train. The average stack gas pressure, moisture content, and oxygen content during the 

test were 760 millimeters of mercury, 12 percent moisture, and 10 percent oxygen, respectively. 

No problems were reported for the test, and all data points will be used for the correlation. 

Table Y-15 summarizes the correlation test data. Figure Y-3 shows a plot of the data. 

Y-29 

VES009737 
Table Y-15. Summa ry ofCorrc.latioo Test Data for FacllJty C 

Ru.o no. PM CEMS rt-sponsc, mA PM co.ocentralion, mg/ocm 

I 5.8 3.5 

2 7.6 9.0 

3 7.8 7.8 

4 16.4 25.2 

5 i 

' 10.1 16.6 

6 11.1 14.3 

7 12.0 19.6 

8 13.2 20.1 

9 15.7 23.4 

10 17.6 26.7 

I I 21.7 26.8 

12 ' 21.0 28.9 

13 I 22.3 25.7 

14 24.1 26.2 

15 24.7 26.2 

------------------------------------. 

~ 30 

a, 25 

E 

g 20 

g 15 c 

8 

c 10 

8 

::E 5 ... 

• • 

• 

• • •• • • • • 

• • 

0+-----r---~~--~----~----~----~ 

0.00 5.00 10.00 15.00 20.00 25.00 30.00 

PM CEMS Response, mA 

Figure Y-3. Plot of correlation test data for Facility C. 

Y-30 

VES009738 
Do I need to con yen the emjssjop ljmit to other units for detennining conmliance wjth PS-l l? 

As indicated in Section 13.2 of PS-ll, you must couven the emission limit to units that are 

consistent with the measurement conditions of your PM CEMS. For this example, you Jnust 

conven the emission limit, which is specified in units of mgfdscm at 7 percent o:>:ygen, to match 

the conditions under which the PM CEMS is operating. The following calculations show this 

conversion: 

where 

~ T, 

T"' 

p•d 

P, 

me 

%0, 

T. P,ld 1 13.9 

= Co X I'.rd X P,, X ( ~-;:) X (z0.9 - %0z) 

= PM concentration at standard conditions, corrected to 7 percent oxygen 

• stack gas temperature at actual conditions, °K 

• ~tack gas temperature at standard conditions; 293°K 

= s~dc k gas pressure at standard conditions= 760 mm of Hg 

• stack gas pressure at actual conditions, mm ofHg 

"' stack gas moisture content, % 

• stack gas oxygen contem, %. 

(121 + 273) ( 760) [ I ] 13.9 

34.0 • c. X 293 X 760 X 1- /! X {20.9- 10) 

34.0 = 1.95C. 

c.= 17.45 

The emission limit in units of concentration at the operating conditions of the PM CEMS is 

17.45 mg!acm. This is the value that must be used for determining the confidence and tolerance 

interval half range percentages. 

What are the rqu!ts of the correlation analysjs? 

Table Y -16 summarizes the results of the correlation analysis. Only the logarithmic model 

satisfies all of the acceptance criteria for PS-11 . The correlation coefficient for the exponential 

model (0.812) f.1ils to meet the minimum allowable value of0.85. The contidence and tolerance 

interval half range percentages for the linear model (11.3 and 35.6 percent, respectively) exceed 

the limits of I 0 and 25 percent, respectively. For the power model. the confidence interval 

(1 4.3 percent) and tolerance interval (46.4 percent) half range percentages also exceed the 

maximum allowable values. 

Y-31 

VES009739 
Table Y-16. Summa.ry of Results for Facility C Corrclutlon Test 

Confidence ToleriUice 

Correlation lnttr>•al half interval half 

Model type Corre.Lttion equation coefficient (r) range range 

Acceptance Criteria for PS.II ~ 0.85 ~ 10"/o £ 25% 

linear ja 2.27 + 1.15x 0.898 11.3% 35.6% - Polynomial j • -1 8.4 + 4.30x · O.lOLx' 0.987 5.56% 13.4% 

Logarithmic ~ -24.4 + 16.8Ln(.<) 0.959 7.32% 23.1% 

Eil<ponential j• 5.26e'"'"" 0.812 19.9% 66.5% 

Power v• o.747x.'·,. 0.908 14.3% 46.4% 

The polynomial model bas the highest correlation coefficient (0. 987) and meets d1e criteria for 

confidence and tolerance interval half range percentages. However, Section 12.4(3) of PS-11 

specifies that the maximum value for a polynomial correlation curve must occur at a PM CEMS 

response value that is al least 1.25 times the highest PM CEMS response value used to develop 

the correlation equation. In this example, the maximum PM CEMS response value is 24.7 rnA, 

and the maximum value for the correlation curve must occur at a PM CEMS response value of at 

least 32.4 rnA (24.7 • 1.25 ~ 30.88). However, the maximum value for the polynomial 

correlation curve occurs at 21.2 rnA, as is evident from Figure Y-4. Therefore, the polynomial 

model carUlOt be used. The complete 011tput for the spreadsheet for tlus ex amp I~ is provided in 

Appendix A.3. 

Plot of Polynomial Regression Curve 

E 35 

-;. " 30 y=.().10tx'•4296x- t8.37t 

E 25 

t! :8 20 i 15 

• 10 I! 8 5 

:I 0 -!--- ~~~~---.....-~~~~~~~---~ ... 0.00 5.00 10.00 15.00 20.00 25.00 30.00 

PM CEMS Response, rnA 

Figure Y-4. Plot of polynomial correlation curve for Facility C. 

Y-32 

VES009740 
Y.4 EXAMPLE 4: WHAT CAN I DO IF MY PM CEMS FAILS THE CORRELATION 

TEST? 

Y.4.1 Correlation Test Scenario 

A correlation test is performed on a boiler at Facility D. The boiler is equipped with an in situ 

PM C.'EMS that responds to PM concentrations at actual conditions. The emission limit, when 

converted to units that are consistent with the PM CEMS sampling conditions, is 18.5 mg/acm. 

The correlation test consists of 15 test runs using a single Method 5 sampling train. Table Y -17 

summarizes the correlation test data. Figure Y -5 shows a plot of the data. 

Table Y-17. Summary of Test Data for Facility D 

PM CEMS response, PM concentration, 

Run no. mA mglacm 

I 6.7 3.1 

2 7.6 7.4 

3 7.8 2. 1 

4 9.2 3.7 ·-- ... -·- 5 10.1 6.1 

6 11.1 2.8 -· -- 7 12.0 4.9 

8 18.2 8.3 

9 13.7 4.5 

10 14.6 8.2 

I I 19.7 13.0 

12 18.0 11.1 

13 20.3 11.9 

14 20.2 10.6 

15 19.7 9.2 

Y-33 

VES009741 
14 

• E 12 

w • • Dl 10 • E • .: 

0 8 • • 

~ • c: 6 • c: 

B 

4 • • 

0 • (,) • • :I! 2 • ... 0 

o.o 5.0 10.0 15.0 20.0 25.0 

PM CEMS Response, mA 

Fi2u re Y -5. Plot of correlation test data for Facility D. 

The results of the correlation analysis are sumnmrized in Table Y -18. None of the models satisfy 

a ll of the correlation criteria specified in PS- 11. Although all of the models meet tlte criteria for 

confidence and tolerance interval half range percentages, only the polynomial model has an 

acceptable correlation coefficient. However, the polynomial correlation curve has a minimum x 

value of 7.05 that is greater than the minimum PM CEMS response (6. 7 mA for Run I) value 

used to develop the correlation curve. As stated in Section 12.4(3Xiii) ofPS-1 I, a polynomial 

correlation curve cannot be used if it has o minimum value that exceeds the minimum PMCEMS response value used to detennine the correlation curve. Consequently, the polynomial 

correlation also is unacceptable. 

Table Y-18. Results of l nltlal Correlation for Facility D 

I Correlation 

Confidence l Tolcroncc 

Interval half intervnl half 

Model type Correlation equation ' coefficient (r) range l range 

Acceptance Criteria for PS-1 1 ~ 0.85 ,; 10% ~ 25% 

Linear I j•-1.20+0.598.• 0.841 5.80"/o 18.3% 

Polynomial ' j• 6.19 . 0.615 + 0.043!U' 0.855 7.68% 18.2% 

Logarithmic j• -11.8 + 7 .37Ln(.r) 0.801 6.41% 20.2% 

Exponential j= 1.69e~"'"' 0.806 6.41% 2 1.3% 

Power j= 0.308x'·" 0.782 6.76% 22.6% 

Y-34 

VES009742 
Several options are available for meeting the requirements of PS-I I, including the following: 

• Re,peat dte correlation test: If you suspect problems with the manual sampling or the 

operation of the PM CEMS, you may want to discard all of the test data and conduct 

a new correlation test. ln such a case, there must be a valid reason for rejecting the 

entire data set. For example, you determine that several of the PM samples were 

contaminated. You still must report all of the rejected data. 

• Conduct additional test runs. You may want to conduct a limited number of 

additional test runs (e.g., five additional rwts) if it appears that some of the data are 

outliers, problems are suspected for some of the runs, or the range of PM 

concentrations measured was not adequate to fully characterize the correlation. In 

this case, you can develop the correlation based on all of the data (i.e., the initial IS 

runs plus the additional runs), or you can reject some of the test rwts, provided you 

meet the requirements of Section 8.6(3) ofPS-11, which allows you to reject up to 

five test runs without explanation and additional runs if the basis for rejecting them is 

specified in the test method, in PS-1 1, or in your quality assurance plan. 

• Relocate the PM CEMS and conduct a new correlation test: If the sampling location 

is not ideal (e.g., PM concentrations are stratified across the stack at the sampling 

point), you may want to consider relocating the PM CEMS to another location that is 

more suitable. In this case, you must conduct a full correlation test (i.e., at least 15 

valid test runs). 

• Replace the PM CEMS: Certain types of PM CEMS may not be suitable for specific 

stack gas conditions (e.g., an in siru light scattering instrument in a wet stack). If that 

is the case, you may decide to install a PM CEMS that uses a different measurement 

principle. If so, you must conduct a full correlation test with the new instrument. 

• Consider alternative solutions: Section 12.4( 4) allows you to petition the regulatory 

authority for approval of alternatives to the con-elation requirements of PS-11. 

For this example, we will assume that you decide to conduct five additional test runs. These 

rwtS are identified as Runs 16 to 20. Table Y-19 shows the results of the additional runs. Figure 

Y -6 shows the results of all20 test runs. 

Table Y-19. Results of Additional Test Runs 

i PM CEMS response, PM concentration, 

Run no. ' mA mglacm 

16 5.2 2.4 ..... ,... .. -- 17 8.6 4.8 

18 14.1 6.2 ·-- .......... ·- 19 14.5 7.0 

20 17.3 8.6 

Y-35 

VES009743 
14 

E 

:;: 12 

a, 

E 10 

s· s 

; 

"' ~ i 6 

~ 4 

0 

(.) 

::E 2 

Q. 

0 

• 

• • 

Run2 • -- ~. • -- -· • - • 

. • • 

• 

• . • .. -· ···-

0 5 10 15 20 25 

PM CEMS Response, mA 

Figure Y-6. Plot of original 15 runs plus 5 additional r uns for Facility D. 

Table Y -20 shows the results of the correlation analysis on aU 20 test rutlS. Tbe analysis shows 

that the linear and polynomial models meet the con-elation criteria ofPS- 11 , and no further 

analyses or testing is required. With the additional data included, the minimum value for the 

polynomial correlation curve now occurs at 4.2 mA, which is less than t.he minimum PM CEMS 

value used«> develop the curve (5.2 mA). However. a review of Figure Y-6 indicates that Run 2 

(7.4 mA, 7.4 mgfocm) may be on outlier. You may decide to repeat the correlation analysis with 

Run 2 excluded. Table Y-21 shows the results of the onalysis without the dat<l from Run 2. In 

the revised analysis, all five models meet the criteria ofPS-11. The correlation coefficients for 

the linear and polyt1omial models also are considerably greater for the revised analysis. You 

have the option of using the results of either analyses. In either case, the polynomial model 

provides the best fit; however, any of the models that satisfy aU of the correlation criteria in PS· 

II can be used. 

Table Y-20. Results ofRevtsed Correl.ation Analysis with All20 Runs 

Confidence Tolerance 

Correlation Interval balf interval half 

Model type Correlation equation coefficient (r) range range 

Acceptance Criterio for PS-11 > 0.85 s 10% s 25% 

Linear f • .().905 + 0.573x 0.861 4.25% 14.7% 

Polynomial I J• 3.94-0.263x + 0.031 4X' O.S15 5.47% 14.4% 

Logarithmic J• -9.79 + 6.S7Lo(x) 0.816 4.82% 16.7% 

Exponential js 1.70c .. "1

"' i 0.840 .52'~ 16.3% 

Power ja OJ67.rl.ll I 0.82.4 4.73% 17.0% 

Y-36 

VES009744 
Table Y-21. Results of Revised Correlation Analysis with Run 2 Excluded 

I I Correlation 

Confidence Tolerance 

I interval half Interval half 

' ~~del type i Correlation equatlo~ .. I coefficient (r) range range 

1 - Acceplance Criteria for PS-11 2 0.85 s I 0"/o !25% 

Linear f= -1.88 + 0.629.< 0.912 3.63% ' 12.4% 

Polynomial f= 2.19 - 0.067x + 0.026x' 0.92 1 4.54% 12.1% 

Logarithmic f= - 11.7 + 7.24Ln(x) 0.867 4.40% 15.0"/o 

Exponential j = 1.41ea11"' 0.909 3.63% 12.6% 

Power f= 0.25Jx'·" 0.894 3.90% 13.6% 

The complete output of the spreadsheet for lhe initial correlation, the correlation with all five 

additional test runs, and the correlation with Run 2 excluded are provided in Appendices A.4 

through A.6. 

Y-37 

VES009745 
Appendix A.l 

Complete Results of Example No. 1 -

Initial Correlation Test 

VES009746 
CORRELATION TEST PM CEMS AND REFERENCE METHOD TEST DATA 

PMCEMS PM concentration 

resoonse ,;,alaan 

Run X y 

1 4.9 2.1 

2 7.6 2.9 

3 7.8 5.1 

4 9.2 4.8 

5 10.1 2.9 

6 11.1 6.3 

10 12.0 8.4 

11 18.2 8.4 

12 15.7 4.9 

13 13.9 6.3 

14 19.7 13.0 

15 21.0 9.1 

16 22.3 13.9 

17 24.1 8.7 

19 24.7 12.6 

20 26.4 15.9 

21 29.1 16.2 

22 29.3 13.5 

23 30.5 15.9 

24- 33.1 16.5 

PS-11 Example 1-R 1D-27-04Jds 

FaciiHy: 

Location: 

Emission Unit: 

Test Datos: 

Emi ssion limit: 

20 

"I 16 c 

.2 14 

e 12 

1: 

! "l 0 8 

v 

l 6 

4 

2 

0 

Faci!HyA 

Anytown USA 

Boiler No. 3 

March 6 to 15, 2004 

• 

Value 

20 

•• 

• • 

• 

• 

UnHs 

mglacm 

• • 

• • 

• • 

0.0 5.0 10.0 15.0 20.0 

PM CEMS Response 

Test Data 

• 

• 

• 

25.0 

• 

• • 

• 

30.0 35.0 

Pago 1 o1 1 

10/2912004 

VES009747 
CALCULATIONS FOR LINEAR CORRELATION 

Facility Emission Unit: Boiler No. 3 : Facl!!ty A 

Location: Anytown USA Test Dates: March 6 to 15, 2004 

PM CEMS I PM """"""'"" 

,...Ar ,.... 11.ATED STAl 

y I (.,xl2 I CYrYl2 I (x;x-)(y,ylj Y" (y'\-

-1-- 4.9 2.1 185.86 54.32 100.48 2.047 0.003 

2 7.6 2.9 119.53 43.16 71.83 3.517 0.381 

3 7.8 5.1 115.20 19.10 46.90 3.626 2.172 

4 9.2 4.8 87.10 21.81 43.59 4.388 0.169 

5 I 10.11 2.91 71.121 43.16) 55.401 4.8781 3.914 

1 6 I 11.06 6.3 55.85 10.05 23.69 5.401 0.808 

10 ..l 12 8.4 42.68 1.14 6.99 5.913 6.18" 

18.2 8.4 0.11 1.14 0.36 9.289 0.79u 

15.7 4.9 8.03 20.88 12.95 7.928 9. 6~ 

6.31 21.461 10.051 14.691 6.9471 0.41' 

131 1.361 12.461 4.1 21 10.1051 8.37! 

21.02 9.1 0.14 -0.92 10.8241 2.97: 

16 I 22.3 13.9 19.62 16.69 11.521 

1 17 24.1 8.7 30.99 0.59 -4.29 12.5011 1"·""" I 

19 24.7 12.6 38.03 9.80 19.30 12.8281 0.052 I 

20 26.4 15.9 61.89 41.34 50.58 13.753 • ono I 

21 29.08 16.2 111 .24 45.29 70.91 

22 29.3 13.5 115.93 16.24 43.3! 

23 30.5 15.9 41 .34 76.9' 

24 33.1 18.5 81.54 131.54 

PS-11 Example 1-R 1G-27-04.JCis Unearcalcs 

15.9861 

17.401 

( 

1.207 

Page 1 of 1 

1012912004 

VES009748 
~SULTS OF LINEAR CORRELATION 

Facility: Facility A 

Location: Anytown USA 

Variable Equation 

n Number of data points = 

X x- - 1/n"(Sum of (X,)) • 

s .. Su = Sum((X1·x-)'2) = 

y y- = 1/n"(Sum of (y1)) = 

Stt s,. = Sum((y,y-)•2) = 

s,.. S")' = Sum((xrx-Xyry-)) = 

bo bo = y--b,x- -

b, t>, = s..;s .. = 

s._ s._ = SORT(1/(n-2XSum(y;'-y.J"2)) -

Y'- y' at mean x value • 

" t, from table • 

Cl Cl = 1(5,. "SQRT(1/n) = 

EL Emission Limn = Cl% Ct% •CIIEL'100% = 

n' n'• n = 

v, v1 from table = 

u •. Un• from table = 

kr kl = Un'• vf = 

Tt Tl = ktS.. = 

TI% TI% • TIIEL"IOO% • 

Sy S,. = SQRT(S,.,i(n-1)) - ,.. r - 1-(s._•21Sy•2J = 

r r = SORT((1-$L•21Sy'2)) = . lndk:8ies c:onelation coefficient is undefined . 

P5-11 Example 1-R 10-27.Q4.xls 

Value 

20 

18.53 

1,442 

9.47 

493 

785.2 

.Q.621 

0.544 

1.910 

9.470 

2.101 

0.897 

20.0 

4.49% 

20 

1.384 

1.177 

1.629 

3.111 

15.6% 

5.095 

0.859 

0.927 

Emission Unit: Boiler No.3 

Test Dates: March 6 to 15, 2004 

(CorrelatlOil equation: - r=-- .0.621 + 0.544 X I 

s ....-voonoou • -fA• · ........ -................. Criteria for PS ··- .. Criterion Actual Allowable Acceptable? 

Correlation coefficient 0.927 ?. 0.85 yes 

Confidence interval 4.49% .:;. 10% yes 

Tolerance interval 15.6% .:;_25% yes ___ - ----·-

• Indicates correlation coefficient Is undefined. 

Plot of Unear Regreuton Una 

y = 0.5445x • 0.6207 j :,- - ---:-------- -

! 10 

~ u 5 

~ 

• 

0~--~--~--~----------~--~ 0.0 5.0 10.0 16.0 20.0 25.0 30.0 35.0 

PM CEMS Rupon10 

PS-t1~Test 

~SpflvWIIon 2-6 10125/04 

Linear Summary Linear Summary 

VES009749 
Fa clllty: Facility A 

Loeatioo : Anytown USA 

I PM CEMS I Pt.looo · -.. I!Sf)C!!!S! !!lJjocm 

Run X y 7 

2.ii 

CALCULATIONS FOR PO!.YNOMIAl CORRELATION 

E.mlufon Unft: Boiler No. 3 

Test Dates: March 6 to 15, 2004 

.:AlCUI.ATBl: 

.£ x• "" --;?; IY"·vl' lvt..-l'Tc. 

118 576 10 ! 17 

439 3.336 22 1681 ..).1171 UOOf U, , 

475 3,702 40 310 I 3.809 I 1567 0,11 

779 7.164 .. • • ... - ••• - .. 

1.030 10,406 1 1 353 14.963 70 771 5.358 U .OOI U..Ul 

I 1u I 121 ~ 144 1 728 20 736 101 1.210 5.819 6.660 0.01 

I 11 I 18.21 8 .41 331 6029 109.720 153 2.782 9.024 0.390 0.12( 

•• ... • • 2<16 3870 60757 77 1,208 7.693 7.828 0.1084 

26.4 

.&.1 29.00 

22 29.3 ..... 30.6 

33.' 

PS-11 E""mple 1-R 10.21~ m 

193 2.686 37,330 88 1,217 6.771 0.222 0.0960 

388 7,645 150.614 256 5045 9.842 9.9n 0.1204_ 

... ---- ·-- 223 191 4,021 10.576 2.178 

"'1"8; 

--.. j 

[iii 

' ..... ,., ..... 865,365 

1,200,361 

~<:ales 

1.vvo v ,..,..,d I n.g\JI , ,,.,. I 

1.&49 0.0960 41 .341 1.271 

>, .;J.;J.;J V,~52 0.14",. AI' ""' t. r:r 

·.469 3.879 0.1492 16.241 1.5! 

1.791...1. 16.221 0.103 0.2046 41 ."' • ' 

17.884 0.379 0.43121 81 .! 

Pogo1oll 

1012912004 

VES009750 
BQU! D Of PQLXNOMLAI COB8ELADON 

Facility: Fael!llty A 

Loc.a:Uon; Anytown USA 

Eminion Unit: Sol., Ho. S 

Tut Oatu:: MMch tto 11-, lOOt 

IV- £4tAi1'on ___y~ (c:_o.ntdon equation: y • D.A20 • 0..401 x • o..tt311 x1 I 

s 

s 

•• 

---c;-

c 

c 

c 

c 

t • &.m. CJI) • 

! • ,..., (>d2) = 

I •S... 

S$•Slm 

SO -~ 

~ n• 

del A• ~S18382·~s:l 

., (S5S2S4•S·tS3S7•S2SOS3-$7S2S2.S3S3S5- A• 

1>1. 

... 0•, s,• 

!I: A• 

CO•( •t·~ )10 

C1 •IS352.S1 ~li0 =-

C2 

C3' 

!S1S2.nS3WO =-

~ .. (nS2· 0= 

371 

1.312 

208.377) 

~ 295 

.., 

O.QOI 

0.40E 

o.0036a 

t.951 

2 120E•09 

1.37S 

.o.1s1e 

•e "rtom !able • 4·. •v 

EL Emla lon Lhlh • 

,.... CI•I,'S, 'SC¥1(dona~ln}• 

v, 

"• 

-.;-" 

s., 

, , __ ? 

.., __ _ 

I 1.251<,.. 

"' 

Sr • IQfi(Sum , 

i'· ,. 

v1tom 

.. -

kl • un' '\11' • 

Tl • kl'\t' 

·II'Ei 1100 

'SwndiY'}) 

' u '"'-- •• 

.. 

1.40( 

1.19e 

1. 

5.0! 

085J 

0.024 

"**"'afT 

·!55.c!! 

41.38 

PSo-11 ~ 1-R to.v-oc..., ..._,.&........,. 

Stnmaly of ~ce CtfteM tot PS..11 - cno.mo ....... - . 1; Zl? 

Cor:rell5on ooefticin 092t ~= ... Coae<lercc ~ 812% ~'"""' ... T~~ 16.$% ~ ..... ... lrd:::all::es cortii!Miion ~is~ 

Check tct COft'efadon Curve Wnfmtft/Ma:dllltn 

ComaCion cunot rinrnum pdnt ~ .. t.lnimt.Jnol slfow:abte x .,.. 

1!'1 ccrrelstion ane miniMum < mlrlhft!n'l • "Value? _}'Of ·~ 

CorrWltion curo,e maxifl'Ull l'l*tt NA 

ExtraDOia1ion litrit for x (125% of ~m ll value NA 

ts correlatiOn curve maximum > extrapolaion !liM? NA 

Ptot ot PotynomJal Regresalon Cur.te 

20 

.. y • O.OCX37~ • 0.4057X • 0 "204 ! ~~ • 

• • • 12 

• 10 

! : • • • 

• l •· • • 

2 • 

0 

0.0 6.0 10.0 15<> 200 25.0 :10.0 35.0 

PM CEMS Rtii)Onte 

" H COrNIIIIIm l C'lll I 

c..v-.~~r~ Sp!Nidlli-HI I 

v~u «11Zto94i 

111\12004 

VES009751 
CALCULATIONS FOR LOGARITHMIC CORRELATION 

Facility: Facility A Emission Unit: Boiler No. 3 

Location: Anytown USA Test Oates: Mareh 6 to 15, 2004 

PMCEMS PM 

Run • 

-- 4.9 

7.6 

11 

1 

2.754 4.9 0.001 20.8851 0.1741 9.1521 18.0831 

I 13 I 13.9 2.632 6.3 0.026 10.0491 0.507 8.139f 3.3821 

14 j 19.7 2.981 13 0.036 12.461 - --- . . - .. - ---

21.02 3.045 9.1 0.064 0.137 . 

22.3 3.105 13.9 0.098 19.625 

I " I 24.1 3.182 8.7 0.152 0.5931 ..(] 

I 19 I 24.7 3.207 12.6 0.112 9.7971 1.2991 M M. O 273 15.9 0.232 41.3451 ° ~--

14 

2 

30.5 

33.1 

370 16.2 0.334 

378 13.5 0.34: 

418 15.9 0.39 

500 18.5 ~ 

;:;a; 

16.241 

41.345 

81.541 6.391 

;.867 

1.67 

).714 

14.6791 1.490 

i.858 

PS.11 Example 1·R 10·27-04.xls Logarithmic Cales 

Page 1 of 1 

1012912004 

VES009752 
RESULTS OF LOGARITHMIC CORRELA TlON 

Facility: Facility A 

Location: Anytown USA 

Variable Eauation 

n Number of data points • 

x' .:- = 1/n' (Sum of (x'J) • 

s,.. S,.. a Sum(()(r x'-)'2) = 

y y- = 1/n' (Sum of (yJ) = 

s ,.. S,.. c Sum((yr y-)"2) = 

s ,., s .., = Sum((x'r X'-)(yry- }) = 

bo bo = y--b,x'-= 

b, b, • s,.;s, • 

s.. S.. = SQRT(11(n-2)(Sum(yfl-y1)"2}) = 

Y"'meen Y"' a t mean x value • 

~ from table= 

Cl Cl = rs,"SQRT(1/n) = 

EL Emission Limit = 

Cl% Cl% =CIIEL'100% • 

n' n' = n = v, v1 from table = 

u •. u. from table = 

kr kc e. ·v • 

Tl Tl = k,'SL = 

Tl% Tl% = TIIEL'100% = 

Sy S., = SQRT(S,/(n-1}) = 

r r. 1-(s..•vs., •21 = 

r r • SQRT((1-S,•21Sy'2})= 

'indicates correlation coefficient is undefined. 

PS-11 "-1-R 11).21.o4.lllt 

Value 

20 

2.79 

5.74 

9.5 

493 

47.8 

·13.766 

8.323 

2.305 

9.470 

2.101 

1.083 

20.0 

5.41% 

20 

1.384 

1.1 n 

1.629 

3.75 

18.8% 

5.095 

0.795 

0.892 

Emission Unit: Boiler No. 3 

Test Oates : March 6 to 15, 2004 

!Correlation equation: y = ·13.766 + 8.323 Ln(x) I 

_ ... ,............ -· ,__"'_ --··--·····--·-· . -.. Criterion Actual Allowable Acceptable? 

Correlation ooertident 0.892 ~0 85 yes 

Confidence interval 5.41% ! 10% yes 

Tolerance interval 18.8% !25% yes 

• Indicates correlation coefficient is undefined. 

Pl.ot of LogarithmiC Rtgl't•tlon Curve 

& :r------ y~=~~~:22=9~.=,.7}~.:.:76 ----------------

1 " t! • . . • -

i 'o _!. o ' • • ~ 5 

l ol ~/ I 

·5 

OjO 5.0 10.0 15.0 20.0 25.0 30.0 s:j.O 

PM CEMS Reaponae 

PS-11 Corntlalicn Tt)st 

Calelil'ltlotl Sc;lr&ldthMII 

v.,_ 2-e 10125104 

Lt>garilhmlc Slnmlary 1 or.!!II2004 

VES009753 
CALCULATIONS FOR EXPONENTIAL CORRELATION 

Facility: facility A Emission Unit: Boiler No. 3 

Location: Anytown USA Test Oates: March 6 to 15, 2004 

PM 

PMCEMS 

Run X y' = Ln(y) 

4.9 0.742 1.164 

~ 7.6 2.9 1.065 11.065 

• 7.8 5.1 1.629 4.804 

0.178 

0.078 

0.074 

4 

5 

9.2 4,8 1.569 87.10 0.258 4.743 1.452 v.v ~ 

10.1 2.9 1.065 71.12 1.024 8.535 1.512 ~ ~ 

~6 

8.4 2.128 42.68 0.003 ·0.336 1.639 0.239 

I II I 10 • .< 8.4 2.128 0.11 0.003 ·0.017 2.055 0.005 

I 12 I 15.71 4.9 1.589 8.03 0.238 1.381 1.887 0.089 

13 .l 13.!!J... 6.3 1.841 21.46 0.056 1.095 1.767 0.005 

14 13 2.565 0.570 

15 9.1 2.208 0.327 

16 I 22.31 13.91-- 2.6321 14.191 o.3081 2.0911 2.3291 o.o92 

17 l -· 24.1 1 8.71 2.1631 30.991 0.0071 0.4821 2.4501 0.082 

19 24.7 12.6 ' . -· 2.534 ~63 0.209 2.818 2.490 0.0021 

20 26.4 15.9 2.766 61.89 0.475 5.424 2.604 0.02f• I 

21 29.08 16.2 2.785 111.24 0.502 7.469 2.783 0.00! 

~· 29.3 13.5 2.603 115.93 0.277 5.66Z 2.791l U.UJ 

23 30.5 15.9 2.766 143.21 0.475 8.251 2.878 0.01 

24 33.1 18.5 2.918 212.20 0.707 12.250 3.052 0.018 

P8-11 Example 1·R 10·27.<J4.xls Exponential Cales 

Page 1 of 1 

1012912004 

VES009754 
RESULTS OF EXPONENTJAL CORRELA TtoN 

ac:illty~ FacWily A 

L.oc-1tJon: Anytown USA 

v- EaUIIIcln 

n -of data • 

X .-• t~nsum of CXJl • 

s,. S,. • Sl.m(("<x-)'2) • 

y '(· • 1/n'(Sum ol ('(~) = 

s,., S,., • Sum(('( .-1-)'2) = 

s.,. Sw • Sl.m(("<x•X'f.-1•)) • 

.... .,. .• f'-b_x- • 

"" ~· "'ttu·· b, b,•Sq~S..• 

s.. S, • SORT(11(,...2)(5um(y',"'\",)'2)) = -r-~r. "{"" 8t rnoiln .r:. value = 

\ \-from t.able • 

c r Ct' • t.•SL•SQRT(11n) • 

LCL' LCL' =v-· . c r 

UCL' UCt.' • +Ct' 

Cl a • c"UCL' • e~~oLCl' 

El Enis-sion Umll = Cl% Cl% •CI/EL'100% • 

n' n· • n • 

v, v, from tDble = 

u, IJ.\· from table = 

k, k, = ~··v,;; 

Tr Tr• k,"SL c 

LTl' LTL'•v-'-Tr 

UTl' u·•t.· .-v .. • ... n· Tl l1 = o'UTL' • e'l Tl'V2 

TI% TI% • TIIEL" 11Xl% .. 

s,. S,. • SORl(S,.,I(<>- 1)) = 

i'•HSL~2) • 

r r • SORT((1·SL •2JS,I'2)) • ··-1 ' ............... !Aft .. .-..lfl .. l ....... ............ ...-~ 

' [:q_~ '.-R to-274'. 

v20 

18.53 

1,442.2 

z.on 

7.86 

96.58 

0.836 

2..'!06 

0.087 

0.278 

2 .077 

2 .101 

0.1307 

1.946 

2 .207 

1.()45 

20.0 

5.23% 

20 

1,384 

1.177 

1.629 

0.453 

1.624 

2.630 

3.740 

18.7% 

0.643 

0.813 

0.902 

Emis5ion Unit Boiler No.3 

Test. Datu: March &to 15,2004 

I correlation equation: y = 2.306 e .. ,.,. ' I 

.......... ,_ ... "'"""""" '""""" ................... --· . Qll!rion Adual AI- ~? 

eon-. cooflidoot 0.902 ~35 yeo 

Cmfodena!- 5.23% .,._ yeo 

T cjera,nc:e "erval 18.7% ~25-% yeo 

• ln<fic:a1es """""'~ coefficlenllo unclellnod. 

Plot of Exponentia.l Regression Cui'Ve 

>•r-------------------------------~ 

20 

c 

~ 1S 

3 

§ 10 

l s 

y: 2.3064eOC6l't 

• ·~--~-------------------- --~ •• •• 10 .0 

~-

15.0 20.0 2&.0 30.0 

PM CEMS RtltPOtiM 

PS-I 1 Corr.lllllon r •tt 

c.ltul.wioo Si1'$1dlt!DIIt 

v.,bf, :toe ~~ 

3&.0 

·-

VES009755 
CALCULATIONS FOR POWER CORRELATION 

Facility: Facility A Emission Unit: Boiler No. 3 

Location; Anytown USA Test Dates: March 6 to 15, 2004 

PM concemraue> 

PM OEMS -r;,;~;j~ I Measure<i'lr..;~ormed I CALCULATED ftlii:::JU OTERS 

Run x t: = Ln(x) y y' - Ln(y) (x'rfl' ('f'rY'f (t:r><'l<Y'rY ) ' ,. I , I 4.9 t .5S9 2.1 0.742 1.446 1.762 1.605 0.773 0.001< 

~ 7.6 2.026 2.9 1.065 0.5S3 1.024 0.773 1.249 0.0341 

7.6 2.054 5.1 1.629 0.544 0.200 0.330 1.277 0.123' 

I " I 9.2 2.219 4.6 1.569 0.326 0.256 0.291 1.456 0.012ti I 

- 10 1 2.313 2.9 1.065 0.230 1.024 0.465 • - - - ~ ~·~-· 

11.06l- 2.4031 s .31 1.641 1 0.151 1 o.o5sl 0.0921 1 

2.4851 8.41 2.1281 0.094] 0.0031 .().0161 1 

2.901 6.4 2.128 0.012 0.003 0.006 2.196 0~ 

I I< I oa.71 2.754 4.9 1.569 0.001 0.236 0.019 2.035 0.1991 I 

·- ·- ~ 2.632 6.3 1.641 0.026 0.056 0.036 1.903 0.0040 

)-- 2 

2.961 13 2.565 0.036 0.236 0.092 2.261 0.0604 

3.045 9.1 2.206 0.064 0.017 0.033 2.352 0.0206 

1.105 13.9 2.632 0.096 0.306 0.174 2.416 0.04671 

I " .... , 3.182 8.7 2.163 0.152 0.007 0.034 2.500 0.1133 I 

19 24.7 3.207 12.6 2.534 0.172 0.209 0.190 2.527 0.0001 

I 20 26.4 3.273 15.9 2.766 0.232 0.475 0.332 2.599 0.0261 

21 29.08 3.370 16.2 2.785 0.334 0.502 0.410 2.704 0.0066 

22 29.3 3.378 13.5 2.603 0.343 0.277 0.306 2.712 0.0119 

23 30.5 3.418 15.9 2.766 0.392 0.475 0.432 2.755 0.0001 

I 24 33.1 3.500 16.5 2.918 0.501 0.707 0.595 2.844 0.0055-1 

P5-11 Example 1·R 10.27.Q4.>ds Power Cales 

Page1ol 1 

10/2912004 

YES009756 
BeSUL TS OF PQWEB CORRELATION 

Foc:ilty: fKility A 

Location: Anytown USA 

Vanabla EaUIIion v-. n N..-oldetl ... 20 

>! 'If-< IM'(S....ol(l(J)& 2.792 

s.., s,., • Sum((IC' r'lf-r2l. 5.74 

y' Y'-; t/n'(S.... ol (Y'J). 2.077 

Sw Sw· s..m<<Y'd-r2l = 7.B8 

s,., S,v • Sum((IC', 'If-XY'r'l-ll • 6.22 

.... t>a'• yo-o,x- = -o.949 

... ~. o"'b," .. 0.387 

b, b, • S,.,,S..,= 1.064 

s, S.. = SQRT n-2XS..m(( "-Y' r2)) • 0249 

>/"- ("' 81 mean x value = 2.077 

~ t1 from table • 2.101 

cr CI• I'Sl 'SQRT(IIn)• 0.117 

LCl' LCL' = '·CI' 1.960 

UCl' UCL' • V""' • Cl' 2.194 

Cl Cl• lo'UCl' • o'LCI/ V2 0936 

El Emlaakln l lrrit • 20.0 

Cl% I% • CIIEL.100% = 4.68% 

n' n• • n = 20 

v, v1 from table • 1.384 

"- u"' from table a 1.177 

k, kt • u.--·v,,.. 1.629 

Tl' Tr • k,Sl ll 0.406 

LTL' l .. , • Tl' 1.671 

UTL' UTL' • ... • + 'f'l' 2.41!3 

Tl •(e~urL·- e~LTL V 3.327 

n% 'f ~ • Tl1El•1o0% • 16.6% 

s. S. • SQRT{S:Jiflol)) • 0.643 

f i' • 1-(SL '2/S,/'2) • 0.850 

r r • SQRT((I.Sl"2/S;I'2)) = 0.922 • I_ ..... ,_ - ... - .. .........Mt'..i-.. t loo urul.all.-.t 

i'$11 ~ 1-R 1.1)-ll oO'.JJs 

EmiS-Sion Unit: Boiler No. 3 

Tnt Oates: March 6 to 15, 200<l 

!correlation eguation: y • 0.3870 x ,.,. I 

...... _ ......... .,._, ,.. "''""""' """'""-.... ..... '' 

Cri1erion Actual 

-- ~:ca"leblo? 

Cotre.ta!!ion ooeftlc:iet ( 0.922 ~0 85 )M 

Confidence- 4.68% <11110 yeo 

ToleranCe interval 16.6% ~25% yeo 

• lndi::stes cxwrelaUon ooeffic:ienl is . 

Plot of Power Function Regression Cutve 

20 ~---------------------------------, 

18 

16 

i 14 i y = 0.387x'·""' • " 12 

§ 10 

g 8 I • ~ • u 

l 8 

4 

2 

0 

0.0 5.0 10.0 150 20.0 

PM CEMS R"ponee 

~$'~ 

• 

• 

2U 300 3&.o 

P$-\1 CclnlllfiiiOn '"' 

Ceb.llllton SPf91d1Mic 

VMion :H "'""'"' 

1-

VES009757 
SELECTION Of BEST MODEL 

Facility: 

Location; 

Model 

li1ear 

PolYnomial 

Logarithmic 

Exponential 

Powor 

Facility A 

Anytown USA 

Correlation 

ooefflclent (a) 

0.927 

0.924 

0.892 

0.902 

0.922 

?. 0.85 7 Confidence lntOMII 

haff range pe~t&ge 

Yea 4.49% 

Yea 6.12% 

Yes 5.A1% 

Yes 5.23% 

Yes 4.68% 

Emission Unit Boiler No. 3 

Test Datos: March 6to 15, 2004 

!: 10%? 

Tolerance inlefval hal !:25%? Minimax willlln 

range percentage allowable range? 

Yes 15.6% Yes lbl 

Yes 16.3'/o Yes Yes 

Yes 18.8% Yes b 

Yes 18.7% Yes 

Yes 16.6% Yes 

(a) • indicales correlation ooel(rcientls I.Widelned; model does not satisfy critelion. 

(b) Not appHcable; cri1erion appios only to polynomial model. 

PS~ 11 Corretaticn Test 

Cab.Miion SP'l!adsheet 

Version2-6 10/251001 

PS-11 EJ<ample 1-R 1~ -o4 .xls 

Bast model: 

Best Model 

Ooca model meet al 

cnlarta? 

Yes 

Yes 

Yes 

Yes 

Yea 

LlnNr 

10/2912004 

VES009758 
Appendix A.2 

Complete Results of Example No.2-

Initial Correlation Test 

VES009759 
CORRELATION TEST PM CEMS AND REFERENCE METHOD TEST DATA 

PMCEMS PM concentration 

lliSDOnSG m lacm 

Run X v 

!A 6.4 1.3 

16 6.4 12 

2A 7.5 2.3 

26 7.5 2.5 

3A 11.2 2.5 

3B 11.2 2.8 

4A 16.4 7.7 

46 16.4 7.9 

5A 29.9 12.9 

58 29.9 12.5 

6A 22.2 7.8 

68 22.2 7.3 

7A 19.5 5.8 

78 19.5 5.9 

8A 27.4 11.7 

68 27.4 12 

9A 32.8 20.4 

98 32.8 20.9 

10A 26.5 20.9 

108 26.5 21.3 

11A 33.5 22.3 

118 33.5 22.6 

12A 28.1 18.4 

128 28.1 17.8 

13A 9.6 3.7 

136 9.6 3.9 

14A 5." 4.2 

148 5.3 4.3 

!SA 18.1 6.8 

158 18.1 ,_l.! 

PS-11 Example 2·R 10·27-04.xls 

FaeUity: 

Location: 

Emission Unit: 

Test Dates: 

Emission limit: 

25 I 

.2 20 1 ~ 15 

3 I c: 8 10 

:;;; 

sJ ll. 

0 

Facility B 

Anytown USA 

Incinerator 

April 3 to 6, 2003 

• 

• 

Value 

24.3 

• • • 

Unitt 

mglaem 

•• • 

• 

• 

0.0 5.0 10.0 15.0 20.0 25.0 

PM CEMS Response 

Test Data 

• 

• 

• • 

• 

30.0 35.0 40.0 

Page 1 o1 1 

1012912004 

VES009760 
CALCULATIONS FOR LINEAR CORRELATION 

Facility: Facility B Emission Unit. Incinerator 

location: Anytown USA Test Dates: April 3 to 8, 2003 

Run 

A 

8 

28 

PM CEMS I PM con<>onu-d<Klll 

X 

6.4 

6.4 

7. 

7 

y 

C'.A.l~ULATEn C::T.&TI~T•ra.. • DADJ 

(x,-x 1• I (Y;Y Y I (x,-x l<Y;Y l I y" I (Y"r Y• 

1.31 174.941 74.941 114.501 1.0651 __ 0.055_ 

1.;![ 174.94r - n 76.681 115.821 1.0651 0.018 

2.3 58.62 92,85 0.245 

2.5 55.60 90.42 ~83 

>.60 

"" I o o ..ll ~.II 11.U1 ~1 22 60.31 4.292 2.226 

4A I 16.4( 7.71 10.411 5.09 7.28 7.788 0.008 1 

48 16.4 7.9 10.41 4.23 6.64 7.788 0.013 I 

29.9 12.9 8.66 30.24 15.703 

29.9 12.5 6.47 26.13 19.034 

6.621 4.651 -5.551 11.6871 15.105 

8 I 22.21 7.31 6.62] 7.06] -6.84] 11.687) 19.242 

5.81 0.021 17.281 0.53L _ nn9.872l 16.577 

78 I 19.51 5.9i 0.02i 16.461 0.511 9.872i 15.n3 

I SA 27.4 11.7 60.42 3.04 13.55 15.182 12.126 

I 88 27.4 12 60.42 4.18 15.88 15.182 10.126 

9A 32.8 20.4 173.54 __ 109.06 _137_.57 ___ 18.812 2.521 

98 32.8 20.9 173.54 119.76 144.16 18.812 4.359 

lOA 26.5 20.9 47.24 119.76 75.22 14.577 39.978 

108 26.5 21.3 47.24 128.61 ____ 7L97 __ 14.577 45.li1A I 33.51 22.31 192.47) 152.36) 171.24) 19.283) 9.1 

18 33.5 22.6 192.47 159.85 175.41 19.2831 11 

12A 28.1 18.4 71.80 71.29 71.54 15.6531 7 

128 28.1 17.8 !:.U!Q 61.52 68.46 

13A- 9.6 3.7 100.53f ;,~.1!> b<.IJ J .<l 

138 9.6 3.9 100.531 36.68 60.73 1~1 -·,-· 1 

14A 5.3 4.2 205.2~ 33.14 82.47 0.3~~ 1!' n<n 

148 5.3 4.3 32.00 81.04 

15A 18.1 6.8 9.96 4.82 

158 I 18.11 7.11 2.331 8.16) 4.36) 8.930) 3.350 

PS-11 Example 2-R 10-27-04.xls linear caJcs 

Pago 1 of 1 

10/2912004 

VES009761 
RESULTS Of LINEAR CORRELATION 

Facility: Facility B 

Location: Anytown USA 

Variable Eauation 

n Number of data oolnls a 

X - x- ~ 1/n'(Sum of (xJ) ~ 

s., S.. ~ Sum((x,-x-)"2) = 

y y- • 1/n'(Sum of (yJ) = 

s.r S,. = Sum((yr y-)"2) ~ 

s.,. s., ~ Sum((XrX-XYrY-)) = 

bu b,• y-b,x- = 

b, b1 ~ S.,.IS.. " 

s.. S, = SQRI(1/(n-2)(Sum(y1"-y1)'2)) = 

Y"....., y• al mean x value = 

t, ~ from table = 

Cl Cl ~ t,"S,'SQBT(1/n) = 

EL Emission Limit = 

Clo/o Cl% =CIIEL'100% ~ 

n' n· = n = 

v, v1 from table = 

u.,. u,. from table = 

kT k1 1! u!\''Vr c 

Tl Tl • k1'Sc = 

I I% Tl% = TI/EL' 100% = 

Sy S, = SORT(Sy/(n-1)) = 

r r' = 1-(S, •215, •21 = 

r r = SORT(( 1-S, •2/Sy •2)) = 

• Indicates com~lation coefficient Is undefined. 

PS-11 El<ample 2-R 10-27-04.xls 

Value 

30 

19.63 

2.738 

9.96 

1,531 

1.840.8 

-3.237 

0.6721 

3.238 

9.957 

2.048 

1.211 

24.3 

4.98% 

30 

1.286 

1.168 

1.503 

4.866 

20.0% 

7.266 

0.801 

0.895 

Emission Unit: Incinerator 

Test Dates: April 3 to 6, 2003 

)Correlation equation: y = -3.237 + 0.672 x) 

-

S-• ......... _, -otA· ~ • .... - _ ..... _ Criteria ~ - -- f-~or PS-11 

Criterion Actual Allowable Acoaptable? 

Correlation coefficient 0.895 ~0 85 yes 

Confidence inte<val 4.98% ~ 10% yes 

Tolerance interval 20.0% 25% yes 

• Indicates oorrelaUon coelflcien1 1s undefined. 

PSot ol Linear Rogn:~sston Llno 

25.,-----------------, e 

j 20 

l 1s 

g 10 

" !: 5 

0 

0 .0 

y = 0.6722x - 32371 •• 

• 

10 .0 20.0 30.0 •o.o 

PM CEMS Response 

PS-11 CorreiOiion T ... 

CltQ.tlation Spteadsheet 

Vcnlon U 1MSI04 

Unear Summary Unear Summary 

VES009762 
Facility: Faeility 8 

Lo<:atlon: Anytown USA 

RlJn 

I PM CEMS I PM ~~ 

molocm 

• I v 

CALCUI.AT!ONS FOR POl. YNOMIAL CORRELATION 

Emission Unit: lncine,rator 

Test Oates: April 3 to s. 2003 

I"' AI l"'ltl 

7 x'i •• ixYix'viY"itV'-vl'i-·itoA. 

262 1.678 8 53 2.649 1.819 0.1326 f4.!04T ' · '" l 

I 262 1.678 8 49 2.649 2.0911 0.13211 78 ~1_ 2. 15 

422 3 164 17 129 2.795 0 245 0 1009 58 -- . --

422 3,164 19 141 2.795 0.087 0.1009 •• 

125 1.405 15,735 28 314 ~ 

~ 

I 4A I 1 11),,. 1.1 269 4,411 72.339 126 2.071 5.942 

L 48 16.4 7.9 269 4411 72,339 130 2.125 5.942 3.833 o.o8< •• "Q.U 12.9 894 26.731 799 254 386 11.533 17.379 20.081 0.07 

1.9 12.5 894 28.731 799.254 374 11.175 17.379 23.804 0.07 

!.2 7.8 493 10 941 242.891 173 3 844 9.872 4.292 0.0823 6~ I. 

QD .2.2 7 3 493 10.941 242.891 162 3,593 9.872 6.614 0.0823 7.06 1, .. 

7A 19.5 5.8 380 7.-4 15 144.590 113 2.20s 7.858 4 236 o 0906 17.28 t.n 

78 19.5 s.u 380 1 A15 144,590 115 2.243 7.858 3.834 o.0906 18.46 1.n 

8A 27.4 11.7 751 20,571 563,641 321 8.784 14.655 8.732 0.0593 3.04 1.44 

88 27.4 12 761 20.671 563,641 329 9.009 14.655 7.049 0.0693 . .. ' •• 

9A 32.8 20.4 1.078 35,288 1,157,432 669 21,947 

"" "" " ••• , 1 076 35288 1 157432 686 22.485 20.884 0.000 0.1573 ... .,iiT . ..... -, ' 7M 18 810 493 155 554 14 677 13.742 51.239 0.0604 1' ' •• . . • 

18.610 493 155 564 14 958 13.742 57.125 0.0604 • 

37.595 1,259,445 747 25.026 21.785 0.265 0.1920 1- I li D I ••. , ... 1,122 37 ~95 1,259,445 757 25.363 21.785 0664 0.1920 •••.• ~ • . odl 

I 12A I 28.11 18.4 700 22.188 623.484 517 14,529 15.300 9.060 0.0805 71." · .. 

·· - -· . ·-. -- " 2.188 623,484 500 14,055 15.300 5.808 0.0605 81. 

885 8 493 36 341 3.223 0.227 0.07()() 39. I 0 

885 8 493 37 359 3.223 0.458 0.0700 36,68 . 2U 149 789 2 118 .... • ... • .... 

••o I o.o • , .a~ I '""I tool 23 121 2.556 3.043 0.17851 32.001" z.4ii 

15A 18.1 6.8 328 5 930 107.328 123 2 228 6.940 0.020 0.0901 I 9.11e 1.Til 

tSB I 18.1 7.1 328 5,930 107.328 129 2,326 6.940 o.ozs 0.0001.1 t~ Q 

PS.11 Elc811"9e 2-R 10·27-0C ldo ~Coles 

Page 1 of 1 

1012912004 

VES009763 
8C$ULU OF PO!._XHQMlAL CORftAADOH 

Fac:ltlty: Fac:lky 8 

l oeetion: Anytown USA 

V-arbb'e ~lion 

• tU!tet ol data = 

s, S1 = SUm(•) = 

s, S2 • S4.tn (dZ). 

s, S>•9un('l3)• 

s. S< = Sun (>dC) = 

s. S5=S<n(y!)= 

s. S6 ::0 S'u'n (dy!) = 

s, S1 = Sum (l<i2yi) = 

O..A -~·n • I •SlS1S~S1S' = 

•• b'O = ($6$2$ot+$1$3S7•S2S883-57S2S2 5353$5 S458$1)fdllll A= 

., bl = (..s6$4.SSSlS2•S2StS7-szses2.S753r>SIStSSYdoiA = 

•• b2,. (nS2$7+$ SI82•SSS183-S29~7StS1Vttet A= 

s. s.. 5Qlt(( ~n.!)Soold(Y"Y)2) = 

0 0 • n1S2S..Sl'WitiS3S2.SIS4!+S2 S IS3-S2 • 

c. 00 • (S:!S'•$3')1) • 

c, Ct • (S3S2.S1S4)i'O• 

c, C2 • (SIS3·SZ )10 • 

c, C3 • ii&•SZ )1) = 

C, C4 • ($ 1$2-nSS};'O • 

c. C5 • (f\S2.S 1 )10 • 

\ ., frOm .able = 

EL Emlssloo UITlrl • 

Cl Cl • 't"S11 ·sqn{oetlil•mln) • 

I I • 1r':i."'100• 

v, v1 fl"om table '"' 

"" l4f' from \able = 

n• n' = 1kkltt.3 = 

k, un ~= 

Tl Tl=kt = 

n% T • 1\IEL "tOO%= 

• 1M• Sun at I • 

s Sy • .~qn(Sumof ( yl-)'-)"2/(n·!) = 

r' • t -(s,•·2JS)""2) = 

' • • ""((I·S,.•>Jl!y"2)) • 

r..I8X~.h? b, > 01 

"'-- y" ·b,12~ · 

1.25><-

• ll'dtaK co"·•r.aoon ~~• Wldelhcd 

-. JO 

5811 

••.m 

585.603 

tG.996.e38 

29$ 

7.703 

216,158 

98E+ ~O 

2.654 

~·73 

0.02203 

2.873 

1198E+10 

0.698 

0199 

1.8951:-03 

1 04SE.02 

·2630E-04 

6.855E.()6 

2.052 

2~ 3 

~36 

5.91% 

1.293 

1.181 

16.86 

1.527 

4.385 

16.0% 

9.96 

1.21 

0.844 

0.919 

minirrun 

3.i3 

41,88 

EmiuSon Unit: lnc:Stler.lllor 

Tut o.tet: April 3 to &. 2003 

leonetatlon ~K~uation: y• 2..154 • .0.1n x .. 1.02203 _ __) 

Summary of Aue~ Critari:a tor PS..1t 

a.- _,., - ..., CJtliiH? 

Correlebon c::oe~ 0.1'19 !. U,_., ,.. CCJr6jence irWval 5..tt .. ,Si'"" ,.. 

Tolerance irr'~ 18.0% ~- ... lndll:::8ts~~ti5 ~ 

W'!IK:III> l~ " '-'UI'Y• ai!Jl""UI'Tll"-.num 

Com Irion eunoe mi':imurn CIOir'4 3.11 

Mii'WftJm 4ikmablt • '4\18 63 

Is ~lftion CUI'Ye trihliiTU'I'I < l'lti'rlum "viNe? ,.. ~lion curve ITQ(lrrun I)Oh1 NA 

Ex!la IJrrj; for )( tlMt: of maxlrrun X YAIU8 NA 

Is c::onelalion CliVe m; .. invm _ > .U.Opej8\0'1 ll'na? NA 

Plot of Potyoomlal Reg I'" l ion Curve 

25 

g 20 

) .. •0 022~ · 0 1731ll •• t86J9 • 

f 15 • 

1.. • • <.> • • l 5 • • ....... 0 1--·· ' 

0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 

PM CEMS RuponH 

P.IICtneltt!orl'f ... 

CIICUollloll lj.wUII,._I 

40.0 

~latiN 0/2~ 

I 

I 

' 

i 

P$.1 1 ~2-R 10.27-0t .• Ptf);o• i8l su~ 11111201M 

VES009764 
CAlCULATIONS FOR LOGARITHMIC CORRELAn ON 

Facility: Facility B Emission Unit; Incinerator 

Locotlon: Anytown USA Test Oates: Aptil3to 6, 2003 

PM 

PM ",;;;;J.;;:,;;,---- I CALCULATEO . 

3 3551 

1---.," I """' ,...,...., _ ..,. .().604 12,'!01 

[742

~ ] 

~ 

10! 

101 

1 

... 

1 

3.33tll I <i 3.336 

1.262 

i.2s2 

1.668 I ~ I 10 , I I 21..6681 896 

156 18.1 2.896 

_l_ 

PS-11 Example 2·R 10-27-04.xls 

6 

-0.743 12.7' ft .:-... A 4" " 

I .328 33.139 6.635 . ', . .,)lltg ,)U, , 

1.328 31.998 6 519 -1.346 ••• 

.006 9 965 -0.239 10.699 

.006 8.161 -0.216 10.699 

I I I 

!1QQJ 

f,951] 

I 

Logarilhmic calcs 

Page 1 o11 

11/1/2004 

VES009765 
RESULTS OF LOGARITHMIC CORRELATION 

Facility: Facility B 

Location: Anytown USA 

Variable Equa!lon 

n Number or data POints = 

'/( '1(- = 1/n"(Sum of '~) = s~ ... S,.. = Sum((X'rx'-r2) • 

y y- = 1/n"(Sum of (y~) = 

Sw Sw • Sum((y; y-)•2) = 

s.-, s .., = Sum((X'rx'-Xy;.y-J) = 

bo bo • y-t>,x'- a 

b, b, = S...,JS,., = 

s, s, = SQRT(1/(n-2)(Sum(yf-y1

)"2)) = 

y" .. .,.. y• at mean x value = ~ ~f om table = 

Cl Cl = ·s,·sOR /n ) = 

EL Emission Limit = 

Cl% Cl% =CIIEL"100% = 

n' n· = n = 

v, v, from table = 

u •. un' from table c 

k, k, s: tv, ~ 

Tl Tl = k,"SL = 

Tl% Tl% = TI/EL"100% = 

Sy Sy = SQRT(S,,t(n-1 )) = ... r" = 1·(S,'2/S,'2) = 

r r ~ SQRT((1.S, ' 2/Sy '2)) = . Indicates correlation coefficient is undefined . 

PS-11 ~ 2-R 10.27·04.J<Is 

Emission Unit: Incinerator 

Test Dates: A pril 3 to 6, 2003 

!Correlation equation: y • ·17.700 + 9.807 Ln(x) I 

Value 

30 --······-· - -.--. - -J· -··--· ·- ·-· . - .. 2.82 Cri1erion Actual Allowable Acceptable? 

10.93 Correlation coellident 0.822 !:,0.85 no 

10.0 Confidence interval 6.37% ~10% yes 

1,531 T oleranoe interval 25.6% 25% no 

107.2 • Indicates con-elation ooelllcient Is undefined. 

- 17.700 

9.807 Ptot of Logarithmic R•grnslon Curv. 

4.141 

9.957 25 

2.048 

1.549 

24.3 

6.37% 

30 

1.266 

1.168 

!j 20 • •• .. y • 9.80671.n(x) • 17.7 

]; 15 • ~ • 10 • 0 5 • " • l 0 

·50 0 5.0 10.0 15.0 20. • . o 

1.503 PM CEMS Ru pon&a 

6.22 

25.6% 

7.266 PS-1 1 Co1re1a1Jon Test 

0.675 Calcul ullon Sprvodsheet 

0.822 Vof'liOn 2·6 1()(.15104 

l.og011thmoc SOn-mary 1012912004 

VES009766 
CALCULATIONS FOR EXPONENTIAL CORRELATJON 

FaciMty: Facltlty 8 Emission Unit: lnclner1tor 

Loc1tlon: Anytown USA Test OatH: Aprl13 to 80 2003 

PM concootra!Scn 

PMCEMS macrn 

Mea!ILnld Translormed CALCULATED STAllSTICAI. PARAMETERS 

Run X y ll' = Ln(y) ,,.,..-y Wd1' ,,.,. ... )(l(d1 '(" CY"rll'iY 

1A 6.4 1.3 0.262 174.94 2.957 22.744 0 ,888 0.3< 

18 8A 1.2 0 .182 174.94 3.239 23.803 0888 Oil! 

2A 7.5 2.3 0.833 147.06 1.320 13.934 0.978 o.a 

28 7.5 2.5 0.916 147.06 1.136 12.923 0.978 0 .0 

3A 11.2 2.5 0.916 71.01 1.136 8.900 1.28< 0 .1: 

38 11.2 2.8 1.030 71.01 0.907 8.025 1~ 0 .0 

4A 16,4 7.7 2.041 10.41 0.004 -4.191 1.715 0.1i 

48 uu 7.9 2.067 10.41 0.007 .0.274 1 715 0 .1: 

5I\ 29.9 12.9 2.557 105.54 0.331 5.910 2833 0 .0 

58 29.9 12.5 2.526 105.54 0.296 5.587 2.833 0.0! 

8A 22.2 7.8 2.054 6.62 0.005 0.166 2.105 0.0: 

811 22.2 7.3 1.988 6.62 0 .000 O.Q15 2.195 0 .0 

7A 19.5 5.8 1.758 0.02 0.050 0.028 1.971 0.0 

78 19.5 5.9 1.775 0.02 0.043 0.026 1.971 0.0 

SA 27.4 11.7 2.460 60.42 0.226 3.713 2.626 0.0: 

88 27.4 12 2.485 60.42 0.253 3.910 2.626 0.0 

9A 32.8 20.4 3.016 173.54 1.068 13.616 3 .073 0,01 

98 32.8 20.9 3.040 173.54 1.119 13.935 3 .073 0,01 

lOA 26.S 20.9 3.040 47.24 1.119 7.271 2.551 0.2 

lOB 26.5 21.3 3.059 47.24 1.159 7.401 2.5!>1 0.2 

IIA 33.5 22.3 3.105 192.47 1.260 15.575 3 .131 0,01 

118 33.~ 22.6 3.118 192.47 1.291 15.760 3 .131 O,QI 

12A 28. 1 18.4 2.912 71.80 O.Be6 7.884 2.504 Q.Q 

128 28.1 17.8 2.879 71.80 0 .805 7.603 2.684 0.0 

IJA 9.8 3.7 1.308 100.53 0 .454 6.754 1.151 0.0 

138 9.6 3 .9 1.361 100.53 0 .386 6226 1.151 0.0 

14A 5.3 4.2 1.435 205.25 0.299 7.835 0 .795 0,41 

148 5.3 4.3 1.459 205.25 0 .274 7.497 0 .795 0.4 

15A 18.1 6.8 1.917 2.33 0.004 0.099 1.855 0.0 

158 18.1 7.1 1.960 2.33 0.000 0 .033 1.855 0 .0 

PS-11 E>amp!e 2·R 10.27-44Jds Exponootial Cales 

Pogo 1 of 1 

10/2912004 

VES009767 
R£SUL'Ci Of EXI'OHENTIAL CORRELATION 

Fadllty: Facility 8 

Location: Anytown USA 

van 

.-

s.. 

y 

~ 

~ 

b,' 

b, 

b, 

-s:-

'(A-

--~ Cl' 

s 

S. • 

-x- • 

s._.•: 

~ 

Sw~_ 

S.,. • 

Voluo 

I = 30 

1'1 ot (xt)).. 19.63 

. 2.738.· 

n ol (y'~) = 1.98l 

!) :. 22.04 

II • 226.81 

b,' • y'-l>,x- • 0.35E 

t., • - · •I 1.4281 

b, • $.,1&.. • O.Oa:l 

1"2)1 = 0.340 

'/" 01 moan x vM!o = 1.9 ...... 

Ct• · 

~from lablo •1 ~ 

0.1270 

1-- 1 ~~--~ 24.3 

3.80% 

30 

1.286 

~ 

v, 

Uw 

k;" 

n· 

L 

~ -

r 

· rndie$ttt 

P$1 ~2~to.n~ 

~-

v, from tobte = 

u,. from tabte • 

k, • u,"v1 u 

Tl' •k,' Sl • 

1.161l 

1.503 

),SIC 

¥C· ~-· + TI' 2492 

Tl • {e"UTL' • e"LR 'V: 3.866 

n% • TIIEL.100% • 15,_9% 

~ • SORT(S,..J(n·1))• 0.871 

(' • I.(Sl "215./'2) = 0.848 

' • SORT({I·Sl•2JS{'2)) =I 0.921 

"" 

Emission u·nlt: k1el .... r•tor 

Test Dotes: April 3 to 8, 2003 

[Comtlatlon equation: y • 1.428 o .... ,. • I 

S• "' __ ,,.,,_ -· 

-· Criteria for PS.. - " 

Crilerion ...,..... Correlation c:oe:l&cient 0.921 

-~0 8.5 -

.,... 1 

Conlldonce i- 3.80% ~10% .,... Tole<ala! intenr.>l 15.9')(, ~25'Jf. )'08 

lndicat .. oorrela5on ooeffident Is 101delln<>d 

Plot of Exponential Regr'f!sslon CLHVe 

~.----------------------------, 

20 

& 

~ tS 

! 

~ 10 

l 

5 

0 . • 

Y = 1.4281eo0121ll 

o.o 5.0 10.0 15.0 20.0 MO 30.0 35.0 

PM CEM.S R61J)Onl6 

---

PS-11 Comii!MionleM 

C*-'laion S~ 

v.nbnU 1~ 

400 

"'""""' 

VES009768 
CALCULATIONS FOR POWER CORRELATION 

FKIII!y: Faelllty B Emi.uion Unit lncinetator 

LoQUon: Anytown USA THt Dates: April 3 to 6, 2003 

PS-11 e-npte 2·R 10.27-()oi.Jds Flower Coles 

Pago1d1 

1~00< 

VES009769 
Facility: FaCJUity 8 

Locallon: Anytown USA 

P$11 ~24t IO.V.()o& . 

RESULTS OF POWER CORREiA TtON 

Emission Unit: lneinetatOr 

Test D1tes: Apri 3 to 6, 2003 

lcornlation equation: y- 0.2007 x um--1 

Plot of Power Function Regreasiotl Curve 

25 ~---------------------------------, 

j 

20 • • ~ IS 

§ Y = 0.2007x'·2122 

~ 10 

s 

.. 5 

0 • 

0.0 s.o 

--

10,0 16-0 zo.o 25.0 30 0 3!5,0 400 

PM CI!:MS Ro•po,.ll 

PS-I 1 CQf'fi!l-*'n T ... 

lion SpotNIH!oec 

··.-nU 11)12511)0 

·-

VES009770 
Facility: 

Location: 

Model 

Unear 

Polynomial 

Logarilhmie 

Exponential 

POWGr 

FacWity B 

Anytown USA 

Correlation 

coefficient (a) 

0.895 

0.919 

0.822 

0.921 

0.892 

~0.85? 

Yes 

Yes 

No 

Yes 

Yes 

SELECTION OF BEST MODEL 

Emission Unit: Incinerator 

Test Dates: Apr113to 6, 2003 

Confidence inte<val ~ 10%? Tolerance interval half ~25%? Minimax wilhin 

hall range peteen!age range percentage ellowallle range? 

4.98% Yes 20.0% Yes b) 

5.91% Yes 18.0% Yes Yes 

6.37% Yes 25.6% No bl 

3.80% Yes 15.9% Yes b) 

1% Yes 18.7% Yes b) 

(a) • indicates comllation cooffiQontls undcmod; model does not satisfy critenon. 

(b) Not applicable; c::ntenon applies only to polynomial model. 

PS-1 1 Co'tt13llotl Test 

CalCulation Sprtad$hMt 

Vorticn 2~ 11l"25.'04 

PS-11 Example 2·R 10.27-04.xls 

Bcalmodol: 

Best Modal 

Oocs model moot all 

cri!ena? 

Yes 

Yes 

No 

Yes 

Yes 

Exponenllol 

1012912004 

VES00977 1 
Appendix A.3 

Complete Results of Example No. 3 -

Initial Correlation Test 

VES009772 
CORRELATION TEST PM CEMS AND REFERENCE METHOD TEST .DATA 

PMCEMS PM concentraUon 

response mg/ocm 

Run X v 

1 58 3.5 

2 7.6 9 

3 7.8 7.8 

4 16.4 25.2 

5 10.1 16.6 

6 11.1 14.3 

7 12.0 19.6 

8 13.2 20.1 

9 15.7 23.4 

10 17.6 26.7 

11 21 .7 26.8 

12 21 .0 28.9 

13 22.3 25.7 

14 2.4.1 26.2 

15 24.7 26.2 

PS..11 Example 3-R 10·27 .Q4.xls 

Facility: 

Location: 

Emission Unit: 

Test Datos: 

Emission limit: 

35 

30 

g 25 

~ t' 20 .. 

u 

s 15 

u 

l 1: l 

0 

0.0 

Test Data 

Facility C 

AnytownUSA 

HWI 

November 1 to 4, 2002 

Value 

17.45 

\ 

• 

• 

Units 

mglacm 

•• 

• 

• 

5.0 10.0 15.0 

• • • 

20.0 

PM CEMS Response 

•• •• 

25.0 30.0 

Page 1 of 1 

1012912004 

VES009773 
CALCULATIONS FOR LINEAR CORRELATION 

Facility: Facility C Emission Unit: HWI 

Location: Anytown USA Test Dates: November 1 to 4, 2002 

- PM-CEMS j PM_, __ , __ ,,_, I "A'"'" ATED STA' 

Run x y (x,-xl' (y,-yl' (x, x-~)(y yl 

I 1 5.8 3.5 92.29 272.25 158.51 

2 7.6 9 60.94 121.00 85.87 

3 7.8 7.8 57.86 148.84 92.8(] 

4 16.4 25.2 0.99 27.04 5.1 

iO 

11 

i2 

13 

14 

15 

PS-11 E.xample 3-R 10-27-04.xls 

10.1( 16.61 28.161 11.561 18.04 

11.11 14.31 18.551 32.491 24 5~ 

2 

22. 

24. 

24. 

19.61 11.611 0.161 1.3€ 

20.11 4.871 O.D11 ·0.22 

23.41 0.091 11.561 1.0C 

26.7 4.81 44.89 14.7C 

26.81 39.611 46.241 42. 

31.29 79.21 49. 

25.71 47.521 32.491 39. I 75.57

1 

38 44l 53.9 

26.2 86.37 38.44 57.6 

Linear Cales 

~-~ 

~ 8.942 

11 01~ 

11.245 

13.892 

15.043 

16.079 

7.460 

20.338 

~ 

<rrd 

29.621 

4.o--

1T.ii 

16.4 

7.3 

0. 

12 

6. 

9. 

17. 

17 

)I 

j3 

14.487 

20.222 

Page 1 ol1 

10/29/2004 

VES009774 
RESULTS OF LINEAR CORRELATION 

Facility: Fa.cility C 

Location: Anytow n USA 

Variable Eauation 

n Number of data points = 

X x-= 1/n"(Sum of (x.)) = 

s... S,. = Sum((x,-x-)"2) -

y y- • 1/n"(Sum of (yJ) = 

s,. S,. = Sum((y, y-)"2) = 

s... s., = Sum((>.rx-)(Yrrll = 

bo bo = y--b,x-= 

b, b1 = S..,tS,.,. = 

s, Sc = SQRI(1/(n·2)(Sum(yh)}"2)) = 

y•,.., Y" at mean x value = 

~ from table • 

Cl Cl = ·s :s /n) = 

EL Emission Limit = 

Cl% Cl% =CIIEL' 100% = 

n' n' = n= 

v, v1 from table = 

"" u •. from table = 

kT kl = u .. ··v, = 

Tl Tl = J<.'S.. = 

I I% Tt% = TI/EL' 100% = 

~ s, = SQRT(Sy,/(n-1)) = 

(' r< = (5 "21~"2) = 

r r = SQRT((1·S, "21~"2)) • 

• Indicates correlation coefficienlls undefined. 

P5-11 Example 3-R 1 0·27 44.xls 

Value 

15 

15.41 

561: 

20.00 

905 

645.2 

2.267 

1.151 

3.530 

20.000 

2.160 

1.969 

17.5 

11.29% 

15 

1.485 

1.184 

1.759 

6.210 

35.6% 

8.038 

0.807 

____Q,_898 .. 

Emission Unit HWI 

Test Oates: November 1 f.o 4, 2002 

!Correlation equation: y • 2.267 + 1.151 xI 

--····· -- . ---·- ··-- --··-··- ·-· . - .. Criterion Actual Allowable Acceplable? 

Conelation coefficient 0.898 ~0 85 yes 

Confidence interval 11.29% 10% no 

Tole ranee intarval 35.6% ~25% no 

Indicates correlation coefficient Is undefined. 

Plot of Linear Regression Une 

.§ 30 35 "----=-~----:- I 

1! 25 

'f ::ll 1.161x + 2.2665 

- 2(1 

~ 15 

8 10 

l ~ • 

0.0 5.0 10.0 15.0 20.0 2M 30.0 

PM CEMS Response 

PS.11 ConuiNlh'l Tat 

Catwlation Sp.-udlhlel 

Vcnion 2-6 1()'2.5104 

LJnear Summary Linear Summary 

VES009775 
Facility: Facility C 

Location: Anytown USA 

Run 

I PN CEMS I ""' ""·-···· malaan 

X I v 

3.! 

CALCULATIONS FOR pQLYp!OM!Al CQRRELA!JQN 

Emission Unit: HWI 

Test Datal: Hovomber 1 to 4, 2002 

r.AI r..111 ATEO: 

>~ 1 ) >~ 1 -tt .. :h .. tyA.v"f- oo~~a (yJ • ..-f- 1 Cl 

M...l. 195...1. I 132 2.0 118 3.138 0.131 0.4679 772-25 I 91 

3.336 68 52.0 8.429 0.328 0 2223 121 00 1: 

I ~ I , .81 , .81 --e1"i" 475j 3.702 61 475 8.976 1.384 0.2.060 1(8 84 1 : 

L 4 16.4 25.2 269 4.4111 n.339 413 6.778 24.851 0.121 ().1650 2704 IJ 

• ·- . •• • 102...1. 1.030...1. 10,406 168 1 693 14.689 3.653 0.1214 

123 1368 15181 159 1.762 16.838 6A44 0.1224 32491 0.981 

I , I •21 111M 144 1 728 2.0 736 235 2.822 18.600 1 000 0.132.0 0. 18 1.01 

8 13.2 20.11 174 2300 30360 265 3502 20,69<1 0.352 0.1488 AA• • AB 

• ••• •• • ... 3.870 so757 367 5768 24.119 o.517 o.1&n 

; 452 95 951 470 8.271 25.876 0.679 0.1541 ., 

1,218 

l ;t t'l ~o tt «1 .... ..... .. ..... ... 497 

.• 581 

... 610 

PS.II E>camjlle 3-R 10·77.0.1.>1!o Polynomoal cates 

Page loll 

IM9/2004 

VES009776 
&ESUL IS Of POL YNQUIAL CQRR£LAJ!ON 

h cii1y. FacUlty C 

l oedon: Anytown USA 

fmiuton Unit HWI 

Test Date~ November 1 to 4, 2002 

Vll'lfl<1tl .. e ... .,. ~ !comil:l.5on esu-atiOn: x • ·11.37t .. 4.211 • • -4.1012l ..:' I 

s 

.. 

"' 

I = S-ii'YI ()i) = 

! •&l'n C 

a• s-.mc 

~ SS•l 

S6·~ 

S7 • S... ()12y~ = elM A • ns2SJiOS2S2S2•S1SSS2:S!iS~·SZs1SSOS4Sis1 • 

""' 4,121 

~ -. 

~ tot.ossl 

bO = (56S254•S1S3S7•S.z:s&S3.$7$.2$2-$3$3$5·$4$6S1)141M A.. I · 18,)71 1 

1)1 •ln:>G::i4•·:i5S3S2•S2S1S1·S2S6S2·S7S3n-S4-S1S5)'det A= 4.296 

b2 • {n&2S1•S 1S&S2•S.SS1SS·S"S2S5·SSSin-S7S 1S 1)to.t A • .0.10123 

I ...,. I S: •SQI1(t11(n.,)Sumd(y""'Y)2)• 1.279 

0 0• n S2$6..S3 •S1 S3S2..S1S4 S St = 1.16BE•06 

• UJ = (~YVD = 3.481 

:, 

c. 

v, 

"" 

s, 

~ 

r 

1--11 .... 

1.2$ltPS-1 1 ~ 3-R 10.27-0t .. 

Sy•t 

J lft:ltlhld 

: 1 • (S3S2.S1S4}.10. ..0.4833 

C2• 1 

""C37(1'S4·S2'VD' 

C4 ;o (S1S2~).10 · I 

"'C601 

Emltib 

Cl• ·s,. -:;;;We 

i'• 

' .. 

~~' tom labte =I 

kl -= oo"''Yf = I 

Tl -= kt'$p = 'rilE 

~ ··MJ~ I 

b,, 011 

't -= .OJ2bt' 

7.124C.o:2l 

2.1 

OJ 

1.515 

1.2001 

1.831 

~ 

~ 8 04 

).975 

""0:981 

1.22 

~88 1 

_ ... Su ....... 

SUIIIMM'J' of Acceptt.nee Cthtt. for P $.11 

Oi:erion 

- - • ' ... , Contllltion~ OJI87 ! .... ,.. 

Cor-Mencel- 5.56,. s 13 .... T-- 13o4Y. ~ ,... lncfQ.tes c:otfNfon ~ i5 ....-red 

ctMICk for COITelation Curw Minimu:ml'lrla:d -m 

Cc:lt'N.lAon CUI"4 miirrun POi'1t w. 

l.i.rimum allowable x wbe NA 

II ~atim OUNe mini'Tun < mlnlrrun • v1luo? w. 

Ccrrel&':':lon curve madnun point 21 2 

atlon lirnll ror,.. 125% ot man-rum x vttut 300 

ts corre!a~ caiiVe Mral(lmiJtn,. tlhi)OIJiiOn Uml? no 

P~ of Po~omlal Regreu lon C!.K'Ve 

.. e 30 l2$ 

" ~ 20 . 

g 15 

u 10 

il! • 

0 

0.0 

.. 7 • ..... 

5.0 10.0 15.0 20 0 25.0 

PM CEMS RetponN 

,..., ecw,..,.,. '"' ~·~ ... 

~., 

30.0 

I 

Mbol 2>e IGI2W 

11/1/201)1 

VES009777 
CALCULATIONS FOR lOGARITHMIC CORRELATION 

Facility: Facility c Emission Unit: HWl 

location: Anytown USA Test Dates: November 1 to 4, 2002 

PM CEMS response I PM • 

Transformed 111Q/acrn CALCULATED STA ltltK:S 

Run x x'; In (x) y {x,- x'")" {yr yl" {x'r x' l<YrY y" ('f"rYIJ" 

5.8 1.758 3.5 0.785 272.250 14.620 i.106 2.578 

7.6 2.028 9 0.379 121.000 6.774 1.849 0.421 

I ., 7.8 2.054 7.8 0.348 148.840 7.196 1.088 5.225 • 18.4 2.797 - · - - A- • A. A • A A •-• 578 8.877 

10.11 2.3131 16.61 0.110) 11.560 

1 1.1 2.407 14.3 0.056 

r 12 2.485 19.6 0.025 0.160 u.u<>« " .... A · - A A rAA - - • 0,004 0,010 - -AA • A AO 

3.1 

1.045 

. 105 

18f 

•.207 

1.9 

--::; 

11.560 

l.2' 

)1 ~ 8.616 

.. 9471 . l-565 : ,,..,.,.. . 0.235 •MI 4.6931 . ... 4.175 

8.11 

10.63' 

PS-11 Example J.R 1 0.27-<l4.xls Logarithmic Cales 

Page 1 of 1 

1012912004 

VES009778 
RESULTS OF LOGARITHMIC CORRELATION 

Facility: Facility C 

Location: Anytown USA 

Variable Eauation 

n Number ol data polnlS -

.: )(- = 1/n"(Sum ol ()(J)-

s,... s ..... = Sum(()(rx'- )"2) " 

y y- = 1/n"(Sum of (YI)) = 

s,., S,., a Sum((yrY-)"2) = 

s.., s.., = Sum(()(r x'-)(YrY-)) = 

bo bo = y--b,)(- " 

b, b, = s,.,;s('( -

~ S.. = SQRT(1/(n· 2)(Sum(y,'·y1)"2)) = 

Y"'mun Y" at mean x value = 

~ ~ trom table = 

Cl C l = ~·s ,'SQRT(1/n) = 

EL Emission Limit = 

Cl% Cl% • CIIEL'100%-

n' n' = n= 

v, v1 from table = 

u~ u,, from table = 

kr k1 = u"·•vl = 

Tl Tl = k1"SL = 

Tl% Tl% = TIIEL "100%" 

s, S, = SQRT(Sy,/(n·1))-

r r • 1-(Sc '2/S, •2) " 

r ra SQRT((1.Sc•21Sy•2))• --

• lndlcales cotr&lallon coefficient is undefined. 

PS-11 EJoJinjl!O 3-R 10.27-()l..ldS 

Value 

15 

2.64 

2.96 

20.0 

905 

49.8 

·24.443 

16.809 

2.290 

20.000 

2.160 

1.277 

17.5 

7.32% 

15 

1.485 

1.184 

1.759 

4.03 

23.1% 

8.038 

0.919 

0.959 

Emission Unit: HWI 

Test Dates: November 1 to 4, 2002 

!Correlation equation: y • ·24.443 + 16.809- [n(x) I 

--•n•••-• -· ··- ··-- ~-· ··- ·-· . - .. cntenon Actual Allowable Aoceptable? 

Correlation coeffident 0.959 ~0 85 yes 

Confidence interval 1-32% ~1 0% yes 

Tole ranee Interval 23.1% .::, 25% yes 

• Indicates oorrelallon coefficient is undefined. 

Plot of LogarithmiC: Regreaslon Curvo 

& : ) y • 1M09lntx) • l4.«3 I 

g2S •--.- .. 

C2Q 

j ~ ~ 5 

0~--~------------~--~--~ 0.0 

Lcgarilllmic &.n-ma<y 

&.0 10.0 15.0 20.0 25.0 30.0 

PMCEMSRaaponsa 

PS. t1 Cotrtkulon T&at 

CaJQJIIIlon Spte8dlhtet 

Voralon 2-6 10/25104 

10/2912004 

VES009779 
CALCULATIONS FOR EXPONENTIAL CORRELATION 

Facility: Facility C Emission Unit: HWI 

Loeallon: Anytown USA Test Dates: November 1 to 4, 2002 

PM• 

PMCEMS 

..,AI "''".A TEC 

Run 

1 

x y y' = Ln{y) (J<r£ )" (y' rYl" (x,-x )(y' ,y) Y" 

5.8 3.5 1.253 92.29 2.614 15.533 2.115 0.744 

L 7.6 9 2.197 60.94 0.452 5.249 2.256 0.004 I 

1 3 1 7.8 7.8 2.054 57.86 0.665 6.203 2.272 0.048 

I • ~ • "" " 3.227 0.99 0.128 0.355 2.948 

2.809 28.16 0.004 0.320 2.453 

!.6601 18.551 0.0441 0.9021 2.531 

I 13 I 22.31 25.71 - - 3.2461 47."LI v. ·~ 14 l4.1 26.21 3.2661 75.5"1 A •• 

15 -- 24.71- - 26.21 3.2661 86.371 0.1571 3.681 

I 

I 

P$-11 Example 3-R 10-27-04.xls Exponential Cales 

Page 1 of 1 

10/2912004 

VES009780 
RESULTS OF EXPQNEHDAl CORRELADON 

fla.cllfty: f .c:IIHy C 

Loca~on: Anyt..., USA 

Variatlle: 

n ~ 

• 

s:-.. 

~ s., 

bo' 

.. 

b, 

s. I s..•l 

"A-· L 

;r· 

v, 

""' k, 

Tl' 

n% 

s,. 

,.--

' 

P$ 1~)-ft t0.21.()4 •• 

~ ... t~t~ 

Nurnbet ot dlsa po!ro ; I 

x-• 11n' (S..mof(ll.))•l 

~-: 

'/-•1 

Sw • s .... <<Y,., 

SC)'•l 

" 

. =: 

b, • O"'tJ,g'•: 

b, .. s~~~ 

'/'" at rnoWl x velve = I 

~ 1rom u.bfo •I 

crsl 

Cl • (e"UCL' .. i 

:mission Llmi 

~00%=: 

v, from t.abte =; 

u,. from table=: 

kr • o,.,"v, *: 

n • ttc'SL•, 

tTL'· Tl~ ~t,."1 oo%--s:· 

Sy• ~ 

" • 1-(Sl "2/Sy'2) • 

l iS\ 

v .... 

~ 

~ Te70I 

j 03 

659 

256 

0.0791 

D.35il 

2.870 ! 

I 0. 196< 

3.• 

1.4851 

1.1841 

1.7591 

~591 

0.812 

Emission Unit: HWI 

Test Data: November 1 t.o 4, 2002 

I correlation !<!Uation: y : 5.256 e ... ,.. • I 

s..nm.ryof A Crit.na for PS..11 

CrMrion AcWof 

Correlo1ion ooellldoi. 0.812 

-~065 

Confidence- 19.93% !~ 

T 01e1anc:e irterval 66.5% !25% . Indicates COTelation c:oefficienC is undefined . 

.. 

35 

g "" 

11 2S 

i J: l,. 

~ 

0 

0.0 

........ ............ 

Plot of EXponenti•l ~r.atiOn Cur'Ve 

• 

5.0 

Y • 5.2565e00785ol 

10.0 15.D ... o ,.. EM H~• 

PS-11 Cclrntl*" T.sa 

C.u!Wion S~Q~cllltl• 

V.r.tbn 2·8 1~ 

Accc~able? 

no 

no 

no 

:tOO 

'""""""' 

VES00978 1 
CALCULAI!ONS FOR POWER CORRELATION 

FacUlty: Facility c Emission Unit: HWI 

Location: Anytown USA Test Dates: November 1to 4, 2002 

Run 

1 

PMCEMS 

X 

PM• 

''"'"""""" 1 m""""'"" 1 Transfonned 1 CALCULATED STA1 

'I( = Ln(x) u " _ ' "'u' Y"\2 I 

•n 

y• 1 I - ' \ I I 

1.758 3.5 1253 0.785 2.614 1.810 310~ 

)28 n "ll ,. ,.,., n "ll~ n .. ~.., n .._ ., .._ .,. • .,.,. n ......,.._., 

2.054 0.6E 

.. '.... 3.227 0.024 0.1< ).055 

i1i20 

50 

r- ..... .. ... ........ ...... .... ... ......... ... .. .. ... 0.0( 

2.4851 19.61 2.9761 0.0251 0.011 [ -0.0171 2.6791 0.081 

13.21 2580L 20.11 3.001 [ 0.0041 0.0111 -0.0081 2.7931 0.0430 

15.7 2.754 23.4 3.153 0.012 0.080 0.031 3.001 0.023 

17.6 2.668 26.7 3.265 0.050 0.172 0.093 3.137 0.021. 

21.7 3.077 26.8 3.288 0.188 0.175 0.181 3.388 0.0099 

21 3.045 28.9 3.364 0.160 0.244 0.198 3.349 0.0002 

!.3[ 3.1051 25.71 3.246[ 0.2121 0.1421 0.1741 3.421 1 0.0303 

I 14 I 24.1

1 

3.182

1 

26.2

1 

3.266

1 

o.290I o.157l o.213l 3.513

1 

o.0613I 

15 24.7 3.207 26.2 3.266 0.317 0.157 0.2.23 3.543 0.0767 

P$-11 Example J.R 1D-27-04.XIs Power Cales 

Page 1 of 1 

1012912004 

VES009782 
RESULTS Of PQW£8 COBBELADON 

Facility: F1clllty C 

L.oulion: Atlytown USA 

Variable e .... ..., value I 

n Nun'mOrof data~ · 15' 

f 'K- • 11n"(&m cl (X', I)~ 2.644; 

s..., S.., • Sum((JI'rll'-)'2} • 2.96 

'(-. 1/n'(St.m of (y'J}. 2.870 

s..,. Sum(('(r'(-}"2}. 5.06 

s,. s,... Sum((JI', IC"-)(y'r'(•}}. 3.54 

., .. DQ' • V'-·b~·- : -0.292 

.,. bo• tf'Oo'• 0.747 

b, bt • S.ryJS ..... : 1.196 

s,_ s,_ • SORT(11(,...2XS""'(y'('-'(.)'2}} = 0.253 

'/"'- yM at mean x value : 2.870 

~ from t.able • 2 .160 

cr Cl • 'Sl'SCIRT(1hl} • 1 

LCl' LCL'• · CI' 2 .729 

UCl' UCL' • yo.' • Cr 3.011 

C1 Cl • (rUCl" • o"l Cl 'V2 2 .492 

El Emission Lirril = 17.5 

Cl% Cl'% • CtiEL "1001 = 14.28'14 

n• n'= n • 15 

v, v1 from table • 1.485 

"" un' rrM'I table • 1.184 

k, k, a u..-·v, • 1.759 

Tl' Tl' • k,S, • 0444 

LTL' Ln' • .. n- 2.425 

un· UTL' • Y .... ... n· 3.3 14 

n Tl • e"'TL' • e"l Tl 'l/2 8 .093 

TO% Tl% • TI/EL'100% = 46.4% 

S.· ~ • SORT(S.,I(r>-1}) = 0601 

, r' • 1-(Sl •21S,1'2} • 0 .824 

, r • SORT((1-5L •21SyA2)} = 0.906 

• lndieal4!s eotre ~ ooefflcl«<t It undeftnecl 

P$ol1~)oR lW7~...at 

Emission Unit; H't'Yl 

Test Dates: No'lerN:loer 1 to 4. 2002 

!correlation equation: y- 0.7466 x ..,,.,~ 

........... ... .._..... -·--.. ..-·-·-. ......- .. Criterion 

- ..,_ -·?I 

Conelatbn ooefticior It 0.908 ~0 65 yes ' 

Confidence intaval 14.28% :!1"" no I 

Tc:Rtance interVal <46.4% ~25% no _j 

lncfgle$ CO<TEiolion ooellic:iool is - -

Plot of Power Function Regreuion Curve 

40 

35 

:5 30 

! 25 c 8 20 c s 15 

l 10 

s 

0 

0.0 

--

y = 0 .7466x1 1959 

• 

s.o tO.O tS.O 20.0 

PM CEMS Ro•pon .. 

2&0 30.0 

S.t t Coi'NIIIon Te1t 

iciAtiUOn ~lhi'Gl 

ICo'2MM 

'"""""" 

VES009783 
SELECTION Of BEST MODEL 

Facility: Emioslon Unlt: HWI 

location: 

Facility C 

Anytown USA Tnt Dates: November 1 to 4, 2.002 

Mode! 1 Conelalion I > 0 .

85 7 I Confidence Interval I < 10%? 

eoellidenl (a) - half range percentage -

Unear I 0.898 Yes - 1t.29% No 

1.987 Yes 5.56% Yes 

1.959 Yea 7.32% Yes 

1.812 No 19.93% ••-

Power I 0 .. 908 Yes 14.28% 

(a) • indlcates correlalion coefficient Is undofiOOd; model does not sa~sf 

(b) Not applicable; criterion applies only to polynomial model. 

PS-11 Cmlllal:ion Tnt 

CaloAabon Spreacbheet 

Vet&ion 2-6 10125104 

I Tolerance inle<Val half 

range pen:enlage 

35.6' 

PS-11 Example 3-R 10.27-0ol.xls Best Model 

!;:25% 7 

No 

YeS 

Yes 

No 

No 

M1nlmax w~hin 

allowallle range? 

...i!!l 

N< 

Boot modo!: 

Docs model mool all 

cnlerla7 

No 

No 

Yos 

No 

N 

Logorllhmlc 

1009/2004 

VES009784 
Appendix A.4 

Complete Results of Example No.4 

Original Test Data 

VES009785 
CORRELATION TEST PM CEMS AND REFERENCE METHOD TEST DATA 

PMCEMS PM ooncentratioo 

re'monse mota em 

Run X v 

1 6.7 3.1 

2 7.6 7.4 

3 7.8 2.1! 

4 9.2 3.7' 

5 10.1 6.1 

6 11.1 2.8 

7 12.0 4.9 

8 182 8.3 

9 13.7 4.5 

10 14.6 8.2 

11 19.7 13.0 

12 18.0 11.1 

13 20.3 11.9 

14 202 10.6 

15 19.7 9.2 

PS-11 Example 4A-R 1().27-04.xls 

Facility: 

Location: 

Emission Unit: 

Teat Dates: 

Emlulon limit: 

14 

12 

g 10 . 

i 

~ 8 

• u 

g 

0 

6 

:I! 4 . ... 2 

0 

0 .0 

Test Data 

Facility 0 

Anytown USA 

Boiler 

November 1 to 4, 2002 

Value 

18.5 

• 

5.0 

• 

• 

llnl!l 

mg/acm 

• 

• 

• • 

10.0 

• 

• 

15.0 

PM CEMS Response 

• • • • 

• • 

20.0 25.0 

Page 1 of 1 

10/29/2004 

VES009786 
Facility: Facility D 

Location: Anytown USA 

PM CEMS 

reSDOnse 

Run X 

1 6.7 

2 7.6 

3 7.8 

4 9.2 

5 10.1 

6 11.1 

7 12 

8 18.2 

9 13.7 

10 14.6 

11 19.7 

12 18 

13 20.3 

14 20.2 

15 19.7 

P5-11 Exemple 4A·R 10-27-04.xls 

CALCULATIONS FOR UN EAR CORRELATION 

Emission Unit: Boiler 

Test Dates: November 1 to 4, 2002 

PM concentration 

mg/acm CALCULATED STATISTICAL PARAMETERS 

y (x,-xl• <YdY (x.-x"J{yrYl Y" <Y"rYY 

3.1 52.22 16.21 29.10 2.808 0.085 

7.4 40.03 0.07 -1.73 3.346 16.437 

2.1 37.54 25.27 30.80 3.465 1.864 

3.7 22.34 t1.74 16.20 4.302 0.362 

6.1 14.64 1.05 3.93 4.840 1.588 

2.8 7.99 18.72 12.23 5.437 6.956 

4.9 3.71 4.96 4.29 5.975 1.156 

8.3 1826 1.38 5.01 9.680 1.906 

4.5 0.05 6.90 0.60 6.991 6.206 

8.2 0.45 1.15 0.72 7.529 0.450 

13 33.33 34.50 33.91 10.577 5.872 

11.1 16.59 15.79 16.18 9.561 2.369 

11.9 40.62 22.78 30.42 10.935 0.930 

10.6 39.35 12.06 21.79 10.876 O.D76 

9.2 33.33 4.30 11.97 10.577 1.896 

linear Cales 

Page 1 of 1 

1012912004 

VES009787 
Facility: Facility D 

Location: Anytown USA 

P$-11 Example4A-R 1o-27-o4.xls 

RESULTS OF LINEAR CORRELATION 

Emission Unit: Boiler 

Test Da.tes: November 1 to 4, 2002 

!Correlation equation: y = -1.196 + 0.598 xI 

s 

_ ... , ..... _.. -· of .-.-- A· -·~~ Criteria for PS-11 --- - -. 

Criterion Actual Allowable Acceptable? 

Correlation coefficient 0.841 ~0 65 no 

Confidence interval 5.80% ~10% yes 

Tole ranee interval 16.3% 25% yes 

• Indicates correlation coeffiCient Is undefined. 

j ~ t ~ 10 

i 8 u 6 8 4 

:1 

.. 2 ~ 

0 

0.0 

Plot or Unoar Rogrosslon Une 

. • 

y = 0.5976x · 1.196 • 

• 

-. 

5.0 10.0 t5.0 20.0 2.:5.0 

PM CEMS Response 

PS·1t Correlation Test 

C..lwi•Uoo Sl>roodsheet 

Veroloo 2-ll 1012510< 

Linear Summary Linear Summary 

VES009788 
Facility: Facility D 

l.ocaUon: Anytown USA 

Run X 

~ 

v 7 

CALCULATIONS FOR POLYNOMIAL CORRELATION 

Emission Unit: Boiler 

Test Dates: Novemb9t 1 to 4 , 2002 

CALCUlATED 

7 1 •' !xviOir' l!'('yl'ldollal!"-v- 301 2._o1s ____ 2~_ _ 139 , _024 o.854 o.• 111 i Cl 

2.'i 

439 I 3.3361 56 I 427 4.032 1 1.348 0.2412 I U .Uf '""' I 

128 4.043 3,775 0 2158 25.27 1.87 

1 • 1 .. .., .,. 1rni 7.1641 34j 313 •.219 0.210 0.1296 •• • • • •• 

5 10.1 8.1 102 1.030 10A06 62 622 4_.23 2812 0.13 

• • • • "" ... 1,3681 15,1811 31 I 345 4.733 • ••• • •• 

1l 144 I 1 728 20 736 59 706 5.085 

I 029 109 720 151 2 749 9,437 

I • I '·"' ... o - 188) - 2.571 35.228 62 84-5 5.945 

10 14.8 8.2 213 3112 45,437 120 1.748 8.502 ... .... • ... 388 "r .... _., .,........ .. "'"'~ I:: n.&tt: .OA nn<> ~ I u I 10 11.1 324 5 832 1~ 976 200 I "'·~ l:l.i'q.q I .J.444 I u. 

.... 203 11.9 412 8365 169.818 242 .. ........ .... ....... ........... ... ' 

20.2 10.6 408 8 242 166.497 214 I 4, 

19.7 9.2 388 7,645 150,614 181 I 3.5 ~ u.vu 1 u.,;c.; "0~ .............. . 1. 

P& 11 Example 4"-R 11).27.()4 ... Polynomial Cates 

"-1ol1 

10129/2004 

VES009789 
RESlJL!& Of POLYN(M!! C068f' e !IOH 

Feclity: F'aeUity 0 

Location: Anytown USA 

EmiUion Unit BOiltr 

Test Dates: Nowmber 1 to 4, 1002 

v..- _ ..... Va ue 

IU'nbetddau 

lcom&atton eguauon: y • t.llt • .o.e:ts a • O.Mllt I? I 

" • IS s, St•Sum<•O = 2($ 

s, S2 • Slm {d2). 3,270 

s, S3 • Smt (1113). 55.562 

s. S4. Sum ():141)• 995.443 

s. SS•&m6") • ,07 

s, se • s..n !><1!'1 • 1.1'0< 

s, $1• Sl.l'n (l'i2yt1 = 29.463 

...... ....... ' I •S2St I I • 2 107E.o7 

.. btl• (~• 9\$)S7 S2SSS3-S7S2S2.S3SJS5.S.4S6St~ A = 6.189 

b, bl • (nSSS4•S5S3S2• S2SIS7-s2SeS:NI7S3n-S<S1S5YOoiA • 5 

b, b2 • (nS287•S1S8&2• S5SUa).S2S.'S5·S3Sen.S7$1$ tYO .. A-= 0.()<361 

s, S. • -(1((1>3- d ~YIZI • 1.843 

D D • n(82S4-SJ')tSt SlS2-StS4)•·S2(StSl·S2') • 2.107E..07 

Co 00=(•<>•····,.,. 7.963 

c, Ct • (SSS2·S1S4)1t) • · 12473 

c, C2 • (SIS3·92")'1) • 4 J4EE.02 

c, C3-. nS4·ST)ID • 01 -01 

c, C4 • (S1S2-nS3)'1) • ·7. 138E-03 

c, C5 • S2·~ ')/D • 2.561E-04 

\ 11 fl'om llitbllt • 2.179 

EL Emh:Wn Limit • 18,5 

Cl Cl• lf"S., "t(J1{~U.·I1Wl•• 1.421 

Cl% =CIIEL"tOO= 7.68% 

,, v1 from lai*t = 1.515 

"" uff from table = 1.208 

.. rf = 1/delta = 7.98 

•• I« = m'"\tf= 1.831 

Tl Tl = kl's = 3.373 

Tl n% = ·wa. "100% • 18.2% 

- • 1/h' Sl.rn d • 7.13 

Sy • IQI1 Sum of ( yi·y-)"2~n-l) • 3.M 

= 1-(S "2/Sr"2} = 0.7l1 

r r • SQO(( I.S,'2/SY'2)) • O.S55 

-..-..? .. >01 mtrimum ,.,_ y • -b,/2 ... 705 

1.25l<._ 25.38 

• l'ldiC::*t~~nt•~

PS-1 t Exetrde ofA.R 10..27-Cot..dl ""'''"' .... Summory 

Sunwnaty ol ~ Oott.n. fot P$-11 

Ctllll'!ion - - kc 3 I ? """""""'_.... .... 0805 ~085 

Conidlr<e- 7.1SI .. .! 10'1. - . 

Tcltlwloe inttwll 18.2% .f2!N .... - -corraladoo-~-

Cbed for CorTebtion Cwve wtnimUII'IIMUtm1 nn 

Careta»n CUM! rt'lf'l1mum po" 705 

MirWoom allowabae • ¥8iJe 6.7 

Ito oorrelation CIJl"ie mhl100m < ,...,um x Ylllue? ,. Correralio" tu'WJ maxim.m ~~ NA 

imil fof x f125')(. of m;uinum ave!UI NA 

Is ccnelation o.nve maxhum > exlr9polatlon lrnlt? NA 

Plol of Polynom.lat Regl'enkm Curve 

.. e 12 

! 10 

1 8 

~ • 

" • " .. 2 · 

0 

o.o 

)' . 0.4)436~'. 0615lb .. 6 18\l • 

• • • 

• • • 

• 

• • • • • 

5.0 10.0 15.0 20.0 

PM CEMS RupoMe 

,..,, Oori'Wib'l r• 

Clf~·~~~Mtl 

! 

25.0 

........ bfl2>4 t012 .... 

11/11'20(M 

VES009790 
CALCVLA TIONS FOR LOGARITHMIC CQRRELATION 

Facility: Facility 0 

location: Anytown USA 

Emission Unit: Boiler 

Test Dates: November 1 to 4, 2002 

PM · PM · 

CALCULATED STATISTICAL PARAMETERS 

Run x I x' = In (x) I y I (x,.·x'l' I (y,.y-)' l<x'rx'')(y,y')l y• I (Y"r Yil2 

1 6.71 1.9021 3.11 o.439l 16.2141 2.6691_ .. 2.2441 o.< 

2 1:6( 2.0281 ···- J AI 0.2881 0075( ·0.147( 3.1731 17. 

7.8 2.054 2.1 0.261 25.267 2.568 3.364 1.5 

4 9.2 2.219 3.7 0.120 11.742 1.185 4.580 0.774 

I ~ I 10.1 2.313 6.1 0.064 1.054 0.259 5.267 0.694 

~ 11.1 2.407 2.8 0.025 18.720 0.684 5.962 10.000 

I , I 12 2.485 4.9 0.006 4.958 0.178 6.536 2.678 

1 8 1 18.2 2.901 8.3 0.113 1.377 0.395 9.604 1.700 

I ft •~ ~ ft ~n 4.5 0.003 6.899 .0.138 7.512 9.073 

8.2 0.013 1.152 0.124 7.981 0.048 

11 I 19.71 2.9811 131 0.1131 34.4961- n2.441I 10.1871 7.9121 

12 18 - 2.890 11.1 0.106 15.787 1.293 9.523 2.4881 

13 20.3 3.011 11.9 0.199 22.785 2.127 10.408 2.2261 

14 20.2 3.006 10.6 0.194 12.064 1.530 10.372 0.0521 

15 19.7 2.981 9.2 0.173 4.299 0.862 10.187 0.975 

PS-11 Example 4A·R 10·27·04.X1s Logarilhmic Cales 

Page 1 of 1 

1012912004 

YES009791 
RESULTS OF LOGARITHMIC CORRELATION 

Facility: Facility D 

Locafton: Anytown USA 

Variable Equation 

n Number of data points ., 

x' x'- : 1/n'(Sum of (x'

1)) = 

slt·x· s,,. = Sum((x',-x'-)"'2) = 

y y- : 1/n'(Sum of (y,)) = 

s,. S.,, - Sum((yr y-)'2) = 

s,~ ~: Sum((x',x'-)(y,-y-)) = 

bo bo- y-·b,x'- " 

b, b,=~: 

s.. S.. = SQRT(1/(n·2XSum(yi'-y,)"'2)): 

Y'- y" at mean x valUe -

~ ~f om tableCl Cl = ~ ·s,· soRT(1/n) = 

EL Emissk>n Limit = 

Clo/o Cl% =CI/EL'100% -

n' n' =n = 

v, v1 from table = 

u, u., from table = 

k, k. = Un'-•v, = 

Tl n = kt'SL = 

TI% TI%- TI/EL' 100% = 

s.,. S.,. = SQRT(S,,(n· 1 )) = 

,. r" : 1-(S.. "215.,. '2) : 

r r = SQRT((1..S,"21S,'2)) = . lndicales correlaUon ooefficient is undeflMd . 

P$~1 1 EK:amplt ~A R 10 27~4.xts 

Emission Unit: Boiler 

Test Dates: November 1 to 4, 2002 

[Correlation equation: y • -11.764 • 7~3~Ln(x) I 

Value 

15 ..,.,.,,,,,,.,, y '" ..,._.,..,....,, ,......, -•••v• ou •~• ~ ~- • • 

2.57 Criterion Aclual Allowable Acceplable? 

2.18 Correlation ooeff-.cient 0.801 >0.85 no 

7.1 Confidence Interval 6.41% !:,10% yes 

177 Tole ranee interval 20.2% ~25% yes 

16.0 • Indicates correlation coeffident is undefined. 

· 11.764 

7.365 PlOt of L.c91rfthm~ Re~aton Cui'Yt: 

2.127 

7.127 

2.160 

1.187 

18.5 

6.41% 

15 

~------------------------------~~------, 

g 12 i • 

y = 7 .3641ln(l() • 11 . 711' • e 10 

1! 8 I • • ~ 6 

• " 2 I . 

.. • 

1.485 0 

1.184 OJ) 5.0 10.0 15 .0 20.0 2M 

1.759 PM CEMS Rtti)O(Ite 

3.74 

20.2% 

3.555 

0.842 

0.801 1012S/0o1 

&..ooarilhmie Surrf1'\ary 10/2912004 

VES009792 
CALCULAIJONS FOR EXPONENTlAL CORRELATION 

Facility: Facility 0 Emission Unit: Boiler 

Location: Anytown USA Test Dates: November 1 to 4, 2002 

PM < 

PM CEMS 

CALCULATEO! "'· 

' I y y I y' = Ln(y) I <x.-x1 I (l"ri"•J' I (X,·X·)(y'rYJ I '/" I IY ml' 

1 6.7 3.1 1.131 5.007 0.000 

~ ~ ~ 7.4 2.001 -1.121 0.590 

~ I ,\,J "-• I 0.742 6.631 0.260 ... "'., <'ll ~ "' <'lll'lol:l 2.438 0.006 

<J 1\1, 1 U , l I,QVQ 14,64 0.061 0.1f7 

~ • • • ., 8 • "'"'n 7 99 2.246 0.281 

0.0551 0 4531 1.6441 0.003 

18.26 0.085 1.248 2.223 0.011 

I a ,..,., ~ • .., 1.504 0.05 0.102 0.073 1.803 0.089 

10 14.6 8.2 2.104 0.45 0.078 0.188 1.887 0.047 

11 19.7 13 2.565 33.33 0.549 4.276 2.364 0.041 

12 18 11.1 2.407 16.59 0.340 2.374 2.205 0.041 

_10.3 11.9 2.4n 40.62 0.426 4.157 2.420 o.oo3 

2.361 39.35 0.288 3.367 2.410 0.002 

1.7 2.219 33.33 0.156 2.280 2.364 0.021 

P$-11 Example 4A·R 10.27-04.xls Exponential Cales 

Pago 1 or 1 

1012912004 

VES009793 
RESULTS Of EXPONEH!!AL CORRELATlOH 

F•eillty: hcllfty 0 

loatlon: Anylown USA 

I Vatiable EQUI..., VW. 

n ~oi<SIQ nts: 1 

x x- • tm•(sum ot ()rJ) • 13.93 

S.. S., • Sum((lt.·x-)'2) c 360.5 

...:h.. 

5.t 

... 

b, 

b, 

s.c I s... • ~~ 

~- CI' 

..!!... 

T, ""' 

TI' 

+ 

' 

P$11 ~ 4A-R 10.V<OtB 

y-. 1/n' (Sum of((~)· 1.82A 

• Sum((y'r'/'•)"2)• 4.66 

S"' • Sum({JCrx·)(Yr'/'-ll = 33.68 

b,; = f~,x- =I o.sZ3 

ba • • ...,..=I 1.687 

b1 = SJS.. =1 0.093, ) = 0.341 

'I'" at mean x vt*Jc • 1.824 

rom table = 2.160 

cr • SI. SORT(IInJ • 0.1~03 

LCe = Y-' • Cl' 1.634 

UCL' • v-' • Cl' 2.015 

iiiC'G 

~ Jmil = 18.5 

Cl% • CUEL' IOO% = 6.41% 

· = n = 15 

v1 lrom tabte = 1.485 

\Iff from tabte = 1.164 

kr• u,.,"v1= ~ 

Tl' • k."SL = 0.600 

L TL' • v-' • TI' 1.224 

lJTL' :;; ... • + 'r1' 2.424 

UTL' • e"LTI,'Ja 3.~7 

n% • TIIEL' IOO% • 21.3% 

• SORT(S /(n-1)) = o.sn • I·(SI. "2/Sy1'2) = 0.650 

r • SORT{{1·SI.•21S,•2)) = 0.606 

E,mission Unk: Boiler 

Test OatH: Novombt<t 1 to 4 , 2002 

[correlation equation: y ~ 1.687 e o.••• • I 

-· .. ·- -~ UIH"'w "'"'""'• ouo r,;ro o 1 

Crierion ........ - -1>10? 

CorreJation coeffic:iet It 0.808 ~O &S no 

Cor1f1denoe In- 6.41% !~ ,... 

Toleranceintefval 21.3% !25 ... ,... . Indicates~ ooet6donlls .-• 

P&ot of Exponentia.t Regteulon CutW 

••r------------------------------------, • 12 • 

Y s 1.6872aP·ot34• i 10 

i : 

" 

• 

l • 

• 2 • 

----------~----~----------~ 0.0 ... 

E~$u'ml;arf 

10.0 ,., :20.0 

PM CEMS AN~~C nte 

PS·1 1 Contlation r•• 

C.rculallon Sotcllldii!Ht 

versbft 2:-e 101'Z61CM 

0 

,..,.,.. 

VES009794 
Facility: Faelllty D 

Location: Anytown USA 

Run 

1 

PMCEMS1 

X 

6.7 

7. 

7. 

CALCULATIONS FOR POweR CORRELATION 

Emission Unit: Boiler 

Test Dates: November 1 to 4, 2002 

PM• 

I ranslotmed 

)( = Ln(x) y = Ln(y) 

1.902 1.131 0.0068 

2:028 7.4 2.001 0.288 0.031 -0.0951 1.1961 0.6486 I 

2.054 2.1 0.742 0.261 1.111 0.5531 1.2271 0.2348 1 9 3.7 1.308 0.120 ~ ..... .,ft . .. "V\. ,.. .., ......... 

. 3131 6.11 1.8081 0.0641 0.000 

!.4071 2.81 1.o:lol o.o25l o.s31l __ o.126l 1.63!11 _ _ o.3h7 

' 12 2.485 1.589 0.019 

ft 18.2 2.901 2.116 0.098 

I "' 13.7 2.617 4.5 1.504 0.003 0.102 -0.017 1.885 0.1454 

• A 14,6 2.681 8.2 2.104 0.013 0,078 0,032 1.960 0,0208 

'' 19.7 2.981 13 2.565 0.173 0.549 0.308 2.310 0.0648 

12 18 2.890 11.1 2.407 0.106 0.340 0.190 2.205 0.0409 

13 20.3 3.011 11.9 2.477 0.199 0.426 0.291 2.345 0.0172 

• • AA A 3,006 10.6 2,361 0,194 0,288 0,236 2,340 0,0004 

2.981 9.2 2.219 0.173 0.156 0.164 2.310 0.0083 

P5-11 Examp!e4A-R 10-27-04.xls Power Cales 

Page 1 of 1 

10/2912004 

VES009795 
RESULTS Of POWER CORRELADON 

Fadllty: Facl~ ty 0 

Loc-o~tion: Anyt.own USA 

Varlalllo 

.r 

s.v 

~ 

~ 

bo' 

... 

b, 

s, 

fA-, 

-"-

cr 

v, 

"

k" , 

u 

~ 

r 

' I ndicatel I 

P$11 £);aqllle fA.R 11).27.Qii.Q 

Vakle 

----:;s ,.._: lln'(Sum of (IC'J) = 2.565 

s,.=&.m<<"•x<-1"2>= 2.1s 

y'-. 1/n'(Sum of (f,)). 1.824 

• &.m((y',-y'-)"2). 4 .66 

s,.,. &.m((IC', IC'•)fY,-y'-)) = 2.55 

S..• SORl 

bo' • t-<>,x- = -1 1n 

...... ,,.., 0, t>,•S~= 1 ml I 0~01 'I" e1 me8n x value = 1 824 

Cl • l _ltlrcm table =j ~~~~ 

:misalon %~""'' I ... 

v, from table =I 1.4i 

u, ~om table •I 1.184 1 

K, • u,.,:v, •I t.759l 

Tl' = k,St =! 

··00% "I '"'""'I :S.,./(~1)) ~ "' ~n 

I" • 1·(SL"2/lV21 •J 0.612! 

r • 50RTI(t-Sl.•21S(2ll •I 0.762 

t is 1 (i8ri 

Emin~ Uni1: Boiler 

Tes-t Oates: Ncwember 1 to 4t 2002 

!correlation equation: y • 0.3084 x '·'

001 I 

--·· .. ··- -· -· -·-·-,__, . __.. .. Crilerico1 

- - A<cot>WH? 

Com!lation coefficienl 0.782 ~o as no 

Confidence interval 6.76% ~1~ l/00 

T o1eranoe irr.ervaJ 22.6% ~25" yeo 

lncicates ""'"""lion coe111aent Is IMldef*lod. 

Plot of Power Function Regress Jon Curve 

t214~ j ----==- • i 10 

" 8 ~ 5 6 

• • 

Y = 0.3084x' ,. 

" l 4 

2 

• 

• • 

00:~~-----;~----~~------~~----~-------- 50 10.0 1S,O 20.0 1tl0 

PM CEMS Ro1pon•o 

--

8-11 Comt~Mkw\ Ttil 

leullliofl $pNCletlftt 

IC/2&01 

·-

VES009796 
Facility: 

LDcation: 

Model 

Li1ear 

Potyr[omlal 

Logarillvnic 

tial 

Power 

Facility D 

Anytown USA 

Correlation 

coeffioont(a) 

0.841 

0.855 

0.801 

0.806 

0.782 

?. 0.85 7 

No 

Yes 

No 

No 

No 

SELECTION OF BEST MODEL 

Emission Unit Boiler 

Test Datos: November 1to 4, 2002 

Confidence interval ~ 10%? Tolerance i1terval haH !:_ 25%7 Mln/max~ln Does model meet au 

haW range percentage range percentage alloWable range? crit.oria? 

5.80% Yes 18.3% Yes (b) No 

7.68% Yes 18.2% Yes No No 

6.4 1% Yes 20.2% Yes b No 

6A1% Yes 21.3% Yes No 

6.76% Ye.s 22.6% Yes No 

(a) • indicates corro!allon coomctonl la undetlned; model does not satisfy criterion. 

(b) Not applicable: criterion applteo only to polynomial model. 

PS-t 1 COtrell.tion T V$l 

Ge"~liOn ~.;h!Q9t 

Version 2.& 

·~ 

Best modo I: Nono eccoptablo 

P5-11 Examp!e4A-R 10.27-04.>do Best Model 10129/2004 

VES009797 
Appendix A.5 

Complete Results of Example No. 4 

With Five Additional Test Runs 

VES009798 
CORRELATION TEST PM CEMS AND REFERENCE METHOD TEST DATA 

PMCEMS PM concentration 

msoonse mg/acm 

Run X y 

1 6.7 31 

2 7.6 7A 

3 7.8 2.1 

4 9.2 3.7 

5 10.1 6.1 

6 11.1 2.8 

7 12.0 4.9 

8 18.2 8.3 

9 13.7 4.5 

10 14.6 8.2 

11 19.7 13.0 

12 18.0 11.1 

13 20.3 11.9 

14 20.2 10.6 

15 19.7 9.2 

16 5.2 2.4 

17 8.6 4.8 

18 14.1 6.2 

19 14.5 7.0 

_ 20_ 17.3 ----- 8.6 

PS.11 Example 4B-R 20 Runs 10-27~.><18 

Facility: 

Locatlon: 

Emi$Sion Unit 

Test Dates: 

Emi$Sion limit: 

14 

12 

.§ 10 

j;j 

~ 8 .. 

u 

g 6 

0 

ll 4 ... 2 

0 

0.0 

Test Dara 

Facility D 

AnytownUSA 

Boiler 

1111-111.U02,1n-118103 

Vafue 

18.5 

• • 

5.0 

• 

• 

• 

Units 

mglacm 

• 

• 

• • 

10.0 

• 

• • 

• 

15.0 

PM CEMS Response 

• 

• • 

• 

• 

• 

• 

20.0 25.0 

Page 1 ol 1 

10/2912004 

VES009799