© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2: Waste Sludge Incineration: Steady-State Nonlinear DR and Detection

© Universidad de Guanajuato, Mexico © University of Ottawa, Canada, 2004

Case Study #2: Waste Sludge Incineration:

Steady-State Nonlinear DR and Detection of Gross Errors Through Analysis

*Flowsheet taken from Felder & Rousseau “Elementary Principles of Chemical Processes”, page 502.


Case Study #2

Dryer Incinerator

Boiler

Sludge F1

Sat. Vap. F2

Conc. Sludge F3 Waste Gas F14

Sat. Liq. F4

Sat. Vap. F5

(p=4 bar)

Waste Gas F8

Boiler Feed Water F6

#6 Fuel Oil F7

Air F10

Cool Water F16Hot Water F15

Preheated Air F9

Natural Gas F11

Hot Water F17

Cool Water F18

Air F13

Preheated Air F12


Case Study #2

Incinerator Problem Engineering Problem

Accounting ProblemManagement Problem


Case Study #2

Stream # Flow Rate (kg/day)

Temperature (ºC)

Mass Fraction Solid

1 23800 23 0.36

2 17100 N/A N/A

3 ? N/A ?

4 ? N/A N/A

5 ? N/A N/A

6 30300 N/A N/A

7 3500 N/A N/A

8 ? N/A N/A

9 ? 127 N/A

Table 2.1a: List of measured and unmeasured variables


Case Study #2

Stream # Flow Rate (kg/day)

Temperature (ºC)

Mass Fraction Solid

10 60000 24 N/A

11 1000 N/A N/A

12 ? 124 N/A

13 160000 26 N/A

14 ? N/A N/A

15 14600 135 N/A

16 14500 35 N/A

17 35600 135 N/A

18 35700 39 N/A

Table 2.1b: List of measured and unmeasured variables


Case Study #2

Gross Errors?

Need to change the problem from

NONLINEAR to BILINEAR!!

See the end of this module.


Case Study #2

#6 Fuel Oil:

87% C, 10% H, 0.84% S, and 2.16% inert (weight percent)

HHV = 3.75 x 104 kJ/kg

Natural Gas:

90% CH4, 10% C2H6 (mole percent)

Boiler:

Efficiency = 62%

25% Excess Air

Sludge:

Cp of solids = 2.5 kJ/kg·ºC

Liquid ~ Water


Case Study #2

Dryer:Efficiency = 55%Pressure at 1 bar on sludge side, and 4 bar on steam sideAll steam and condensate flows are saturated

Incinerator:Sludge must enter at higher than 75% consistencyHHV of concentrated sludge = 19000 kJ/kg dry solids195 SCM natural gas/tonne wet sludge2.5 SCM air/10000 kJ of sludge HHV for complete combustion100% Excess Air for sludge and natural gas

Standard Deviation:Flows 500 kg/day, Temperature 2ºC, Composition

0.03


Case Study #2

1) Define the y and z matrices for the measured variables and unmeasured variables.

2) Set up the V matrix, scaling down if necessary.3) Fill in the matrix with the measured values and the matrix with

guesses for the unmeasured variables, scaling down appropriately.4) Determine the mass, component, and heat balances for the process, f(

, ) = 0, scaling down parameters where necessary.

5) Determine the Jacobian matrices, Ay and Az, and solve them using the values from the and matrices.

6) Calculate b0 = Ay + Az – f ( , )

7) Carry out QR Factorization of the Az matrix, separating the Q matrix into Q1 (m x n) and Q2 (m x m-n) matrices, and the R matrix into an R1 (n x n) matrix.

0y

0y

0y

y

0y

0z

0z

0z

0z

z


Case Study #2

8) Calculate = – V(Q2TAy)T[(Q2

TAy)V(Q2TAy)T]-1(Q2

TAy – Q2Tb0)

9) Calculate = R1-1Q1

Tb0 – R1-1Q1

TAy - R1-1R2

(the last term is not necessary if R2 is a zero matrix)

10) Replace , , and b0 with , , and b1.

11) Repeat steps 5 to 10 until the difference between both and , and and are very small.

0y

0y

0y

0z

1y

1y

1y

1z

1z

ˆN rz

ˆny1

ˆny

ˆnz

1nz


Step 1:

1 1

2 2

3 6

4 7

5 10

6 11

7 13

8 15

9 16

10 17

11 18

12 1

13 1

14 9

15 10

16 12

17 13

18 15

19 16

20 17

21 18

y F

y F

y F

y F

y F

y F

y F

y F

y F

y F

y y F

y X

y T

y T

y T

y T

y T

y T

y T

y T

y T

,

1 3

2 4

3 5

4 8

5 9

6 12

7 14

8 1

9 3

10 3

z F

z F

z F

z F

z Fz

z F

z F

z y

z y

z x


Step 2:

V(1,1), V(2,2), V(3,3), … , V(11,11) = 0.000025 (after scaling)

V(12,12) = 0.0009 (not necessary to scale)

V(13,13), V(14,14), … , V(21,21) = 0.0004 (after scaling)

Assumed that Covariance = 0


Step 3:

0

0.238

0.171

0.303

0.035

0.600

0.010

1.600

0.146

0.145

0.356

ˆ 0.357

0.36

0.23

1.27

0.24

1.24

0.26

1.35

0.35

1.35

0.39

y

, 0

0.11

0.30

0.30

0.64

0.60ˆ

1.60

1.73

0.64

0.22

0.78

z


Step 4:F3 + F11 + F12 - F14 = 0

F4 - F5 = 0

F15 – F16 = 0

F5 - F6 = 0

Dryer Incinerator

Boiler

Sludge F1Sat. Vap. F2

Conc. Sludge F3 Waste Gas F14

Sat. Liq. F4Sat. Vap. F5

(p=4 bar)

Waste Gas F8


#6 Fuel Oil F7

Air F10


Preheated Air F9

Natural Gas F11

Hot Water F17

Cool Water F18

Air F13

Preheated Air F12

F12 - F13 = 0

F9 - F10 = 0

F7 + F9 - F8 = 0

F17 – F18 = 0

Mass Balances:


Step 4:

Component Balances:

DryerSludge F1

Sat. Vap. F2

Conc. Sludge F3

Solids:

F1x1 - F3x3 = 0

Water:

F1y1 - F3y3 - F2 = 0

Normalization Constraints:

x3 + y3 - 1 = 0

x1 + y1 - 1 = 0


Step 4:

Balances on Heat Exchangers:

Air F10


Preheated Air F9

Hot Water F17

Cool Water F18

Air F13

Preheated Air F12

1.046F9T9 – 1.046F10T10 + 4.18F16T16 – 4.18F15T15 = 0

1.046F12T12 – 1.046F13T13 + 4.18F18T18 – 4.18F17T17 = 0


Step 4:

Heat Flow Through Boiler:

(0.62)(HHVOIL)(F7) = (h5)(F5) – (h6)(F6)

23250F7 – 2737.6F5 + 83.9F6 = 0

2.325F7 – 0.27376F5 + 0.00839F6 = 0

Boiler

Sat. Vap. F5

(p=4 bar)

Waste Gas F8


#6 Fuel Oil F7

Preheated Air F9


Step 4:

Air / Oil Ratio in Boiler:

Boiler

Sat. Vap. F5

(p=4 bar)

Waste Gas F8


#6 Fuel Oil F7

Preheated Air F9

For 1 kg of oil:

0.87 kg C 0.0724 kmoles C 0.10 kg H 0.0992 kmoles H 0.0084 kg S 0.000262 kmoles S

O2 required (25% excess):

For C (0.0724)(1.25)(1) For H (0.0992)(1.25)(0.25) For S (0.000262)(1.25)(1)

Total 0.1218 kmoles O2

16.83F7 – F9 = 0

0.5801 kmoles Air

16.83 kg Air


Step 4:

Heat Flow Through Dryer:DryerSludge F1Sat. Vap. F2

Conc. Sludge F3

Sat. Liq. F4Sat. Vap. F5

(p=4 bar)

(x1)(F1)(Cps)(TB-T1) + (y1)(F1)(CpH2O)(TB-T1) + (F2)(LvH2O) – (0.55)(h5-h4)(F5) = 0

250x1F1 – 2.5x1F1T1 + 418y1F1 – 4.18y1F1T1 + 2257F2 – 1173.1F5 = 0

0.25x1F1 – 0.0025x1F1T1 + 0.418y1F1 – 0.00418y1F1T1 + 2.257F2 – 1.1731F5 = 0


Step 4:

NG / Sludge Ratio:IncineratorConc. Sludge F3

Natural Gas F11

195 SCM/tonne wet sludge

8.7054 kmol NG/tonne F3

7.8348 kmol CH4 125.67 kg 0.1257 tonnes

0.8706 kmol C2H6 26.18 kg 0.0262 tonnes

0.1519 tonnes NG/tonne F3

0.1519F3 – F11 = 0


Step 4:

Air / Sludge + NG Ratio:

Therefore for 1 kg of NG: 0.8275 kg CH4 0.0516 kmoles CH4 0.1725 kg C2H6 0.0057 kmoles C2H6

O2 required (100% excess):

For CH4 (0.0516)(2)(2) For C2H6 (0.0057)(2)(7/2)

Total 0.2464 kmoles O2

1.1728 kmoles Air

32.01 kg Air/kg NG

0.9 kmole CH4 14.44 kg (82.75%) 0.1 kmole C2H6 3.01 kg (17.25%)

Since NG is 10 mol% C2H6 and 90 mol% CH4:

IncineratorConc. Sludge F3

Natural Gas F11

Preheated Air F12

To burn NG:


Step 4:

Air / Sludge + NG Ratio:

IncineratorConc. Sludge F3

Natural Gas F11

Preheated Air F12

To burn sludge:

(2.5 SCM air/10000 kJ)(19000 kJ/kg solid) (x3 kg solid/kg sludge) = 4.75x3 SCM air/kg sludge

0.2121x3 kmol air/kg sludge

6.1496x3 kg air/kg sludge

100% Excess:

12.3x3 kg air/kg sludge

12.3F3x3 + 34.01F11 – F12 = 0

Combining the air needed for the both the sludge and the NG:


Step 4:

Summary of Constraints:

1) 2.325y4 – 0.27376z3 + 0.00839y3 = 0

2) 0.25y12y1 – 0.0025y12y1y13 +0.418z8y1 – 0.00418z8y1y13 + 2.257y2 – 1.1731z3 = 0

3) y1y12 – z1z10 = 0

4) y1z8 – z1z9 – y2 = 0

5) z10 + z9 – 1 = 0

6) y12 + z8 – 1 = 0

7) z2 – z3 = 0

8) z1 + y6 + z6 – z7 = 0

9) z6 – y7 = 0


Step 4:

Summary of Constraints:

10) z5 – y5 = 0

11) z3 – y3 = 0

12) y4 + z5 – z4 = 0

13) 16.83y4 – z5 = 0

14) 0.1519z1 – y6 = 0

15) 12.3z1z10 + 34.01y6 – z6 = 0

16) 1.046z5y14 – 1.046y5y15 + 4.18y9y19 – 4.18y8y18 = 0

17) 1.046z6y16 – 1.046y7y17 + 4.18y11y21 – 4.18y10y20 = 0

18) y8 – y9 = 0

19) y10 – y11 = 0


Step 5:

0 0 0.0084 2.325 0 0 0 0 0 0 0

0.3567 2.257 0 0 0 0 0 0 0 0 0

0.36 0 0 0 0 0 0 0 0 0 0

0.64 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0

0 0 0 16.83 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 34.01

yA

0 0 0 0 0

0 0 0 0 0.251 0 0 5.643 1.463 0 0

0 0 0 0 0 0 0.272 0 0 5.643 1.6302

0 0 0 0 0 0 0 1 1 0 0

0 0 0 0 0 0 0 0 0 1 1


Step 5:

0 0 0 0 0 0 0 0 0 0

0.0594 0.0009 0 0 0 0 0 0 0 0

0.238 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0.6276 0.6276 0 0 0.6103 0.6061 0 0

0 0 0

0 1.6736 1.6736 0 0 1.4881 1.4923

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0


Step 5:

0 0 0.2738 0 0 0 0 0 0 0

0 0 1.1731 0 0 0 0 0.0993 0 0

0.78 0 0 0 0 0 0 0 0 0.11

0.22 0 0 0 0 0 0 0.238 0.11 0

0 0 0 0 0 0 0 0 1 1

0 0 0 0 0 0 0 1 0 0

0 1 1 0 0 0 0 0 0 0

1 0 0 0 0 1 1 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0

0 0 0 1 1 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0

0.1519 0 0 0 0 0 0 0 0 0

9.594 0 0 0 0 1

zA

0 0 0 1.353

0 0 0 0 1.3284 0 0 0 0 0

0 0 0 0 0 1.297 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0


Step 6:

0 0 0 0 0

0.0000

0.0847

0.0001

0.1281

1.0000

1.0000

0

0.0000

0

ˆ ˆ ˆ ˆ( , ) 0

0

0.0000

0

0

1.0553

0.0347

0.2132

0

0

y zb A y A z f y z


Step 7:

1

0 0 0.1749 0 0 0.0000 0.0000 0.0126 0.0003 0.0029

0 0 0.7493 0 0 0.0000 0.0000 0.0423 0.0011 0.0099

0.0806 0 0.0000 0 0 0.0362 0.0343 0.0006 0.0003 0.0027

0.0227 0 0.0000 0 0 0.0102 0.0097 0.2309 0.1035 0.9554

0 0 0 0 0 0 0 0 0.994

Q

3 0.1051

0 0 0 0 0 0 0.0000 0.9709 0.0245 0.2264

0 1.0000 0 0 0 0 0 0 0 0

0.1033 0 0.0000 0 0 0.5532 0.8266 0.0000 0.0000 0.0000

0 0 0 0 0 0.5067 0.3391 0.0020 0.0009 0.0090

0 0 0 0 0.5154 0 0 0 0 0.0000

0 0 0.6387 0 0 0.0000 0.0000 0.0461 0.0012 0

.0108

0 0 0 1.0000 0 0 0 0 0 0

0 0 0 0 0.5154 0 0 0 0 0.0000

0.0157 0 0.0000 0 0 0.0071 0.0067 0.0001 0.0001 0.1547

0.9910 0 0.0000 0 0 0.0610 0.0830 0.0052 0.0024 0.0241

0 0 0 0 0.6847 0 0 0 0 0.0000

0 0 0 0 0 0.6573 0.4399 0.0025 0.0011 0.0117

0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0


Step 7:

2

0.1473 0.0977 0.0648 0.0024 0.9469 0.0860 0.1728 0 0

0.4942 0.2036 0.1180 0.0219 0.1140 0.1567 0.3147 0 0

0.5864 0.3539 0.2125 0.0438 0.2713 0.2823 0.5670 0 0

0.0001 0.0211 0.0017 0.1485 0.0021 0.0023 0.0014 0 0

0.0000 0.

Q

0023 0.0002 0.0163 0.0002 0.0003 0.0002 0 0

0.0490 0.0252 0.0113 0.0332 0.0108 0.0150 0.0309 0 0

0 0 0 0 0 0 0 0 0

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0 0

0.0220 0.3567 0.2968 0.0413 0.0041 0.3942 0.5051 0 0

0.0080 0.

6453 0.3305 0.1029 0.0082 0.4391 0.0725 0 0

0.6201 0.2121 0.1206 0.0263 0.1255 0.1603 0.3218 0 0

0 0 0 0 0 0 0 0 0

0.0029 0.2334 0.7578 0.0372 0.0030 0.3217 0.0262 0 0

0.0007 0.1389 0.0114 0.9777 0.0136 0.0151 0.0090 0 0

0.0477 0.030

5 0.0171 0.0085 0.0222 0.0228 0.0460 0 0

0.0038 0.3100 0.3217 0.0494 0.0039 0.5727 0.0348 0 0

0.0198 0.2515 0.2156 0.0384 0.0140 0.2864 0.4249 0 0

0 0 0 0 0 0 0 1.0000 0

0 0 0 0 0 0 0 0 1.0000


Step 7:

1

9.6812 0 0 0 0 0.8877 0.1033 0.0054 0.0025 1.3497

0 1.0000 1.0000 0 0 0 0 0 0 0

0 0 1.5656 0 0 0.0000 0.0000 0.0744 0.0000 0.0000

0 0 0 1.0000 1.0000 0 0 0 0 0

0 0 0 0 1.9403 0 0 0 0 0

0 0 0 0 0 1.9734 0.5532 0.0024 0.0011 0.0785

0 0 0 0 0 0 0.8266 0.0

R

023 0.0011 0.1161

0 0 0 0 0 0 0 1.0300 0.0254 0.0072

0 0 0 0 0 0 0 0 1.0057 0.9911

0 0 0 0 0 0 0 0 0 0.1380

2

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

R


Step 8:

1

1 0 2 2 2 2 0 2 0

0.2293

0.1277

0.3147

0.0359

0.6045

0.0154

1.5988

0.1482

0.1482

0.3590

ˆ ˆ ˆ( ) [( ) ( ) ] ( ) 0.3590

0.3798

0.2303

1.2570

0.2530

1.2069

0.2931

1.3627

0.3374

1.3794

0.3605

T T T T TT T

y y y yy y V Q A Q A V Q A Q A y Q b


Step 9:

1 1

1 1 1 0 1 1 1

0.1017

0.3147

0.3147

0.6404

0.6045ˆ ˆ

1.5988

1.7159

0.6202

0.1475

0.8525

T T

yz R Q b R Q A y


Steps 10 & 11:

0.2294

0.1277

0.3147

0.0359

0.6045

0.0155

1.5988

0.1482

0.1482

0.3590

ˆ 0.3590

0.3803

0.2303

1.2571

0.2529

1.2070

0.2930

1.3626

0.3375

1.3793

0.3606

y

0.1017

0.3147

0.3147

0.6404

0.6045ˆ

1.5988

1.7160

0.6197

0.1424

0.8576

z


Results:

VariablesRaw

MeasurementReconciled

MeasurementAdjustment

(%)F1 23800 22940 -3.61

F2 17100 12770 -25.32

F3 ? 10170 N/A

F4 ? 31470 N/A

F5 ? 31470 N/A

F6 30300 31470 3.86

F7 3500 3590 2.57

F8 ? 64040 N/A

F9 ? 60450 N/A

F10 60000 60450 0.75

Table 2.2a: Measured and reconciled data


Results:

VariablesRaw



(%)F11 1000 1550 55.00

F12 ? 159880 N/A

F13 160000 159880 -0.08

F14 ? 171600 N/A

F15 14600 14820 1.51

F16 14500 14820 2.21

F17 35600 35900 0.84

F18 35700 35900 0.56

x1 0.36 0.3803 5.64

x3 ? 0.8576 N/A

Table 2.2b: Measured and reconciled data


Results:

VariablesRaw



(%)y1 ? 0.6197 N/A

y3 ? 0.1424 N/A

T1 23 23.03 0.13

T9 127 125.71 -1.02

T10 24 25.29 5.38

T12 124 120.70 -2.66

T13 26 29.30 12.69

T15 135 136.26 0.93

T16 35 33.75 -3.57

T17 135 137.93 2.17

T18 39 36.06 -7.54

Table 2.2c: Measured and reconciled data


Interpretation:

Dryer IncineratorF1 = 23800

F2 = 17100

F3 = 23800 – 17100 = 6700

F11 = 6700(0.1519) = 1018

Raw Measurements:


Interpretation:

Dryer IncineratorF1 = 22940

F2 = 12770

F3 = 22940 – 12770 = 10170

F11 = 10170(0.1519) = 1545

!Reconciled Measurements:


Interpretation:

Control natural gas feed based on

incinerator temperature

Incorrect measurement of vapor flow from dryer (F2)

Insufficient natural gas fed to incinerator (F11)

Incinerator temperature too low

Install more measurement devices

throughout the process


Case Study #2

MATLAB code used to solve Case Study #1:

f=[0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0];V=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];

for I=1:11,V(I,I)=0.005^2;end;


Case Study #2

for I=12,V(I,I)=0.03^2;end;

for I=13:21,V(I,I)=0.02^2;end;

y=[0.238; 0.171; 0.303; 0.035; 0.600; 0.010; 1.600; 0.146; 0.145; 0.356; 0.357; 0.36; 0.23; 1.27; 0.24; 1.24; 0.26; 1.35; 0.35; 1.35; 0.39];

z=[0.11; 0.3; 0.3; 0.64; 0.6; 1.6; 1.73; 0.64; 0.22; 0.78];

flag=1;

while flag>0,SSEy=0;SSEz=0;


Case Study #2

Ay=[0 0 0.00839 2.3250 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0.25*y(12)-0.0025*y(12)*y(13) +0.418*z(8)-0.00418*z(8)*y(13) 2.257 0 0 0 0 0 0 0 0 0 0.25*y(1)-0.0025*y(1)*y(13) -0.0025*y(12)*y(1)-0.00418*z(8)*y(1) 0 0 0 0 0 0 0 0;y(12) 0 0 0 0 0 0 0 0 0 0 y(1) 0 0 0 0 0 0 0 0 0;z(8) -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 16.83 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 34.01 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 -1.046*y(15) 0 0 -4.18 *y(18) 4.18*y(19) 0 0 0 0 1.046*z(5) -1.046*y(5) 0 0 -4.18*y(8) 4.18*y(9) 0 0;0 0 0 0 0 0 -1.046*y(17) 0 0 -4.18*y(20) 4.18*y(21) 0 0 0 0 1.046*z(6) -1.046*y(7) 0 0 -4.18*y(10) 4.18*y(11);0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0];

Az=[0 0 -0.27376 0 0 0 0 0 0 0; 0 0 -1.1731 0 0 0 0 0.418*y(1)-0.00418*y(1)*y(13) 0 0; -z(10) 0 0 0 0 0 0 0 0 -z(1); -z(9) 0 0 0 0 0 0 y(1) -z(1) 0; 0 0 0 0 0 0 0 0 1 1; 0 0 0 0 0 0 0 1 0 0; 0 1 -1 0 0 0 0 0 0 0; 1 0 0 0 0 1 -1 0 0 0; 0 0 0 0 0 1 0 0 0 0; 0 0 0 0 1 0 0 0 0 0; 0 0 1 0 0 0 0 0 0 0; 0 0 0 -1 1 0 0 0 0 0; 0 0 0 0 -1 0 0 0 0 0; 0.1519 0 0 0 0 0 0 0 0 0; 12.3*z(10) 0 0 0 0 -1 0 0 0 12.3*z(1); 0 0 0 0 1.046*y(14) 0 0 0 0 0; 0 0 0 0 0 1.046*y(16) 0 0 0 0;0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0];


Case Study #2

f(1)=2.3250*y(4)-0.27376*z(3)+0.00839*y(3);f(2)=0.25*y(12)*y(1)-0.0025*y(12)*y(1)*y(13)+0.418*z(8)*y(1)-0.00418*z(8)*y(1)*y(13) +2.257*y(2)-

1.1731*z(3);f(3)=y(1)*y(12)-z(1)*z(10);f(4)=y(1)*z(8)-z(1)*z(9)-y(2);f(5)=z(10)+z(9)-1;f(6)=y(12)+z(8)-1;f(7)=z(2)-z(3);f(8)=z(1)+y(6)+z(6)-z(7);f(9)=z(6)-y(7);f(10)=z(5)-y(5);f(11)=z(3)-y(3);f(12)=y(4)+z(5)-z(4);f(13)=16.83*y(4)-z(5);f(14)=0.1519*z(1)-y(6);f(15)=12.3*z(1)*z(10)+34.01*y(6)-z(6);f(16)=1.046*z(5)*y(14)-1.046*y(5)*y(15)+4.18*y(9)*y(19)-4.18*y(8)*y(18);


Case Study #2

f(17)=1.046*z(6)*y(16)-1.046*y(7)*y(17)+4.18*y(11)*y(21)-4.18*y(10)*y(20);f(18)=y(8)-y(9);f(19)=y(10)-y(11);b=Ay*y+Az*z-f;[Q,R]=qr(Az); for I=1:19,for J=1:10,Q1(I,J)=Q(I,J);end;end; for I=1:19,for J=11:19,Q2(I,J-10)=Q(I,J);end;end;


Case Study #2

for I=1:10,for J=1:10,R1(I,J)=R(I,J);end;end; yhat=y-V*(Q2'*Ay)'*inv((Q2'*Ay)*V*(Q2'*Ay)')*(Q2'*Ay*y-Q2'*b);zhat=inv(R1)*Q1'*b-inv(R1)*Q1'*Ay*yhat; for I=1:21,SSEy=SSEy+(y(I)-yhat(I))^2;end; for I=1:10,SSEz=SSEz+(z(I)-zhat(I))^2;end;


Case Study #2

if and(SSEy<1.0e-6,SSEz<1.0e-6),

flag=-1;

end;

y=yhat;

z=zhat;

end


Case Study #2Bilinear Method with Gross Error Detection

y12 = x1F1 = 0.0857

y13 = T1F1 = 0.0547

y14 = T10F10 = 0.1440

y15 = T13F13 = 0.4160

y16 = T15F15 = 0.1971

z8 = y1F1

z9 = y3F3

z10 = x3F3

z11 = T9F9

z12 = T12F12

z13 = T1F1y1

y17 = T16F16 = 0.0507

y18 = T17F17 = 0.4806

y19 = T18F18 = 0.1392

y20 = T1F1x1 = 0.0197

Altered Bilinear Variables:


Case Study #2

1) 2.325y4 – 0.27376z3 + 0.00839y3 = 0

2) 0.25y12 – 0.0025y20 + 0.418z8 – 0.00418z13 + 2.257y2 – 1.1731z3 = 0

3) y12 – z10 = 0

4) z8 – z9 – y2 = 0

5) z10 + z9 – z1 = 0

6) y12 + z8 – y1 = 0

7) z2 – z3 = 0

8) z1 + y6 + z6 – z7 = 0

9) z6 – y7 = 0

Altered Bilinear Constraints:


Case Study #2Altered Bilinear Constraints:

10) z5 – y5 = 0

11) z3 – y3 = 0

12) y4 + z5 – z4 = 0

13) 16.83y4 – z5 = 0

14) 0.1519z1 – y6 = 0

15) 12.3z10 + 34.01y6 – z6 = 0

16) 1.046z11 – 1.046y14 + 4.18y17 – 4.18y16 = 0

17) 1.046z12 – 1.046y15 + 4.18y19 – 4.18y18 = 0

18) y8 – y9 = 0

19) y10 – y11 = 0


Case Study #2Variables

Raw Measurement

Reconciled Measurement

Adjustment (%)

F1 23800 24370 2.39

F2 17100 16530 -3.33

F3 ? 7840 N/A

F4 ? 31030 N/A

F5 ? 31030 N/A

F6 30300 31030 2.41

F7 3500 3540 1.14

F8 ? 63160 N/A

F9 ? 59620 N/A

F10 60000 59620 -0.63

Table 2.3a: Measured and reconciled data using a bilinear approach


VariablesRaw



(%)F11 1000 1190 19.00

F12 ? 159880 N/A

F13 160000 159880 -0.13

F14 ? 168910 N/A

F15 14600 14550 -0.34

F16 14500 14550 0.34

F17 35600 35650 0.14

F18 35700 35650 -0.14

x1 0.36 0.3984 10.67

x3 ? 1.2387 N/A

Case Study #2

Table 2.3b: Measured and reconciled data using a bilinear approach


VariablesRaw



(%)y1 ? 0.6016 N/A

y3 ? -0.2387 N/A

T1 23 22.45 -2.39

T9 127 122.29 -3.71

T10 24 24.15 0.62

T12 124 111.35 -10.20

T13 26 26.02 0.08

T15 135 135.46 0.34

T16 35 34.85 -0.43

T17 135 134.81 -0.14

T18 39 39.05 0.13

Case Study #2

Table 2.3c: Measured and reconciled data using a bilinear approach


Case Study #2

592.12)6(0463.12 95.02

NO GROSS ERROR?!?

Need to impose inequality constraints on mass fractions

IMPOSSIBLE WITH BILINEAR DR!


Case Study #2

0349.0mb

Calculating the bias in measurement y2:

Adjusting the measurement:

y2 = 0.171 - 0.0349 = 0.1361

Faulty measurement is known from previous analysis(bilinear DR can’t properly detect error).



Raw Measurement


F1 23800 23800

F2 17100 13610

F3 ? 10190

F4 ? 31030

F5 ? 31030

F6 30300 31030

F7 3500 3540

F8 ? 63160

F9 ? 59620

F10 60000 59620

Table 2.4a: Bilinear reconciliation without bias



Raw Measurement


F11 1000 1550

F12 ? 159980

F13 160000 159980

F14 ? 171720

F15 14600 14550

F16 14500 14550

F17 35600 35650

F18 35700 35650

x1 0.36 0.3667

x3 ? 0.8566

Table 2.4b: Bilinear reconciliation without bias



Raw Measurement


y1 ? 0.6333

y3 ? 0.1434

T1 23 22.98

T9 127 122.29

T10 24 24.15

T12 124 111.28

T13 26 26.00

T15 135 135.46

T16 35 34.85

T17 135 134.81

T18 39 39.05

Table 2.4c: Bilinear reconciliation without bias


Case Study #2

NONLINEAR DR!!!


Case Study #2

y=[0.238; 0.171; 0.303; 0.035; 0.600; 0.010; 1.600; 0.146; 0.145; 0.356; 0.357; 0.0857; 0.0547; 0.144; 0.416; 0.1971; 0.0507; 0.4806; 0.1392; 0.0197];

V=[0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0.000025 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0.00002 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0.00002 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0.0001 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.001 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.00005 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.00001 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.00001 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.00005 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.000004];

MATLAB code used for bilinear DR:


Case Study #2Ay=[0 0 0.00839 2.325 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 2.257 0 0 0 0 0 0 0 0 0 0.25 0 0 0 0 0 0 0 -0.0025;0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0;0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;-1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 16.83 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 34.01 0 0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 -1.046 0 -4.18 4.18 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.046 0 0 -4.18 4.18 0;0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0 0 0];

Az=[0 0 -0.27376 0 0 0 0 0 0 0 0 0 0; 0 0 -1.1731 0 0 0 0 0.418 0 0 0 0 -0.00418;0 0 0 0 0 0 0 0 0 -1 0 0 0;0 0 0 0 0 0 0 1 -1 0 0 0 0;-1 0 0 0 0 0 0 0 1 1 0 0 0;0 0 0 0 0 0 0 1 0 0 0 0 0;0 1 -1 0 0 0 0 0 0 0 0 0 0;1 0 0 0 0 1 -1 0 0 0 0 0 0;0 0 0 0 0 1 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0 0 0 0;0 0 1 0 0 0 0 0 0 0 0 0 0;0 0 0 -1 1 0 0 0 0 0 0 0 0;0 0 0 0 -1 0 0 0 0 0 0 0 0;0.1519 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 -1 0 0 0 12.3 0 0 0;0 0 0 0 0 0 0 0 0 0 1.046 0 0;0 0 0 0 0 0 0 0 0 0 0 1.046 0;0 0 0 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 0 0 0 0 0 0];


Case Study #2[Q,R]=qr(Az);

for I=1:19,

for J=1:13,

Q1(I,J)=Q(I,J);

end;

end;


Case Study #2for I=1:19,

for J=14:19,

Q2(I,J-13)=Q(I,J);

end;

end;

for I=1:13,

for J=1:13,

R1(I,J)=R(I,J);

end;

end;


Case Study #2G=Q2'*Ay;

yhat=y-V*G'*inv(G*V*G')*G*y;

zhat=-inv(Az'*Az)*Az'*(Ay*yhat);

yhat2=yhat;

yhat2(12)=yhat2(12)/yhat2(1);









Case Study #2zhat2=zhat;

zhat2(8)=zhat2(8)/yhat2(1);

zhat2(9)=zhat2(9)/zhat2(1);




Vr=G*V*G';

r=G*y;

tau=r'*inv(Vr)*r;

Id=[1 0 0 0 0 0;0 1 0 0 0 0;0 0 1 0 0 0;0 0 0 1 0 0;0 0 0 0 1 0;0 0 0 0 0 1];


Case Study #2for I=1:20,

for J=1:6,

Ac(J,1)=G(J,I);

end;

Vri=inv(Vr)*(Id-Ac*inv(Ac'*inv(Vr)*Ac)*Ac'*inv(Vr));

O(I)=r'*Vri*r;

end;

B=[0;1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0];

P=G*B;

m=inv(P'*inv(G*V*G')*P)*P'*inv(G*V*G')*G*y;

Documents

© Universidad de Guanajuato, Mexico© University of Ottawa, Canada, 2004 Case Study #2: Waste Sludge Incineration: Steady-State Nonlinear DR and Detection