1. 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.

2. Case Study #2 Incinerator Dryer 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 F16
Hot Water F15
Preheated Air F9
Natural Gas F11
Hot Water F17
Cool Water F18
Air F13
Preheated Air F12
A pulp and paper plant has chosen to dispose of its waste sludge by incineration. Before the sludge can be incinerated however, it must first be dried in order to lose its excess water. The heat required for the dryer comes from saturated steam at a pressure of 4 bars. This steam is produced in the boiler, where #6 fuel oil is burned with air. After being dried, the concentrated sludge is sent to the incinerator. Natural Gas is used as a supplementary fuel in order to keep the incinerator temperature high enough for complete combustion.

3. Case Study #2 Incinerator Problem Engineering Problem
For a few weeks now, the incinerator has not been completely incinerating the sludge, resulting in a continuous build-up of residue that must be removed frequently, costing the company time and money. Management (which has thus far ignored complaints from the engineers) has begun to see reports from the accounting department on the new costs. The problem is now recognized, and heavy pressure is put on the plant engineers to solve it. Based on measurements taken of sludge entering and water vapor leaving the dryer, the sludge entering the incinerator should have more than a high enough concentration, and based on the flow of concentrated sludge, the amount of natural gas and air entering the incinerator should be more than enough as well. Increasing the heat in the dryer (lowering the sludge flow and raising its concentration) does not solve the problem, and since the models developed that are used in the control systems are known to be correct, the engineers are stumped. One engineer finds some free time and decides to try reconciling the data.
Management Problem
Accounting Problem

4. Case Study #2 Stream # Flow Rate (kg/day) Temperature (ºC)
Mass Fraction Solid 1
23800 23
0.36 2
17100
N/A 3 ? 4 5 6
30300 7
3500 8 9
127
The following two slides are a list of all the measured and unmeasured variables for each stream in the incineration process. A number indicates a measured variable and a question mark indicates an unmeasured variable. N/A does not necessarily indicate that a measurement does not or should not exist, but simply that it is an unnecessary measurement for the mass or energy balances required, and so has not been included. Listed here are the variables for the first nine streams.
Table 2.1a: List of measured and unmeasured variables

5. Case Study #2 Stream # Flow Rate (kg/day) Temperature (ºC)
Mass Fraction Solid 10
60000 24
N/A 11
1000 12 ?
124 13
160000 26 14 15
14600
135 16
14500 35 17
35600 18
35700 39
And here are the variables for the next nine streams.
Table 2.1b: List of measured and unmeasured variables

6. Case Study #2 NONLINEAR to BILINEAR!! Gross Errors?
Need to change the problem from
NONLINEAR to BILINEAR!!
The best thing to do first is to check the data for gross errors. However, this requires changing the problem from a nonlinear one to a bilinear one. This will be shown briefly at the end, but mainly this case study will focus on nonlinear steady-state data reconciliation and the logical detection of a gross error. The purpose here is to show that while gross error detection is a powerful tool, sometimes it is still better to use nonlinear DR.
See the end of this module.

7. 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
The following two slides show essential information required to make the mass and energy balances describing the process. The fuel oil composition by weight is 87% carbon, 10% hydrogen, 0.84% sulfur, and the remaining oxygen, nitrogen, and ash, and has a higher heating value of kJ/kg. The natural gas composition on a molar basis is 90% methane and 10% ethane. 62% of the oil heating value fed to the boiler is used to create saturated steam at 4 bar, and air is fed to the boiler 25% in excess of that required for complete combustion of the oil. The heat capacity of the solids in the sludge is 2.5 kJ/kg·ºC, and the water in the sludge can safely be approximated as water.

8. Case Study #2 Dryer: Efficiency = 55%
Pressure at 1 bar on sludge side, and 4 bar on steam side
All steam and condensate flows are saturated
Incinerator:
Sludge must enter at higher than 75% consistency
HHV of concentrated sludge = kJ/kg dry solids
195 SCM natural gas/tonne wet sludge
2.5 SCM air/10000 kJ of sludge HHV for complete combustion
100% Excess Air for sludge and natural gas
Standard Deviation:
Flows kg/day, Temperature ºC, Composition
In the dryer, 55% of the heat lost by the condensing steam is transferred to the sludge. The pressure is constant in the dryer at 1 bar on the sludge-drying side and at 4 bar on the condensing steam side. In the incinerator, the concentrated sludge (which must have a concentration higher than 75% solids) has a higher heating value of approximately kJ/kg of dry solids standard cubic meters of natural gas are required per metric ton of wet sludge, and 2.5 standard cubic meters of air per kJ of sludge heating value are required for complete combustion of the sludge. Air is fed to the incinerator in 100% excess of that required for complete combustion of both the sludge and natural gas. The standard deviation of all flow measurements is 500 kg/day, all temperature measurements 2ºC, and all composition measurements 0.03.

9. Case Study #2
Define the y and z matrices for the measured variables and unmeasured variables.
Set up the V matrix, scaling down if necessary.
Fill in the matrix with the measured values and the matrix with guesses for the unmeasured variables, scaling down appropriately.
Determine the mass, component, and heat balances for the process, f( , ) = 0, scaling down parameters where necessary.
Determine the Jacobian matrices, Ay and Az, and solve them using the values from the and matrices.
Calculate b0 = Ay + Az – f ( , )
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.
Since this problem involves not only mass flow and temperature reconciliation, but composition reconciliation as well, it is a nonlinear problem and will require a method such as successive linearization in order to solve it. The method used for this problem is very similar to the example seen in Tier 1 Chapter 5. The solution method is broken down into 11 steps, shown in this and the following slide.

10. Case Study #2
8) Calculate = – V(Q2TAy)T[(Q2TAy)V(Q2TAy)T]-1(Q2TAy – Q2Tb0)
9) Calculate = R1-1Q1Tb0 – R1-1Q1TAy - R1-1R (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.

11. Step 1: ,
As shown here, the measured and unmeasured variables (including the mass fraction of liquid in the sludge) have been arbitrarily assigned to different y and z values.

12. Step 2:
V(1,1), V(2,2), V(3,3), … , V(11,11) = (after scaling)
V(12,12) = (not necessary to scale)
V(13,13), V(14,14), … , V(21,21) = (after scaling)
Assumed that Covariance = 0
At this point, it is already seen that most of the variables will have to be scaled down. The diagonal elements in the V matrix up to V(11,11) are the variance for the flow measurements, scaled down by a factor of The single diagonal element in V(12,12) is the variance for the composition measurement, and was not necessary to scale. The diagonal elements from V(13,13) to V(21,21) are the variance for the temperature measurements, scaled down by a factor of Since the covariance is assumed to be zero, all non-diagonal elements equal zero.

13. Step 3: ,
Here, the y hat zero matrix has been filled with the appropriate measured values, and the z hat zero matrix has been filled with approximations for the appropriate unmeasured values. As seen in the previous slide, the flow measurements have been scaled down by a factor of and the temperature measurements have been scaled down by a factor of 100.

14. Step 4: F3 + F11 + F12 - F14 = 0 F4 - F5 = 0 F15 – F16 = 0 F5 - F6 = 0
Mass Balances:
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 F16
Hot Water F15
Preheated Air F9
Natural Gas F11
Hot Water F17
Cool Water F18
Air F13
Preheated Air F12
There are eight independent mass balances for this process, all very self-explanatory and listed here. Since the parameters are all equal to one, there is no need to scale these equations.

15. Step 4: Component Balances: Solids: F1x1 - F3x3 = 0 Water:
F1y1 - F3y3 - F2 = 0
Dryer
Sludge F1
Sat. Vap. F2
Conc. Sludge F3
Normalization Constraints:
x3 + y3 - 1 = 0
x1 + y1 - 1 = 0
There are only two streams which require component balances, F1 and F3. There is one mass balance for each component, and a normalization constraint for each stream. Xi represents the mass fraction of solids in Stream i, and Yi represents the mass fraction of water in Stream i.

16. Step 4: Balances on Heat Exchangers:
Air F10
Cool Water F16
Hot Water F15
Preheated Air F9
1.046F9T9 – 1.046F10T F16T16 – 4.18F15T15 = 0
Hot Water F17
Cool Water F18
Air F13
Preheated Air F12
Here are the heat balances on the two exchangers. Since there are two different specific heat capacities involved with each, the specific heat can not be factored out of the equations. Again, there is no need to scale these equations, as the parameters are sufficiently small.
1.046F12T12 – 1.046F13T F18T18 – 4.18F17T17 = 0

17. Step 4: Heat Flow Through Boiler:
(0.62)(HHVOIL)(F7) = (h5)(F5) – (h6)(F6)
Boiler
Sat. Vap. F5
(p=4 bar)
Waste Gas F8
Boiler Feed Water F6
#6 Fuel Oil F7
Preheated Air F9
23250F7 – F F6 = 0
Doing a heat balance on the boiler, the change in enthalpy of the two water streams is equal to 62% of the heating value of the oil stream (since the boiler has a 62% efficiency). Plugging in all the numbers, rearranging, and scaling results in the bottom equation.
2.325F7 – F F6 = 0

18. Step 4: For 1 kg of oil: Air / Oil Ratio in Boiler:
0.87 kg C kmoles C kg H kmoles H kg S kmoles S
O2 required (25% excess):
Boiler
Sat. Vap. F5
(p=4 bar)
Waste Gas F8
Boiler Feed Water F6
#6 Fuel Oil F7
Preheated Air F9
For C (0.0724)(1.25)(1) For H (0.0992)(1.25)(0.25) For S ( )(1.25)(1)
Total kmoles O2
To find the air/oil ratio in the boiler, the amount of O2 required to completely combust the oil is calculated. After adding 25% excess and using various conversion factors, the bottom equation results.
16.83F7 – F9 = 0
kmoles Air
16.83 kg Air

19. Step 4: Heat Flow Through Dryer:
Sludge F1
Sat. Vap. F2
Conc. Sludge F3
Sat. Liq. F4
Sat. Vap. F5
(p=4 bar)
Heat Flow Through Dryer:
(x1)(F1)(Cps)(TB-T1) + (y1)(F1)(CpH2O)(TB-T1) + (F2)(LvH2O) – (0.55)(h5-h4)(F5) = 0
250x1F1 – 2.5x1F1T y1F1 – 4.18y1F1T F2 – F5 = 0
Since 55% of the heat lost by the vapor in condensing is used in the dryer, this amount is equal to the heat required to bring the solids and water in stream F3 to the boiling point plus that required to vaporize the water in stream F2. After simplifying and scaling, the final equation shown results.
0.25x1F1 – x1F1T y1F1 – y1F1T F2 – F5 = 0

20. Step 4: NG / Sludge Ratio: 195 SCM/tonne wet sludge
Incinerator
Conc. Sludge F3
Natural Gas F11
NG / Sludge Ratio:
195 SCM/tonne wet sludge
kmol NG/tonne F3
kmol CH kg tonnes
kmol C2H kg tonnes
tonnes NG/tonne F3
Assuming that there are 22.4 kmol per cubic meter of natural gas, the natural gas/sludge ratio is calculated as shown.
0.1519F3 – F11 = 0

21. Step 4: To burn NG: Air / Sludge + NG Ratio:
Since NG is 10 mol% C2H6 and 90 mol% CH4:
0.9 kmole CH kg (82.75%) kmole C2H kg (17.25%)
Therefore for 1 kg of NG:
kg CH kmoles CH kg C2H kmoles C2H6
Incinerator
Conc. Sludge F3
Natural Gas F11
Preheated Air F12
O2 required (100% excess):
For CH4 (0.0516)(2)(2) For C2H6 (0.0057)(2)(7/2)
Total kmoles O2
Here is shown the calculation of air in 100% excess of that needed for complete combustion of the natural gas entering the incinerator.
kmoles Air
32.01 kg Air/kg NG

22. Step 4:
To burn sludge:
(2.5 SCM air/10000 kJ)(19000 kJ/kg solid) (x3 kg solid/kg sludge) = 4.75x3 SCM air/kg sludge
Air / Sludge + NG Ratio:
0.2121x3 kmol air/kg sludge
6.1496x3 kg air/kg sludge
100% Excess:
Incinerator
Conc. Sludge F3
Natural Gas F11
Preheated Air F12
12.3x3 kg air/kg sludge
Here is shown the calculation of air in 100% excess of that needed to burn the sludge. The final equation is the mass balance for the air entering the incinerator, taking into account both the sludge and natural gas.
Combining the air needed for the both the sludge and the NG:
12.3F3x F11 – F12 = 0

23. Step 4: Summary of Constraints: 2.325y4 – 0.27376z3 + 0.00839y3 = 0
0.25y12y1 – y12y1y z8y1 – z8y1y y2 – z3 = 0
y1y12 – z1z10 = 0
y1z8 – z1z9 – y2 = 0
z10 + z9 – 1 = 0
y12 + z8 – 1 = 0
z2 – z3 = 0
z1 + y6 + z6 – z7 = 0
z6 – y7 = 0
In total, there are 19 independent mass, energy, and component balance equations. They are presented on this and the following slide, using the appropriate y’s for the measured variables, and z’s for the unmeasured variables.

24. Step 4: Summary of Constraints: 10) z5 – y5 = 0 11) z3 – y3 = 0
12) y4 + z5 – z4 = 0
13) y4 – z5 = 0
14) z1 – y6 = 0
12.3z1z y6 – z6 = 0
1.046z5y14 – 1.046y5y y9y19 – 4.18y8y18 = 0
1.046z6y16 – 1.046y7y y11y21 – 4.18y10y20 = 0
y8 – y9 = 0
y10 – y11 = 0

25. Step 5:
Here are the first 11 columns of the Jacobian matrix for the measured variables. Since this is the first iteration, it is solved using the raw measurements. As shown in Tier 1, each row represents an equation (for example, the first row applies to the first constraint equation) and each column represents the first derivative with respect to the column number for each constraint equation). Therefore, the value in row 2 column 2, which is 2.257, represents the derivative of the second constraint equation with respect to y2, solved using the raw measurement values in the first iteration.

26. Step 5:
Here are columns 12 to 21 for the Jacobian matrix corresponding to the measured variables.

27. Step 5:
Finally, the Jacobian matrix for the unmeasured variables, solved using the initial guesses.

28. Step 6:
With the Jacobian matrices known, the calculation of the right-hand vector, b0, can be carried out.

29. Step 7:
Here is Q1, the first n columns (10 in this case) of the Q matrix from the QR factorization of the Jacobian for the unmeasured variables.

30. Step 7:
And here is Q2, the last m-n columns (9 in this case) of the Q matrix.

31. Step 7:
The R matrix resulting from the QR factorization is shown here, split into it’s respective R1 and R2 matrices.

32. Step 8:
The calculation of the reconciled values for the measured variables in the first iteration.

33. Step 9:
And the calculation of the reconciled values for the unmeasured variables in the first iteration. Notice the last term in the equation is neglected, since the R2 matrix was composed entirely of zeroes.

34. Steps 10 & 11:
After a few more iterations, using the new reconciled data for the measured and unmeasured variables each time, the results converge to those shown on the slide. Most of the values are not significantly different from the values obtained in the first iteration.

35. Reconciled Measurement
Results:
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%) F1
23800
22940
-3.61 F2
17100
12770
-25.32 F3 ?
10170
N/A F4
31470 F5 F6
30300
3.86 F7
3500
3590
2.57 F8
64040 F9
60450
F10
60000
0.75
A table of results is presented over the next three slides. The table shows both raw and reconciled measurements, in order to observe any large differences or anomalies. One such anomaly can be observed with the flowrate for stream 2, whose reconciled measurement is over 25% smaller than the raw measurement.
Table 2.2a: Measured and reconciled data

36. Reconciled Measurement
Results:
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%)
F11
1000
1550
55.00
F12 ?
159880
N/A
F13
160000
-0.08
F14
171600
F15
14600
14820
1.51
F16
14500
2.21
F17
35600
35900
0.84
F18
35700
0.56 x1
0.36
0.3803
5.64 x3
0.8576
Again, something strange is happening with the flowrate for stream 11, whose reconciled value is a very large 55% larger than its raw measurement.
Table 2.2b: Measured and reconciled data

37. Reconciled Measurement
Results:
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%) y1 ?
0.6197
N/A y3
0.1424 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
137.93
2.17
T18 39
36.06
-7.54
At a glance, nothing seems out of the ordinary for the last 11 variables shown here.
Table 2.2c: Measured and reconciled data

38. Interpretation: Raw Measurements: Incinerator Dryer F1 = 23800
As seen in the results, there are 2 major anomalies, the flowrate for stream 2 (the saturated vapor leaving the dryer) and the flowrate for stream 11 (the natural gas entering the incinerator). The flow with the larger anomaly, stream 11, is inspected first, and the measurement device is found to be operating correctly. However, while inspecting the control system for stream 11, the engineer finds that its flow is regulated by the flow measurements for streams 1 and 2. Based on the raw measurements for streams 1 and 2, the flow in stream 3 should be equal to around 6700 kg/day. The natural gas (F11) should be fed to the incinerator at a ratio of times that amount, which is 1018 kg/day. The raw measurement for stream 11 was 1000 kg/day, so it appears as though there is nothing wrong with the control system. In other words, the incinerator is being fed the correct amount of gas based on the the measurements taken of streams 1 and 2.

39. ! Interpretation: Reconciled Measurements: Incinerator Dryer
F1 = 22940
F2 = 12770
F3 = – = 10170
F11 = 10170(0.1519) = 1545
If the reconciled flows for streams 1 and 2 are used, however, stream 3 should have a flow of around kg/day, which means that natural gas should be fed to the incinerator at a rate of 1545 kg/day in order to keep the temperature high enough for complete combustion. If the reconciled flows for streams 1 and 2 are correct then, the sludge not being completely incinerated can be explained by the fact that lower than the required amount of natural gas is being used.

40. Interpretation:
Incorrect measurement of vapor flow from dryer (F2)
Insufficient natural gas fed to incinerator (F11)
Incinerator temperature too low
Based on this knowledge, the dryer is shut down for a few hours while the measurement device for the saturated vapor stream is changed and reconnected to the control system. After the dryer is restarted, the new raw measurement for stream 2 is kg/day. As a result, the control system for stream 11 allows 1640 kg/day of natural gas to be fed to the incinerator. The temperature in the incinerator is increased dramatically, and the sludge going through is incinerated completely. In order to prevent this problem from reoccurring, it is suggested that perhaps there should be a temperature indicator for the incinerator, and the natural gas flow be regulated based on that.
Control natural gas feed based on incinerator temperature
Install more measurement devices throughout the process

41. 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=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ];
for I=1:11,
V(I,I)=0.005^2;
end;
The following 7 slides contain the MATLAB code used to solve case study #1. It is essentially the same as the steps followed earlier to solve the problem. Look at the code carefully, and associate each bit of code with each step.

42. Case Study #2 for I=12, V(I,I)=0.03^2; end; for I=13:21,
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;

43. Case Study #2
Ay=[ ;0.25*y(12) *y(12)*y(13) *z(8) *z(8)*y(13) *y(1) *y(1)*y(13) *y(12)*y(1) *z(8)*y(1) ;y(12) y(1) ;z(8) ; ; ; ; ; ; ; ; ; ; ; ; *y(15) *y(18) 4.18*y(19) *z(5) *y(5) *y(8) 4.18*y(9) 0 0; *y(17) *y(20) 4.18*y(21) *z(6) *y(7) *y(10) 4.18*y(11); ; ];
Az=[ ; *y(1) *y(1)*y(13) 0 0; -z(10) z(1); -z(9) y(1) -z(1) 0; ; ; ; ; ; ; ; ; ; ; 12.3*z(10) *z(1); *y(14) ; *y(16) ; ; ];

44. 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) *y(12)*y(1)*y(13)+0.418*z(8)*y(1) *z(8)*y(1)*y(13) *y(2) *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);

45. 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;
for J=11:19,
Q2(I,J-10)=Q(I,J);

46. Case Study #2 for I=1:10, for J=1:10, R1(I,J)=R(I,J); 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;
SSEz=SSEz+(z(I)-zhat(I))^2;

47. Case Study #2 if and(SSEy<1.0e-6,SSEz<1.0e-6), flag=-1; end;
y=yhat;
z=zhat;
end

48. Case Study #2 Bilinear Method with Gross Error Detection
Altered Bilinear Variables:
y12 = x1F1 =
y13 = T1F1 =
y14 = T10F10 =
y15 = T13F13 =
y16 = T15F15 =
y17 = T16F16 =
y18 = T17F17 =
y19 = T18F18 =
y20 = T1F1x1 =
z8 = y1F1
z9 = y3F3
z10 = x3F3
z11 = T9F9
z12 = T12F12
z13 = T1F1y1
As promised, it will now briefly be shown how to solve this problem using the bilinear method with gross error detection, and the drawbacks involved. The nonlinear problem must first be changed to a bilinear one by a change of variables, as shown here. The variables not shown are unaltered.

49. Case Study #2 Altered Bilinear Constraints:
2.325y4 – z y3 = 0
0.25y12 – y z8 – z y2 – z3 = 0
y12 – z10 = 0
z8 – z9 – y2 = 0
z10 + z9 – z1 = 0
y12 + z8 – y1 = 0
z2 – z3 = 0
z1 + y6 + z6 – z7 = 0
z6 – y7 = 0
Of course, the changing of the variables results in new balance equations, shown on the following 2 slides.

50. Case Study #2 Altered Bilinear Constraints: 10) z5 – y5 = 0
12) y4 + z5 – z4 = 0
13) y4 – z5 = 0
14) z1 – y6 = 0
12.3z y6 – z6 = 0
1.046z11 – 1.046y y17 – 4.18y16 = 0
1.046z12 – 1.046y y19 – 4.18y18 = 0
y8 – y9 = 0
y10 – y11 = 0
Notice how all of the equations are now linear.

51. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%) F1
23800
24370
2.39 F2
17100
16530
-3.33 F3 ?
7840
N/A F4
31030 F5 F6
30300
2.41 F7
3500
3540
1.14 F8
63160 F9
59620
F10
60000
-0.63
Reconciling the data using the altered constraints gives the results listed in Table The reconciliation seems to be good, as the adjustments are smaller than they were with using the nonlinear method.
Table 2.3a: Measured and reconciled data using a bilinear approach

52. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%)
F11
1000
1190
19.00
F12 ?
159880
N/A
F13
160000
-0.13
F14
168910
F15
14600
14550
-0.34
F16
14500
0.34
F17
35600
35650
0.14
F18
35700
-0.14 x1
0.36
0.3984
10.67 x3
1.2387
However, there is a serious problem. Observe that the mass fraction of solid in stream 3 is given as greater than 1.
Table 2.3b: Measured and reconciled data using a bilinear approach

53. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled Measurement
Adjustment (%) y1 ?
0.6016
N/A y3 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
134.81
-0.14
T18 39
39.05
0.13
And here the mass fraction of water in stream 3 is given as a negative number, which is also impossible.
Table 2.3c: Measured and reconciled data using a bilinear approach

54. Case Study #2 NO GROSS ERROR?!?
Need to impose inequality constraints on mass fractions
The problem here is clearly that bilinear DR has found the optimum method of reconciliation by violating mass fractional constraints. Indeed, the reconciliation works so well that a global test shows no gross error at all! The only way to correct this problem is by imposing inequality constraints on the mass fractions, something which bilinear DR is incapable of doing.
IMPOSSIBLE WITH BILINEAR DR!

55. Case Study #2 Faulty measurement is known from previous analysis
(bilinear DR can’t properly detect error).
Calculating the bias in measurement y2:
Adjusting the measurement:
y2 = =
Since we already know where the gross error is located, we can calculate the bias, correct the faulty measurement, and illustrate that bilinear DR does work. However, it would have been extremely difficult to locate the gross error in this case study by using bilinear DR alone.

56. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled Measurement F1
23800 F2
17100
13610 F3 ?
10190 F4
31030 F5 F6
30300 F7
3500
3540 F8
63160 F9
59620
F10
60000
Performing Bilinear DR on the data after adjustment of the faulty measurement gives the results shown in Table Notice that the results are very similar to those given by performing nonlinear DR without adjusting for gross error.
Table 2.4a: Bilinear reconciliation without bias

57. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled Measurement
F11
1000
1550
F12 ?
159980
F13
160000
F14
171720
F15
14600
14550
F16
14500
F17
35600
35650
F18
35700 x1
0.36
0.3667 x3
0.8566
Table 2.4b: Bilinear reconciliation without bias

58. Reconciled Measurement
Case Study #2
Variables
Raw Measurement
Reconciled 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
134.81
T18 39
39.05
Table 2.4c: Bilinear reconciliation without bias

59. NONLINEAR DR!!! Case Study #2
In conclusion then, this case study was meant to show how nonlinear DR can be the preferred method of reconciliation, even though standard gross error detection can not be used.

60. Case Study #2 MATLAB code used for bilinear DR:
V=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ];
Even though bilinear DR did not work for this case study, the MATLAB code used is shown in the remaining slides, for interest’s sake.

61. Case Study #2
Ay=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ];
Az=[ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ];

62. Case Study #2 [Q,R]=qr(Az); for I=1:19, for J=1:13, Q1(I,J)=Q(I,J);
end;

63. Case Study #2 for I=1:19, for J=14:19, Q2(I,J-13)=Q(I,J); end;
R1(I,J)=R(I,J);

64. Case Study #2 G=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);
yhat2(13)=yhat2(13)/yhat2(1);
yhat2(14)=yhat2(14)/yhat2(5);
yhat2(15)=yhat2(15)/yhat2(7);
yhat2(16)=yhat2(16)/yhat2(8);
yhat2(17)=yhat2(17)/yhat2(9);
yhat2(18)=yhat2(18)/yhat2(10);
yhat2(19)=yhat2(19)/yhat2(11);

65. Case Study #2 zhat2=zhat; zhat2(8)=zhat2(8)/yhat2(1);
Vr=G*V*G';
r=G*y;
tau=r'*inv(Vr)*r;
Id=[ ; ; ; ; ; ];

66. Case Study #2 for 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;
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;