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Case Study #2: Waste Sludge Incineration:

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. Sat. Vap. F2. Incinerator. Sludge F1. Conc. Sludge F3. Waste Gas F14. Dryer.

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Case Study #2: Waste Sludge Incineration:

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  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. Sat. Vap. F2 Incinerator Sludge F1 Conc. Sludge F3 Waste Gas F14 Dryer Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) Preheated Air F12 Hot Water F17 Boiler Boiler Feed Water F6 Natural Gas F11 Waste Gas F8 #6 Fuel Oil F7 Cool Water F18 Preheated Air F9 Air F13 Air F10 Hot Water F15 Cool Water F16 Case Study #2

  3. Case Study #2 Incinerator Problem Engineering Problem Management Problem Accounting Problem

  4. Case Study #2 Table 2.1a: List of measured and unmeasured variables

  5. Case Study #2 Table 2.1b: List of measured and unmeasured variables

  6. Case Study #2 Gross Errors? Need to change the problem from NONLINEAR to BILINEAR!! See the end of this module.

  7. #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

  8. 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 = 19000 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 500 kg/day, Temperature 2ºC, Composition 0.03 Case Study #2

  9. 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. Case Study #2

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

  11. Step 1: ,

  12. 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 2:

  13. Step 3: ,

  14. Sat. Vap. F2 Incinerator Sludge F1 Conc. Sludge F3 Waste Gas F14 Dryer Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) Preheated Air F12 Hot Water F17 Boiler Feed Water F6 Boiler Natural Gas F11 Waste Gas F8 #6 Fuel Oil F7 Cool Water F18 Air F10 Preheated Air F9 Air F13 Hot Water F15 Cool Water F16 Step 4: F3 + F11 + F12 - F14 = 0 F4 - F5 = 0 F15 – F16 = 0 F5 - F6 = 0 F12 - F13 = 0 F9 - F10 = 0 F7 + F9 - F8 = 0 F17 – F18 = 0 Mass Balances:

  15. Component Balances: Sat. Vap. F2 Sludge F1 Conc. Sludge F3 Dryer Step 4: Solids: F1x1 - F3x3 = 0 Water: F1y1 - F3y3 - F2 = 0 Normalization Constraints: x3 + y3 - 1 = 0 x1 + y1 - 1 = 0

  16. Balances on Heat Exchangers: Preheated Air F12 Hot Water F17 Air F10 Preheated Air F9 Cool Water F18 Cool Water F16 Hot Water F15 Air F13 Step 4: 1.046F9T9 – 1.046F10T10 + 4.18F16T16 – 4.18F15T15 = 0 1.046F12T12 – 1.046F13T13 + 4.18F18T18 – 4.18F17T17 = 0

  17. Heat Flow Through Boiler: Sat. Vap. F5 (p=4 bar) Boiler Feed Water F6 Boiler Waste Gas F8 #6 Fuel Oil F7 Preheated Air F9 Step 4: (0.62)(HHVOIL)(F7) = (h5)(F5) – (h6)(F6) 23250F7 – 2737.6F5 + 83.9F6 = 0 2.325F7 – 0.27376F5 + 0.00839F6 = 0

  18. Air / Oil Ratio in Boiler: 0.87 kg C 0.0724 kmoles C 0.10 kg H 0.0992 kmoles H 0.0084 kg S 0.000262 kmoles S 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 Sat. Vap. F5 (p=4 bar) Boiler Feed Water F6 Boiler 0.5801 kmoles Air Waste Gas F8 #6 Fuel Oil F7 16.83 kg Air Preheated Air F9 16.83F7 – F9 = 0 Step 4: For 1 kg of oil: O2 required (25% excess):

  19. Heat Flow Through Dryer: Sat. Vap. F2 Sludge F1 Conc. Sludge F3 Dryer Sat. Liq. F4 Sat. Vap. F5 (p=4 bar) Step 4: (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

  20. NG / Sludge Ratio: Incinerator Conc. Sludge F3 7.8348 kmol CH4 125.67 kg 0.1257 tonnes 0.8706 kmol C2H6 26.18 kg 0.0262 tonnes Natural Gas F11 0.1519 tonnes NG/tonne F3 Step 4: 195 SCM/tonne wet sludge 8.7054 kmol NG/tonne F3 0.1519F3 – F11 = 0

  21. Air / Sludge + NG Ratio: 0.9 kmole CH4 14.44 kg (82.75%) 0.1 kmole C2H6 3.01 kg (17.25%) Incinerator Conc. Sludge F3 Preheated Air F12 Natural Gas F11 Step 4: To burn NG: Since NG is 10 mol% C2H6 and 90 mol% CH4: 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

  22. Air / Sludge + NG Ratio: Incinerator Conc. Sludge F3 Preheated Air F12 Natural Gas F11 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 0.2121x3 kmol air/kg sludge 6.1496x3 kg air/kg sludge 100% Excess: 12.3x3 kg air/kg sludge Combining the air needed for the both the sludge and the NG: 12.3F3x3 + 34.01F11 – F12 = 0

  23. Summary of Constraints: Step 4: • 2.325y4 – 0.27376z3 + 0.00839y3 = 0 • 0.25y12y1 – 0.0025y12y1y13 +0.418z8y1 – 0.00418z8y1y13 + 2.257y2 – 1.1731z3 = 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

  24. Summary of Constraints: Step 4: • 10) z5 – y5 = 0 • 11) z3 – y3 = 0 • 12) y4 + z5 – z4 = 0 • 13) 16.83y4 – z5 = 0 • 14) 0.1519z1 – y6 = 0 • 12.3z1z10 + 34.01y6 – z6 = 0 • 1.046z5y14 – 1.046y5y15 + 4.18y9y19 – 4.18y8y18 = 0 • 1.046z6y16 – 1.046y7y17 + 4.18y11y21 – 4.18y10y20 = 0 • y8 – y9 = 0 • y10 – y11 = 0

  25. Step 5:

  26. Step 5:

  27. Step 5:

  28. Step 6:

  29. Step 7:

  30. Step 7:

  31. Step 7:

  32. Step 8:

  33. Step 9:

  34. Steps 10 & 11:

  35. Results: Table 2.2a: Measured and reconciled data

  36. Results: Table 2.2b: Measured and reconciled data

  37. Results: Table 2.2c: Measured and reconciled data

  38. F2 = 17100 Raw Measurements: Incinerator F1 = 23800 F3 = 23800 – 17100 = 6700 Dryer F11 = 6700(0.1519) = 1018 Interpretation:

  39. ! F2 = 12770 Incinerator F1 = 22940 F3 = 22940 – 12770 = 10170 Dryer F11 = 10170(0.1519) = 1545 Interpretation: Reconciled Measurements:

  40. Incinerator temperature too low Interpretation: Incorrect measurement of vapor flow from dryer (F2) Insufficient natural gas fed to incinerator (F11) Control natural gas feed based on incinerator temperature Install more measurement devices throughout the process

  41. 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

  42. 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

  43. 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

  44. 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

  45. 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(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

  46. 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

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

  48. Case Study #2 Bilinear Method with Gross Error Detection Altered Bilinear Variables: y12 = x1F1 = 0.0857 y13 = T1F1 = 0.0547 y14 = T10F10 = 0.1440 y15 = T13F13 = 0.4160 y16 = T15F15 = 0.1971 y17 = T16F16 = 0.0507 y18 = T17F17 = 0.4806 y19 = T18F18 = 0.1392 y20 = T1F1x1 = 0.0197 z8 = y1F1 z9 = y3F3 z10 = x3F3 z11 = T9F9 z12 = T12F12 z13 = T1F1y1

  49. Case Study #2 Altered Bilinear Constraints: • 2.325y4 – 0.27376z3 + 0.00839y3 = 0 • 0.25y12 – 0.0025y20 + 0.418z8 – 0.00418z13 + 2.257y2 – 1.1731z3 = 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

  50. Case Study #2 Altered 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 • 12.3z10 + 34.01y6 – z6 = 0 • 1.046z11 – 1.046y14 + 4.18y17 – 4.18y16 = 0 • 1.046z12 – 1.046y15 + 4.18y19 – 4.18y18 = 0 • y8 – y9 = 0 • y10 – y11 = 0

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