LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES & THERMAL PROCESS EXAMPLE
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In this presentation is shown the linearization of functions with two of more variables with an example for a thermal process.
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LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES & THERMAL PROCESS EXAMPLE
- 1. LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES & THERMAL PROCESS EXAMPLE 1 CONTROL PROCESS FIRST-ORDERDYNAMICSYSTEMS "Good, better, best. Never let it rest. 'Til your good is better and your better is best." - St. Jerome
- 2. LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES & THERMAL PROCESS EXAMPLE IRIS BUSTAMANTE PÁJARO* ANGIE CASTILLO GUEVARA* ALVARO JOSE GARCÍA PADILLA * KARIANA ANDREA MORENO SADDER* LUIS ALBERTO PATERNINA NUÑEZ* CHEMICAL ENGINEERING PROGRAM UNIVERSITY OF CARTAGENA 2 CONTROL PROCESS FIRST-ORDERDYNAMICSYSTEMS
- 3. 3 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION FUNCTIONS OF TWO OR MORE VARIABLES Rate of a chemical reactionEquation of state Raoult’s law 𝑘 𝑐 𝐴 2 𝑡 𝑐 𝐵(𝑡) 𝑟 𝑐 𝐴 𝑡 , 𝑐 𝐵(𝑡) ) 𝑦 𝑇 𝑡 , 𝑝 𝑡 , 𝑥(𝑡) 𝑝0 𝑇 𝑡 𝑝 𝑡 𝑥(𝑡) 𝜌 𝑝 𝑡 , 𝑇(𝑡) 𝑀 𝑝(𝑡) 𝑅 𝑇(𝑡) Smith & Corripio, 2005
- 4. 4 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION Smith & Corripio, 2005 LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES Taylor series expansion 𝑓 𝑥1 𝑡 , 𝑥2 𝑡 , … ≈ 𝑓 𝑥1, 𝑥2, … + 𝜕𝑓 𝜕𝑥1 𝑥1 𝑡 − 𝑥1 + 𝜕𝑓 𝜕𝑥2 𝑥2 𝑡 − 𝑥2 + ⋯ 𝜕𝑓 𝜕𝑥 𝑘 = 𝜕𝑓 𝜕𝑥 𝑘 𝑥1, 𝑥2,… Where, 𝑥1, 𝑥2, … basic values of each variable.
- 5. 5 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION Smith & Corripio, 2005 EXAMPLE 2-6.2 FUNCTION 𝑎 𝑤 𝑡 , ℎ(𝑡) = 𝑤 𝑡 ℎ(𝑡) Area of a rectangle 𝑤 𝑡 ℎ 𝑡 𝑎 𝑤 𝑡 , ℎ(𝑡) ≈ 𝑎 𝑤, ℎ + 𝜕𝑎 𝜕𝑤 𝑤 𝑡 − 𝑤 + 𝜕𝑎 𝜕ℎ ℎ 𝑡 − ℎ How to linearize? 𝑎 𝑤 𝑡 , ℎ(𝑡) ≈ 𝑎 𝑤, ℎ + ℎ 𝑤 𝑡 − 𝑤 + 𝑤 ℎ 𝑡 − ℎ 𝑎 𝑤, ℎ 𝑤 ℎ 𝑤 ℎ 𝑡 − ℎ ℎ𝑤𝑡−𝑤 Error Small error
- 6. 6 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION Smith & Corripio, 2005 EXAMPLE 2-6.3 T p Density of an ideal gas as function of pressure and temperature: 𝜌 𝑝 𝑡 , 𝑇(𝑡) = 𝑀 𝑝(𝑡) 𝑅 𝑇(𝑡) Linear approximation? Additional information 𝑀 = 20 𝑘𝑔 𝑘𝑚𝑜𝑙 𝑇 = 300 𝐾 𝑃 = 101.3 𝑘𝑃𝑎 𝑅 = 8.314 𝑘𝑃𝑎 ∙ 𝑚3 𝑘𝑚𝑜𝑙 ∙ 𝐾
- 7. 7 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION Smith & Corripio, 2005 How to linearize? 𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈ 𝜌 𝑝, 𝑇 + 𝜕𝜌 𝜕𝑝 𝑝 𝑡 − 𝑝 + 𝜕𝜌 𝜕𝑇 𝑇 𝑡 − 𝑇 𝜕𝜌 𝜕𝑝 = 𝜕𝜌 𝜕𝑝 𝑝, 𝑇 = 𝜕 𝜕𝑝 𝑀 𝑝(𝑡) 𝑅 𝑇(𝑡) 𝑝, 𝑇 = 𝑀 𝑅 𝑇(𝑡) 𝑝, 𝑇 = 𝑀 𝑅 𝑇 𝜕𝜌 𝜕𝑇 = 𝜕𝜌 𝜕𝑇 𝑝, 𝑇 = 𝜕 𝜕𝑇 𝑀 𝑝(𝑡) 𝑅 𝑇(𝑡) 𝑝, 𝑇 = − 𝑀𝑝 𝑡 𝑅 𝑇2 𝑡 𝑝, 𝑇 = − 𝑀 𝑝 𝑅 𝑇2 𝜌 𝑝, 𝑇 = 𝑀 𝑝 𝑅 𝑇
- 8. 8 MATHEMATICALTOOLSFORCONTROLSYSTEMS LINEARIZATION Smith & Corripio, 2005 Linearized function 𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈ 𝑀 𝑝 𝑅 𝑇 + 𝑀 𝑅 𝑇 𝑝 𝑡 − 𝑝 − 𝑀 𝑝 𝑅 𝑇2 𝑇 𝑡 − 𝑇 Numerically 𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈ 1.178 + 0.01163 𝑝 𝑡 − 101.3 − 0.00393 𝑇 𝑡 − 300 𝜌 = 𝑘𝑔 𝑚3 ; 𝑇 = 𝐾; 𝑝 = 𝑘𝑃𝑎 Where,
- 9. 9 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝐹𝑖, 𝑇𝑖 𝐹𝑜, 𝑇𝑜 Assumptions Control volume Liquid is well mixed Tank is well insulated Energy input by the stirrer is negligible Constant and equal inlet and outlet volumetric flow, liquid densities and heat capacity Question Mathematical model, 𝑇𝑜 response to changes in 𝑇𝑖 Case 1: Adiabatic
- 10. 10 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS Energy balance: 𝐹𝑖 𝜌𝑖ℎ𝑖 𝑡 − 𝐹𝑜 𝜌 𝑜ℎ 𝑜 𝑡 = 𝑑 𝑉 𝜌 𝑢 𝑡 𝑑𝑡 Rate of energy into control volume Rate of energy out of control volume Rate of change of energy accumulated in control volume 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 Replacing internal energy 𝑢(𝑡) and enthalpy ℎ(𝑡) 𝑢 𝑡 = 𝑐 𝑣 𝑇 𝑡 − 𝑇𝑟𝑒𝑓 ℎ 𝑡 = 𝑐 𝑝 𝑇 𝑡 − 𝑇𝑟𝑒𝑓 Eq. 1
- 11. 11 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝐹𝜌𝑐 𝑝 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇𝑠𝑠 = 0 Deviation Subtracting Eq. 1 & 2 Stable State (SS): Eq. 2 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑇𝑠𝑠 = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 𝑇𝑖(𝑡) = 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 𝑇(𝑡) = 𝑇 𝑡 − 𝑇𝑠𝑠 𝐹𝜌𝑐 𝑝 𝑇𝑖(𝑡) − 𝐹𝜌𝑐 𝑝 𝑇(𝑡) = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 𝑇𝑖(𝑡) = 𝑉 𝜌𝑐 𝑣 𝐹𝜌𝑐 𝑝 𝑑 𝑇 𝑡 𝑑𝑡 + 𝑇(𝑡) 𝝉
- 12. 12 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS Use of Laplace transform yields ℒ 𝑇𝑖(𝑡) = 𝜏 ℒ 𝑑 𝑇 𝑡 𝑑𝑡 + ℒ 𝑇(𝑡) ℒ 𝑑 𝑦 𝑡 𝑑𝑡 = 𝑠𝑦 𝑠 − 𝑦(0) 𝑇𝑖(𝑠) = 𝜏 𝑠 𝑇 𝑠 − 𝑇 0 + 𝑇(𝑠) 𝑇 0 = 𝑇 0 − 𝑇𝑠𝑠 = 0 𝑇 𝑠 = 1 (𝜏𝑠 + 1) 𝑇𝑖 (𝑠) Transfer function first-order processes Assuming inlet temperature increases of M degrees in magnitude 𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 + 𝑀 𝑡 < 0 𝑡 ≥ 0
- 13. 13 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑇𝑖 𝑡 = 𝑀 𝑢(𝑡) 𝑇𝑖 𝑠 = 𝑀 𝑠 Unit step function 𝑇 𝑠 = 1 (𝜏𝑠 + 1) 𝑀 𝑠 Partial fractions method 𝑇 𝑠 = 𝑀 𝑠 𝜏𝑠 + 1 = 𝐴 𝑠 + 𝐵 𝜏𝑠 + 1 𝐴 𝜏𝑠 + 1 + 𝐵𝑠 = 𝑀 𝑇 𝑠 = 𝑀 𝑠 + −𝑀𝜏 𝜏𝑠 + 1 ℒ−1 1 𝑠 + 𝑎 = 𝑒−𝑎𝑡 ℒ−1 1 𝑠 = 𝑒−0𝑡 = 1ℒ−1 𝑇 𝑠 = ℒ−1 𝑀 𝑠 + ℒ−1 −𝑀𝜏 𝜏𝑠 + 1 Laplace transform inverse
- 14. 14 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑇 𝑡 = 𝑀 (1 − 𝑒−𝑡/𝜏) 𝑇 𝑡 = 𝑇𝑠𝑠 + 𝑀 (1 − 𝑒−𝑡/𝜏)or 0 M 𝑇 𝑡 , °C 𝑇 𝑇 + 𝑀 𝜏 Time 0.632 𝑀 Figure. Response of a first-order process to a step change in input variable
- 15. 15 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS Case 2: Non adiabatic 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑞 𝑡 = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 𝑞 𝑡 = 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠(𝑡)Assumption: U constant 𝐹𝜌𝑐 𝑝 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇𝑠𝑠 − 𝑈𝐴 𝑇𝑠𝑠 − 𝑇𝑠,𝑠𝑠 = 0 Stable State (SS): Eq. 2 Eq. 1𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠(𝑡) = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 Subtracting Eq. 1 & 2
- 16. 16 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑇𝑠𝑠 − 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠𝑠 − (𝑇𝑠 𝑡 − 𝑇𝑠,𝑠𝑠) = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 𝑇𝑖(𝑡) = 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 𝑇(𝑡) = 𝑇 𝑡 − 𝑇𝑠𝑠 𝑇𝑠(𝑡) = 𝑇𝑠 𝑡 − 𝑇𝑠,𝑠𝑠 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑈𝐴 𝑇(𝑡) − 𝑇𝑠(𝑡) = 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 𝑉 𝜌𝑐 𝑣 𝑑 𝑇 𝑡 𝑑𝑡 + 𝐹𝜌𝑐 𝑝 + 𝑈𝐴 𝑇 𝑡 = 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 + 𝑈𝐴 𝑇𝑠(𝑡)
- 17. 17 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑉 𝜌𝑐 𝑣 𝐹𝜌𝑐 𝑝 + 𝑈𝐴 𝑑 𝑇 𝑡 𝑑𝑡 + 𝑇 𝑡 = 𝐹𝜌𝑐 𝑝 𝐹𝜌𝑐 𝑝 + 𝑈𝐴 𝑇𝑖 𝑡 + 𝑈𝐴 𝐹𝜌𝑐 𝑝 + 𝑈𝐴 𝑇𝑠 (𝑡) 𝝉 𝑲 𝟏 𝑲 𝟐 Use of Laplace transform yields 𝜏 𝑠 𝑇 𝑠 − 𝑇 0 + 𝑇 𝑠 = 𝐾1 𝑇𝑖 𝑠 + 𝐾2 𝑇𝑠 𝑠 𝑇 𝑠 = 𝐾1 𝜏𝑠 + 1 𝑇𝑖 𝑠 + 𝐾2 𝜏𝑠 + 1 𝑇𝑠 𝑠 𝐾 = ∆𝑂 ∆𝐼 = ∆ output variable ∆ input variable
- 18. 18 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑇 𝑠 = 𝐾2 𝜏𝑠 + 1 𝑇𝑠 𝑠 Constant: surrounding temperature Constant: liquid temperature 𝑇 𝑠 = 𝐾1 𝜏𝑠 + 1 𝑇𝑖 𝑠 𝑇𝑠 𝑡 = 𝑇𝑠,𝑠𝑠; 𝑇𝑠 𝑠 = 0 𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠; 𝑇𝑖 𝑠 = 0 Assuming inlet temperature increases of M degrees in magnitude 𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 + 𝑀 𝑡 < 0 𝑡 ≥ 0
- 19. 19 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑇𝑖 𝑡 = 𝑀 𝑢(𝑡) 𝑇𝑖 𝑠 = 𝑀 𝑠 Unit step function 𝑇 𝑠 = 𝐾1 (𝜏𝑠 + 1) 𝑀 𝑠 Partial fractions method 𝑇 𝑠 = 𝐾1 𝑀 𝑠 𝜏𝑠 + 1 = 𝐴 𝑠 + 𝐵 𝜏𝑠 + 1 𝐴 𝜏𝑠 + 1 + 𝐵𝑠 = 𝐾1 𝑀 𝑇 𝑠 = 𝐾1 𝑀 𝑠 + −𝐾1 𝑀𝜏 𝜏𝑠 + 1 ℒ−1 1 𝑠 + 𝑎 = 𝑒−𝑎𝑡 ℒ−1 1 𝑠 = 𝑒−0𝑡 = 1ℒ−1 𝑇 𝑠 = ℒ−1 𝐾1 𝑀 𝑠 + ℒ−1 −𝐾1 𝑀𝜏 𝜏𝑠 + 1 Laplace transform inverse
- 20. 20 FIRST-ORDERDYNAMICSYSTEMS THERMAL PROCESS EXAMPLE Smith & Corripio, 2005 THERMAL PROCESS 𝑇 𝑡 = 𝐾1 𝑀 (1 − 𝑒−𝑡/𝜏) 𝑇 𝑡 = 𝑇𝑠𝑠 + 𝐾1 𝑀 (1 − 𝑒−𝑡/𝜏)or 0 M 𝑇 𝑡 , °C 𝑇 𝑇 + 𝐾1 𝑀 Time 𝐾1 𝑀 Figure. Response of a first-order process to a step change in input variable
- 21. THANKS ! 21