Ordinary differentiall equations

Transcription

1 Ordinary differentiall equations

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3 Finite element method Divide macroscopic object into small pieces Can investigate real-size objects in real world time regimes Disregarding of atomic nature of matter and long-range interactions

4 Case study Reactor with reaction

5 More complicated problem Accumulation = inputs - outputs

6 inputs outputs outputs More complicated problem Accumulation = inputs - outputs Q 55 = 2 m3 min Q 15 = 3 m3 min outputs outputs inputs c 5 Q 54 = 2 m3 min Q 25 = 1 m3 min outputs Q 44 = 11 m3 min Q 01 = 5 m3 min Q 12 = 3 c 01 = 20 mg/m 3 min m 3 c 1 c 2 c 4 inputs outputs inputs outputs inputs outputs Q 31 = 1 m3 min outputs Q 23 = 1 m3 min Q 34 = 8 m3 min Q 03 = 8 m3 min c 03 = 20 mg/m 3 inputs c 3 outputs

7 outputs How to solve that problem? Accumulation = inputs - outputs Q 15 = 3 m3 min So? outputs Q 01 = 5 m3 min Q 12 = 3 c 01 = 20 mg/m 3 min inputs c 1 outputs m 3 V 1 dc 1 dt = c 01Q 01 + c 3 Q 31 c 1 Q 12 c 1 Q 15 Q01c01 = 50 mg/min, Q03c03 = 160 mg/min, V1 =50 m3, V2 = 20 m3, V3 = 40 m3, V4 = 80 m3, and V5 = 100 m3 Q 31 = 1 m3 min

8 outputs How to solve that problem? Accumulation = inputs - outputs Q 15 = 3 m3 min So? outputs Q 01 = 5 m3 min Q 12 = 3 c 01 = 20 mg/m 3 min inputs c 1 outputs m 3 V 1 dc 1 dt = c 01Q 01 + c 3 Q 31 c 1 Q 12 c 1 Q 15 Q01c01 = 50 mg/min, Q03c03 = 160 mg/min, V1 =50 m3, V2 = 20 m3, V3 = 40 m3, V4 = 80 m3, and V5 = 100 m3 Q 31 = 1 m3 min V 1 dc 1 dt = 0.12c c 3 + 1

9 outputs How to solve that problem? Accumulation = inputs - outputs Q 15 = 3 m3 min So? outputs Q 01 = 5 m3 min Q 12 = 3 c 01 = 20 mg/m 3 min inputs c 1 outputs m 3 V 1 dc 1 dt = c 01Q 01 + c 3 Q 31 c 1 Q 12 c 1 Q 15 Q 01 c 01 = 50 mg min, Q 03c 03 = 160 mg min, V 1 = 50 m 3, V 2 = 20 m 3, V 3 = 40 m 3, V 4 = 80 m 3, V 5 = 100 m 3 Q 31 = 1 m3 min dc 1 dt = 0.12c c 3 + 1

10 How to solve that problem? Accumulation = inputs - outputs For other reactors dc 1 dt = 0.12c c dc 2 dt = 0.15c c 2 dc 3 dt = 0.025c c dc 4 dt = 0.1c c c 5 dc 5 dt = 0.03c c c 5 Initially the concentration in all reactors will be zero. What will be the evolution of concentrations in time? How to solve this problem?

11 A really simple example Converstion of cyclopropane to propane cyclopropane propane k = s, T = 500o C 1st order dc dt = c t If the initial concentration c(0) of cyclopropane is 0.05 M, what is the concentration after 30 min?

12 How to calculate c(t)? Differential equation dc dt = c t

13 How to calculate c(t)? Differential equation dc dt = c t t 1 න 0 c dc dt dt = න 0 t dt c t dc න c(0) c = න 0 t dt

14 How to calculate c(t)? Differential equation dc dt c t dc න c(0) c = c t = න 0 t dt ln c t ln c 0 = t

15 How to calculate c(t)? Differential equation dc dt c t dc න c(0) c = c t = න 0 t dt ln c t ln c 0 c t c 0 = e t = t

16 How to calculate c(t)? Differential equation dc dt c t dc න c(0) c = c t = න 0 ln c t ln c 0 c t c 0 = e t c t = c 0 e t t dt = t

17 Different approach Euler s scheme df f (t + Dt) - f (t) f (t + Dt) - f (t) dt = lim Dt 0 Dt» Dt df dt = f i+1 - f i Dt

18 Instead of solving function analytically step by step procedure f t + Δt = f t + f t Δt f Analytical solution t

19 Instead of solving function analytically step by step procedure f t + Δt = f t + f t Δt f Analytical solution f Point-by-point solution t t

20 How does it works we start from f(0) 1. Choose step Δt = 1 2. Compute next point using chain rule f t + Δt = f t + f t Δt Example f t = 2 t f t = 2

21 How does it works we start from f(0) f 0 + Δt = 1 = = 2 f f(1) 1 t

22 How does it works we start from f(0) f = = 4 f f(2) f(1) t 2 2

23 And so on point by point we reproduce solution f f(t) = 2t f(2) f(1) t 2 2

24 Problems? - Accuracy Example derivative varies (change of the function s shape) f Wrong value! t

25 Solution? More points, shorter interval Example derivative varies (change of the function s shape) f Problem? Computing efficiency! t

26 How to calculate c(t)? Different approach finite difference method Δc Δt = c t

27 How to calculate c(t)? Different approach finite difference method Δc Δt = c t c t + Δt c t Δt = c t

28 How to calculate c(t)? Different approach finite difference method Δc Δt = c t c t + Δt c t Δt = c t c t + Δt c t = Δtc t

29 How to calculate c(t)? Different approach finite difference method, explicit scheme Δc Δt = c t c t + Δt c t Δt = c t c t + Δt c t = Δtc t c t + Δt = c t Δtc t = c t 1 Δt

30 c t + Δt = c t Δtc t = c t 1 Δt c n = c n 1 1 Δt c t

31 c t + Δt = c t Δtc t = c t 1 Δt Δt = 0. 1 c 0 = c 0 = 1 c n = c n 1 1 Δt c 1 t

32 c t + Δt = c t Δtc t = c t 1 Δt Δt = 0. 1 c 0 = c 0 = 1 c n = c n 1 1 Δt c 1 = c 0 1 Δt = = 0. 9 c t

33 c t + Δt = c t Δtc t = c t 1 Δt Δt = 0. 1 c 0 = c 0 = 1 c n = c n 1 1 Δt c 1 = c 0 1 Δt = = 0. 9 c c 2 = c 1 1 Δt = = t

34 c t + Δt = c t Δtc t = c t 1 Δt Δt = 0. 1 c 0 = c 0 = 1 c n = c n 1 1 Δt c 1 = c 0 1 Δt = = 0. 9 c 1 c 2 = c 1 1 Δt = = c 3 = c 2 1 Δt = = t

35 c t + Δt = c t Δtc t = c t 1 Δt Δt = 0. 1 c 0 = c 0 = c n = c n 1 1 Δt c 1 = c 0 1 Δt = = 0. 9 c 2 = c 1 1 Δt = = c 3 = c 2 1 Δt = = c explicit exp(-kt)

36 Problems? c t + Δt = c t Δtc t = c t 1 Δt Say Δt = 1 then c 1 = c = 0 c 2 = c = 0 Say Δt > 1 Δt = 2 then c 1 = c = c 0 < 0 negative concentration. Unphysical!!! c t + Δt = c t 1 kδt 1 kδt > 0 Δt > 1/k c exp(-kt)

37 Solution? Implicit scheme f f t Δt 0 df dt t Δf Δt t or Δf Δt t + Δt or anywhere inbetween Δf f t + Δt Δt Slope in point t Δf tgβ = lim Δt 0 Δt = df dt t α t + Δt Average slope on Δt interval β t tgα = Δf Δt

38 Solution? Implicit scheme Before Now Δc Δc = c t = c t + Δt Δt Δt c t + Δt c t c t + Δt c t = c t = c t + Δt Δt Δt c t + Δt = c t Δtc t + Δt c t + Δt = c t Δtc t

39 Solution? Implicit scheme c t + Δt = c t Δtc t + Δt c t + Δt + Δtc t + Δt = c t c t + Δt 1 + Δt = c t c t + Δt = c t 1 + Δt Always positive

40 Solutions, comparison dc = c t dt c t c t + Δt = c t Δtc t c t + Δt = 1 + Δt Δt = 0. 1 c t = c 0 e t Δt = c explicit c implicit exp(-kt)

41 Problems? - Accuracy Example derivative varies (change of the function s shape) f Wrong value! t

42 But what about set of ODE? Same approach combined with linear equations dc 1 dt = 0.12c c dc 2 dt = 0.15c c 2 dc 3 dt = 0.025c c dc 4 dt = 0.1c c c 5 dc 5 dt = 0.03c c c 5 dc 1 dt dc 2 dt dc 3 dt dc 4 dt dc 5 dt = c 1 c 2 c 3 c 4 c

43 dc 1 dt dc 2 dt dc 3 dt dc 4 dt dc 5 dt = But what about set of ODE? Same approach combined with linear equations c 1 c 2 c 3 c 4 c d dt c 1 c 2 c 3 c 4 c 5 = c 1 c 2 c 3 c 4 c

44 But what about set of ODE? Same approach combined with linear equations d dt c 1 c 2 c 3 c 4 c 5 = c 1 c 2 c 3 c 4 c തc A തc ഥb dതc dt = Aതc + ഥb

45 What next? Finite difference approach but on vectors! dതc dt = Aതc Δതc Δt = Aതc

46 What next? Finite difference approach but with vectors! Δതc Δt = Aതc തc t + Δt ҧ c(t) Δt = Aതc തc t + Δt = തc t + ΔtAതc

47 What next? Finite difference approach but with vectors! Explicit scheme തc = തc(t) തc t + Δt = തc t + ΔtAതc(t) തc t + Δt = I + ΔtA cҧ t തc t + Δt = B ҧ c t B = I + ΔtA

48 What next? Finite difference approach but with vectors! Explicit scheme തc = തc(t) തc t + Δt = B ҧ c t B = I + ΔtA Define c(0) Compute B = I + ΔtA Problem? Unstable! Compute c in the next time frame c nδt = Bc( n 1 Δt)) n + 1

49 What next? Finite difference approach but on vectors! Implicit scheme തc = തc(t + Δt) തc t + Δt = തc t + ΔtAതc(t + Δt) തc t + Δt ΔtAതc(t + Δt) = തc t I ΔtA തc t + Δt = തc t

50 What next? Finite difference approach but on vectors! Implicit scheme തc = തc(t + Δt) I ΔtA തc t + Δt = തc t Bതc t + Δt = തc t തc t + Δt = B 1 തc t B = I ΔtA

51 Implicit scheme തc = തc(t) തc t + Δt = B 1 ҧ c t Define c(0) B = I ΔtA Problem? Computing B 1 Compute B = I + ΔtA Compute B 1 Compute c in the next time frame c nδt = B 1 c( n 1 Δt)) n + 1

52 Working example implicite scheme A k 1=1 B k 2 =1 C da dt = A db = A B dt dc dt = B d dt Initially A = 1, B = 0, C = 0 A B C cҧ 0 = = A B C

53 Working example 1. Initiate, calculate B d dt A B C = A k 1=1 B k 2 =1 C A B C cҧ 0 = B = I ΔtA Δt=1 cҧ 0 = B = =

54 Working example 2. Calculate B 1 B = B 1 =

55 Working example 3. Iterate n=1 cҧ nδt = B 1 cҧ n 1 Δt cҧ 1 = B 1 cҧ 0 = = A B C

56 Working example 3. Iterate n=2 cҧ nδt = B 1 cҧ n 1 Δt cҧ 2 = B 1 cҧ 1 = =

57 Working example 3. Iterate n=3 cҧ nδt = B 1 cҧ n 1 Δt cҧ 3 = B 1 cҧ 2 = =

58 Working example 3. Iterate n=4 cҧ nδt = B 1 cҧ n 1 Δt cҧ 4 = B 1 cҧ 3 = =