Reservoir modelling Non-linear reservoirs
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The linear reservoir Qin (n) Reservoir modelling Non-linear reservoirs General equations: Paul Torfs dS = Qin (t) − Qout dt Qout = F(S) S(n) linear reservoir: F(S) = k S Qout(n) Sumova 2008 The non-linear reservoir Sumova 2008 The power reservoir 4 I Qin (n) but many other formula’s are “thinkable” 3 Q = a Sb F(S) = k1 S + k2 S 2 Q2 (a = 1) F(S) = k S b S(n) F(S) = k .. . Qout(n) dS = Qin (t) − Qout dt Qout = F(S) (eb S − 1) .. . 1 0 0 1 2 3 S F(0) = 0?, F increasing? I b = 1.2 b = 1.0 b = 0.8 many other formula’s are used in practice Sumova 2008 I b < 1: more than linear for small S, less then linear for big S I b > 1: less than linear for small S, more then linear for big S Sumova 2008 The broken linear reservoir The broken linear reservoir Q Q Sc Sc Sc if S ≤ Sc S Sc if S ≥ Sc Q = k1 S Sumova 2008 S Q = k1 Sc + k2 (S − Sc ) Sumova 2008 The finite reservoir The finite reservoir Qin Smax Qin Smax Qout Qov but should defined with the help of an overflow flux: ( 0 if S < Smax Qov = max(Qin − Qout , 0) if S = Smax Qout reservoir with maximal content is not a special form Sumova 2008 Sumova 2008 2.0 The shape of a reservoir I typical “outflow” is given by function S ↔ Q 1.5 The shape of a reservoir I linear reservoir: straight cylinder R(h) h h = water level in reservoir I let the reservoir by cylindrical R(h) = radius for level h I “change” formula: 0.5 S(n) Qout(n) 0.0 I 0.0 0.2 0.4 0.6 0.8 1.0 1.2 outlet with fixed diameter 1 Q=h 1.0 Q Qin (n) I what is the “form” corresponding to an arbitrary function as the one on the left? π R2 (h) = dS dS 1 = = dQ dh dQ dS S Sumova 2008 Numerics: explicit scheme dS = Qin (t) − F(S(t)) dt S[n + 1] − S[n] = Qin [n] − F(S[n]) ∆t 1.0 S[n + 1] = S[n] + ∆t Qin [n] − F(S[n]) 0.0 0.0 0.5 0.5 h 1.0 1.5 1.5 2.0 2.0 The shape of a reservoir Q Sumova 2008 0.0 0.2 0.4 0.6 S 0.8 1.0 1.2 −0.4 −0.2 0.0 0.2 0.4 I non-linearity of F causes no problem whatsoever I works as well and as easily for system of many reservoirs R Sumova 2008 Sumova 2008 Numerics: implicit scheme Numerics: implicit scheme S[n + 1] − S[n] = Qin [n + 1] − F(S[n + 1]) ∆t dS = Qin (t) − F(S(t)) dt S − S[n] ∆t Qin − F (S) Q S[n + 1] − S[n] = Qin [n + 1] − F(S[n + 1]) ∆t F(S) S[n + 1] =? S a “general” non-linear relation non-linear problem to solve! S[n + 1] Sumova 2008 Numerics: implicit scheme S Sumova 2008 Numerics: implicit scheme S[n + 1] − S[n] = Qin [n + 1] − F(S[n + 1]) ∆t dS = Qin (t) − F(S(t)) dt S[n + 1] − S[n] = Qin [n + 1] − F(S[n + 1]) ∆t S[n + 1] =? S − S[n] ∆t Q Qin − F (S) F(S) S a non-linear (and non-continuous) relation with an “if” statement S Sumova 2008 I non-linear problem to solve! I when with “if”-statements even more difficult/impossible I for system of many reservoirs: very difficult/impossible in general Sumova 2008 Recession analysis I Emptying a broken linear reservoir ∂S (t) = −k S(t)b ∂t in general: 1.0 0.9 0.8 0.7 0.6 S 0.5 0.4 0.3 0.2 0.1 0.0 ∂S (t) = −F(S) ∂t is difficult to solve I some examples can be done: ∂S (t) = −k S(t)b ∂t S(0) = S0 i1/(1−b) h S(t) = S0 1 − k (1 − b) S0b−1 t b = 1.4 b = 1.0 b = 0.6 0 1 2 3 4 5 t Sumova 2008 Non-linear recession analysis Sumova 2008 Non-linear recession analysis Selecting the recession limbs 2.5 The data: only Q Qout ● ● ● ● ● 2.5 ● ● ● ● 1.5 ● ● 1.5 ● ● ● ● ● ● ● ● ● ● ● 0.5 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●●●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●●● ●● ●●● ●●●●● ●● ●●●●●●●●● ● ● ● ● ● ●●● ● ● ●● ● ● ●● ●● ● ● ●● ●● ●● ●●● ●●● ●●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ●● ●●●● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●● ●●●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ●● ●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ●● ●● ●●●● ●●●●●● ●●●●●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● 0 0 200 ● ● 400 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ●● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 0.5 ● ● ● 1.0 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● 0.0 ● ● ● ● ● ● ● ● ● ● ● 1.0 ● ● 0.0 results[[s]] ● 2.0 2.0 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● 100 ● ● ● ● ● ● ● ●● ● ● ●● ● ● ● ● ● ● ●● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● 200 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 300 600 Sumova 2008 Sumova 2008 Non-linear recession analysis Non-linear recession analysis 1.0 ● ● ● ● 0.8 ● ● ● ● ● I we want S-Q scatter plots to fit non-linear relations I but only Q given I for each recession limb i we do have: dS = 0 − Q(t) dt ● ● ● 0.6 ● ● ● ● ● ● ● ● ● ● 0.4 ● ● ● ● ● ● ● ●● ●● ●● ● ● ● ● ● ● ●● ● ●● ● ● ● ● ●● ● ● ●● ●● ● ● ● ● ●● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ●● ●● ● ● ● ● ● ● ●● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● 0.0 0.2 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● Q 4 3 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 2 0 1 −log(Q)+log(Q(t)) 5 6 “Classical” linear recession: no succes 0.0 ●● ●● ● ● ●● ● ● ● ● I 0.5 1.0 or discrete: Si [n] = Si [n − 1] − Q[n] 1.5 S 5 10 15 I t so we do known most of S but only do not know Si [0] for each recession limb Sumova 2008 Sumova 2008 Non-linear recession analysis Non-linear recession analysis what we know with Si [0] = 0 for each limb (so each curve should be shifted to the right) 1.0 ● 0.8 (so each curve should be shifted to the right) so assume we want to fit: 1.0 ● Q(t) ● 0.0 ● ● ● ● ● ●● ● ● ● ●● ● ● ● ●● ● ● ● ● ● ● ● ● ●●● ● ● ● ● ● ●● ● ● ● ● ●● ● ● ● ● ● ● ●●● ● ● ●● ●● ●● ●● ●●● ●● ● ●● ● ● ● ● ●● ●●● ●●● ● ● ●●● ●● ● ●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● 0.2 ● ● −2.5 −2.0 −1.5 −1.0 Si(t)−Si(0) −0.5 ● ● ● ● ● ● ● ● ● ●● ● ● ● ●● ● ● ● ●● ● ● ● ● ● ● ● ● ●●● ● ● ● ● ● ●● ● ● ● ● ●● ● ● ● ● ● ● ●●● ● ● ●● ●● ●● ●● ●●● ●● ● ●● ● ● ● ● ●● ● ●●● ●●● ● ● ●●● ●● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● Qi [n] ∼ F(Si [n] + Si [0], p1 , p2 , . . .) −2.5 ● −2.0 −1.5 −1.0 Si(t)−Si(0) ● ● ● ● ● ● ● ● ●● ●●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● 0.2 0.4 ● ● do solve fitting problem: ● ● ● 0.4 ● ● ● Q ∼ k Sb ● 0.6 0.6 ● ● ● ● Q ∼ F(S, p1 , p2 , . . .) ● 0.8 ● 0.0 Q(t) make shift part of the fitting what we know ● −0.5 ● ● ● ● ● ● ● ● ● ●● ●●●● ● 0.0 i = number of recession limb n = pos in limb 0.0 Sumova 2008 Sumova 2008 Non-linear recession analysis Non-linear recession analysis after the fitted shifting fitted Q-S relation 0.0 ●● 0.8 ● ● ● ● 0.2 ●● ● ● ●● ● ● ● ● ● 0.4 0.4 ● ● ●● ● ●● ● ● ● ● ●● ● ● ●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ●● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ●● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● 0.0 0.2 0.0 ● ● ●● ● ●● ● ● ● ● ●● ● ● ●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ●● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ●● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● Q ● ● ● ● ● 0.6 0.8 ● ● ● 0.6 Q ● 1.0 1.0 ● ● 0.5 1.0 1.5 0.0 S ● ● ● ●● ●● ●● ● ● ● ● ●● ●● ● ● ●● ● 0.5 ● ● 1.0 1.5 S Sumova 2008 Sumova 2008 Non-linear recession of Isel river 6 Non-linear recession of Isel river ● ● ● 6 ● ● 5 ● 5 ● ● ● ● ● ● ● ● ● ● ● ● 4 ● ● ● ● ● ● ● ● ● ● 2 ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ●● ●● ● ● ● ● ● ● ●●● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ●● ●● ● ● ● ● ● ● ● ●●● ● ● ● ● ● 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