urban heat islands: an optimal control approach › 2015 › 09 › alvarezvazquez-icia… · urban...

Urban heat islands:An optimal control approach

L.J. Alvarez-Vazquez, F.J. Fernandez,

N. Garcıa-Chan, A. Martınez,

M.E. Vazquez-Mendez.

Departamento de Matematica Aplicada II

Universidad de Vigo

The Urban Heat Island: a metheorological phenomenon

(published on website. 30 June, 2010)http://blog.rtve.es/eltiempo/2010/06/isla-de-calor.html

Connected problems: “Beret polution”

http://elpais.com/diario/2011/10/07/catalunya/1317949650_850215.html (published on website. 7 October, 2011)

Improvement Strategy: Build green zones

http://www.construyeargentina.com/wp-content/uploads/2013/06/green_roof3b.jpg

Objective: Control UHI by building green zones and parks

http://www.fincasfiol.com/wp-content/uploads/2013/09/VERDEVERTICAL_0721.jpg

Numerical Simulation Optimal Control Numerical Results

Table of contents

1 Numerical Simulation: air temperature and velocityMathematical modelNumerical Resolution

2 Optimal Control: location of green zonesMathematical FormulationNumerical Resolution

3 Numerical ResultsNumerical simulation without green zonesOptimal location of two green zones

Table of contents

Domain Ω (3D)

Domain Ω (2D)

State System

Air velocity u(x, t) (m/s) and pressure p(x, t)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t+ u.∇u−∇.(Km∇u) +∇p =

θrefg

−cdf

NGZ∑

LADk1Ωk∥u∥u in Ω× (0, T ),

∇.u = 0 in Ω× (0, T ),u.n = 0 on (Γr ∪ Γw ∪ Γs)× (0, T ),u.n = −u∗ on Γ1 × (0, T ),u.n = 0 on Γ2 × (0, T ),u.n = u∗ on Γ3 × (0, T ),u(0) = u0 in Ω,

State System

Air temperature θ(x, t) (K)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∂t+ u.∇θ −∇.(Kh∇θ) =

NGZ∑

LADk1Ωk

θkf − θ

rin Ω× (0, T ),

θ = θin on Γ1 × (0, T ),∇θ.n = 0 on (Γ2 ∪ Γ3)× (0, T ),Kh∇θ.n = γ1(T

4rw − θ4) on Γw × (0, T ),

Kh∇θ.n = γ1(T4rr − θ4) on Γr × (0, T ),

Kh∇θ.n =NGZ∑

(σk1γ1(T

4rp − θ4) + σk

2γ2(θkf

4− θ4)

)1Γgk

+γ1(T4r − θ4)

NGZ∑

)on Γs × (0, T ),

θ(0) = θ0 in Ω,

State System

Foliage temperature θkf (x, t) (K) at parkland Ωk, k = 1, . . . , NGZ

θkf − θ

r= σk

1γ1(T4rf − θkf

4) + σk

2γ2(θk4− θkf

4) in Ωk × (0, T )

Numerical Resolution of the State System

Time discretization: the method of characteristics

We choose a natural number N ∈ N, define the time step ∆t = TN , and

consider the discrete times tnNn=0 ⊂ [0, T ] given by tn = n∆t, forn = 0, . . . , N. The characteristic method is based on

∂t+ u.∇c ≃ 1

∆t(cn+1 − cn Xn),

where Xn(x) = X(x, tn+1, tn) is given by

⎧⎨

dτ= u(X(x, t, τ), τ),

X(x, t, t) = x.

Semi-discrete problem

Then, given initial fields u0 and θ0, we are interested in finding, for eachn = 0, . . . , N − 1, the fields un+1, θn+1, pn+1, and θkf,n+1, k = 1, 2,solving the following system of partial differential equations:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αun+1 −∇.(Km∇un+1) +∇pn+1 =θn+1

θrefg

−cdf

LADk1Ωk∥un+1∥un+1 + α(un Xn) in Ω,

∇.un+1 = 0 in Ω,un+1.n = 0 on Γr ∪ Γw ∪ Γs,un+1.n = −u∗ on Γ1,un+1.n = 0 on Γ2,un+1.n = u∗ on Γ3.

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αθn+1 −∇.(Kh∇θn+1) =2∑

LADk1Ωk

θkf,n+1 − θn+1

rn+1+ α(θn Xn)

θn+1 = θin on Γ1

∇θn+1.n = 0 on Γ2 ∪ Γ3,Kh∇θn+1.n = γ1(T

4rw,n+1 − θ4n+1) on Γw,

Kh∇θn+1.n = γ1(T4rr,n+1 − θ4n+1) on Γr,

Kh∇θn+1.n =2∑

(σk1γ1(T

4rp,n+1 − θ4n+1) + σk

2γ2(θkf,n+1

4− θ4n+1))1Γgk

+γ1(T4ra,n+1 − θ4n+1)(1− 1Γg1

− 1Γg2) on Γs

θkf,n+1 − θn+1

rn+1= σk

1γ1(T4rf ,n+1 − θkf,n+1

4) + σk

2γ2(θkn+1

4− θkf,n+1

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

αθn+1 −∇.(Kh∇θn+1) =2∑

LADk1Ωk

θkf,n+1 − θn+1

rn+1+ α(θn Xn)

θn+1 = θin on Γ1

∇θn+1.n = 0 on Γ2 ∪ Γ3,Kh∇θn+1.n = γ1(T

4rw,n+1 − θ4n+1) on Γw,

Kh∇θn+1.n = γ1(T4rr,n+1 − θ4n+1) on Γr,

Kh∇θn+1.n =2∑

(σk1γ1(T

4rp,n+1 − θ4n+1) + σk

2γ2(θkf,n+1

4− θ4n+1))1Γgk

+γ1(T4ra,n+1 − θ4n+1)(1− 1Γg1

− 1Γg2) on Γs

θkf,n+1 − θn+1

rn+1= σk

1γ1(T4rf ,n+1 − θkf,n+1

4) + σk

2γ2(θkn+1

4− θkf,n+1

Space discretization: the finite element method

We consider a mesh τh of Ω and, associated to this mesh, we define thefollowing finite element spaces:

Wh = v ∈ [C(Ω)]n : v| !T ∈ [P1b(T )]n, ∀T ∈ τh, v · n = 0 on ∂Ω,Mh = q ∈ C(Ω) : q| !T ∈ P1(T ), ∀T ∈ τh,Vh = z ∈ C(Ω) : z| !T ∈ P1(T ), ∀T ∈ τh, z|Γ1

= 0,Xh = z ∈ L2(Ω) : z| !T ∈ P0(T ), ∀T ∈ τh.

Fully discrete problem

For each n = 0, . . . , N − 1, we look for(uh,n+1, ph,n+1, θh,n+1, θ1f,h,n+1, θ

2f,h,n+1) ∈ Wh ×Mh × Vh ×Xh ×Xh,

solution of the variational formulations of the previous semi-discreteproblems −→ Freefem++.

[1] F.J. Fernandez, et al. Optimal location of green zones in metropoli-tan areas to control the urban heat island. J. Comput. Appl.Math., in press (2015)

= 0,Xh = z ∈ L2(Ω) : z| !T ∈ P0(T ), ∀T ∈ τh.

Control 2D

Control variable

We asumme that Ωk = Γgk × [0, Zk], where Zk is known. If Ω ⊂ R2,then

Γgk = [p1, p1 + lk]

We take NGZ = 2 and suppose that l1 + l2 = L is given. Thenl2 = L− l1 and the control variable is

b = (p1, p2, l1) ∈ R3

Control 2D

Control variable

We asumme that Ωk = Γgk × [0, Zk], where Zk is known. If Ω ⊂ R2,then

Γgk = [p1, p1 + lk]

We take NGZ = 2 and suppose that l1 + l2 = L is given. Thenl2 = L− l1 and the control variable is

b = (p1, p2, l1) ∈ R3

Control 3D

Control variable

We asumme that Ωk = Γgk × [0, Zk], where Zk is known. If Ω ⊂ R3 andΓgk is a rectangle, then Γgk = [pk1 , p

k1 + lk1 ]× [pk2 , p

k2 + lk2 ]

We take NGZ = 2 and suppose thatNGZ∑

lk1 lk2 = L is given. Then

l22 = (L− l11l12)/l

21 and the control variable is

b = (p11, p12, p

21) ∈ R7

Optimal Control

Objective function

J(b) =

(Γs\(Γg1∪Γg2 ))×[a,b]θ(x, t) dx dt

T (b− a)µ (Γs \ (Γg1 ∪ Γg2))

Optimal Control

Admissible set

Uad = b ∈ R4(n−1)−1 : ∀k = 1, 2, Γgk ⊂ Γsjk, for any jk ∈ 1, . . . ,M,

with Γsj1∩ Γsj2

= ∅ and µmin ≤ µ(Γgk) ≤ µmax.

Optimal Control

Optimal Control Problem

minb ∈ Uad

Numerical Resolution of the Optimal Control Problem

The discrete control problem

minb∈Uad

J∆th (b) =

Ω(1− 1Γg1

− 1Γg2)1[a,b]θh,n(x)dx

Ω(1− 1Γg1

− 1Γg2)1[a,b] dx

Equivalent Formulation: The MINLP Problem

(MINLP)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

minb,y

f(b,y)

s.t. h(b,y) ≤ 0Ay ≤ cy ∈ 0, 1p

where p = 2M , y ∈ R2M , b ∈ R4(n−1)−1, f : R4(n−1)−1+2M → R,

h : R4(n−1)−1+2M → Rm1 , A ∈ Mm2×2M and c ∈ Rm2

Numerical Resolution of the Optimal Control Problem

The discrete control problem

minb∈Uad

J∆th (b) =

Ω(1− 1Γg1

− 1Γg2)1[a,b]θh,n(x)dx

Ω(1− 1Γg1

− 1Γg2)1[a,b] dx

Equivalent Formulation: The MINLP Problem

(MINLP)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

minb,y

f(b,y)

s.t. h(b,y) ≤ 0Ay ≤ cy ∈ 0, 1p

where p = 2M , y ∈ R2M , b ∈ R4(n−1)−1, f : R4(n−1)−1+2M → R,

h : R4(n−1)−1+2M → Rm1 , A ∈ Mm2×2M and c ∈ Rm2

Numerical procedure

(MINLP)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

minb,y

f(b,y)

s.t. h(b,y) ≤ 0Ay ≤ cy ∈ 0, 1p

We define Y =y ∈ 0, 1p : Ay ≤ c

. For each y∗ ∈ Y we

denote by b∗ ∈ R4(n−1)−1 the numerical solution of

f(b,y∗)

s.t. h(b,y∗) ≤ 0−→ IPOPT code

(MINLP) is solved by solving the following problem

(IP) miny∗∈Y

f(b∗,y∗) −→ Exhaustive search

Numerical procedure

(MINLP)

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

minb,y

f(b,y)

s.t. h(b,y) ≤ 0Ay ≤ cy ∈ 0, 1p

We define Y =y ∈ 0, 1p : Ay ≤ c

. For each y∗ ∈ Y we

denote by b∗ ∈ R4(n−1)−1 the numerical solution of

f(b,y∗)

s.t. h(b,y∗) ≤ 0−→ IPOPT code

(MINLP) is solved by solving the following problem

(IP) miny∗∈Y

f(b∗,y∗) −→ Exhaustive search

Numerical simulation without green zones

Trw = 352.60K, Trr = Tra = 368.28K, Trp = 358.90K,Trf = 309.71K

Rsw,net(1− am) +Rlw,dow − ϵm σB T 4r = 0

Results

Numerical simulation without green zones

Trw = 352.60K, Trr = Tra = 368.28K, Trp = 358.90K,Trf = 309.71K

Rsw,net(1− am) +Rlw,dow − ϵm σB T 4r = 0

Results

Optimal location of two green zones

Z1 = Z2 = 6m

L = 8m, µmin = 1m, and µmax = 5m

LAD1(z) = LAD2(z)

Results

Without green zones

Results

Without green zones

Work in progress

3D approach

Preliminary numerical results

Work in progress

3D approach

Preliminary numerical results

urban heat islands: an optimal control approach › 2015 › 09 › alvarezvazquez-icia… · urban...

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