]> 3.3.1 Example 1: Nuclear reactor temperature control II

3.3.1 Example 1: Nuclear reactor temperature control II

The performance index (3.4) has a form (3.16) with

P = ccT ,cT = 10 ,Q = R = 0 L(2) .

Treating a and r as fixed constants, we seek for a pair (b,d) belonging to the domain of stability which minimizes the performance index. The stability domain of the characteristic quasipolynomial (s2 as) + (bs d)esr, in the plane 0bd is an open bounded set with boundary Γ1 Γ2 where the curve Γ1 is given parametrically

b(ω) = a cos ωr ω sin ωr d(ω) = ω2 cos ωr+aω sin ωr ,ω [0,ω0] .

Here ω0 denotes the smallest positive root of the equation ω = a tan ωr and Γ2 is an interval of 0b axis with ends coinciding with the ends of Γ1.

Taking into account the nature of initial conditions we conclude that

J(x0)(b,d) = x0,x0H = v0T αv 0 = z20 2α 22,α = αT = α11α12 α12α22 .

But, since C = R = 0 L(2) to find α22 from (3.33) we ought to put γ = 0 in its right hand side. To determine the system (3.33) we first calculate the polynomial (3.29),

[λ4 + λ2(b2 a2) d2]λ2(λ2 a2) .

It has eight roots but we write down only four of them because others may differ of signs,

λ1 = 1 2 a2 b2 + (b2 a2 )2 + 4d2 , λ2 = jK,K = 1 2 b2 a2 + (b2 a2 )2 + 4d2 , λ3 = 0,λ4 = a .

From (3.31) we determine the corresponding eigenvectors of (3.28)

Li = d b λi2(λ i + a) d bλi bλi2(λ i + a) d(d bλi) ,Mi = dλi(λi + a) d bλi λi bλi(λi + a) d bλi bλi d

for i = 1, 2 and

L3 = db 00 ,M3 = 0 0 a 1 ,L4 = 0 0 a1 ,M4 = d0 b0 .

The system (3.33) takes the form

0 0 0 0s1s8 2d 2dear 2 2a 0 0s2s9 a + ba bear 2 02a 0s2s9 a + ba bear 0 2 2 4as3s10 2 2 0 2d 0 0s4s11 d d 0 2b 0 0s5s12a + b b 0 0 0 2ds6s13 0 aear 0 0 0 2bs7s14 1 ear α11 α12 α12 α22 2κ1 2κ2 2κ3 2κ4 = 2 0 0 0 0 0 0 0

with

s1 = 2d 2eλ1rdλ1(λ1 + a) d bλ1 ,s8 = 2d 2eλ2rdλ2(λ2 + a) d bλ2 ,

s2 = b + λ12(λ 1 + a) d bλ1 + eλ1r λ 1 bλ1(λ1 + a) d bλ1 ,

s9 = b + λ22(λ 2 + a) d bλ2 + eλ2r λ 2 bλ2(λ2 + a) d bλ2 ,

s3 = 2bλ12(λ 1 + a) d(d bλ1) + 2bλ1 d eλ1r,s 10 = 2bλ22(λ 2 + a) d(d bλ2) + 2bλ2 d eλ2r ,

s4 = eλ1rd dλ1(λ1 + a) d bλ1 ,s11 = eλ2rd dλ2(λ2 + a) d bλ2 ,

s5 = beλ1r bλ1(λ1 + a) d bλ1 ,s12 = beλ2r dλ2(λ2 + a) d bλ2 ,

s6 = eλ1rλ12(λ 1 + a) d bλ1 + λ1,s13 = eλ2rλ22(λ 2 + a) d bλ2 + λ2 ,

s7 = bλ12(λ 1 + a) d(d bλ1) eλ1r + bλ1 d ,s14 = bλ22(λ 2 + a) d(d bλ2) eλ2r + bλ2 d .

The third equation is redundant. Adding both sides of the fifth equation multiplied by b and the sixth equation multiplied by d one obtains adκ3 = 0, whence κ3 = 0. Next adding both sides of the seventh equation multiplied by b and the eighth equation multiplied by d one obtains dκ3 + (ab d)earκ 4 = 0, whence κ4 = 0, provided that ab d0. Since the common part of the straight line ab = d and the stability domain is of planar measure zero we can neglect the possible degeneration getting a preliminary reduction of the linear system to be solved,

0 0 01 2s1 1 2s8 2 2a 0 s2 s9 0 2 2a1 2s31 2s10 0 2 01 ds41 ds11 0 0 2d s6 s13 α11 α12 α22 2κ1 2κ2 = 1 0 0 0 0 .

The final reduced form arises by elimination of α11 and α12,

0 1 eλ1rλ1(λ1 + a) d bλ1 1 eλ2rλ2(λ2 + a) d bλ2 2ad eλ1r(d + bλ 1) λ1(λ1 + a)eλ2r(d + bλ 2) λ2(λ2 + a) 2d eλ1rλ12(λ 1 + a) d bλ1 + λ1 eλ2rλ22(λ 2 + a) d bλ2 + λ2 α22 2κ1 2κ2 = = 1 d 0 0 .

This yields

1 z02 2J(x0)(a,b) =α22 = G3F3 F2G5 2d2 aF3G1 F1G3 + G1F2 aG5F1 = = G4F3 F2G6 2d2 aF3G2 F1G4 + G2F2 aG6F1 (3.34)

with

F1 = d bλ1 eλ1rλ 1(λ1 + a) F2 = eλ1r(d2 b2λ 12) λ 1(λ1 + a)(d bλ1) F3 = eλ1rλ 12(λ 1 + a) + λ1(d bλ1) G1 = d + K2 cos Kr aK sin Kr G2 = bK K2 sin Kr aK cos Kr G3 = (d ab)K2 + (d2 + b2K2) cos Kr G4 = adK bK3 (d2 + b2K2) sin Kr G5 = bK2 aK2 cos Kr K3 sin Kr G6 = Kd K3 cos Kr + aK2 sin Kr .

The results of minimization of the performance index (3.34) for a = 5, r = 0.5 are presented at Figure 3.4 and they agree with calculations obtained in [36, Formula (44), p. 1049] with the use of the frequency–domain method. The minimal value of z202J(x 0) is 0.01947 and it is achieved at b = 3.78, d = 6.897.


PIC

Figure 3.4: The level curves of the performance index z202J(x 0) as a function of b and d