Колесников А.А. Синергетические методы управления сложными системами: энергетические системы
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δ
1
z ≡ D
x
0
4
z
˙z(t)=δ
2
(x
4
− x
0
4
); D − D = δ
1
z.
dx
1
dt
=
1
F
1
∆
e
((˜e
22
− ˜e
1
e
21
)l
1
− (˜e
12
− ˜e
1
e
11
)l
2
)+
1
e
33
∂ ¯ϕ (x
2
,x
3
)
∂x
3
V l
3
;
dx
2
dt
=
1
∆
e
(e
11
l
2
− e
21
l
1
);
dx
3
dt
=
e
32
∆
e
e
33
(e
21
l
1
− e
11
l
2
)+
1
e
33
l
3
;
dx
4
dt
=
1
e
44
(
˜
D
(x
2
,x
4
) − δ
1
z);
dz
dt
= δ
2
(x
4
− x
0
4
),
l
i
ψ
1
= β
11
x
1
+ β
12
(
˜
D
(x
2
,x
4
) − δ
1
z + e
44
α
3
(x
4
− x
0
4
));
ψ
2
= β
21
x
1
+ β
22
(
˜
D
(x
2
,x
4
) − δ
1
z + e
44
α
3
(x
4
− x
0
4
)).
ψ
i
=0
˙
ψ
1
(t)+α
1
ψ
1
=0,
˙
ψ
2
(t)+α
2
ψ
2
=0,
u
1
u
2
u
1
=
1
∆
F
⎛
⎜
⎜
⎝
f
3
⎛
⎜
⎜
⎝
˜
D
(x
2
,x
4
) − δ
1
z
∂
˜
D
(x
2
,x
4
)
∂x
4
e
44
∂
˜
D
(x
2
,x
4
)
∂x
4
+ e
44
α
3
− δ
1
δ
2
(x
4
− x
0
4
)
⎞
⎟
⎟
⎠
− f
1
f
6
+
+ f
3
f
4
+
1
∆
β
∂
˜
D
(x
2
,x
4
)
∂x
2
k
1
f
3
+ k
2
f
6
∂
˜
D
(x
2
,x
4
)
∂x
2
⎞
⎟
⎟
⎠
,
u
2
= −
1
∆
F
⎛
⎜
⎜
⎝
f
2
⎛
⎜
⎜
⎝
˜
D
(x
2
,x
4
) − δ
1
z
∂
˜
D
(x
2
,x
4
)
∂x
4
e
44
∂
˜
D
(x
2
,x
4
)
∂x
4
+ e
44
α
3
− δ
1
δ
2
(x
4
− x
0
4
)
⎞
⎟
⎟
⎠
− f
1
f
5
+
+ f
2
f
4
+
1
∆
β
∂
˜
D
(x
2
,x
4
)
∂x
2
k
1
f
2
+ k
2
f
5
∂
˜
D
(x
2
,x
4
)
∂x
2
⎞
⎟
⎟
⎠
.
u
1
u
2
ψ
1
=0
ψ
2
=0
z
˙x
4
(t)=−α
3
(x
4
− x
0
4
),
α
3
> 0 x
4
x
0
4
D =0, 85D
0
D =0, 9D
0
D =0, 95D
0
β
11
=1 β
12
=1 β
21
=40 β
22
=1 α
1
=1/30 α
2
=1/32 α
3
=3/2 δ
1
=1 δ
2
=1/x
0
5
z
1
z
2
˙z
1
(t)=δ
2
(x
4
− x
0
4
);
˙z
2
(t)=γ
2
x
1
;
D
− D = δ
1
z
1
;
D
= γ
1
z
2
,
D
D
dx
1
dt
=
1
F
1
∆
e
((˜e
22
− ˜e
1
e
21
)l
1
− (˜e
12
− ˜e
1
e
11
)l
2
)+
1
e
33
∂ ¯ϕ (x
2
,x
3
)
∂x
3
V l
3
+ γ
1
z
2
;
dx
2
dt
=
1
∆
e
(e
11
l
2
− e
21
l
1
);
dx
3
dt
=
e
32
∆
e
e
33
(e
21
l
1
− e
11
l
2
)+
1
e
33
l
3
;
dx
4
dt
=
1
e
44
(
˜
D
(x
2
,x
4
) − δ
1
z
1
);
dz
1
dt
= δ
2
(x
4
− x
0
4
);
dz
2
dt
= γ
2
x
1
u
1
u
2
f
1
˜
f
1
˜
f
1
= f
1
+ γ
1
z
2
.
β
11
=1 β
12
=1 β
21
=40 β
22
=1 α
1
=1/30 α
2
=1/32
α
3
=3/2 δ
1
=1 δ
2
=1/x
0
5
γ
1
=0.01 γ
2
=1/3
0, 95D
0
D γ
2
= −1/3
D D D
D D
z
2
k
k
D D D
D D
Q D
Q
Q
D
D
D
Q D
Q D
˙x
7
(t)=k
1
x
8
;
˙x
9
(t)=k
2
x
10
;
D
= u
1
= F
1
(x
7
);
T
1
˙x
8
(t)=−x
8
+˜u
1
;
T
2
˙x
10
(t)=−x
10
+˜u
1
;
Q = u
2
= F
2
(x
9
).
x
7
x
9
x
8
x
10
˜u
1
˜u
2
T
1
T
2
F
1
F
2
D =
ξρ ∆pf
ξ ρ ∆p
f
f = f
x
i
+
1
4
sin(πx
i
)
,i=7, 9.
D = D
0
f
f
0
x
i
+
1
4
sin(πx
i
)
,i=7, 9.
f /f
0
> 1
F
1
(x
7
)=1, 2D
0
x
7
+
1
4
sin(πx
7
)
.
B
Q
Q =(1+˜α )Vc (T −T
)B,
˜α Vc
T
÷ T T
T
T − T
= T
⎡
⎢
⎢
⎢
⎣
1 −
1
M
σ
ψ F a T
3
ϕ
BV c
0,6
+1
⎤
⎥
⎥
⎥
⎦
.
M σ
F a ψ
ϕ
Q =
Q
0
T
T − (T
)
0
⎡
⎢
⎢
⎢
⎣
1 −
1
T − (T
)
0
(T
)
0
B
0
B
0,6
+1
⎤
⎥
⎥
⎥
⎦
B
B
0
,
B(x
9
)
B(x
9
)=1, 2B
0
x
9
+
1
4
sin(πx
9
)
.
u
1
u
2
˜u
1
˜u
2
ψ
4
= F
1
(x
7
) − u
1
(x);
ψ
6
= F
2
(x
9
) − u
2
(x);
ψ
5
= x
8
− ϕ
1
(x
7
, x);
ψ
7
= x
10
− ϕ
2
(x
8
, x),
ψ
i
=0 i =4...7
˙
ψ
i
(t)+α
i
ψ
i
=0,i=4...7
˜u
1
˜u
2
x =
#
x
1
x
2
x
3
x
4
z
1
z
2
$
˜u
1
= x
8
+ T
1
⎡
⎢
⎢
⎣
∂ϕ
1
∂x
7
k
1
− α
5
x
8
− α
4
α
5
1
k
1
∂F
1
∂x
7
(F (x
7
) − u
1
(x))
⎤
⎥
⎥
⎦
;
˜u
2
= x
10
+ T
2
⎡
⎢
⎢
⎣
∂ϕ
2
∂x
9
k
2
− α
7
x
10
− α
6
α
7
1
k
2
∂F
2
∂x
9
(F (x
9
) − u
2
(x)),
⎤
⎥
⎥
⎦
∂ϕ
1
∂x
7
=
α
4
k
1
⎡
⎢
⎢
⎢
⎣
1
∂F
1
∂x
7
2
∂
2
F
1
∂x
2
7
(F
1
(x
7
) − u
1
(x)) − 1
⎤
⎥
⎥
⎥
⎦
;
∂ϕ
2
∂x
9
=
α
6
k
2
⎡
⎢
⎢
⎢
⎣
1
∂F
2
∂x
9
2
∂
2
F
2
∂x
2
9
(F
2
(x
9
) − u
2
(x)) − 1
⎤
⎥
⎥
⎥
⎦
.
D =0, 85D
0
D =0, 9D
0
D =0, 95D
0
x
1
=∆h x
2
= p
x
3
= x x
4
= p x
5
= p x
6
= ω x
7
= s x
8
= ω u
1
= D u
2
= Q u
3
= U
˙x
1
(t)=f
1
+ f
2
u
1
+ f
3
u
2
;
˙x
2
(t)=f
4
+ f
5
u
1
+ f
6
u
2
;
e
44
˙x
4
(t)=
˜
D
(x
2
,x
4
)+D − D (x
4
,x
5
,F(x
7
));
e
55
˙x
5
(t)=
x
5
x
0
5
m
(D (x
4
,x
5
,F(x
7
)) − D (x
5
));
e
66
x
6
˙x
6
(t)=∆H ·D (x
5
) − N ;
˙x
7
(t)=k x
8
;
T
˙x
8
(t)=−x
8
+ u
3
;
...,
e
55
= J e
66
=
V
ρ
0
γp
0
f
i
i =1...6
D
x
4
F (x
7
)
g =
D
D
0
p =
x
4
x
0
4
f =
F
F
0
g = p
−ε
2
+2ε ε +1− 2ε
−ε
2
0
+2ε ε
0
+1− 2ε
f.
∂g
∂p
∂g
∂f
g
p f
∂g
∂p
=
ε
ε +1− 2ε
−ε
2
0
+2ε ε
0
+1− 2ε
f;
∂g
∂f
= p
−ε
2
+2ε ε +1− 2ε
−ε
2
0
+2ε ε
0
+1− 2ε
,