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I HC THI NGUYN

TRNG I HC K THUT CNG NGHIP--------------------------------------

LUN VN THC S K THUTNGNH: TNG HA

thit k b quan st v iu khin nhit trong phi tm

NG MINH C

THI NGUYN 2009


2/93

I HC THI NGUYN

TRNG I HC K THUT CNG NGHIP--------------------------------------

LUN VN THC S K THUT

NGNH: TNG HO

thit k b quan st v iu khin nhit trong phi tm

Hc vin: Ng Minh cNgi HD Khoa Hc: PGS.TS. Nguyn Hu Cng

THI NGUYN 2009


3/93

LI CAM OAN

Tn ti l: Ng Minh c

Sinh ngy 19 thng 08 nm 1982

Hc vin lp cao hc kho 9 Ngnh Tng ho - Trng i hc k thut Cng

nghip Thi Nguyn.

Hin ang cng tc ti khoa in - Trng i hc K thut Cng nghip Thi

Nguyn.

Xin cam oan: ti Thitkb quan st v iu khin nhit trong phi

tmdo thy gio, PGS.TS. Nguyn Hu Cng hng dn l cng trnh nghin cu

ca ring ti. Tt c cc ti liu tham kho u c ngun gc, xut x r rng.

Tc gi xin cam oan tt c nhng ni dung trong lun vn ng nh ni dung

trong cng v yu cu ca thy gio hng dn. Nu c vn g trong ni dung

ca lun vn th tc gi xin hon ton chu trch nhim vi li cam oan ca mnh.

Thi Nguyn, ngy 04 thng 4 nm 2009


4/93

Lun vn thc s k thut

Sha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

LI CM N

Li u tin em xin by t lng bit n chn thnh, li cm n su sc ti

thy gio - PGS,TS Nguyn Hu Cng, ngi trc tip ch bo v hng dn

em trong sut thi gian qua.Em xin by t lng cm n i vi cc thy c gio trong Khoa , b mn

cng ng o bn b, ng nghip c v rt nhiu cho vic thc hin lun vn

ny.

Mcdc s ch bo st sao ca thy hng dn, s n lc c gng ca

bn thn. Song v kin thc cn hn ch, nn chc chn lun vn ny khng trnh

khi nhng thiu st nht nh. Em rt mong c s ch bo ca cc thy c gio

v s gp chn thnh ca cc bn.

Em xin chn thnh cm n!


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Lun vn thcskthut


2

LI NI U

Hin nay t nc ta ang trong thi k i mi, thi k cng nghip ho,

hin i ha cng vi s pht trin ca cng ngh thng tin, ngnh k thut in t

l s pht trin ca k thut iu khin v t ng ha. Trong lnh vc gia cngnhit ta thng gii quyt bi ton l iu khin nhit trong cc l nung theo mt

ch tiu no , tuy nhin cht lng ca cc sn phm trong qu trnh gia cng

nhit li ph thuc vo trng nhit trong phi. Nh vy t ra bi ton phi

iu khin c nhit trong phi nung theo ch tiu cht lng t ra, tc l phi

iu khin mt thng s m khng th dng sensor o c. Tt ra bi ton

Bit v tm li

Trong khun kh lun vn em i vo nghin cu tm hiu mt s phng

php tnh ton trng nhit trong phi tm. Nghin cu xy dng m hnh quan

st nhit di dng m hnh hm truyn. Sau khi c m hnh hm truyn v

trng nhit trong tm, thit k biu khin bng phng php kinh in

v iu khin m.Nh vy c th iu khin trng nhit trong phi tho mn

yu cu cng ngh t ra (Trc kia ta ch iu khin c nhit trong khng

gian l).

Sau thi gian tm hiu v nghin cu v c bit di s hng dn caThyPGS.TS Nguyn Hu Cnglun vn c hon thnh.

Trong qu trnh thc hin lun vn, chc chn khng trnh khi nhng thiu

st. Em rt mong c s ch bo ca cc thy c gio v s gp chn thnh ca

cc bn.

Em xin chn thnh cm n!

Thi nguyn, ngy 4/4/2009

Hc vin

Ng Minh c


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3

MC LCNi dung Trang

Li cmn ............................................................................................................ ... 1Li ni u..................................................................................................................2

Mc lc...................................................................................................................... 3CHNG 1: GII THIU MT S PHNG PHP TNH TON

TRNG NHIT TRONG PHI TM .....................................51.1. Thnh lp phng trnh truyn nhit................................................................51.2. iu kin ban u v iu kin bin................................................................71.3. Tnh ton trng nhit trong phi tm bng phng php gii tch ..........81.4 Tnh ton trng nhit trong phi tm bng phng php s ...................10

1.4.1. Phng php sai phn gii bi ton c tr ban u ................................. 11

1.4.1.1. M hnh bi ton ...............................................................................111.4.1.2. Li sai phn ................................................................................... 111.4.1.3. Hm li ...........................................................................................111.4.1.4. o hm li ....................................................................................111.4.1.5. Lin h gia o hm v o hm li .............................................121.4.1.6. Phng php Euler hin.....................................................................131.4.1.7. Phng php Euler n........................................................................131.4.1.8. Phng php Crank Nicolson ........................................................14

1.4.2. Phng php sai phn gii bi ton truyn nhit mt chiu ...................141.4.2.1. M hnh bi ton ...............................................................................141.4.2.2. Li sai phn v hm li ................................................................151.4.2.3. Xp x cc o hm ...........................................................................171.4.2.4. Phng php sai phn hin (c in) ................................................181.4.2.5. Phng php n (c in) .................................................................191.4.2.6. Phng php Crank -Nicolson (6 im i xng) ...........................20

1.5. Kt lun chng 1..........................................................................................22

CHNG 2: XY DNG M HNH HM TRUYN XC NH

NHIT TRONG PHI TM.......................................................23

2.1. t vn .....................................................................................................23

2.2. Nghin cu i tng iu khin...................................................................23

2.3. Xy dng m hnh hm truyni vi vt mng .........................................24

2.4. Xy dng m hnh hm truyn khiphi c chia lm 2 lp (n=2) ............25




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2.7. Xy dng m hnh hm truyn khi phi c chia thnh n lp ...................31

2.8. V d tnh ton hm truyn tng lp khi chia phi thnh 1 lp v 3 lp .....33

2.9. Kt qu m phng cho b quan st nhit ..................................................35

2.10. Kt lun........................................................................................................38CHNG 3: THIT K B IU KHIN NHIT TRONG PHI TM ....39

3.1. Gii thiu mt s phng php thit k ........................................................39

3.1.1. Phng php a thc c trng c h s suy gim thay i c...........39

3.1.2. Phng php b hng s thi gian tri ...................................................42

3.1.3. Thit k b iu chnh cho h c hnh vi tch phn ..............................46

3.1.4. Phng php thit k b b ....................................................................50

3.1.5. B iu khin m....................................................................................51

3.1.6. Thit k b iu khin m......................................................................67

3.2. Thit k..........................................................................................................75

3.2.1. Thit k b iu khin PID iu khin nhit cho lp 2 khi chia

phi lm 3 lp .........................................................................................75

3.2.2. Thit k b iu khin m iu khin nhit cho lp 2 khi chia

phi lm 3 lp .........................................................................................77CHNG 4: CC KT QU M PHNG...........................................................83

4.1. Kt qu m phng khi thit k b iu khinPID iu khin nhit

cho lp 1 v lp 2 khi phi c chia thnh 3 lp ...........................................83

4.2. Kt qu m phng khi thit k b iu khin m iu khin nhit

cho lp 1 v lp 2 khi phi c chia thnh 3 lp ...........................................84

4.3. Kt lun v kin ngh nghin cu tip theo......................................................85

4.3.1 Kt lun.......................................................................................................85

4.3.2 Nhng kin ngh nghin cu tip theo........................................................85

TI LIU THAM KHO.........................................................................................86

PH LC..................................................................................................................87


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CHNG 1GII THIU MT SPHNG PHPTNH TON TRNG

NHIT TRONG PHI TM

1.1. Thnh lp phng trnh truyn nhitXt mt vt rn truyn nhit ng hng, ),,,( tzyxu l nhit ca n ti

im ),,( zyx thi im t. Nu ti cc im khc nhau ca vt nhit khc nhau

th nhit s truyn t im nng hn ti im ngui hn. S truyn nhit tun

theo nh lut sau:

Nhit lng Q i qua mt mnh mt kh b S cha im ),,( zyx trong mt

khong thi gian t t l vi S , t v o hm php tuyn nu

. Tc l

n

uStzyxkQ

= ),,( (1.1)

Trong 0),,( >zyxk l h s truyn nhit ( ),,( zyxk khng ph thuc vo hng

ca php tuyn vi S v s truyn nhit l ng hng), n

l vect php ca S

hng theo chiu gim nhit .

Gi q l dng nhit, tc l nhit lng i qua mt n v din tch trong mt n v

thi gian. T )1.1( ta suy ran

ukq

= .

By gi ta ly trong vt mt th tch tu V gii hn bi mt mt kn trn S v xt

s bin thin ca nhit lng trong th tch trong khong thi gian t 1t n

2t .T )1.1( ta suy ra nhit lng qua mt S vo trong t thi im 1t n thi im

2t l

=

2

1

),,(1

t

t S

ds

n

uzyxkdtQ .

Trong n

l vecvt php hng vo trong ca mt S . p dng cng thc

Ostrogradsky i t tch phn trn mt S sang tch phn ba lp ta c

( ) =2

1

1

t

t V

dxdydzkgradudivdtQ

Gi s rng trong vt c cc ngun nhit, gi ),,,( tzyxF l mt ca chng tc l

nhit lng sinh ra hay mt i trong mt n v th tch ca vt v trong mt n v

thi gian.


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6

Nhit lng sinh ra hay mt i trong th tch V t thi im 1t n thi im 2t l

=2

1

),,(2

t

t V

dxdydzzyxFdtQ

Mt khc ta li bit rng nhit lng cn cho th tch V ca vt thay i nhit t),,,( 1tzyxu n ),,,( 2tzyxu l

[ ] =V

dxdydzzyxzyxCtzyxutzyxuQ ),,(),,(),,,(),,,( 123 .

Trong ),,( zyxC l nhit dung, ),,( zyx l mt ca vt.

V

=2

1

),,,(),,,( 12

t

t

dtt

utzyxutzyxu nn c th vit

=

2

1

3

t

t V

dxdydzt

uCdtQ .

Mt khc 213 QQQ += nn ta c

( ) =

2

1

0),,,(

t

t V

dxdydztzyxFkgradudivt

uCdt

V khong thi gian ),( 21 tt v th tch V c chn tu , nn ti mi im ),,( zyx

ca vt v mi thi im t biu thc di du tch phn u bng khng

( ) ),,,( tzyxFkgradudivt

uC +=

.

Hay ),,,( tzyxFz

uk

zy

uk

yx

uk

xt

uC +

+

+

=

(1.2)

Phng trnh gi l phng trnh truyn nhit trong vt ng hng khng ng

cht. Nu vt ng cht th kC ,, l nhng hng s v phng trnh c dng

),,,(2

2

2

2

2

22 tzyxf

z

u

y

u

x

ua

t

u+

+

+

=

(1.3)

Trong C

ka =2 ,

C

tzyxFtzyxf ),,,(),,,( = . l phng trnh truyn nhit khng

thun nht. Nu trong vt khng c ngun nhit th 0),,,( tzyxF ta s c

phng trnh truyn nhit thun nht:

+

+

=

2

2

2

2

2

22

z

u

y

u

x

ua

t

u )4.1(


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7

Nu ta xt s truyn nhit trn mt mt vt ng cht rt mng (ch kho st s

truyn nhit theo hai phng) t trn mt phng Oxy th nhit ),,( tyxu ti

im ),( yx thi im t tho mn phng trnh truyn nhit:

),,(2

2

2

22

tyxfy

u

x

ua

t

u+

+

=

)5.1(

Cn phng trnh truyn nhit trn mt vt ng cht rt mng t dc theo trc x

l:

),(2

22

txfx

ua

t

u+

=

)6.1(

1.2. iu kin ban u v iu kin bin

Trong vt l ta bit rng mun xc nh c nhit ti mi im trong vt mi

thi im, ngoi phng trnh )3.1( ta cn cn phi bit phn b nhit trong vt

thi im u v ch nhit bin S ca vt.

iu kin bin c th cho bng nhiu cch

* Cho bit nhit ti mi im P ca bin S ),(| 1 tPu S = )7.1(

* Ti mi im ca bin S cho bit dng nhitn

ukq

= vy ta c iu kin bin

),(2 tPn

u

S

=

)8.1(

Trong k

tPqtP

),(),(2

= l mt hm chotrc.

* Trn bin S ca vt c s trao i nhit vi mi trng xung quanh, m nhit

ca n l 0u th ta c iu kin bin sau:

0)( 0 =

+S

uuhnu )9.1(

Nu bin S cch nhit th 0=h suy ra )9.1( tr thnh 0=

Sn

u

Nh vy bi ton truyn nhit trong mt vt rn, ng cht truyn nhit ng hng

t ra nh sau: Tm nghim ca phng trnh )3.1( tho mn iu kin u

),,(0

zyxut

==

v mt trong cc iu kin bin )9.1)(8.1)(7.1( .


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8

1.3. Tnh ton trng nhit trong phi tm bngphng php gii tch

Gii hn bi ton l mt tm phng c chiu dy 2s, h s dn nhit , c h s to

nhit t b mt ti mi trng l . Gi thit t gc to tm ca tm phng, ta

c phng trnh truyn nhit nh sau:2

2

u ua

x

= (1-10)

Trong :u(x,=0) = uo=const

0

0; ( )w fx x S

u ut t

x x

= == =

Trong cng thc trn:

a- l h s dn nhit

u- hm nhit ca vt

Vi tf l nhit trong khng gian l nung. gii phng trnh (1 -10) ta dng

phng php phn ly bin s:

t: u(x,) = ().(x) ta c :

,

2,,

2

( ). ( )

( ). ( )

ux

ux

x

=

=

, ,,( ). ( ) . ( ). ( )x a x =

Phng trnh (1-10) s tng ng vi:

, ,,( ) ( )

( ) ( )

x

a x

= (1-11)

Phng trnh (1-11) v tri l mt hm theo thi gian , v phi l mt hm theo to

khng gian x, do ch tho mn khi c hai v u l hng s. Ta k hiu hng

s ny l k2, vy t (1-11) ta c :

,() =ak2() (1-12)

() = k2(x) (1-13)

Nghim tng qut ca (1-12) l :

() = B1exp(ak2)

Nghim tng qut ca (1-13) l:(x) = B2exp(kx) + B3exp(-kx)


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Vy nghim ca (1-10) l:

u(x, ) = () . (x) = B1exp(ak2).[B2exp(kx) + B3exp(-kx)] (1-14)

Ta thy nhit khng th tng v hn theo thi gian nn k2< 0.

t k2

=-q2

hay k= iq. (1-14) tr thnh .u(x,) = B1exp(-aq

2)[B4cosqx +B5isinqx) (1-15)

C phn thc v phn o ca (1-15) u l nghim ca phng trnh vi phn v tng

cc nghim cng l 1 nghim. Vy nghim ca phng trnh c dng:

u(x,) = C1exp(-aq2)[C2cosqx +C3sinqx] (1-16)

V:

00

x

u

x

= = nn C3 = 0 . Vy nghim tr thnh:

u(x,) = Aexp(-aq2)cosqx (1-17)

Hn na t iu kin bin ( )w fx s

ut t

x

== ta nhn c phng trnh c

trng:

coti

qsgqs

B= hay cot

i

gB

= (1-18)

Trong qs = v tiu chun Bii

sB

=

Phng trnh (1-18) c hng lot nghim 1, 2, ... n... cc nghim ny tho mn:

1


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Khi s dng cc iu kin u cho,ta xc nh c n s cn li trong phng

trnh (1-17) bng cch nhn hai v ca phng trnh vi cosn

x

s = , sau ly tch

phn theo cn t x= - s n x= + s ta c :

0

2sin

sin cosn

n

n n n

A u

=

+(1-20)

Tm li nghim ca (1-10) l :

2

21

2 sincos( )exp( )

sin cos

o nn n

n n n n

u x au

s s

=

= + (1-21)

Khi ta s dng cc t hp khng th nguyn ( h tng i )

,

o

u= :uu l nhit khng th nguyn

xX

s= v h s khng th nguyn n =

o

nD

u

Thi gian khng th nguyn (theo tiu chun Fourier)0 2

aF

s

= ,

Phng trnh (1-21) c vit

, 2

n n n o

n=1

= D cos( x)exp(- F )u

(1-22)

Thc t cho thy khi Fo ln, s hng ca chui (1-22) gim rt nhanh. Khi F o

0,3 ta ch cn ly s hng u tin ca chui m sai s khng vt qu 1%.

Ngoi phng php gii tch ngi ta cn dng phng php s gii bi ton dn

nhit tc l dng phng php sai phn

1.4. Tnh ton trng nhit trong phi tm bng phng php sNh bit vic s dng cc cng c gii tch tnh ton cc bi ton k thut c

nhiu hn ch, do ngi ta tm cch tnh gn ng bng cc phng php s.

y xin gii thiu phng php sai phn, trc ht ta xt mt s bi ton n gin

i vi phng trnh vi phn thng.


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1.4.1. Phng php sai phn gii bi ton c tr ban u

1.4.1.1. M hnh bi ton

Cho khong [x0, X]. Tm hm u = u(x) xc nh ti [x0, X] v tha mn:

,

0( , )u f x u x x X = < < (1.23)

0( )u x = (1.24)

Trong f(x,u) l mt hm s cho trc v l mt s cho trc.

Gi s bi ton (1.23), (1.24) c nghim u = u(x) trn, ngha l n c o

hm lin tc n cp m ta cn.

1.4.1.2. Li sai phn

Ta chia on [x0, X] thnh N on con bng nhau, mi on con di ( ) /h b a N =

bi cc im 0 , 0,1,..,ix x ih i N= + = (hnh 1.1). Tp cc im x i gi l mt li sai

phn trn [x0, X] k hiu l ,h mi im xi gi l mt nt ca li, h gi l bc i

ca li.

Ta s tm cch tnh gn ng gi tr ca nghim u(x) ti cc nt x i ca li ,h .

l tng u tin ca phng php sai phn, cn gi l phng php li.

1.4.1.3.Hm li

l nhng hm s xc nh ti cc nt ca li ,h . Gi tr ca hm li v ti nt

xi vit l vi.

Mt hm s u(x) xc nh ti mi x [a,b] s to ra hm li u c gi tr ti

nt xi l ui = u(xi).

1.4.1.4.o hm li

Xt hm s v. o hm li tin cp mt ca v, k hiu l v x, c gi tr ti nt xi l:

1i ixi

v vv

h

+ =

o hm li li cp mt ca v, k hiux

v , c gi tr ti nt xi l:

x

x0

x1 x2

xi xN=Xxi+1

Hnh 1.1 Li sai hn


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12

1i ixi

v vv

h

=

Khi h b th o hm li xp x c o hm thng.

1.4.1.5. Lin h gia o hm v o hm li

Gi s hm u(x) trn theo cng thc Taylor ta c:

' 2

1( ) ( ) ( ) ( ) ( )i i i iu x u x h u x h u x O h+ = + = + +

Ta suy ra:

'1( ) ( ) ( ) ( )i ixi iu x u x

u u x O hh

+ = = + (1.25)

Cng c:

' 2

1( ) ( ) ( ) ( ) ( )i i i iu x u x h u x h u x O h = = + Do :

'1( ) ( ) ( ) ( )i i ixiu x u x

u u x O hh

= = + (1.26)

Ngoi ra vi quy c:

1/ 2 1/ 2 1/ 2, ( )2

i i i i

hx x u u x+ + += + =

Ta cn c:' 2 '' 3

1 1/ 2 1/ 2 1/ 2 1/ 2

1( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2! 2i i i i i

h h hu x u x u x u x u x O h+ + + + += + = + + +

' 2 '' 3

1/ 2 1/ 2 1/ 2 1/ 2

1( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2! 2i i i i i

h h hu x u x u x u x u x O h+ + + += = + +

Ta suy ra:

' 3

1 1/ 2( ) ( ) ( ) ( )i i iu x u x h u x O h+ + = +

Do :' 21

1/ 2

( ) ( )( ) ( )i ixi i

u x u xu u x O h

h

++

= = = + (1.27)

ng thi:

211/ 2

( ) ( )( ) ( )

2

i i

i

u x u xu x O h+ +

+= + (1.28)


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1.4.1.6.Phng php Euler hin

Trong (1.23) thay ' ( )iu x bi xiu th (1.25) cho:

'1( ) ( ) ( ) ( ) ( , ( )) ( )i ixi i i iu x u x

u u x O h f x u x O h

h

+ = = + = +

Ta suy ra:

21( ) ( ) ( , ( )) ( )i i i iu x u x h f x u x O h+ = + + (1.29)

B qua v cng b 2( )O h v thay ( )i

u x bi vi xem l gn ng ca ( )iu x , ta c:

1 ( , )i i i iv v hf x v+ = + (1.30)

Cng th c (1.30) cho php tnh 1iv + khi bit vi. Da vo (1.24) ta t thm iu kin:

0v = (1.31)Th hai cng thc (1.30), (1.31) cho php tnh ra tt c cc v i. Phng php tnh vi

bng (1.30), (1.31) g i l phng php Euler.Sau khi c vi ta xem vi l gn ng ca

u(xi).

Phng php Euler l phng php sai phn n gin nht gii gn ng

bi ton (1.23), (1.24).

y khi bit vi mun tnh vi+1 ta ch phi tnh gi tr ca biu thc v

phi ca (1.30), ch khng phi gii mt phng trnh i s no. V l phng

php sai phn (1.30), (1.31) thuc loi phng php sa i phn hin. N cng c tn

l phng php Euler hin.

1.4.1.7. Phng php Euler n

Nu trong (1.23) thay '( )i

u x bixi

u th (1.26) cho:

'1( ) ( ) ( ) ( ) ( , ( )) ( )i i i i ixiu x u x

u u x O h f x u x O hh

= = + = +

Ta suy ra:

2

1( ) ( ) ( , ( )) ( )i i i iu x u x h f x u x O h= + + (1.32)

B qua v cng b 2( )O h v thay u(xi) bi vi xem l gn ng ca u(xi), ta c:

1 ( , )i i i iv v hf x v= + (1.33)

Cng thc (1.33) cho php tnh v ikhi bit v i-1. Thm iu kin (1.31) th

cc cng thc (1.31), (1.33) cho php tnh ra tt c cc v i. Phng php tnh vi bng

(1.33), (1.31) li l mt phng php sai phn khc. y khi bit v i -1 mun


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tnh ra vi ta phi gii phng trnh i s (1.33) i vi n s v i. V l phng

php sai phn ny thuc loi phng php sai phn n. N cng c tn l phng

php Euler n.

1.4.1.8. Phng php Crank- NicolsonNu p dng (1.27) ta c:

' 2 211/ 2 1/ 2 1/ 2

( ) ( )( ) ( ) ( , ( )) ( )i i i i i

u x u xu x O h f x u x O h

h

++ + +

= + = +

Theo (1.28) ta li c:

21 11/ 2 1/ 2

( , ( )) ( , ( ))( , ( )) ( )

2

i i i ii i

f x u x f x u xf x u x O h+ ++ +

+= +

Ta suy ra:

21 1 1( ) ( ) ( , ( )) ( , ( )) ( )2

i i i i i iu x u x f x u x f x u x O hh

+ + + += +

Do :

3

1 1 1( ) ( ) [ ( , ( )) ( , ( ))]+O(h )2

i i i i i i

hu x u x f x u x f x u x+ + += + + (1.34)

B qua v cng b 3( )O h v thay u(xi) bi vi xem l gn ng ca u(xi), ta c:

1 1 1[ ( , ) ( , )]2i i i i i i

hv v f x v f x v+ + += + + (1.35)

Cng thc (1.35) cho php tnh v i+1khi bit v i. Thm iu kin (1.31) th cng

thc (1.35), (1.31) cho php tnh ra tt c cc v i. y khi bit v i mun tnh ra

vi+1 ta phi gii phng trnh i s (1.35) i vi n s v i+1. V l phng php

tnh vi bng (1.35), (1.31) thuc loi phng php sai phn n. N c tn l phng

php Crank - Nicolson.

1.4.2. Phng php sai phn gii bi ton truyn nhit mt chiu1.4.2.1. M hnh bi ton

Cho cc s a, b; a < b v T > 0. Xt:

( , ) (0, ]; [a,b] [0,T]T TQ a b T Q= =

Xt bi ton bin th nht i vi phng trnh truyn nhit:

Tm hm s u(x, t) tha mn:

2

2

( , ), ( , )T

u uLu f x t x t Q

t x

=

(1.36)


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( , 0) ( ),u x g x a x b= < < (1.37)

( , ) ( ), ( , ) ( ), 0a b

u a t g t u b t g t t T = = < (1.38)

Trong f(x, t), g(x), ga(t), gb(t) l nhng hm s cho trc.

Phng trnh (1.36) l phng trnh Parabol ic v gi phng trnh (1.36) lphng trnh truyn nhit mt chiu. Bin x gi l bin khng gian, cn bin t l

bin thi gian.

Bi ton (1.36) - (1.38) l mt bi ton va c iu kin ban u ( l iu

kin (1.37)), va c iu kin bin ( l iu kin (1.38)); l bi ton bin loi

mt i vi phng trnh (1.36).

Gi s bi ton (1.36) - (1.38) c nghim duy nht trn trongTQ .

1.4.2.2. Li sai phn v hm li

a. Li sai phn

Chn hai s nguyn N > 1 v M 1 v t:

, , 0,1,2,...,i

b ah x a ih i N

N

= = + =

, , 0,1,2,...,j

Tt j j M

M = = =

Ta chia min Q T thnh bi nhng ng thng x = x i, t = tj (Hnh 1.2). Mi im

(xi, tj) gi l mt nt, nt im (x i, tj) cn c vit gn l (i, j); h gi l bc i

theo khng gian, gi l bc i theo thi gian.

Tp tt c cc nt to thnh mt li sai phn trnTQ .

Li trn [a,b] (li vi khng gian): Tp:

{ }1,2,..., 1h ix i N = =

gi l tp cc nt trn [a, b]. Tp:

{ }0,h ix i N = = gi l tp cc nt bin trn [a, b]; nt 0 v nt N l hai nt

bin. Tp:h h h

= gi l mt li sai phn trn [a,b].


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Li trn [0, T] (li thi gian): Tp:

{ }1,2,...,jt j M = = gi l mt li sai phn trn (0, T]. Tp:

{ } { }00,1,..., 0jt j M t = = = = gi l mt li sai phn trn [0, T]; nt

t0= 0 l nt ban u.

Tp:h h

= l tp cc nt trong trnTQ .

Tp: { }0h x a = = gi l tp cc nt bin tri.

Tp: { }h Nx b +

= = gi l tp cc nt bin phi.

Tp: { }0

0 0hh

t

= = gi l tp cc nt ban u.

Nh vy tp:

0

h h h hh h

+

= = chnh l li sai phn trn TQ .

Ta phn li sai phn TQ thnh nhiu lp: Lp th j to bi cc nt ng cng mt

gi tr thi gian tj l:

{ }( , ), 0,1,..., ;jh i jx t i N = = nt (x0, tj) = (a, tj) v (xN, tj) = (b, tj) l hai nt bin.

tM =T

tj

xx0 = a xN = bxi

t

0

Hnh 1.2 Li sai phn v hm li


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b. Hm li

Hm s xc nh ti cc nt ca mt li no gi l mt hm li.

Gi tr ca hm li v ti nt (i, j) vit l jiv . Cc gi tr ca hm li v ti cc nt

ca lp jh to thnh hm li jv xc nh trn h . Ta c:1

0 1( , ,..., )j j j j N

Nv v v v R+=

Trong tp cc hm li ny ta xt hai loi chun:

{ }ji0 i N

ax vjv m

= ; 2 2 20 12 ( ) ( ) ... ( )j j j j

Nv v v v= + + +

Mi hm s u(x, t) xc nh trnTQ c gi tr ti (i, j) l u(xi, tj) v to ra hm li u

xc nh bi ( , )ji i j

u u x t = .

1.4.2.3. Xp x cc o hm

p dng cng thc Taylor

2' '' ( ) 1( ) ( )( ) ( ) ( ) ( ) . . . ( ) (( ) )

1! 2! !

mm mx x x

F x x F x F x F x F x O xm

+ + = + + + + +

Ta c:

1( , ) ( , )

( , ) ( )i j i j

i j

u x t u x t ux t O

t

+ = +

(1.39)

1

1

( , ) ( , )( , ) ( )

i j i j

i j

u x t u x t ux t O

t

++

= +

(1.40)

1 2( , ) ( , )

( , / 2) ( )i j i j

i j

u x t u x t ux t l O

t

+ = + +

(1.41)

21 1 2

2 2

( , ) 2 ( , ) ( , )( , ) ( )

i j i j i j

i j

u x t u x t u x t ux t O h

h x

+ + = +

(1.42)

21 1 1 1 1 2

12 2

( , ) 2 ( , ) ( , ) ( , ) ( )i j i j i j

i j

u x t u x t u x t u x t O hh x

+ + + ++

+ = +

(1.42a)

1 1 1 1 1 1 1

2 2

22 2

2

( , ) 2 ( , ) ( , ) ( , ) 2 ( , ) ( , )1

2

( , / 2) ( )

i j i j i j i j i j i j

i j

u x t u x t u x t u x t u x t u x t

h h

ux t O h

x

+ + + + + + + +

= + + +

(1.43)

Nh vy, ta c nhiu cch xp x phng trnh o hm ring (1.36). T ta suy ra

nhiu phng n khc nhau thay th bi ton vi phn bi bi ton sai phn.


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1.4.2.4. Phng php sai phn hin (c in)

Mc ch ca phng php l tm cch tnh ( , )ji i j

v u x t ti mi nt

( , ).i jx t S dng (1.39), (1.42) ta suy ra:

1 11

2

22

2

( , ) 2 ( , ) ( , )( , ) ( , )

( , ) ( , ) ( )

ji i j i ji j i j

i j i j

u x t u x t u x t u x t u x t

h

u ux t x t O h

t x

+ + +

= + +

(1.44)

Do c ( , )ji i j

v u x t , da vo (1.44), (1.37), (1.38) ta vit bi ton sai phn sau

y thay th cho bi ton vi phn (1.36), (1.37), (1.38):

1

1 1

2

2

( , )

j j j j j

i i i i i

h i j

v v v v v

L v f x th

++ +

= (1.45)0 ( ), 0,1,...,i iv g x i N = = (1.46)

0 ( ), ( ), 0,1,...,j j

a j N b jv g t v g t j M = = = (1.47)

Mi phng trnh (1.45) cha mt n 1ji

v + lp trn j + 1 v ba n 1 1, ,j j j

i i iv v v + lp

di j theo s Hnh 1.3.

t 2/h = ta gii (1.45) ra n 1jiv+ :

1

1 1(1 2 ) ( ) ( , )j j j j

i i i i i jv v v v f x t + + = + + + (1.48)

iu kin (1.46) cho 0i

v lp 0.

iu kin (1.47) cho0

jv v jNv hai nt bin (0,j) v (N, j) ca

j

h .

Nh vy phng trnh (1.45) tc (1.48) v iu kin bin (1.47) cho php tnh 1jiv

+

lp trn j + 1 khi bit jiv lp di j m khng phi gii mt h phng trnh i s no.

Cho nn phng php (1.45), (1.46), (1.47) gi l phng php sai phnhin; n cn c tn l phng php sai phn hin c in gii bi ton (1.36) -

(1.38).N c s hnh 1.3. S ny gi l s hin bn im.


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1.4.2.5. Phng php n (c in)

p dng (1.40), (1.42) ta c:

1 1 1 1 1 1

2

22

1 12

( , ) ( , ) ( , ) 2 ( , ( , ))

( , ) ( , ) ( )

i j i j i j i j i j

i j i j


h

u ux t x t O h

t x

+ + + + +

+ +

+

= + +

(1.49)

c ( , )j

i i jv u x t , ta vit bi ton sai phn sau y thay cho bi ton vi phn:1 1 1 1

1 112

2( , )

j j j j j

i i i i i

h i j

v v v v vL v f x t

h

+ + + ++

+

+ = (1.50)

0 ( )i iv g x= (1.51)

0 ( ), ( )j j

a j N b jv g t v g t = = (1.52)

t

j-1

j

j+1

xi-1 xi xi+1 x

Hnh 1.3 S hin bn im

t

j-1

j

j+1

xi-1 xi xi+1 xHnh 1.4 S n bn im


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20

Mi phng trnh (1.50) cha ba n 1 1 11 1, ,j j j

i i iv v v+ + +

+ lp trn j +1 v mt nj

iv lp

di j theo s hnh 1.4.

Cng nh trn t 2/h = , khi (1.50) vit:1 1 1

1 1 1(1 2 ) ( , )j j j j

i i i i i jv v v v f x t + + +

+ + + + = (1.53)

Tc dng ca cc iu kin (1.51), (1.52) cng nh phng n hin: Chng cho

0

0, ,j j

i Nv v v . Nhng y khi bit j

iv lp j mun tnh1j

iv+ lp j + 1 ta phi gii h

i s tuyn tnh (1.53) i vi 1 1 11 2 1, ,...,j j j

Nv v v+ + +

. Theo ngha ta ni phng php

sai phn (1.50), (1.51), (1.52) l mt phng php n. N cn c tn l phng

php n c in. N c s hnh 1.4. S ny gi l s n bn im.

H (1.53) l mt h ba ng cho c th gii bng phng php truy ui.

1.4.2.6. Phng php Crank-Nicolson (6 im i xng)

p dng (1.41), (1.44) ta c:

1 1 1 1 1 1

2

1 1

2

22 2

2

( , ) ( , ) ( , ) 2 ( , ( , )1[

2

( , ) 2 ( , ) ( , )]

( , / 2) ( , / 2) ( )

i j i j i j i j i j

i j i j i j

i j i j


h

u x t u x t u x t

h

u ux t x t O h

t x

+ + + + +

+

+ +

+=

= + + + +

(1.54)

c ( , )ji i j

v u x t , ta vit bi ton sai phn thay th cho bi ton vi phn:

1 1 1 1

1 1 1 1

2 2

2 21[ ] ( , / 2)

2

j j j j j j j j

i i i i i i i ih i j

v v v v v v v vL v f x t

h h

+ + + ++ + + + + = + (1.55)

0 ( )i iv g x= (1.56)

0 ( ), ( )j j

a j N b jv g t v g t = = (1.57)

Mi phng tnh (1.55) cha ba n 1 1 11 1, ,

j j j

i i iv v v+ + +

+ lp trn j + 1 v ba n 1 1, ,j j j

i i iv v v +

lp di j theo s hnh 1.5


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21

S ny gi l s 6 im i xng hay s Crank - Nicolson.

t 2/h = , phng trnh (1.55) vit:

1 1 1

1 1

1 1(1 )

2 2

j j j j

i i i iv v v F + + + + + + = (1.58)

trong :

1 1

1( ) (1 ) ( , / 2)

2

j j j j

i i i i i jF v v v f x t += + + + + (1.59)

Cc iu kin (1.56), (1.57) cho 00

, ,j j

i N

v v v .

Khi bit jiv lp j, phng trnh (1.55) tc (1.58 ) cho php tnh

1j

iv + nhng phi

gii mt h i s tuyn tnh i vi 1 1 11 2 1, ,...,j j j

Nv v v

+ + + . y l mt phng php n.

p dng phng php sai phn tnh ton trng nhit trong phi tm vi

thng s c th: Mt tm phng ( bng thp) c chiu dy s=0,2 m c nung trong

mt l nung c nhit l 10000C, h s dn nhit = 55,8 W/m.K, nhit dung

ring c=460 J/Kg.K ; khi lng ring =7800 Kg/m

3

; h s to nhit t b mt timi trng l =335W/m2. Ta s tnh ton c trng nhit trong phi phn b

nh hnh v sau: (Chng trnh tnh km theo phn ph lc)

xi-1 xi xi+1 x

t

j-1

j

j+1

Hnh 1.5 S Crank - Nicolson


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22

1.5. Kt lun chng 1

Trong chng ny ta i thnh lp phng trnh truyn nhit trong phi tm.

Phng trnh truyn nhit trong phi t m chnh l mt phng trnh vi phn o

hm ring (partial differential equations). Vic tnh ton trng nhit trong phi

chnh l ta phi i gii phng trnh trn vi cc iu kin c th . chng ny gii thiu cng c ton hc vi hai phng php l gii tch v phng php s

gii bi ton.

Hn ch ca cc phng php gii thiu l kh khn cho vic thc hin cc

bi ton iu khin v vi cc phng php thit k hin nay, khi thit k b iu

khin, ta phi bit hm truyn ca i tng,.....

Hnh 1.6 Trng nhit trong phi


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23

CHNG 2

XY DNG M HNH HM TRUYN XCNH NHIT

TRONG PHI TM

2.1.t vn

Nh bit vi cc h thng iu khin muniu khin mt thng s no

ta phi o lngc thng s v ly c tn hiu phn hi . Tuy nhin trn

thc t c nhiu thng s cng ngh ca i tng cn iu khin m ta khng th

o trc tipc, v vy t ra vn xy dng m hnh Bit v tm Li

2.2. Nghin cu i tng iu khin

Xt mt l gia nhit t mt pha nh hnh v (hnh-2.1). Gi thit th tch

bung l nh, coi nhit trong l l nh nhau. Nu b qua s truyn nhit qua u

v cnh ca tm kim loi phng, rng ln vi cc thng s sau:

H s dn nhit ca tm : W/m.K

H s truyn nhit ca tm : W/ 2m

Chiu di a (mt)Chiu rng b (mt)

Chiu dy d (mt)

Khi lng ring : Kg/ 3m

Nhit dung ring c: J/kg.K

Din tch b mt tip xc A=a*b ( 2m )

Ta coi phi l mt i tng ng hc v c chia thnh n lp.i tng nghc ny c lng vo l nhit trong khng gian l; lng ra l nhit ca lp

d, T(t)

Tf(t)Heat source

Hnh-2.1 M hnh phi 1 lp


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di cng. Vic chn n bng bao nhiu tu thuc Dy ca tm v chnh xc

yu cu.

2.3. Xy dng m hnh hm truyni vi vt mng

Vt mng l vt c h s BIO


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2.4. Xy dng m hnh hm truyn khi phi c chia thnh 2 lp (n=2)

Dng nhit chy vo lp 1 l:

1( )

11

T Tf

Q A T T f

R

= = (2.5)

Vi1

1

.R

A=

Dng nhit chy ra lp 1 hay cng l dng nhit chy vo lp 2

1

2

1

1 2( )1 2/ 2

( )2

/ 2

A T TQ T T

d R

RA

d

= =

=

(2.6)

Vy phng trnh cn bng nhit l:

11 1 2

11 2

2 1 22

2

T TdT T T fC

dt R R

dT T T C

dt R

=

=

(2.7)

Xut pht t phng trnh2 1 2

22

dT T T C

dt R

= ta c:

1

2 2 2 1 2 2 2 2

( 1)C R T s T T C R s T T = + =

Suy ra hm truyn ca lp th 2:

Tf(t)Heat source

d/2

d/2

1, T1(t)

2, T2(t)



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26

( ) 12( )2

( ) 11 2 2

T sW s

T s R C s= =

+(2.8)

Xut pht t phng trnh11 1 2

11 2

T TdT T T f

Cdt R R

= ta c :

1

( ) 1 12

2 1 1 1 2 1 1 1

( )( )

f sW ss s

R C R C R C R C

TT + +

=

Suy ra hm truyn lp 1

( ) 11( )1

( ) 1 1

1

( ) 1 12

2 1 1 1 2 1

T sW s

T s R C fW s

sR C R C R C

= =

+ +

(2.9)

( )

1( )1

11

1 1 2

1 W (s)2

W sR

R C s

R

=

+ + (2.10)


Dng nhit chy vo lp 1 l:

11

( ) ( )1 11

;

T Tf

Q A T T Rf R A

= = = (2.11)

Tf(t)Heat source

d/3d/3

d/3

1, T1(t)2, T2(t)3, T3(t)



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Dng nhit chy ra lp 1 (cng chnh l dng nhit chy vo lp 2)

1 2( )1 1 2

/ 32

/ 3

2

1

1

A T TT T

d R

dR

A

Q

= =

=(2.12)


2 3( )

2 2 3/ 3

3

/ 3( )3

2

2

A T TQ T T

d R

dRA

= =

=

(2.13)

Do khng c nhit chy ra lp 3 nn t (2.11), (2.12), (2.13). Ta c phng trnh

cn bng nhit:

11 1 21

1 2

2 32 1 2

22 3

3 2 33

3

T TdT T T fC

dt R R

T TdT T T

Cdt R R

dT T T C

dt R

=

=

=

11 1 2

1 1 2 1

2 32 1 2

2 2 3 2

3 2 3

3 3

(a)

(b)

(c)

T TdT T T f

dt R C R C

T TdT T T

dt R C R C

dT T T

dt R C

=

=

=

(2.14)

Xut pht t phng trnh (2.14c) ta xy dngc hm truyn ca lp th 3

( ) 13( )3

( ) 12 3 3

T sW s

T s R C s= =

+(2.15)

Xut pht t phng trnh (2.14b) ta xy dngc hm truyn ca lp th 2


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( )

( )12

( )2 ( )

1 2 2

1( )2

21 2 2

3

1

( ) 1 13

3 2 2 2 3 2

1 W ( )3

T s

W sT s R C

W sR

R C sR

W ss

R C R C R C

s

= =

=

+ +

+ +

(2.16)

Xut pht t phng trnh (2.14a) ta xy dngc hm truyn ca lp th 1

( )

( ) 11( )1( ) 1 1

1( )1

11 1 1

2

1( ) 1 12

2 1 1 1 2 1

1 W ( )2

T sW sT s R C

f

W sR

R C sR

W ss

R C R C R C

s

= =

=

+ +

+ +

(2.17)


Dng nhit chy vo lp 1 l

11( ) ; ( )

1 11

T Tf

A T T Rf

R AQ

= = = (2.18)


Tf(t)Heat source

d/4

d/4

d/4d/4

1, T1(t)2, T2(t)

3, T3(t)4, T4(t)



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/ 41 2( ) ; ( )

1 1 2 2/ 4

2

1

1

A T T dT T R

d R AQ

= = = (2.19)


/ 42 3( ) ; ( )2 2 3 3

/ 4 3

2

2 2

A T T l dT T R

d R A AQ

= = = = (2.20)

Dng nhit chy ra lp 3(cng chnh l dng nhit chy vo lp 4)

3

3

/ 43 4( ) ; ( )

3 3 4 4/ 4 4Q

A T T dT T R

d R A

= = = (2.21)

Do khng c nhit chy ra lp 4 nn t (2.18), (2.19), (2.20), (2.21) ta c h

phng trnh cn bng nhit:

11 1 21

1 2

2 32 1 22

2 3

3 2 3 3 43

3 4

3 444

4

T TdT T T fC

dt R R

T TdT T T C

dt R R

d TT T T T C

dt R R

T TdTC

dt R

=

=

=

=

(a)

(b)

(c)

11 1 2

1 1 2 1

2 1 2 2 3

2 2 3 2

3 2 3 3 4

3 3 4 3

4 3 4

4 4

T TdT T T f

dt R C R C

dT T T T T

dt R C R C

dT T T T T

dt R C R C

dT T T

dt R C

=

=

=

= (d)

(2.22)


33/93

Lun vn thcskthut


30

Xut pht t phng trnh (2.22d) ta xy dngc hm truyn ca lp th 4

( ) 14( )4 ( ) 1

3 4 4

T sW s

T s R C s= =

+(2.23)

Xut pht t phng trnh (2.21c) ta xy dngc hm truyn ca lp th 3

( )

1

( ) 1 14

4 3 3 3 4 3

1 W ( )4

1( )

33 3

1( )

33

1 3 34

W ss

R C R C R C

s

W sR C

W sR

R C s R

+ +

=

=

+ +

(2.24)

Xut pht t phng trnh (2.22b) ta xy dngc hm truyn ca lp th 2

( )

1

( ) 1 13

3 2 2 2 3 2

1 W ( )3

1( )

22 2

1( )2

212 2

3

W ss

R C R C R C

s

W sR C

W sR

R C sR

+ +

=

=

+ +

(2.25)

Xut pht t phng trnh (2.22a) ta xy dngc hm truyn ca lp th nht:

( )

1

( ) 1 12

2 1 1 1 2 1

1 ( )2

( ) 11( )1( ) 1 1

1( )1

11 1 12

W ss

R C R C R C

W s

T sW s

T s R C f

W sR

R C sR

+ +

= =

=

+ +

(2.26)


34/93

Lun vn thcskthut


31

2.7. Xy dng m hnh hm truyn khi phi c chia thnh n lp

1 1W ( )

n 1s

R C sn nR Cn n

=+

n-1

1 1 n

1 1 1 1

1

W ( ) 1 1

1W ( ) n n

n n n n n n

ss

R C R C R C

s R C

+ +

=

.

.

.

3

3 3

4

4 3 3 3 4 3

1

W ( ) 1 11W ( )

ss R C R C R C

sRC

+ +

=

2

2 2 3

3 2 2 2 3 2

1

W ( ) 1 11W ( )

ss

R C R C R C

sR C

+ +

=

Tf(t) Heat source

d/n

d/n

d/n

1, T1(t)2, T2(t)

n, Tn(t)

.

.

.

.

.

.

Hnh-2.5 M hnh phi n lp


35/93

Lun vn thcskthut


32

2

2 1 1 1 2 1

1W ( )

11 1

1

W ( ) 1 1s

R C ss

R C R C R C

=

+ +

Hay:

( )

( )4

( ) 1( )

1( )1

1( )

11

1 1 1

1( )

331

3 34

1( )

21

2 2

1 ( )

1 ( )

n

T snW sn R C sT s n nn

W sn R

nR C sn n Rn

W sR

R C sR

W s

R C s

W s

W s

= =+

= + +

=+ +

=+

( )

( )

3

2

;

1 2 1

2

3

1( )

111

1 12

1 d/n d/n d/n(R = ; R = R = ... R = )n1 2 3A A A A

1 ( )

1 ( )

; ;n

R

R

W sR

R C sR

W s

W s

+

=+ +


36/93

Lun vn thcskthut


33

2.8. V d tnh ton hm truyn tng lp khi chia phi thnh 1 lp v 3 lp

Ly vt liu l thp tm vi cc thng s nh sau :

H s dn nhit ca tm =55.8 w/m.K (y coi h s dn nhit ca tm l

hng s)Khi lng ring: =7800kg/ 3m

Nhit dung ring c=460 j/kg.K

H s truyn nhit =335 w/ 2m

Chiu di tm a=40 cm=0.4 m

Chiu rng tm b =25 cm =0.25 m

Chiu dy tm d =5 cm =0.05 m

Din tch b mt tm:A=a*b =0.4*0.25 =0.1 2m

- Gi scoi tm thp l 1 lp :

Khi s truyn nhit qua tm l truyn nhiti lu:

V=0.4*0.25*0.05 = 0.005 3m

m=V* =0.005*7800 =39 kg

C =m*c =39*460 =17940

R=0.0299

Hm truyn i tng l

1( )

1W s

RCs=

+

1( )

536.406 1W s

s=

+

- Gi s coi tm thp l 2 lp:Khi chiu dy mi lp l d/2=0.05/2 m

V1=V2=.4*0.25*0.025=0.00253

m

m1 =m2 =V1* =0.0025*7800 =19.5 kg

C1 =C2 =m1*c =19.5 *460 = 8970

1

1 10.0299

0.1*335R

A= = =


37/93

Lun vn thcskthut


34

2

/ 2 0.0250.00448

55.8*0.1

dR

A= = =

Hm truyn tng lp ca i tng l:

22

1 2 2

1 21

1 1 2

2

( ) 1 1 1( )( ) 1 0.00448*8970 1 40.1856 1

1 40.1856 1( )

10762 575.7127 11 (1 W (s))

T sW sT s R C s s s

SW s

R s sR C s

R

= = = =+ + +

+= =

+ ++ +

- Gi s coi tm thp l 3 lp

Khi chiu dy mi lp l d/3=0.05/3 m

V1=V2=V3 =0.4*0.25*(0.05/3)3

m m1=m2=m3 =V1*=0.4*0.25*(0.05/3)*7800=13kg

C1=C2=C3 =m1*c =13*460 =5980

1

1 10.0299

0.1*335R

A= = =

2 3

/ 3 0.05/ 30.00299

55.8*0.1

l dR R

A A = = = = =

Hm truyn tng lp ca i tng l :

33

2 3 3

2 22

2 2 3

3

2

1 3 21

1 1 22

( ) 1 1( )

( ) 1 17.88 1

1 17.88 1( )

318.85 53.55 11 (1 W (s))

1 318.85 53.55 1( )

57449 13127 589.05 1

1 (1 W (s))

T sW s

T s R C s s

sW s

R s sR C s

R

s sW s

R s s s

R C s R

= = =+ +

+= =

+ ++ +

+ += =

+ + +

+ +


38/93

Lun vn thcskthut


35

2.9. Kt qu m phng cho b quan st nhit

- Khi coi tm phi l 1 lp:

Hnh -2.6 B quan st phi 1 lp v kt qu m phng


39/93

Lun vn thcskthut


36

- Khi coi tm phi l 2 lp:



40/93

Lun vn thcskthut


37

- Khi coi tm phi l 3 lp ta c :



41/93

Lun vn thcskthut


38

2.10. Kt lun

Da trn cc nh lut v truyn nhit, cc phng trnh cn bng nhit ta xy

dng c m hnh hm truyn cho phi 1 lp, 2 lp, 3 lp, 4 lp, t tng q ut

ha ta xy dng c m hnh hm truyn ca phi khi c chia thnh n lp.

y chnh l nhng m hnh quan st nhit c m t ton hc di dng hm

truyn. Nhng m hnh quan st ny s cho ta xc nhc nhit ti mt im

bt k mt thi im bt k. y cngchnh l c s cho vi ciu khin trng

nhit trong phi tha mn mt cng nght ra.


42/93

Lun vn thcskthut


39

CHNG 3

THIT K B IU KHIN NHIT TRONG PHI TM

3.1. Gii thiu mt s phng php thit k

3.1.1.Phngphp a thc c trng c h ssuy gim thay i c

Phng php h s suy gim ( Phng php a thc c trng c h s suy gim

thay i c) da vo a thc chun bc 2 c n ghin cu y tng qut

cho bc cao hn.

- Xt h bc 2 :

Gi s h bc 2 c hm truyn

( )2

0 02 2 2

0 1 2 0 0. 2 . .aW s

a a s a s s s

= =+ + + +

(3.1)

: h s suy gim

0 : tn s ring

Khi h s suy gim thay i s lm cht lng ca h thay i, kho st cht lng

ca h khi thay i, c th cng nh qa iu chnh cng tng ln.

Ta c :2

2 1

0 2

4 aa a

=

-Phng php a thc c trng c h s suy gim thay i c cho h bc

cao

Gi s hm truyn ca h c dng:

( )0

1

1

0

... asasa

asW

n

n

n

n +++=

(3.2)

Ta dng h s c trngnh sau:

2 2 2 2

1 2 11 2 1

0 2 1 3 2 1 1

0 1 10 1 1

1 2 1

2

1 1 1 1 21 2

0 0 2 0 2 1 1

, ;........; ;

; ;..............; ;

; ;.......;

n nn n

n n n n

n nn n

n n

nn

n

a a a a

a a a a a a a a

a a a a

a a a a

a a a

a a a a

+

+

= = = =

= = = =

= = = = =


43/93

Lun vn thcskthut


40

Mta thc cho c xc nh bng cch cho mt tn s c trng th nht

0 v h s suy gim ly c nh. Vy ta tnhc cc thng s khc c xc

nh bng cch nhn k tip nhau vi.

2 2 20 1 0 1 010 1 1 0 3 0

1 2 0 2 0 1 2

; ; ;a a a aaa a a a a a a

= = = = = =

Tng t nh vy ta xc nh: 2 30 0 0 0, , , , ...

Thng thng ta chn a0 = 1 v a1=1

30

33

2012

10

0

011 1

==

===

a

a

aaa

Vy ta c:

( ) kkkk

kk

a=

=

02/1

0

Ch

1%

0 0

2,22,2

at

a

=

: Khi cho cng 1 s h s cho cc gi tr n khc th cht lng ca h

thng thay i, n cng ln th thi gian hm qu ln u tin t xc lp cngnh.

H s c tnh cht ca h s suy gim, khi cng b h dao ng cng

mnh, < 1,5 h tr ln mt n nh, nh qu iu chnh % ln

Lng qu iu chnh quan h vi theo cng thc kinh nghim

Lg(%)=4,8 2 (3.3)

Thi gian qu t cc i

(3.4)

Ngita thng chn > 1,6

Bng - 3.1 Bng tnh sn mt s gi tr % theo

1,6 1,75 2 2,4

% 40 20 6 1


44/93

Lun vn thcskthut


41

- Xt nh hng ca t s hm truyn

Gi s hm truyn kn ca h c dng:

( ) 011

0

1

1

...

'...''

asasa

asasasW

nn

nn

m

m

m

m

+++

+++=

(3.5)

Khi m tng th % tng v t gim, c cht lng % cho trc ngi ta

dng h s hiu chnh nh sau:

Xt khi t s hm truyn c dng bc 1

( )0

1

1

01

...

''

asasa

asasW

n

n

n

n +++

+=

(3.6)

( )

1

00

0

0

'

''

5,1'

45,1'

a

a=

+=

(3.7)

Khi thit k c xc nh theo mu s ca (3.6) sau dng cng thc (3.7)

xc nh li ri xc nh lng qu iu chnh theo cng thc (3.3)

Thi gian qu c tnh:

=

00 '4

112,2

t (3.8)

Khi t s hm truyn c dng bc 2

( )0

1

1

01

2

2

...

'''

asasa

asasasW

n

n

n

n +++

++=

(3.9)

Ta c :


45/93

Lun vn thcskthut


42

( )

=

==

+=

00

20

12

1

00

20

203

'

112,2

''

'4,

'

''

5,1'

6,15,1'

t

aa

a

a

a

3.1.2.Phng php b hng s thi gian tri

- Khi nim chung

Trong cc h thng iu khin i tng cng nghip ta thng gp cc i

tng c 1 hoc 2 hng s thi gian ln, trong khi c cu iu khin chng li c

hng s thi gian rt b

Khi i tng iu khin c 1 hoc 2 hng s thi gian ln nu ta thit k b

iu khin c kh nng b c nhng hng s thi gian ln a h kn ca h

thng v dng bc 2 chun c dng:

( )2

00

2

2

0

2

++=

sssWk (3.10)

Cc i tng cng nghip ni chung thng lm vic trong c ng 1 tn s

thp, mong mun ( ) 1jWk khi 0 (3.11)

Khi 0 hm tuyn tnh s hWh(j ) , nn trong h phi c khu

tch phn

Vi tn s cao, iu kin (3.11) khng tho mn c

Vy khi 0 th ( ) 0jWk do tn s ct cng ln cng tt

- Xc nh thng s ca biu chnh theo tiu chun phng

Theo tiu chun phng h c hnh vi tch phn

( )kW j

Hnh-3.1c tnh bin-tn ca hm mdun ti u

1

c


46/93

Lun vn thcskthut


43

xt trng hp tng qut:

( )1

1

1 1

1 1

bsnn

dtk

jsk bj

W s KT s T s= =

= + + (3.12)

Tsk: L cc hng s thi gian ln ca i tng

Tbj : L cc hng s thi gian b ca i tng

Ch

( ) ( )1

11

dn

dc k

ki

d s

k sk

W s T sT s

n n

T T

=

= +

=

=

: i tng phi a v phn hi -1

Nguyn tcchung l b cc hng s thi gian tri trong mch h. Do vy

trong mch ch cn li hng s thi gian b. Khi h c 1 hng s thi gian ln chn

b iu chnh l PI, khi h c 2 hng s thi g ian tri chon b iu chnh l PID,

nu i tng c nhiu hn 2 hng s thi gian tri th dng phng php ni tip

cc b iu chnh, hoc dngphng php khc.

Chn b iu khin:

(3.13)

Tuy trng hp c nhiu hng s thi gian b, th hng s thi gian b tng

ng c tnh:

1

bn

b bj

j

T T=

= (3.14)

Sau khi b , h h c dng:

( )1

1

1

bn

h

ji bj

KW s

T s T s==

+ (3.15)

Ti l hng s tch phn ca biu chnh cn cxc nh

(-)

xWh(s)

y

Hnh 3.2


47/93

Lun vn thcskthut


44

Khi b h kn c hm truyn :

( )

( )( )

1

1 1

11 1 1

bk n

ibj

h j

W sT s

T s

W s k =

= =+ + +

(3.16)

Bnh phng modul c tnh tn h kn

( ) ( ) ( )( ) ( )

( )

= =

= = =

= =+ + +

=

+ + + +

2

1 1

2 6

2 2 4 2

2 21 1 1

1 1.

1 1 1 1

1

1 2 2 1

bb b

n nk k kb b

i i

bj bj j j

nn n

i i ibj bj i bj j j j

W s W j W j Ts Ts

T s T sK K

T T TT T T Tk K K K

(3.17)

tho mn iu kin (3.11)ngi ta thng thit k sao cho:

( ) ( )1 1

2 1 0 2 1 2b bn n

ibj i bj b

j j

TT T K T KT

K = = + = = + =

Hm truyn ca h kn sau khi chn b iu chnh c dng:

( )22221

1

sTsTsW

bb

k++

=

Tiu chun phngc tng kt theo bng 3.2.

Im

b

1J.

2T

b

1-J.

2T

b

1-

2T

.Re

Hnh-3.3


48/93

Lun vn thcskthut


45

ttB iu chnh Tn Tv Tv2 Ti

1 PI :sT

sT

i

n 1+ T1 - - 2KTb

2( )( )+ +1 1

:T s T sn v

PIDT si

T1 T2 - 2KTb

3( )( )( )21 1 1

2 :n v v

i

T s T s T sPID

T s

+ + + T1 T2 T3 2KTb

B iu chnh PID2 t dng, v kh thc hin c phn cng.

Hm qu i vi tn hiu t:

Hm truyn kn ca h sau khi chn b iu chnh :

( )22221

1

sTsTsW

bb

k++

= (3.18)

bT2

1707,0

2

10 ===

Hm qu :

( ) ( ) ( )[ ]bbTt

TtTteth b 2/sin2/cos12/ = (3.19)

Tc ng qu vi tc ng ca nhiu:

Hm nhiu f vit dng:

( )( )( )

( )( ) ( )

( ) ( )

( )sTsTsWsW

sWsW

sW

sE

sYsW

bb

dcdt

dcdt

dt

f

+

=

+==

12

1

1(3.20)

dcW ( )s dtW ( )s x e

f

y

(-)

Hnh-3.4

Bng - 3.2 La chn b iu khin theo tiu chun phng


49/93

Lun vn thcskthut


46

Xt i tng c 2 hng s thi gian ln :

( )( )( )

( )

( )

( )( )( )2 21 2 1 22 11 1

11 1 1 1 1 2 21

2 1

b b

f

b b

b b

KT s T sW s

T s T s T s T s T s T s

sT T s

+= =

+ + + + + ++

+

(3.21)

Hm truyn c im khng: 0 v -1/Tb

Xt trng thi ca h khi c nhiu trng thi xc lp

Gi s f(t) = 1(t) 1

( )F ss

= Y(s) = Wf(s)F(s)

( ) ( )( )

( )( )( )2 20 0

1 2

2 1lim lim 0

1 1 1 2 2

b b

s s

b b

KT s T sy sY s s

T s T s T s T s

+ = = =

+ + + +

ch xc lp nh hng ca nhiu khng cn na

gi s i tng c 1 hng s thi gian tri

( )

( )( )

( )( )

1

2 2

1

1

2 1

1 2 2 1

dt

b b

f

b b

KW s

T s

KT T sW s

T s T s T s

=+

+=

+ + +

Ta cng chng minh tng t :

( ) ( ) 0== fhy

3.1.3.Thit k b iu chnh cho h c hnh vi tch phn

-t vn

Ta xt i tng bc 1

( )( )( )1 1 1b

KW s

T s T s=

+ +(3.22)

Theo tiu chun phng, chn b iu chnh l PI:

( ) 11

2dc

b

T sW s

KT s

+= (3.23)

Gi s hng s thi gian T1 rt ln th b iu chnh PI c tc dng nh b

iu chnh P do thnh phn tch phn khng cn na, tng t b iu chnh l PIDkt qu c hiu qu nh PD, nhng vn cn sai lch tnh


50/93

Lun vn thcskthut


47

Khi T1 rt ln ta c:

( )( )

( )( )( )

( )2 2 2 21 1

1 2 1

1 2 2 1 2 2 1

b b b b

f

b b b b

KT s T s KT T sW s

T s T s T s T T s T s

+ += =

+ + + + + (3.24)

Ta thy ch xc lp s0 nhng Wf(s) khng th bng khng.

Nh vy khi h c hnh vi tch phn hay c hng s thi gian qu ln m

dng tiu chun i xng , th s dnn sai lch tnhi vi tn hiut v vi

nhiu.

- Thit k b iu chnhc hnh vi tch phn theo tiu chun i xng

c tc ng nhanh i vi nhiu, cn c h s khuch i ln khi tn s

b, c th chn hng s thi gian ca biu chnh nh sau:

1 2...d d dn d T T T T = = = =

B iu chnh c dng: ( )( )1 d

n

d

dc

i

T sW s

T s

+= (3.25)

Hm truyn h h:

( )( )

( ) ( )1

1

1 1s

n

h n

i b k

k

K T s

W sT s T s T s

=

+=

+ +(3.26)

Khi hng s thi gian ca i tng l rt ln

( )( )

( )1

1

1

d

s

s

n

n

h nn

i b k

k

K T sW s

T s T s s T =

+

+ (3.27)

Cng nh tiu chun phng, iu kin trc tin l: ns = nd. n gin ta dng khiu:

0

1

d

d

n

d

n

k

k

KTK

T=

=

Suy ra (3.27) c dng: ( )( )

0 1

1

sn

dh

i b d

K T sW s

T s T s sT

+

+ (3.28)

Dng php bin i gn ng:


51/93

Lun vn thcskthut


48

1 1 11 1

s sn n

d s de

d d de de

d

des

T s n T s

T s T s T s T s

TT

n

+ += + + =

=

(3.29)

Vy ta c :

( )( )

( )

0

2

0

1

1

1( )

1 1

deh

i b de

dek

de ide b

K T sW s

sT T s T s

T sW s

T TT s s T s

K

+=

+

+=

+ + +

(3.30)

Bnh phng modul c tnh tn h kn c dng

( ) ( ) ( )( )

( )

2 22

2

24 2

2 2 6

0 0 0 0

1

1 2 2

dek k k

i de i i de b ide de b

TW W j W j

D

T T T T T T T D T T T

K K K K

+=

= + + +

cho ( ) ( ) 11lim 220

DjW

Ta rt ra :

02

4

i b

de b

T K T

T T

=

=(3.31)

Thng s ca b iu chnh c chn theo:

0

1 1

4

2,

s s

s s

dde d s b

sn n

d ds d i bn n

k k

k k

TT T n T

n

KT KT n n K T T

T T= =

= =

= = =

(3.32)

Vy ta c hm truyn ca h h:

( )( )

1 4 1

4 2 1

bh

b b b

T sW s

T s T s T s

++

(3.33)


52/93

Lun vn thcskthut


49

c tnh tn s logarit ca h h ( )sWh i xng nhau qua tn s ct

bc

T2

1= nn

gi l tiu chun i xng

tt B iu chnh Tn Tv Tv2 Ti

1sT

sTPI

i

n+1

: bT4 - -2

1

8 bK

TT

2( )( )

sT

sTsTPID

i

vn ++ 11: bT8 bT8 -3

1 2

128 bK

TTT

3( )( )( )21 1 1

2 :n v v

i

T s T s T sPID

T s

+ + + bT12 bT12 bT12

4

1 2 3

3456 bK

TT T T

Biu thc (3.33) l biu thc x p x khi h l bc 1 v c hnh vi tch phn.

Trong trng hp h bc 1 vi khu qun tnh th biu thc qun tnh:

( )( )( )

1

2

1

1

1 4

8 1 1

b

bbb

T sTW s

TTs T s T s

T

+=+ +

(3.34)

Hm truyn kn vi tn hiu t x(t) = 1(t)

( )

( )( )

1

2

11

1 4

1 4 8 1 1

b

bb

b b

T sTW s

TT

T s s T s T sT

+=

+ + + +

(3.35)

Hm truyn kn ca h thng c thit k theo tiu chun i xng:

( )( )

( )

2 2

2 2 3 3

1 4

8 1

1 4

1 4 8 8

bh

b b

bk

b b b

T sW s

T s T s

T sW s

T s T s T s

+=

+

+=

+ + +

Bng- 3.3 Quy tc xc nh b iu chnh theo tiu chun i xng


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50

Vy khi T1 cng ln so vi Tb, s tng qu iu chnh gim thi gian p

ng T0, tc ng nhanh ch yu ph thuc vo Tb. gim lng qu iu

chnh, dng b lc u vo vi mc ch l b tr im 0

( ) sTsW bl

41

1

+=

3.1.4.Phng php thit k b b

Xc nh b iu khin Wk(s) da trn c s bit trc hm truyn ca i tng

v bin hm truyn ca c h thng W *(s), W*(s) c xc nh t y u cu cht

lng ca bi ton iu khin

Gi s i tng c hm truyn dng:

( ) ( )( )

( ) ( )( )

( )( ) ( )

( ) ( )

( )( )

( )( )

( )( )

( )( ) ( )

*

*

*

&

.

1

1`. .

1

dt

dc dt

dc dt

dc

dt

A s C sW s W sB s D s

W s W sW s

W s W s

W s B s C sW s

W s W s A s D s C s

= =

=+

= =

iu kin : D(s) C(s) phi l a thc Hurwist (h n nh: tt c cc im

khng v im cc phi nm bn tri trc o)Gi nA l bc ca A(s)

Gi nB l bc ca B(s)

Gi nC l bc ca C(s)

Gi nD l bc ca D(s)

Vy

DCAB

DACB

nnnn

nnnn

Mun tch hp c b iu khin b th bc ca i tng ca h kn tng

i khng nh hn b tng i ca i tng

-xt trng hp W*(p) c dng:

( )* 0 10 1

. ... .

. ... .

m

m

n

n

c c s c sW s

d d s d s

+ + +=

+ + +


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Lun vn thcskthut


51

Mun cho h khng c sai lch tnh :

( ) ( ) 00*

011lim dcsWimlth

st===

xt

( ) ( ) ( ) ( )11.211121 ......... ++++++= mmnn scsccsdsddssCsD

m ( )( )( )

( )( ) ( )

.dcB s C s

W sA s D s C s

=

Vy h kn khng c sai lch tnh b iu khin thit k theo phng php

b cha thnh phn tch phn nu i tng c ha c thnh phn , ngc li khi

i tng c sn thnh phn tch phn th b iu khin s khng cha thnh

phn tch phn na.3.1.5. B iu khin m

-Cu trc ca b iu khin m

Cc b iu khin m c thit k da trn logic m c gi l b iu

khin m (FLC : Fuzzy Logic Control)

-B iu khin m c bn

B iu khin m c bn c dng nh hnh -3.5 gm 3 khi:

Khi 1: lm m ho

Khi 2: xc nh lut hp thnh

Khi 3: Gii m

B iu khin m c bn gm ba khu chnh l khu m ho, thit b thc

hin lut hp thnh v khu gii m.

Hnh -3.5 B iu khin m c bn


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52

- B iu khin m ng

Do b iu khin m c bn ch c

kh nng x l cc gi tr tn hiu hin thi

nn n thuc nhm cc b iu khin m

tnh. Tuy vy, m rng min ng dng

ca chng vo cc bi ton iu khin

ng, cc khu ng hc cn thit s c

ni thm vo b iu khin m c bn

hnh -3.6. Cc khu ng c nhim v cung cp thm cho b iu khin m c bn

cc gi tr o hm hay tch phn ca tn hiu. Cng vi cc khu ng b sung ny,

b iu khin m c bn s c gi l b iu khin mng.

- u im nhc im ca iu khin m

- Khi lng cng vic thit k gim i nhiu do khng cn s dng m hnh

i tng trong vic tng hp h thng.

- B iu khin m d hiu hn so vi cc b iu khin khc v d dng

thay i.

- i vi cc bi ton thit k c phc tp cao, gii php dng b iu

khin m cho php gim khi lng tnh ton v gim gi thnh sn phm.

- Trong nhiu trng hp b iu khin m lm vic n nh hn, bn vng

hn v cht lng iu khin cao hn.

- iu khin m c th s dng cho cc h thng khng cn bit chnh xc

m hnh i tng.

- V h thng iu khin m gn vi nguyn liu khin ca con ngi

(con ngi khng c cc cm bin cm nhn chnh xc i tng), do cc b

cm bin s dng c th khng cn chnh xc cao.

+ Vic nghin cu v l thuyt i vi l thuyt m cha tht hon thin

(tnh n nh, tnh phi tuyn, ti u).

+ Cho n nay cha c nguyn tc chun mc cho vic thit k cng nh

cha th kho st tnh n nh, tnh bn vng, cht lng, qu trnh qu cng

nh qu trnh nh hng ca nhiu cho cc b iu khin m.

- Khng thit k h iu khin m cho cc bi ton m h iu khin kinh

in c th d dng thc hin c nh cc b iu khin P, PI, PD, PID.

Hnh -3.6. B iu khin m ng


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53

- Hn ch s dng iu khin m cho cc h thng cn m bo an ton

cao do nhng yu cu v cht lng v mc ch ca h thng iu khin m ch c

th xc nh v t c qua thc nghim.

- H thng iu khin m l h thng iu khin mang tnh chuyn gia, gn

vi nguyn l iu khin ca con ngi, do ngi thit k phi hon ton

hiu bit v kinh nghim v h thng cn iu khin mi c th thit k c h

iu khin m.

- M ho

M ho c nh ngha nh l s nh x ( s lm tng ng), t tp m cc gi

tr thc x* U thnh cc gi tr m A U, nguyn tc chung vic thc hin m

ho l:

- T tp gi tr thc x u vo s to ra tp m A vi hm lin thuc c gi tr

rng ti cc im r x

- Nu c nhiu u vo th vic m ho s gp phn kh c nhiu

- Vic m ho phi to iu kin n gin tnh ton cho sau ny

- C 3 phng php m ho:

+ M ho n v (Singleten fuzzifier) l t cc im gi tr thc x U ly cc

gi tr n v ca tp mA

==

'0

'1)('

xxkhi

xxkhixA

ngha l hm lin thuc dng:

+ M ho Gaus (Gaussian fuzzifier) : l t cc im gi tr thc x* U ly cc

gi tr trong tp mA vi hm lin thuc dng hnh tam gic hoc vung

- Quy lut suy din v c ch suy din m

- Mnh hp thnhLut m c bn l lut m t bi quan h: Nu ... Th...(IF....THEN....), mt

cch tng qut c dng:

NU TH

Mt mi quan h Nu.... Th ..... gi l mt mnh hp thnh, trong mt

mnh hp thnh c th c mt mnh iu kin hoc nhiu mnh iu kin

v mt hoc nhiu mnh kt lun.

Mt s dng mnh m:


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x = A v x1 = A1 v x2 B.

x1 = A1 v x2 = A2 v ... v xn = An

x1 = A1 hoc x2 = A2 hoc ... hoc xn = An (3.36)

(lu rng cc php logic v (and), hoc (or), Ph nh (not) trong log ic m tngng cc php giao, hp, b).

Trong h m lut m l b no ca n, ngi thit k phi da vo kinh

nghim ca mnh m pht biu v xy dng cho c mt tp m dng ny lm c

s cho vic trin khai thit k tip theo.

- Qui tc hp thnh

T mt gi tr u vo x0 hay c th hn l ph thuc A(x0) ta phi xc

nh c u ra hay ph thuc ca u ra. ph thuc u ra s l mt tp

m gi l tp mB'(y), tp m B' cng c s vi tp m kt lun B.

Nh vy, biu din h s tha mn mnh kt lun nh mt tp m B' cng

c s vi B th mnh hp thnh chnh l nh x.

A(x0) B'(y). (3.37)

M t mnh hp thnh chnh l m t nh x trn, c ngha l phi tm

c hm lin thuc AB

(x,y) cho mnh hp thnh A B, c nhiu cch m t

mnh hp thnh gi l cc qui tc hp thnh l:

1- Cng thc Zadeh: (qui tc hp thnh Zadeh)

AB(x,y) = MAX{MIN{A(x), B(y)}, 1 - A(x)}. (3.38)

2- Cng thc Lukasiewicz: (qui tc hp thnhLukasiewicz)

AB(x,y) = MIN{1, 1 - A(x) + B(y)}. (3.39)

3- Cng thc Kleene-Dienes: (qui tc hp thnh Kleene-Dienes)

AB(x,y) = MAX{1 - A(x), B(y)}. (3.40)

Theo nguyn tc ca Mandani " ph thuc ca kt lun khng c ln

hn ph thuc ca iu kin" ta c cch xc nh hm lin thuc AB(x,y) cho

mnh hp thnh AB nh sau.

4-Cng thc MIN: (qui tc hp thnh MIN ca Mandani,sch gi l qui

tc hp thnh MAX-MIN)

AB(x,y) = MIN{A(x), B(y)}. (3.41)


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5-Cng thc PROD: (qui tc hp thnh MIN ca Mandani, sch gi l qui

tc hp thnh MAX-PROD)

AB(x,y) = A(x)B(y). (3.42)

Cc cng thc (3.38, ..., 3.42) cho mnh hp thnh AB c gi l ccquy tc hp thnh. Hai quy tc hp thnh theo Mamdani l MIN (MAX -MIN) v

PROD (MAX-PROD) hay c s dng hn c.

Xt mnh hp thnh mt iu kin: Nu x = A th y = B, (x c th l tc

xe, y l bn p ga, A l chm, B l tng) x c xc nh bi cc hm lin

thuc A(x), v y c xc nh bi cc hm lin thuc B(y) th hm lin thuc

AB(x,y) s dng quy tc MIN v quy tc PROD ti mt gi tr r

0xx = c ch ra trn hnh 3.6 : a v b.

Hnh -3.7: Hm lin thuc ca lut hp thnh A B(x,y)a, Hm lin thucb, Vi qui tc MAX-MINc, Vi qui tc MAX-PROD

b,A(x)

B(y)

x0

H

AB(x0,y)

x y

A(x)

B(y)

x0

H

AB(x0,y)

x

y

c,

a,A(x)

B(y)

x y


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56

- Luthp thnh

n gin ngi ta k hiu mnh hp thnh AB ti mt gi tr r

0xx = l R. Tn gi chung ca m hnh R (ma trn) l lut hp thnh.

Hm lin thuc AB(x,y) ca m hnh R c biu din theo cch t hpcc mnh hp thnh no, theo quy tc hp thnh no th lut hp thnh c tn gi

l tn ghp ca cch t hp v tn quy tc hp thnh .

+ Hm lin thuc AB(x,y) c t hp theo php hp A B(x) =

MAX{A(x), B(x)} v quy tc MIN th ta c lut hp thnh MAX-MIN.

+ Hm lin thuc AB(x,y) c t hp theo php hp A B(x) =

MAX{A(x), B(x)} v quy tc PROD th ta c lut hp thnh MAX-PROD.

+ Hm lin thuc AB(x,y) c t hp theo php hp Lukasiewier: A

B(x) = min{1, A(x) + B(x)} v quy tc MIN th ta c lut hp thnh SUM-MIN.

+ Hm lin thuc AB(x,y) c t hp theo php hp Lukasiewier: A

B(x) = min{1, A(x) + B(x)} v quy tc PROD th ta c lut hp thnh SUM -

PROD.

Ch

Nh vy:

: Nu lut hp thnh ch c mt mnh hp thnh (khng ph i t hp) th

thc cht cha th hin c khi nim MAX hoc SUM, khi lut hp thnhMAX-MIN tng ng SUM-MIN, MAX-PROD tng ng SUM-PROD.

K hiu gi tr m u ra l B' th hm lin thuc ca B' ti mt gi tr r x 0

vi quy tc MAX-MIN s l:

B'(y) = R(x0,y) = MIN{A(x0) B(y)} (3.43)

T cng thc (3.39) ta thy khi cao ca tp m B l 1 th cao ca tp

m B' s chnh l cao ca tp m A ti x0, hnh-3.6b.

)x()x(H 0A0 =

Ta gi )x(H 0 l tha mn mnh iu kin hay gi tt l tha mn.

Th hai lut hp thnh MAX-MIN v MAX-PROD c vit nh sau:

1- Lut hp thnh MAX-MIN:

B'(y) = R(x0,y) = MIN{ )x(H 0 , B(y)}. (3.44)

2- Lut hp thnh MAX-PROD:

B'(y) = R(x0,y) = )x(H 0 B(y). (3.45)


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57

Do xc nh hm lin thuc B'(y) ta phi xc nh tha mn

)x(H 0 sau c th s dng cc cng thc (3.44) hoc (3.45).

* Cch xc nh tha mn )x(H 0

Cch xc nh tha mn )x(H 0 c ch ra trn hnh -3.8.

+ Khi tn hiu vo l mt gi tr r x0 hnh 3.8a.

+ Khi tn hiu vo l mt gi tr m vi hm lin thuc A'(x) hnh 3.8-b.- Lut hp thnh mt iu kin

T cc khi nim v lut hp thnh v tp m u ra B'(y) nh trn ta c th

xy dng thut ton xc nh lut hp thnh v tp m u ra.

- Thut ton xy dng lut hp thnh R

Lut hp thnh R chnh l m hnh ma trn R ca mnh hp thnh AB,

ng vi mi cng thc tnh hm lin thuc AB(x,y) khc nhau ta c cc lut hp

thnh khc nhau. Nhng nhn chung xy dng lut hp thnh R (mt iu kin)ta c th tin hnh theo cc bc sau:

Bc 1

ni21 x,...,x,...,x,x

: Ri rc ha cc hm lin thuc A(x), B(y), s im ri rc ha vi tn s

ln sao cho khng b mt tn hiu. Chng hn ri rc hm A(x) vi n im

, hm B(y) vi m im y1, y2 ... yj ...ym .

Bc 2 )x(TA

: Xc nh hm lin thuc ri rc v )y(TB

l: (T l chuyn v)

)}x(,...),x(),x({)x( nA2A1ATA =

Hnh -3.8: Xc nh tha mn H(x0)a, Vi gi tr vo r x0b, Vi gi tr vo m c hm lin thucA'(x)

A(x)

x0

)x(H 0

xa,

A(x)

xb,

A'(x)

)x(H 0


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58

)}y(,...),y(),y({)y( mB2B1BT

B= (3.46)

Bc 3

=

=

nm1n

m111

mnR1nR

m1R11R

r...r

r...r

)y,x(...)y,x(

)y,x(...)y,x(

R

: Xy dng ma trn hp thnh R, ma trn ny c n hng v m ct:

(3.47)

trong : rij = R(xi, yj) c tnh theo cc cng thc (3.38) n (3.42). Thc t hay

dng hai cng thc MIN v PROD ca Mandani (3.41) v (3.42) l:

- Theo cng thc MIN (vi lut hp thnh MAX-MIN):

rij = R(xi, yj) = MIN {A(xi), B(yj)}. (3.48)

- Theo cng thc PROD (vi lut hp thnh MAX-PROD):rij = R(xi, yj) = A(xi).B(yj). (3.49)

* Cng thc tng qut xy dng lut hp thnh R

T cc cng thc (3.46) n (3.49) ta thy c th a ra cng thc tng qut

(cng thc dyadic) tnh ma trn hp thnh R nh sau:

)y().x(RT

BA= (3.50)

Trong cng thc (3.50 ) nu p dng quy tc MAX-MIN th php nhncthay bng php ly cc tiu (min), vi quy tc MAX-PROD th thc hin php nhn

nh bnh thng.

- Xc nh hm lin thuc u ra B'(y) khi c lut hp thnh

T ma trn R ta thy hm lin thuc u ra B'(y) ng vi mt gi tr u vo

x0 chnh l mt hng ca ma trn R.

n gin ta gi a l vector xc nh v tr ca gi tr r x 0, vector xc nh

v tr ch c mt gi tr bng 1 ti v tr c x0 cn cc gi tr khc u bng 0. Do vycho mt gi tr r bt k }x,...,x,...,x{Xx ni1= ta s c mt vector chuyn v a

T

vi:

aT = (a1, a2, ... ai ..., an)

trong ch c mt phn t a i duy nht c ch s i l v tr ca x0 trong x c gi tr

bng 1, cc phn t cn li u bng khng. Nh vy hm lin thuc B'(y) s c

xc nh:


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B'(y) = aT.R = (a1, a2, ... ai ..., an)

nm1n

m111

r...r

r...r

= (l1, l 2, ..., l j, ..., l m)

vi: ==n

1iijij ral (3.51)

Trong thc t trnh phi s dng thut ton nhn ma trn (tng tc x

l) th php nhn ma trn kiu (3.51 ) c thay bi lut max-min ca Zadeh vi

max (ly cc i) thay vo v tr php cng, min (ly cc tiu) thay vo v php

nhn.

}r,amin{max ijini1

j

=l (3.52)

Kt qu ca hai php tnh (3.51) v (3.52) vi u vo l gi tr rl hon ton nhnhau.

*Ch : Khi lng vo l tp m A' vi hm lin thuc A'(x), th vector xcnh

v tr a gm cc gi tr ri rc ca hm lin thuc A'(x) ti cc im

}x,...,x,...,x{Xx ni1= khi ny khng s dng cn g thc (3.52) c, phi s

dng cng thc (3.51).

- . Lut hp thnh nhiu iu kin

Thut ton xy dng lut hp thnh R:

+ Ri rc ha min xc nh cc hm lin thuc ca cc mnh iu kin

v mnh kt lun.

+ Xc nh tha mn H cho tng vector cc gi tr r u vo l vector t

hpd im mu thuc min xc nh ca cc hm lin thuc )x( iAi ,1,...,i d= .

Chng hn vi mt vector cc gi tr r u vo

1c

x

cd

=

M trong ci, i = 1, ..., d l

mt trong cc im mu min xc nh ca ( )xiAi

, th:

)}c(),...,c(),c({MINH dA2A1A d21 = (3.53)

+ Lp m hnh ma trn R gm cc hm lin thuc gi tr m u ra cho tng

vector cc gi tr u vo theo nguyn tc:

)}y(,H{MIN)y( B'B = nu quy tc s dng l MAX-MIN (3.41).


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60

)y(.H)y( B'B = nu quy tc s dng l MAX-PROD (3.42).

Khng nh lut hp thnh mt iu kin, lut hp thnh R ca d mnh

iu kin khng th biu din di dng ma trn c na m thnh mt li trong

khng gian d +1 chiu.- Lut ca nhiu mnh hp thnh

Trong thc t t c h m no ch lm vic vi mt mnh hp thnh m

thng vi nhiu mnh hp thnh, hay cn gi l mt tp cc mnh hp thnh

Rk.

Vy ta phi lin kt cc lut hp thnh ring r li, c hai kiu lin kt l lin

kt theo kiu "cc i" (MAX-MIN, MAX-PROD) v kiu "tng" (SUM -MIN,

SUM-PROD) tng ng vi hai php hp l php hp bnh thng v php hpLukasiewicz.

- Lin kt lut hp thnh kiu "cc i" (MAX)

Khi c cc lut hp thnh thnh phn R1, R2 ,... , Rp ta c lut hp thnh

tng hp:

==

}r,...,r,rmax{...}r,...,r,rmax{

}r,...,r,rmax{...}r,...,r,rmax{

R...RRR

pnm

2nm

1nm

p1n

21n

11n

p

m1

2

m1

1

m1

p

11

2

11

1

11

p21 (3.54)

*Ch : tng mnh thnh phn nn c m hnh ha thng nht theo mt quy

tc chung, cng theo quy tc MAX-MIN hoc cng theo quy tc MAX-PROD... khi

lut hp thnh chung s c tn l lut hp thnh MAX-MIN hoc lut hp thnh

MAX-PROD...

Lut hp thnh MAX-MIN mt iu kin c th hin trn hnh -3.6c

- Lin kt lut hp thnh kiu "tng" (SUM)

Lut hp thnh chung lin kt theo kiu "cc i" (MAX) khng c tnh

thng k. Chng hn khi a s cc mnh hp thnh thnh phn c cng mt gi

tr u ra nhng v khng phi l gi tr lp nht nn s khng c n v b

mt trong kt qu chung.

C nhiu cch khc phc nhc im ny, mt trong cc cch l s dng

php HocLukasiewicz lin kt cc mnh thnh phn.


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==

==

==

=}r,1min{...}r,1min{

}r,1min{...}r,1min{

R,1minRp

1k

k

nm

p

1k

k

1n

p

1k

k

m1

p

1k

k

11p

1kk (3.55)

Vi cch lin kt ny ta c lut hp thnh SUM-MIN v SUM-PROD.

Lut hp thnh SUM-MIN mt iu kin c th hin trn hnh-3.8d.

- Thut ton xy dng lut hp thnh chung ca nhiu mnh

Thut ton xy dng lut hp thnh chung ca nhiu mnh ni chung

tng t nh ca mt mnh , ch thm bc tng hp cc mnh .

Xt mnh hp t hnh chung cho p mnh hp thnh mi mnh hp

thnh c 1 iu kin gm:

R1: NU 1A= ,, TH 1By = Hoc

R2: NU 2A= ,, TH 2By = hoc

...

Rp: NU Ap = ,, TH y Bp= hoc

Hnh -3.9:Cch k t hp cc mnh a, b, Lut hp thnh ca mt mnh .c, Lut hp thnh kt hp kiu MAX-MINd, Lut hp thnh kt hp kiu SUM-MIN

)y(1B

y

a,

)y(2B

y

b,

y

c,

)y(B

y

d,

)y(B


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62

Trong cc gi tr m A1, A2, , Apc cng c s X

B1, B2, , Bpc cng c s Y

Gi hm lin thuc Ak v Bk l )x(kA

v )y(kB

vi 1, 2, ...,k p=

Cc bc thut ton:

Bc 1

)x(kA

: Ri rc ha cc hm lin thuc iu kin X v kt lun Y, s im ri rc

ha vi tn s nh sao cho khng b mt tn hiu. Chng hn ri rc hm

vi n im ni21 x,...,x,...,x,x , hm )y(kB vi m im y1, y2 ... yj ...ym .

Bc 2 )x(T

Ak: Xc nh hm lin thuc ri rc v )y(T

Bk l:

)}x(,...),x(),x({)x( nA2A1AT

A kkkk = )}y(,...),y(),y({)y( mB2B1B

T

B kkkk= (3.56)

Bc 3

: Xy dng ma trn hp thnh R, (theo cng thc cng thc dyadic)

)y().x(RT

BAk kk= , 1, 2,...,i n= v 1,2,...,j m=

ma trn ny c n hng v m ct:

=kk

kk

k

nm1n

m111

r...r

r...rR (3.57)

trong : -php nhn c gi nguyn nu s dng nguyn tc MAX-PROD hoc

SUM-PROD.

- php nhn c thay bng php ly cc tiu khi s dng nguyn tc

MAX-MIN hoc SUM-MIN.

Bc 4:Theo MAX-PROD v MAX-MIN (cng thc 3.53)

Xc nh lut hp thnh chung

==

}r,...,r,rmax{...}r,...,r,rmax{

}r,...,r,rmax{...}r,...,r,rmax{

R...RRRp

nm

2

nm

1

nm

p

1n

2

1n

1

1n

p

m1

2

m1

1

m1

p

11

2

11

1

11

p21

Theo SUM-PROD v SUM-MIN (cng thc 3.48)


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Lun vn thcskthut


63

=

=

==

==

=}r,1min{...}r,1min{

}r,1min{...}r,1min{

R,1minRp

1k

k

nm

p

1k

k

1n

p

1k

k

m1

p

1k

k

11p

1kk

- Xc nh hm lin thuc u ra ti cc u vo

Vi cc gi tr u vo c xc nh bi vecto v tr a ta cng c:

R.a)y(T

'B = (3.58)

Ch

Gii m l qu trnh xc nh mt gi tr

r y' no c th chp nhn c t hm lin

thuc

: Thut ton trn vit cho p mnh hp thnh vi 1 iu kin, c th m

rng cho p mnh hp thnh vi q iu kin.

- Gii m

Vi b iu khin m th u ra l mt tp m, vy a cho cc b iu

khin thc t cha lm vic c. Cn phi gii m tc l cn r ho tp m u ra

B.

)y('B ca gi tr m B'.

Phng php cc i

gii m theo phng php cc i

phi tin hnh theo hai bc:

Xc nh min cha gi tr r y': Min

cha gi tr r y' l min m ti hm lin thuc

t gi tr cc i:

G = { yY, )y('B

= H} (3.59)

Min cha gi tr r 21 y'yy trn

hnh -3.10

Xc nh gi tr r y c th chp nhn c

trong min G theo mt trong ba nguyn l:

+Nguyn l trung bnh

Theo nguyn l trung bnh cho kt qu y l honh ca im trung bnh

gia cn tri y1 v cn phi y2 ca min G:

Hnh -3.11 Nguyn l trung bnh

y

y1

)y(B

y2y'

Hnh -3.10 Xc nh min cha gi tr r

y

y1

)y(B

y2


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