Báo Cáo Thí Nghiệm Điều Khiển Quá Trình

8/10/2019 Bo Co Th Nghim iu Khin Qu Trnh

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BO CO TH NGHIM IU KHIN QU TRNH

H v Tn: L Thi n c

SHSV: 20125469

Nhm: 12

Hiu chnh:

+ start time: 0.0s , Stoptime: 13700s, Type: Fixed step, solve: Ode 4, Fixed step size: 0.2s

Cc Tham s trong kh i single-tank:

+ course number: 57

+ class number : 12

+ name list number: 69


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BI 1

XY D NG H THNG IU KHIN MT BNHN C

I. XY DNG M HNH CHO I T NG BNH M C

1. Xc nh Cc Tn Hiu Vo/ra v nhiu h thng.

- Tn hi u vo: m van vo (In value ) - F1

- Tn hi u ra : M c n c trong bnh (Level) - h

- Nhiu h thng: m van ra (Out value) F2

- S bin vo l 2

- S bin ra l 1

=> s bc t do l : 3-1=2 v b ng s bin vo . Nn h c kh nng iu khin c.

H Th ng

Nhi u F2

Bi n iu Khi n

F1

Bi n C n iu Kh

h


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2. Xy d ng m hnh ton hc cho i t ng v i cc tham s hnh th c:

- Phng trnh cn bng vt cht:

1 2dV dh

A F F dt dt

(1)Trong : A l tit din ct ngang c a bnh ch a (coi nh u t trn xu ng).

Phng trnh m hnh tr ng thi xc l p:

0 1 2d h

A F F dt

(2)

Tr v ca (1) (2) ta c:

1 2d h

A F F dt

t: ; 2; 1 y h u F d F - Phng trnh tr thnh:

1 ( )dy d udt A

- Ti tr ng thi ban u tt c cc bi n chnh l ch u y, u, d v dy dt u bng 0.-Laplace 2 v ta c

1 1( ) ( ) ( ) sy s u s d s

A A

- Do van l khu qun tnh b c nht nn hm truy n ca h thng s c dng:

()( ) ( )

3. S dung Simulink xc nh tham s m hnh (nhn dng h thng):

- To hm p ng u ra c a h thng bng cch t o mt tn hi u bc thang u vo:

M hnh trong Simulink


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Ta c p ng qu nh sau:

- Xc nh cc h s k, T:

+ T th trn ta k tim cn ct tr c honh t t i im ng v i ta (1.95,0). Ta ch n T=1.95.

- T ta tnh h s gc K= tan = v i a= 16.7 v b=10 1.95 = 8.05

Suy ra: K= = 2.07Vy, ta c hn truy n t sau: G(s) = ( )


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- Kim tra l i p ng qu bng m hnh simulink trong matlab:

Hai th p ng ca hm truy n v c a h thng m ph ng


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- Nhn Xt:

+ th p ng ca hm truy n va tm c cha st vi p ng ca qu trnh.Vy, ta ph i iu ch nh cc thng s T v K p ng ca hm truy n st v i qu trnh hn.

- sau khi iu ch nh ta c cc thng s mi T= 2.5 v K = 2.8, th p ng m i

- p ng ca hm truy n sau khi iu ch nh v c a h thng m ph ng:


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II. CC SCH LC IU CHNH

1. Sch lc iu chnh c th s dng:

+ Sch lc iu khin phn hi: V sch lc iu khin phn hi c tc d ng n nh mt h thngkhng n nh c th l h thng bnh m c.

+ Sch lu c iu khin tng: V sch lc iu khin tng cng c tc dng n nh mt h thng khngn nh ngoi ra cn gi m ti thiu c nh h ng ca nhiu.

2. sch lc iu chnh khng th s dng:

- Sch lc iu khin truyn thng:

+ iu khin truyn thng khng c tc d ng n nh mt h thng khng n nh, c th l h thng bnh m c trong bi th nghi m c thnh ph n tch phn.

+ iu khin truyn thng p ng r t nhanh v i nhiu, do v y nu cm bin o khng chnh xc hc sais m hnh l n s lm m c n c trong bnh b trn qu ho c cn ht.

- Sch lc iu khin t l: Do ng uyn l iu khin ca sch lc iu khin t l l duy tr quan h giahai bi n iu khin nhm iu khin gin ti p bin th 3. Nhng iu khin bnh m c ch c mt biniu khin l F1 nn khng th p dng

- Sch lc iu khin la chn: D o iu khin la chon cng yu cu t nh t hai bi n iu khin m h thng bnh m c ch c mt bin iu khin nn khng xc nh c tn hi u iu khin ln t ln khngth p dng sch lc iu khin ny.

III. THIT K B IU KHIN PID CHO I T NG:

- S dng phng phpZiegler Nichol 1 v phng php ny p dng cho i tng c c tnh l m tkhu qun trnh b c nht hoc khu tch phn c th i gian tr tng i nh.

Dng hm truy n ca b iu khin PID:

(s) = Kp( 1 + + )

Kp

P = 0.4 - -

PI = 0.36 3.3T=9.24 -

PID = 0.48 2T=5 0.5T= 1.4


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2. M Phng B iu khin PID trn Simulink

Nh iu ch nh cc thng s , , ta s c c nhng b iu khin P, PI v PID tng ng nhhnh v trn.

IV. M HNH CC SCH LC IU KHIN.

1. Sch Lc iu Khin Truyn thng.

1.1. M Hnh Sch L c Truyn Thng :

F2

SP F1 hB iu khi n

PID

Qu Trnh


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1.2. Lu P&ID Ca Sch L c Truyn Thng.

Kp Ti TdPID 0.48 5.6 1.4

1.3. M Hnh Trn Simulink.


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- Ta c p ng ca h thng:

- v i hai b iu khin PI v P ta cng c k t qu sai khc r t t so v i 2 k t qu trn.

1.4. Nhn Xt

+ khi b t u qu trnh m ph ng, khi t mc n c trong bnh th m c nc trong bnh tng lin tc v

nhanh chng lm trn bnh trong khi th i gian m ph ng vn cn.+ Sau khi m van 2, v n gi nguyn gi tr t th m c n c trong bnh l i gim lin tc n cn bnh.

=> h thng khng n nh.

=> khng th p dng c sch lc ny iu khin h thng.

2. SCH LC IU KHIN PHN HI VNG N.

2.1. M Hnh Sch Lc iu Khin Vng n

F2

SP F1 hQU TRNHB IU KI N


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2.2. Lu P&ID Cho Sch Lc iu Khin Phn Hi Vng n.

2.2. M Hnh M Phng Trn Simulink.


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2.3. S Dng Sch Lc iu Khin Phn Hi Vng n V i Ba B iu Khin P, PI, PID.

2.3.1. Tr ng h p s dng b iu khin P v i Kp=0.4 .

- p ng qu

- Nhn Xt:

+ H thng hot ng n nh, qu iu ch nh nh.+ Thi gian qu ngn.

+ Lu l ng ra n inh

+ tn hi u ra b lch so v i tn hi u t nh hnh d i


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Nh Vy, s dng b iu khin P r t ngin (c bit h m ph ng bnh n c bn thn h c thnh phn tch phn), tc ng r t nhanh m c cht lng gn st v i gi tr t v c tnh n nh cao.

2.2.2. Tr ng h p s dng b iu khin PI v i Kp = 0.36 v = 9.24.

- p ng ca h thng

- Hin T ng: Khi ti n hnh m ph ng th x y ra hi n t ng mc nc tng vt trong th i gian ng n kodi th i gian qu . y l hin t ng bo ha tch phn (Reset Windup) x y ra khi c thnh ph n tch

phn trong b iu khin.

-Nhn Xt:

+ qu iu ch nh l n.

+ Thi gian qu l n.

+ d tr n nh gim do c thnh ph n I trong b iu khin.

Nn khc phc tnh tr ng ny t c n them m t khu ch ng bo ha tch phn vo b iu khin.


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2.2.3. Tr ng h p s dng b iu khin PID v i Kp = 0.48 , =5.6 v = 1.4

- p ng ca h thng

- Hin T ng: Gn ging v i khi s dng b iu khin PI nhng thi gian qu di hn

- Nhn Xt: + qu iu ch nh l n

+ Thi gian qu di

+ Tn ti sai l ch tnh l n


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3. S dng sch lc iu khin phn hi vng ndng cc b iu khin P, PI-RW v PID-RW.

3.1. S b iu khin:

- V i = 0.5 v thng s cn t cho kh i Saturation l : Upper limit :1

Lower limit: 0

3.2. K t qu m phng.

3.2.1. i v i b iu khin PI- chng bo ha tch phn v i Kp= 0.36, Ti=9.24, Tt=4.12

- p ng ca h thng


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- Nhn Xt:

+ gim c hin t ng baoc ha tch phn , tn hi u ra nhanh chng bm t i tn hi u t m khnggiao ng qu nhi u , qu trnh nhanh chng i n n nh.

+ qu iu ch nh nh..

+ Thi gian qu nhanh.

+ Gim c sai l ch tnh xung mc th p.

+ p ng ra bm st v i tn hi u t v i sai l ch tnh rt nh nh hnh d i c hai b iu khin


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3.2.2. i v i b iu khin PID- chng bo ha tch phn v i = 0.48, = 5.6, = 2.8 v

=1.4 .

- th p ng qu :

- Nhn Xt:

+ qu iu ch nh nh.

+ Thi gian qu nh v nh hn thi gian qu khi dng b PI- ch ng bo ha tch phn

( do c thnh ph n vi phn trong b iu khin).

+ sai l nh tnh gim xung mc th p.


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4. S dng sch lc iu khin tng.

4.1. L do cn s dng iu khin tng

- vng iu khin phn hi vng n cha p ng c yu c u cht l ng.

- iu khin tng c cc u im nh: Ci thin kh nng loi b nhiu cc b, gim qu iuchnh, ci thin tnh n nh ca ton h kn, nng cao tnh b n vng ca h kn.

- sch lc iu khin tng c t c p ng nhanh hn so vi iu khin phn hi. Do iu khintng c vng ngoi iu khin thng, ph n ng nhanh v i nhiu v tri t tiu nhi u ny.

4.2. Xc nh cc vng iu khin, nhim v v c im ca t ng vng.

+ Vng 1 (vng ngoi): o mc ca bnh r i phn hi li so snh v i SP1 cho ta lu l ng vo mongmun. ( gi tr SP2 c a vng 2).

+ Vng 2 (vng trong): o hiu lu l ng Out Flow & In Flow, ph n hi & so snh v i gi tr uvo. (ch n Kp=10000).

c im:

+ H s khuch i K p ca vng ngoi r t ln nn p ng nhanh v i s thay i ca nhiu u vo.(Chn K p = 10000 cho b iu khin P vng ngoi).

+ Cc thng s trong b PID-2 (vng trong) d a theo tnh ton trn.

4.3.iu khin tng khng o lu l ng ra.

4.3.1. Cu trc:

h

- -

B iu khi n sc p

B iu khi n th c p

Qu trnh

In Flow


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4.3.2. Lu P&ID.

4.3.3. M phng trn Simulink.


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4.3.4. S dng 2 b iu khin P v i : Kp = 10000; : Kp = 0.4

- p ng ca h thng:


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- Nhn Xt: H thng n nh vi qu iu ch nh v th i gian qu nh. Tuy nhin v n cn s chnh l ch gia Set Point & Level; gi a Out Flow v In Flow.

4.3.5. S dng b iu khin P/PI-RW v i : Kp = 10000; : Kp = 0.36, = 8.34 v

.( hiu chnh)

- p ng ca h thng:


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- Nhn Xt:

+ H thng n nh v i thi gian qu nh

+ qu iu ch nh nh.

+ Sai l ch tnh tng i l n.

+ Vn cn s chnh l ch kh ng k gia Set Point v i Level, gi a Out Flow v i In Flow .

+ p ng ra b nh h ng khi nhi u qu trnh( m van 2) l n.

.


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4.4. iu khin tng o lu l ng ra.

4.4.1. Cu trc b iu khin.

SP + + h

_ + _

4.4.2. Lu P&ID.

B iu Khi nS C p

B iu Khi nTh C p

Qu Trnh

In F Out F


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3.4.3. M phng Trn Simulink

4.4.4. S dng b iu khin P/P.

- p ng ca h thng:


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-Nhn xt:

+ p ng ra lun bm st gi tr t vi qu iu ch nh th p, thi gian qu nh, sai l ch tnh nh

ng k, sai l ch gia lu l ng vo v i lu l ng ra nh . Ngoi ra khi nhi u c gi tr l n nht ( m van 2 b ng 1) m c n c tr ng bnh v n n nh gi tr t (gi tr t y trong khong t 0 n 400,nu ln hn 400 th mc n c s khng cn n nh na).


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4.4.6. S dng b iu khin P/PI-RW v i : Kp = 10000; : Kp = 0.36, = 8.34 v. ( hiu chnh)

- p ng ca h thng.


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- Nhn Xt: +H thng n nh.

+ qu iu ch nh nh, thi gian qu nh nhng vn lu hn so vi tr ng h p s dng

b iu khin P/P.

+ Sai l ch gia lu l ng vo v i lu l ng ra nh .

+ H khng b nh h ng nhi u b i nhiu.


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4.4.7. So snh hai tr ng hp khng i v c o lu l ng ra.

- Tr ng hp khng o lu l ng ra:

+ khi h n nh, nu thay i l ng ra th m c n c trong bnhthay i lm cho ch t l ng ca h thng thay i (do khng

chit tiu c nhiu).

+ qu iu ch nh tng i l n, sai l ch gia Out Flow v

In Flow l ng k.

- Tr ng hp c i lu l ng ra: + Khi h thng n nh d thay i lu l ng ra th h thng vn

n nh (hn ch ti a c nhiu)

+ qu iu ch nh nh, sai l ch gia Out Flow v In Flow nh .


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BI 2

XY D NG H THNG IU KHIN HAI BNH N CTHNG NHAU

I. XY D NG M HNHCHO I T NG.

1. Xc nh cc tn hiu vo, tn hiu ra, nhiu ca h thng v s bc t do.

- Tn hi u vo: m van F1, m van F2 v m van F3 F3

- Tn hi u ra: M c ca hai bnh m c: level 1: h1 v level 2: h2 Nhi u

- Nhiu: m van F3

F1 h1F2 h2

Bin iu khin Bin cn K

- S bc t do: 5 2 = 3 v ng bng s bin vo.

Nn h c kh nng iu khin c.

QU TRNH


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2. Xy d ng m hnh ton hc cho i t ng v i cc tham s hnh th c.

- Phng trnh cn bng vt cht:

- Bin i Laplace:

() () ( ) () () ( ) - Nu khng tnh n c tnh qu tnh b t nht ca cc van, th:

{( )( ) Do c tnh van 1 l m t khu c hm truy n l m t khu qu tnh b c nht v do hai bnh thng nhaunn c tnh c a van 2 l m t khu c hm truy n l m t khu qu tnh b c nht c tr .

V th ta c 2 hm truy n t ca 2 bnh nh sau:

{ ( )( )( )( ) trong :{ 3. S dng Simulink xc nh cc tham s m hnh.

M hnh trong Simulink


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3.1.Xc nh

- p ng qu ca bnh 1 ( )

- K tim cn ca ( ) ct tr c honh t i im ng v i ta (2.06;0) . Ch n Suy ra ta tnh c hng s gc c a ti p tuyn l: = tan 4.12

Vy hm truy n t ca bnh 1 c d ng: ( )( )


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3.2. Xc nh

- p ng qu ca bnh 2 ( )

- K tim cn ca ( ) ct tr c honh t i im ng v i ta (4.3;0) . Ch n - th i gian tr : = 1.2(s).

Suy ra ta tnh c hng s gc c a ti p tuyn l: = tan 0.56

Vy hm truy n t ca bnh 1 c d ng: ( )( )


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3.2. So snh khim ch ng ta dng m hnh sau trn Simulink.

- so snh p ng ca hm truy n G1(s) v i p ng ca bnh 1:


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- So sanh p ung ca hm truy n G2(s) v i p ng ca bnh 2

- Nhn xt: Hai hm truy n tm c cha st vi p ng qu ca hai bnh m c nn ta ph i hiu

chnh li nh sau:

( )( ) () ( )

-


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- t ta c hai p ng qu gn st v i p ng qu ca hai bnh :


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II. CC SCH LC IU CHNH

1. Sch lc iu chnh c th s dng:

+ Sch lc iu khin phn hi: v sch lc iu khin phn hi c tc d ng n nh mt h thngkhng n nh c th l h thng bnh m c.

+ Sch lu c iu khin tng: v sch lc iu khin tng c tc d ng n nh mt h thng khng nnh, ngoi ra con gi m c nh h ng ca nhiu, trit tiu c sai l ch tnhcho ch t l ng tt hniu khin phn hi.

2. sch lc iu chnh khng th s dng:

- Sch lc iu khin truyn thng:

+ iu khin truyn thng khng c tc d ng n nh mt h thng khng n nh, c th l h thng bnh m c trong bi th nghi m c thnh ph n tch phn.

+ iu khin truyn thng p ng r t nhanh v i nhiu, do v y nu cm bin o khng chnh xc hc sais m hnh l n s lm m c n c trong bnh b trn qu ho c cn ht.


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III. THIT K B IU KHIN PID CHO I T NG.

- Ta s dng phng php: Ziegler Nichol

Dng hm truy n ca b iu khin PID:

(s) = Kp( 1 + + )

1. Thit k b iu khin PID cho value1

( )( ) Kp

P = 0.27 - -

PI = 0.24 3.3T=9.47 -

PID = 0.32 2T=5.74 0.5T= 1.44

2. Thit k b iu khin PID cho value2

() ( ) Kp

P = 1.43 - -

PI = 1.3 3.3T=14.2 -

PID = 1.71 2T=8.6 0.5T= 2.15


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IV. SCH LC IU KHIN PHN HI VNG N S DNG HAIB IU KHIN P V PI-RW .

1. Lu P&ID.

2. M phng bng Simulink.


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3. S dng b iu khin P.

PID valve 1: 0.27 v PID valve 2: = 1.43


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- Nhn Xt: + B iu khin P lm cho h thng n nh v p ng r t nhanh v i thi gian qu nh

v qu iu ch nh nh, khng c hi n tng giao ng quanh gi tr t.

+ Tuy nhin sai l ch tnh vn cn ng k do khng c thnh ph n tch phn gim sai l ch

Tnh trong b iu khin.4. S dng b iu khin PI-RW.

PID valve 1: ; PID valve 2:


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- Nhn Xt: +H thng hot ng n nh nhng vi d tr n nh nh hn tr ng h p dng b iu

khin P.

+ qu iu ch nh mc ch p nhn c ( do c thnh ph n I lm tng qu iu ch nh).

+ Thi gian qu nh.+ Gim c sai l ch tnh xung mc thp hn so v i khi dng b iu khin P.

V.SCH LC IU KHIN TNG O LU L NG RA.

1. Lu P&ID.


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2. Xc nh cc vng iu khin cn xy d ng.

- Vng 1 (vng ngoi): o mc ca bnh ch a(level 1 v level 2) r i phn hi li so snh v i SP1 v

SP2.

- Vng 2(vng trong): o hiu lu l ng ra Flow 1 v i Flow 2, Flow 2 v i Flow 3, ph n hi r i sosnh v i gi tr u ra c a vng 1 cho ta tn hi u iu khin( m van ph h p).

- c im ca b iu khin :

+ H s khuch i ca vng ngoi r t ln ln p ng nhanh v i s thay i ca nhiu u vo

( chn ) .+ Cc thng s trong b iu khin PID valve 1 v PID valve 2 ( vng trong) d a theo ci tnh ton

Trn.

3. M phng trn Simulink.


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4. S dng b iu khin P/P v i thng s PID valve 1: 0.27 v PID valve 2: = 1.43.

- p ng bnh 1.


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- p ng bnh 2.

- p ng lu l ng ra flow 1.


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- Nhn Xt: +H n nh. Khi tng SP1 th mc n c bnh 2 n nh hn sai lch tnh rt nh, nhng

Khi gi m SP1 th sai l ch tnh bnh 2 l i tng ln ng k.

+ qu iu ch nh nh, thi gian qu nh.

+ Lu lng n c ra van 3 n nh.

5. S dng b iu khin P/PI-RW.

- V i cc thng s PID valve 1: 0.27, v PID valve 2: = 1.43, .- p ng bnh 1


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- p ng ca bnh 2.



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- p ng u ra flow 3.


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- Nhn Xt: H hot ng n inh, khi tng SP1 th mc n c trong bnh 2 n nh hn v khi gim thmc nc trong bnh 2 giao ng vi bin nh. Nu gim mc nc trong bnh 1 n gn bng mcn c trong bnh 2 th xu t hin sai l ch tnh tng i l n bnh 2.

+ Tc p ng nhanh, qu iu ch nh nh, thi gian qu nh.

+ gim c nh h ng ca nhiu, sai l ch tnh nh.

+ lu lng n c ra van 3 n nh khng giao ng nhi u.

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Báo Cáo Thí Nghiệm Điều Khiển Quá Trình