Van hanh he thong dien

Mn hoc: Vn hanh H thng in

Chng 2

TNH TOAN PHN B TI U CNG SUT TRONG H THNG IN BNG PHNG PHAP LAGRANGE

2.1. M U Cn phai xac nh s phn b ti u cng sut gia cac nha may in trong h thng in ( co th ch co cac nha may nhit in , hoc co ca nhng nha may thuy in ) u ap ng mt gia tr phu ta tng cho trc (k ca cac tn tht) nhm nng cao tnh vn hanh kinh t cua h thng in . y la bai toan a ch tiu: - Chi ph nhin liu tng trong toan h thng la nho nht (min) - am bao tin cy hp ly - Cht lng in nng am bao... Giai quyt bai toan a ch tiu nh vy hin nay cha co mt m hnh toan hoc cht che, ma thng ch giai quyt cac bai toan ring bit, sau o kt hp lai. V vy bai toan phn b ti u cng sut gia cac nha may in thng ch xet at muc tiu quan trong la chi ph nhin liu tng trong toan h thng la nho nht. 2.2. BAI TOAN LAGRANGE: Bai toan c phat biu nh sau: Cn phai xac nh cac n s x1, x2,..., xi,........ ,xn sao cho at cc tr ham muc tiu :

F(x1, x2,..., xj,........ ,xn) min (max) (2-1)

va thoa man m iu kin rang buc: (m


Bai giai :

T 132

21 =+xx suy ra

236 1

2xx =

Thay vao ham muc tiu F :

min236

),(2

121

22

2121

+=+=

xxxxxxF

iu kin cc tr :

01

=xF

hoc la : 0)2(4

182 111

== xxxF

giai ra c : x1 = 18/13 va x2 = 12/13 Xet ao ham cp 2 :

04

264

18221

2

>=+=xF

nn ham F at cc tr tai : 1318*

1 =x va 1312*

2 =x

va khi o gia tr ham muc tiu la :

1336* =optF

Phng phap thay th trc tip trn y ch tin li khi h phng trnh rang buc la tuyn tnh va s lng m khng ln lm. Trong trng hp chung giai bai toan xac nh cc tr co rang buc la ng thc va tuyn tnh thng s dung rng rai phng phap nhn t Lagrange . Ni dung chu yu cua phng phap Lagrange nh sau: Cn phai xac nh cac n s x1, x2,..., xj,........ ,xn sao cho: F(x1, x2,..., xj,........ ,xn) min (max) (2-3) va thoa man

g1(x1, x2,..., xj,........ ,xn) = 0 g2(x1, x2,..., xj,........ ,xn) = 0

........................................ (2-4) gm(x1, x2,..., xj,........ ,xn) = 0

trong o m


Nghim ti u X*opt cua ham muc tiu F cung chnh la nghim ti u cua ham Lagrange L(X) va ngc lai v gi(x1, x2,..., xi,........ ,xn) = 0 vi moi i=1..m. V vy ta cn tm li giai ti u cho ham L(x1, x2,..., xi,........ ,xn) Bai toan Larange phat biu nh sau: Hay xac nh (x1, x2,..., xi,........ ,xn) va (1, 2,.........., m ) sao cho :

0)()()(1

=+= =

m

i j

ii

jj xXg

xXF

xXL

(2-6)

vi j=1..n va thoa man cac ieu kin rang buc : vi 0),.....,,( 21 =ni xxxg mi ,1= (2-7)

T (2-6) ta co n phng trnh va t (2-7) co m phng trnh nn se giai c (n+m) n s xj va i xac nh ham L(X) at cc tiu hay cc ai ta cn phai xet thm ao ham cp hai cua F(X) hay L(X) tai cac im dng a giai ra c trn: Nu d2L< 0 th ham F(X) ( hoc L(X) ) at cc ai va ngc lai nu d2L > 0 th ham muc tiu se at cc tiu. Ta se giai lai bai toan v du 1 theo phng phap Lagrange : Tm cac nghim s x1 , x2 sao cho : min),( 22

2121 += xxxxF

vi rang buc 132

21 =+xx

Thanh lp ham Lagrange :

=

=

+=1

1212121 ),(.),(),(

m

iii xxgxxFxxL

)132

(),( 21122

2121 +++=

xxxxxxL

Xac nh cac im dng bng cach giai cac phng trnh :

02

2)( 111

=+=

x

xXL

03

2)( 122

=+=

x

xXL

0132

21 =+xx

Giai h 3 phng trnh trn c :

1318*

1 =x va 1312*

2 =x

va khi o gia tr ham muc tiu la :

1336* =optF

( nh kt qua a nhn c bng phng phap th )

Nhom Nha may in - B mn H thng in - HBK a Nng . 16


Xet cac ao ham bc hai tai im dng:

02)(21

2

>=x

XL

02)(22

2

>=x

XL

nn ham L(X) va ham muc tiu F(X) at cc tiu tai im X* (18/13 ; 12/13). Trong trng hp ham muc tiu F(X) va cac rang buc g(X) la nhng phim ham ( tn tai tng quan gia nhng ham ) khi o tm cc tr cua cac phim ham phai s dung cac bai toan bin phn. V du nh trng hp tnh phn b ti u cng sut i vi cac nha may thuy in v khi o phai xet ti u trong ca chu ky iu tit. Bai toan c phat biu nh sau : Cn phai xac nh cac ham s x1, x2,..., xi,........ ,xn cua thi gian t sao cho ham muc tiu la phim ham at cc tr: (2-8) min(max).)',....,',',,....,,,(1

02121 = dtxxxxxxtFV

t

t nn

va thoa man m iu kin rang buc : g1(t,x1, x2,..., xj,........ ,xn) = 0 g2(t,x1, x2,..., xj,........ ,xn) = 0

............................................. (2-9) gm(t,x1, x2,..., xj,........ ,xn) = 0

Trong o : dtdx

x jj =' vi nj ,1= (2-10)

Thanh lp ham Lagrange :

(2-11) =

+=m

iii xtgtxtFxtL

1

)],().([),(),(

sau o tm cc tr cua phim ham:

(2-12) min(max).),(1

0

** = dtxtFVt

t

vi (2-13) =

+=m

iii xtgtxtFxtF

1

* )],().(),(),(

Cac gia tr xj(t) vi j = [1..n] va cac h s nhn i(t) vi i = [1..m] co th nhn c bng cach giai h phng trnh ao ham ring cua ham Lagrange va vit trong dang h phng trnh Euler nh sau :



=

=

=

0)'()(

......................................

0)'()(

0)'()(

**

2*

2*

1*

1*

nn xfdtdxf

xfdtdxf

xfdtdxf

(2-14)

Trong o :

nj

xFxf

njxFxf

jj

jj

,1 ; '

)'(

,1 ; )(

**

**

==

==

(2-15)

Kt hp n phng trnh cua h (2-14) va m phng trnh rang buc (2-9) ta se giai c (m+n) gia tr ham xj(t) va i(t) vi j = [1..n], i = [1..m]. Ngoai ra xac nh 2n hng s tch phn ta se s dung cac iu kin u :

njxtxxtx jjjj ,1 )( ; )( 1100 === (2-16)

2.3.- PHN B TI U CNG SUT GIA CAC NHA MAY NHIT IN: Xet bai toan : Co n nha may nhit in cung cp cho phu tai tng Ppt c nh. Bit nhng s liu v c tnh tiu hao nhin liu tng nha may. Cn phai xac nh cng sut phat ti u cua mi nha may Pj vi j = [1...n], sao cho chi ph nhin liu tng trong h thng at cc tiu, vi rang buc v iu kin cn bng cng sut. M ta dang toan hoc: Cn xac inh b nghim ti u P*(P*1,P*2,......,P*n) sao cho ham muc tiu v chi ph nhin liu tng at cc tiu :

(2-17) min)(),...,,....,,(1

21 == =

n

jjjnj PBPPPPfB

thoa man iu kin rang buc v cn bng cng sut :

(2-18) 0.......)(1

21 ==+++++= =

pt

n

jjptnj PPPPPPPPPPg

vi const= P const;=P ; n1,=j 0 ptjP (2-19)

Ta giai bng phong phap Lagrange : Thanh lp ham Lagrange : )()()( PgPBPL += (2-20)



iu kin ham s L(P) at cc tr :

=+=

=+=

=+=

0)()()(.............................................

0)()()(

0)()()(

222

111

nnn PPg

PPB

PPL

PPg

PPB

PPL

PPg

PPB

PPL

(2-21)

Gia thit : )(.........)()()( 21 PBPBPBPB n+++= (2-22) Khi o :

jj

j

j

n

j

j

jjj PB

PB

PB

PB

PB

PPB

==+++++= .............)( 21 (2-23)

vi gia thit j k ; 0 =j

k

PB

ngha la chi ph nhin liu nha may th k khng phu

thuc vao cng sut phat ra cua nha may th j .

Ta t jj

j

PB

= va goi la sut tng tiu hao nhin liu cua nha may th j, noi ln

nhp tng tiu hao nhin liu khi tng cng sut phat Pj , j phu thuc vao c tnh cua lo hi va turbin. T iu kin rang buc :

(2-24) 0.......)(1

21 ==+++++= =

pt

n

jjptnj PPPPPPPPPPg

ta tnh c :

1)(

.............)(

1

1

111

2

1

1

1

==+

+++=PP

PPP

PP

PP

PP

PPg ptn

(2-25)

Tng quat :

1)(

.............)( 21 ==+

+++++=j

j

j

pt

j

n

j

j

jjj PP

PPP

PP

PP

PP

PP

PPg

(2-26)

Thay vao iu kin cc tr (2-21 ) ta co h phng trnh :



=+=+=

=+=+=

=+=+=

0)()()(...........................................................

0)()()(

0)()()(

2222

1111

nnnn PPg

PPB

PPL

PPg

PPB

PPL

PPg

PPB

PPL

(2-27)

Do o iu kin cc tr la:

0...............21 =+==+==+=+ nn (2-28) hay : )( ...............21 ====== nn (2-29)

y chnh la nguyn ly phn b ti u cng sut gia cac nha may nhit in trong HT. Khi xem Ppt = const , P = const th chi ph nhin liu tng trong h thng nho nht th cac nha may phai phat cng sut Pj* ti u khi thoa man nguyn ly cn bng sut tng tiu hao nhin liu j = const. Vi c tnh sut tng tiu hao nhin liu j cua cac t may phat la ham khng giam khi tng cng sut phat Pj (thc t nh vy) ta co th chng minh ham muc tiu B(P) at cc tiu bng cach xet thm cac ao ham cp hai va co c:

0)(dhay 0)( 222

PLP

PL

j (2-30)

Nu xet tn tht cng sut phu thuc vao cng sut phat Pj ngha la: P = P(P1,P2,.....,Pn)

iu kin cc tiu cua ham Lagrange co th vit :

=

+=+=

=

+=+=

=

+=+=

0)1()()()(.........................................................................

0)1()()()(

0)1()()()(

22

222

11

111

nn

nnn PP

PPg

PPB

PPL

PP

PPg

PPB

PPL

PP

PPg

PPB

PPL

(2-31)



Khi o, nguyn ly phn b cng sut ti u la :

n

n

PP

PP

PP

==

=

1

..............11

2

2

1

1 (2-32)

i

i

PP

1 goi la sut tng tiu hao NL khi co xet n tn tht P

Qua o cho thy khi P = const th cho ta kt qua iu kin phn b ti u cng sut nh a trnh bay trn. T nguyn ly cn bng sut tng tiu hao nhin liu nay, ta co th tm ra c nghim ti u P* = (P*1,P*2,.......,P*n). 4.4. THU TUC PHN PHI TI U CNG SUT : Vic phn phi ti u cng sut gia cac nha may nhit in c tun theo nguyn ly cn bng v sut tng tiu hao nhin liu . Sut tng th hin nhp tiu tn nhin liu khi tng cng sut P phat ra. V vy theo nguyn ly phn phi trn y at cc tiu nhin liu tiu hao trong toan h thng, nha may co nho se nhn phat nhiu cng sut va nha may co ln (ngha la lam vic khng kinh t) se phai phat t cng sut. Nguyn ly nay th hin tnh cng bng trong phn phi ti u. Cn quan tm nhng c im sau: 4.4.1. Sut tng tiu hao nhin liu va sut tiu hao nhin liu :

Cn phai phn bit ro sut tng tiu hao nhin liu va sut tiu hao nhin liu .

ng vi mi nha may nhit in co th xy dng c ng c tnh tiu hao nhin liu B phu thuc cng sut phat ra P nh hnh 2-1. Gia s t may phat ang lam vic im a :

BP

tgaa

a= = (2-33)

a: goi la sut tiu hao nhin liu cua nha may ng vi im a [kg n.lieu/KWh ]

]n.lieu/KWh [kg tgdPdB

aa == (2-34)

a: goi la sut tng tiu hao nhin liu.



Hnh 2-1 T O ve tip tuyn Ob, im b goi la im lam vic kinh t , tai im lam vic nay cng sut phat la Pkt ng vi chi ph nhin liu la Bkt . Khi P > Pkt th theo c tnh ta thy sut tng tiu hao nhin liu tng nhanh, cang tiu hao nhin liu. V vy theo quan im kinh t tit kim nhin liu ch vn hanh vi P


dPdQ

L = - goi la sut tng tiu hao nhin liu cua tucbin [Kcalo/KWh]

ng c tnh sut tng tiu hao nhin liu cua lo hi L thng co dang ng cong (hnh 2-3a) tuy thuc cac loai lo hi khac nhau.

Hnh 2-3

ng c tnh tiu hao nhit lng Q cua turbin trong nhiu trng hp co dang

o nhit lng cua turbin T la gia tr ao ham cua

ng phap gia cng toan hoc, chng han phng phap bnh phng cc tiu xy dng c quan h giai tch B = B(P). T o xac nh c c tnh s

t trng hp tn tht cng sut la hng s, khng phu thuc vao cng sut phat

i mi nha may ta xy dng c quan h sut tng tiu hao nhin liu phu

ng cong j ta xy dng c ng cong (P) cua toan h ng g

,12 Ppt ), nh cach lam m ta trn hnh ve ta xac

gn tuyn tnh (hnh 2-3b). ng c tnh co ch gay khuc ng vi gia tr Pkt, iu o giai thch khi van qua tai m, nhit lng tng nhanh va tnh kinh t giam t ngt. ng c tnh sut tng tiu hang Q theo P. T cac ng T va L xy dng c ng c tnh sut tng tiu hao nhin liu cua t may nh hnh 2-3c. Ngoai ra xy dng c tnh sut tng tiu hao nhin liu cua t may hoc nha may in co th thc hin bng cach thng k cac tp s liu B va P trong cac ch vn hanh khac nhau va nh cac ph

ut tng tiu hao nhin liu. 4.4.3.Thu tuc phn phi ti u cng sut : Xecua cac nha may. Gia s ta cn phai phn phi cng sut Ppt cho n nha may, ta tin hanh nh sau: - Vthuc vao cng sut phat j = j(Pj) vi j = [1..n] bng dang giai tch hoc bng s cho theo bang . - Da trn cac th m n nha may, bng cach gi nguyn tr s trn truc tung, cng n gia tr cng sut P trn truc hoanh. - Cn c vao phu tai tng cng Ppt cn cung cp k ca tn tht cng sut P (trong tnh toan s b co th ly bng 0,07 - 0



nh c cac g yia tr ti u cng sut phat ra t cac nha ma in Pj* thoa man iu kin cn bng sut tng tiu hao nhin liu: 1 2= = = = = = ....... ........ ( )n n

tuc phn phi nh trn cn phai chu y: 1. Khi gia nhin liu nha may th i nao o khac gia nhin liu tiu chun th cn hiu chnh i thanh i theo :

va thoa man iu kin cn bng cng sut. ptNJ PPPPPP +=+++++

***2

*1 .......

Ta nhn thy nha may nao co sut tng tiu hao nhin liu cang nho th nhn cang nhiu cng sut. Khi tin hanh thu

0

.'aai

ii =

Trong o : ai la gia nhin liu cua nha may th i va a0 la gia nhin liu tiu chun, t o ta thy rng nha may nao co gia nhin liu cang t th ch nn phat t cng sut. 2. Co th xay ra trng hp tm ra nho hn ng vi cng sut cc tiu Pmin

y phn phi ti u ta se nha may co nho nhn thm cng sut trc, nhng ui cung cung phai am bao i bng nhau vi moi nha may th i va phai ap ng y u

g sut trong h thng gm cac nha may thuy

t la thi gian gia 2 ln thao nc va tr nc k tip nhau. Tuy

y li, thi tit v.v.... V vy ch lam

hoc ln hn ng vi cng sut cc ai cho phep Pmax th khi o ch cho nha may nhn cng sut Pmin hoc Pmax v o la gi han kha nng phat cng sut cua nha may. 3. Thng trong thc t vn hanh ngi ta ch cho bang sut tng tiu hao nhin liu va Pi thay cho ng c tnh d phn b hn. Khi phu tai tng ln th theo nguyn lcphu tai. 4.5. PHN B CNG SUT TI U GIA NHIT IN VA THUY IN: Trong vn hanh khng phai nha may thuy in lun lun phat ht cng sut la ti u mc du no co nhiu u im la gia thanh in nng re, khng tiu hao nhin liu... Ch tiu ti u cua s phn b cnin va nhit in la lam cc tiu chi ph nhin liu nhit in, ng thi phai thoa man iu kin thuy nng nha may thuy in. Ch ti u ch xet i vi nhng thuy in co h cha nc, ngha la co kha nng iu chnh dong chay vao tuc bin ( goi la kha nng iu tit ) Chu ky iu titheo dung tch h cha thng phn nha may thuy in iu tit theo ngay, tun, mua, nm hoc nhiu nm. Trong mt chu ky iu tit lng nc tiu ph cho nha may thuy in la khng i va c xac nh bi nhng iu kin v thuvic ti u cua thuy in phai xet trong toan b chu ky iu tit va iu kin rang buc y chnh la lng nc tiu hao a qui nh.



Ngoai ra co nhng thi gian nha may thuy in buc phai lam vic theo ch gii han va vn phn b cng sut ti u khng cn t ra. Chng han i vi thuy in ch phat in khng co yu cu v giao thng, thuy li... thi im phu tai cao

nh ( cn phai tit kim nc mua nc can ), hoc thuy

p :

a may nhit in ng tr theo iu kin cn bng sut tng tiu hao nhin liu .

(2-35) V xet trong chu ky iu tit nn ta phai xet B con phu thuc vao t va xet ca s

im phai am nhn phu tain khng co h cha, h cha nho phai tn dung ht thuy nng nn phai phat ht cng sut ngha la nhn phn phu tai nn (xem giao trnh Nha May in ). Ta xet trng h Co n nha may thuy in lam vic trong h thng cung vi mt s nha may nhit in ma ta xem nh mt nh

Goi B la lng tiu hao nhin liu nha may nhit in ng tr trong mt n v thi gian. ( n v la tn/h )

),,( ,NDND PPtBB =

thay i cua PN theo thi gian t :

dtdP

P NDND =,

Goi Qi la lu lng nc tiu hao trong mt n v thi gian nha m3

ay thuy in th i [ m /s ].

n1,=i voi),,( ,TDiTDiii Lng nc qui nh i vi thuy in th

PPtQQ = (2-36) u ti

Khi o bai toan c phat biu nh sau : c nha may in liu:

(2-37)

thoa man cac rang buc v lng nc tiu hao i vi cac nha may thuy in:

,222 WdtPPtQ

T

TT = (2-38)

i trong chu ky i t T:

=T

ii dtQW 0 .

Xac nh cng sut phat cua nha may nhit in ng tr PN va cua cathuy in PT1, PT2,.........., PTn sao cho at cc tiu ham muc tiu v chi ph nh

min).,,( ,0

dtPPtB NT

N

.),,( 10,

111 WdtPPtQT

TT = .),,( 20

.....................................

.),,(0

,n

T

TnTnn WdtPPtQ = va thoa man rang buc v iu kin cn bng cng sut:

0.....),( 21 =++++= PPptPPPPPtg TDnTDTDND (2-39)



Ta giai bai toan ti u nay theo phng phap Lagrange nh a trnh bay muc t ta lp phim ham Lagrange:

+++TT

PtgdtPtQdtPtQ ),().,(........).,(

1, 2,......, n : la nhng h s khng xac nh a vao cac phng trnh rang buc theo iu kin lu lng nc.

s khng xac nh a vao phng trnh rang buc cn bng cng sut. T y tm cc tiu cua phim ham L(t,P) :

01

++= =

T

i

n

ii PtQPtB

ii

=

th

tm nghim cua bai toan ta lp h phng trnh Euler di dang:

2.2. Trc h

+=T

dtPtBPtL ).,(),( tnn 00 110 Trong o:

t : h

min)].,(,( t dtPtgPtL ),(),([)

t ),(),(),(),(* PtgPtQPtBPtFn

++= ti1

min).,(*),(0

= (2-40)

TdtPtFPtL

0'** = PiPi fdf dt

(2-41)

o : Pi la cng sut cu ay nhit in n ,

i N T1 T2 Tn

Trong a nha m g tr PN va cac nha may thuy in PT1PT2,...,PTn. P la cac ao ham P ,P , P ,........,P

i

PiP

PtFf

),(** = va i

PiP

PtFf'

),(*'

*

= (2-42)

c h phng trnh Euler dang : Ta

..............................

1TD

=

+

=

+

=

+

0)1('

(

......................................

0)1('

(

0)1('

11

111

tnn

n

TDt

TD

NDt

NDND

PP

PQ

dtd

PQ

PP

PQ

dtd

PQ

PP

PB

dtd

PB

(4-43)

vi gia

TDnTDnTDn

thit : Ppt = hng s



Ta k hiu :

=NDPB - sut tn

trong ch c lp.

goi la g tiu hao nhin liu nha may nhit in

xa

11 qQ = , 2

2

2 qPQ

TD

= ,....... t tng tiu hao nc nha may thuy in

1,2,.... trong ch xac lp. - la su

1PTD

Nhn thy cac thanh phn :

'. '

=

Bd va iTD

qQid '. =Pdt '

NDPdtxut hin trong qua trnh bin i ch lam vic cua h i itc bin i theo thi gian cua cng sut nha may in.

h phng trnh (4-43) kh t ta co :

thng va , q phu thuc vao

Thng ta gia thit i = 0, qi = 0 ; khi o t

TDnTDND QP

QP

PP

1111

1

1

Nu xem tn th

nq (4-44) q === ...........1

t cng sut khng i th: nn qqq ............ 2211 ==== (4-45)

y la nguyn ly cng bng cua vic phn b ti u cng sut gia cac nha may in theo sut tng tiu hao nhin liu, trong o i vi thuy in i co ai din la sut tng ng tr la i.qi. Nhng ga tr cua i la nhng hng s ng vi nha may thuy in i va c chon

n ti u cua bai toan a nu. Sau y ta se x s va xy dng thu tuc phn phi cng sut ti u gia

4.6. C IM VA THU TUC PHN PH4.6.1. Y ngha cua h s

trong chu ky iu tit nhm thoa man iu kiet thm y ngha cua cac h i

nhit in va thuy in.

:

Trong trng hp n gian khi khng xet n s thay i cua cng sut trong mang in, t biu thc (4-45) ta co :

tdindi dPdQi

dPdB

qi :== (4-46)

Gia thit rng s thay i cng sut phat r i cng sut phat ra nha may nhit in, chng han khi nhit in phat cng sut giam i th t

a nha may thuy in th i la do thay

huy in i phai phat cng sut tng ln. Mt cach gn ung v gia tr tuyt i ta xem nh : dPt = dPn. Nh vy tng quat ta co th vit :



1,2...n=i voidQidBi = (4-47)

Nh vy i c nh ngha la s bin i cua tiu hao nhin liu nha may

i thuy in lam vic vi ln th nhin liu tit kim c nhit

t cua thuy in i tng ln, thng gia tr cua sut tng tiu hao nc qi cua no se tng, khi o do

c i cn phai

va lai dn n gia tr khng i ban u.

c (4-45).

n liu, xy dng ng c tnh cho nha may nhit in ng tr (hnh 2-4).

t la moi gia tr i la nhng hng s a cho, xy dng cac ng c tnh iqi cho cac nha may thuy in i=1,2,...,n (hnh 2-

4 tn

nhit in theo s thay i cua lu lng nc nha may thuy in i. Th nguyn cu i la [ tn nhin liu/m3 nc ] va chnh i la ch tiu phan anh hiu qua s dung nc nha may thuy in i. Khin trn 1m3 nc cang nhiu, do o goi la h s hiu qua s dung nng lng cua thuy in. Ngoai ra cn chu y rng co ch lam vic ti u ga tr i cua mi nha may thuy in sau khi xac nh cn gi khng i trong sut chu ky iu tit. iu o c giai thch nh sau : Gia thit thi im nao o ga tr i c chon tng ln. Khi o tit kim nhin liu nhit in cn tng cng sut phat thuy in i. Nhng v lng nc trong chu ky iu tit a xac nh nn khi tng cng sut thuy in se tng lng nc tiu hao va bt buc phai giam cng sut thi im khac. Mt khac, cng sut phang sut phat cua nhit in gam i nn gia tr cua giam, v vy = /q la

chon giam i. Tom lai, khi tng i ta cn phai tng Pti, nhng khi Pti tng ( Pn gam ) se lam giam i va khi i giam tit kim nhin liu ta lai cn phai giam Ptitng i. Qua trnh tip tuc cho n khi i tr v 4.6.2. Thu tuc phn phi ti u cng sut gia nhit in va thuy in: Vic phn phi ti u cng sut gia nha may nhit in va thuy in trong HT da trn nguyn ly cn bng sut tng tiu hao nh trn biu th Thu tuc phn phi tin hanh nh sau: - i vi cac nha may nhit in cn c vao nguyn ly cn bng sut tng tiu hao nhi - i vi tng nha may thuy in, cn c vao lng tiu hao nc Qi va cng sut phat Pti ta xy dng ng c tnh sut tng tiu hao nc qi. - Trc ht khao sat trng hp n gian nh

).

- T gia tr ph ng cua h th pt k ca tn tht trong mang trn th sut

ng tiu hao nhin liu tng HT (hnh 2-4 ) ta xac nh cac gia tr ti u v cng sut cua hit in va cac thuy in P*n,P*t1,P*t2,.....,P*tn.

u tai t ng P



1q1 nqn

P*N PN P*T1 PT1 P*Tn PTn

Hnh 2-4 Tuy nhin trong thc t thng cac ga tr cua i cua thuy in phai xac nh theo iu kin ti u ma khng bit trc, v vy thu tuc phc tap hn. Nh a phn tch, ch lam vic ti u cua cac nha may thuy in phai am bao 2 muc tiu : - at cc tiu tiu hao nhin liu trong cac nha may nhit in. - at lng tiu hao nc Wi trong chu ky iu tit nh qui nh. T y thy rng phai chon cac gia tr i mt cach hp ly.



Hnh 2-6

T hnh 2- 4 ta thy rng nu nha may thuy in i nao o nu chon gia tr i ln th ng c tnh iqi nng cao ln do o cng sut phat cua thuy in th i se giam i va dn n lng nc trong chu ky iu tit nho hn qui nh. V vy trong trng hp tng quat thu tuc phn phi ti u cng sut gia nhit in va n nha may thuy in c tin hanh gn ung theo thut toan trn s hnh 2-5. Trong mt s trng hp do kho d bao chnh xac lng nc trong chu ky iu tit dai nn thng xac nh ch lam vic cua thuy in theo lng nc tiu hao trung bnh trong mt ngay m Qtb . Vi

Hnh 2-5




nhng gia tr chon khac nhau, gia tr cua QBtb B ta co th xy dng theo ng c tnh nh hnh 2-6, da theo th phu tai cua thuy in. T y cung thy rng khi chon ln, cng sut PBTB se nho, dn n QBtb B nho . Trong trng hp co mt nha may thuy in, vic xac nh gia tr co th n gan suy t gia tr QBtb B qui nh. Khi co nhiu thuy in vic xy dng cac ng QBtb B cung phc tap, luc o thng chon cac h s Bi B theo phng phap dn ung nh a nu . Cn chu y rng cac gia tr c chon co tuy thuc vao tnh thi tit. Chng han vao mua nc ln khi h khng cha ht toan b lng dong chay, cn chon nho, co th dn n Bq B nho hn ca gia tr cc tiu cua nhit in, nh vy QBTB se ln, thuy in se phat toan b cng sut, nhit in ch am bao phn phu tai con lai. Tng t khi nc can co th thc hin chon ln . Trn y khi xet ch lam vic ti u cua nhit in va thuy in ch nhm thoa man ch tiu cc tiu chi ph nhin liu va am bao cng sut phu tai h thng. Trong thc t vic chon cac tham s con phai thoa man nhng ch tiu khac nh mc nc qui nh ha lu phai am bao, cac ch tiu v cht lng in nng nh in ap v.v...

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