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Ngy 27 thng 9 nm 2014 TNH TON NGN MCH PHN II - VOER

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TNH TON NGN MCH PHN II

KHOA HC V CNG NGH

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Ngn mch 1 pha chm t.

Ma trn tng dn ngn mch 1 pha chm t pha a thu c t bng 7.1.

Dng ngn mch v in p nt thu c bng cch thay th

t phng trnh (7.24) vo trong (7.16), (7.18) v (7.20). Dng ngn mch ti nt p l:

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Bin i n gin ta c:

Thnh phn pha ca dng ngn mch ti nt p c th thu c bng cch nhn c hai v ca phng trnh

(7.25) bi T . Dng thu c l:

in p ngn mch ti nt p l:

Bin i n gin ta c:

Thnh phn pha ca in p ngn mch.

s



in p ti cc nt khc nt p l:

Bin i n gin ta c:

Cc thnh phn pha l:

Cc cng thc thu c trong cc mc trn tng kt trong bng 7.3. in p ca mt pha i vi t xem nh

mt n v so vi gc qui chiu. Cng thc trong bng 7.2 bao gm in p mt pha i vi t, n c th xem

nh mt n v. Dng lc ngn mch trong cc nhnh ca mng in c th tnh ton t cng thc (7.21).



7.4. TNH TON NGN MCH BNG CCH DNG Z VNG

Dng v p lc ngn mch c th tnh ton bng cch dng ma trn tng tr vng cho h thng n gin trnh

by trong hnh 7.2. Dng in vng ca h thng in n gin l bng 0 trc lc ngn mch khng ch

n tt c cc dng nt. l cn thit v vy kt qu tnh ton dng in vng trong tng dng ngn mch

xc nh dng v p ngn mch. Tnh ton ngn mch c th thc hin c bng cch tnh theo h thng 3

pha hoc l tnh theo cc thnh phn i xng. Phng php sau y s biu din bng cch dng h thng 3

pha.

S nhnh ca h thng 3 pha n gin bng s nhnh ca mng in cng vi s my pht tng ng. S nt

bng s nt n cng vi t, ngha l bng n+1. S nhnh cy hay s vng c bn ca h thng n gin l:

l = (e + e ) - (n + 1) + 1

Hay

l = e + e + n

Vi e l s nhnh ca h thng 3 pha v e l s my pht tng ng 3 pha.

Ngn mch ti nt p tng ng vi cng thm mt nhnh cy t nt n t.

Dng din t h thng trong hnh 7.3, in p lc ngn mch l:

n q

n q

q



(7.26)

Trong vect

biu din thay i trong in p nt thu c t in p nt ngun

lc ngn mch.

Phng trnh t tnh ca mng in trong dng vng nh sau.

Cho h thng ngn mch trnh by trong hnh 7.3, vect in p vng bit l:

Kch thc ca ma trn tng tr vng, n bao gm c vng ngn mch l 3(l + 1) x 3(l + 1). Vevt dng in

vng cha bit trong ngn mch l:

Trong

l dng in lin kt vi vng ngn mch. Dng in vng c th tnh ton t.

Dng in trong tt c cc nhnh ca mng in lc ngn mch c th tnh nh sau:

(7.27)

Vi C l ma trn vng hng c bn trn 3 pha. Vect dng c th phn chia nh sau:

n n



Vi

: L vect dng in trong nhnh b cy

: L vect dng in trong nhnh cy

Do vect in p thay i l:

Vi K l ma trn ng dn - nhnh b cy c bn trn 3 pha.

: L ma trn tng tr gc i vi nhnh b cy

in p nt lc ngn mch thu c bng cch cng thm in p thay i vi in p trc lc ngn mch.

Phng trnh (7.26) tr thnh.

(7.28)

Dng ti nt ngn mch l ging nh dng trong vng ph l

. Phng php c th l mt cng vic biu din tnh ton ngn mch ti nhiu v tr trong h thng bng cch cng thm

cc nhnh cy, ti mi thi im, gia nt ngn mch vi t. Yu cu hnh thnh v nghch o ma trn tng tr vng

cho mi v tr ngn mch khc nhau. Php ton ma trn cn thit i hi cung cp d liu ngn mch cho mt s ln v tr

v vy n tn rt nhiu thi gian.

Phng php tng bc, mi nhnh cy ng thi c cng thm vo gia mi nt v t, yu cu hnh

thnh ma trn tng tr vng n v ch nghch o mt ma trn con. Trong phng php ny, dng in trong

vng ph l thay i ng vi tng v tr ngn mch khc nhau. Dng in pha c xem nh l lin kt trong

vng ph vi nt ngn mch p n ph thuc vo dng ngn mch. Xem dng in pha l mt n v, dng in

trong vng ph th p l:

i vi ngn mch 3 pha.

i vi ngn mch 1 pha chm t (trn pha a).



i vi ngn mch hai pha (gia pha b v pha c).

Dng trong tt c cc vng ph khc l xem nh bng 0.

Vect in p, dng in v ma trn tng tr vng trong phng trnh biu din cho ton mng in, bao gm

vng ph c th phn chia nh sau:

Trong phng trnh (7.29) vect

v

c xem nh l cc vect dng in v in p vng trong h thng n gin v

v

c xem nh l cc vect dng in v in p vng ph.

Vect

c th tnh ton cho ngn mch ti nt p t phng trnh (7.29) bng cch xem dng in trong vng ph l:



Trong

c xem nh vect dng 3 pha ca vng ph th p. T phng trnh (7.29) ta c.

(7.31)

T

phng trnh (7.31) tr thnh.

i vi h thng n gin dng in vng tm c l:

(7.32)

in p vng ph t phng trnh (7.29) l:

Th

vo trong phng trnh t (7.32) ta c .

(7.33)

Phng trnh (7.33) xc nh ngun in p vng ph, t dng in vng ph tnh bi phng trnh (7.30).

Thc t xc nh dng ngn mch vi ngun in p trong vng ph th p

phi bng in p nt th p trc ngn mch. Tnh ton ngun in p ca vng ph th p



thu c t phng trnh (7.33) dng tng ng tnh ton dng in. Dng ngn mch thc t ti nt p l:

i vi pha a:

(thc t) =

(tng ng)

i vi pha b:

(thc t) =

(tng ng)

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

Dng in vng

ca h thng n gin c th thu c t phng trnh (7.32) dng dng in vng hin ti. Dng nhnh b cy c th

tnh ton t phng trnh (7.27) v in p nt, sau c th xc nh t phng trnh (7.28).

Trong phng trnh (7.33) xem dng in vng ph

trong cc nhnh cy ph kt ni cc nt ca mng in vi t v v vy n c xem l dng nt. in p vng ph

l in p nt thu c t dng in hin ti. Trong phng trnh (7.33).

V vy trong phng php ma trn tng tr vng dng xc nh ma trn tng tr nt cho tnh ton ngn

mch.

7.5. CHNG TRNH M T TNH TON NGN MCH

Phn ln nghin cu ngn mch l ch cn tnh cho ngn mch 3 pha v 1 pha chm t. Tnh ngn mch

trong h thng in l nhm mc ch tnh ton nhng s c, dng ma trn tng tr nt th t thun v th t

khng biu din nh mc 7.3 v h thng in n gin trnh by trong mc 7.2.



D liu nhp vo din t h thng l thuyt thnh lp ngun nng lng v cc bin i trung gian. D liu

cho my pht, b t, s im ni v in khng th t thun th t khng. V l thuyt 1 pha gm 2 thnh

phn, thnh phn th nht l cho mi mt im ni dc theo chiu di ng dy l mt in khng ng

dy, thnh phn th hai ca ng dy l in khng tng h i hi gia hai dy vi nhau. My bin p v

l thuyt c xem nh mt im ni ti mi trm vi s cun dy, s kt ni v in khng th t thun th t

khng ca n.

Chng trnh tnh ton u tin gn cho mt dy s nt v sp xp h thng d liu cho thun li, hnh thnh

cc ma trn tng tr nt th t thun, th t khng. Kim tra v biu din d liu trong mi pha. Tip theo,

thit lp cc ma trn tng tr nt th t thun, th t khng. Ma trn c lu tr tm thi trong vng nh ph

cung cp cho chng trnh tnh tip theo. Sau cc ma trn tng tr nt th t thun v th t khng

c gi ra dng trong tnh ton ngn mch. T ma trn i xng ch lu tr cc thnh phn trn ng

cho. Chng trnh tnh ngn mch th t tng bc c trnh by trong hnh 7.6.

S thut ton tnh ton ngn mch trong h thng in:

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