gt cung cap dien

Gio trnh Cung cp in http://www.ebook.edu.vn

Bin son: Nh Trng 3

LI NI U

t nc Vit Nam trong cng cuc cng nghip ho - hin i ho, nn kinh t ang trn pht trin. Yu cu s dng in v thit b in ngy cng tng. Vic trang b kin thc v h thng in nhm phc v cho nhu cu sinh hot ca con ngi, cung cp in nng cho cc thit b ca kh vc kinh th, cc khu ch xut, cc x nghip l rt cn thit.

Vi mt vai tr quan trng nh vy v xut pht t yu cu, k hoch o to, chng trnh mn hc ca Trng Cao ng Ngoi ng - Cng ngh Vit Nht. Chng ti bin son cun gio trnh Cung cp in gm 5 chng vi nhng ni dung c bn sau:

- Chng 1: Tnh ton ph ti in

- Chng 2: Tnh ton tn tht in p, tn tht cng sut.

- Chng 3: La chn cc thit b in trong li cung cp in.

- Chng 4: Nng cao h s cng sut.

- Chng 5: Tnh ton chiu sng.

Gio trnh cung cp in c bin son phc v cho cng tc ging dy ca gio vin v l ti liu hc tp ca hc sinh.

Do chuyn mn v thi gian c hn nn khng trnh khi nhng thit st, vy rt mong nhn c kin ng gp ca ng nghip v bn c cun sch t cht lng cao hn.

TC GI


Bin son: Nh Trng 4

GII THIU CHUNG V CUNG CP IN 1. LI IN V LI CUNG CP IN H thng in bao gm ba khu: ngun in, truyn ti in v tiu th in Ngun in l cc nh my in (thu in, nhit in, in nguyn t ) v cc trm pht in (izen, in mt tri).

Tiu th in bao gm tt c cc i tng s dng in nng trong cc lnh vc kinh t v i sng: cng nghip, nng nghip, lm nghip, giao thng vn ti, thng mi, dch v, phc v sinh hot

truyn ti in t ngun pht n cc h tiu th ngi ta s dng li in. Li in bao gm ng dy ti in v trm bin p. Li in nc ta hin c nhiu cp in p: 0,4KV, 6KV, 10KV, 22KV, 35KV, 110KV, 220KV v 500KV. Mt s chuyn gia cho rng trong tng lai li in Vit nam ch nn tn ti nm cp in p: 0,4KV, 22KV, 110KV, 220KV v 500KV.

C nhiu cch phn loi li in: Cn c vo tr s ca in p, chia ra li in siu cao p (500KV), li

in cao p (220KV, 110KV), li trung p (35KV, 22KV, 10KV, 6KV) li in h p (0,4KV).

Cn c vo nhim v, chia ra li cung cp (500KV, 220KV, 110KV), li phn phi (35KV, 22KV, 10KV, 6KV, 0,4KV).

Ngoi ra cn nhiu cch chia khc, V d cn c vo phm vi cp in, chia ra li khu vc, li a phng: cn c vo s pha, chia ra li mt pha, hai pha, ba pha; cn c vo i tng cp in chia ra li cng nghip, li nng nghip, li th

2. NHNG YU CU I VI PHNG N CUNG CP IN Bt k mt phng n (hoc d n) cung cp in no cng phi tho mn

4 yu cu c bn sau:

2.1. tin cy cung cp in

l mc m bo lin tc cung cp in tu thuc vo tnh cht ca h dng in.


Bin son: Nh Trng 5

H loi 1: l nhng h rt quan trng khng c mt in, nu xy ra mt in s gy ra hu qu nghim trng.

- Lm mt an ninh chnh tr, mt trt t x hi. l sn bay, cng hang hi, khu qun s, khu ngoi giao on, cc i s qun, nh ga, bn xe, trc giao thng chnh trong thnh ph

- Lm thit hi ln i vi nn kinh t quc dn. l khu cng nghip, khu ch xut, du kh, luyn kim, nh my c kh ln, trm bm nng nghip ln Nhng h ny ng vai tr quan trng trong nn kinh t quc dn.

- Lm nguy hi n tnh mng con ngi. H loi 2: bao gm cc x nghip ch to hang tiu dng (nh xe p, vng bi, bnh ko, nha ) v thng mi, dch v (khch sn, siu th, trung tm thng mi ln). Vi nhng h ny nu mt in s b thua thit v kinh t nh dn cng,gy th phm, ch phm ph v hp ng cung cp nguyn liu hoc sn phm cho khch hang, lm gim st doanh s v li xut H loi 3: l nhn h khng quan trng cho php mt in tm thi khi cn thit. l h nh sang sinh hot th v nng thn.

2.2. Cht lng in

Cht lng in c th hin hai ch tiu: tn s (f) v in p (U). Mt phng n cp in c cht lng tt l phng n m bo tr s tn s v in p nm trong gii hn cho php. C quan Trung tm iu Quc gia chu trch nhim iu chnh tn s chung cho h thng in. Vic m bo cho in p ti mi im nt trn li trung p v h p nm trong phm vi cho php l nhim v ca k s thit k v vn hnh li cung cp in. m bo cho cc thit b dng in (ng c, n, qut, t lnh, ti vi) lm vic bnh thng yu cu in p t vo cc cc thit b dng in khng c chnh lch qu 5% so vi tr s in p nh mc. chnh lch in p so vi tr s nh mc gi l chnh lch in p, k hiu l U.

U = U Um Yu cu: U 5%Um

2.3. Kinh t

Tnh kinh t ca mt phng n cp in th hin qau hai ch tiu: vn du t v ph tn vn hnh.


Bin son: Nh Trng 6

Vn u t mt cng trnh in bao gm tin mua vt t, thit b, tin vn chuyn, tin th nghim, th nghim, tin mua t ai, b hoa mu, tin kho st thit k, tim lp t, nghim thu. Ph tn vn hnh bao g cc khon tin phi chi ph trong qu trnh vn hnh cng trnh in: Tin lng cn b qun l, cn b k thut, cng nhn vn hnh, tin bo dng nh k, tin sa cha, trung i tu, tin th nghim, th nghim, tin tn tht in nng trn cng trnh in.

Thng th hai khon kinh ph ny lun mu thun nhau, nu vn u t ln th ph tn vn hnh nh v ngc li. V d: nu chn tit din dy dn nh th tin mua t i nhng tin tn tht in nng li tng ln do in tr dy ln hn. V d: nu mua thit b in loi tt th t nhng gim c ph tn vn hnh do t phi sa cha, bo dng

2.4. An ton

An ton l vn quan trng, thm ch phi t ln hang u khi thit k, lp t, vn hnh cng trnh in. An ton cho cn b vn hnh, an ton cho thit b, cng trnh in, an ton cho ngi dn v cc cng trnh dn dng ln cn.

Ngi thit k v vn hnh cng trnh in phi nnghim chnh tun th trit cc quy nh, ni quy an ton, v d khon cch an ton gia cng trnh in v cng trnh dn dng, khong cch an ton t dy dn ti mt t

3. MT S K HIU TRN S CP IN

TT Thit b in K hiu trn bn v My pht in

F, Trm bin p Trm phn phi

My bin p 2 cun dy, 3 cun dy

My ct in

Dao cch ly, cu dao


Bin son: Nh Trng 7

Dao ct ph ti, my ct ph ti

Cu ch

My bin dng in

Dy dn

Dy dn ghi r s dy

p t mt

Khi ng t

ng c in

n si t

n tup

Ni t

ng h ampe, vn

A V


Bin son: Nh Trng 8

Chng 1: TNH TON PH TI IN 1.1. KHI NIM CHUNG

Ph ti in l s liu u tin v quan trng nht tnh ton thit k h thng cung cp in. Xc nh ph ti in qu ln so vi thc t s dn n chn thit b in qu ln lm tng vn u t. Xc nh ph ti in qu nh dn ti chn thit b in qu nh s b qu ti gy chy n h hi cng trnh, lm mt in.

Xc nh chnh xc ph ti in l vic lm kh. Cng trnh in thng phi c thit k, lp t trc khi c i tng s dng in. V d: cn thit k v lp t trm bin p trung gian cp in cho khu ch xut ngay t giai on xy dng c s h tng (ng giao thng, in, nc), sau mi mi cc x nghip vo mua t xy dng nh my. Khi thit k lp t ng dy cao p v trm bin p trung gian cp in cho khu ch xut ngi thit k ch bit thng tin rt t: din tch khu ch xut v tnh cht ca cc x nghip s xy dng ti (cng nghip nng, nh).

Ph ti cn xc nh trong giai on tnh ton thit k h thng cung cp in gi l ph ti tnh ton. Cn lu phn bit ph ti tnh ton v ph ti thc t khi cc nh my i vo hot ng. Ph ti tnh ton l ph ti gn ng ch dng tnh ton thit k h thng cung cp in cn ph ti thc t l ph ti chnh xc c th xc nh c bng cc ng h o in trong qu trnh vn hnh.

C nhiu phng php xc nh ph ti in. Cn cn c vo lng thng tin thu nhn c qua tng gai on thit k la chn phng php thch hp. Cng c nhiu thng tin v i tng s dng cng la chn c phng php chnh xc.

2.2. XC NH PH TI IN KHU VC NNG THN Nng thn c nhiu i thng s dng in, ph bin nht vn l trm

bm, trng hc v nh sng sinh hot.

2.2.1. Ph ti in trm bm

Cc my bm nng nghip thng c cc thang cng sut 14KW, 20KW, 33KW, 45KW, 55KW, 75KW, 100KW. Vi my bm cng sut nh s dng


Bin son: Nh Trng 9

in h p, my bm cng sut ln 100KW tr ln thng dng in 6KV hoc 10KV.

Trm bm chia lm 2 loi: trm bm ti v trm bm tiu. Trm bm ti lm vic hu nh quanh nm. Trm bm tiu ch lm vic t ngy vo nhng dp ng lt.

Ph ti trm bm c xc nh theo cng thc sau:

.tgPQ

Pk.kP

tttt

n

1idmitidttt

=

= =

Trong : Ptt, Qtt - ph ti tc dng v phn khng tnh ton ca trm bm

kt - h s ng thi, ly theo tc t

n

nk lvdt =

vi n - tng s my bm t trong trm nlv - s my bm lm vic. Vi trm bm ti t nhiu my bm ngi ta thng cho mt my bm

thay phin nhau ngh bo dng. Vi trm bm tiu, do tnh cp bch ca vic chng l lt bo v hoa mu,

cn cho 100% my bm lm vic Kt - h s ti vi trm bm ti ly theo thc t. vi trm bm tiu cho my ti 100% cng sut. Nh vy vi trm bm tiu trong nhng ngy lm vic phi cho 100%

my bm vn hnh y ti, ngha l: Kt = kt = 1 Khi ph ti in ca trm bm tiu s l:

Ptt = =

n

1idmik

Tr s cos ca trm bm ly nh sau: vi trm bm tiu cos = cos m 0,8 (kt = 1) vi trm bm ti cos = 0,6 0,7 tu theo kt


Bin son: Nh Trng 10

2.2.2. Ph ti in trng hc

Hin nay nng thn trng hc pht trin mnh m v u khp, mi x c trng hc tiu hc, trng ph thng c s, mi huyn c 1, 2 thm ch 3, 4 trng ph thng trung hc.

Vi cc trng ph thng, in ch dng chiu sang v qut mt, v th ph ti in c xc nh theo din tch.

thit k cp in cho trng cn xc nh ph ti in tng phng hc, c nh hc v ton trng.

Ph ti in mt phng hc xc nh theo cng thc: Ptt = Po.S Trong : S - Din tch phng hc (m2). Mt phng hc ca trng ph thng

thng c din tch S = 8x10 = 80m2. Po - Sut ph ti trn n v din tch Po = (15 20) (W/m2). Qp = Pp.tg h s cng sut cos ca phng hc ly nh sau: Nu l n tup + qut: cos = 0,8 Nu l n si t + qut: cos = 0,9 Ph ti tnh ton mt tng nh gm n phng hc:

PT = kt.=

n

1i

pP

Trong : kt - h s ng thi. Nu cc phng hc thng xuyn s dng ht th kt = 1.

V d 2.2: Yu cu xc nh ph ti tnh ton ca mt trng ph thng c s ca x bao gm nh hc 2 tng, mi tng 6 phng hc mi phng c din tch 80m2 v khu nh thng trc, hiu trng, phng hp, gio vin c tng din tch 100m2.

GII Ph ti mt phng hc vi Po = 15 (W/m2) Pp = Po.S = 15.80 = 1200 (W) = 1,2 (kW) Ph ti tng gm 6 tng hc ging nhau: PT = 6.Pp = 6.1,2 = 7,2 (kW) Ph ti c nh hc 2 tng:


Bin son: Nh Trng 11

PN = 2.7,2 = 14,4 (KW) Ph ti khu nh thng trc, phng hp: PH = 20.100 = 2000 (W) = 2 (kW) Tng ph ti in ton trng P = PN + PH = 14,4 + 2 = 16,4 (kW) Gi thit dng n tup, cos = 0,8, xc nh c ph ti ton phn

5,208,0

4,16

cos==

=

P

S (kVA)

2.2.3. Ph ti nh sng sinh hot

y l ph ti in ca cc h gia nh. nng thn, cc gia nh dng in khng chnh lch nhau lm.

Ph ti tnh ton ca mt thn, xm hoc lng c xc nh nh sau: Ptt = Po.H Qtt = Ptt.tg Trong : H - s h dn trong thn, lng Po - sut ph ti tnh ton cho 1 h, thng ly Po = (0,5

0,8) (kW/h). vi 0,5 dnh cho khu vc thun nng 0,6 0,8 dnh cho khu vc c ngh ph hoc lng xm ven ng. phc v sinh hot cc h thng dng nhiu loi thit b in gia dng

khc nhau nh: n, qut, tivi, radio, bn l, t lnh v.vtrong tnh ton cung vp in thng ly h s cng sut chung l cos = 0,85.

Ph ti tnh ton ton x bao g cc thng xm, trng hc, trm bm v.v l:

PX = kt. =

n

1i ttiP

QX = kt. =

n

1i ttiQ

2x

Q2xP

xS +=

kt - h s ng thi vi n = 1, 2 kt =1 n = 3, 4 kt =0,9 0,95


Bin son: Nh Trng 12

n = 5, 6, 7 kt =0,8 0,85 V d 2.3. Yu cu xc nh ph ti in cho 1 x nng nghip bao gm: Thn 1: 300 h dn, thun nng Thn 2: 200 h dn, thun nng Thn 3: 120 h dn, bm mt ng lin x Trng PTCS: 12 lp hc + 100m2 khu hnh chnh Trm bm: 1x33 (kW)

GII xc nh ph ti in ton x cn xc nh ph ta cho tng khu vc: Ph ti in thn 1: l thn thun nng ly Po = 0,5 (kW/h) P1 = 0,5.300 = 150 (kW) Q1 = 150.0,527 = 79 (kVAr) (cos = 0,85 tg = 0,527) Ph ti thn 2: P2 = 0,5.200 = 100 (kW) Q2 = 100.0,527 = 52,7 (kVAr) Ph ti thn 3 vi Po = 0,8 (kW/h) P3 = 0,8.120 = 96 (kW) Q3 = 96.0,527 = 50,59 (kVAr) Ph ti trng hc tnh v d 2.2 PT = 16,4 (kW) QT = 16,4.0,75 = 24,75 (kVAr) Ph ti trm bm vi kt = 1 PB = 33 (kW) QB = 33.0,75 = 24,75 (kVAr) Ly h s ng thi kt = 0,8, xc nh c ph ti in ton x PX = kt.(P1 + P2 + P3 + PT + PB) PX = 0,8.(150 + 100 + 96 + 16,4 + 33) = 316 (kW) QX = 0,8.(79 + 52,7 + 50,6 + 12,5 + 24,7) = 219 (kVAr) SX = 22 219316 + = 385 (kVA)


Bin son: Nh Trng 13

2.3. X C NH PH TI IN KHU VC CNG NGHIP 2.3.1. Trong giai on d n kh thi

Trong giai oanny cc nh my, hoc khu cng nghip cha xy dng. Cn xc nh ph ti in chun b ngun in, thit k v xy dng ng dy cao p v trm bin p trung gian. Thng tin thu nhn c giai on d n kh thi rt t, ch l din tch hoc sn lng. Cng thc xc nh ph ti in cho khu ch xut hoc khu cng nghip thng cn c vo din tch: Stt = s0.D Trong : s0 - sut ph ti trn mt n v din tch (ha) D - din tch khu ch xut hoc khu cng nghip (ha). Tr s ca so ly nh sau:

- Vi khu cng nghip nh (dt, may, giy dp, ko bnh) so = 100 200 (kVA/ha)

- Vi khu cng nghip nng (luyn kim, c kh, ho cht) so = 300 400 (kVA/ha) Vi mt s x nghip, trong giai on d n kh thi thng bit sn lng, cng thc xc nh ph ti in nh sau:

max

ttT

a.MP =

Trong : a - sut in nng chi ph sn xut 1 sn phmm (kWh/1sp) M - sn lng, ngha l s sn phm mt nm Tmax - thi gian s dng cng sut ln nht Tr s ca a v Tmax tra bng. V d 2.4. Yu cu xc nh ph ti in cho mt khu ch xut trong giai on d n kh thi, d nh s xy dng sau 5 nm, bit rng khu ch xut c xy dng trn din tch 80 (ha) v l khu cng nghip nng.

GII V ch bit duy nht thng tin l din tch, ph ti in ca khu ch xut xc nh theo cng thc Stt = s0.D. Gi thit cc nh my trong khu u c trang b my mc hin i, cng ngh cao, dy chuyn sn xut tin tin, chn sut ph ti s0 = 400 (kVA/ha), ph ti in ca khu ch xut l:


Bin son: Nh Trng 14

Stt = s0.D = 400.800 = 32000 (kVA). V d 2.5. Yu cu xc nh ph ti in cho xi nghp sn xut xe p, sn lng mt vn chic/nm, d nh xy dng sau 3 nm.

GII Thng tin v nh my tng lai l sn lng, ta phi p dng cng thc

max

ttT

a.MP =

Tra cm nang vi nh my sn xut xe p ao = 200 (kWh/xe) v Tmax = 5000 (h), xc nh c ph ti in;

4005000

10.200

T

a.MP

4

max

tt === (kW)

Tip tc tra cm nang c cos = 0,6 tg = 1,33 Qtt = Ptt.tg = 400.1,33 = 533 (kVAr)

6676,0

400

cos===

ttPS (kVA)

2.3.2. Trong giai on xy dng nh xng

giai on ny, thng tin m ngi thit k in nhn c l cng sut t ca tng phn xng v din tch ca cc phn xng. Ph ti in ca tng phn xng xc nh theo cng thc:

Ptt = knc.P Qtt = Ptt.tg

Trong : knc - h s nhu cu P cng sut t ca phn xng (kW)

P =

=

n

1iP mi

vi Pm cng dut nh mc ca tng my (ng c) n - s my (ng c) t trong phn xng. Hai cng thc trn xc inh ph ti in ca cc my mc t trong phn xng, cn gi l ph ti ng lc. Ph ti in chiu sng phn xng c xc nh theo din tch

Pcs = Po.D Trong : D - din tch phn xng (m2) Po Cng sut chiu sang trn n v din tch (W/m2)


Bin son: Nh Trng 15

Tu theo yu cu,tnh cht lm vic ca cc phn xng m ly tr s P0 thch hp: vi cc phn xng c kh, luyn kim P0 = 12 15 (W/m2) vi cc phn xng dt, may, ho cht P0 = 15 20 (W/m2) vi kho bi P0 = 5 15 (W/m2) vi xng thit k P0 = 25 30 (W/m2) vi nh hnh chnh P0 = 20 25 (W/m2) Ph ti chiu sng phn khng ca phn xng xc nh theo cng thc:

Qcs = Pcs.tg Nu dng n si t cos = 1 tg = 0 Qcs = 0

Nu dng n tup cos = 0,8 tg = 0,75 Ph ti in ton phn xng PPX = Ptt + Pcs QPX = Qtt + Qcs

2

PX

2

PXPX QPS +=

PX

PX

PXQ

Pcos =

Sau khi ln lt tnh ton ph ti in cc phn xng, ta xc nh c ph ti in ton x nghip bao gm n phn xng:

PXN = kt.=

n

1i

PXiP

QXN = kt.=

n

1i

PXiQ

T y s tnh c ph ti ton phn v cos ca x nghip:

2

PX

2

PXPX QPS +=

PX

PX

PXQ

Pcos =

V d 2.6. Mt xng sa cha t c cng sut t 200kW, din tch xng 2040 = 800m2. Yu cu xc nh ph ti in

GII Xng sa cha t t cc my cng c nh phay, tin, dp, khoan,

mi thc cht l mt xng c kh.


Bin son: Nh Trng 16

Tra PL 2 vi xng c kh knc = 0,2 0,3. V l xng sa cha, my mc lm vic khng lin tc, khng ti c nh nh sn xut dy chuyn ly knc = 0,3, ph ti tnh ton ng lc:

Ptt = knc.P = 0,3.200 = 60 (kW) Ph ti chiu sng vi P0 = 12 (W/m2) Pcs = Po.D = 12.800 = 9600 (W) = 9,6 (kW) Ph ti ton xng PX = Ptt + Pcs = 60 + 9,6 = 69,6 (kW) Xng dng n si t Qcs = 0 QXN = Qtt = Ptt.tg Tra s tay c cos = 0,5 0,6. V l xng sa cha ly cos = 0,5.

Xc nh c ph ti in phn khng QX = Qtt = 60.1,73 = 103,8 (kVAr) Ph ti in ton phn SX = 69,6 + j103,8 (kVA) SX = 22 103,869,6 + = 125 (kVA) cos = 56,0

125

6,69

S

P

X

X=

2.3.3. Trong giai on thit k chi tit

y l cng on cui cng trong qu trnh thit k cp in cho x nghip cng nghip. giai on ny bit ht thng tin v i tng s dng in: cng dut, chng loi ng c, v tr t trong phn xng v c tnh k thut, cng ngh ca chng. Nhim v ca ngi thit k l phi ra phng n cp in hp l cho cc phn xng v thit k mng h p phn xng a in n tng ng c.

xc nh ph ti in phn xng, ta chia ra thnh cc nhm my cho cc ng c t gn nhau, mi nhm khong 8 12 my, sau xc nh ph ti in cho tng nhm my v cui cng cho c phn xng.

Ph ti tnh ton cho mt nhm n my xc nh theo cng thc cn c vo cng sut trung bnh Ptb v h s cc i kmax.

Ptt = kmax.Ptb = kmax.ksd.=

n

i 1

Pmi


Bin son: Nh Trng 17

Qtt = Ptt.tg Trong : Ptb cng sut trung bnh ca nhm my trong thi gian kho st, thng ly l 1 ca hoc 1 ngy m. Pm cng sut nh mc ca my ksd - h s s dng cos - h s cng sut ca nhm my cng c, tra PL1 vi nhm my cng c cos = 0,5 0,6. kmax - h s cc i, tra PL5 (theo ksd v nhq) nhq - s thit b dng in hiu qu. Nhq l s thit b gi tng c cng sut bng nhau, c cng ch lm vic v gy ra mt ph ti tnh ton ng bng ph ti tnh ton do nhm thit b thc t gy ra.

ngha ca nhq l ch: mt nhm my bt l bao gm nhiu my c cng sut khc nhau, c tnh k thut khc nhau, ch lm vic, qu trnh cng ngh khc nhau rt kh tnh chnh xc ph ti in Ngi ta a vo i lng trung gian nhq nhm gip cho vic xc nh ph ti in ca nhm my d dng tin li m sai s phm phi l cho php.

Trnh t xc nh nhq nh sau: 1. Xc nh n1 - s ng c c cng sut ln hn hay bng mt na cng

sut ng c c cng sut ln nht.

2. Xc nh Pn1 cng sut ca n1 ng c trn. =

=

n

1i

n1P Pmi

3. Xc nh cc t s:

=

==

==

n

1i

dmi

n

1i

dmin

*1

*

P

P

P

PP,

n

nn 1

4. Tra ph lc 4 (theo n* v P*) tm c nhq* 5. Xc nh nhq theo biu thc

nhq = n.nhq*

Ghi ch: 1. Nu trong nhm my c thit b mt pha th phi quy i v 3 pha theo

cc biu thc: - Dng in p pha:

Pq = 3P*m - Dng in p dy


Bin son: Nh Trng 18

Pq = 3 Pm 2. Nu trong nhm ny c ng c lm vic ch ngn hn lp i lp li

th phi quy i v di hn: Pq = Pm. %kd

kd% - h s ng in phn trm, ly theo thc t k% = Thi gian lm vic (ng my)/Thi gian kho st Cc ng c lm vic ch ngn hn lp li l cn cu, my nng, cu trc, bin p hn. Ring bin p hn thng ch to mt pha u vo in p dy, khi xc nh ph ti in phi quy i. V d: Yu cu xc nh ph ti in ca nhm my cng c c cc s liu cho theo bng:

TT Tn my Pm(kW) c im S lng 1. Cu trc 14 kd% = 36% 1 2. Bin p hn 12 Ud, kd% = 49% 1 3. My mi th 10 2

4. My mi tinh 7 2

5. My tin 5,5 3 6. My khoan 4,5 3 7. Qut gi 1,7 Uf 1

GII Trc ht phi quy i cc thit b v 3 pha v di hn Cu trc: Pq = Pm. %kd = 14. 36% = 8,4 (kW) Bin p hn Pq = 3 Pm. %kd = 3 12. 49% = 14,57 (kW) Qut gi Pq = Pm = 3.1,7 = 5,1 (kW) Sau khi quy i, cn sp xp li th t cc my theo ln cng sut

TT Tn my S lng Pm(kW) 1. Bin p hn 1 14,57


Bin son: Nh Trng 19

2. My mi th 2 10

3. Cu trc 1 8,4

4. My mi tinh 2 7

5. My tin 3 5,5 6. Qut gi 1 5,1 7. My khoan 3 4,5

Xc nh s thit b dng in hiu qu: 1. Thit b c cng sut ln nht l bin p hn 14,57 (kW), mt na cng

sut l 7,28 (kW). Vy c 4 thit b c cng sut ln hn tr s ny l bin p hn (1), my mi th (2) v cu trc (1). n1 = 4

2. Tng cng sut ca n1 my: Pn1 = (14,57 + 2.10 + 8,4) = 42,97 (kW) 3. Xc nh n*, P*

0,313

4

n

nn 1* ===

0,473.4,55,13.5,52,742,97

42,97P* =++++

=

4. Tra s tay vi n* = 0,3 v ksd = 0,88 5. Tnh c nhq = n.nhq* = 13.0,88 = 11,44 Tip theo, tra s tay vi nhq = 11,44 v ksd = 0,2 c kmax = 1,8 T y xc nh c ph ti in ca nhm

Ptt = kmax.Ptb = kmax.ksd.=

n

i 1

Pmi = 1,8.0,2.92,37= 33,25 (kW)

Tra cm nang c cos = 0,6 tg = 1,33 Qtt = 33,25.1,33 = 44,22 (kVAr) Vy ph ti in ca nhm my l: Stt = 33,25 + j44,22 (kVA)

BI TP CHNG 1 Bi tp 1. Yu cu xc nh ph ti in cho mt trng dy ngh, bao gm:

1 nh ging ng 3 tng 5 lp hc 1 nh 2 tng: - tng di hi trng 200m2


Bin son: Nh Trng 20

- tng trn th vin 200m2 1 xng thc tp c kh 300m2, cng sut t P = 150kW 1 k tc x 3 tng 12 phng. Mi phng 10 hc sinh.

Bi tp 2. Mt khu vn phng i din gm: 1 nh 4 tng, mi tng 8 phng hc 24m2

1 nh 2 tng, mi tng 6 phng hc 40m2 Yu cu xc nh ph ti in cn cp cho khu vn phng.


Bin son: Nh Trng 21

Chng 2: TNH TON TN THT IN P, TN THT CNG SUT, TN THT IN NNG

2.1. S THAY TH LI CUNG CP IN Thng dng 2 loi s : s nguyn l v s thay th.

S nguyn l l s chp ni cc phn t ca li cung cp in (MBA, ng dy, my ct, attomat, cu dao, cu ch) nhm m t cch thc cp in t ngun n cc ph ti.

S thay th l s dng trong qu trnh tnh ton li cung cp in, trn ngi ta thay th cc phn t ca li bng cc i lng c trng cho qu trnh truyn ti in.

2.1.1. S thay th ng dy ti in

S thay th y ca mt on ng dy ti in l s hnh 2.1

Hnh 2.1. S nguyn l v s thay th on ng dy ti in di l (km) tit din F

Ba i lng c trng cho qu trnh truyn ti in trn ng dy l Z, G v B

Trong : Z - tng tr ca on dy, l i lng phc: Z = R + jX vi R - in tr on ng dy:

Fl

R =

- in tr sut ca vt liu lm dy C 3 loi vt liu lm dy: nhm (A), ng (M) v thp (C), trong A,

M dn in, C lm tng bn c. A = 31,5 (mm2/km), M = 18,8 (mm2/km).

l, F

1S&

A 1

2G

2G

2B

2B

Z A 1

1S&


Bin son: Nh Trng 22

R tng trng cho tn tht cng sut tc dng do pht nng dy dn. X tng trng cho tn tht cng sut phn khng do t ho dy dn. Trong tnh ton thc t ngi ta lp sn cc bng tra ro (/km) v xo

(/km) trong Ph lc, khi tng tr on ng dy l (km) l: Z = rol + jxol Mun tra xo, ngoi bit tit din dy cn bit cch treo dy trn x xc

nh khong cch trung bnh hnh hc D gia cc dy. Trong tnh ton s b, c th cho php ly xo = 0,4 (/km) Vi cp, nu khng c bng tra, ly gn ng xo = 0,08 0,1 (/km). G - in dn ca on ng dy, tng trng cho tn tht cng sut tc

dng do r in qua s, ct v do vng quang in. Vng quang in l hin tng khi l cng in trng trn b mt dy dn ln lm ion ho lp khng kh xung quanh to nn mt vng sng xung quanh dy dn, mt thg c th nhn thy c vo nhng m m t cui thng ti tri, lm tn hao cng sut.

G = 2vq

UP

Tn tht do cng sut tc dng do vng quang thc t ch xy ra ng dy trn khng in p >220kV.

B Dung dn ca on ng dy. Khi dy dn ti in, gia cc dy t gn nhau v gia dy vi t hnh thnh nhng bn cc, kt qu l to ra mt cng sut phn khng Qc phng ln ng dy. Vi ng dy cao p (110, 220KV) nhiu khi hin tng ny c li v n b li lng cng sut Q tn tht trn in khng X ca ng dy, nhng li rt nguy him nhng ng dy siu cao c bit khi khng ti v non ti, lm cho in p cui ng dy tng cao vt qu tr s cho php.

B = bo.l vi: bo l dung dn trn 1km ng dy; l l chiu di ng dy. Lng Qc do ng dy sinh ra t l vi bnh phng in p ti in,

viin p n dy U 35 (kV) lng Qc ny nh, c th b qua. Cng v in p trung v h p tn tht vng quang v r in rt nh,

ngi ta cho php b qua i lng G trn s thay th.

Tng dn ng dy 2Bj+

2G

=

2Y


Bin son: Nh Trng 23

Tm li, vi li cung cp in cho php s dng s thay th n gin ch bao gm tng tr cc on ng dy.

Hnh 2.2. S nguyn l v s thay th ng dy trung p v h p

3.1.2. S thay th my bin p

My bin p l thit b in lm nhim v bin i in p v truyn ti cng sut.

Hnh 2.3. S thay th my bin p hai cun dy a) s thay th chnh xc my bin p; b) s thay th gn ng my bin p

My bin p lm vic theo nguyn tc cm ng in t, gm 3 b phn chnh l cun dy 1, cun dy 2 v li thp non c dn t cao. c trng cho cc i lng tn tht trn 3 phn t trong qu trnh ti in ngi ta dng s thay th hnh T vi 3 phn i lng Z1, Z2, Zo. S ny tnh ton kh. Ngi ta thng s dng s thay th gn ng hnh .

Tng tr MBA: ZB = RB + jXB Trong : RB - in tr hai cun dy, tng trng cho tn tht cng sut

tc dng do pht nng 2 cun dy. XB - in khng hai cun dy, tng trng cho tn tht cng

sut phn khng do t ho hai cun dy. Vi my bin p nh ch to cho 4 thng s sau: Po (W, kW) - tn hao khng ti

1S&

A

A

1 2

1 2

1S& 2S&

2S&

lA1, FA1 l12, F12

ZA1 Z12

1

Z1 Z2

Z0

1

oS&

Zo

a) b)


Bin son: Nh Trng 24

PN (W, kW) - tn hao ngn mch, chnh l tn hao nh mc trong 2 cun dy.

Io(%) dng in khng ti (%) UN(%) - in p ngn mch (%) T 4 thng s sny ta c th xc nh c cc i lng trn s thay

th my bin p:

.10S

UUj.10S

UPZ

dmB

2dmBN3

2dmB

2dmBN

B += ()

Trong cng thc ny: UdmB(kV) - inp nh mc ca bin p. Nu tnh ZB v pha cao p th

ly UdmB pha cao, nu tnh ZB v pha h th ly UdmB pha h. SdmB cng sut nh mc ca MBA UN(%) nh ch to So - tn tht cng sut trong li thp cn gi l tn tht khng ti v

khng ph thuc vo tr s ca cng sut ti qua bin p. Tr s So khng i trong sut ti gian ng my vo li in.

So = Po + jQo Trong : Po nh ch to cho, tng trng cho tn tht cng sut tc

dng do pht nng li thp. Qo - tn tht cng sut phn khng do t ho li thp, xc

nh theo cng thc:

100SI

Q dmBoo = .

Nu hai my bin p lm vic song song

.102S

UUj.102S

UPZdmB

2dmBN3

2dmB

2dmBN

B += ()

V So = 2Po + j2Qo.

2.2. TNH TON TN THT IN P Tn tht i p l i lng phc (vc t phc) UjUU +=& . Trong li

cung cp in, ngi ta ch quan tm n tr s ca tn tht in p, tr s sny c ln xp x ln ca thnh phn thc U.


Bin son: Nh Trng 25

Nhn trn hnh 2.4 nhn thy tr s ( ln) ca vct U&: U&=OA OB (tr s ca thnh phn thc U). V th, n gin trong tnh ton, c th tnh tn tht in p theo tr s ca hnh phn thc.

Hnh 2.4. Vc t tn tht U& v thnh phn U Tn tht in p (thnh phn thc) l do cng sut tc dng gy nn in

tr R v cng sut phn khng gy trn X.

dmdmdm UQXPR

UQX

UPR

U +=+=

Nu P(kW), Q(kVAr), R, X(), Udm(kV) th U(V)

2.2.1. ng dy 1 ph ti

Hnh 2.5: S nguyn l v s thay th ng dy 1 ph ti Trn s thay th, tnh tn tht in p theo cng thc

dmdmdm UQXPR

UQX

UPR

U +=+= dng Scos v dng P + jQ

Tn tht in p trn on ng dy A1 l:

dm

A11A11A1 U

XQRPU +=

Trong : ZA1 = RA1 + jXA1 = rolA1 + jxolA1 V sin1jScos1S1SA1S +== && .

0

A

B

U&

U

jU

l, F

1S&

A 1

ZA1

1S& P1 + jQ1

A 1


Bin son: Nh Trng 26

V d 2.1: ng dy trn khng 10(kV) (Vit tt l DK 10 kV) cp in cho x nghip c cc s liu ghi trn hnh v. Yu cu xc nh tn tht in p trn ng dy.

Hnh 2.6. DK 10 (kV) cp in cho x nghip GII

Trc ht cn v s thay th ng dy. Tra Ph lc 25 vi dy AC 50 c ro = 0,64 (/km), xo = 0,4 (/km)

ZA1 = 0,64.5 + j0,4.5 = 3,2 + j2 () j600800j1000.0,61000.0,80,810001SA1S +=+=== && (kVA)

Hnh 2.7. S thay th ng dy ca v d 2.1 in p tn tht trn ng dy cp cho x nghp:

37610

600.2800.3,2UA1 =

+= (V)

2.2.2. ng dy n ph ti

Vi ng dy lin thng cp in cho 3 ph ti, tn tht in p bng tng tn tht in p trn 3 on ng dy

U = Umax = UA123 = UA1 + U12 + U23

Hnh 2.8. S nguyn l v s thay th ng dy lin thng cp in cho 3 ph ti

AC 50, 5km A 1

1000 0,8

3,2 + j2 () A 1

800 + j600 (kVA)

A 1 2

1S& 2S&

lA1, FA1 l12, F12

3S&

3 l23, F23

ZA1 Z12 A 1 2 3 Z23

P1+ jQ1 P2+ jQ2 P3+ jQ3 S1 S2 S3

S2 S3

S3


Bin son: Nh Trng 27

Vi li in trung v h p, tnh ton tn tht in p cho php coi inp ti mi im trn ng dy bng Um v cho php coi dng cng sut chy trn cc on ng dy bng cng sut ph ti, ngha l cho php b qua tn tht in p v tn tht cng dut trn cc on ng sau khi tnh tn tht trn on ng dy trc. V d khi tnh ton on 12, l ra cng sut chy trn on 12 bao gm ph ti 2, 3 ( 2S& , 3S& ) v tn tht trn on 2, 3 nhng cho php b qua lng tn tht ny, 3S2S12S &&& +=

in p tn tht trn cc on nh sau:

( ) ( )

( ) ( )dm

A1321A1321A1

dm

1232123223

dm

23323323

UXQQQRPPP

U

UXQQRPP

U

UXQRP

U

+++++=

+++=

+=

T y xc nh c tn tht in p trn ton b tuyn dy ( ) ( )

( ) ( )dm

233233

dm

12321232

dm

A1321A1321123

UXQRP

UXQQRPP

UXQQQRPPP

++

++++

++++++

== AUU

Tng qut dm

n

1

n

1ijijijij

U

XQRPU

+=

Trong : n - s on ng dy Pij, Qij cng sut tc dng v phn khng chy trn on

ng dy ij. V d: Cng sut chy trn on A1: PA1 = P1 + P2 + P3 QA1 = Q1 + Q2 + Q3 Cng sut chy trn on 1, 2: P12 = P2 + P3 Q12 = Q2 + Q3 Cng sut chy trn on 2, 3: P23 = P3 Q23 = Q3 V d 2.2. DK 10kV cp in cho 2 x nghip, ton b ng dy dng

AC 50, cc s liu khc cho trn hnh v. Yu cu: 1. Kim tra tn tht in p 2. Bit U1 = 10,250 (kV), cn xc nh U2, UA


Bin son: Nh Trng 28

BI GII S thay th ng dy:

Hnh 2.9. S nguyn l v s thay th DK 10(kV) ca v d 2.2 j11,6

2j0,4.50,64.5ZA1 +=+= ()

Z12 = 0,64.2 + 0,4.2 = 1,28 + j0,8 () 1S&= 1000 0,8 = 800 + j600 (kVA) 2S&=5000,7 = 350 + j350 (kVA)

1. Kim tra tn tht in p Cn xc nh U ca ng dy v so snh vi tr s cho php xem c

tho mn hay khng. Cc biu thc so snh nh sau: - Khi ng dy lm vic bnh thng: U Ucp = 5%Um - Khi ng dy c s c: U Ucp = 10%Um

Vi ng dy trong v d ny, tn tht in p khi ng dy lm vic bnh thng l:

( ) ( ) 351,872,827910350.0,8350.1,28

1013506001,6350800U =+=+++++= V

Khi s c 1 ng dy trn on A1, ng dy l kp ch cn l n, tng tr tng gp i nn U cng tng gp i. Usc = 2.279 + 72,8 = 630,8 (V) Kt qu kim tra: U = 351,8 (V) < Ucp = 5%.10 000 = 500 (V) Usc = 630,8 (V) < Ucp = 10%.10 000 = 1000 (V) Vy ng dy m bo yu cu (hoc tho mn yu cu) v tn tht in p.

AC 50, 5km A 2

500 0,7(kVA)

1 AC 50, 2km

1000 0,8(kVA)

1,6 + j1 () A 2

350 + j350 (kVA)

1,28 + j0,8() 1

800 + j600 (kVA)


Bin son: Nh Trng 29

2. Xc nh in p cc im khi bit U1 = 10,250 (kV) U2 = U1 - U12 = 10,250 - 0,073 = 10,177 (kV) UA = U1 +UA1 = 10,250 + 0,279 = 10,529 (kV)

2.2.3. ng dy phn nhnh

l,F

2S&

3S&

1S&

Hnh 2.10. ng dy phn nhnh

Trn li cung cp in nhiu khi gp ng dy phn nhnh, ngha l n 1 nt no th ng dy r ra thnh 2, 3 tuyn theo hng khc nhau. kim tra tn tht in p trn ng dy phn nhnh cn lu rng: tn tht in p l tn tht trn tng tuyn dy t ngun n i nt xa nht ca tuyn. V d vi ng dy phn nhnh trn hnh 2.10 cn kim tra U theo 2 tuyn dy: tuyn A12 v tuyn A13, tuyn c tr s U ln phi nh hn Ucp.

cpA13

A12max UU

UMAXU

=

2.3. TNH TON TN THT CNG SUT 2.3.1. Tn tht cng sut

Tn tht cng sut trn ng dy l mt i lng phc QjPS +=&

Trong : P - tn tht cng sut tc dng do pht nng trn in tr ng dy Q - tn tht cng sut phn khng do t ho ng dy.


Bin son: Nh Trng 30

Tn tht cng sut trn ng dy xc nh theo biu thc:

jX)(RU

QPZUSZIS

dm

22

2dm

22 +

+===

&

Nu S (kVA), P(kW), Q(KVAr), Z, X, Y () v Udm (kV) th S& (VA). n v cng sut v tn tht cng sut thng dng li cung cp in l (kVA) cn phi nhn vi 10-3. 1. ng dy 1 ph in

Hnh 2.11. ng dy 1 ph ti v s thay th Vi ng dy 1 ph ti, cng sut chy qua tng tr Z12 chnh l ph ti

S1. Vy: A1A1A12dm

21

A12dm

2A1

A1 jPZUSZ

USS +===&

V d 2.3. DK 10 (kV) cp in cho x nghip c kh c ph ti in 2000 (kV), cos = 0,6. Dy dn AC 70, di 5km. Yu cu xc nh tn tht cng sut trn ng dy.

GII S thay th ng dy

Hnh 2.12. ng dy 10 (kV)cp in cho x nghip v s thay th Tra bng c ro = 0,46 (/km), xo = 0,4 (/km) ZA1 = 0,46.5 + j0,4.5 = 2,3 + j2 () Tn tht cng sut trn ng dy:

l, F

1S&

A 1

ZA1 A 1

AC 70, 5km A 1

2000 0,6

2,3 + j2 () A 1

92000 + j80000 (kVA)


Bin son: Nh Trng 31

( ) ( ) ( )kVAj8092VAj8000092000j22,310

2000S 22

A1 +=+=+=&

2. ng dy n ph ti

Hnh 3.13. S nguyn l v s thay th ng dy cp in cho 2 ph ti

Cng tng t nh khi tnh ton U, khi tnh S& coi in p cc im bng Um v coi cng sut gy S& trn cc on ch l cng sut ti (b qua S& ca on sau)

( ) ( )12Z2

dmU

22S

A1Z2dmU

22Q1Q

22P1PS

SSSS 12A1A12

++++

+==

=

&

&&&&

Tng qut vi ng dy n ti

( )2dm

n

1ij

2ij

2ij

2dm

n

1ij

2ij

U

ZQP

U

ZSS

+==

&

Trong : n - s on ng dy hoc s ph ti Sij, Pij, Qij - cng sut S, P, Q chy trn on ng dy ij Zij - tng tr ca on ng dy ij Udm - in p nh mc ca ng dy. V d 2.4. ng dy trn khng 10 (kV) cp in cho 3 ph ti, ton b

dng dy AC 50. Chiu di cc on ng dy v s liu ph ti cho trn hnh v. Yu cu xc nh tng tn tht cng sut trn ng dy.

GII

1S&

A

A

1 2

1 2

1S& 2S&

2S&

lA1, FA1 l12, F12

ZA1 Z12


Bin son: Nh Trng 32

Trc ht ta cn v s thay th ng dy, dng cng thc tnh tng tr quen thuc v cc cng thc bin i cng sut S sang P, Q c cc thng s ca s thay th.

Hnh 2.14. S nguyn l v s thay th ng dy Tng tn tht cng sut trn ng dy l:

( ) ( )13Z2

dmU

23S

12Z2dmU

22S

A1Z2dmU

23Q2Q1Q

23P2P1PSSSS 1312A1 ++

+++++=++=

&&&&

Thay s vo ta c:

( ) ( ) ( ) ( )

( )( ) ( ) ( )

( )kVAj39,4663,56j1,923,07j23j35,5456,86

j1,21,923102102400

j0,81,283102102500j1,6310210

22403007002320400700S

+=

+++++=

=++

+++++++++

=&

2.3.2. Tn tht cng sut trong tm bin p

1. Trm 1 bin p

Hnh 2.15. S nguyn l v s thay th trm bin p

AC 50, 5km A 2

500 0,8(kVA)

1 AC 50, 2km

400 0,8(kVA)

1,6 + j1 () A 2

400 + j300 (kVA)

1,28 + j0,8() 1

1,92+j1,20 (kVA)

AC 50, 3km

320 + j240 (kVA)

320 + j240 (kVA)

3

3

1S& 1S

&

0S& SdmB


Bin son: Nh Trng 33

Tn tht cng sut trong trm bin p ch l tn tht trong cc my bin p t trng trm, cc thit b in khc nh my ct, dao cch ly c tng tr nh gn nh bng 0, tn tht cng sut trn chng l khng ng k.

Tn tht cng sut trong my bin p bao gm tn tht trong li thp v tn tht trn hai cun dy. Tn tht trong my bin p l mt i lng phc:

BQBPcuS0SS jB +=+= &&&

Trong : 100

dmBSoIjo

PBQjoPoS +=+=&

cuS& - tn tht trn 2 cun dy c th xc nh theo 2 cch Theo tng tr bin p:

B2dm

2

cu ZUSS =&

Theo PN v UN 2

dmBSS

100dmSNUj

2

dmBSS

NPcuQjcuPcuS

+

=+=&

Trong : PN, UN l s liu nh ch to cho vi ti nh mc, cn quy i v ti S bt k bng cch nhn vi bnh phng h s ti.

Nh vy, nu tnh ton tn tht cng sut trn 2 cun dy theo tng tr bin p th:

++

+= 2

2

2

2

100 dmBdmBo

Bdm

oB USSIjR

USPS&

Nu tnh cuS& theo PN, UN th

++

+= 2

2

2

2

100100 dmBdmBNdmBo

dmNoB U

SSUSIjUSPPS&

V d 3.5. Trm bin p cp in cho x nghip c kh t 1 bin p 1000(kV) 10/0,4 (kV) c cc s liu k thut Po = 5 (kW), PN = 12 (kW), Io% = 3 (%), UN (%) = 5 (%). Ph ti nh my l 8000,6 (kVA) hnh v. Yu cu xc nh tn tht cng sut trong trm.

GII

1x1000(kVA)

10(kV) 0,4(kV)

8000,6 (kVA) 5 + j30 (kVA)

1,2 + j5 (kVA) 8000,6 (kVA)


Bin son: Nh Trng 34

Hnh 2.16. S nguyn l v s thay th trm bin p ca x nghip c kh

( )

( )+=+=+=

+=+=+=

52,1101000

210.531021000

210.12102

3102

2

3051001000.35

100dmBSoIj

oPoS

jjdmS

dmBUNUjdmBS

dmBUNPBZ

kVAjj&

Xc nh cuS& theo ZB:

( ) ( )kVAjjZUSScu 3268,71052,110

800 32

2

2

2+=+== &

Trng hp ny tn tht trong trm bin p

cuS& = (5 + 7,68) + j(30 + 32) = 12,68 + j62 (kVA) 2. Trm bin p t 2 my

Hnh 2.17. S nguyn l v s thay th trm bin p t 2 my Vi trm bin p t 2 my, so vi trm 1 my, tng tr gim i mt na cn 0S& tng gp i.

100dmBSo2IjoP2BQj2oP2oS +=+=&

102

231022

2

dmSdmBUNUj

dmBSdmBUNP

BZ +

=

Tn tht cng sut trong trm 2 my:

++

+= 2

2

2

2

10021

1002

212

dmB

dmBNdmBo

dmNoB U

SSUSIjUSPPS&

1S& 1S

&

0S&

2SdmB

ZB


Bin son: Nh Trng 35

V d 2.6. X nghip luyn kim t hai my bin p do Cng ty thit b in ng anh ch to 2x1000(kVA) 22/0,4 (kV). Ph ti x nghip S = 1500 (kVA), cos = 0,9. Yu cu xc nh tn tht trong 2 trm bin p. GII Tra ph lc 6 vi bin p 1000 (kVA) 22/0,4 (kV) do cng ty thit b in ng Anh ch to ta c: Po* = 1570 (W), PN = 9500 (W), Io (%) = 1,32 %, UN % = 5% p dng cng thc ta c:

( )kVAj82,2513,38S10001500

1005.1000

21

1002.1,3.1000j

100015009,5

212.1,57S

US

100SU

21

100S2Ij

US

P212S

B

2

2

2

2

B

2dmB

2dmBNdmBo

2dm

2

NoB

+=

++

+=

++

+=

&

&

&

BI TP CHNG 2 Bi tp 1. ng dy 10kV cp in cho 3 ph ti, ton b dng dy AC 70 cc s liu cho trn hnh v. Yu cu:

1. Kim tra tn tht in p 2. Cho bit U2 = 10.18kV, hy xc nh tr s in p UA, U1, U3.

Bi tp 2. ng dy phn nhnh cp in cho 4 ph ti, ton b dng dy AC 50, c s liu cho trn hnh v. Yu cu:

1. Kim tra tn tht in p 2. Bit U3 = 9,850kV, Yu cu xc nh UA, U1, U3, U4.

4km A 2

500 0,7(kVA)

1 3km

800 0,8(kVA) 300 0,8(kVA)

2km 3


Bin son: Nh Trng 36

Chng 3: LA CHN CC THIT B IN TRONG LI CUNG CP IN

H thng in bao gm cc thit b in c chp ni vi nhau theo mt nguyn tc cht ch to nn mt c cu ng b, hon chnh. Mi thit b in cn c la chn ng thc hin tt chc nng trong s cp in v gp phn lm cho h thng cung cp in vn hnh m bo cc ch tiu k thut, kinh t v an ton.

3.1. LA CHN MY BIN P Trong s cp in, my bin p c vai tr rt quan trng, lm nhim v bin i in p v truyn ti cng sut. Ngi ta ch to ra my bin p rt a dng, nhiu kiu cch, kch c, nhiu chng loi. Ngi thit k cn c vo c im ca i tng dng in (khch hng) la chn hp l my bin p. Thng k hiu my bin p nh sau: Kiu my cng sut U1/U2 V d: 4JB 5444 3LA 250 24/0,4 l my bin p phn phi do Siemens ch to, kiu 4JB 4LA, cng sut 250 kVA, in p U1=24 kV, U2 = 0,4 kV. Cng c khi k hiu n gin hn: Kiu Cng sut/in p V d: 8CB8 400/35 l bin p phn phi kh, cng sut 400 kVA, in p 35/0,4 kV do ChongQing ch to. La chn my bin p bao gm la chn s lng, cng sut, chng loi, kiu cch v cc tnh nng khc ca bin p.

S lng bin p t trong mt trm ph thuc vo tin cy cung cp in cho ph ti ca trm .

- Vi ph ti quan trng khng c php mt in, phi t hai bin p.

- Vi cc x nghip hng tiu dng, khch sn, siu th (h loi 2) thng t 1 bin p cng vi my d phng.

- Vi c h nh sng sinh hot thng ch t trm 1 my. Cng sut my bin p c chn theo cc cng thc sau: Vi trm 1 my: SdmB Stt

Vi trm 2 my: 1,4SS ttdmB


Bin son: Nh Trng 37

Trong : SdmB cng sut nh mc ca my bin p, nh ch to cho Stt cng sut tnh ton, ngha l cng sut yu cu ln nht ca ph ti 1,4 - h s qu ti k = 1,4 Cn lu rng h s qu ti ph thuc thi gian qu ti. Ly kqt = 1,4 l ng vi iu kin thi gian nh sau: qu ti khng qu 5 ngy m, mi ngy qu ti khng qu 6 gi. Nu khng tho mn iu kin thi gian trn phi tra th tm kqt trong s tay cung cp in hoc khng cho qu ti. Hai cng th trn ch dng chn my bin p ch to trong nc hoc vi my bin p ngoi nhp nhit i ho.

Khi s dng my ngoi nhp cha nhit i ho cn phi a vo cng thc h s hiu chnh nhit khc k n s chnh lch nhit gia mi trng ch to v mi trng s dng my.

1001k 21hc

=

Trong : 1 - nhit mi trng s dng (oC) 2 - nhit mi trng ch to (oC) V d: Nu dng my bin p Nga Vit nam th

81,0100

5241k hc =

=

Vi 24 - nhit trung bnh H Ni 5 - nhit trung bnh Mtcva. Khi cng sut my bin p c tnh theo cng thc:

hc

ttdmB k

SS v hc

ttdmB 1,4k

SS

Cng cn lu l my bin p rt t xy ra s c, nu nh kho st thng k c trong h loi 1 c mt s phn trm no h loi 3 c th ct in khi cn thit vi thi gian k trn th khi 1 bin p s c, bin p cn li ch cn cp in cho h loi 1. Kt qu l s la chn c c my nh hn, hp l hn. Cng thc chn cng sut my cho 2 trm my s l:

1,4SS 1dmB

V d 3.1. Yu cu chn my bin p cho khu chung c c ph ti in Stt = 300 kVA, in p trung p 22 KV.


Bin son: Nh Trng 38

GII V cp in cho khu chung c, trm t 1 my Ta c: SdmB 300 kVA Chn my 315 kVA do ABB ch to: 315 22/0,4 V d 3.2. Yu cu chn my bin p cho trm bin p nh my luyn kim

c ph ti in SB = 1200 (kVA) trong hai trng hp: khng bit s % ph ti loi 3 v bit ph ti loi 3 l 20%.

GII Trm cp in cho nh my luyn kim phi t 2 my bin p. Khi khng bit s ph ti loi 3 ca nh my, khi s c 1 my bin p,

my cn li phi cp cng sut 1200 (kVA). ( )kVA857

4,11200

1,4SS ttdmB ==

Chn dng 2 my bin p do Cng tu thit b in ng Anh ch to, cng sut 1000 (kVA), 2x1000 22/0,4

Trng hp kho st thng k c tron gnh my c 20% ph ti loi 3 (v d nh kho, nh hnh chnh, phn xng sa cha)

( )kVA6854,11200%.80

1,4SS 1dmB ==

Chn dng 2 my bin p do Cng ty thit b in ng Anh ch to, cng sut 750 (kVA), 2x750 11/0,4

So snh hai phng n chn my: Phng n chn 2 my 1000 (kVA) c li l khi 1 my s c khng phi

ct in ph ti loi 3 nhng c hi l vn u t ln, h s ti nh.

6,020001200

21===

dmB

tt

SS

K

Phng n chn 2 my 750 (kVA) c li l vn u t nh, h s ti cao 8,0

15001200

21===

dmB

tt

SS

K

Tuy nhin, khi c s c 1 my, my cn li cho php ti 1,4 s phi ct 1 lng ti loi 3 l:

1200 1,4.750 = 150 (kVA) S % phi ct:


Bin son: Nh Trng 39

%5,12100.15001200

=

Ngha l khng cn ct ht 20% ph ti loi 1.

3.2. LA CHN MY CT IN My ct in, k hiu l MC l thit b ng ct mch in cao p (trn 1000V). Ngoi nhim v ng, ct in ph ti phc v cho cng tc vn hnh, my ct cn c chc nng ct dng in ngn mch bo v cc phn t ca h thng cung cp in.

My ct cng c ch to nhiu chng loi, nhiu kiu cch, mu m. C my ct t du, my ct nhiu du, my ct khng kh, my ct chn khng, my ct kh SF6.

My ct hp b (MCHB) l loi my ct ch to thnh t, trong t sn my ct v hai dao cch ly, loi my dng rt tin li cho cc trm bin p hoc trm phn phi kiu trong nh.

My ct ph ti (MCPT) bao gm dao ct ph ti dng kt hp vi cu ch, trong dao ct ph ti dng ng ct dng ph ti cn cu ch (CDPT - CC) ct dng ngn mch. My ct ph ti r tin hn nhng lm vic khng chc chn, tin cy bng my ct.

My ct in c chn v kim tra theo cc iu kin ghi trong bng Bng 3.1. CC IU KIN CHN V KIM TRA MY CT

Cc iu kin chn v kim tra iu kin

in p nh mc (kV) UdmMC UdmL Dng in nh mc (A) IdmMC Icb Dng ct nh mc (kA) ICdm IN Cng sut ct nh mc (MVA) SCdm SN Dng in n nh ng (kA) Iod ixk Dng in n nh nhit (kA)

Ionh nh.dm

qd

t

t.I


Bin son: Nh Trng 40

Bng 3.2. IU KIN CHN V KIM TRA MY CT PH TI Cc iu kin chn v kim tra iu kin

in p nh mc (kV) UdmMC UdmL Dng in nh mc (A) IdmMC Icb Dng n nh ng (kA) Iod ick Dng n nh nhit (kA)

Ionh nh.dm

qd

t

t.I

Dng in nh mc ca cu ch (A) Imcc Icb Dng ct nh mc ca cu ch (kA) Icn I Cng sut ct nh mc ca cu ch (MVA) Scm S

Trong 2 bng trn: UmL - in p nh mc ca li in (kV) Icb dng in cng bc, ngha l dng in lm vic ln nht i qua

my ct, xc nh theo s c th. I, I dng ngn mch v cng v siu qu trong tnh ton ngn mch

li cung cp in, coi ngn mch l xa ngun, cc dng ny bng nhau v bng dng ngn mch chu k.

Ixk dng in ngn mch xung kch, l tr s tc thi ln nht ca dng ngn mch.

Ixk = 1,8. 2 IN S cng sut ngn mch S = 3 UtbI tnh. m thi gian n nh nhit nh mc, nh ch to cho tng ng vi

Inh.m (Ionh). tq - thi gian quy i, xc nh bng cch tnh ton v tra th. Trong

tnh ton thc t li trung p, ngi ta cho php ly tq bng thi gian tn ti ngn mch, ngha l bng thi gian ct ngn mch.

Vy: Icm = Inh.dm nh.dm

c

tt


Bin son: Nh Trng 41

Cc thit b in c Im > 1000 (A) khng cn kim tra n nh nhit. V d 3.1. Trm bin p phn phi 1000 (kVA) 22/0,4 (kV) cp in cho

khch sn dng my ct ph ti (DCPT - CC) 22 (kV). Bit dng ngn mch sau cu ch trung p I = 8 (kA), yu cu la chn my ct ph ti cho TBAPP.

GII Dng cng bc qua my ct chnh l dng nh mc ca bin p vi gi

thit khng cho bin p qu ti th Iqt = 1,25IdmB (trm 1 my), Iqt = 1,4IdmB (trm 2 my)

Icb = IdmB = ( )A27,2622.3

1000=

Chn dng dao ct ph ti ca ABB kt hp vi b cu ch ng ca Siemens c cc thng s k thut ghi trong bng

Bng 3.3. THNG S K THUT CA DAO CT PH TI DO ABB CH TO

Loi dao ct ph ti

Udm (kV) Idm (A) INmax(kA) IN3s (kA)

NPS 24 B1/A1 24 400 40 10

Bng 3.4. THNG S K THUT CA CU CH NG DO SIEMENS CH TO

Loi cu ch ng Udm (kV) Idm (A) INmax(kA) IN3s (kA) 3GD1 406 4B 24 32 31,5 27

Cn c vo dng ngn mch cho, lp bng kim tra b DCTT CC Bng 3.5. BNG KIM TRA DCTT CC CHO Cc iu kin chn v kim tra iu kin

in p nh mc (kV) UdmMC = 24 UdmL = 22 Dng in nh mc (A) IdmMC = 400 Icb = 16,27 Dng n nh ng (kA) Iod = 40 ick= 2 .1,8.8 = 20,3 Dng n nh nhit (kA)

Ionh = 10 13,438,08

t

t.I

nh.dm

qd==


Bin son: Nh Trng 42

Dng in nh mc ca cu ch (A) Imcc = 32 Icb = 26,27 Dng ct nh mc ca cu ch (kA) Icn = 40 I = 8 Cng sut ct nh mc ca cu ch (MVA)

Scm = 3 .24.40 S = 3 .8.23

3.3. LA CHN CU CH, DAO CCH LY Cu ch l phn t yu nht trong h thng cung cp in do ngi thit

k to ra nhm ct t mch in khi c dng in ln qu tr s cho php i qua. V th chc nng ca cu ch l bo v qu ti v ngn mch. Cn lu l dy ch ch to rt kh ng nht tit din v kh kh ht tp cht nn lm vic khng c tin cy lm, khng ct dng tht chnh xc, v th chc nng ch yu l bo v ngn mch, cu ch ch lm d phng bo v qu ti cho ptmt hoc khi ng t.

Dao cch ly (cn gi l cu dao) c nhim v ch yu l cch ly phn c in v phn khng c in to khong cch an ton trng thy phc v cho cng tc sa cha, kim tra, bo dng. S d khng cho php dao cch ly ng ct mch khi ang mang ti v khng c b phn dp h quang. Tuy nhin, c th cho php dao cch ly ng, ct khng ti bin p khi cng sut my khng ln (thng nh hn 1000 kVA)

Cu ch v dao cch ly c ch to vi mi cp in p. Trong li cung cp in, cu ch c th dng ring r, nhng thng

dng kt hp vi dao cch ly hoc dao ct ph ti. Dao cch ly cng c th dng ring r, nhng thng dng kt hp vi my ct v cu ch.

3.2.1. La chn dao cch ly, cu ch cao p

Trong li in cao p, cu ch thng dng cc v tr sau: - Bo v my bin in p - Kt hp vi dao ct ph ti thnh b my ct ph ti trung p bo

v cc ng dy. Cu ch c ch to nhiu loi, nhiu kiu, in p trung ph bin nht l cu ch ng. in p trung ngi ta cn dng cu ch t ri (CCTR) thay cho b cu dao - cu ch (CD - CC).


Bin son: Nh Trng 43

Trong li cung cp in trung v cao p, dao cch ly t dng ring r, thng dng kt hp.

- Kt hp vi my ct trong t my ct hoc trong b MC DCL - Kt hp vi cu ch trung p t ti cc trm BAPP

Cu ch v cu dao cch ly trung, cao p c chn v kim tra theo cc iu kin ghi trong bng 3.6, 3.7. Bng 3.6. CC IU KIN CHN V KIM TRA DAO CCH LY


in p nh mc (kV) UdmDCL UdmL Dng in nh mc (A) IdmDCL Icb Dng in n nh ng (kA) Id. ixk Dng in n nh nhit (kA)

Inh.m nh.dm

qd

t

t.I

Bng 3.7. CC IU KIN CHN V KIM TRA CU CH Cc iu kin chn v kim tra iu kin

in p nh mc (kV) UdmCC UdmL Dng in nh mc (A) IdmCCL Icb Dng ct nh mc (kA) Im I Cng sut ct nh mc (MVA) Scm S

V d 3.2. Trm bin p ca mt x nng nghip t 1 my bin p 320 (kVA) in p 110/0,4 (kV), Bit rng trm c cp in t TBATG 35/10 (kV) ca huyn cch 3km bng DK 10, dy AC 35. My ct du ng dy l ca Lin x (c) mt Catalog. Yu cu chn DCL CC 10 (kV) cho trm.

GII Dng in ln nht qua dao cch ly v cu ch chnh l dng qu ti my bin p. cc my bin p cp in cho nng thn, do non ti sut ngy, bui ti c th cho php qu ti vi krt = 1,25.


Bin son: Nh Trng 44

Dng cng bc qua DCL CC l:

Icb = IqtB = 1,25IdmB = 1,2510.3

320= 27,75 (A)

Cn c vo Icb = 27,75 (A) chn dao cch l 3DC in p 12kV v cu ch ng 3GD do Siemens ch to c cc thng s k thut ghi trong bng: BNG CHN DAO CCH LY

Loi DCL Udm (kV) Idm (A) INmax(kA) IN3 (kA) 3DC 12 400 40 10

BNG CHN CU CH Loi CC Udm (kV) Idm (A) INmax(kA) IN3 (kA)

3GD1 - 120 - 2B 12 100 80 40

kim tra n nh in ng dao cch ly cn tnh tr s dng ngn mch xung kch: ixk = 1,8 2 IN = 1,8. 2 .1,9 = 4,79 (kA). KT QU KIM TRA DAO CCH LY


in p nh mc (kV) UdmDCL =12 UdmL = 10 Dng in nh mc (A) IdmDCL = 400 Icb = 27,75 Dng in n nh ng (kA) Id. = 40 ixk = 4,79 Dng in n nh nhit (kA)

Inh.m = 10 38,09,1

t

t.I

nh.dm

qd=

BNG KT QU KIM TRA CU CH Cc iu kin chn v kim tra iu kin

in p nh mc (kV) UdmCC = 12 UdmL = 10 Dng in nh mc (A) IdmCCL = 100 Icb = 27,75 Dng ct nh mc (kA) Im = 80 I = 1,9 Cng sut ct nh mc (MVA) Scm = 3 .12.80 S = 3 .10,5.1,9


Bin son: Nh Trng 45

3.3.2. La chn cu dao, cu ch h p

li h p thng gi dao cch ly l cu dao. Ngi ta ch to cu dao 1 pha, 2 pha, 3 pha vi s cc khc nhau: 1 cc, 2 cc, 3 cc, 4 cc. V kh nng ng ct, cu dao c ch to gm hai loi:

- Cu dao (thng, khng ti) ch lm nhim cch ly, ng ct khng ti hoc dng nh

- Cu dao ph ti lm nhim v cch ly v ng ct dng ph ti. Cu ch h p cng c ch to gm 3 loi:

- Cu ch thng thng (khng lm nhim v cch ly, ct ti). - Cu ch cch ly c mt u c nh, 1 u m ra c nh dao cch ly

lm nhim v cch ly nh cu dao. - Cu ch ct ti l cu ch cch ly c th ng ct dng ph ti nh cu

dao ph ti. Ngi ta cng ch to b cu dao cu ch theo loi:

- B cu dao cu ch thng thng - B cu dao ph ti - cu ch

Bng 3.5 ch ra k hiu, s v chc nng ca tng loi cu dao, cu ch h p. Cu ch h p c c trng bi 2 i lng: Idc - dng nh mc ca dy chy cu ch (A) Iv - dng nh mc ca v cu ch (bao gm c v np). Khi la chn cu ch h p phi la chn c Idc v Iv. Thng chn Iv

Idc vi cp khi dy chy t v qu ti, ngn mch hoc khi cn tng ti ta ch cn thay dy chy ch khng cn thay v. K hiu y cu ch h p cho trn hnh v.

Hnh 3.1. K hiu y cu ch h p v cc v d Cng cn lu l: Khi ni cu ch 100 (A) phi hiu cu ch c Iv = 100 (A) Khi ni b cu dao - cu ch 100 (A) phi hiu l ICD = IvCC = 100 (A)

( )( )AIAI

dc

vo 15

30100200

200500


Bin son: Nh Trng 46

Trong li h p cu ch v cu dao thng c t kh xa ngun (TBAPP) v th dng ngn mch qua chng d nh, nn khng cn kim tra cc i lng lin quan n dng ngn mch.

Bng 3.8. CC LOI CU DAO, CU CH H P Loi K hiu Chc nng Cu dao (dao cch ly)

Cch ly, ng ct dng nh

Cu dao ph ti (dao ct ti)

Cch ly, ng ct dng ph ti

Cu ch

Bo v qu ti v ngn mch

Cu ch cch ly

Bo v qu ti ngn mch, cch ly

Cu ch ct ti

Bo v qu ti v ngn mch, ng ct dng in ph ti

B cu dao - cu ch

Bo v qu ti v ngn mch, cch ly

B cu dao ph ti - cu ch

Bo v qu ti v ngn mch, ng ct dng in ph ti

1. La chn cu dao h p Cu dao h p c chn theo 2 iu kin: UdmCD UdmL IdmCD Itt Trong : UdmCD - in p nh mc ca cu dao, thng ch to 220V,

230V, 250V, 380V, 400V, 440V, 500V, 690V. UdmL - in p nh mc ca li in h p, c tr s 220V

(in p pha), hoc 380V (in p dy)

CD(DCL)

CDPT

CC

CCCL

CCCT

CDPT - CC

CD - CC


Bin son: Nh Trng 47

Ngoi ra, cn phi ch n s pha, s cc, kh nng ct ti, trong nh, ngoi tri 2. La chn cu ch h p

a) Trong li in thp sang, sinh hot Cu ch c chn theo 2 iu kin UdmCD UdmL Idc Itt

Trong : UdmCC - in p nh mc ca cu ch, ch to cc c in p nh vi cu dao. Idc dng in nh mc ca dy chy (A) nh ch to cho Ib dng in tnh ton, l dng lu di ln nht chy qua dy chy cu ch (A) - Vi ph ti 1 pha (v d cc thit b gia dng)

cosUPII

pdm

dmdmtt ==

Trong : Updm - in p pha nh mc, bng 220V cos - h s cng sut vi n si t, bn l, bp in, bnh nng lnh, cos = 1 vi qut, t lnh, iu ho, n tup, cos = 0,8 vi cn h gia nh cos = 0,83 vi lp hc dng qut + n si t cos = 0,9 vi lp hc dng qut + n tup cos = 0,8 - Vi ph ti 3 pha:

cosU3PIdm

tttt =

Trong : Udm - in p dy inh mc, bng 380V cos - ly theo ph ti b) Trong li in cng nghip Ph ti ch yu ca li cng nghp l cc my mc cng c, cc ng c. S cp in cho cc ng c gii thiu trn hnh v.


Bin son: Nh Trng 48

Hnh 3.2. Cu ch bo v 1 ng c (CC2, CC3), bo v 2 ng c (CC1) v c nhm ng c (CCT)

- Cu ch bo v 2, 3 ng c Trong thc t, cm 2, 3 ng c nh hn hoc 1 ng c ln cng 1, 2

ng c nh gn c th c cp in chung 1 ng dy v c bo v chng bng 1 cu ch (nh 1, 2 trn hnh 3.2)

Trng hp ny cu ch c chn theo 2 iu kin sau:

IkII

IkI

1n

1dmitimmmax

dc

n

1idmitidc

=

+

ly theo tnh cht ca ng c m my. - Cu ch tng bo v nhm ng c Cu ch tng c chn theo 3 iu kin

Idc Itt

II mmdc

Hai iu kin trn l iu kin chn lc, ngha l CTT ch chy khi c ngn mch trn thanh ci t in, cn khi xy ra ngn smch ti ng c no hoc dy dn no th ch cu ch nhnh chy, m bo cho c nh khng b mt in. Mun vy, ngi ta quy c phi chn Idc ca cu ch tng ln hn t nht l 2 cp so vi Idc ca cu ch nhnh ln nht.

V d 3.3. Yu cu la chn cu ch bo v bng n si t 100 (W) GII

1

2

3

4

CCT

CC1

CC2

CC3

KT

KT

KT

KT


Bin son: Nh Trng 49

Bng n si t dng in p 220V v cos = 1, cu ch chn theo cng thc:

0,45A220100

cosU3P

IIdm

ttttdc ===

Vy chn cu ch h p c Idc = 2A, Iv = 5A

3.4. LA CHN PTMT ptmt l thit b ng ct h p c chc nng bo v qu ti v ngn

mch. Do c u im hn hn cu ch l kh nng lm vic chc chn, tin cy, an

ton, ng ct ng thi 3 pha v c kh nng t ng ho cao nn ptmt mc d c gi t hn vn ngy cng c dng rng ri trong li in h p cng nghip, dch v cng nh li in sinh hot.

ptmt c ch o vi in p khc nhau: 400V, 440V, 500V, 600V, 690V.

Ngi ta cng ch to cc loi ptmt 1 pha, 2 pha, 3 pha vi s cc khc nhau: 1 cc, 2 cc, 3 cc, 4 cc. K hiu ptmt cho bng di dy:

K hiu

S cc 1 cc 1 cc +TT 2 cc 3 cc 3 cc + TT 4 cc

Ngoi ptmt thng thng, ngi ta cn ch to loi ptmt chng r in. ptmt chng r t ng ct mch in nu dng r c tr s 30mA, hoc 300mA tu loi.

ptmt c chn theo 3 iu kin: UdmA UdmL IdmA IB IcmA IN


Bin son: Nh Trng 50

V d 3.4. Yu cu chn ptmt tng cho cn h gia nh c cng sut t l 6kW.

GII Ph ti tnh ton cn h Ptt = kt.P = 0,8.6 = 4,8 kW Cn h dng in p 220V, cos = 0,85

25,66A0,22.0,85

4,8IB ==

C th chn ptmt 32A, y d phng pht trin ph ti, chn pttmt 1 pha G4CB 1040 do Clipsal ch to c Idm = 40A, Icm = 6 kA.

3.4. LA CHN DY DN V CP 3.4.1. Gii thiu chung v cc phng php v phm vi p dng.

C 3 phng php la chn tit din dy dn v cp. 1. Chn tit din theo mt kinh t ca dng in Jkt (A/mm2) l s ampe ln nht trn 1mm2 tit din chn theo phng php ny s c li v kinh t. Phng php chn tit din dy theo Jkt p dng vi in c in p U 110kV, bi v trn li ngy khng c thit b s dng in trc tip u vo, vn in p khng cp bch, ngha l yu cu khng cht ch. Li trung p th v x nghip ni chung khong cch ti in ngn, thi gian s dng cng sut ln cng c chn theo Jkt. 2. Chn tit din theo in p cho php Ucp

Phng php la chn tit din ny ly ch tiu cht lng in lm iu kin tin quyt. Chnh v th n c p dng la chn tit din dy cho li in nng thn, thng ng dy ti in kh di, ch tiu in p rt d b vi phm. 3. Chn dy dn theo dng pht nng lu di cho php. Phng php ny tn dng ht kh nng ti ca dy dn v cp, p dng cho li h p th, cng nghip v sinh hot. Bng 3.9. PHM VI P DNG CC PHNG PHP LA CHN TIT DIN DY DN V CP.

Li in Jkt Ucp Icp


Bin son: Nh Trng 51

Cao p Mi i tng - - Trung p th, cng nghip Nng thn - H p - Nng thn th, cng

nghip

Tit din d chn theo phng php no cng phi tho mn cc iu kin k thut sau y: Ubt Ubtcp Usc Usccp Isc Icp Trong : Ubt, Usc l tn tht in p lc ng dy lm vic bnh thng v khi ng dy b s c nng n nht (t 1 ng dy trong l kp, t on dy trong mch kn). Ubtcp, Usccp - tr s U cho php lc bnh thng v s c. vi U 110kV: Ubtcp = 10%Um Usccp = 20%Um Vi U 35kV: Ubtcp = 5%Um Usccp = 10%Um

Isc, Icp dng in s c ln nht qua dy dn v dng in pht nng lu di cho php.

Ngoi ra, tit din dy dn ng dy trn khng phi tho mn cc iu kin bn c hc v tn tht vng quang in.

Ring vi cp mi cp in p phi tho mn iu kin n nh nhit dng ngn mch.

qdt..IF

Trong : - h s, vi nhm = 11, vi ng = 6 tq - thi gian quy i, vi ngn mch trung, h p cho php ly tq = tc (thi gian ct ngn mch), thng tc = (0,5 1)s.

3.4.2. La chn tit din theo Jkt Trnh t la chn tit din theo phng php ny nh sau:


Bin son: Nh Trng 52

1. Cn c vo loi dy nh dng (dy dn hoc cp) v vt liu lm dy (nhm hoc ng) v tr s Tmax tra bng chn tr s Jkt.

Bng 3.10. TR S Jkt THEO Tmax V LOI DY DN Tmax (h) Loi dy

5000 Dy ng 2,5 2,1 1,8 Dy A, AC 1,3 1,1 1 Cp ng 3,5 3,1 2,7 Cp nhm 1,6 1,4 1,2

Nu ng dy cp in cho nhiu ph ti c Tmax khc nhau th xc nh tr s trung bnh ca Tmax theo biu thc:

=

=

=

=

=n

1ii

n

1imaxii

n

1ii

n

1imaxii

maxtb

P

TP

S

TST

Trong : Si, Pi l ph ti in (ph ti tnh ton) ca h tiu th 2. Xc nh tr s dng in ln nht chy trn cc on dy

cosU3nP

U3nS

Idm

ij

dm

ijij ==

Vi n - s l ng dy (l n n = 1, l kp n = 2) 3. Xc nh tit din kinh t tng on

ij

ijktij J

IF =

Cn c vo tr s Fij tnh c, tra s tay tm tit din tiu chun nht nht b hn.

4. Kim tra tit din chn theo cc iu kin k thut. Nu khng tho mn phi nng tit din ln 1 cp v th li.

V d3.5: Yu cu la chn dy dn cho ng dy t 10kV cp in cho 2 x nghip nh hnh v. Cc s liu ph ti cho trong bng.

Ph ti S (kVA) cos Tmax (h) X nghip 1 2000 0,8 5200 X nghip 2 1000 0,7 4000


Bin son: Nh Trng 53

Hnh 3.3. ng dy 10kV cp in cho 2 x nghip GII

V ng dy cp in cho x nghip (c chiu di ngn, Tmax ln) chn tit din theo Jkt, dy AC.

( )h4800100020001000.40002000.5200Tmaxtb =+

+=

T Tmax = 4800h v dy AC tra bng c Jkt = 1,1 (A/mm2) Tr s dng in trn cc on:

( )( ) ( ) ( ) ( ) ( )A86,4

103.271012007001600

U3.2QQPP

I

A57,8.103

1000U3

SI

22

dm

221

221

A1

dm

212

=

+++=

+++=

===

Tit din kinh t mi on:

( )( )21

212

5,781,14,86

5,521,18,57

mmF

mmF

A ==

==

Tra bng tit din tiu chun chnj dy gn nht b hn on 12 chn AC 50 c xo = 0,65 (/km); ro = 0,368 (/km) on A1 chn 2AC 70 c xo = 0,46 (/km); ro = 0,36 (/km) Tn tht in p ln nht lc bnh thng (khng dy no b t) Umax = UA12 = UA1 + U12

( ) ( ) ( )VU 8,56110

368,0.71064,0.70010.2

36,0710120046,07001600max =

++

+++=

Umax = 561,8V > Ubtcp = 5%Udm = 500V Cn tng tit din on A1 ln 95mm2 c xo = 0,33 (/km); ro = 0,34 (/km)

( ) ( ) ( )V494,510

710.0,368700.0,642.10

0,3471012000,337001600Umax =

++

+++=

Umax = 494,5V < Ubtcp = 5%Udm = 500V

A 1 2 4km 3km


Bin son: Nh Trng 54

Khi s c mt dy trn dy kp A1, dng in trn ng cn li tng gp i. Isc = 2IA1 = 2.86,4 =172,8A Isc = 172,8A < Icp = 325A Vy chn dy dn cho ton b ng dy nh sau: on A1: 2AC 95 on 12: AC 50

3.4.3. Chn tit din dy dn theo Ucp

Xut pht t nhn xt: khi tit din dy dn thay i th in tr thay i theo cn in khng rt t thay i, tra s tay thy xo (/km) c gi tr xo = 0,33 0,45 bt k c dy dn v khong cch gia cc pha. V th cho mt tr s xo ban u nm trong khong gi tr trn th sai s l khng ln. Tn tht in p c xc nh theo biu thc bit

UUUQX

UPR

UQXPR

Udmdmdm

+=+=+

=

Khi cho gi tr xo tnh c:

dmo

dm UQl

xUQXU ==

T y xc nh c U = Ucp - U

Mt khc dm

o

dm F.U.lP.

xUPRU ==

Suy ra: dm

o.UU

.lP.F

=

Vy trnh t xc nh tit din dy theo phng php ny nh sau: 1. Cho mt tr s ln cn 0,4 (/km), trng hp tng qut ng dy n ti,

tnh c: ln

i.Q

UxU

1ij

dm

o =

=

2. Xc nh thnh phn U : U = Ucp - U 3. Xc nh tit din tnh ton theo Ucp


Bin son: Nh Trng 55

l.P.UUF

dmo

=

Chn tit din tiu chun gn nht ln hn, 4. Kim tra li tit din chn theo cc tiu chun k thut

Trong cc cng thc trn: Q(kVAr), P(kW), l(km), U(V), Um(kV)

V d 3.6. DK 10 kV cp in cho 2 ph ti (hnh v). Cho bit tn tht in p cho php t im r A n ph ti 2 l 3%Um. Yu cu xc nh tit din dy dn cho ng dy.

Hnh 3.4. ng dy cp in cho 2 ph ti GII

Dy c chn theo Ucp, loi dy AC Chn xo = 0,35 (/km),

( )[ ] ( )V75,6200.0,6.3.5200.0,65400.0,610

0,35U =++=

U = 3%Um - U = 300 75,6 = 224,4 (V) Tit din tnh ton:

( )[ ] 240,43mm200.0,8.35200.0,8400.0,810.224,4

31,5F =++=

Chn tit din tiu chun 50mm2 AC 50 Kim tra li: Tra s tay vi dy AC 50, treo trn nh tam gic u cch 2m c xo = 0,393 (/km); ro = 0,64 (/km).

Hnh 3.5. S thay th ng dy 10kV ZA1 = 064.5 + j0,393.5 = 3,2 + j1,965 () Z12 = 064.3 + j0,393.3 = 1,92 + j1,179 ()

A 1 2 5km 3km

4000,8 kVA 2000,8 kVA

A 1 2 3,2 + j1,965() 1,92 + j1,179()

320 + j240()

160 + j120()


Bin son: Nh Trng 56

( ) ( ) ( )V,269210

120.1,179160.1,19210

1,9651202403,2016320UA12 =

++

+++=

UA12 = 269,2V < Ucp = 300V Vy chn dy AC 50 cho ton tuyn l hp l. Cng thc xc nh tit din theo Icp rt n gin

K1K2Icp Itt (*) Trong : K1 - h s hiu chnh nhit k n s chnh lch nhit mi trng ch to v mi trng t dy, tra s tay. K2 - h s hiu chnh nhit , k n s lng cp t chung mt rnh, tra s tay. Icp dng pht nng cho php, nh ch to cho ng vi tng loi dy, tng tit din dy, tra s tay. Itt dng in lm vic ln nht (di hn) qua dy. Tit din dy sau khi chn theo (*) phi th li mi iu kin k thut, ngoi ra cn phi kim tra iu kin kt hp vi cc thit b bo v. Nu bo v bng cu ch:

K1K2Icp

Idc

= 3 vi mch ng lc (cp in cho cc my) = 0,8 vi mch sinh hot Nu bo v bng ptmt

K1K2Icp 1,51,25IdmA

Vi 1,25IdmA l dng khi ng nhit (Ik.nh) ca ptmt, trong 1,25 l h s ct qu ti ca ptmt.

V d 3.7. Yu cu la chn dy dn cp in cho ng c my mi c s liu k thut cho theo bng di dy, bit rng dy dn i chung mt rnh vi 5 dy khc, nhit mi trng +30oC, dy c bo v bng cu ch c Idc = 50A. ng c Pm(kW) cos kmm My mi 10 0,8 5 0,9

GII Dng in lu di ln nht qua my mi l dng nh mc:


Bin son: Nh Trng 57

21,125A0,9.0,38.0,8.3

10II dmDtt ===

Tra bng chn cp ng 4 li PVC (3.2,5 + 1.2,5) c Icp = 36A T nhit mi trng tra s tay c K1 = 0,94 Vi 6 cp i chung mt rnh tra s tay c K2 = 0,75.

K1K2Icp = 0,94.0,75.36 = 25,38A > 21,125A Th li iu kin kt hp cu ch bo v:

K1K2Icp = 25,38A > A67,16350

3Idc

==

Khng cn kim tra U v ng dy ngn. Khng cn kim tra n nh nhit dng ngn mch v xa ngun.

BI TP CHNG 3 Bi tp 1. Yu cu la chn b CD CCT v cc cu ch nhnh t trong t ng lc cp in cho mt nhm my cng c c cc s liu cho trong bng di y ng c Pm (kW) cos Kmm Kt My mi 12 0,8 5 0,8 0,9 My khoan 7,5 0,8 5 0,8 0,9 My tin 1 5 0,8 5 0,8 0,9 My tin 2 4 0,8 5 0,8 0,9 Qut gi 1,7 0,8 5 0,8 0,9

Bi tp 2. Cho mt ng dy 10kV cp in cho ph ti cho trn hnh v. Bit khong cch trung bnh gia cc pha l 1m. Yu cu la chn tit din dy dn theo Jkt.

Bi tp 3. Mt ng dy 10kV cp in cho 3 ph ti (hnh v), bit khong cch trung bnh gia cc pha l 0,8.Yu cu la chn tit din dy dn theo hao tn in p cho php. Ucp = 5%.

3km A 2

500 0,8(kVA)

1 2km

800 0,6(kVA) 200 0,8(kVA)

2km 3

3km A 2

18 0,85(kVA)

1 4km

20 0,8(kVA) 15 0,8(kVA)

6km 3


Bin son: Nh Trng 58

Chng 4. NNG CAO H S CNG SUT 4.1. H S CNG SUT V NGHA CA VIC NNG CAO H S CNG SUT 4.1.1. Khi nim chung

in nng l nng lng ch yu ca cc x nghip. Cc x nghip ny tiu th khong 70% lng in nng xn xut ra. V th vn s dng hp l v tit kim in nng trong cc x nghip c ngha rt ln.

V mt sn xut in nng, vn t ra l phi tn dng ht kh nng ca c nh my sn xut ra c nhiu in nng, ng thi v mt dng in phi ht sc tit kim in nng, gim tn tht in nng n mc thp nht, phn u 1kWh in nng ngy cng lm ra nhiu sn phm hoc chi ph in nng trn mt n v sn phm ngy cng gim.

Tnh ton trong ton h thng in, tn tht in nng trong qu trnh truyn ti v phn phi chim khong 10 15%, trong tn tht in nng cho mng 1 10kV (mng in x nghip) chim ti 64,4% tng tn tht in nng. Nguyn nhn l mng in x nghip thng dng in p tng i thp, ng dy ti v phn tn gy ra tn tht ln. V th cc bin php tit kim in nng trong x nghp c ngha rt quan trng khng nhng c li cho bn than x nghip m cn c li cho nn kinh t quc dn.

H s cos l mt ch tiu nh gi x nghip dng in c hp l v tit kim khng. Do nh nc ban hnh cc chnh sch khuyn khch cc x nghip nng cao h s cos . H s cos ca cc x nghip ni chung l thp khong 0,6 0,7%, chng ta phi phn u nng cao h s cos ln 0,9.

Tit kim in nng v nng cao h s cos khng phi l bin php tm thi i ph m l bin php lu di vi mc ch pht huy hiu qu cao nht trong qu trnh truyn ti, phn phi v s dng in nng.

4.1.2. Cc bin php tit kim in nng

- To thi quen tt cc thit b in khi ra khi phng - Bo tr tt cc thit b in

i vi thit b in, cc ng c in phi c bo ng theo nh k


Bin son: Nh Trng 59

i vi thit b chiu sng phi thng xuyn lau chi bi bn bm vo b ngoi tng hiu qu pht sng.

- Nng cao h s cos - S dng cc thit b in c hiu sut cao: n compact, ng c in

c hiu sut cao.

4.1.3. ngha ca vic nng cao h s cos 4.1.3.1 Cc nh ngha ca vic nng cao h s cos

* H s cos tc thi: l h s cng sut ti mt thi im no o c nh dng c o cos hoc nh dng c o cng sut, in p.

IUP

..3cos =

Do ph ti lun bin i nn cos tc thi lun bin i. V th h s cos thc thi khng c gi tr trong tnh ton.

* H s cos trung bnh: l h s cos trong mt khong thi gian no (mt ca, mt ngy em, mt thng)

tb

tb

PQ

artgcoscos =

H s cos dng nh gi mc s dng in tit kim v hp l ca x nghip.

* H s cos t nhin: l h s costb tnh cho c nm khi khng c thit b b. H s cos t nhin dng lm cn c tnh ton, nng cao h s cng sut b v cng sut phn khng. 4.1.3.2. ngha ca vic nng cao h s cos

Nng cao h s cos l mt trong nhng bin php quan trng tit kim in nng. Phn ln cc thit b in u s dng cng sut phn khng Q v cng sut tc dng P. Nhng thit b tiu th nhiu cng sut phn khng l:

- ng c khng ng b: chng tiu th khong 60 65% tng cng sut phn khng ca mng.

- My bin p tiu th khong 20 25% - ng dy trn khng, in khng v cc thit b khc tiu th

khong 10%.


Bin son: Nh Trng 60

Cng sut tc dng P l cng sut bin thnh c nng hoc nhit nng trong cc thit b b in.

Cng sut phn khng: l cng sut t ho trong cc my in xoay chiu, n khng sinh ra cng. Qu trnh trao i cng sut ph khng gia my in v h dng in l mt qu trnh dao ng. Mi k ca dng in Q i chiu 4 ln, gi tr trung bnh trong mt na chu k ca dng in bng khng. Cho nn vic to ra cng sut phn khng khng i hi tiu tn nng lng ca lng c s cp quay my pht in. Mt khc cng sut phn khng cung cp cho cc h tiu th khng nht thit phi ly t ngun. V vy trnh vic truyn ti mt lng cng sut phn khng ln trn ng dy, ngi ta t gn cc h tiu th cc thit b sinh ra Q (t b, my b ng b) cung cp trc tip cho ph ti. Lm nh vy gi l b cng sut phn khng. Khi b cng sut phn khng th gc lch pha l:

PQ

artg=

Khi lng P khng i nh c b cng sut phn khng lng Q truyn ti trn ng dy gim gim cos tng.

H s cos tng s a n hiu qu sau: - Gim c tn tht cng sut trn mng in

( ) ( )QQRRR +=+=+

= P2

2

2

2

2

22

PUQ

UP

UQP

P

Khi Q gim s gim c P(Q) - Gim c tn tht in p trong mng in

( ) ( ) ( )Qp UUUQX

UPRXR

UQP

U +=+=++=

Khi Q truyn ti trn ng dy gim th U(Q) gim. - Tng kh nng truyn ti ca ng dy v my bin p. Kh nng

truyn ti ca ng dy v my bin p ph thuc vo iu kin pht nng tc l ph thuc vo ICP. Dng in truyn ati trong dy dn v trong my bin p c tnh theo cng thc:

UQP

USI

33

22 +==


Bin son: Nh Trng 61

Biu thc trn cho ta thy nu cng mt iu kin pht nng ca dy dn v ca my bin p, chng ta c th tng kh nng truyn ti cng sut tc dng P ca chng bng cch gim cng sut Q. V nhng l do trn m vic nng cao h s cos, b cng sut phn khng tr thnh vn quan trng cn phi quan tm ng mc khi thit k cng nh vn hnh h thng cung cp in.

4.2. CC BIN PHP NNG CAO H S cos T NHIN Nng cao h s cos t nhin l tm cc bin php cc h dng in

gim bt c lng cng sut phn khng Q nh: p dng cc qu trnh cng ngh tin tin, s dng hp l cc thit b in.

Nh vy nng cao h s cos t nhin rt c li v a li hiu qu kinh t m khng phi t thm cc thit b b. V th ki xt n cc vn nng cao h s cos bao gi cng xt n c bin php nng cao h s cos t nhin trc sau mi xt n bin php b.

1. Thay i v ci tin quy trnh cng ngh cc thit b lm vic ch hp l nht.

Cn c vo iu kin c th cn sp xp quy trnh cng ngh mt cch hp l nht. Vic gim bt cc ng tc, nhng nguyn cng tha v p dng cc bin php gia cng tin tin u a n hiu qu tit kim in nng gim bt in nng tiu th cho mt n v sn phm. Trong cc x nghip, cc thit b cng sut ln thng l ni tiu th nhiu in nng. V th nghin cu cc thit b vn hnh ch kinh t v tit kim in nht.

V d: cc nh my c kh, my nn kh thng tiu th khong 30 40% in nng cung cp ton nh my. V vy vic nh ch vn hnh hp l cho my nn kh c nh hng ln n vn tit kim in nng. Qua kinh nghim vn hnh khi h s ca my nn kh xp x bng 1 th in nng tiu hao cho mt n v sn phm s gim ti mc thi thiu. V vy cn b tr cho cc my nn kh lun lm vic ch y ti, lc ph ti ca x nghip nh c th ct bt my nn kh.

2.Thay th ng c khng ng b lm vic non ti bng ng c c cng sut nh hn

Khi lm vic ng c khng ng b tiu th lng cng sut phn khng bng:


Bin son: Nh Trng 62

( ) 2pt.koQdmQOQQ += Trong : Qo lc ng c lm vic khng ti Qdm cng sut phn khng lc ng c lm vic ch

nh mc Kpt - h s ph ti Cng sut phn khng Qo = (60 70)% Qdm H s cng sut ca ng c c tnh theo cng thc sau:

( )( )*

.

1

1cos

1

1cos

220

222

+

+

=

+

=

+==

dm

ptOdm

PkQQQ

PQQP

PSP

Biu thc (*) cho ta thy: nu ng c lm vic non ti th cos s thp. V d: Nu ng c c kpt = 1 th cos = 0,8 Kpt = 0,3 th cos = 0,51 Nhn xt: thay th ng c lm vic non ti bng ng c c cng sut

nh hn s tng h s mang ti do nng cao c h s cos . iu kin kinh t cho php thay th ng c l vic thay th ng c

phi gim c cng sut tc dng trong mng v trong ng c. Nu m bo c iu kin trn th vic thay th ng c mi c li.

- Nu kpt < 0,45 th vic thay th bao gi cng c li - Nu 0,45 < kpt < 0,7 th phi so snh kinh t k thut mi xc nh

c vic thay th ng c c li hay khng. iu kin k thut cho php thay th ng c l: vic thay th phi m bo nhit, m bo iu kin m my v lm vic n inh. 3. Gim in p ca nhng ng c lm vic non ti.

Bin php ny c dng khi khng c iu kin thay th ng c lm vic non ti bng ng c c cng sut nh hn.

Trong thc t ngi ta thng dng cc bin php sau y gim in p t vo ng c lm vic non ti:

- i ni dy qun Stato t Y. Khi thay i dy qun t tam gic sang sao th in p t ln mt pha ca ng c s gim i 3 ln, do cos


Bin son: Nh Trng 63

v hiu sut ca ng c cng tng ln. ng thi mmen cc i ca ng c s gim i 3 ln so vi trc v vy phi kim tra li kh nng m my v iu kin lm vic n nh ca ng c. Bin php ny thng dng cho cc ng c c in p nh hn 1000V v kpt nm trong khong 0,35 0,4.

- Thay cch phn nhm ca dy qun stato. Bin php ny thng c dng vi ng c c cng sut ln v c nhiu mch nhnh song song trong mt pha. Bin php ny kh thc hin v phi tho ng c ra mi thay i c cch u cc cun dy ca stato.

- Thay i u phn p ca my bin p h thp in p ca mng in phn xng. Bin php ny ch c thc hin khi tt c cc ng c trong phn xng u lm vic ch non ti v phn xng khng c cc thit b yu cu cao v mc in p. Trong thc t bin php ny t c s dng.

4. Hn ch ng c chy khng ti Bin php ny c thc hin theo hai hng: - Vn ng cng nhn hp l ho cc thao tc, hn ch n mc thp

nht thi gian my chy khng ti. - t b hn ch chy khng ti trong s khng ch ng c. Thng

thng nu ng c chy khng ti qu thi gian chnh nh no th ng c b ngt ra khi mng in.

5. Dng ng c ng b thay th ng c khng ng b V ng c ng b c mt s u im r rt sau y so vi ng c

khng ng b. - H s cng sut cao, khi cn c th lm vic ch qu kch thch

tr thnh my b cung cp cng sut phn khng cho mng. - Mmen quay t l bc nht vi in p, v vy t ph thuc vo ph ti

do nng sut lm vic ca my cao hn. Khuyt im: cu to phc tp, gi thnh cao. V vy ng c ng b ch chim khong 20%. 6. Nng cao cht lng sa cha ng c

Do cht lng sa cha ng c khng tt nn sau khi sa cha tnh nng ca ng c thng km hn trc: tn tht trong ng c tng ln, h s cos gim V vy cn ch n khu nng cao cht lng sa cha ng c gp phn ci thin vn h s cos ca x nghp.


Bin son: Nh Trng 64

4.3. CC BIN PHP NNG CAO H S cos NHN TO 4.3.1. Khi qut

Bng cch t cc thit b b gn cc h tiu th cung cp cng sut phn khng, ta gim c cng sut phn khng truyn ti trn ng dy v do cos ca mng tng ln. Bin php b khng gim c cng sut phn khng ca c h m ch gim c cng sut phn khng truyn ti. V vy sau khi thc hin cc bin php nng cao h s cos t nhin m vn khng t yu cu th mi xt n cc bin php b.

Cn ch thm b cng sut phn khng Q ngoi mc ch l nng cao h s cng sut cos tit kim in, n cn c tc dng khng km phn quan trng l iu chnh v n nh in p cung cp.

B cng sut phn khng a li hiu qu kinh t nhng tn km thm v mua sm thit b b v chi ph vn hnh cho chng. V vy quyt nh phng n b phi da trn c s tnh ton v so snh kinh t, k thut.

4.3.2. Chn cc thit b b

1. T in L loi thit b in tnh, lm vic vi dng in vt trc nh p c th

sinh ra cng sut phn khng cung cp cho mng. * u im: - Tn tht cng sut b, khng c phn t quay nn lp rp bo qun d

dng - T in c ch to thnh tng n v cng sut nh, v th c th

tu theo s pht trin ca ph ti trong qu trnh sn xut l chng ta ghp dn t in vo mng do hiu sut s dng cao v khng phi b nhiu vn u t ngay mt lc. * Nhc im:

- Rt nhy cm vi inp t ln t: Qc = 3U2..C.10-3 - T in c cu to km chc chn, d b ph hng khi b ngn mch.

Khi in p tng ln 110%Um th t in b chc thng. - Khi ng t in vo mng s c dng in xung, cn khi ngt t ra

khi mng trn cc ca t in vn cn in p d c th gy nguy him cho ngi vn hnh.


Bin son: Nh Trng 65

* ng dng: T in c s dng rt rng ri trong cc x nghip c cng sut trung bnh v nh. Thng thng nu dung lng b 5000kVr th ngi ta dng t, cn dng lng > 5000kVr th phi so snh gia dng t v my b ng b. 2. My b ng b L mt ng c ng b lm vic ch khng ti. Do khng c ph ti nn my b ng b c ch to gn nh v r hn my my b cng cng sut. ch qu kch thch my b ng b sn xut ra cng sut phn khng cho mng. cn ch thiu kch thch my b l thit b rt tt iu chnh in p trong mng. * Nhc im:

- Gi thnh cao - My b c phn t quay nn lp rp, bo qun, vn hnh kh khn.

* ng dng: - c s dng nhng ni quan trng hoc nhng ni tp trung vi

lng b ln. 3. ng c khng ng b dy qun c ng b ho Khi cho dng in mt chiu vo rto, ng c s lm vic nh ng c ng b v sinh ra cng sut phn khng. * Nhc im: Hao tn cng sut kh ln, kh nng qu ti ln v vy ng c ch lm vic 75% cng sut nh mc. * ng dng: ng c khng ng b dy qun c ng b ho rt t c s dng. Ngoi ra c th to ra cng sut phn khng bng cch:

- Dng ng c ng b lm vic ch qu kch thch - Dng my pht lm vic ch b.

4.4. XC NH DUNG LNG B Dung lng b c xc nh theo cng thc:

Qb = P(tg1 - tg2). Trong : P cng sut tc dng ca ph ti tnh ton 1 gc tng ng vi costb1 trc khi b. 2 gc tng ng vi costb2 sau khi b


Bin son: Nh Trng 66

costb2 c ly bng h s cng sut do c quan qun l h thng in quy nh, thng ly bng 08 0,95. : H s tnh n kh nng nng cao h s cos bng phng php t nhin, thuc khong 0,9 1,0. * Thng ngi ta xc nh theo dung lng b ti u Do khi b c th tit kim c dung lng cng sut tc dng:

Ptk = kkt.Qb kb.Qb = Qb (kkt kb) Trong : kkt - ng lng kinh t ca cng sut phn khng kW/kVAr Kb - sut tn tht cng sut tc dng trong thit b b, kW/kVAr kkt c xc nh nh sau:

Trc khi b Q gy ra tn tht cng sut tc dng:

RUQ

P 22

1 =

Sau khi b mt lng Q gy ra tn tht cng sut tc dng l:

RU

)Q-(QP 2

2bu

2 =

Vy lng cng sut tc dng tit kim c l:

RU

)Q-(Q-R

UQ

PPP 22

bu2

2

21 ==

Theo nh ngha ta c:

== Q

Q2UQR

QPk bu2

bukt (Kw/kVAr)

Nu dung lng Qb nh hn nhiu so vi cng sut phn khng truyn ti trn ng dy Q (iu ny thng xy ra trong thc t), tc l c th coi

0=QQbu

, lc ny ng lng kinh t ca cng sut phn khng c tnh theo

cng thc n gin sau:

2kt UQR2k =

Nu Q v R cng ln th kkt cng ln, ngha l nu ph ti phn khng cng ln v cng xa ngun th vic b cng c hiu qu kinh t. Dung lng b ti u

Qb t.u = Q - 2RU2 kb (*)


Bin son: Nh Trng 67

Dung lng b ti u

Qb. t.u = Q - 2RRU2 (**)

T (*) v (**) ta c: Qb t.u

=

ktkbuk1Q

4.5. PHN PHI DUNG LNG B TRONG MNG IN 4.5.1. V tr t thit b b

1. t t b pha cao p ca x nghip: t ti v tr ny c li l gi t cao p thng r hn t h p, tuy nhin ch lm gim tn tht in nng t v tr t t b tr ln li in, khoong gim c tn tht in nng trong mng bin p v li h p x nghip.

2. t t b ti thanh ci h p ca TBA x nghip. T in t ti v tr ny lm gim tn tht in nng trong trm bin p v cng khng lm gim tn tht in nng trn li h p x nghip.

3. t t b ti cc t ng lc. t t b ti cc im nu lm gim c tn tht in nng trn cc ng dy t t phn phi ti cc t ng lc v trong trm bin p x nghip.

4. t t b ti cc ca tt c ng c. t b ti cc ng c c li nht v gim tn tht in nng, tuy nhin vn u t s cao v tng chi ph qun l, vn hnh, bo dng t.

Trong thc t b cos cho x nghip, tu theo quy m ca x nghip v kt cu ca li in x nghip, ngi ta tin hnh b nh sau:

1. Vi 1 xng sn xut hoc x nghip nh nn t tp trung t b ti thanh ci h p trm bin p x nghip

2. Vi x nghip loi va c 1 trm bin p v mt s phn xng vi cng sut kh ln v kh xa trm bin p, gim tn tht in nng trn cc ng dy t TBA n cc phn xng c th t phn tn t b ti cc t phn phi phn xng v ti cc ng c c cng sut ln (50 70kW).

3. Vi x nghip quy m ln bao gm hng chc phn xng, thng li in kh phc tp gao gm trm phn phi trung tm v nhiu trm bin p phn xng, khi xc nh v tr v cng sut b thng tnh theo 2 bc:


Bin son: Nh Trng 68

- Bc 1: xc nh cng sut b t ti thanh ci h p tt c cc TBA phn xng.

- Bc 2: phn phi cng sut b ca tng trm ( xc nh c t bc 1) cho cc phn xng m trm bin p cp in. 4. Cng c th xt t b ton b pha cao p, hoc 1 phn b bn cao, 1 phn b bn h p tu thuc vo chnh lch gi t cao v h p.

Trong trng hp b t nhiu im (trng hp 2 v 3), cng sut b ti u ti im i no xc nh theo biu thc:

Qb i = Qi (Q - Qb )iRtdR

Trong : Qi cng sut phn khng yu cu ti nt i Q - tng cng sut phn khng yu cu, Q =

=

n

1i iQ

Qb - tng cng sut b, xc nh theo bc 1 ca trng hp 3 Ri - in tr nhnh n v tr nt i Rt - in tr tng ng ca li in

Rt

n21 R1

...

R1

R1

1

+++=

V d 4.1. Mt xng c kh nng nghip cng sut 100kW, cos = 0,6. Yu cu xc nh cng sut b t b nng cos = 0,9.

GII Vi cos = 0,6 tg = 1,33 Vi cos = 0,9 tg = 0,48 Tng cng sut b t b l:

Qb = P(tg1 - tg2) = 100(1,33 0,48) = 85kVAr V d 4.2. X nghip c kh gm 3 phn xng c mt bng v s liu ph ti cho trn hnh v. Yu cu t t b bn cnh cc t phn phi ca 3 phn xng cos = 0,95.

3S& = 50 + j70kVA 2S& = 50 + j50kVA 1S& = 50 + j120kVA

PX1 PX2 PX3


Bin son: Nh Trng 69

Cc ng cp t TBA v 3 phn xng c cc s liu cho trong bng ng dy Loi cp l(km) ro(/km) R() TBA PX1 PVC(3x50+1,35) 50 0,387 0,0194 TBA PX2 PVC(3x25+1,16) 70 0,727 0,0509 TBA PX3 PVC(3x16+1,10) 100 1,15 0,115

GII Tng cng sut tnh ton ca x nghip

j240kVA180SSS 321 +=++ &&&

0,33tg0,95cos

1,33180240

PQ

1tg

22 ==

===

Tng cng sut phn khng cn b ti 3 phn xng nng cos ca x nghip ln 0,95 l:

Qb = P(tg1 - tg2) = 180(1,33 0,33) = 180kVAr S thay th v s tng ng li in h p dng xc nh Qb i

in tr tng ng ca li in h p x nghip

Rt ( )=++

=

+++= 0126,0

115,01

0505,01

0194,01

1

R1

...

R1

R1

1

n21

p dng cng thc: Qb i = Qi (Q - Qb )

iRtdR

Xc nh c cng sut cc t t b t ti 3 phn xng:

TBA

Q1 Qb1

Q2 Qb2

Q3 Qb3

R1

R2

R3 Q Qb

Rt TBA


Bin son: Nh Trng 70

Qb1 = 120 (240 - 180 ) 0,01940,0126

= 81kVAr

Qb2 = 50 (240 - 180 ) 0,05090,0126

= 35kVAr

Qb3 = 70 (240 - 180 ) 0,1150,0126

= 64kVAr.

BI TP CHNG 4 Bi tp 1. Mt siu th ln, c ph ti tnh ton P = 600kW, cos = 0,9. Yu cu la chn t b t ti thanh ci h p TBA siu th nng cos = 0,9. Bi tp 2. Mt trm bm cao p 10kV t 5 my bm 200kW, cos = 0,7. Yu cu la chn t b 10kV cho trm bm cos = 0,9. Bi tp 3. Mt mng hnh tia c 4 nhnh, in p 6kV, in tr v ph ti phn khng ca tng nhnh nh sau: r1 = 0,1, Q1 = 400KVAr r2 = 0,05, Q1 = 400KVAr r3 = 0,06, Q1 = 500KVAr r1 = 0,2, Q1 = 200KVAr Dung lng b ca mng Qb = 1200 kVAr. Hy tnh dung lng b ca tng nhnh.


Bin son: Nh Trng 71

Chng 5. TNH TON CHIU SNG 5.1. KHI NIM CHUNG V CHIU SNG

Chiu sng ng vai tr ht sc quan trng trong i sng sinh hot cng nh trong sn xut cng nghip. Nu nh sng thit s gy hi mt, hai sc kho, lm gim nng sut lao ng, gy ra th phm ph phm, gy tai nn lao ng c bit, c nhng cng vic khng th tiwns hnh c nu thiu nh sng hoc nh sng khng tht (ngha l khng ging nh sng ban ngy) nh b phn kim tra cht lng my, b phn pha ch ho cht, b phn nhum my

C nhiu cch phn loi c hnh thc chiu sng. - Cn c vo i thng cn chiu sng chia ra chiu sng dn dng v

chiu sng cng nghip. Chiu sng dn dng bao gm chiu sng cho cn h gi nh, cc c quan, trng hc, bnh vin, khch sn Chiu sng cng nghip nhm cung cp nh sng cho cc khu vc sn xut nh nh xng, kho bi

- Cn c vo mc ch chiu sng chia ra chiu sng chung, chiu sng cc b, chiu sng s c, chiu sng chung to nn sng ng u trn ton b din tch cn chiu sng (phng khch, hi trng, nh hng, phn xng). Chiu sng cc b l hnh thc tp trung nh sng vo 1 im hoc 1 din tch hp (bn lm vic, chi tit cn gia cng chnh xc nh tin, khoan, ng ch my khu). Chiu sng s c l hnh thc chiu sng d phng khi xy ra mt in li nhm mc ch an ton cho con ngi trong cc khu vc sn xut hoc ni tp trung ng ngi (nh ht, hi trng).

- Ngoi ra cn chia ra chiu sng trong nh, ngoi tri, chiu sng trang tr, chiu sng bo v Mi hnh thc chiu sng c yu cu ring, c im ring, dn ti phng php tnh ton, cch s dung loi n, b tr n khc nhau.

5.2. MT S I LNG DNG TRONG TNH TON CHIU SNG. 5.2.1. Quang thng

Nng lng do mt ngun sng pht ra qua mt in tch trong mt n v thi gian gi l thng lng ca quang nng. Nhng nh sng ca ngun quang pht ra gm nhiu song in t c di song khc nhau, do nng lng ca ngun quang in biu th bng biu thc:


Bin son: Nh Trng 72

=2

1

21

deE

Trong : e - hm phn b nng lng - bc sng E12 thng lng ca quang nng t 1 n 2 Thng lng ton phn:

=

0deE

Trong ngun quang c cng sut kh ln, nhng c cc bc song khc nhau, s gy cho mt ta cm gic nhau. Do , ngi ta a thm vo khi nim r, k hiu V.

Cui cng ngi ta nh ngha quang thng l tch phn ca thng lng quang nng v hm r V:

=

0 deVF

n v ca quang thng l Lumen (lm).

5.2.2. Cng nh sng

Nu c mt ngun sng S bc x theo mi phng, trong gc c d n

truyn i mt quang thng dF th i lng ddF

gi l cng nh sng ca

ngun sng trong phng : ddFI =

Nu dF tnh bng Lumen, gc c tnh bng st-ra-i-an th cng nh sng tnh bng nn quc t, gi tt l Nn, k hiu l Cd, 1Cd = 1lm/1sr.

5.2.3. trng v ri

Mt ngun sng c kch thc gii hn, trn ly mt din tch dS, quang thng bc x theo mi phng ca gc c 2pi gi l dF th trng ca ngun sng c nh ngha:

dSdFR = (*)


Bin son: Nh Trng 73

Nh vy trng l quang thng bc x trn mt n v din tch ca ngun.

Ngc li, ri l phn quang thng thi trn mt n v din tch dS. ri k hiu l E.

Hnh 5.1. Hnh minh ho xc nh ri Gi thit c ngun sng C, din tch c chiu sng dS c phng php

tuyn N (hnh v), thng lng ca ngun C i qua din tch dS l dF = ld.

2r

dScosd =

r l khong cch t C n tm dS Thay vo cng thc (*) ta c:

dSrldScos

dSdFE 2

==

Vy ri ca ngun sng t l thun vi cng nh sng v t l nghch vi bnh phng khong cch t ngun ti tm din tch c chiu sng, ngoi ra cn ph thuc vo hng ti ca ngun.

Tm li, ngi ta nh ngha mt quang thng ri trn mt b mt gi l ri, n v l lux (vit tt l lx)

SFE =

Trong : F quang thng

gt cung cap dien

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