giáo trình cung cấp Điện

Gio Trnh Cung Cp in

LI NI Ut 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

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Gio Trnh Cung Cp in

GII THIU CHUNG V CUNG CP IN1. 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.

2

Gio Trnh Cung Cp in 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.3

Gio Trnh Cung Cp in 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 My pht in Trm bin p Trm phn phi My bin p 2 cun dy, 3 cun dy K hiu trn bn v F,

My ct in Dao cch ly, cu dao4

Gio Trnh Cung Cp in

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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, vnA V

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Gio Trnh Cung Cp in


Chng 1: TNH TON PH TI IN1.1. 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 dng6

Gio Trnh Cung Cp in


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: Ptt = k dt . k ti Pdmii =1 n

Q tt = Ptt .tg Trong : bm kt - h s ng thi, ly theo tc t k dt = n lv n 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 = k dmii =1 n

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

vi

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

7

Gio Trnh Cung Cp in 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 (m ). Mt phng hc ca trng ph thng thng c din tch S = 8x10 = 80m . Po - Sut ph ti trn n v din tch Po = (15 20) (W/m ). 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. Ppi =1 n

2

2

2

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 80m v khu nh thng trc, hiu trng, phng hp, gio vin c tng din tch 100m . GII Ph ti mt phng hc vi Po = 15 (W/m ) 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:82 2 2

Gio Trnh Cung Cp in 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 phnS = P 16,4 = = 20,5 (kVA) cos 0,8

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: n PX = kt. P i=1 tti n QX = kt. Q i=1 tti P 22+ Q x x kt - h s ng thi vi n = 1, 2 kt =1 n = 3, 4 kt =0,9 0,95 x9

S

=

Gio Trnh Cung Cp in 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 + 100m 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 = 316 2 + 219 = 385 (kVA)2

2

10

Gio Trnh Cung Cp in


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:Ptt = a.M T max

-

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:11

Gio Trnh Cung Cp in


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 thcPtt = a.M T max

Tra cm nang vi nh my sn xut xe p ao = 200 (kWh/xe) v Tmax = 5000 (h), xc nh c ph ti in;Ptt = a.M 200.10 4 = 400 (kW) = Tmax 5000

Tip tc tra cm nang c cos = 0,6 tg = 1,33 Qtt = Ptt.tg = 400.1,33 = 533 (kVAr)S= Ptt cos = 400 = 667 (kVA) 0,6

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) n P = P mi i=1 Pm cng dut nh mc ca tng my (ng c)

vi

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 (m ) 2 Po Cng sut chiu sang trn n v din tch (W/m )122

Gio Trnh Cung Cp in


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/m ) 2 vi cc phn xng dt, may, ho cht P0 = 15 20 (W/m ) 2 vi kho bi P0 = 5 15 (W/m ) 2 vi xng thit k P0 = 25 30 (W/m ) 2 vi nh hnh chnh P0 = 20 25 (W/m ) 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 + Qcs2 2 S PX = PPX + Q PX

2

cos PX =

PPX Q PX

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:n

PXN = kt. n

PPXi i=1PXi i=1

QXN = kt. Q T y s tnh c ph ti ton phn v cos ca x nghip:2 2 S PX = PPX + Q PX

cos PX =

PPX Q PX2

V d 2.6. Mt xng sa cha t c cng sut t 200kW, din tch xng 2040 = 800m . 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.

13

Gio Trnh Cung Cp in


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/m ) 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 = 69,6 2 +103,8 SX2

2

= 125 (kVA)

cos = PX = 69,6 0,56125

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. Pmii=1 n

14

Gio Trnh Cung Cp in 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. Pn = Pmi1

n

i=1

3.

Xc nh cc t s: n =*1

2n 1

,P =*

P n

=

Pdmii =1 n

n

n

P

Pdmi i=1

4. 5.

Tra ph lc 4 (theo n* v P*) tm c nhq* 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 dy15*

Gio Trnh Cung Cp in 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. k % d 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 1. 2. 3. 4. 5. 6. 7. Cu trc Bin p hn My mi th My mi tinh My tin My khoan Qut gi Pm(kW) c im 14 12 10 7 5,5 4,5 1,7 Uf kd% = 36% Ud, kd% = 49% S lng 1 1 2 2 3 3 1

GII Trc ht phi quy i cc thit b v 3 pha v di hn Cu trc: Pq = Pm. k d % = 14. 36% = 8,4 (kW) Bin p hn Pq = 3 Pm. k d % = 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 1. Bin p hn S lng 1 Pm(kW) 14,5716

Gio Trnh Cung Cp in 2. 3. 4. 5. 6. 7. My mi th Cu trc My mi tinh My tin Qut gi My khoan 2 1 2 3 1 3

http://www.ebook.edu.vn 10 8,4 7 5,5 5,1 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* n 4 = 0,3 n* = 1 = n 13 42,97 P* = = 0,47 42,97 + 2,7 + 3.5,5 + 5,1 + 3.4,5 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. Pmi = 1,8.0,2.92,37= 33,25 (kW)i=1 n

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 1Bi 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

17

Gio Trnh Cung Cp in - tng trn th vin 200m2 2


1 xng thc tp c kh 300m , 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 24m 2 1 nh 2 tng, mi tng 6 phng hc 40m Yu cu xc nh ph ti in cn cp cho khu vn phng.2

18

Gio Trnh Cung Cp in


Chng 2: TNH TON TN THT IN P, TN THT CNG SUT, TN THT IN NNG2.1. 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 2.1A l, F 1& S1

AG 2 B 2

ZG 2 B 2

1& S1

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: R = l F

- 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 (mm /km), M = 18,8 (mm /km).192 2

Gio Trnh Cung Cp in


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. U 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 Y G B = +j 2 2 220

G=

Pvq2

Gio Trnh Cung Cp in


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. A lA1, FA1 1 & S1 A ZA1 1 & S1 Z12 l12, F12 2 & S2 2 & S2

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.1 Z1 Z2 1 Zo

Z0

& So

a)

b)

Hnh 2.3. S thay th my bin p hai cun dya) 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 ti21

Gio Trnh Cung Cp in


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:ZB =2 PN U dmB U U 3 + j N dmB .10 () .10 2 SdmB SdmB 2

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:Q o = I o S dmB . 100

Nu hai my bin p lm vic song songZ B = PN UdmB .103 + j U NU dmB .10 () 2 2SdmB 2SdmB2 2

V So = 2Po + j2Qo. 2.2. TNH TON TN THT IN P Tn tht i p l i lng phc (vc t phc) U& = U + jU . 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.

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Gio Trnh Cung Cp in


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.A U& jU 0 U B

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.QX PR + QX U = PR + U = U dm U dm dm

Nu P(kW), Q(kVAr), R, X(), Udm(kV) th U(V) 2.2.1. ng dy 1 ph tiA l, F 1& S1

A

ZA1

1& S1

P1 + jQ1 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 thcU = PR + QX = PR + QX dng Scos v dng P + jQ U dm U dm U dm

Tn tht in p trn on ng dy A1 l:U A1 = P1R A1 + Q1X A1 U dm

Trong : ZA1 = RA1 + jXA1 = rolA1 + jxolA1 & V S = & = S cos + jS sin . S A1 1 1 1

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Gio Trnh Cung Cp in


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.A AC 50, 5km 1 1000 0,8

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 () & S = & =10000,8 = 1000.0,8 + j1000.0,6 = 800 + j600 (kVA) S A1 1A 3,2 + j2 () 1 800 + j600 (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:U A1 = 800.3,2 + 600.2 = 376 (V) 10

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 A lA1, FA1 1 & S1 AS1 S2 S3

l12, F12

2 & S2

l23, F23

3 & S3

ZA1

1 P1+ jQ1S2 S3

Z12

2 P2+ jQ2S3

Z23

3 P3+ jQ3

Hnh 2.8. S nguyn l v s thay th ng dy lin thng cp in cho 3 ph ti24

Gio Trnh Cung Cp in


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 ( &2 ,&3 ) v tn tht trn on 2, 3 nhng cho php S S& S S b qua lng tn tht ny, S12 = & 2 +& 3 in p tn tht trn cc on nh sau:U 23 = U 23 = U A1 = P3 R 23 + Q 3 X 23 U dm

(P2

(P1 + P2

+ P3 )R 12 + (Q 2 + Q 3 )X12 U dm

+ P3 )R A1 + (Q1 + Q 2 + Q 3 )X A1 U dm

T y xc nh c tn tht in p trn ton b tuyn dyU + = U A123 =

(P 1 + P 2

(P2

+ P3 )R 12 + (Q 2 + Q 3 )X12 P3 R 23 + Q 3 X 23 + U dm U dmn

+ P3 )R A1 + (Q1 + Q 2 + Q 3 )X A1 + U dm

Tng qut U Trong :

PijR ij + QijX ij1

n

=

1

U dm

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, UA25

Gio Trnh Cung Cp in BI GII S thay th ng dy:A AC 50, 5km 1


AC 50, 2km

2

1000 0,8(kVA)

500 0,7(kVA)

A

1,6 + j1 ()

1

1,28 + j0,8()

2

800 + j600 (kVA)

350 + j350 (kVA)

Hnh 2.9. S nguyn l v s thay th DK 10(kV) ca v d 2.2Z A1 = 0,64.5 + j0,4.5 = 1,6 + j1 () 2

Z12 = 0,64.2 + 0,4.2 = 1,28 + j0,8 () S&= 1000 0,8 = 800 + j600 (kVA)1

S& =5000,7 = 350 + j350 (kVA) 2 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: (800 + 350)1,6 + (600 + 350)1 350.1,28 + 350.0,8 U = + = 279 + 72,8 = 10 10 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.26

351,8 V

Gio Trnh Cung Cp in


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

& S2

& S1

& S3

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.Umax

= MAX

U A12 U A13

U cp

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 phcS&= P + jQ

Trong : ng dy

P - tn tht cng sut tc dng do pht nng trn in tr Q - tn tht cng sut phn khng do t ho ng dy.27

Gio Trnh Cung Cp in


Tn tht cng sut trn ng dy xc nh theo biu thc:2 2 + Q &= I 2 Z = Z = P 2 S (R + jX) S 2 U dm U dm

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 . 1. ng dy 1 ph inA l, F 1& S1-3

A

ZA1

1

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: S& = A1S2 1 A2 U dm

Z A1 =

S2 12 U dm

Z A1 = PA1 + j A1

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 dyA AC 70, 5km 1 2000 0,6 A 2,3 + j2 () 1 92000 + j80000 (kVA)

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:28

Gio Trnh Cung Cp in2000 & = A1 2 102

http://www.ebook.edu.vn= 92000 + j8 0 0 0 0 ) = 92 + j8 0 kVA ) (VA (

(2,3 + j2)

2. ng dy n ph ti A lA1, FA1 1 & S1 A ZA1 1 & S1 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 S& qua ca on sau)& & & & S = S A12 = S A1 + S12 & S =

l12, F12

2 & S2

Z12

2 & S2

(P1 + P2 )2 + (Q 1U2 dmn n 1

+Q 2

)2)

Tng qut vi ng dy n ti& S =

S2 2 Z + Z A1 U 2 12 dm

Sij2 Z ij1

U2 dm

=

2 ij

+ Q2ij Z ij

U2 dm

Trong : n Sij, Pij, Qij Zij Udm

- s on ng dy hoc s ph ti - cng sut S, P, Q chy trn on ng dy ij - tng tr ca on ng dy ij - 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

29

Gio Trnh Cung Cp in


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.A AC 50, 5km AC 50, 3km 3 A 1,6 + j1 () 1 400 0,8(kVA) 1,28 + j0,8() 320 + j240 (kVA) 320 + j240 (kVA) 1 AC 50, 2km 2 500 0,8(kVA)

2 400 + j300 (kVA)

1,92+j1,20 (kVA) 3

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

(P + PS = S A1 + S12 + S13 = 1 2 & & & &

+ P 2+ Q + Q 3 1 2 Udm 2

) (

+Q 3

)2Z A1 +

S

2

S2

2 3 2 Z12 + 2 Z 13 U dm U dm 3

Thay s vo ta c:S = &

(700 + 400 + 320 )2 + (700 + 300 + 240 )2102

10 +

3

(1,6 + j) 50010

2

2

10

(1,28 + j0,8)

+

2 400 3 + 10 (1,92 + j1,2) = 2 10 = (56,86 + j35,54 ) + (3 + j2 ) + (3,07 + j1,92 ) = 63,56 + j39,46(kVA )

2.3.2. Tn tht cng sut trong tm bin p 1. Trm 1 bin p SdmB

& S1S& 0

& S1

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

30

Gio Trnh Cung Cp in


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:+ S& = B & 0 & S S cu = P B + jQ B

Trong :

I S &o = Po + S = Po + j o dmB jQ B 100 & Scu - tn tht trn 2 cun dy c th xc nh theo 2 cch

Theo tng tr bin p:2 & = S ZB S cu U2dm

Theo PN v UN2 2 S U NS dm S +j S&cu = Pcu + jQ cu = P NS 100 S dmB dmB

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 S & S B = Po + RB + o 2 U dm 2 I S S j dmB + 2 U dmB 100

Nu tnh & theo PN, UN th cu S& = B P S2 + + P N 2 Udm I S U S j o dmB + N 100 100 dmBS

o

2 U dmB 2

31 Gio Trnh Cung Cp in

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 10(kV) 0,4(kV)8000,6 (kVA) 1x1000(kVA) 5 + j30 (kVA) 1,2 + j5 (kVA) 8000,6 (kVA)

32

Gio Trnh Cung Cp in


Hnh 2.16. S nguyn l v s thay th trm bin p ca x nghip c khS = P + j IoS dmB = 5 + j 3.1000 = 5 + j30(kVA) o & 100 100 o 2 2 2 2 P UdmB U U 3 + j N dmB 10 = 1 2.10 10 3 + j 5.10 10 = 1,2 + j5( ) N Z = 10 B 2 S 1000 S2 1000 dm dmB& Xc nh Scu theo ZB: S& = cu S2 U2

Z=

800 2 102

(1,2 + j5)10 3

= 7,68 + j3 2(kVA)

Trng hp ny tn tht trong trm bin p S& = (5 + 7,68) + j(30 + 32) = 12,68 + j62 (kVA) cu 2. Trm bin p t 2 my ZB

& S12SdmB

& S1

S& 0

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 S& tng gp i. 02I S S&o = 2Po+ 2 jQ B = 2P + j o dmB o 1002 2 U U P U N dmB 103 + j N dmB 10 ZB = 2 2S 2S dmB dm

Tn tht cng sut trong trm 2 my: 1 S2 o S& = 2P + P N 2 B 2 Udm + 2I S2 1U S S o dmB N dmB j + 2 2 100 U dmB 100 33

Gio Trnh Cung Cp in


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: S&B

2IoSdmB 1 U NSdmB S2 j P 2 N 100 + = 2 + o + U 2 2 2 100 U dmB dm 1

2 S

2 2 2.1,3.1000 1 5.1000 1500 & = 2.1,57 + 1 + j S 1500 + 9,5B 2 10002 100 2 100 10002 & SB = 13,38 + j82,25(kVA )

BI TP CHNG 2Bi 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.A 4km 1 800 0,8(kVA) 3km 500 0,7(kVA) 2 2km 3 300 0,8(kVA)

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.

34

35

Gio Trnh Cung Cp in


Chng 3: LA CHN CC THIT B IN TRONG LI CUNG CP INH 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: SdmB S tt 1,4 36

Gio Trnh Cung Cp in Trong : ph ti 1,4 - h s qu ti k = 1,4


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

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.k hc = 1 1 2 100o

1 - nhit mi trng s dng ( C) o 2 - nhit mi trng ch to ( C) V d: Nu dng my bin p Nga Vit nam th Trong :k hc =1 24 5 =

0,81 100

Vi

24 - nhit trung bnh H Ni 5 - nhit trung bnh Mtcva. Khi cng sut my bin p c tnh theo cng thc:S dmB S tt S v S dmB tt k hc 1,4k hc

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:S dmB S1 1,4

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.

37

Gio Trnh Cung Cp in


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).S dmB S tt 1200 = = 857(kVA) 1,4 1,4

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)S dmB S1 80%.1200 = 685(kVA) = 1,4 1,4

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.K1 = 1200 S tt = = 0,6 2S dmB 2000 1200 S tt = = 0,8 2S dmB 1500

Phng n chn 2 my 750 (kVA) c li l vn u t nh, h s ti caoK1 =

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:

38

Gio

Trnh

Cung

Cp

in

http://www.ebook.edu.vn1200 1500 .100 = 12,5%

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 in p nh mc (kV) Dng in nh mc (A) Dng ct nh mc (kA) Cng sut ct nh mc (MVA) Dng in n nh ng (kA) Dng in n nh nhit (kA) iu kin UdmMC UdmL IdmMC Icb ICdm IN SCdm SN Iod ixk Ionh I .t qd t nh.dm

39

Gio Trnh Cung Cp in


Bng 3.2. IU KIN CHN V KIM TRA MY CT PH TI Cc iu kin chn v kim tra in p nh mc (kV) Dng in nh mc (A) Dng n nh ng (kA) Dng n nh nhit (kA) iu kin UdmMC UdmL IdmMC Icb Iod ick Ionh I . Imcc Icb Icn I Scm St qd t nh.dm

Dng in nh mc ca cu ch (A) Dng ct nh mc ca cu ch (kA) Cng sut ct nh mc ca cu ch (MVA) 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 tc t nh.dm

40

Gio Trnh Cung Cp in


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 =1000 = 26,27( A) 3.22

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 NPS 24 B1/A1 Udm (kV) 24 Idm (A) 400 INmax(kA) 40 IN3s (kA) 10

Bng 3.4. THNG S K THUT CA CU CH NG DO SIEMENS CH TO Loi cu ch ng 3GD1 406 4B Udm (kV) 24 Idm (A) 32 INmax(kA) 31,5 IN3s (kA) 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 in p nh mc (kV) Dng in nh mc (A) Dng n nh ng (kA) Dng n nh nhit (kA) iu kin UdmMC = 24 UdmL = 22 IdmMC = 400 Icb = 16,27 Iod = 40 ick= 2 .1,8.8 = 20,3 Ionh = 10 I .t qd t nh.dm =8 0,8 = 4,13 3

41

Gio Trnh Cung Cp in Dng in nh mc ca cu ch (A) Dng ct nh mc ca cu ch (kA) Cng sut ct nh mc ca cu ch (MVA)

http://www.ebook.edu.vn Imcc = 32 Icb = 26,27 Icn = 40 I = 8 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).42

Gio Trnh Cung Cp in


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 Cc iu kin chn v kim tra in p nh mc (kV) Dng in nh mc (A) Dng in n nh ng (kA) Dng in n nh nhit (kA) iu kin UdmDCL UdmL IdmDCL Icb Id. ixk Inh.m I .t qd t nh.dm

Bng 3.7. CC IU KIN CHN V KIM TRA CU CH Cc iu kin chn v kim tra in p nh mc (kV) Dng in nh mc (A) Dng ct nh mc (kA) Cng sut ct nh mc (MVA) iu kin UdmCC UdmL IdmCCL Icb Im I 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 = 1,25.43

Gio Trnh Cung Cp in Dng cng bc qua DCL CC l:


Icb = IqtB = 1,25IdmB = 1,25 320 = 27,75 (A)3.10

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 3DC BNG CHN CU CH Loi CC 3GD1 - 120 - 2B Udm (kV) 12 Idm (A) 100 INmax(kA) 80 IN3 (kA) 40 Udm (kV) 12 Idm (A) 400 INmax(kA) 40 IN3 (kA) 10

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 Cc iu kin chn v kim tra in p nh mc (kV) Dng in nh mc (A) Dng in n nh ng (kA) Dng in n nh nhit (kA) iu kin UdmDCL =12 UdmL = 10 IdmDCL = 400 Icb = 27,75 Id. = 40 ixk = 4,79 Inh.m = 10 I .t qd t nh.dm = 1,9 0,8 3

BNG KT QU KIM TRA CU CH Cc iu kin chn v kim tra in p nh mc (kV) Dng in nh mc (A) Dng ct nh mc (kA) Cng sut ct nh mc (MVA) iu kin UdmCC = 12 UdmL = 10 IdmCCL = 100 Icb = 27,75 Im = 80 I = 1,9 Scm = 3 .12.80 S =3 .10,5.1,9

44

Gio Trnh Cung Cp in 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.I vo ( A) I dc ( A) 30 15 200 100 500 200

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)45

Gio Trnh Cung Cp in


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 Cu dao (dao cch ly) Cu dao ph ti (dao ct ti) Cu chCC CDPT CD(DCL)

K hiu

Chc nng Cch ly, ng ct dng nh Cch ly, ng ct dng ph ti Bo v qu ti v ngn mch Bo v qu ti ngnCCCL

Cu ch cch ly

mch, cch ly Bo v qu ti v ngn

Cu ch ct tiCCCT

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

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

CD - CC

CDPT - CC

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)

46

Gio Trnh Cung Cp in


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 : 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)I tt = I dm = Pdm U pdm cos

UdmCC - in p nh mc ca cu ch, ch to cc c in p nh

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:I tt = Ptt 3U dm cos

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.47

Gio Trnh Cung Cp in

http://www.ebook.edu.vn1

CC1 CC2 CCT CC3

KT 2 KT 3 KT 4 KT

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:n

I dc

k ti I dmi i=1

I mmmax + I dc

n 1

k ti

I dmi1

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 IttI dc I mm

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

48

Gio Trnh Cung Cp in


Bng n si t dng in p 220V v cos = 1, cu ch chn theo cng thc:I dc I tt = Ptt 3U dm cos = 100 220 = 0,45A

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 IN49

Gio Trnh Cung Cp in


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 = 0,85IB = 4,8 = 25,66A 0,22.0,85

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 1mm 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 Icp502

Gio Trnh Cung Cp in Cao p Trung p H p Mi i tng th, cng nghip -

http://www.ebook.edu.vn th, cng nghip

Nng thn Nng thn

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. F .I . t qd 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:

51

Gio Trnh Cung Cp in


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 Loi dy Dy ng Dy A, AC Cp ng Cp nhm Tmax (h) 5000 1,8 1 2,7 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: Tmaxtb = Trong :

Si Tmaxii =1

n

Sii=1

n

Pi Tmaxii =1

n

Pii=1

n

Si, Pi l ph ti in (ph ti tnh ton) ca h tiu th Sij n 3U dm Pij n 3U dm cos

2. Xc nh tr s dng in ln nht chy trn cc on dy I ij = =

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

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 X nghip 1 X nghip 2 S (kVA) 2000 1000 cos 0,8 0,7 Tmax (h) 5200 400052

Gio Trnh Cung Cp in


A

4km

1

3km

2

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. 2000.5200 + 1000.4000 Tmaxtb = = 4800(h ) 2000 + 1000 2 T Tmax = 4800h v dy AC tra bng c Jkt = 1,1 (A/mm ) Tr s dng in trn cc on: I12 = S2 = 1000 = 57,8(A ) 3U dm 3.10 I

)A1

( P 1 + P2=2

1 2 + (Q + Q ) 700) 2

(1600 +2

+ (1200 + 710)2

86,4(A ) =

2 3.U dm Tit din kinh t mi on:

=

2 3.10

57,8 = 52,5(mm 2 ) 1,1 86,4 = 78,5(mm 2 ) FA1 = 1,1 F12 = 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

( + ) + (1200 + U max = 1600 700 0,46710)0,36 +

700.0,64 + 710.0,368 = 561,8(V ) 10

2.10 Umax = 561,8V > Ubtcp = 5%Udm = 500V 2 Cn tng tit din on A1 ln 95mm c xo = 0,33 (/km); ro = 0,34 (/km) U =

(1600 + 700)0,33 + (1200 + 710)0,342.10

+

700.0,64 + 710.0,368 10

=

494,5(V ) max

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

53

Gio Trnh Cung Cp in


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 PR + QX PR QX U = = = U + U + U dm U dm Udm

Khi cho gi tr xo tnh c: U = T y xc nh c U = Ucp - U Mt khc Suy ra: U = F =o PR P. .l = xo U dm F.U dm QX U dm =xo Ql U dm

P. .l U.U dm

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:n U = x o Q ij .l U dm i =1

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

54

Gio Trnh Cung Cp in F =o P.l U.U dm


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. 5km A 1 3km 22000,8 kVA

4000,8 kVA

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), 0,35 [(400.0,6 + 200.0,65).5 + 200.0,6.3] = 75,6(V ) U = 10 U = 3%Um - U = 300 75,6 = 224,4 (V) Tit din tnh ton: 31,5 [(400.0,8 + 2 0 00,8)5 + 2 0 00,8.3] = 40,43mm 2 . . F= 10.224,4 Chn tit din tiu chun 50mm 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). A 3,2 + j1,965() 1 1,92 + j1,179() 22

320 + j240()

160 + j120()

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 ()55

Gio Trnh Cung Cp in U =

http://www.ebook.edu.vn + 160.1,192 + 120.1,179 =

(320 + 160)3,2 + (240 + 120)1,965

10 10 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: I dc K1K2Icp = 3 vi mch ng lc (cp in cho cc my) = 0,8 vi mch sinh hot Nu bo v bng ptmt 1,25I dmA 1,5 K1K2Icp 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 o khc, nhit mi trng +30 C, dy c bo v bng cu ch c Idc = 50A. ng c My mi Pm(kW) 10 cos 0,8 kmm 5 0,9

269,2(V ) A12

56

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

Gio Trnh Cung Cp in I tt = I dmD =

http://www.ebook.edu.vn 10 = 21,125A 3.0,38.0,8.0,9

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: 50 I = 16,67A K1K2Icp = 25,38A > dc = 3 3 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 3Bi 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 My mi My khoan My tin 1 My tin 2 Qut gi Pm (kW) 12 7,5 5 4 1,7 cos 0,8 0,8 0,8 0,8 0,8 Kmm 5 5 5 5 5 Kt 0,8 0,8 0,8 0,8 0,8 0,9 0,9 0,9 0,9 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.A 3km 1 800 0,6(kVA) 2km 500 0,8(kVA) 2 2km 3 200 0,8(kVA)

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%.A 3km 1 20 0,8(kVA) 4km 18 0,85(kVA) 2 6km 3

15 0,8(kVA)

57

Gio Trnh Cung Cp in


Chng 4. NNG CAO H S CNG SUT4.1. 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

58

Gio Trnh Cung Cp in


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.cos = P 3.U . I

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)cos = co sartg Qtb Ptb

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%.

59

Gio Trnh Cung Cp in


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:Q = artg P

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 in2 P +Q Q P2 R = 2 R + 2 R = P(P ) + Q(Q ) P = 2 U U U 2 2

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

(R + X )

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:I= S 3U = P +Q2 2

QX = U (p ) + U (Q ) =P + U R U

3U

60

Gio Trnh Cung Cp in


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:61

Gio Trnh Cung Cp in

http://www.ebook.edu.vn 2 Q = QO + Qdm Qo .kpt

(

)

Trong : nh mc

Qo lc ng c lm vic khng ti Qdm cng sut phn khng lc ng c lm vic ch

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:cos = P = S P P2

+Q

2

=

1 Q 1+ P2

cos =

1 + 1 Q0 + (Qdm Pdm2 2

(* )

QO ).k pt

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 cos62

Gio Trnh Cung Cp in


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.63

Gio Trnh Cung Cp in


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 = 3U ..C.10 T in c cu to km chc chn, d b ph hng khi b ngn mch. Khi ng t in vo mng s c dng in xung, cn khi ngt t ra2 -3

Khi in p tng ln 110%Um th t in b chc thng. khi mng trn cc ca t in vn cn in p d c th gy nguy him cho ngi vn hnh.64

Gio Trnh Cung Cp in


* 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 b65

Gio Trnh Cung Cp in


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:P1 = Q2 R U2

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

P 2 =

(Q - Q U

bu 2

)2 R2

Vy lng cng sut tc dng tit kim c l:P = P1 P2 Q2 = U2R

(Q - Q ) Ubu 2

R

Theo nh ngha ta c:k kt = P QR Q = 2 2 bu (Kw/kVAr) Q bu U Q

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 coiQbu Q = 0 , lc ny ng lng kinh t ca cng sut phn khng c tnh theo

cng thc n gin sau:QR k kt = 2 U 2

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 u2 Qb t.u = Q - U kb (*)

2R

66

Gio Trnh Cung Cp in Dung lng b ti u2 Qb. t.u = Q - U R (**)


2R

k T (*) v (**) ta c: Qb t.u = Q1 bu k kt

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:

67

Gio Trnh Cung Cp in


- 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:R Qb i = Qi (Q - Qb ) td R i

Trong :

Qi cng sut phn khng yu cu ti nt i Q - tng cng sut phn khng yu cu, Q = Q i i=1 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 =1 1 1 1 + + ... + R1 R 2 Rn n

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.PX1 PX2 PX3

S& = 50 + 1 j120kVA

& S 2 = 50 + j50kVA

S& = 50 + j70kVA3

68

Gio Trnh Cung Cp in


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

GII Tng cng sut tnh ton ca x nghip & S & S1 + &2 + S3 = 180 + j240kVA Q 240 tg = = 1 P 180 = 1,33 cos 2 = 0,95 tg 2 = 0,33 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 iR1 TBA R2 R3 Q3 Qb3 Q1 Qb1 Rt Q2 Qb2 TBA Q Qb

in tr tng ng ca li in h p x nghip Rt = 1 1 1 + R1 R 2 + ... + 1 Rn = 1 = 0,0126( ) 1 1 1 + + 0,0194 0,0505 0,115

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

Xc nh c cng sut cc t t b t ti 3 phn xng:69

Gio Trnh Cung Cp in Qb1 = 120 (240 - 180 ) Qb2 = 50 (240 - 180 ) Qb3 = 70 (240 - 180 ) 0,0126 = 81kVAr 0,0194


0,0126 = 35kVAr 0,0509 0,0126 = 64kVAr. 0,115

BI TP CHNG 4Bi 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.

70

Gio Trnh Cung Cp in


Chng 5. TNH TON CHIU SNG5.1. 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:71

Gio Trnh Cung Cp in E = e d1 2


2 1

Trong :

e - hm phn b nng lng - bc sng E12 thng lng ca quang nng t 1 n 2 Thng lng ton phn: E = e d0

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: F = V e d0

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 dF gi l cng nh sng ca d dF ngun sng trong phng : I = d 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 2 gi l dF th trng ca ngun sng c nh ngha: dF R= (*) dS

72

Gio Trnh Cung Cp in


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. N dS

d C

r

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. dScos d = r2 r l khong cch t C n tm dS Thay vo cng thc (*) ta c: dF ldScos E= = 2 dS r dS 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) F E=S Trong : F quang thng ca ngun sng (lm) S - din tch chiu sng (m ) 5.3. CC LOI N to ngun sng ngi ta thng dng cc loi n in, thng dng nht l n si t v n tup.2

73

Gio Trnh Cung Cp in 5.3.1. n si t


n si t cn gi l n dy tc c dng rng ri trong cc lnh vc do cu to n gin, lp t d dng. Nguyn tc lm vic ca n si t da trn c s bc x nhit. Khi dng in i qua si dy tc, dy tc s pht sng v pht quang. Vt liu lm dy tc l Vnfram, Tungsten vn xon c hoc thng mc dch dc trn cc cc ph v hai cc chnh trong bng n. Nhit ca dy tc trong bng c th ln ti 200 3000 C. Trong bng c th cha kh tr hoc chn khng. Thng bong cng sut nh th ht chn khng, bng cng sut ln (trn 75W) th np kh tr. u im ca n si t: - Ni trc tip vo li in - Kch thc nh - R tin - Bt sng ngay - To ra mu sc m p - cos cao (bng 1) Nhc im ch yu l tn in v pht nng. Ngoi ra, tnh nng ca n thay i ng k theo bin thin in p ngun. Mi bin thin in dn ti bin thin dng in v do bin thin si t nng l nh hng n quan thng v tui th ca n. V d, nu in p gim i 10% tui th n l 3700h, khi qu in p 5% tui th ca n nh hn 500h. 5.3.2. n tup n tup cn gi l n hunh quang. Nguyn tc pht quang ca loi n ny da trn c s phng in ca cc cht kh. Sau khi rt chn khng ngi ta np vo trong bng mt t kh argon v thu ngn. Pha mt trong ng bi mt t bt hunh quang. Hai in cc t hai u ng. Hai in cc t hai u ng. S ni dy n tup cho trn hnh 5.2. Khi dng in hai u cc ca Stc-te c in th kh ln lm cho Stcte phng in, mch in c ni lin. Hai in cc A v B ca bng c t nng. S t nng ny l cn thit cho s phng in trong n. Khi stcte phng in th in th trn hai cc ca n gim xung, nhit lng trn stcte cng gim, tip im ca stcte m ra. Hin tng qu trong mch in xy ra lm cho n phng in t cc A sang cc B. Cc sng in t phng t A sang B v ngc li c tn s ln, cc74o

Gio Trnh Cung Cp in


song ny p vo mn hunh quang vch bng v pht ra cc tia bc x th cp ln 2 cc bc sng ny mt ngi ta mi cm thy c.

Hnh 5.2. S ni dy n tup 1. bng n; 2. chn lu; 3. Stcte; 4. t in b cos a) Vai tr ca cun dy chn lu (chn lu in cm) Chn lu in cm cho php san bng dng song dng in. Dng in khng cn hnh sin na b) Chn lu in t Do s pht trin ca k thut in t nn ngi ta nghin cu v thay th cc chn lu kiu cun dy li thp bng mt mch bn dn nh hn v tiu th in nng t hn. Vn c bn ddaay l bin i tn s t 50Hz ln khong 20kHz bng b chnh lu - nghch lu. Chn lu in t c kch thc nh hn nhiu so vi cc loi chn lu cun dy li thp v loi tr c hiu ng nhp nhy c) u im n tup: - Hiu sut nh sng ln, dng ni cn ri ln. - Tui th cao - Din tch pht quang ln - Khi in p thay i trong phm vi cho php, quang thng gim t (1%) d) Nhc im: Ch to phc tp, gi thnh cao, cos thp

75

Gio Trnh Cung Cp in o


Quang thng ph thuc vo nhit , pham vi pht quang cng ph

thuc nhit . Khi nhit di 15 C th stcte lm vic kh khn. - Khi ng in n khng sng ngay. Ngoi hai loi n si t v hunh quang cn dng cc n khc nh: n kh natri p sut thp, n kh natri p sut cao, n halogen kim loi 5.3.3. Cc loi chao n Chao n l b phn bao bc ngoi bng n. N c dng phn phi li quang thng ca bng n mt cch hp l v theo yu cu nht nh. Chao n cn c tc dng lm cho mt khi b chi, bo v cho bng khi va p, bi bn hoc khi ph hu bi cc kh n mn Theo cch phn b quang thng, chao n c chia ra lm 3 loi: chao n chiu trc tip, chao n phn x v chao n khuch tn. Chao n chiu trc tip c th tp trung hn 90% quang thng ca ngun sng xung pha di. Ngc li, chao n phn x tp trung ghn 90% quang thng ca ngun sng ln pha trn ri phn x tr xung. Chao n khuch tn to ra nh sng khuch tn ch khng thiu nh sng trc tip. 5.4. THIT K CHIU SNG DN DNG Chiu sng dn dng bao gm chiu sng cho cc khu vc nh sng sinh hot nh nh , hi trng, trng hc, c quan, vn phng i din, ca hng, siu th, bnh vin v.v nhng khu vc ny yu cu chiu sng chung, khng i hi tht chnh xc tr s ri cng nh cc thng s k thut khc. Trong chiu sng dn dng, tu theo kh nng kinh ph, ttu theo mc yu cu m quan c th s dng mi loi n: n si t, n tup, n halogen, n natri cao, thp p. Trnh t thit k chiu sng dn dng nh sau: 1. Cn c vo tnh cht ca i tng cn chiu sng, chn sut ph ti chiu sng Po (W/m ) thch hp, t y tnh c tng cng sut chiu sng cho khu vc c din tch S (m ).2 2

Pcs = Po.S

2. Chn loi n, cng sut n P, xc nh tng s bng n cp lp trong P khu vc: n = cs Pd

76

Gio Trnh Cung Cp in


3. Cn c vo din tch cn chiu sng, vo s lng bng n, vo tnh cht yu cu s dng nh sng m chn cch b tr n thch hp (b tr di u hay thnh rnh, thnh cm, s lng bng trong mi cm) 4. V s u dy t bng in n tng bng n. l bn v mt bng cp in chiu sng. 5. V s nguyn l li in chiu sng. 6. La chn v kim tra cc phn t trn s (ptomat, cu ch, thanh ci, dy dn). Ghi ch: Trong tnh ton chiu sng dn dng th bao gm c tnh ton thit k cho qut. Trong trng hp ny c 2 cch lm: - Ly sut ph ti cung cho c chiu sng v qut, sau tr i cng sut qut (ly thoe thc t) tm c cng sut chiu sng. - Ly ring sut ph ti cho chiu sng tnh ton thit k chiu sng, cn qut ly theo thc t, tnh ton ring. V d 5.1. Yu cu thit k chiu sng cho 1 siu th nh, din tch 10x10m. GII Siu th in thoi yu cu mc chiu sng cao. Chn sut chiu sng; Po = 30W/m . Tng cng sut cn cp cho chiu sng siu th P = Po.S = 30(10x10) = 3000W Chn dng n tup di 1,2m, cng sut 40W. S lng bng n cn dng l 3000 n= = 75 bng 40 s lng ny c b tr thnh 5 dy, mi dy 15 bng chia lm 3 cm, mi cm 3 bng. 5.5. THIT K CHIU SNG CNG NGHIP Vi cc nh xng sn xut cng nghip thng l chiu sng chung khi cn tng cng nh sng ti im lm vic c chiu sng cc b. V l phn xng sn xut, yu cu kh chnh xc v ri ti mt bn cng tc, nn thit k chiu sng cho khu vc ny thng dng phng php h s s dng. Trnh t tnh ton theo phng php ny nh sau:772

Gio Trnh Cung Cp in 1. Xc nh treo cao n H = h h1 h2 Trong :


h - cao ca nh xng h1 - khong cch t trn n bng n h2 - cao mt bn lm vich1

L

L L

L L

L

h

H

h2

Hnh 5.2. B tr n trn mt bng v mt ng 2. Xc nh khong cch gia 2 n k nhau (L) theo t s hp l L/H tra theo bng 5.1 Bng 5.1. T S L/H HP L CHO CC I TNG CHIU SNG L/H b tr nhiu dy Tt Max cho nht php 2,3 3,2 Chiu rng gii hn ca Max cho nh xng b tr 1 dy php 2,5 1,3H

L/H b tr 1 dy Tt nht 1,9

Loi n v ni s dng

Chiu sng nh xng dng chao m hoc st trng men Chiu sng nh xng dng chao vn nng Chiu sng c quan vn phng

1,8 1,6

2,5 1,8

1,8 1,5

2,0 1,8

1,2H 1,0H

3. Cn c vo s b tr n trn mt bng, mt ct xc nh h s phn x ca tng (tg%), trn (tr%). 4. Xc nh ch s ca phng (c kch thc ab) ab H(a b )78

=

Gio Trnh Cung Cp in 5. T tg, tr, tra bng tm h s s dng ksd 6. Xc nh quang thng ca n k E.S.Z F = dt (lm) n.k sd Trong : kdt - h s d tr, tra bng 5.2


E - ri (lx) theo yu cu ca nh xng S - din tch nh xng (m ) Z - h s tnh ton Z = 0,8 1,4 n - s bng n c xc nh chnh xc sau khi b tr n mt bng2

trn

7. Tra s tay tm cng sut bng c F F tnh ton tnh theo cng thc bc 6 8. V s cp in chiu sng trn mt bng 9. V s nguyn l cp in chiu sng 10. La chn cc phn t trn s nguyn l. Bng 5.2. H S D TR S ln lau bng t nht mt thng 4 3 2 H s d tr n tup 2 1,8 1,5 n si t 1,7 1,5 1,3

Tnh cht mi trng Nhiu bi khi, tro, m hng Mc khi, bi, m hng trung bnh t bi khi, tro, m hng

V d 5.2. Yu cu thit k chiu sng cho phn xng c kh c a = 20m, b = 50m, cao = 4,5m, h2 = 0,8m, h1 = 0,7m GII 1. Xc nh s lng, cng sut bng Ni dung phn ny bao gm cc hng mc t 1 n 7 trong trnh t tnh ton nu trn. V l xng sn xut, d nh dng n si t cos = 1 Chn ri cho chiu sng chung l E = 30lx treo cao n79

Gio Trnh Cung Cp in


H = 4,5 - 0,8 - 0,7 = 3m Tra bng n si t, bng vn nng c L/H = 1,8. Xc nh c khong cch gia cc n L = 1,8.H = 1,8.3 = 5,4m Cn c vo b rng phng (20m) chn L = 5m n s c b tr lm 4 dy cch nhau 5, cch tng 2,4m tng cng 36 bng, mi dy 9 bng. Xc nh ch s phng

==

ab 20.50 5 H(a b ) 3(20 + 50)

Ly h s phn x tng 50%, trn 30%, tra s ta tm c h s s dng ksd = 0,48. Ly kdt = 1,3, h s tnh ton Z = 1,1 xc nh c quang thng mi n l: F= k dt E.S.Z 1,3.30.1000.1,1 = = 2483 (lm) n.k sd 36.0,48

Tra bng chn bng si t 200W c F = 2528lm. Tng cng sut chiu sng ton xng: P = 36.200 = 7200W = 7,2kW.

BI TP CHNG 5Bi tp 1. Yu cu thit k chiu sng v qut cho 1 lp hc kch thc 810m. Bi tp 2. Yu cu thit k chiu sng cho 1 phng lm vic ca vn phng i din nc ngoi kch thc 46m. Bi tp 3. Yu cu thit k chiu sng cho mt hi trng c din tch 1220m. Bi tp 4. Yu cu thit k h thng chiu sng chung cho phn xng may xut khu kch thc 1530m. Yu cu ri 100lx.

80

Gio Trnh Cung Cp in


1. 2. 3. 4. 5.

TI LIU THAM KHO Cung cp in (NXB KHKT), Nguyn Xun Ph, Nguyn Cng Hin, Nguyn Bi Kh... Gio trnh cung cp in (NXB i hc v THCN), Nguyn Cng Hin, ng Ngc Dinh... Mng cung cp v phn phi in (NXB KHKT), Bi Ngc Th Li in v H thng in (NXB KHKT), Trn Bch Gio trnh cung cp in (NXB GD), Ng Hng Quang

81

Gio Trnh Cung Cp in PH LC


PL1 Tr s trung bnh ksd v cos ca cc nhm thit b in Nhm thit bNhm my gia cng kim loi (tin, ca, bo, khoan...) - Phn xng c kh - Phn xng sa cha c kh Nhm my phn xng rn Nhm my ca phn xng c Nhm ng lm vic lin tc (qut gi, my bm, my nn kh) Nhm ng c lm vic ch ngn hn lp li (cu trc, cn cu...) Nhm my vn chuyn lm vic lin tc (bng ti, bng chuyn) Nh

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