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8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
1/78
Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
chng I
gii thiU CHUNG V NH MY
Danh sch cc phn xng trong nh my:
Theos mt
bng
Tn phn xng v ph ti Slngthit b
in
Tngcng
sut t,kW
H snhu cu,
Ksd
H scngsut,cos
1 Phn xng trm t 280 500 0.34 0.68
2 Phn xng vt liu hn 200 800 0.35 0.56
3 Phn xng nha tng hpPlasmace 100 1100 0.37 0.67
4 Phn xng tiu chun 70 250 0.37 0.78
5 Phn xng kh c in 100 700 0.37 0.72
6 Phn xng dp 100 800 037 0.67
7 Phn xng xi mng aming 50 850 0.4 0.72
8 Kho thnh phm 15 85 048 0.87
9 Kho ph liu kim loi 15 70 0.48 0.8110 Phn xng m in 50 70 0.48 0.76
11 Xem d liu phn xng
12 Trm trung ha 10 100 0.52 0.66
13 Ra knh thot axit 3 30 0.7 0.68
14 Trm bm 8 260 0.55 0.68
15 Nh n 30 70 0.43 0.56
16 Phn xng in 25 150 0.44 0.72
17 Nh iu hnh 20 50 0.46 0.78
18 Phn xng lm ngui 2 30 0.79 0.77
19 Kho axit 2 20 0.79 0.67
20 My nn s N0 1 15 200 0.48 0.72
- 1 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Cp in theo tiu chun loi 1Gi thnh tn tht in nng C=1000/kWhSut thit hi do mt in Gth=7500/kWhHao tn in p cho php trong mch tnh t ngun ( im u in) l Ucp
= 5%Cc s liu khc ly trong ph lc v s tay thit k in
Ngun in c ly t li cch nh my cch 468.72m, dng dy ACDung lng ngn mch v phia h p ca trm bin p : 250MVA, nh mylm vic 3 ca
Nhim v thit kI. Tnh ton ph ti Xc nh ph ti tnh ton phn xng Xc nh ph ti ca cc phn xng khc Tng hp ph ti ca ton x nghip, xy dng biu ph
ti trn mt bng x nghip di dng cc hnh trn bn knhr
II Xc nh s ni mng in ca nh my Xc nh v tr ca trm bin p Chon cng sut v s lng nh my bin p Chn dy dn t ngun ti trm bin p La chn s ni in t trm bin p n phn xng
( so snh t nht 2 phng n)III Tnh ton in Xc nh tn hao in p trn ng dy v trong nha my Xc nh tn hoa cng sut Xc nh tn tht in nng
IV Chn v kim tra thit b Tnh ton ngn mch ti cc im c trng Chn v kim tra thit b
Kim tra thit b m my ca ng cV Tnh ton b h s cng sut Tnh ton b cng sut phn khng nng cao h scng sut ln cos 2 = 0.9 nh gi hiu qu b
VI Tnh ton ni t v chng stVII Hch ton cng trnh
Lit k cc thit b Xc nh cc ch tiu kinh t
- 2 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
S mt bng x nghip vi s b tr cc thiwwts v vbiu ph ti
S mng in trn mt bng x nghip( gm c s cacc phng n so snh)
S trm bin p ( s nguyn l, s mt bng v s mt bng v mt ct trm bin p)
S ni t Bng s liu tnh ton so snh cc phng n
chng II
Xc nh ph ti tnh toncho cc phn xng v cho ton nh my
Ph ti in l cng sut tc dng P v cng sut phn khng
Q yu cu ti mt im no ca li. Khi thit k cung cp
in cho mt cng trnh tc l tnh ton lng cng sut yu
cu ca ph ti cung cp vi cng sut in m ph
ti yu cu. Do , vic u tin khi thit k cung cp incho mt cng trnh l xc nh ph ti in ca cng trnh
, tc l phi d bo ph ti ngn hn hoc di hn. y,
trong gii hn v thit k v kh nng ta ch xt bi ton xc
nh ph ti ngn hn.
D bo ph ti ngn hn l xc nh ph ti ca cng trnh
ngay sau khi cng trnh i vo vn hnh. Ph ti ny c gi
l ph ti tnh ton. Ph ti tnh ton l mt s liu cn thit
v quan trng. Bi n chnh l s liu la chn cc thit
b in nh: my bin p, thit b bo v, thit b ng ct,
dy dn, tnh ton cc tn tht cng sut, tn tht
in p, la chn cc thit b b.
Tu thuc vo qui m, c im ca cng trnh, tu
thuc vo giai on thit l s b hay k thut thi cng m- 3 - n
mn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
ngi thit k phi cn nhc, la chn phng php tnh ton
ph ti cho thch hp.
I. Cc i lng c bn v h s tnh ton
1. Cng sut nh mc(Pm):
Cng sut nh mc ca mt thit b tiu th in l cng
sut ghi trn nhn hiu ca my hoc ghi trong l lch my.Vi
ng c cng sut nh mc ghi trn nhn hiu my chnh l
cng sut c trn trc ng c.
2. Cng sut t ( P ):
ng c c hiu sut c th cng sut t ca ng c:
c
m
P=P
Trong thc t th c = 0,8 0,9 c gi tr kh cao nn c ta
th xem P = Pm.
Vi thit b chiu sng cng sut t chnh l cng sut t-
ng ng vi s ghi trn hay bu n.
Vi ng c in lm vic ch lp li v d nh: cu trc,my hn th khi tnh ton ph ti in ca chng phi thc
hin qui i v cng sut nh mc lm vic ch di
hn tc l phi qui i v ch lm vic c h s tip in:
% = 100% .
Gi Pm l cng sut nh mc qui i v ch lm
vic di hn th Pm c xc nh:
Pt = Pm = Pm. m
Vi Pm, m: Cho trong l lch my.
Vi my bin p ca l in cng sut t l :
Pm = Sm. cos m
Trong :
Sm- cng sut biu kin nh mc c trong l
lch my.- 4 - n
mn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
cos m- h s cng sut ca l in khi ph ti
ca n t n nh mc.
Vi my bin p hn: Pm = Sm. cos m . m
3. Ph ti trung bnh(Ptb):
L mt c trng tnh ca ph ti trong khong thi gian
no . Tng ph ti trung bnh ph ti ca cc thit b cho
ta cn c nh gi gii hn di ca ph ti tnh ton. Hin
ti, ph ti trung bnh c tnh:
ptb = tP
; qtb = tQ
Trong :
P, Q - in nng tiu th trong thi gian kho st,
kW, kVAR.
t - khong thi gian kho st, h
Nu l mt nhm thit b th ph ti trung bnh c tnh:
Ptb = =
n
1i
ip ; Qtb = =
n
1iiq
Ph ti trung bnh l mt s liu quan trng nh gi
mc s dng in cu thit b, xc nh ph ti tnh
ton v tnh tn hao in nng.
4. Ph ti tnh ton (Ptt):
L mt ph ti gi thit lu di khng i, tng ng vi
ph ti ph ti thc t (bin i) v mt hiu ng nhit ln
nht.Tc l ph ti tnh ton l ph ti c tc dng lm nngdy dn ln ti nhit ln nht do ph ti thc t gy ra.
Ph ti tnh ton c gi tr:
Ptb Ptt Pmax
5. H s s dng (ksd): L t s gia ph ti tcdng trung bnh vi cng sut t trong khong thi gian
xem xt.
- 5 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
i vi mt thit b:
ksd =dm
tb
P
P
i vi mt nhm thit b:
ksd =dm
tb
P
P=
=
=n
i
dmi
n
i
tbi
P
P
1
1
H s s dng c ngha l h s ni ln mc s dng,
mc khai thc cng sut ca thit b trong khong thi
gian xem xt.
6. H s ph ti (hay cn gi l h s mang ti kpt):
L t s gia cng sut ca thit b in trong thc t vi
cng sut nh mc trong khong thi gian ang xt. N c
th hin bi cng thc:
kpt =dm
thct
P
P
H s ph ti c ngha tng t nh h s s dng l n
ni ln mc s dng, mc khai thc ca thit b trong
khong thi gian xem xt.
7. H s nhu cu (knc):
L t s gia cng sut tnh ton hoc cng sut tiu th vi
cng sut t (cng sut nh mc) ca nhm h tiu th:
knc =dm
tb
tb
tt
dm
tt
P
P.
P
P
P
P= = kmax . ksd
H s nhu cu thng dng tnh cho ph ti tc dng. Trong
thc t h s nhu cu thng do kinh nghim vn hnh tng
kt li. N c gi tr nh hn 1: knc1
8. H s cc i (kmax ) :
L t s gia ph ti tnh ton ph ti trung bnh trong mt
khong thi gian ang xt:
- 6 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
kmax =tb
tt
P
P
H s ny c gi tr ln hn 1 (kmax 1).H s cc i ph
thuc vo nhiu yu t: ph thuc vo sthit b hiu qu,ph thuc vo h s s dng, cc ch c trng cho ch vn hnh ca thit b. N thng oc tnh vi ca lm vicc ph ti ln nht. 9. S thit b tiu th in nng hiu qu (nhq):
Xt mt nhm gm c n thit b c cng sut nh mc vch lm vic khc nhau. Th nhq c gi l s thit b tiuth in nng hiu qu ca nhm, l mt s qui i gm
c nhq thit b c cng sut nh mcv ch lm vic nhnhau to nn ph ti tnh ton bng ph ti tiu th thc t bin thit b ca nhm ang xt.
Khi s thit b n 5 th s thit b hiu qu c xc nh:
nhq =( )
=
=
n
i dmi
n
i
dmi
P
P
1
2
2
1 (*)
Nu m nhm c tt c cc thit b cng c cng sut nhmc nh nhau th:
nhq =( )
2
2
.
.
dm
dm
Pn
Pn= n
Nu n > 5 th vic tnh nhq theo cng thc (*) kh phc tpnn thng s dng phng php n gin ho tnh nhq vi
sai s cho php trong phm vi 10%. Trnh t tnh ton Xc nh s thit b trong nhm n v tng cng sut nhmc dmnP
Chn nhng thit b c cng sut m cng sut nh mcca mi thit b ny nh hn hoc bng mt na cng sutca thit b c cng sut ln nht trong nhm thit b xt.
Xc nh s n1: l s thit b c cng sut khng nh hnmt na cng sut ca thit b cng sut ln nht, v ng vi
s thit b n1 xc nh tng cng sut nh mc: 1dmnP- 7 - n
mn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Tm gi tr n* =n
n1 v p*=
dmn
dmn
P
P1
Sau khi xc nh c ( n*, p*) da vo cm nang tra ta s c
nhq*, t rt ra nhq* nhq = nhq* . nII. Cc phng php xc nh ph ti tnh Ton
1. Xc nh ph ti tnh ton theo sut ph ti trn mtn v din tch sn xut:
Cng thc tnh: FpPtt .0=
Trong :po - sut ph ti trn 1 m2 din tch sn xut,
KW/m2(thng c tra trong cc s tay k thut).
F - din tch sn xut, m2.Sut ph ti tnh ton trn mt n v din tch sn xut
ph thuc vo dng sn xut v theo s liu thng k.2. Xc nh ph ti tnh ton theo sut tiu hao
in nng cho mt n v sn phm:
Cng thc tnh:max
0catt
T
W.MP =
Trong :Mca - s n v sn phm sn xut ra trong
mt ca.Wo - sut tiu hao in nng cho mt n v sn phm
(KWh/nsn phm).Tmax- thi gian ca ca ph ti ln nht, h.
3. Xc nh ph ti tnh ton theo cng sut t
v h s nhu cu:Ph ti tnh ton ca nhm thit b c cng ch lm vic
c tnh theo cng thc :
tgPQ
PkP
tttt
n
i
dminctt
.
.1
=
= =
Trong :
- 8 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
knc - h s nhu cu (thng c tra trong cc s tayk thut).n - l s thit b trong nhm.
tg - ng vi cos c trng cho nhm thit btra trong cn
Nu h s cng sut cos ca cc thit b trong nhmkhng ging nhau th phi tnh h s cng sut trung bnhtheo cng thc:
cos tb =n
n
PPP
PPP
++++++
...
cos....cos.cos.
21
21
Phng php tnh ph ti theo cng sut t v h s nhu
cu c u im l n gin, tnh ton n gin. Nhng cnhoc im l chnh xc khng cao v knc tra trong stay.
4. Xc nh ph ti tnh ton theo h s cc ikmax v cng sut trung bnh Ptb(hay phng php sthit b hiu qu nhq):
nng cao chnh xc ca ph ti tnh ton hoc khi
khng c cc s liu p dng cc phng php n ginth s dng phng php xc nh ph ti tnh ton theo hs cc i v cng sut trung bnh. Theo phng php nyph ti tnh ton c tnh:
=
==n
i
dmisdtbtt PkkPkP1
maxmax ...
Trong :
kmax - h s cc i c tra bng da theonhq v ksd.
Phng php ny cho kt qu tng i chnh xc v khi xc
nh s thit b hiu qu th xt n mt lot cc yu t
quan trng nh: nh hng ca s lng thit b trong nhm, s
thit b c cng sut ln nht, s khc nhau v ch lm
vic ca cc thit b ...- 9 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Khi tnh ph ti theo phng php ny, trong mt s trng
hp c th dng cc cng thc gn ng sau xc nh ph
ti tnh ton :
Trng hp n 3 , nhq < 4 :
=
=
=
=
n
i
dmitt
n
i
dmitt
PP
1
1
Trng hp n > 3 v nhq < 4 :
dmi
n
i
ptitt PkP .1==
Trong : Kpti - h s ph ti ca thit b th i.
Nu nhq > 300 v ksd < 0,5 th h s cc i
kmax s ly ng vi n hq= 300. Cn nu nhq> 300
v ksd 0,5 th:
Ptt = 1,05.ksd.Pdm
III Xc nh ph ti tnh ton c th
A. Xc nh ph ti tnh ton phn xng
Tu theo thng tin thu nhp c ca ph ti in m ta
chn phng php xc nh ph ti tnh ton cho hp l.
Do mi ch bit cng sut t ca tng phn xng nn ta
xc nh ph ti tnh ton theo phng php cng sut t
v h s nhu cu.
i vi ph ti chiu sng th ta xc nh ph ti tnh ton
theo phng php sut ph ti trn mt n v din tch sn
xut
- 10 - nmn hc : SV : Ng xun Li H in 1_K3
8/3/2019 Do an Mon Cung Cap Dien Cua Ngo Xuan Loi
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Din tch thc t: din tch bn v x 10002 ( t l : 1/10002)
Theo t l ta c thng s din tch theo bng s liu sau:
theos mt
bng
Tn Phn Xng Din tchbn v(mm2)
Din tchthc t(m2)
1 Phn xng trm t 33*19 627
2 Phn Xng vt liu hn 36*29 1044
3 Phn xng nha Tng hpPlasmace
34*12 408
4 Phn xng tiu chun 33*11 363
5 Phn xng kh c in 34*7 238
6 Phn xng dp 38*16 608
7 Phn xng Xi mng aming 12*22 264
8 Kho thnh phm 12*22 264
9 Kho ph liu kim loi 10*8 80
10 Phn xng m in 6*22 13211 Xem d liu phn xng 15*10 150
12 Trm trung ha 14*9 126
13 Ra knh thot axit 11*8 88
14 Trm bm 9*10 90
15 Nh n 8*12 96
16 Phn xng in 15*10 150
17 Nh iu hnh 13*8.5 110.5
18 Phn xng lm ngui 28*9 252
- 11 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
19 Kho axit 10*8 80
20 My nn s No 1 12*10 120
Phn xng trm t
Cng sut tnh ton ng lc:
Pdl = knc.P = 0.34*500=170kW
Cng sut tnh ton chiu sng:
Pcs = p0.F = 15*627 = 9.405kW
Cng sut tnh ton tc dng ca phn xng:
Ptt= Pdl + Pcs =170 + 9.405= 179.405kW
Cng sut tnh ton phn khng ca phn xng:
Qtt = Qdl = Ptt.tg =179.405*1.078= 193.4kW- 12 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Cng sut tnh ton ton phn ca phn xng:
Stt = Ptt/cos = 179.405/0.68 =263.83kW
Cc phn xng khc tnh tng t, kt qu c ghi trong bng
sau:
- 13 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- 14 - nmn hc : SV : Ng xun Li H in 1_K3
peo
t
ng
Tn Phn Xng Pd,kW
cos
Ksd Po,W/m2
Pdl,kW
F(m2)
Pcs,kW
Ptt,kW
Qtt,kWAr
Stt,KVA
1 Phn xng trm t 500 0.68 0.34 15 170 627 9.405
179.4
193.4
263.8
2 Phn Xng vt liu hn 800 0.56 0.35 15 280 1044
15.66
295.7
437.5
528
3 Phn xng nha Tng hpPlasmace
1100
0.67 0.37 15 407 408 6.12 413.1
457.7
616.6
4 Phn xng tiu chun 250 0.78 0.38 15 95 363 5.445
100.4
80.55
128.8
5 Phn xng kh c in 700 0.72 0.37 15 259 238 3.57 262.6
253.1
364.7
6 Phn xng dp 800 0.67 0.37 15 296 608 9.12 305.1
338.1
455.4
7 Phn xng Xi mngaming
850 0.72 0.4 14 340 264 3.696
343.7
331.3
477.4
8 Kho thnh phm 85 0.87 0.48 15 40.8 264 3.96 44.76
25.37
51.45
9 Kho ph liu kim loi 70 0.81 0.48 14 33.6 80 1.12 34.72 25.14 42.8610 Phn xng m in 120
00.76 0.4 14 480 132 1.84
8481.
8412 634
11 Xem d liu phn xng 150 0 012 Trm trung ha 100 0.66 0.52 12 52 126 1.51
253.5
160.9
181.0
813 Ra knh thot axit 30 0.68 0.7 10 21 88 0.88 21.8
823.6 32.1
814 Trm bm 260 0.68 0.55 15 143 90 1.35 144.
4155.
7212.
315 Nh n 70 0.56 0.43 15 30 96 1.44 31.4
446.5
156.1
416 Phn xng in 150 0.72 0.44 15 66 150 2.25 68.2
565.7
894.7
917 Nh iu hnh 50 0.78 0.46 20 23 110.
52.21 25.2
120.2
332.3
2
18 Phn xng lm ngui 30 0.77 0.79 14 23.7 252 3.528 27.23 22.56 35.3619 Kho axit 20 0.67 0.79 14 15.8 80 1.12 16.9
218.7
525.2
520 My nn s No 1 200 0.72 0.48 10 96 120 1.2 97.2 93.6
8135
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Ph ti tnh ton tc dng ton nh my:
Pttnm=kt.20
1
Ptti
Pttnm = 0.8*(179.4 + 295.7+ 413.1+ 100.4+ 262.6+ 305.1+ 343.7+44.76+ 34.72+ 481.8+ 53.51+ 21.88+ 144.4+ 31.44+ 68.25+25.21+ 27.23+ 16.92+ 97.2)
Pttnm= 0.8*2947 =2357.6 (kW)Ph ti tnh ton phn khng ton nh my:
Qttnm= kt.20
1
Qtti
Qttnm = 0.8*3062 = 2449.6 (kW)Ph ti tnh ton ton phn ca nh my:
Sttnm= )6.2449*6.2449()6.2375*6.2357( + =3399.826(kW)
H s cng sut ca nh my:cos nm= Pttnm/Sttnm =2357.6/3399.826 =0.6934
Vy Pttnm = 2357.6(kW)Qttnm = 2449.6 (kW)Sttnm = 3399.826(kW)
Xy dng bn ph ti
Xc nh biu ph ti:
-Vic phn b hp l cc trm bin p trong phm vi x nghip l mt vn quan trng xy dng s cung cp in c cc ch tiu kinh t, k thut
cao, m bo c chi ph hng nm nh. xc nh c v tr t cc trm
bin p ta xy dng biu ph ti trn mt bng tng ca x nghip.
- Biu ph ti l mt vng trn c din tch bng ph ti tnh ton ca
phn xng theo mt t l la chn.
- 15 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- mi phn xng c mt biu ph ti. Tm ng trn biu ph ti
trng vi tm ca ph ti phn xng, tnh gn ng c th coi ph ti ca phn
xng ng u theo din tch ca phn xng.
- Biu ph ti cho php hnh dung c r rng s phn b ph ti trong
x nghip
- Mi vng trn biu ph ti chia ra thnh hai phn hnh qut tng ngvi ph ti ng lc v ph ti chiu sng
1.2.3/ Xc nh bn knh vng trn ph ti:
- ng knh t l ca biu ph ti:
gip cho vic t trm phn phi v cc trm bin p
phn xng mt cch hp l t hiu qu kinh t nht v gim
c tn tht n mc thp nht ta phi xc nh trung tm
ph ti ca tng trm. Mun xc nh trung tm ph ti ta
cn phi da vo bn ph ti ca cc phn xng. Tm ca
nhng ng trn chnh l trung tm ph ti ca cc phn x-
ng.
+Cng thc tnh:
RX=m
SttX.
Tron :
+ S ttX :Ph ti tnh ton ca phn xng th X ( kva)
+ Ri : Bn knh vng trn biu ph ti ca phn xng th X (mm)
+ m : T l xch ( kva/mm 2 )
*/ Gc chiu sng ca biu ph ti :
cs =tt
cs
P
P.360
xc nh biu ph ti ta chn m=50 kva/cm 2
1.2.4. Biu xc nh tm ph ti.- 16 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Trn s mt bng x nghip v mt h to xoy, c
v tr to trng tm ca cc phn xng l ( xi, yi ) ta xc
nh c to ti u M0(x0, y0 )
Vng trn ph ti :
1545
1
Tn phn xng; Cng sut tnh ton ca phn xng
Phn Xng trm t:
- 17 - nmn hc : SV : Ng xun Li H in 1_K3
Gc ph ti chiusng.
Ph ti nglc.
cs
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
theos mt
bng
Tn Phn Xng Pcs,kW
Ptt,kW
Qtt,kWAr
Stt,KVA
R(mm)
0cs
1 Phn xng trm t 9.405 179.4 193.4 263.8 8.2 18.872 Phn Xng vt liu hn 15.66 295.7 437.5 528 10.5 19.063 Phn xng nha Tng hp Plasmace 6.12 413.1 457.7 616.6 6.6 5.334 Phn xng tiu chun 5.445 100.4 80.55 128.8 6.2 19.55 Phn xng kh c in 3.57 262.6 253.1 364.7 5 4.896 Phn xng dp 9.12 305.1 338.1 455.4 8 10.7
7 Phn xng Xi mng aming 3.696 343.7 331.3 477.4 5.3 3.878 Kho thnh phm 3.96 44.76 25.37 51.45 5.3 31.89 Kho ph liu kim loi 1.12 34.72 25.14 42.86 3 11.6
10 Phn xng m in 1.848 481.8 412 634 3.75 1.38
11 Xem d liu phn xng 0 0
12 Trm trung ha 1.512 53.51 60.91 81.08 3.65 10.1713 Ra knh thot axit 0.88 21.88 23.6 32.18 3 14.47
14 Trm bm 1.35 144.4 155.7 212.3 3.1 3.36615 Nh n 1.44 31.44 46.51 56.14 3.2 16.4816 Phn xng in 2.25 68.25 65.78 94.79 4 11.8617 Nh iu hnh 2.21 25.21 20.23 32.32 3.4 31.5518 Phn xng lm ngui 3.528 27.23 22.56 35.36 5.2 46.6419 Kho axit 1.12 16.92 18.75 25.25 2.9 23.8320 My nn s N0 1 1.2 97.2 93.68 135 3.57 4.44
- 18 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- 19 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
21
10
53 4
18
15
8 7 611
16
149 19 20 13 12
17
- 20 - n mn hc : SV : Ng xun LiH in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
CHNG III
Thit k mng cao p cho ton nh my
Theo s liu ban u th Nh my c c cung cp in
t trm bin p trung gian (BATG) cch nh my 507,05m. ch-
ng 2, ta xc nh c cng sut tnh ton ca cc phn
xng v ton nh my. Vi s liu d nh t trm phn
phi trung tm (PPTT) nhn in t trm BATG v ri phn
phi cho cc trm bin p phn xng (BAPX).
I. V tr trm phn phi trung tm.
gim chi ph u t cho dy dn v gim tn tht in
nng tc m bo v tiu chun kinh t th trm phn phi
trung tm phi t trung tm ph ti ca ton nh my.
Trn mt bng nh my gn mt h to vung gc XOY.
V tr tm ph ti ca ton nh my M(X,Y) c xc nh
theo cng thc:
X =
=
=11
1
11
1
.
i
i
i
i
i
S
xS
; Y =
=
=11
1
11
1
.
i
i
i
i
i
S
yS
(3.1)
Trong :
xi ,yi : To tm ph ti ca phn xng th i.
X,Y : To tm ph ti in ton nh my.
Si : Cng sut tnh ton ph ti phn xng
th i,kVA.
- 21 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Bng 3.1 Ta tm ph ti ca cc phn xng, trm
BA trung tm v ngun.
theos mt
bng Tn Phn Xng X Y1 Phn xng trm t 77.5 54.52 Phn Xng vt liu hn 83 21.53 Phn xng nha Tng hp Plasmace 84.5 102.54 Phn xng tiu chun 84.5 102.55 Phn xng kh c in 84.5 84.256 Phn xng dp 51.5 104.57 Phn xng Xi mng aming 45.5 64.58 Kho thnh phm 45.5 419 Kho ph liu kim loi 14 58.510 Phn xng m in 103 5311 Xem d liu phn xng 45.5 141.512 Trm trung ha 13 124
13 Ra knh thot axit 15.3 105.514 Trm bm 11.5 42.515 Nh n 71.75 144.7516 Phn xng in 30.5 141.517 Nh iu hnh 11 42.7518 Phn xng lm ngui 91.5 143.519 Kho axit 14 69.520 My nn s No 1 14.75 87.5
- 22 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Tm ph ti ca ton nh my c to l : M
(67.57;73.47)
trong :
X=67,57
Y=73.47
Cc phng n t trm bin p
Phng n 1: Ch dng mt trm bin p t ngay trmphn phi trung tm. Trm ny c ta 00O(67,57;73,47)
Phng n 2: : Dng 4 trm bin p : TBA1,TBA2,TBA3,TBA4
Trm 1 cung cp cho phn xng:1, 2, 10
Trm 2 cung cp cho phn xng: 3, 4, 5, 18, 15
Trm 3 cung cp cho phn xng: 7, 8, 9, 14, 19
Trm 4 cung cp cho phn xng: 6, 11, 12, 13, 16, 17, 20
Ta xc nh to ca cc trm bin p theo cng thc:
+ X BAi =
i
ii
S
.xS; + Y BAi =
i
ii
S
.yS
-Trm 1:
X 1BA = 87.908.1425
5.129570
5286348.263
103*52883*6348.263*5.77==
++++
Y1BA
=27.39
8.1425
1.55992
5286348.263
53*5285.21*6348.263*5.54==
++
++
-Trm 2:
X 2BA = 11.846.1201
101067
36.3514.567.3648.1286.616
101067==
++++
Y 2BA = 14.1006.1201
120330
36.3514.567.3648.1286.616
16.5074265.812630726132025.63201 ==++++++++
-Trm 3:
X 3BA = 25.253.21286.4245.514.477 14*25.253.212*5.1114*86.425.45*45.515.45*4.477 ++++ ++++ = 93.3326.809 66.27457 = - 23 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Y 3BA =25.253.21286.4245.514.477
5.69*25.253.212*5.425.58*86.4241*45.515.64*4.477
++++++++
= 57.07
-Trm 4:
X 4BA = =+++++ +++++ 13532.3279.9418.3208.814.455135*75.1432.32*1179.94*5.3018.32*3.1508.81*134.455*5.51 36.4
Y 4BA =
5.10513532.3279.9418.3208.814.455
135*5.8732.32*75.4279.94*5.14118.32*5.10508.81*1244.455*5.104=
++++++++++
Ta xc nh c to ca cc trm bin p trn l:
Trm 1 O 1 (90.87;39.27)
Trm 2 O 2 (84.11 ; 100.14)Trm 3 O 3 (33.93 ; 57.07)
Trm 4 O 4 (36.4; 105.5)
Phng n 3: chia thnh 7 nhm:
Trm 1 cp in cho Phn xng 1; 5
Trm 2 cp in cho Phn xng 2
Trm 3 cp in cho Phn xng 10Trm 4 cp in cho Phn xng 3
Trm 5 cp in cho Phn xng 4; 15; 18
Trm 6 cp in cho Phn xng 7; 8; 9; 14; 19
Trm 7 cp in cho Phn xng 6; 11; 12; 13; 16; 17; 20
S ni dy t trm bin p n cc phn xng:
m bo an ton v m quan trong x nghip cc tuyn dy s c xy dngbng ng cp.
C th so snh theo cc phng n sau:
Phng n 1: Ta dng 1trm bin p t ngay trm phn phi trung tm.
T trm bin p ni dy n cc phn xng theo ng thng, cc t phn phi s
c t ngay ti u cc phn xng cung cp in cho cc thit b trong xng
- 24 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Phng n ny c tng chiu di nh nht nhng khng thun tin cho vic thi cng,
vn hnh v pht trin mng in nn khng c tnh kh thi. V vy ta loi b phng
n ny.
Phng n 2 : Cng ko trc tip t trm bin p n cc phn xng ,
nhng theo ng b gc , Cc ng cp s c xy dng dc theo cc mp
ng v nh xng , nh vy s thun tin cho vic xy dng , vn hnh v
pht trin mng in , tuy nhin chiu di ca cc tuyn dy s tng hn so vi
phng n 1.
Phng n 3: Dng 2 trm bin p c ni t trm phn phi trung
tm. Phng n ny ni dy t cc trm bin p n cc phn xng theo
ng thng, cc t phn phi s c t ngay ti u cc phn xng cung
cp in cho cc thit b trong xng.
Phng n ny c tng chiu di nh nht nhng khng thun tin cho vic
thi cng , vn hnh v pht trin mng in nn khng c tnh kh thi. V vy ta
loi b phng n ny.
Phng n 4: Dng 2 trm bin p c ni t trm phn phi trung tm.
Phng n ny cng ko dy trc tip t trm bin p n cc phn xng nhng theo
ng b gc, cc ng cp s c xy dng dc theo cc ng mp v nh xng.
Nh vy s thun tin cho vic xy dng, vn hnh v pht trin mng in, tuy nhin
chiu di tuyn dy s tng ln so vi phng n 3.
Phng n 5: Dng 4 trm bin p c ni t trm phn phi trung tm.
Phng n ny c ni dy trc tip t cc trm bin p n cc phn xng theo
ng cho. Phng n ny c chiu di dy ngn nhng kh lp t v vn hnh.
Do vy phng n ny cha c ti u.
Phng n 6: Dng 4 trm bin p c ni t trm phn phi trung tm.- 25 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Phng n ny c ni dy trc tip t cc trm bin p n cc phn xng theo
cc ng b vung gc. Phng n tuy chiu di dy ln nhng d lp t v vn
hnh, pht trin nh my.
I./ LA CHN MY BIN P.
1. Cc phng n ni dy bin p.
1.1. Phng n 1: s ni dy hnh tia.
H1: S ni dy hnh tia
a. u - nhc im:
+ u im: tin cy cung cp in cao, thun tin cho qu trnh thi cng, vn
hnh
V sa cha.
+ Nhc im: Vn u t kh ln.
b. Thnh phn:
B: Trm bin p phn xng.
1 : Thanh ci trm bin p phn xng.
2 : Thanh ci t phn phi ng lc.
3 : Ph ti dng in.
- 26 - nmn hc : SV : Ng xun Li H in 1_K3
B
22
2 23
1
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
1.2. Phng n 2: S ni phn nhnh.
H 2: S ni phn nhnh.
a. u_nhc im:
-u im: Gi thnh r, lp rp nhanh v tit kim c t phn phi.
-Nhc im: tin cy cung cp in thp, phc tp khi bo v.b. Thnh phn: trm bin p trn gm c:
B : Trm bin p phn xng.
1 : Thanh ci trm bin p phn xng.
2 : Thanh ci t phn phi ng lc.
3 : Ph ti dng in.
Cn c kt qu trn chn dng cc b t 3 pha do Lin X ch to, b t c
bo
v bng ptmt, trong t c t cc bng n lm in tr phng in.
1.3. Phng n 3: S ni dy hn hp.- 27 - n
mn hc : SV : Ng xun Li H in 1_K3
B
122
23
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
H 3: S ni dy hn hp.a. c im:
- tin cy cung cp in cao, thun li cho qu trnh vn hnh v sa cha.
- Gi thnh thp, lp rp nhanh, tit kim t phn phi.
b. Thnh phn:
B : Trm bin p phn xng.
1 : Thanh ci trm bin p phn xng.2 : Thanh ci t phn phi ng lc.
3 : Ph ti dng in.
Trong cc phng n trn ta thy phng n 3 l kh thi nht v n kt hp
c c 2 ch tiu kinh t v k thut.
1. Xc nh s lng, dung lng trm bin p.
2.1./ Gii thiu chung.
Trong h thng in bin i t cp in p ny sang cp in p khc
ngi
ta dng my bin p. Vn u t cho my bin p trong tng vn u t l kh
ln,
- 28 - nmn hc : SV : Ng xun Li H in 1_K3
B
21
2
2
211
1
3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
v vy vic la chn s lng v cng sut cc my bin p trn c s kinh t,
k
thut cho cc trm bin p phn xng c ngha quan trng xy dng hp
l s
cung cp in cho nh my.
1.1.1. S lng trm bin p.
H tiu th loi 1 cn c cung cp t hai ngun in c lp trong
c th m bo cung cp cho tt c cc h tiu th khc. m bo tin
cy cung cp in cho
h tiu th loi 1 t mt trm bin p, t nht mi phn on thanh ci phi ni
vi mi my bin p v cng sut ca my bin p c chn sao cho khi hng
mt trong
chng th my cn li vi kh nng qu ti ca mnh, phi m bo cung cp
cho tt
c cc h tiu th loi 1.
Cn lu rng, ti cc trm bin p phn xng vi hai my bin p,
hp l nht l cc phn on thanh gp in p thp lm vic ring r. Khi
dng in ngn mch s gim bt hai ln v iu kin lm vic ca cc kh c
in, in p thp s c
gim nh. Khi ct mt my bin p my cn li s m nhn c ph ti ca my
kianh ng ptmt phn on. Vic ng ngun d tr cho h loi 1 c thc
hin t ng. Cc phn xng loi 3 ly in t pha h p ca trm bin p
phn xng
cnh hoc t ring trm bin p c mt my bin p.
2.1.2.V tr cc trm bin p.- 29 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Trm c t cnh cc phn xng, v kt cu trm c xy kn mt
tng l
tng ca phn xng. Cc trm bin p p dng s ni my bin p qua
dao cch
ly-cu ch. i vi trm mt my b tr thnh hai phng: phng cao p t cc
thit b
cac p, phng h p t my bin p v cc thit b phn phi h p. Trm hai
my bin p b tr 4 phng, mt phng t hai t cao p, mt phng t cc t
h p v hai phng cn li mi phng t mt my bin p.
2.1.3. Chn dung lng my bin p.
Trong h thng in cung cp in cho nh my, cng sut ca my bin
p in
lc trong iu kin bnh thng phi cung cp cho tt c cc h tiu th. Khi
chn cng sut MBA cn phi m bo ch lm vic v mt kinh t, m
bo d tr r nt hoc khng r rt thch hp cung cp cho h tiu th khi
ct mt my bin p. Thm vo ph ti ca MBA trong iu kin bnh
thng khng c lm gim tui th ca MBA (theo iu kin pht nng).
tin cy cung cp in ca nh my t c nh trm hai MBA. Cn
lu rng trong khi s c mt MBA my cn li phi m bo ton b cng
sut yu cu ca h tiu th loi 1. Vic m bo cng sut yu cu thc hin
c khng ch do s dng cng sut danh nh ca MBA m phi k n khnng qu ti ca n (vi mc ch gim cng sut t ca MBA). Kh nng qu
ti ca MBA tu thuc vo th ph ti ca h tiu th nhn in t MBA. C
hai trng hp cho php qu ti MBA: qu ti s c v qu ti thng xuyn
khi vn hnh.
Qu ti s c: l ch qu ti cho php trong mt s trng hp ngoi l
(s c) vi mt thi gian hn ch khng gin on vic cung cp in nng.- 30 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Trong iu kin s c, cho php MBA (vi bt k h thng lm mt no)
khng ph thuc nhit mi trng lm mt, c th lm vic trong thi gian 5
ngy m theo th ph ti hai bc, vi ph ti bc hai khng qu 0,93 v ph
ti bc hai t n 1,4 cng sut nh mc nhng thi gian bc hai khng qu 6
gi mt ngy (tng thi gian gin on hay lin tc). Lc ny cn tng cng
lm mt.
Qu ti thng xuyn: l ch lm vic xt trong mt khong thi gian
no (ngy, thng, nm), trong c mt khong thi gian MBA lm vic
non ti v khong thi gian cn li ca chu k kho st MBA mang ti nh hn
nh mc.Mc qu ti c tnh ton sao cho hao mn cch in trong
khong thi gian xt khng vt qu nh mc, tng ng vi nhit cun
dy 98 .
C hai kh nng qu ti cho php cng li nhng khng c vt qu 50% tc
l
S qt 1,5.S dd
Tm li chn cng sut MBA ta s dng CT sau:
Vi trm t 1 MBA: iu kin S mB ttS
Vi trm t 2 MBA: iu kin S mB 2
ttS v S mB 4,1
SCS
Trong :
S mB - cng sut nh mc MBA (kVA).
Stt
- cng sut tnh ton ca ph ti (kVA).S SC - cng sut yu cu lc s c (kVA).
Chn MBA loi do ABB sn sut, khng cn hiu chnh nhit , ng thi
gim c chi ph u t do tit kim ngoi t. Sau y tin hnh tnh ton la
chn MBA cho cc trm bin p.
2.2. Tnh ton v la chn.
- 31 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Ph ti loi 1
Gi thnh tn tht in nng C = 1000 / kwh
Sut thit hi do mt in g th = 7500 / kwh.
2.2.1. Phng n 1:
Dng mt trm bin p t ngay trm phn phi trung tm. Trm ny cung cp
cho tt c cc phn xng nn ta chn t 2 MBA lm vic song song.
Chn MBA c: S mBA 4,1
ttS = 30474,1
4267 = (kVA)
Ta chn 2 MBA c cng sut = 3200 (kVA) do ABB ch to
2.2.4. Phng n 2:Ta dng 4 trm bin p:
Trm 1 cung cp in cho cc phn xng : 1, 2, 10
Trm 2 cung cp in cho cc phn xng : 3, 4, 5, 18, 15
Trm 3 cung cp in cho cc phn xng : 7, 8, 9, 14, 19
Trm 4 cung cp in cho cc phn xng : 6, 11, 12, 13, 16, 17, 20
Vi trm 1 ta c:
S 1ttN = Stt1+ Stt2+ Stt10 = 1398
Chn MBA c : S 1mBA 4,1
1ttS = 10004,1
1398= (kVA)
Nh vy ta chn 2 MBA c cng sut S 1mBA =1000 (kVA).
Hoc 1 MBA c cng sut 2000 (kVA).
Vi trm 2 ta c:
S 2ttN = Stt3+ Stt4+ Stt5 + Stt15+ Stt18 = 1202
Chn MBA c : S 1mBA 4,1
1ttS = 2.8584,1
1202= (kVA)
Nh vy ta chn 2 MBA c cng sut S 1mBA =1000 (kVA).
Hoc 1 MBA c cng sut 2000 (kVA).- 32 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Vi trm 3 ta c:
S 3ttN = Stt7+ Stt8+ Stt9 + Stt14+ Stt19 = 809.26
Chn MBA c : S1mBA
4,11ttS
=578
4,1
26.809
= (kVA)Nh vy ta chn 2 MBA c cng sut S 1mBA =630 (kVA).
Hoc 1 MBA c cng sut 1000 (kVA).
Vi trm 4 ta c:
S 1ttN = Stt6+ Stt16+ Stt12 + Stt13+ Stt17 + Stt20= 830.77
Chn MBA c : S 1mBA 4,1
1ttS = 407.5934,1
77.830= (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =630 (kVA).
Hoc 1 MBA c cng sut 1000 (kVA).
Phng n 3: chia thnh 7 nhm:
Trm 1 cp in cho Phn xng 1; 5
Trm 2 cp in cho Phn xng 2
Trm 3 cp in cho Phn xng 10
Trm 4 cp in cho Phn xng 3
Trm 5 cp in cho Phn xng 4; 15; 18
Trm 6 cp in cho Phn xng 7; 8; 9; 14; 19
Trm 7 cp in cho Phn xng 6; 11; 12; 13; 16; 17; 20
Vi trm 1 ta c:
S 1ttN = Stt1+ Stt5 = 263,8 + 364,7 = 628,5 (kVA)
Chn MBA c : S 1mBA 4,1
1ttS = 4494,1
5.628= (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =500(kVA).
Hoc 1 MBA c cng sut 1000 (kVA).
Vi trm 2 ta c:
SttN2 = Stt2 = 528 (kVA)- 33 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Chn MBA c : S 1mBA 4,1
1ttS = 3774,1
528 = (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =400(kVA).
Hoc 1 MBA c cng sut 800 (kVA).Vi trm 3 ta c:
SttN3 = Stt10 = 528 (kVA)
Chn MBA c : S 1mBA 4,1
3ttS = =4,1
634453(kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =500(kVA).
Hoc 1 MBA c cng sut 1000 (kVA).
Vi trm 4 ta c:
SttN4 = Stt3 = 528 (kVA)
Chn MBA c : S 1mBA 4,1
4ttS = =4,1
3.616440(kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =500(kVA).
Hoc 1 MBA c cng sut 1000 (kVA).
Vi trm 5 ta c: SttN5 = Stt4+ Stt15 + Stt18 = 128.8 + 56.14 + 35.35= 220.3(kVA)
Chn MBA c : S 1mBA 4,1
1ttS = 35.1574,1
3.220 = (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =160(kVA).
Hoc 1 MBA c cng sut 315 (kVA).
Vi trm 6 ta c:
SttN6 = Stt7+ Stt8 + Stt9 + Stt14+ Stt19 = 477.4+ 51.45+ 42.86+ 212.3+ 25.25
= 810(kVA)
Chn MBA c : S 1mBA 4,1
6Stt= 578
4,1
810= (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =600(kVA).
Hoc 1 MBA c cng sut 1200 (kVA).
- 34 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Vi trm 7 ta c:
SttN7 = Stt6+ Stt12+ Stt13+ Stt16+ Stt17+ Stt20=455.4+ 81.08+ 32.18+ 94.79+
135+32.32= 830.77(kVA)
Chn MBA c : S 1mBA 4,1
7SttN= 4.593
4,1
77.830 = (kVA)
Nh vy ta chn 2 MBA do ABB sn xut c cng sut S 1mBA =600(kVA).
Hoc 1 MBA c cng sut 1200 (kVA).
Ta c bng tnh ton sau:
Phngn
S trmv
S trng hp
Cngsut
nh
mc
Cngsut
tnh
ton
Hao tncng
sut
inp
NM
Vnu t
Trm Trng
hp
S mBA S tt 0P KP U K K.10 6
VN1 1 1 23200 4267 1,75 13 7 220.5
2
1 1 21000 1398 1,75 13 5 220.52 12000 2,7 22,5 6 731.5
2 1 2 1000 1202 1,75 13 5 204.35
2 12000 2.7 22.5 6 731.53 1 2630 809,26 1,3 8,2 4,5 204.35
2 11000 1.9 13 6.5 220,5
4 1 2630 830.77 1,3 8,3 4,5 204.352 11000 1,75 13 5 220,5
3 1 1 2500 628.5 1 7 4 187.352 11000 1.75 13 5 220.5
2 1 2400 528 0.84 5.75 4 167.652 1800 1.4 10.5 5
3 1 2630 809.26 1.3 8.3 4,5 204.35- 35 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
2 11000 1,75 13 5 220.54 1 2500 528 1 7 4 187.35
2 11000 1,75 13 5 220.55 1 2160 220.3 0.5 2.95 4
2 1315 0.72 4.85 4 158.256 1 2630 810 1.3 8.3 4.5 204.35
2 11000 1.9 13 5 220.57 1 2630 830.77 1.3 8.3 4.5 204.35
2 11000 1.9 13 5 220.5
Bng cc tham s ca my bin p do ABB Vit Nam sn xut
So snh cc phng n:
By gi ta s i so snh 4 phng n la chn trn tm ra phng n ti
u.
Phng n 1 :
t 1 trm trung tm 2 my bin p c cng sut 3200 KVA
Phng n 2 :
Trm 1 :t 2 my bin p c cng sut = 2000 (KVA)
Trm 2 :t 2 my bin p c cng sut = 2000 (KVA)
Trm 3 :t 2 my bin p c cng sut = 630 (KVA)
Trm 4 :t 2 my bin p c cng sut = 630 (KVA)
Phng n 3 :
Trm 1 :t 2 my bin p c cng sut = 500 (KVA)
Trm 2 :t 2 my bin p c cng sut = 400 (KVA)
Trm 3 :t 2 my bin p c cng sut = 630 (KVA)
Trm 4 :t 2 my bin p c cng sut = 500 (KVA)
Trm 5 :t 2 my bin p c cng sut = 160 (KVA)
Trm 6 :t 2 my bin p c cng sut = 630 (KVA)
Trm 7 :t 2 my bin p c cng sut = 630 (KVA)
Ta c:
= ( 0,124+ T M .10 4 ) 2 . 8760- 36 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
= ( 0,124 + 4500.10 4 ) 2 . 8760 = 2886,21 (h)
a. Phng n 1:
S SC = m12 . S tt = 0,7.1679,89= 1175,9 (kVA)
Kqt = 1759,11000
9,1175==
mBA
SC
S
S
Xc nh tn tht do mt in :
Do phng n ny chu c qu ti khi gp s c nn khng mt in
Vy Y 2 =0
Tn tht trong MBA c xc nh theo biu thc:
A 1BA = (n.P 01 .8760 + ..2 2 1
21
nBA
K
S
SP )
= 2.1,75.8760 + 21,2886.1000
89,1679.
2
132
2
= 83602,32 (kWh).
Chi ph b tn tht l:
C 1 =C . 1BAA = 1000. 83602,32 = 83,6.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Suy ra: Z 1 = 0,185.220,5.10 6 + 83,6.10 6 =124,4.10 6 (/nm).
b. Phng n 2:
- Trm 1:
Tn tht trong MBA c xc nh theo CT:
A 21BA =(n. 01P .8760+ ..2 2 1
21
nBA
K
S
SP)
= 2.1,75.8760 + 21,2886.1000
1398.
2
132
2
= 67325.33 (KWh)
Chi ph b tn tht l:
C 21 =C. 21BAA =1000. 67325.33= 67,325.10 6 (/nm).
- 37 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z 21 = 0,185.220,5.10 6 + 67,325.10 6 =108,117.10 6 (/nm).
Trm 2:
Tn tht trong MBA c xc nh theo CT:
22BAA =(n.P 02 .8760 + ..2 2 2
221
nBA
K
S
SP)
= 2.1,75.8760 + 21,2886.1000
1202.
2
132
2
= 57765,05 (KWh).Chi ph b tn tht l:
C 22 = C . 22BAA = 1000 . 57765,05 =57.765.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z 22 = 0,185.220.5.10 6 + 57,765.10 6 =98,557.10 6 (/nm).
Trm 3:
Tn tht trong MBA c xc nh theo CT:
A 31BA =(n. 01P .8760+ ..2 2 1
21
nBA
K
S
SP)
= 2.1,3.8760 + 21,2886.630
5.628.
2
2,82
2
= 42301.7(KWh)
Chi ph b tn tht l:C23=C. A BA23 =1000. 42301.7= 42,301.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z23= 0,185.204,35.10 6 + 42,301.10 6 =80,105.10 6 (/nm).
Trm 4:
Tn tht trong MBA c xc nh theo CT:- 38 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
ABA24 =(n.P 02 .8760 + ..2 2 2
2
nBAS
SP)
= 2.1,3.8760 + 21,2886.630
77.830.
2
2,82
2
= 43353.48 (KWh).
Chi ph b tn tht l:
C24 = C .ABA24= 1000 . 43353.48 =43,354.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z24= 0,185.204.35.106
+ 43,354.106
=72,158.106
(/nm).Vy tng chi ph qui i l: Z2= Z21+ Z22+Z23+Z24= 108,117.10 6 +80,105.10 6
+98,557.10 6 +72,158.10 6 = 358,973.10 6 (/nm).
c. Phng n 3:
Trm 1:
Tn tht trong MBA c xc nh theo CT:
ABA31=(n.P 03 .8760 + ..2 2 3
231
nBA
K
S
SP)
= 2.1.8760 + 21,2886.500
5.628.
2
72
2
= 33481,23 (KWh).
Chi ph b tn tht l:
C31 = C
.ABA31= 1000 . 33481,23 =33,481.106
(/nm).Chi ph qui i :
Z = (a vh + a tc ).V + C
Z31= 0,185.187,35 6 + 33,481.10 6 =58,44.10 6 (/nm).
Trm 2:- 39 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Tn tht trong MBA c xc nh theo CT:
A 41BA =(n. 01P .8760+ ..2 2 1
21
nBA
K
S
SP)
= 2.0,84.8760 + 21,2886.400
528.
2
75,52
2
= 14716.85(KWh)
Chi ph b tn tht l:
C 32 =C. 32BAA =1000. 14716.85 = 14,716.10 6 (/nm).
Chi ph qui i :
Z = (avh
+ atc
).V + C Z 32 = 0,185.167,65.10 6 + 14,716.10 6 =45,731 .10 6 (/nm).
Trm 3:
Tn tht trong MBA c xc nh theo CT:
ABA33=(n.P0.8760 + ..2 2 2
233
nBA
K
S
SP)
= 2.1,3.8760 + 21,2886.63026.809
.2
3,8
2
2
= 42540 (KWh).
Chi ph b tn tht l:
C 42 = C. 42BAA = 1000 . 42540= 42,54.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z33= 0,185.204,35.106
+ 42,54.106
=80,345.106
(/nm).Trm 4:
Tn tht trong MBA c xc nh theo CT:
34BAA =(n.P 03 .8760 + ..2 2 3
234
nBA
K
S
SP)
= 2.1.8760 + 21,2886.500
528.
2
72
2
= 28785 (KWh).- 40 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Chi ph b tn tht l:
C 43 = C . 43BAA = 1000 . 28785= 28,785.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z34= 0,185.187,35.10 6 + 28,785.10 6 = 63,444.10 6 (/nm).
Trm 5:
Tn tht trong MBA c xc nh theo CT:
ABA35=(n.P 04 .8760 + ..2 2 4
235
nBA
K
S
SP)
= 2.0,5.8760 + 21,2886.160
3.220.
2
95,22
2
= 16830,65 (KWh).
Chi ph b tn tht l:
C35 = C . 44BAA = 1000 . 16830,65 =16,83.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z 44 = 0,185.100,25.10 6 + 16,83.10 6 = 35,376.10 6 (/nm).
Trm 6:
Tn tht trong MBA c xc nh theo CT:
ABA36=(n.P0.8760 + ..2 2 2
233
nBA
K
S
SP)
= 2.1,3.8760 + 21,2886.630
810.
2
3,8
2
2
= 42576 (KWh).
Chi ph b tn tht l:
C36 = C . 36BAA = 1000 . 42575= 42,576.10 6 (/nm).
Chi ph qui i :
Z = (a vh + a tc ).V + C
Z33= 0,185.204,35.106
+ 42,576.106
=80,380.106
(/nm).- 41 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Trm 7:
Tn tht trong MBA c xc nh theo CT:
ABA37=(n.P0.8760 + ..2 2 2
233
nBA
K
S
SP
)
= 2.1,3.8760 + 21,2886.630
77,830.
2
3,82
2
= 43604,4 (KWh).
Chi ph b tn tht l:
C37 = C . 42BAA = 1000 . 43604,4= 43,604.10 6 (/nm).
Chi ph qui i :Z = (a vh + a tc ).V + C
Z37= 0,185.204,35.10 6 + 43,604.10 6 =81,408.10 6 (/nm).
Vy tng chi ph qui i l:
Z3 = Z31+ Z32+Z33+Z34+ Z35+Z36+ Z37
= 58,44.10 6 + 45,731.10 6 + 80,345.10 6 + 63,444.10 6 +35,376.10 6
+80,380.10 6 + 81,408.10 6 = 445,124 (/nm).
T cc phng n trnh by trn, ta c th thy l phng n 2(Phng n
s dng 2 my bin p c cng sut 1000 (KVA) ,2 my bin p c cng sut
630 (KVA) c chi ph nh nht ng thi cng m bo cc yu cu k thut.
Phng n ny cn t 2 trm bin p ngay trm phn phi trung tm vi 2
my bin p lm vic song song c cng sut l 1000(KVA). 2 my bin p c
cng sut 630 (KVA)
- 42 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
H 1: S ni trm bin p phn xng t 1 MBA
H 2: S ni cc trm phn xng t 2 MBA.
- 43 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Hnh 3- 6: S nguyn l mng cao p ton x nghip
- 44 - nmn hc : SV : Ng xun Li H in 1_K3
35KV
CSV
MC
BATT
MCBU
TG 10 KV MCLL TG 10 KV
0,4 KV
B1
CC
CD
B2 B3 B4 B7 B8 B9
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
II. La chn dy dn.
Khi la chn phng n c th chn chi tit dy dn theo phng
php n gin nht theo dng in t nng cho php, nhng sau khi xc
nh c phng n ti u th tit din dy dn phi c kim tra li theo hao
tn in p cho php. Do dy l mng in h p nn cht lng in p dc
t ln hng u, do ta chn tit din dy dn ln hng u, do ta ch tit
din dy dn theo hao tn in p, ly gi tr hao tn in p cho php l cpU
= 5% (i vi mng in cp in p 380V, = cpU 19%). D nh t cp
trong cc rnh ngm di lng t . Do s b chn gi tr in tr khng l X0
= 0,07/km.
1. Phng n .
S ni cc dy ca mng in c th hin hnh trn. Chiu di dy dn
c ni t trm bin p n cc phn xng theo ng b gc c xxc
nh bng biu thc:
Li = ( ) ( ) 22 iBAiBA yYxX +
Vi phn xng 1:
Chiu di dy dn c ni t trm bin th n phn xng 1 theo ng b
gc c xc nh :
L1 = ( ) ( ) ( ) ( ) 22212
1 5.5457,735,7747,67 +=+ yYxXBABA
= 29.1 (m)
Thnh phn phn khng ca hao tn in p:
)(036.138,0
10.1,29.07,0.4,193.. 311 V
U
LxQU Nx ===
Thnh phn tc dng ca hao tn in p:
11 XcpR UUU = =19 1,036 = 17,97 (V).
- 45 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Tit din dy :
( )21
11
1 2497,1738,032
1,29.4,179mm
UU
LPF
R
=
=
=
Ta chn cp loi ABBG c F Ch = 25mm 2 c r=1,25 / km v x= 0,06
/ km.
Hao tn in p thc t:
3
1
0101
1 101,2938,0
06,0.4,19325,1.4,179 +
=+
= LU
xQrPU = 18,06(V) 1000A do khng phi
kim tra dng n nh nhit.
b. Chn my ct hp b 10kv:
- Cc my ct ni vo thanh ci 10kv chn cng mt loi
SF6, k hiu 8DC11 do SIEMENS ch to c bng thng s sau:
Loi Um,kV Im, A Im.C, 2skA
i, kA
8DC11 12 1250 25 63
- Kim tra :
Im.MC Icb = 364,72 (A)
Im.ct IN = 4,55 (kA)
im. ixk = 11,58 (KA)
2.2. Chn v kim tra dao cch li cp 35 kV:
iu kin chn v kim tra:
- in p nh mc, kV : UmDCL Um.m- 54 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- Dng in lu di nh mc, A : Im.DCL Icb
- Dng n nh ng, kA : im. ixk
- Dng n nh nhit, kA : tm.nh.I2
m.nh tq.I2
Chn dao cch li t ngoi tri, li dao quay trong mt
phng nm ngang loi 3DE do SIEMENS ch to:
Loi Um, kv Im, A INt, kA IN max, kA
3DC 36 1000 25 60
- Kim tra: UmDCL Um.m = 35 kV
Im.DCL Icb =104 A
IN max ixk = 16,8 kA
2.3. Chn t cao p trn b cp 10kv
- Chn t cao p trn b, c dao cch ly 3 v tr, cch in
bng SF6 do SIEMENS ch tao, loi 8DH10
Loi t Um, kV Im, A INt, kA IN max,kA
Thit b
8DH10 12 200 25 25 Dao ct phti
3.Cu ch
2.4 Chn v kim tra cp :
Chn cp ng 3 li, cch in XLPE, ai thp, v PVC
do hng FURUKAWA ch to, cp c t trong hm cp :
ng cp F, 1li Hnh ICP, IN, 1s Um, kV- 55 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
mm2 dng250c
AkA
BATT-B1,2,6 25 Vn
xon
140 3,37 10
BATT-B3,4,5,7
16 Vnxon
110 2,28 10
- Kim tra iu kin n nh nhit:
qdN tI.F
Trong :
- h s nhit , vi ng =7.
tq- thi gian qui i, s.
- Ngn mch trong h thng cung cp in c coi l
ngn mch xa
ngun: I=I do thi gian qui i ly bng thi gian tn ti
ngn mch.
tq = tnm = tbv + tmc .
Ta ly:
+ Thi gian tc ng ca bo v : tbv =0,02 s
+ Thi gian tc ng ca my ct :tmc =0,1 s
Thi gian quy i tq =0,12 s.
- Ta ch cn kim tra cho tuyn cp no c dng ngn
mch ln nht. Tuyn cp BATT-B6 c dng ngn mch ln nht
IN3 = 4,55KA.
Fmin= . IN3 max qdt = 7 . 4,55 . 12,0 =11mm2 < F =16mm2
- Cp c chn vt cp v c di ngn nn khng cn
kim tra iu kin tn tht in p v dng cho php.- 56 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
2.5. Chn v kim tra Aptomat .
- Vi trm 2 MBA ta t 2 t aptomat tng, 2 t aptomatnhnh v 1 t aptomat phn on.
- Vi trm 1MBA ta t 1 t aptomat tng v 1 t aptomat
nhnh.
- Mi t aptomat nhnh t 2 aptomat.
Aptomat c chn theo dng lm vic lu di:
m.dmdmA
dm
tt
ttmax.lvdmA
UU
U*3
SIII
==
- Vi aptomat tng sau my bin p, d tr c th
chn theo dng nh mc ca MBA.
dm
B.dm
B.dmA.dm U.3
S
II =
- Aptomat phi c kim tra kh nng ct ngn mch : ICt
m IN
Dng qua cc aptomat:
- Dng ln nht qua aptomat tng, MBA 1000 kVA
)A(14434,0.3
1000Imax
==
- Dng ln nht qua aptomat tng, MBA 630 kVA
)(32,9094,0.3
630max AI ==
2.6. Chn v kim tra thanh dn:
- 57 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- Thanh dn cp in p 10kv c chn thanh dn ng
cng.
Chn thanh dn theo iu kin pht nng lu dicho php:
K1 . K2 . ICP Icb.
- Thanh dn t nm ngang : K1=0,95.
- K2 : h s hiu chnh theo nhit .
0CP
'
0CP
2K
=
Trong :
cp =700c : nhit cho php ln nht khi lm vic bnh
thng.
0 =250c : nhit mi trng thc t.
K2 = 0,88
- Chn Icb theo iu kin qu ti ca my bin p:
)A(71510.3.88,0.95,0
7500.4,1
U.3.K.K
S.4,1I
U.3
S.4,1I
dm21
B.dmcp
dm
B.dmcb
==
=
Chn thanh dn ng, tit din trn 40x5, c dng Icp
=700(A)
Kim tra iu kin n nh ng: cp tt
- 58 - nmn hc : SV : Ng xun Li H in 1_K3
ah
b
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- Lc tnh ton do tc dng ca dng in ngn mch :
)KG(i.al.10.76,1F 2
XK
8
tt=
Trong :
l =100cm : khong cch gia cc s .
a =50cm : khong cch gia cc pha.
ixk : dng in xung kch ca ngn mch ba pha, A
ta c : ixk = 2 .1,8 . IN2 = 2 .1,8 . 4,6 = 11,71.103 A
Ftt =1,76.10-8.50
100.(11,71.103 )2 = 4,83 KG.
- M men un : )cm.KG(3,4810
100.83,4
10
l.FM tt ===
- ng sut tnh ton : )/(2cmKG
W
M
tt =
6
h.bW
2
= (cm3)
Thanh dn c b =0,5cm ; h = 4cm
)cm/KG(23,364.5,0
3,48.6
h.b
M.6 222tt===
ng sut cho php ca thanh dn ng cp=1400kG/cm2
cp >> tt= 36,23(kG/cm2).
Kim tra n nh nhit : qdtI.S
Ta c: =7; I= 4,6 (KA); tq = 0,12s
. I qdt = 7 . 4,6 . 12,0 = 11,15 mm2 .
- 59 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
S =40 . 5 = 200mm2>> 11,15mm2
2.7. Chn v kim tra s :
Chn s t ngoi tri doLin X ch to loi OH-10-
500(H-10)
C Um =10kv, Fph =500 KG.
- Kim tra n nh ng : Fcp Ftt.
Thanh dn t nm ngang nn lc tc ng cho phptrn u s l Fcp=0,6Fph0,6 Fph Ftt .
)KG(05,86,083,4
6,0
FF ttph
== , s chn tho mn.
2.8. Chn bin dng in BI
- Chn bin dng do SIEMENS ch to loi 4MA72 c thng
s k thut cho bng sau.
K hiu UmkV
Uchung
kV
Uchu pxung
kV
I1 mA
I2.mA
I.ngkA
4MA72 12 28 75 20 -2500
1hoc
5
120
2.9. Chn my bin p BU
- Chn my bin in p 3 pha 5 tr do Lin X ch to loi
HTM-10 c cc thng s k thut sau:
Loi Um, V Cng sut nh mctheo cp chnh xc
VA
SmVA
- 60 - nmn hc : SV : Ng xun Li H in 1_K3
Ftt
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Scp Thcp
0,5 1 3
HTM-10 10000 100 120 120 200 1200
Chng 4.
xc nh cc tham s ca mng in
I. Hao tn in p
1. Trn ng dy
Nh tnh ton, tn tht trn ng dy vi phng n ti u
(Phng n 2) l U = 159,7 (V). Trong tn hao in p ln
nht trn ng dy l on M
2. Trong my bin p
2
2
dmBA
dm
NBAS
UPR =
dmBAdm
nBAS
UUX
=
100%
2
RBA1 = 15 . 322
10.1000
4,0 = 2,4 . 10
-3
XBA1 = 5,5 . 32
10.1000.100
4,0 = 8,8 . 10
-3
Khi my bin p lm vic song th :
RBA =2
BAR = 1,2 . 10-3
XBA =2BAR = 4,4 . 10-3
II. Hao tn cng sut- 61 - n
mn hc : SV : Ng xun Li H in 1_K3
68,11=+
=U
XQRPU BABABA
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
1. Trn ng dy
Hao tn cng sut tc dng trn on dy OV c xc nh
theo biu thc
1mP = lrU
QP.02
22 + = 83,810.76.67,038,0
73,9726,124 62
22
=+ (kw)
Hao tn cng sut phn khng trn on dy OM c xc
nh theo biu MQ0 = lxU
QP.02
22 += 79,010.76.06,0
38,0
73,9726,124 62
22
=+
(kVAR)
Tnh ton tng t cho cc on dy khc ghi trong bng di
y.
NQ P
L0i r0 x0 P Q
KVar KVar m /km /km KW KVarM1 193.4 179.4 29.1 1.25 0,06 8.81 0.79
M2 437.5 295.7 46 0,95 0,06 7.77 0.49M3 457.7 413.1 53 0,95 0,06 10.81 0.68M4 80.55 100.4 146 0,48 0,06 6.54 0.82M5 253.1 262.6 102 0,4 0,06 16.82 2.52M6 338.1 305.1 146 0,18 0,06 10.85 3.62M7 331.3 343.7 158 0,18 0,06 11.15 3.72M8 25.37 44.76 90 0,48 0,06 11.17 1.4M9 25.14 34.72 118 0,4 0,06 15.6 2.34M10 412 481.8M12 60.91 53.51M13 23.6 21.88M14 155.7 144.4M15 46.51 31.44M16 65.78 68.25M17 20.23 25.21M18 22.56 27.23M19 18.75 16.9
M20 93,68 97.2- 62 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
2. Trong my bin p
=
+=
+=
22
03200
4267.
2
159,4.2..
2
1.2
nBA
kBAS
SPPP 32,72 (kW)
Iii. Tn tht in nng
Tng tn tht in nng trong ton mng in
BAdd AAA += = 287187,5 + 61517,8 + 67899,3 =
416604,6 (kWh)
(Gi tr ddA tng tn tht in nng trn ng dy (Mc
III.1), tng tn tht in nng trong cc my bin p BAA -
c tnh theo phng n 2 (Theo 2.2.4.b))
CHNG 5. TNH TON NI T CHO TRM BIN P CA PHN
XNG N
1. Khi nim v ni t.
Dng in c th ngi gy nn nhng tc dng nguy him :gy
bng,git,trng hp s lm cht ngi, do tc dng ca ni t l tn
dng in v gi mc in th thp trn cc thit b c ni t.H thng
ni t c 2 chc nng: ni t lm vic va ni t an ton.
- Ngi b tai nn v in l do chm phi nhng phn t mang in,
song cng c th chm phi nhng b phn ca thit b ih bnh thng khng
mang in nhng li c in p khi cch in b hng.trong trng hp ny,
m bo an ton c th thc hin bng cch ni t tt c nhng b phn bnh
thng khng mang in nhng khi chc in b hng c th c in p.
- 63 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- Khi c ni t ,qua ch cch in b chc thng v thit b ni t s
co dng in ngn mch mt pha vi t v in p i vi u ca v thit b
bng : U =I R
Trong :
+I : dng in mt pha chm t.
+ R : in tr ni t ca cc trang b ni t.
Trng hp ngi chm phi v thit b c in p ,dng in qua ngi xc
dnh theo biu thc : I ng .Rng =I R .
V R
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
in tr ni t theo thi tit. Cc in cc c ni vi nhau bng cch hn
vi thp nm ngang t su 0,5 0,8m. chng n mn cc ng thp
trong t phi c b dy khng nh hn 3,5mm, cc thanh thp dt, gc khng
c nh hn 4mm,v dy ni t cn c tit din ho mn bn c kh n
nh nhit v chu c dng lm vic lu di.
in tr ni t ch yu xc nh bi in tr xut ca t , hnh
dng kch thc in ccv chn su trong t. ni t thng bao gm mt
s in cc ni song song vi nhau v t cch nhau mt khong tng i
nh, v vy khi c dng ngn mch chm t th tch t tn dng in t mi
cc gim i do lm tng in tr ni t, h s s dng in cc ni t s
gim xung khi tng s in cc v gim khong cch gia chng, ngoi ra cn
ph thuc vo dng ni t.
in tr sut ca t ph thuc vo thnh phn, mt , m v
nhit ca t v ch c th xc nh bng o lng. in tr sut khng phi
c nh trong c nm m hay i do s thay i m, v vy trong tnh ton
ni t phi dng in tr sut tnh ton l tr s ln nht trong c nm tt
=kmax. ; kmax l h s tng cao, ph thuc vo iu kin kh hu ni s xy
dng trang b ni t.
2.Xc nh in tr ni t nhn to.
Cng thc tnh
tnndnt RRR
111=
Trong :
+ Rnt : in tr ni t nhn to
+ Rnt
: in tr ni t cho php ca h thng- 65 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
+ Rtn : in tr ni t t nhin
- i vi li h p c trung im ni t trc tip, theo quy phm ca trang b
in th Rnd 4 .
Do khng bit r cc h thng ni t ca cc cng trnh khc nn coi Rtn 0.
Vy ta c :Rnt = Rn = 4.
2. XC NH IN TR TN CA MT IN CC CHN SU.
3.1 Xc nh in tr sut tnh ton
Trang b in s c xy dng ti ni dt thuc loi t vn c in tr sut
=50 m ( bng 8_1sch tra cu cung cp in )
- h s nng cao K cho in cc nm ngang khi chn su 0,8m l K= 1,6 in
trsut
tnh ton ca thanh nm ngang tt.ng =1,6.50 =80 m.
- h s nng cao K cho in cc thng ng 2 3m chn su cch mt tkhong 0,5 0,8m l K=1,4
=>in tr sut tnh ton ca cc thng ng tt =1,4.50 = 70 m
3.2 Cch thc chn su v loi in cc.
- in cc c chn thng ng, u trn cch mt t khoang h = 0,7m c
chiu di l= 2,5m, cc in cc cach nhau khong a= 5m v c ni vi nhaubng cc thanh in cc ngang chn su khong h= 0,8m .
- Chn in cc thng ng bng thp gc c kch thc 70.70mm.
- Chn in cc ngang bng thanh dt c kch thc 40.4mm.
- 66 - nmn hc : SV : Ng xun Li H in 1_K3 t h =0,7 m
l =2,5 m
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
3.3 Tnh in tr ca mt cc in thng ng.
Cng thc tnh cho in cc trn: ( sch tra cu cung cp in)
+
+
=
lt4
lt4ln
2
1
d
l2ln
l.2R d.tt
dc
Trong :
+ l : chiu di in cc, m
+ d : ng knh in cc trn, m
+ t : chn su, m
- tnh cho in cc l thp gc phi quy i v in
cc trn theo
cng thc d = 0,95.b; b = 0,07 m: b rng ca in cc vung
=> d = 0,95.0,07 = 0,0665m
- chn su : t = 0,7 + l/2 = 1,95 m.
Ta tnh c :
=
+
+= 74,185,295,1.4
5,295,1.4ln
2
1
0665,0
5,2.2ln
5,2.14,3.2
70Rdc
4. Xc nh s b s in cc thng ng.
ntd.sd
dc
R.K
Rn=
- 67 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- Ksd. : h s s dng in cc thng ng, khng xt
n nh hng ca thanh ni ngang; in cc t thnh
vng, la
=1chn s b Ksd.=0,58.
cc1097,94.58,0
74,20n ==
5. Xc nh in tr tn ca in cc nm ngang.
Cng thc tnh :
t.bL.2
lnL.2.KR
2
ng.sd
g.tt
ng
=
Trong :
+ L =10.2,5 =25m : tng chiu di ca in cc ngang.
+ b =40mm = 0,04m :chiu rng ca thanh dt.
+ t =0,8m : chn su.
+ Ksd.g =0,34 : h s s dng thanh ni ngang trong vng
in cc thng ng khi chng bng 10.
Thay vo cng thc trn tnh c :
== 85,158,0.04,0
25.2ln
25.14,3.2.34,080
R2
ng
6. Tnh chnh xc in tr ca in cc thng ng.Cng thc tnh :
=
=
= 35,5485,154.85,15
RR
R.RR
ntng
ntng
d
Tnh chnh xc s in cc thng ng c xt n nh
hng ca in cc ngang, h s s dng thanh ng Ksd.
=0,58. (Bng 8-5, sch tra cu cung cp in )- 68 - n
mn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
6n04,635,5.58,0
74,18
R.K
Rn
dd.sd
dc ====
Vy ta ly n =6 in cc thng ng.
Mt bng b tr cc in cc :
- 69 - nmn hc : SV : Ng xun Li H in 1_K3
TBA
11
2
0,7
m
0,8
m
2,5
m
1. Cc2. Thanh ni
a 2,5m
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Chng 6: TNH TON DUNG LNG B
Phn ln h cng nghip trong qu trnh lm vic tiu th t mnh in
c cng sut tc dng P ln cng sut phn khng Q. Cc ngun tiu th cng
sut phn khng l: ng c khng ng b, tiu th khong 60 65% tng
cng sut phn khngca mng in x nghip, my bin p tiu
th khong 20-25%. ng dy v cc thit b khc tiu th
khong 10%,... ty thuc vo thit b in m x nghip c
th tiu th mt lng cng sut phn khng nhiu hay t.
Truyn ti mt lng cng sut phn khng qua dy dn v my
bin p s gy ra tn tht in p, tn tht tn tht in
nng ln v lm gim kh nng truyn ti trn cc phn t ca
mng in do c li v kinh t - k thut trong li in
cn nng cao h s cng sut t nhin hoc a ngun b
cng sut phn khng ti gn ni tiu th tng h s cng
sut cos lm gim lng cng sut phn khng nhn t h
thng in .
- Nng cao h s cng sut t nhin bng cch :
+ Thay cc ng c non ti bng cc ng c cng sut nh
hn.
+ Gim in p t vo ng c thng xuyn non ti.
+ Hn ch ng c khng ng b chy khng ti.
- 70 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
+ Thay ng c khng ng b bng ng c ng b.
- Nu tin hnh cc bin php trn gim lng cng sut
phn khng tiu th m h s cng sut ca x nghip vn chat yu cu th phi dng bin php khc t thit b b
cng sut phn khng.
1. Xc nh dung lng b
1.1. Tnh h s Cos tb ca ton x nghip.Cng thc :
=PXi.tt
PXiPXi.tt
tb P
cos.PCos
Trong :
+ Ptt.Pxi : cng sut tnh ton ca phn xng th i.
CosXN
tb =17,1748
78,0.2448,0.08,11682,0.08,1768,0.19,160786,0.09,158 ++++
+ 17,174877,0.43,21875,0.27,19376,0.89,23775,0.14,244 +++
= 0,776
H s Cos ti thiu do nh nc quy nh l t (0,85 0,9),
nh vy ta phi b cng sut phn khng cho x nghip
nng cao h s Cos .
1.2. Tnh dung lng b tng ca ton x nghip :
Cng thc tnh :
Qb = Ptt.XN ( tg 1 - tg 2 )
Trong :
- 71 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
+ tg 1 : tng ng vi h s Cos 1 trc khi b.
+ tg 2 : tng ng vi h s Cos 2 cn b, ta b n Cos 2 t
gi tr quy nh khng b pht t (0,85 0,95) ta b nCos 2 = 0,9.
Cos 1 = 0,776 tg 1=0,813
Cos 2 = 0,9 tg 2=0,484
Qb = 1003,9. ( 0,813 - 0,484) = 330,28 (KVAR)
Qb = 330,28 KVAR
2. Chn v tr t v thit b b.
2.1. V tr t thit b b .
V nguyn tc c li nht v mt gim tn tht
in p, tn tht in nng cho i tng dng in l t
phn tn cc b t b cho tng ng c in, tuy nhin nut phn tn qu s khng c li v vn u t, lp t v
qun l vn hnh . V vy vic t thit b b tp trung hay
phn tn l tu thuc vo cu trc h thng cp in ca i
tng, theo kinh nghim ta t thit b b pha h p ca
trm bin p phn xng ti t phn phi. V y ta coi gi
tin n v (/KVAR) thit b b h p ln khng ng k sovi gi tin n v tn tht in nng qua my bin p.
2.2. Chn thit b b .
b cng sut phn khng cho x nghip c th dng cc
thit b b sau:
My b ng b :
- 72 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
+ C kh nng iu chnh trn.
+ T ng vi gi tr cng sut phn khng pht ra (c th tiu
th cng sut phn khng).+ Cng sut phn khng khng ph thuc in p t vo,
ch yu ph thuc vo dng kch t
+ Gi thnh cao.
+ Lp rp, vn hnh phc tp.
+ Gy ting n ln.
+ Tiu th mt lng cng sut tc dng ln .
T in :
+ Tn tht cng sut tc dng t
+ Lp t, vn hnh n gin, t b s c
+ Cng sut phn khng pht ra ph thuc vo in p tvo t.
+ C th s dng ni kh ro bt k t b t.
+ Gi thnh r.
+ Cng sut phn khng pht ra theo bc v khng th thay
i c.
+ Thi gian phc v, bn km.
Theo phn tch trn th thit b T b thng c dng
lp t nng cao h s cng sut cho cc x nghip
3. Tnh ton phn phi la chn dung lng b
- 73 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- S nguyn l t thit b b :
- S thay th .
C: Qb = 1003,9. ( 0,813 - 0,484) = 330,28 (KVAR)
Trong trng hp ny, ta s chn t b CEP131A1 c Qbu = 100
(kvar) v t CEP 124M8 c Qbu = 50 (kvar) ng vi cc phn x-
ng c th.
4. nh gi hiu qu b
Cng sut biu kin ca phn xng sau khi b l :SV = PV + j(QV - QbuV ) = 124,26 + j (97,73 50)
= 124,26 + j.47,73
Tn tht sau khi b l:
sbA =2
22
38,0
73,4726,124 +
. 0,67. 76 . 10-6 . 2886 = 18032,2 (kwh)
Lng in nng tch kim c do b cng sut phn khng l :
AAA = = 25433,9 - 18032,2 = 7401,2 (kwh)
S tin tit kim c trong nm
= cAC = 7401,2 . 1000 = 7,401 .106 (/nm)
Vn u t b- 74 - n
mn hc : SV : Ng xun Li H in 1_K3
35KV 10KV
BATT
Qb
CpBAP
Xi
0,4KV P
i+JQ
i
Qbi
10KVR
CiR
Bi 0,4KV
Qb
(Qi- Q
bi)
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Vb = vob . Qbn = 75.103 . 50 = 3,75. 106 (/nm)
Chi ph quy i :
Zb = p.Vb = 0,185 . 3,75 . 106 = 0,694.106 (/nm)
Tnh ton tng t cho cc phn xng khc , kt qu ghi trong
bng sau:
- 75 - nmn hc : SV : Ng xun Li H in 1_K3
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- 76 - nmn hc : SV : Ng xun Li H in 1_K3
Phn
Xng
Qtt Qbn Loi t sbA
Kwh
sbA
Kwh
C
.106
(/n
m)
Vb
.106
(/
n)
Zb
.106
(/
n)
V 97,7
3 50
CEP
124M8
18032.2
3 7401.67 7.4 3,75
0,6
94
U 96,1
1 50
CEP
124M8
16200.2
1 6210.69 6.21 3,75
0,6
94
T 100,
79 100
CEP131
A1
20977.6
1
10221.9
9 10.22 7,5
1,3
88
69,6
5 50
CEP
124M8
12618.4
1 6253.79 6.25 3,75
0,6
94
152,
67 100
CEP131
A1
31804.2
4
16741.5
6 16.74 7,5
1,3
88
N 161,
45 100
CEP131
A1
19595.4
3
11704.2
7 11.7 7,5
1,3
88
A 154,
61 100
CEP131
A1
20274.9
3
11892.7
7 11.89 7,5
1,3
88
O 127,
83 100
CEP131
A1
18807.7
7
13441.1
3 13.44 7,5
1,3
88
H 139,
37 100
CEP131
A1
28149.6
8
16859.1
2 16.86 7,5
1,3
88
Tng 186460,
51 100727
100,7
27
56,2
5
10,
41
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
Tng s tin tit kim c do t t b l :
TK = C - Zb = (100,727 10,41) . 106 = 90,32 .106 /nm
C th thy vic y t b mang li hiu qu kinh t rt ln
Nh vy ta s t mi phn xng mt t b loi CEP 124M8
hoc CEP131A1 ty thuc vo cc phn xng c th
S nguyn l, s lp t c biu din hnh di
y:
- S nguyn l t t b trong trm bin p.
- 77 - nmn hc : SV : Ng xun Li H in 1_K3
X
XX
X
XX X X X
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Trng i Hc Cng Nghip H Ni Mn Cung Cp in---- Khoa i n ----
- S lp t t b trong trm t 2 my.
- S lp t t b trong trm t 1 my:
T
aptomat tng
T b
cos
T b
cos
T
aptomattng
T
phn
phi
cho cc
phn x-
ng
T
aptomat
phn
on
T
phn
phi
cho cc
phn x
ng
T
aptomat
tng
T b
cos
T phn
phi cho
cc phn
xng
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