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M C L C
CH NG I/ GI I THI U CHUNG V NH MY .....3
CH NG II / PHAN CHIA NHOM PHU TAIVA XAC NH PHU TAI TNHTOAN......7
CHNG VIII/ LA CHON MAY BIEN AP LC VABU CONG SUAT PHAN
KHANG..................32
CHNGIV/ LA CHON DAY DAN VA THIET B BAOVE...........33
CHNG V /TNH TOAN SUT AP..42
CHNG VI/TNH TOAN DONG NGAN MACH..50
Ch ng VII/ La chon thiet b baove.61
CHNG VIII/ THIET KE BAO VE AN TOAN IEN CHONGI....79
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L I NI UCng nghi p lun l khch hng tiu th i n l n nh t. Trong tnh hnh k
t th tr ng hi n nay, cc x nghi p l n nh , cc t h p s n xu t u ph i ho
ton kinh doanh trong cu c c nh tranh quy t li t v ch t lng v gi c s n ph m.i n n ng th c s ng gp m t ph n quan tr ng vo l li c a x nghi p. N
thng x y ra m t i n 1, 2 ngy x nghi p khng c li, n u m t i n lu h n xnghi p s thua l . Ch t lng i n x u(ch y u l i n p th p ) nh h ng l n nch t l ng s n ph m. Ch t lng i n p th c s quan tr ng v i x nghi p may, xnghi p ho ch t, x nghi p l p t ch t o c kh, nh my bia, nh my s n xu t
th c n gia sc, nh my xi m ng, nh my luy n gang thep.vv V th , m tin c y c p i n p v nng cao ch t lng i n n ng l m i quan tm hng u
c a n thi t k c p i n cho nh my, x nghi p. Nh m h th ng ho v vn d ng nh ng ki n th c c h c t p trong mn
cung c p i n gi i quy t nh ng v n th c t , em c th y L Minh Ph ggiao th c hi n ti thi t k mn h c v i n i dung: Thi t k h th ng cun
i n cho nh my s n xu t bia. n gi i thi u chung v nh my, v tr a l, c i m cng ngh ,
b ph t i . . . ng th i n c ng xc nh ph t i tnh ton, thi t k m ng cao p, h p . . .
hon thnh t t n ny, em c s gip t n tnh c a cc th y ctrong khoa i n- i n T , c bi t l s gip v ch b o t n tnh c a th y LMinh Ph ng.
Do ki n th c v th i gian c h n, b n n khng trnh kh i sai st, knmong cc th y ,c gp ki n b n n c hon thi n h n.
Em xin chn thnh c m n !
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CH NG IGI I THI U CHUNG V NH MY
I. Tong quan ve thiet ke cung cap ien: Thiet ke cung cap ie n la trn h bay cac ph ng phap tn h toan e
qua o ta co the la cho n n hng thiet b cua he tho ng ien phu hp viyeu cau ho tieu thu. Muc tieu ch n h cua cua viec thiet ke cu ng cap ie n la am bao n hu cau ie n n a ng cho ho tieu thu vi chat lng ien n a ng trong pham vi cho phep. Vi cung mot ho tieu thu, ta co the co n hieuphng an thiet ke. Mot phng an cung cap ie n x ng hiep c xem lahp ly khi thoa man n hng yeu cau :
Von au t n ho, chu y en viec tiet kiem ng oai te qu ( muan hng thiet b ng oai khong ca n thiet hoac n hng thiet b co thethay the ba ng n hng san pham sa n xuat trong n c n hng van am bao yeu cau k thuat va ki n h te ) vat t hiem.am bao o tin cay cung cap ie n cao tuy theo tn h chat ho tieuthu.Chi ph van ha n h, bao tr, bao dng ha ng n am thap.am bao an toan cho ng i va thiet bThuan tien trong van han h va sa cha, .v .v .am bao chat lng ien n a ng , chu yeu la am bao o lechva o dao o ng ien ap be va n am trong pham vi g ia tr chophep so vi g ia tr n h mc.
Tuy n hien , trong thc te thiet ke, kho ng n hat thiet phai thoa ma n tat cac cac tieu ch tre n ma ta phai ca n n hac tn h toan va ket hp haihoa cac tieu ch cho phu hp vi hoa n can h cu the cua t ng ho tieuthu ien .
Co 3 loai ho tieu thu ien :Ho loai 1:La n hng ho rat qua n trong khong c e mat ie n ,n eu xay ra mat ie n se g ay ra hau qua rat ng hiem trong : an n in h, chn h tr, tn h mang con ng i,y Lam mat a n n in h chn h tr, mat trat t xa hoi. o la sa n bay,
hai cang , khu quan s, khu ng oai g iao oan , cac ai s qua n ,be n h vien ,
y Lam thiet hai ln e n n en kin h te quoc da n . o la khu cong ng hiep, khu che xuat. Nh ng ho n ay ong vai tro quan trong trong n en kin h te quoc da n hoac co g ia tr xuat khau cao emlai n hieu ng oai te cho at n c.
y Lam ng uy hai en tn h mang con ng i.Ho loai 2: bao g om cac x ng hiep che tao ha ng tieu dung , luyen kim, n ha may c kh ln va thng mai dch vu ( trung tam thng mai ln , khach sa n ,). Vi n hng ho n ay n eu mat ie n se b
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thua thiet ve kin h te n h dan cong , g ay th pham, phe pham,pha v hp ong cung cap ng uyen lieu hoac sa n pham chokhach ha ng , lam g iam sut doan h so va lai suatHo loai 3: la n hng ho khong quan trong cho phep mat ie n tamthi khi can thiet.
Ng oai ra, ng i thiet ke phai chu y e n yeu cau ca n phat trien phu tai trong tng lai, rutng an thi g ian thi cong , .v .v .Ng ay n ay, trong s phat trien cua n e n kin h te th trng , cac co ng ty,
x ng hiep du ln hay n ho eu phai hoach toa n kin h doan h trong cuoc ca n htran h khoc liet ve chat lng va g ia ca. S mat ie n , chat lng ien xau hay do s co .v.v. .eu a n h hng e n chat lng san pham, g ayphe pham, g iam hieu suat lao o ng . ac biet a n h hng rat ln e n cac x ng hiep det may, hoa chat ie n t oi hoi s chn h xac va va n ha n h lien tuc cao. Do o am bao o tin cay cap ie n , n a ng cao chatlng ien n a ng la moi quan tam ha ng au.
Tom lai: Ng an h ien n oi chung va cung cap ie n n oi rieng ong vai
tro het sc qua n trong trong s phat trien cua n en kin h te cu ng n h cuaxa hoi.Viec thiet ke cung cap ie n phai am bao cac tieu ch ve kythuat cung n h li ch kin h te. e co c mot ph ng an thiet ke toi u taca n ng hien cu ky lng , so sa n h tat ca cac ph ng an co the co vachon phng an tot n hat.
Noi dung cua o a n cung cap ie n n ay la ng hien cu thiet ke, tn htoan , van ha n h li ien ha ap sao cho am bao cac chi tieu ki n h te kythuat va a n toan .
S o mat ba ng : g om ba s o ch n h sau :S o mat ba ng cua pha n xng .S o au day cua pha n xng .
S o ng uyen ly cua he tho ng cung cap ie n cho pha n xng .Ng oai cac yeu cau n eu tre n , ng i thiet ke ca n lu y sao cho hethong cap ie n that n g ian , de thi cong , de va n ha n h, de s du ng , dephat trien .vv
II / Gi i thi u chung v nh my biaTrong nh my bia c nhi u h th ng my mc khc nhau r t a d
phong ph v ph c t p. Cc h th ng my mc ny c tnh cng ngh cao v t
i. Do v y m vi c cung c p i n cho nh my ph i m b o ch t lng v tinc y cao.
ng v m t cung c p i n th vi c thi t k i n ph i m b o s gi ph t i trong t ng lai; v m t k thu t v kinh t ph i ra ph ng n c p i n
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sao cho khng gy qu t i sau vi n m s n xu t v c ng khng gy qu dth adung l ng cng su t d tr .
Nh my 9 lo i thi t b , cc lo i thi t b ny c phn b trn di n tch
5525 , cc thi t b ny c xy d ng t ng i g n nhau va cc thng s t nglo i thi t b c cho trong b ng sau:
S O MAT BANG NHA MAY BIADai:85m Rong:65m
Dien tch:5525 m
BANG LIET KE CONG SUAT THIET B IEN
KHMB Ten thiet b Pm (kw) Ksd cos N 68 May xuc chai 42 0,4 0,667 May g ap rong 9 0,5 0,669 May g ap day 9 0,4 0,670 Motor vi tren 15 0,6 0,6
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71 Motor vi di 15 0,54 0,572 May chiet bia 18 0,7 0,6773 Bm n c n ong
hap4,5 0,7 0,67
74 Motor cau 5 0,7 0,7
75 Quat cong ng hiep 0.75 0,5 0,7
III/ Gi i thi u c i m ph t i i n c a nh my:Ph t i i n trong nh my cng nghi p c th phn ra lm 2 lo i ph t+ Ph t i ng l c.+ Ph t i chi u sng.
Ph t i ng l c th ng c ch lm vi c di h n, i n p yu c u tr
ti p n thi t b l 380/220V, cng su t c a chng n m trong d i t 1 nch c kW v c cung c p b i dng i n xoay chi u t n s f=50Hz.
Ph t i chi u sng th ng l ph t i 1 pha, cng su t khng l n. Ph tchi u sng b ng ph ng , t thay i v th ng dng dng i n xoay chi u t n s f =50 Hz.
CH NG II PHAN CHIA NHOM PHU TAI
VA XAC NH PHU TAI TNH TOAN
I.XAC NH TAM PHU TAI.
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Viec chia n hom ong c trong cac pha n xng g iup ta bo tr s olap at cho hp ly ,am bao t n h my thuat.Co n hieu phng phap chian hom :
1. Pha n n hom theo mat ba ng lap at,v tr a ly
2. Pha n n hom theo cong suat3. Pha n n hom theo chc n a ng lam viec ,theo yeu cau cua xng hiep Viec xac n h tam phu tai ta du ng e xac n h c v tr lapat tu o ng lc, tu phan phoi phan xng , tu phan phoi chn hva may bien ap. ay ta pha n chia theo v tr a ly.
Tam phu tai c tn h theo cong thc sau:
XTDL =
!
!n
i
dmi
n
idmii
P
P X
1
1
YTDL =
!
!n
idmi
n
idmii
P
P Y
1
1
XTPPPX =
!
!4
1
4
1dmTDLjtdlj .PX
jdmTDL j
j
P
YTPPPX =
!
!4
1
4
1
idmTDL j
jdmTDL jt dl j
P
P Y
e at tu o ng lc hay tu phan phoi trc het ta phai xacn h tam phu tai cua t ng n homthiet b(tu ong lc)n ham cung cap ie n vi ton that ie n ap va to n thatcong suat n ho, g iam c chi ph kin h te trong lap at dayda n . ong thi viec la chon v tr cuoi cung con phuthuoc vao yeu to my qua n va thua n tien lu thong , thaotac
T s o mat ba ng cua pha n xng a cho ta co the pha n than h4 n hom hay 4TDL e ta co the de da ng hn trong cong viec quan ly haytien ch trong van e kin h ph
II/ PH T I TNH TON C A NH MY BIA
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1 / Ph t i tnh ton: Ph t i tnh ton l ph t i gi thi t lu di khng i, t ng ng
v i ph t i th c t v m t hi u qu pht nhi t ho c m c hu ho i cch i
cch khc, ph t i tnh ton c ng t nng thi t b ln t i nhi t t ng t nh ph t i th c t gy ra, v v y ch n cc thi t b theo ph t i tnh ton s m ton thi t b v m t pht nng.
Ph t i tnh ton c s d ng l a ch n v ki m tra cc thi t b trong th ng cung c p i n nh: my bi n p, dy d n, cc thi t b ng c t, b o v tnh ton t n th t cng su t, t n th t i n n ng, t n th t i n p; l a ch nl ng b cng su t ph n khng ph t i tnh ton ph thu c vo nhi u y unh : cng su t, s l ng, ch lm vi c c a cc thi t b i n, trnh v ph ngth c v n hnh h th ng N u ph t i tnh ton xc nh c nh h n ph t ith c t th s lm gi m tu i th c a thi t b i n, ngc l i n u ph t i tnh tonxc nh c l n h n ph t i th c t th gy ra dth a cng su t, lm ng v n
u t, gia t ng t n th t c ng v v y c nhi u cng trnh nghin c u ph ng php xc nh ph t i tnh ton, song cho n nay v n cha c c ph ng ph ng php no th t hon thi n. Nh ng ph ng php cho k t qu tinc y th l i qu ph c t p, kh i lng tnh ton v cc thng tin ban u v ph tl i qu l n. Ng c l i nh ng ph ng php tnh n gi n l i c k t qu c chnhxc th p.2 / Cc ph ng php xc nh ph t i tnh tona/ Ph ng php xc nh ph t i tnh ton (PTTT) theo cng su t t v h
nhu c u:Ptt = k nc.P
Trong :k nc : l h s nhu c u , tra trong s tay k thu t .
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P : l cng su t t c a thi t b ho c nhm thi t b , trong tnh c th l y g n ng P } Pd (kW) .b/ Ph ng php xc nh PTTT theo cng su t cng su t trung bnh v h s
hnh dng c a th ph t i :Ptt = k hd . Ptb
Trong :k hd : l h s hnh dng c a th ph t i tra trong s tay k thu
bi t th ph t i .Ptb : l cng sut trung bnh c a thi t b ho c nhm thi t b (kW) .
c / Ph ng php xc nh PTTT theo cng su t trung bnh v l ch c a ph t i kh i gi tr trung bnh :
Ptt = Ptb W Fs . Trong :
W: l l ch c a th ph t i kh i gi tr trung bnh .
F : l h s tn x c aW.
d/ Ph ng php xc nh PTTT theo su t tiu hao i n n ng cho m t n v ph m :
Ptt =max
0
TM.a
Trong :a0: l su t chi ph i n n ng cho m t n v s n ph m, kWh/ vsp.M: l s s n ph m s n su t trong m t n m .
Tmax: l th i gian s d ng cng su t l n nh t , (h)e/ Ph ng php xc nh PTTT theo su t trang b i n trn m t n v di n t
Ptt = p0 . FTrong :
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p0 : l su t trang b i n trn m t n v di n tch , (W/m2) .F : l di n tch b tr thi t b , (m2) .
f / Ph ng php tnh tr c ti p :
L ph ng php i u tra ph t i tr c ti p xc nh PTTT p d ng chotr ng h p:
- Ph t i r t a d ng khng th p d ng ph ng php no xc nh ph t itnh ton.
- Ph t i r t gi ng nhau v l p i l p l i cc khu v c khc nhau nhpht i khu chung c . g/ Xc nh ph t i nh nh n c a nhm thi t b :
Theo ph ng php ny th ph t i nh nh n c a nhm thi t b s xu t khi thi t b c dng kh i ng l n nh t m my cn cc thi t b khc trong nhm
ang lm vi c bnh th ng v c tnh theo cng th c sau:I n= Ik (max)+ (Itt - k sd . I m (max))
Trong :Ik (max): l dng kh i ng c a thi t b c dng kh i ng l n nh t
trong nhm my.Itt: l dng i n tnh ton c a nhm my.I m (max): l dng nh m c c a thi t b ang kh i ng.k sd: l h s s d ng c a thi t b ang kh i ng.
Trong cc ph ng php trn, 3 ph ng php4 ,5,6 d a trn kinh nghi mthi t k v v n hnh xc nh PTTT nn ch cho cc k t qu g n ng tuy
chng kh n gi n v ti n l i. Cc ph ng php cn l i c xy d ng trn c s l thuy t xc su t th ng k c xt n nhi u y u t do c k t qu chnh x n,nh ng kh i l ng tnh ton h n v ph c t p.
Trong bi t p di ny v i phn x ng SCCK ta bi t v tr, cng su t tv cc ch lm vi c c a t ng thi t b trong phn x ng nn khi tnh ton ph t i
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ng l c c a phn x ng c th s d ng ph ng php xc nh ph t i tnh tontheo cng su t trung bnh v h s c c i. Cc phn x ng cn l i do ch bi t di ntch v cng su t t c a n nn xc nh ph t i ng l c c a cc phn ng
ny ta p d ng ph ng php tnh ton theo cng su t t v h s nhu c u. Ph chi u sng c a cc phn x ng c xc nh theo ph ng php su t chi u sngtrn m t n v di n tch s n xu t.h/ Ph ng php xc nh PTTT theo cng su t trung bnh v h s c c i
V c thng tin chnh sc v m t b ng b tr my mc thi t b bi t ccng su t v qu trnh cng ngh c a t ng thi t b nn ta xc nh ph t i tntheo cng su t trung bnh v h s c c i.Theo ph ng php ny ph t i tnh ton
c xc nh nhsau:+ Tnh ton ph t i ng l c
y V i 1 ng c
Ptt = P my V i nhm ng c n 3
Ptt =n
i P mi y V i nhm ng c n 4
Ptt = k max . k sd . n
iP mi
Trong :P mi: cng su t nh m c c a thi t b
k sd :h s s d ng c a nhm thi t b . tra s tayn: S thi t b trong nhm.k max: H s c c i, tra trong s tay k thu t theo quan h :
k max = f(nhq, k sd)
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nhq: S thi t b dng i n hi u qu .
y Tnh nhq Xc nh n1 : s thi t b cos cng su t l n hay b ng m t n a
cng su t thi t b c cng su t l n nh t. Xc nh P1 : cng su t c a n1 thi t b trn
P1 = n
iPdmi
Xc nhn* = n
n1 P* = 7 P P 1
Trong : n : t ng s thi t b trong nhm
P : t ng cng su t m i nhm, P = n
iP mi
T n* v P* tra b ng ta cnhp*+ Khi nhq 4 Tra b ng v i nhqv k sd c k max
+ Khi nhq < 4 Ph t i tnh ton c xc nh theo cng th c
Ptt = n
i( k ti. Pdmi )
Trong :k ti : h s t i c a thi t b i
k ti = 0,9 v i thi t b lm vi c ch di h nk ti = 0,75 v i thi t b lm vi c ch ng n h n l p l i.
+ Ph t i ng l c ph n khng
Qtt = Ptt . tg
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Trong Cos : h s cng su t tnh ton c a nhm thi t b , tra s tay
cos tb=
dmi
dm i J cos.
III: PHN CHIA NHM PH T I V TNH TON CC THNG SC A PH T I
1 .1 Cac thong so ky thuat cua TL 1
STT Ten thiet b KHMB
Pmi(KW)
Xi(m)
Yi(m)
N N
tg Cos ksd
1 Motor cau 74 5 4,5 6,7 17,0 0,7
2 Bm nc nonghap 73 4.5 9 6,7 1,167,0 0,7
3Q uat cong
nghiep 75 0,75 5,2 15,2 17,0 0,5
4 Motor cau 74 5 5,38 20 17,0 0,75 Motor cau 74 5 5,37 24,3 1
7,0 0,7
6 Bm nc nonghap 73 4,5 6,6 29,1 1,167,0 0.7
7 Motor cau 74 5 5,37 33,5 17,0 0.7
8 Motor cau 74 5 5,55 38,7 17,0 0,7
9 Bm nc nonghap 73 4,5 6,41 43,5 1,167,0 0,7
10 B m n c nng h p 73 4,5 15,1 43,5 1,167,0
0,7
11 May chiec bia 72 18 16,3 27,4 1,167,0 0,7
12 May chiec bia 72 18 22,5 27,4 1,167,0 0,7
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13 May gap rong 67 9 30,8 30,1 3,16,0 0,5
14 May gap rong 67 9 30,8 23.4 3,16,0
0,5
15 Motor cau 74 5 40 6,71
7,0
0,7
16 Motor cau 74 5 30,4 6.7 17,0 0,7
Tong cong suat n h mc cuaTL1 la:!
16
1idm i = 107,75(KW)
V tr cua TDL1:XTDL1=
!
!16
1
16
1
idm i
idm ii
= 17,7(m)
YTDL1=
!
!16
1
16
1
idmi
idmii
P
P Y = 25,2(m)
Theo tn h toan ta xac n h c toa o cuaTDL1(17,7 ;0)n hng n h vay th n mqu st v i my chi t bia,khng an ton v kh v n hnhn e n a ve v tr m i c t a l XTDL1 =17,7, YTDL1 =01 .2. Tnh toan phu tai cuaTL 1 .Cong suat trung bn h cuaTDL1 la:
PtbTDL 1=
!16
1i sdidmi K P = 72,05(KW)
He so co ng suat cos NTDL1 la:
cosNTDL1 =
!
!16
1
16
1cos
idm i
iidm i
N = 0,643
T ay ta suy ra: tg N TDL1= 1,189
QtbTDL 1= PtbTDL 1 tgNTDL1 = 1 28,2 (kvar)
He so s du ng KsdTDL1 la:
KsdTDL1=
!
!16
1
16
1
idmi
i sdidmi
P
K P = 0,66
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So thiet b hieu qua cua TDL 1 la:
NhqTDL 1=
!
!16
1
2
216
1)(
i
dmi
idmi
P
P = 10,88
04,115,11max !v! sd
sd
hq K K
n K
Ta co:4< n hq
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TB3-Quat cong ng hiep (75): Im3= 63,17,0*38,0*3
75,0cos..3
!! luoi
dm
U
P (A)
TB4-May chiet bia (72):Im4= 8,4 067,0*38,0*3
18
cos..3!!
N luoidm
U
P (A)
TB5-Mayg ap rong (67):Im5= 8,22.6,0*38,0*3
9cos..3
!! luoi
dm
U
P (A)
Nh vay ta thay trong tat ca cac thiet b cua TDL1 th dong n h mc cuaTB4 la ln n hat.
Suy ra:Immax=ITB4=40,8(A) ;Ksdmax=0,7Ta luon chon Kmm=3,5
T (*) ta tn h c In TDL1=ImmTDL1= IttTDL1+Immax(Kmm Ksdmax)In TDL1=ImmTDL= 242,5 +40,8(3,5-0,7) = 356,75(A)
Kmm:la he so m may cua thiet b :+ Vi ong c rotor long soc:kmm=57+ V i ng c DC ho c K B rotor dy qu n :Kmm=3.54+ i v i ng c cng su t 10kw :ch n Kmm=47+ i v i ng c cng su t 30kw:chon Kmm=34+ i v i ng c cngsu t 30kw :ch n Kmm=2.53
K t qu tnh ton cc thng s TDL1
Ten thietb KHMB
Pmi(KW)
Xi(m)
Yi(m)
N N
tg Cos
ksdPtt(kw)
Qtt(kvar)
Stt(kva)
Itt(A)
Motor cau 74 5 4,5 6,7 17,0 0,7 3,64 3,85 5,3 8
Bm ncnong hap 73 4.5 9 6,7
1,167,0
0,7
3,8 3,9 5,4 8,27
Q uat cong
nghiep75 0,75 5,2 15,2
1
7,0 0,5 0,39 0,41 0,56 0.86
Motor cau 74 5 5,38 20 17,0 0,7 3,64 3,85 5,3 8
Motor cau 74 5 5,37 24,3 17,0 0,7 3.64 3,85 5,3 8
Bm ncnong hap 73 4,5 6,6 29,1
1,167,0
0.7
3,8 3,9 5,4 8,27
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7 May xuc chai 68 42 77,6 22 1,167,0 0,7
8 Q uat congnghiep 75 0,75 79 14,8 17,0 0,5
Tong cong suat n h mc cuaTL2 la:!8
1idm i P =105KW)
V tr cua TDL2:XTDL2=
!
!8
1
8
1
idm i
dm ii
i
P
p X =74,8(m)
YTDL2=
!
!8
1
8
1
i
dmi
idmii
P
P Y =36,1(m)
V tr cua TDL2 la(74,8 ;36,1).Nhng ta thay n eu at n h vay th ta th y khngc kh thi v qu g n v i m y xc chai, gy m t m quan v an ton nn ta a TDL2 v v tr
c t a lXTDL1 =85 YTDL1 =35 2.1 . Tnh toan phu tai cuaTL2.
Cong suat trung bn h cuaTDL2 la:
PtbTDL 2= !
8
1i sdidmi K P = 73,2(KW)
QtbTDL 2= tg K P i
sdidmi **8
1
!Ni = 76,5(Kvar)
He so s du ng KsdTDL2
la:
KsdTDL2=
!
!8
1
8
1
idmi
i sdidmi
P
K P = 0,697
He so co ng suat cos NTDL2 la: cosNTDL2 =
!
!8
1
8
1cos
idmi
iidmi
P
P N = 0,698
T ay ta suy ra: tg N TDL2= 1,02
So thiet b hieu qua cua TDL 2 la:
NhqTDL 2=
!
!8
1
2
28
1)(
i
dmi
idmi
P
P = 3,4
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53,115,11max !v! sd
sd
hq K K
n K
Ta co: n> 3 va n hq
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+ Vi ong c rotor long soc:kmm=57+ V i ng c DC ho c K B rotor dy qu n :Kmm=3.54+ i v i ng c cng su t 10kw :ch n Kmm=47+ i v i ng c cngsu t 30kw:chon Kmm=34+ i v i ng c cng su t 30kw :ch n Kmm=2.53
K t qu tnh ton cc thng s TDL2
STTTenthietb
KHMB Pmi(KW)
Xi(m)
Yi(m) N
N tg
Cos ksd Ptt(kw
Qtt(kvar)
Stt(kva)
Itt Gi
ch
1Q uatcongnghiep
75 0,75 79 52,1 17,0 0,5 0,33 0,33 0,47 0,7
Vtr
TDL2
cr isov itnhton
2 Motor cau 74 5 76,3 60,2 17,0 0,7 3,15 3,15 4,45 6,7
3 Motor cau 74 5 71,1 60,7 17,0 0,7 3,15 3,15 4,45 6,7
4Bmncnong
hap
73 4,5 66 60,2 1,167,0 0,7
3,3 3,2 4,6 7
5 Motor cau 74 5 58 60,2 17,0 0,7 15,1 20,7 25,6 39
6Mayxucchai
68 42 75 39,1 3,16,0 0,7
15,1 20,7 25,6 39
7Mayxucchai
68 42 77,6 22 3,167,0 0,7
15,1 20,7 25,6 39
8Q uatcong
nghiep
75 0,75 79 14,8 17,0
0,5
15,1 20,7 25,6 39
T L2 Thi t BT ng4 Lo i
TB 105 85 35 02,1698,0 0,697
94,5 62,5 113,2 172
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3.1 Cac thong so ky thuat cua TDL3.
STT ten thiet b KHMB Pmi
(KW)
Xi
(m)
Yi
(m) N
N
tg
Cos ksd
1May gap
rong 67 9 67 32,5 3,16,0 0,5
2May gap
rong 67 9 67 24,3 3,16,0 0,5
3 Motor cau 74 5 76,3 6,7 17,0 0,7
4 Motor cau 74 5 71,1 6,7 17,0 0,7
5 Motor cau 74 5 54,8 6,7 17,0
0,7
6 Motor cau 74 5 49,2 6,7 17,0 0,7
7May chiet
bia 72 18 54,5 28 1,167,0 0,7
8May chiet
bia 72 18 61,4 28 1,167,0 0,7
9motor vi
tren 70 15 56 14,8 3,16,0 0,6
10Bm ncnong hap 73 4,5 68 6,7 1,1
67,0 0,7
11
Bm nc
nong hap 73 4,5 44 6,7 1,1
67,0
0,712
May gapday 69 9 10,5 10,5 3,1
6,0 0,4
Tong cong suat n h mc cuaTL3 la:!
8
1idm i P =107(KW)
V tr cua TDL3:XTDL3=
!
!12
1
12
1
idmi
idmii
P
P X =53(m)
YTDL3=
!
!8
1
8
1
idm i
idm ii
P
P Y =18,6(m)
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V tr cua TDL3 la(53 ;16,8).Nhng ta thay n eu at n h vay th ta th y khngc kh thi , gy m t m quan v an ton nn ta a TDL2 v v tr c t a lXTDL1= 53
YTDL1 =0
3.2. Tnh toan phu tai cuaTL3.
Cong suat trung bn h cuaTDL3 la:PtbTDL 3=
!
8
1i sdidmi K P = 67,1(KW)
QtbTDL 3= tg K P i
sdidmi **8
1
!Ni = 101,55(Kvar)
He so s du ng KsdTDL3 la:
KsdTDL3=
!
!8
1
8
1
idmi
i sdidmi
P
K P =0,49
He so co ng suat cos NTDL3 la:
cosNTDL3 =
!
!8
1
8
1cos
idmi
iidmi
P
P N = 0,648
T ay ta suy ra: tg N TDL3= 1,17
So thiet b hieu qua cua TDL 3 la:
NhqTDL 3=
!
!8
1
2
28
1)(
i
dm i
idm i
P
P
= 9,1
51,115.11max !v! sd
sd
hq K
K
n K
Ta co:4< n hq
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67,0151
8,1013
33 !!!
tt TDL
tt TDLTDLtt S
P Cos N
Xac nh dong nh nhon(dong m may)cuaTDL3 .In TDL3=ImmTDL31=IttTDL3+Immax(Kmm Ksdmax) (*)TL3 co8 thiet bn hng ch co 6 loai n e n ta cung ch tn h cho cac
thiet b sau:TB1- Mayg ap rong (67):Im5= 8,22
.6,0*38,0*39
cos..3!!
N luoidm
U
P (A)
TB2- Motor cau(74):Im1= 8,107,0*38,0*3
5cos..3
!!N luoi
dm
U
P (A)
TB3-Motor vitrn(70): Im3= 4 65,0*38,0*3
15cos..3
!!N luoi
dm
U
P (A)
TB4-May chiet bia(72):Im4= 8,4 067,0*38,0*3
18cos..3
!!N luoi
dm
U P (A)
TB5- Bmn c n ong hap(73):Idm2= 2,1067,0*38,0*3
5,4cos..3
!!N luoi
dm
U
P (A)
TB6- Mayg ap day(69):Idm2= 8,226,0*38,0*3
9cos..3
!!N luoi
dm
U
P (A)
Kmm:la he so m may cua thiet b :
+ Vi ong c rotor long soc:kmm=57+ V i ng c DC ho c K B rotor dy qu n :Kmm=3.54+ i v i ng c cngsu t 10kw :ch n Kmm=47+ i v i ng c cngsu t 30kw:chon Kmm=34+ i v i ng c cng su t 30kw :ch n Kmm=2.53
Nh vay ta thay trong tat ca cac thiet b cua TDL3 th dong n h mc cuaTB4 la ln n hat.
Suy ra:Immax=ITB4=40,8(A) ;Ksdmax=0,7Ta luon chon Kmm=3,5
T (*) ta tn h c In TDL3=ImmTDL3= IttTDL3+Immax(Kmm Ksdmax)
InTDL3=ImmTDL=229,6+40,8(3,5-0,7)=343,8 (A).
K t qu tnh ton cc thng s T L 3
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TT ten thiet b KHMB Pmi(KW)
Xi(m)
Yi(m) N
N tg
Cos ksd Ptt(kw
Qtt(kvar)
Stt(kva)
Itt(A) c
1May gap
rong 67 9 67 32,5 3,16,0
0,5 6,8 6,43 6,8 10,3
2May gap
rong 67 9 67 24,3 3,16,0 0,5 6,8 6,43 6,8 10,3
3 Motor cau 74 5 76,3 6,7 17,0 0,7 5,3 3,85 6,5 9,9
4 Motor cau 74 5 71,1 6,7 17,0 0,7 5,5 3,9 7,75 10,3
5 Motor cau 74 5 54,8 6,7 17,0
0,7 5,5 3,9 7,75 10,3
6 Motor cau 74 5 49,2 6,7 17,0
0,7 5.5 3,9 7,75 10,3
7May chiet
bia 72 18 54,5 28 1,167,0 0,7 5,3 3,85 6,5 9,9
8May chiet
bia 72 18 61,4 28 1,167,0 0,7 5,3 3,85 6,5 9,9
9motor vi
tren 70 15 56 14,8 3,16,0 0,6 13,6 12,87 18,7 28,4
10Bm ncnong hap 73 4,5 68 6,7 1,1
67,0 0,7 5,5 3,9 6,75 10,3
11Bm ncnong hap 73 4,5 44 6,7 1,1
67,0 0,7 5,5 3,9 6,75 10,3
12 May gapday 69 9 10,5 10,5 3,16,0 0,4 5,4 5,1 7,42 11,3TDL
3Cc Thi t B
T ng6 lo i
Thi t b 107 53 0 17,1684,0 0,49 101,8 111,7 229,6 229,6
4.1 Cac thong so ky thuat cua TDL4.
STT ten thiet b KHMB Pmi(kw)
Xi(m)
Yi(m) N
N tg
Cos Ksd
1 Motor cau 74 5 5,5 47,8 17,0 0,7
2 Motor cau 74 5 5,5 52,1 17,0 0,7
3 Motor cau 74 5 13,9 52,1 17,0
0,7
4 Motor cau 74 5 14 47,8 17,0
0,7
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5 Motor cau 74 5 21,5 49,7 17,0 0,7
6 Motor cau 74 5 26,7 49,7 17,0
0,7
7 Motor cau 74 5 5,2 60,2 17,0
0,7
8 Motor cau 74 5 25,3 60,7 17,0
0,7
9 Motor cau 74 5 26,3 60,7 17,0
0,7
10 Motor cau 74 5 36,8 60,7 17,0
0,7
11 Motor cau 74 5 42,3 60,7 17,0 0,7
12 Motor cau 74 5 47,5 60,2 17,0 0,7
13Bm nc nonghap 73 4,5 52,7 60,2 1,1
67,0 0,7
14Bm nc nonghap 73 4,5 32 50,6 1,1
67,0 0,7
15Bm nc nonghap 73 4,5 10,4 60,2 1,1
67,0 0,7
16Bm nc nonghap 73 4,5 31,5 60,2 1,1
67,0 0,7
17 Motor vi d i 71 15 47,5 60,7 73,15,0 0,54
18 may gap day 69 9 50 41,6 3,16,0 0,4
Tong cong suat n h mc cuaTL4 la:!
18
1idm i P = 102(KW)
V tr cua TDL4: XTDL4=
!
!18
1
18
1
idmi
idmii
P
P X =30 (m)
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YTDL4=
!
!18
1
18
1
idmi
idmii
P
P Y =55,2(m)
V tr cua TDL4 la(30 ;55,2).Nhng ta thay n eu at n h vay th ta th y khngc kh thi , gy m t m quan v an ton nn ta a TDL2 v v tr c t a lXTDL1= 30
YTDL1 =65
4.2. Tnh toan phu tai cuaTL4 .Cong suat trung bn h cuaTDL4 la:
PtbTDL 4= !
18
1*
i sdidmi K P = 66,3(KW)
QtbTDL 4= tg K P i
sdidmi **18
1
!Ni = 74,55(Kvar)
He so s du ng KsdTDL4 la:
KsdTDL4=
!
!18
1
18
1*
idmi
i sdidmi
P
K P =0,65
He so co ng suat cos NTDL4 la:
cosNTDL4 =
!
!18
1
18
1cos
i
dmi
iidmi
P
P N = 0,656
T ay ta suy ra: tg N TDL4= 1,15
So thiet b hieu qua cua TDL 4 la:
NhqTDL 4=
!
!18
1
2
218
1)(
i
dmi
idmi
P
P = 15,1
28,115,11max !v! sd
sd
hq K
K
n K
Ta co:4< n hq 10 n e n QttTDL4= QtbTDL4= 74,55(kvar)
SttTDL 4 = QP 4ttTDL24ttTDL2 = 22 55,7485 =113(KVA)
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Dong tn h toan cua TDL4 la:
IttTDL4 =luoi
4ttTDL
3 S =
0,383113v
=171,7(A)
75,01138544
4 !!!tt TDL
tt TDLTDLtt S
P Cos N
Xac nh dong nh nhon(dong m may) cuaTDL4.TL4 co18 thiet bn hng ch co 4 loai thiet b n e n ta ch ca n tn h
cho cac thiet b sau:
TB1-Motor cau(74):Im1= 8,107,0*38,0*3
5cos..3
!!N luoi
dm
U
P (A) TB2-
Bm n c n ong hap(73):Idm2= 2,1067,0*38,0*3
5,4cos..3
!!N luoi
dm
U
P (A)
TB3- Motor vi di(71): Im3= 4 65,0*38,0*3
15cos..3
!!N luoi
dm
U
P (A)
TB4-Mayg ap day(69):Idm4= 8,226,0*38,0*3
9cos..3
!!N luoi
dm
U
P (A)
Nh vay ta thay trong tat ca cac thiet b cua TDL4 th dong n h mc cuaTB3 la ln n hat.
Suy ra:Immax=ITB1=46(A) ;Ksdmax=0,54Ta luon chon Kmm=3,5
T (*) ta tn h c In TDL4=ImmTDL4= IttTDL4+Immax(Kmm Ksdmax)InTDL4
=ImmTDL
= 171,7 +46v (3,5-0,54) =307,8(A) Kmm:la he so m may cua thiet b :
+ Vi ong c rotor long soc:kmm=57+ V i ng c DC ho c K B rotor dy qu n :Kmm=3.54+ i v i ng c cng su t 10kw :ch n Kmm=47+ i v i ng c cng su t 30kw:chon Kmm=34+ i v i ng c cng su t 30kw :ch n Kmm=2.53
K t qu tnh ton cc thng s TDL4
STTten thietb KHMB Pmi
(kw)Xi
(m) Yi(m)
N N tg Cos Ksd Ptt(kw Qtt(kvar) S
tt(kva) Itt(A) Gch
1 Motor cau 74 5 5,5 47,8 17,0 0,7
4,48 3,5 5,7 8,6
2 Motor cau 74 5 5,5 52,1 17,0 0,7
4,48 3,5 5,7 8,6
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3 Motor cau 74 5 13,9 52,1 17,0
0,74,48 3,5 5,7 8,6
4 Motor cau 74 5 14 47,8 17,0
0,74,48 3,5 5,7 8,6
5 Motor cau 74 5 21,5 49,7 17,0 0,74,48 3,5 5,7 8,6
6 Motor cau 74 5 26,7 49,7 17,0
0,74,48 3,5 5,7 8,6
7 Motor cau 74 5 5,2 60,2 17,0
0,74,48 3,5 5,7 8,6
8 Motor cau 74 5 25,3 60,7 17,0
0,74,48 3,5 5,7 8,6
9 Motor cau 74 5 26,3 60,7 17,0
0,7
4,48 3,5 5,7 8,6
10 Motor cau 74 5 36,8 60,7 17,0 0,74,48 3,5 5,7 8,6
11 Motor cau 74 5 42,3 60,7 17,0 0,7
4,48 3,5 5,7 8,6
12 Motor cau 74 5 47,5 60,2 17,0 0,7
4,48 3,5 5,7 8,6
13
Bmncnonghap 73 4,5 52,7 60,2 1,1
67,0 0,7
4,67 3,54 5,8 8,9
14
Bmncnonghap 73 4,5 32 50,6 1,1
67,0 0,7
4,67 3,54 5,8 8,9
15
Bmncnonghap 73 4,5 10,4 60,2 1,1
67,0 0,7
4,67 3,54 5,8 8,9
16
Bmncnonghap 73 4,5 31,5 60,2 1,1
67,0 0,7
4,67 3,54 5,8 8,9
17Motor vid i 71 15 47,5 60,7 73,1
5,0 0,5410,33 14 1,4 26,4
18may gapday 69 9 50 41,6 3,1
6,0 0,4
4,6 4,63 6,5 9,9
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TDL4Cc thi t bt ng
4 lo ithi t b 102 30 65 15,1
656,0 0,6585 74,55 113 171,7
III .XAC NH TAM TPPPX HAY TPPC.
XTPPPX =
!
!4
1
4
1dmTDLjtdlj .PX
jdmTDL j
j
P =40(m)
YTPPPX =
!
!4
1
4
1.
idmTDL j
jdmTDL jt dl j
P
P Y =25(m)
V tr cua TPPTT la(40 ;25).Nhng ta thay n eu at n h vay th ta th y khngc kh thi , gy m t m quan v an ton nn ta a TPPTT v v tr c t a lXTDL1= 4 0
YTDL1 =0
IV.TNH TOAN TONG PHU TAI CHIEU SANG VA SINH HOAT.1 .Tnh toan P CS .
Pcs = Po*S trong o: Po= 0,012 (kw/m2)S la die n tch phan xng =85*65 =5525(m2)
Pcs = Po*S =0,012 v 5525=66,3 (kw)Qcs= Pcs.tg c sN ta co:Cos 75,08,0 !! c sc s tg N N
Qcs= Pcs.tg c sN = 66,3 7,4 975,0 !v (kvar)
)(2,817,4 93,66 2222 kV AQ P S c sc sc s !!! 2.Tnh toan P SH .
e xac n h cong suat si n h hoat cua pha n xng ta ca n tn h cong suat cua o cam ay ta ch t n h cong suat cua o cam, kho ng tn h e n quat hay may lan h.
2.1 c m 1 pha
Ta co: I o cam =10(A) ;Upha = 220(V) ;CosN ocam = 0,8 tg N ocam =
0,75 Ksd= 0,4 z 0,8 ta chon Ksd =0,6Po cam = Upha.Io cam . Cos N ocam . Ksd= 220 6,08,010 vvv =1,1(kw)Qo cam =Po cam. tg N ocam =1,1 83,075,0 !v (kvar)
Moi n hom thiet b ta s dung 10 o cam, ma pha n xng bia ta chia ra lam4 n hom thiet b n en :
P1 = )(2,4 26,08,010220104 kw!vvvvv
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Q1 = 2275,01,1104 !vvv (kvar)
2.2 c m 3 phaTa ch n lo i c m n b n ch u ba pha 16A:
Ta co: I o cam =16(A) ;Upha 380(V) ;CosN ocam = 0,8 tg N ocam = 0,75Ksd= 0,4 z 0,8 ta chon Ksd =0,7
Po cam = Upha.Io cam . Cos N ocam . Ksd= 380 7,08,016 vvv =3,4(kw)Qo cam =Po cam. tg N ocam =3,4 55,275,0 !v (kvar)
Moi n hom thiet b ta s dung 10 o cam, ma pha n xng bia ta chia ra lam4 n hom thiet b n en :
P2= )(7,1367,08,010380164 k !vvvvv Q2 = 3375,01,1104 !vvv (kvar)
PSH = P2 + P1 =136,7+42,2=179(kw)QSH = Q2 + Q1 =22+33=55 (kw)
)(18755179 2222 kV AQ P S SH SH SH !!!
V. XAC NH PHU TAI TNH TOAN CUA TPPPX.
Cong suat cua TPPPX la:
PttTPPPX =Pcs +Psh +Kt !
4
1 jTDL jtt P
QttTPPPX =Qcs+Qsh+Kt !
4
1 jTDL jtt Q
Kt: He so o ng thi cua TPPPX.Kt = 0,9z 1 Nen ta chon : Kt = 0,9
PttTPPPX =66,3+179+0,9 8,565858,1015,949,74 !v (kw)QttTPPPX = 49,7+55+0,9 4 9355,747,1115,6214 12,4 2 !v (kvar)
SttTPPPX = 22 4 938,565 = 750(KVA)T o ta suy ra:
Itt TPPPX =luoi
T PPPX tt
U
S
.3=
38,0.3750 = 1140(A)
Cos !!!750
8,565T PPPX tt
T PPPX tt T PPPX tt S
P N 0,75 .
B ng s li u cho TPPPX
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CHNG IIILA CHON MAY BIEN AP LC VA
BU CONG SUAT PHAN KHANG I. LA CHON THIET B BU.
Muc ch cua viec la chon thiet b bu co ng suat pha n khang chophan xng la:
y Nang cao he so co ng suat.y Giam cong suat bieu kie n cua pha n xng .y Giam ton hao ien ap va co ng suat trong li ien cua pha n
xng .Co the la chon n hieu loai thiet b bu : Tu bu hay may bu o ng bo. ay ta chon thiet b bu la tu bu. Va bu tre n than h cai cuaTPPPX.
Cach xac n h dung lng bu:Ta co co ng thc: Q bu = PttTPPPX(tg NTPPPX- tg N chua n)Trong o: !TPPPX tt os N 0,75 suy ra tg NTPPPX =0,87
K hi u
xi (m) yi (m) Pmi(kw)
Ptt(kw)
Q tt
(kvar) Stt
(kva) Itt(A)
N N
tg Cos
TDL1 17,7 0 107,75 74 ,93 14 1 159,6 224 ,54,1
58,0
TDL2 85 35 105 94 ,5 62,5 113 17202,1
69,0
TDL3 53 0 107 101,8 111,7 229,6 229,617.1
68,0
TDL4 30 65 102 85 74 ,55 113 171,715,1
656,0
TPPPX 4 0 0 4 21,75 565,8 4 93 750 114 087,0
75,0
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CosN chua n = 95,085,0 z ta chon CosN chua n= 0,95tg N chua n= 0,3
Vay co ng suat pha n khang ca n bu la: Qbu=565,8 5,3223,087,0 !v (kvar).Ta chon loai tu co:U= 0,38 KV;Qmax=50(kvar).
So tu bu ta ca n chon la: 4,650
5,322 !!n
tra bang 8.18 trang 57 sach hng dan thiet ke cung cap ien. Taca n chon 7 tu bu loai:KC-0.38-50-3Y3.
II .LA CHON MBA LC. So lng may bien ap c la chon phu thuoc vao ac iem cap
ien cho ho tieu thu, pha n xng bia ay thuoc ho tieu thu loai hai n en
ta ch cho ca n mot may cho pha n xng la u.Va khi la chon cong suat cua may bie n ap th ta da vao g ia thStttrc bu hay Sttsaubu.Neu chon may bien ap theo S tttrc bu th sau n ay ta cothe them tai vao pha n xng trong qua trn h phat trien cong ng hiep tuyn hien theo ta n e n chon theo S ttsau bu v chi ph cho mot may bien ap co co ng suat ln la rat to n kem, ma trong qua trn h tn h toan va la chon MBA thta cung a tha mot khoa ng n ao o roi.
ieu kien la chon MBA lc la:SttTPPPX sau bu = 22 b uT PPPX tt T PPPX tt QQ P
= !22 5,3224 938,565 591(KVA)
Tra b ang 8.22 t rang 60 DH TKCC . Ta chon loai MBA phan phoi ba pha hai dayquan do ABB che tao co : Sm=630(KVA); ( Po=1 ,2(KW); K=22/0,4( PN=8,2(KW); UN=4%; UMBA HA =0,4(KV)
III; l a Ch n My Pht i n D Phng
Ta ch n My Pht i n KOHLER 5KVA-630KVA Made in FRANCE c thng s nhsau
model dm(kv)
Pdmkw
S(kva) ng c model u pht
Kv630 410 509 630 volvo TAD1642GE ls
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CHNG IVLA CHON DAY DAN VA
THIET B BAO VEA.LA CHON DAY DAN.I .LA CHON DAY DAN T TDL EN CAC THIET B.1 . La chon day dan t TDL 1 en cac thiet b .
Ta co: Icpdd u hc
lv
K I max
V ay tn h t TDL en thiet b n e n : Ilvmax = Itt thiet b =Im thiet b Co n hieu cach lap at :i trong khong kh hay trong at n hng e bao
am an toan va tn h my quan ta chon cach i trong at n e n :Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 16 n e n tra bang ta tm c:K5= 0,41+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,41.1.0,95=0,39 TDL1 co 16 thiet bn hng ch co 5 loai n e n ta ch ca n tn h cho 5
loai thiet b o ma thoi.
Thiet b 1(Motor cau -74):IcpddTB1 7,2739,0 8,101 !!u
hc
TBdm K
I (A)
T ra bang 8.8 trang 49 sach HDCC ta chon c:IcpddTB1=30(A); S= 4,0(mm2); Ro= 4,61 K m/; .
Thiet b 2( Bmn c ong hap -73)IcpddTB2 2639,02,102 !!u
hc
dm TB
K
I (A)
T ra bang 8.8 trang 49 sach HDCC ta chon c:IcpddTB2= 27(A); S= 3,5(mm2); Ro= 5,3 .
Thiet b 3(Quat cong ng hiep -75)IcpddTB3 2,439,063,13 !!u
hc
dmTB
K I (A)
T ra bang 8.8 trang 49 sach HDCC ta chon c:IcpddTB3=14(A); S= 1,0(mm2); Ro=18,1 .
Thiet b 4(May chiet bia -72)IcpddTB439,08,4 04 !u
hc
dmTB
K I =104,6(A).
K m/;
K m/;
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T ra bang 8.8 trang 49 sach HDCC ta chon c:IcpddTB4=108(A); S= 35(mm2); Ro=0,524 .
Thiet b 5(Mayg ap rong -67)IcpddTB5 5,5839,08,225 !!u
hc
dmTB
K I (A).
T ra bang 8.8 trang 49 sach HDCC ta chon c:IcpddTB5=62(A); S= 14(mm2); Ro=1,33 .
2.La chon day dan t TDL2 en cac thiet b .Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 8 n e n tra bang ta tm c:K5= 0,52+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,52.1.0,95=0,49 TDL2 co 8 thiet bn hng ch co 4 loai n e n ta ch ca n tn h cho 4 loai thietb o.Giong n h cach tn h tren ta se tn h va la chon c day da n choTDL2n h sau:
T ra bang 8.8 trang 49 sach HDCC ta c bang sau:
Ten thiet b KHMB Icpdd (A)Iluachon (A)
Tietdien (mm2)
ien tr
( km/; )Q uat congnghiep 75 3,3 14 1 18,1Motor cau 74 22 22 2,5 5,3Bm ncnong hap 73 20,8 22 2,5 5,3my xc chai 68 216 228 120 0,153
3.La chon day dan t TDL3 en cac thiet b .Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 12 n e n tra bang ta tm c:K5= 0,45
+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,45.1.0,95=0,43Tng t n h cach tn h tren ta, se t n h va la chon c day da n t TDL3e n cac thiet b:
T ra bang 8.8 trang 49 sach HDCC ta c bang so lieu sau:
K m/;
K m/;
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Ten thietb KHMB Icpdd (A)
Ilachon (A)
Tietdien (mm2)
ien tr
k m/; Motor vi trn 70 107 108 35 0,524
May gaprong 67 53 62 14 1,33B m n cnng h p 73 23,7 27 3,5 5,3
Mo t c u 74 25 27 3,5 5,3May chietbia 72 94,8 97 30 0,635May gapday 69 53 62 14 1,33
4.La chon day dan t TDL4 en cac thiet b .Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 18 n e n tra bang ta tm c:K5= 0,38+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,38.1.0,95=0,36Tng t n h cach tn h tren ta se tn h va la chon c day da n t TDL4e n cac thiet b.
T ra bang 8.8 trang 49 sach HDCC ta c bang sau:
Ten thiet b KHMB Icpdd (A)
Ilachon (A)
Tietdien (mm2)
ien tr
K m/; May gap day 69 63 68 16 1,15Motor vi d i 71 127,7 132 50 0387Bm nc nonghap 73 28,3 30 4 4,61
Motor cau 74 30 30 4 4,61
II. LA CHON DAY DAN T TPPPX EN CAC TDL.
Ta co: Icpdd u
V ay tn h t TPPPX e n TDL n e n : Ilvmax = Itt TDL Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.
hc
lv
K I max
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+So cap at ca n h n hau la 4 n e n tra bang ta tm c:K5= 0,65+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,65.1.0,95=0,621 . la chon day dan t TPPPX en TDL 1 .
Icpdd62,0
5,24 21 !uhc
tt TDL
K
I = 391(A)
T ra bang 8.8 trang 49 sach HCC ta la chon c :Icpdd = 400(A) S= 300(mm2); Ro= 0,0601 .
2. la chon day dan t TPPPX en TDL2.
Icpdd62,0
1722 !uhc
tt TDL
K
I = 277,4(A).
T ra bang 8.8 trang 49sach HCC ta la chon c:Icpdd =298A) S= 185(mm2); Ro=0,0991 .
3. la chon day dan t TPPPX en TDL3.
Icpdd62,0
6,2293 !uhc
tt TDL
K I = 370(A).
T ra bang 8.4 trang 45 sach HCC ta la chon c (loai cap 4 loi) :Icpdd =400(A) S= 300(mm2); Ro=0,0601 .
4.la chon day dan t TPPPX en TDL4.
Icpdd62,0
7,1714 !uhc
tt TDL
K I = 278,4(A)
T ra bang 8.8 trang 49 sach HCC ta la chon c : Icpdd =298 (A) S= 185(mm2); Ro=0,0991 .
III. LA CHON DAY DAN T MBA EN TPPPX.
Ta co: Icpdd u
y V ay tn h t MBA en TPPPX n e n :y Ilvmax = A
U
S
luoi
dmM B A 9573*38,0
6303
!!
Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 4 n e n tra bang ta tm c:K5= 0,65
K m/;
K m/;
K m/;
K m/;
hc
lv
K I max
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+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,65.1.0,95=0,62
Icpdd!!u 62,0
957hc
lvT PPPX
K
I 1543(A)
T ra bang 8.7 trang 48 sach HDCC ta chon 4 si cap co: Icpdd =420(A) S= 150(mm2); Ro=0,1 24 .V y t ng dng dy d n ch n l Icpdd=1680(A)LU Y:Tat ca cac day da n ta eu lay Xdd=0.08(; /km) 4 day at so ng song vay n en :RoL1 =Ro /4 =0,1 24/4=0,031 ( )X
oL1 =0,08/4 =0.02( )
IV: L a ch n dy d n cho chiu sng v sinh ho t , t b, my pht i n d phng1: sinh ho t
y Ilvmax = )(2843*38,0
1873
AU
S
luoi
dm sh !!
Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.
+So cap at ca n h n hau la 4 n e n tra bang ta tm c:K5= 0,65+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,65.1.0,95=0,62
Icpdd !!u62,0
284
hc
lv sh
K I 458(A)
T ra bang 8.7 trang 48 sach HDCC ta chon Icpdd =480(A) S= 200(mm2); Ro=0,094 .2: chi u sng
y Ilvmax = )(1233*38,0
2,813
AU
S
luoi
dm c s !!
Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.
K m/;
K m/;
K m/
;
K m/;
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+So cap at ca n h n hau la 4 n e n tra bang ta tm c:K5= 0,65+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,65.1.0,95=0,62
Icpdd !!u62,0
123hc
lvc s
K I 199(A)
T ra bang 8.7 trang 48 sach HDCC ta chon Icpdd =204(A) S= 50(mm2); Ro=0,387 .
3: T b
y Ilvmax = )(5313*38,0
50*73
7 A
U
Q
luoi
dm !!
Khc =K4.K5.K6.K7 trong o:+Cap at trong vat lieu cach ie n n en :K4=1.+So cap at ca n h n hau la 4 n e n tra bang ta tm c:K5= 0,65+ac tn h cua at kho n e n K6=1+Nhiet o cua at khoa ng 25oC vo boc lapolyme n e n K7= 0,95
Khc=K4.K5.K6.K7=1.0,65.1.0,95=0,62
Icpdd !!u62,0
531maxhc
lv
K
I 857(A)
T ra bang 8.7 trang 48 sach HDCC ta chon Icpdd =864(A) S= 500(mm2); Ro=0,0366
4: My pht i n d phngThong s c a my pht i n d phng gi ng nh thong s c a my bin p, nn ta
chon dy d n cho my pht i n gi ng nh cho my bin p t i TPPPX
Icpdd !!u62,0
957hc
lvm pd
K
I 1543(A)
T ra bang 8.7 trang 48 sach HDCC ta chon 4 si cap co: Icpdd =420(A) S= 150(mm2); Ro=0,1 24 .V y t ng dng dy d n ch n l Icpdd=1680(A)
K m/;
K m/;
K m/;
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1 .Tnh sut ap t MBA en TPPPX che o b n h thng :( U1= tt TPPPX Ltt TPPPX Ltt TPPPX X R L I N N sincos13 1010 vvvvv
Trong o:IttTPPPX = 1140(A); L1 = 20(m) =0,02(km) ; R0L1=0,031(; /km)X0L1 =0,02(; /km); cosNttTPPPX =0,75 sin NttTPPPX =0,66
T tren ta tn h c: ( U1 =1 ,44(V)
che o khi o ng :( U1= tt T PPPX Ltt T PPPX LMM TTT PPPX X R L I I N N sincos1)(3 1010' vvvvv
Trong o:IttTPPPX = 1140(A);
IMM= IMM- K sd *Idm =394,6-106*0,7=320(A)L1 = 20(m) =0,02(km) ; R0L1=0,031(; /km)
X0L1 =0,02(; /km); cosNttTPPPX =0,75 sin NttTPPPX =0,66T tren ta tn h c: ( U1 = 1 ,84(V)
2.Tnh sut ap t TPPPX en TDL che o b n h thng :( U2= 3 .IttTDL.L2(R0l2.cosNttTDL+ X0l2.sin NttTDL) che o khi o ng :( U2 tt TDL xtt TDL L MM tt TDL X R L I I N N sincos2)(3 2020 vvvvv Trong o: IttTDL
:dong tn h toan cua TDLL2 :chieu dai day da n t TPPPX e n tng TDL
R0l2 :ien tr cua day da n X0l2 :ien khang cua day da n IMM:dong m may cua TDL
Theo cong thc tn h n h tren ta co c bang so lieu sau:
TDL
ChieuDaiL2
(km)
ien trR0L2
(; /km)
ien khang
X0L2(; /km)
IttTDL IMM cosN sin N ( U2Bn h
thng
( U2Khiong
1 0.0223 0.0601 0.08 224,5 356,75 0.58 0.81 0.53 0,842 0.08 0.0991 0.08 172 394,6 0.69 0.72 2,38 5,43 0.013 0.0601 0.08 229,6 343,8 0.68 0.73 0,31 0,474 0.135 0.0901 0.08 171,7 307,8 0.656 0.75 3,65 6,54
3.Tnh sut ap t TDL en thiet b che o b n h thng :( U3= .IdmT BI.L3(R0l3.cosNT BI +X0L3.sin NT BI) che o khi o ng :
3
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( U3= .ImmT BI.L3(R0l3.cosNm m+X0L3.sin Nm m)Trong o: IdmT BI:dong n h mc cua thiet b
L3 :chieu dai day da n t TDL en thiet bR0l3 :ien tr cua day da n X0l3 :ien khang cua day da n ImmT BI:dong m may cua thiet bcosNm m la he so m may cua tat ca cac thiet b cho n ba ng
0,35 sin Nm m =0,94
3.1 Tnh sut ap t TDL 1 en thiet b Theo cong thc tren ta tn h c sut ap t TDL1 en cac thiet b trong n hom n h sau:
STT KHMB
Chieu dai
L3(Km)
ien trROL3
; /Km
ien khang XOL3 ; /Km
IdmTB ImmTBCosN Thiet b
Sin N Thiet b
U3Bn h
thng
U3Khion
g
1 74 0.025 4,61 0.08 10.8 54 0.7 0.71 4,16 16,62 73 0.025 5,3 0.08 10.2 51 0.67 0.74 1,59 7,93 75 0.027 18.1 0.08 1.63 8,15 0.7 0.71 3,8 15,24 74 0.030 4,61 0.08 10.8 54 0.7 0.71 18 9,25 74 0.034 4,61 0.08 10.8 54 0.7 0.71 2,08 10,46 73 0.039 5.3 0.08 10.2 51 0.67 0.74 2,48 12,47 74 0.043 4,61 0.08 10.8 54 0.7 0.71 2,6 13,28 74 0.048 4,61 0.08 10.8 54 0.7 0.71 2,64 13,29 73 0.053 5.3 0.08 10.2 51 0.67 0.74 3,38 17
10 73 0.048 5.3 0.08 10.8 51 0.67 0.74 3,06 15,311 72 0.037 0,524 0.08 40,8 102 0.67 0.74 1,07 2,712 72 0.035 0,524 0.08 40,8 102 0.67 0.74 0,2 0.513 67 0,048 1,33 0.08 22,8 114 0.6 0.84 1,6 5,714 67 0.048 1,33 0.08 22,8 114 0.6 0.84 1,64 5,715 74 0.021 4,61 0.08 10.8 54 0.7 0.71 1,28 6,4416 74 0.021 4,61 0.08 10.8 54 0.7 0.71 1,28 6,44
3
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3.2 Tnh sut ap t TDL2 en thiet b :
STT KHMB
Chieu dai
L3(Km)
ien trROL3
; /Km
ien khang XOL3 ; /Km
IdmTB ImmTBCosN Thiet b
Sin N Thiet b
U3Bn h
thng
U3Khion
g
1 75 0.023 18.1 0.08 1.63 8,15 0.7 0.7 0.82 4,132 74 0.044 5,3 0.08 10.8 54 0.7 0.7 3,09 12,33 74 0.039 5,3 0.08 10.8 54 0.7 0.7 2,7 10,8
4 73 0.049 5.3 0.08 10.2 51 0.67 0.74 3,12 12,485 74 0.057 5,3 0.08 10.8 51 0.7 0.7 3,2 12,26 68 0.012 0,153 0.08 106 265 0.6 0.8 0,85 2,147 68 0.021 0,153 0.08 106 265 0.6 0.8 1,5 3,748 75 0.027 18,1 0.08 1,63 8,15 0.7 0.7 0,96 4,8
3.3 Tnh sut ap t TDL3 en thiet b :
STT KH
MB
Chieu dai
L3(Km)
ien trROL3
; /Km
ien khang XOL3 ; /Km
IdmTB
ImmTB
CosN Thiet b
Sin N Thiet b
U3Bn h
thng
U3Khion
g
1 670.046
5 1,33 0.08 22,8 114 0.6 0.8 1,58 7,92 67 0.038 1,33 0.08 22,8 114 0.6 0.8 1,3 6,463 74 0.03 5,3 0.08 10,8 54 0.7 0.7 2 104 74 0.025 5,3 0.08 10,8 54 0.7 0.7 1,66 8,35 74 0.007 5,3 0.08 10,8 54 0.7 0.7 0,46 2,3
6 74 0.007 5,3 0.08 10,8 54 0.7 0.7 0.46 2,37 72 0.029 0,635 0.08 40,8
142,8 0.7 0.7 1,02 3,58
8 72 0,037 0.635 0,08 40,8142,
8 0,7 0,7 1,3 4,589 73 0,016 0,635 0,08 10,2 51 0,67 0,74 0,13 0,68
10 73 0,021 0,635 0,08 10,2 51 0,67 0,74 0,19 0,9
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11 70 0,018 0,524 0,08 46 161 0,6 0,8 0,54 1,912 69 0.011 1.33 0.08 22.8 114 0.6 0.8 0,37 1.86
3.4 Tnh sut ap t TDL4 en thiet b:
STT KH
MB
Chieudai
L3(Km)
ien
trROL3; /Km
ien khang XOL3 ; /Km
IdmTB ImmTB
CosN Thie
tb
Sin N Thie
tb
U3Bn h
thng
U3Khion
g
1 74 0.0425 4,61 0.08 10.8 54 0.7 0.7 2,61 10,442 74 0.038 4,61 0.08 10.8 54 0.7 0.7 2,33 9,323 74 0.034 4,61 0.08 10.8 54 0,7 0.7 2,08 8,32
4 74 0.03 4,61 0.08 10.8 54 0.7 0.7 1,84 7,365 74 0.024 4,61 0.08 10.8 54 0.7 0.7 1,47 5,886 74 0.019 4,61 0.08 108 54 0.7 0.7 1,16 4,647 74 0.032 4,61 0.08 10.8 54 0.7 0.7 1,96 7,848 74 0.014 4,61 0.08 10.8 54 0.7 0.7 0,86 3,449 74 0.011 4,61 0.08 10.8 54 0.7 0.7 0,67 2,68
10 74 0.0166 4,61 0.08 10,8 54 0.7 0.7 1,01 4,0411 74 0,0086 4,61 0.08 10,8 54 0,7 0,7 0,52 2,0812 74 0,022 4,61 0.08 10,8 54 0,7 0,7 1,47 5,88
13 73 0,024 4,61 0.08 10,2 51 0,67 0,74 0,88 3,5214 73 0,016 4,61 0.08 10,2 51 0,67 0,74 0,58 2,3215 73 0.006 4,61 0.08 10,2 51 0,67 0,74 0,21 0,8416 73 0,022 4,61 0.08 10,2 51 0,67 0,74 0,8 3,217 71 0,052 0,387 0.08 46 161 0,5 0,86 1,08 4,3218 69 0.043 1.15 0.08 22.8 114 0.6 0.8 0.2 0,8
V.KIEM TRA DAY DAN THEO IEU KIEN SUT APTa co :( U=( U1+( U2+( U3
%( U= luoiU U 100v(
che o b n h thng th: %( U %5e che o khi o n h th: %( U %8e
Theo cong thc tren ta tn h c o sut ap t cac thiet b e n MBA:
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1 .Tu ong lc 1 .CHE O BNH THNG CHE O KHI ONG
STT ( U1 ( U2 ( U3 ( U ( U% ( U1 ( U2 ( U3 ( U ( U%TBdmln 1.44 0.53 3,06 5,03 2,01 1.84 0,84 15,3 17,9 7,2TBcdln 1,44 0.53 3,38 3,56 2,14 1.84 0,84 17 19,6 7,8
Nhan xet : T bang ket qua tre n ta thay %( U che o bn h thng vakhi ong eu n ho hn g ii han cho phep n e n ta chap n han cach chon day da n tren
2. Tu ong lc 2.
CHE O BNH THNG CHE O KHI ONGSTT ( U1 ( U2 ( U3 ( U ( U% ( U1 ( U2 ( U3 ( U ( U%
TBdmln 1.44 2,38 3,12 6,94 2,77 1.84 5,4 12,48 19,7 7,8TBcdln 1,44 2,38 3,2 7.02 2,8 1.84 5.4 12,2 19.4 7.7
Nhan xet : T bang ket qua tre n ta thay %( U che o bn h thng vakhi ong eu n ho hn g ii han cho phep n e n ta chap n han cach chon day da n tren 3. Tu onh lc 3.
CHE O BNH THNG CHE O KHI ONG
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STT ( U1 ( U2 ( U3 ( U ( U% ( U1 ( U2 ( U3 ( U ( U%TBdmln 1.44 0,31 1,3 3,05 1,22 1.84 0,47 4,58 5,9 2,4TBcdln 1,44 0.31 5,8 7,55 3,02 1.84 0,47 7,9 10,02 4,08
Nhan xet : T bang ket qua tre n ta thay %( U che o bn h thng vakhi ong eu n ho hn g ii han cho phep n e n ta chap n han cach chon day da n tren
4. Tu ong lc 4CHE O BNH THNG CHE O KHI ONG
STT ( U1 ( U2 ( U3 ( U ( U% ( U1 ( U2 ( U3 ( U ( U%TBdmln 1,44 3,65 2,08 7.17 2,86 1.84 6,54 8,32 16,7 7,5TBcdln 1.44 3,65 2.61 7,7 3,08 1,84 6,54 10,44 18.8 7,52
Nhan xet : T bang ket qua tre n ta thay %( U che o bn h thng vakhi ong eu n ho hn g ii han cho phep n e n ta chap n han cach chon day da n tren .
CHNG VITNH TOAN DONG NGAN MACH
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S o ngan mach
Theo s o t ng ng tren th ta co the xac n h c 3 iem ng a n mach: K1, K2 va K3
A.Ng n m ch 3 pha
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I. TNH DONG NGAN MACH TAI IEM K1 1 .Tnh ien tr cua cac phan t
May bien ap.Cac thong so cua may bie n ap pha n xng ma ta a la cho n la:
SmMBA= 630(KVA)
( PN =8,2(KW)UNM =4%UMBA ha=0,4(KV)
Ta co: RMBA= !vv(dmM B A
ha N M
S
U P 2
32 10 2
32
4 00104,02,8 vv = 3102,8 v (; )
XMBA= 3222
10100
vv
(
dmMB A
ha
dmMB A
N M N M
S
U S
P U = 3106,9 v (; )
May bien ap at cach pha n xng mot khoang la 20 m vay n e n L1= 0,02(km),day dan t TPPPX e n MBA co Ro =0,124(; /Km)
RL1 =L1*R0L1 =0,00248(;
)XL1 =0,02 (; /Km) .a tn h pha n la chon day t TPPPX e n MBA)(0016,008,002,01011 ;!v!v! L L X L X
Con vi tat ca cac CB ta eu lay R CB = )(1015,0 3 ;v Nh vay ta tn h c tong tr ZK1 t iem ng a n mach K1 tr ve
ng uon ba ng : !! 21
211 LM B A LC BM B A K X X R R R Z 0,0155(; )
Vay do ng ng an mach tai K1 la:IK1= )(14 823
34 00
3 11 A
Z Z
U
K K
ha !v
!v
II. TNH DONG NGAN MACH TAI IEM K2Tong tr Zk2 t iem ng a n mach K2 tr ve ng uon la:ZK2 = 221
2213 L LM B A L LC BM B A X X X R R R R
Trong o:+RL2 = L2 v R0L2(ien tr cua day da n t cac TDL ve TPPPX)+ XL2 = L2 v X0L2(ien khang cua day da n t cac TDL ve TPPPX)Tn h tong tr Zk2 t cac TDL tr ve ng uon theo cong thc tren t
o ta tn h c dong ng a n mach t cac TDL ve ng uon n h sau:
TDL L2(Km)
X0L2 (; /Km)
R0L2(; /Km)
XL2(; )
RL2(; )
Uha(V)
ZK2 (; )
IK2(A)
1 0.0223 0.08 0.0601 0,0017 0,00134 400 0.071 32162 0.08 0.08 0.0991 0,0064 0,0079 400 0,022 10265
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3 0.013 0.08 0.0601 0,00104 0,0078 400 0,015 152884 0.135 0.08 0.0901 0.0108 0,012 400 0,031 7333
III. TNH DONG NGAN MACH TAI IEM K3Tong tr ZK3 t cac thiet b cua cac TDL tr ve ng uon la:ZK3 = 2
321
2
3214
L L L MB A L L LCB MB AX X X X R R R R R
+RL3 = L3 v R0L3(ien tr cua day da n t cac thiet b ve TDL)+ XL3 = L3v X0L3(ien khang cua day da n t cac thiet b ve TDL)Tn h tong tr Zk3 t cac thiet b tr ve ng uon theo co ng thc tren
t o ta tn h c dong ng an mach t cac thiet b ve ng uon n h sau1 .tnh dong ngan mach cac thiet b cua TDL 1 ve nguon
STT L3(Km)
X0L3 (; /Km)
R0L3(; /Km)
XL3(; )
RL3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.025 0.08 4,61 0,002 0.115 400 0,127 1805,72 0.025 0.08 5,3 0,002 0,1325 400 0,145 1589,53 0.027 0.08 18.1 0,00216 0,4887 400 0,5 4614 0.030 0.08 4,61 0,0024 0,1383 400 0,151 1528,45 0.034 0.08 4,61 0,0027 0,156 400 0,168 13676 0.039 0.08 5.3 0,00312 0,206 400 0,218 1056,47 0.043 0.08 4,61 0,0034 0,198 400 0,211 10968 0.048 0.08 4,61 0,0038 0,2212 400 0,233 987,79 0.053 0.08 5.3 0,0042 0,2809 400 0.293 787
1 0 0.048 0.08 5.3 0,0038 0,2544 400 0,266 86511 0.037 0.08 0,524 0,0029 0,019 400 0,034 66331 2 0.035 0.08 0,524 0,0028 0,018 400 0.033 68171 3 0,048 0.08 1,33 0,0038 0,063 400 0.076 30131 4 0.048 0.08 1,33 0,0038 0,063 400 0.076 30131 5 0.021 0.08 4,61 0,0016 0,096 400 0,108 21191 6 0.021 0.08 4,61 0,0016 0,096 400 0,108 2119
2.tnh dong ngan mach cac thiet b cua TDL2 ve nguon
STT L3(Km)
X0L3 (; /Km)
R0L3(; /Km)
XL3(; )
RL3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.023 0.08 18.1 0,00184 0,416 400 0,428 5382 0.044 0.08 5,3 0,00352 0,233 400 0,245 9403 0.039 0.08 5,3 0,00312 0,206 400 0,218 10564 0.049 0.08 5.3 0,0039 0,259 400 0,27 8505 0.057 0.08 5,3 0,00456 0,302 400 0,314 734
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6 0.012 0.08 0,153 0,00096 0,0018 400 0,019 117997 0.021 0.08 0,153 0.0016 0,0032 400 0,021 109868 0.027 0.08 18,1 0,00216 0,488 400 0,5 461
3.tnh dong ngan mach cac thiet b cua TDL3ve nguonSTT
L3(Km)
X0L3 (; /Km)
R0L3(; /Km
)
XL3(; )
RL3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.0465 0.08 1,33 0,0037 0,061 400 0,07 30842 0.038 0.08 1,33 0,003 0,05 400 0,064 36073 0.03 0.08 5,3 0,0023 0,158 400 0,17 13534 0.025 0.08 5,3 0,002 0,132 400 0,144 15955 0.007 0.08 5,3 0,00056 0,037 400 0,05 45436 0.007 0.08 5,3 0,00056 0,037 400 0,05 4543
7 0.029 0.08 0,635 0,00232 0,018 400 0,033 68618 0,037 0.08 0.635 0,0029 0,023 400 0,038 60119 0,016 0.08 0,635 0,00128 0,01 400 0,026 8817
1 0 0,021 0.08 0,635 0,00168 0,013 400 0,028 797511 0,018 0.08 0,524 0,00014 0,0094 400 0,025 92091 2 0.011 0.08 1.33 0,00088 0,014 400 0,029 7843
4.tnh dong ngan mach cac thiet b cua TDL4ve nguon:
STT L3(Km)
X0L3 (; /Km)
R0L3(; /Km)
XL3(; )
RL3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.0425 0.08 4,61 0,0034 0,195 400 0,2 11122 0.038 0.08 4,61 0,003 0,175 400 0,18 12303 0.034 0.08 4,61 0,0027 0,156 400 0,168 13684 0.03 0.08 4,61 0,0024 0,138 400 0,15 15315 0.024 0.08 4,61 0,0019 0,11 400 0,122 18786 0.019 0.08 4,61 0,0015 0.087 400 0,315 7337 0.032 0.08 4,61 0,00256 0,147 400 0,06 35388 0.014 0.08 4,61 0,00112 0,06 400 0,073 31479 0.011 0.08 4,61 0,00088 0,05 400 0,065 351410 0.0166 0.08 4,61 0,00132 0,076 400 0,09 259011 0,0086 0.08 4,61 0,00068 0,039 400 0,017 131341 2 0,022 0.08 4,61 0,00176 0,1 400 0,11 20441 3 0,024 0.08 4,61 0,0019 0,11 400 0,122 1879
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1 4 0,016 0.08 4,61 0,00128 0,073 400 0,086 26791 5 0.006 0.08 4,61 0,00048 0,027 400 0,04 55981 6 0,022 0.08 4,61 0,00176 0,1 400 0,113 20441 7 0,052 0.08 0,387 0,00416 0,02 400 0,036 63651 8 0.043 0.08 1.15 0,00344 0,049 400 0,063 3656
B.Ng n m ch 1 pha
I. TNH DONG NGAN MACH TAI IEM K1 1 .Tnh ien tr cua cac phan t
May bien ap.Cac thong so cua may bie n ap pha n xng ma ta a la cho n la:
SmMBA= 630(KVA)( PN =8,2(KW)UNM =4%UMBA ha=0,4(KV)
Ta co: RMBA= !vv(
dmMB A
ha N M
S
U P 2
32 10 2
32
4 00104,02,8 vv = 3102,8 v (; )
XMBA= 3222
10100
vv
(
dmMB A
ha
dmMB A
N M N M
S
U S
P U = 3106,9 v (; )
May bien ap at cach pha n xng mot khoang la 20 m vay n e n L1= 0,02(km),day dan t TPPPX e n MBA co R0PE =0,124(; /Km)
RPE1 =L1*R0PE =0.144(; )
day da n pha t TPPPX e n MBA co Ro =0,124(; /Km)RL1 =L1*R0L1 =0.00248(; )
XL1 =0,02 (; /Km) .a tn h pha n la chon day t TPPPX e n MBA)(0016,008,002,01011 ;!v!v! L L X L X )(0016,008,002,0011 ;!v!v! PE PE X L X
Con vi tat ca cac CB ta eu lay R CB = )(1015,0 3 ;v Nh vay ta tn h c tong tr ZK1 t iem ng a n mach K1 tr ve
ng uon ba ng : !! 2112111 PE LM B A PE LC BM B A K X X X R R R R Z 0,106(; )
Vay do ng ng an mach tai K1 la:IK1= )(2178
34 00
3 11 A
Z Z
U
K K
ha !v
!v
II. TNH DONG NGAN MACH TAI IEM K2
V SphaTDL1 =300mm2, n n ta ch n l i dy PE c SPE=150mm2
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Tra b ng 8.8 trang4 9 HDDNMHday PE t TPPPX e n TDL1co R0l2 =0,124(; /Km
V SphaTDL2 =185mm2, n n ta ch n l i dy PE c SPE=95mm2
Tra b ng 8.8 trang4 9 HDDNMHday PE t TPPPX e n TDL2 co R0l2 =0,193(; /Km
V SphaTDL3 =300mm2, n n ta ch n l i dy PE c SPE=150mm2Tra b ng 8.8 trang4 9 HDDNMH
day PE t TPPPX e n TDL3 co R0l2 =0,124(; /Km
V SphaTDL4 =185mm2, n n ta ch n l i dy PE c SPE=95mm2
Tra b ng 8.8 trang4 9 HDDNMHday PE t TPPPX e n TDL4 co R0l2 =0,193(; /Km
Tong tr Zk2 t iem ng a n mach K2 tr ve ng uon la:ZK2 = 22121
221213 PE PE L LM B A PE PE L LC BM B A X X X X X R R R R R R
Trong o:+RL2 = L2 v R0L2(ien tr cua day da n t cac TDL ve TPPPX)+ XL2 = L2 v X0L2(ien khang cua day da n t cac TDL ve TPPPX)RPE2 = L2 v R0PR2 (ien tr cua day PE t cac TDL ve TPPPX)+ XPR2 = L2 v X0PE@ (ien khang cua day PE t cac TDL ve TPPPX)
Tn h tong tr Zk2 t cac TDL tr ve ng uon theo cong thc tren to ta tn h c dong ng a n mach t cac TDL ve ng uon n h sau:
TDL L2(Km)
X0L2 (; /Km)
R0L2(; /Km)
RPE2(; /Km)
XPE2 (; /Km)
XL2(; )
RL2(; )
Uha(V)
ZK2 (; )
IK2(A)
1 0.0223 0.08 0,00134 0.0601 0.0017 0,0017 0,00134 400 0.217 10632 0.08 0.08 0,0079 0.0991 0,0064 0,0064 0,0079 400 0,262 8793 0.013 0.08 0,0078 0.0601 0,001 0,00104 0,0078 400 0,223 10334 0.135 0.08 0,012 0.0901 0.01 0.0108 0,012 400 0,259 890
III. TNH DONG NGAN MACH TAI IEM K3Tong tr ZK3 t cac thiet b cua cac TDL tr ve ng uon la:
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1 0.023 0.08 18.1 0,00184 0,00184 0,416 0,416 400 1,45 1592 0.044 0.08 5,3 0,00352 0,00352 0,233 0,233 400 1,06 2163 0.039 0.08 5,3 0,00312 0,00312 0,206 0,206 400 1.03 2234 0.049 0.08 5.3 0,0039 0,0039 0,259 0,259 400 1,14 2025 0.057 0.08 5,3 0,00456 0,00456 0,302 0,302 400 1,22 1886 0.012 0.08 0,153 0,00096 0,00096 0,0018 0,0018 400 0,626 3687 0.021 0.08 0,153 0.0016 0.0016 0,0032 0,0032 400 0,629 3678 0.027 0.08 18,1 0,00216 0,00216 0,488 0,488 400 1,59 144
3.tnh dong ngan mach cac thiet b cua TDL3ve nguonV Spha t TDL1 v cc thi t b nh h n 35mm2 nn ta ch n thng s dy PE b ngthong s dy pha
STT
L3(Km)
X0L3 (; /Km)
R0L3(; /Km
)
XL3(; )
XPE3(; )
Rl3(; )
RPE3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.0465 0.08 1,33 0,0037 0,0037 0,061 0,061 400 0,345 6672 0.038 0.08 1,33 0,003 0,003 0,05 0,05 400 0,323 7133 0.03 0.08 5,3 0,0023 0,0023 0,158 0,158 400 0,539 4284 0.025 0.08 5,3 0,002 0,002 0,132 0,132 400 0,487 4735 0.007 0.08 5,3 0,00056 0,00056 0,037 0,037 400 0.297 7766 0.007 0.08 5,3 0,00056 0,00056 0,037 0,037 400 0,297 7767 0.029 0.08 0,635 0,00232 0,00232 0,018 0,018 400 0,259 8888 0,037 0.08 0.635 0,0029 0,0029 0,023 0,023 400 0,269 8559 0,016 0.08 0,635 0,00128 0,00128 0,01 0,01 400 0,243 947
1 0 0,021 0.08 0,635 0,00168 0,00168 0,013 0,013 400 0,249 92411 0,018 0.08 0,524 0,00014 0,00014 0,0094 0,0094 400 0,242 9521 2 0.011 0.08 1.33 0,00088 0,00088 0,014 0,014 400 0,251 917
4.tnh dong ngan mach cac thiet b cua TDL4ve nguon:
V Spha t TDL1 v cc thi t b nh h n 35mm2 nn ta ch n thng s dy PE b ngthong s dy pha
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STT L3(Km)
X0L3 (; /Km)
R0L3(; /Km)
XPE3(; )
XL3(; )
RL3(; )
RPE3(; )
Uha(V)
ZK3 (; )
IK3(A)
1 0.0425 0.08 4,61 0,0034 0,0034 0,195 0,195 400 0,65 3552 0.038 0.08 4,61 0,003 0,003 0,175 0,175 400 0,069 378.3 0.034 0.08 4,61 0,0027 0,0027 0,156 0,156 400 0,57 4044 0.03 0.08 4,61 0,0024 0,0024 0,138 0,138 400 0,525 4315 0.024 0.08 4,61 0,0019 0,0019 0,11 0,11 400 0,479 4816 0.019 0.08 4,61 0,0015 0,0015 0.087 0.087 400 0,434 5327 0.032 0.08 4,61 0,00256 0,00256 0,147 0,147 400 0,55 4178 0.014 0.08 4,61 0,00112 0,00112 0,06 0,06 400 0,38 607,59 0.011 0.08 4,61 0,00088 0,00088 0,05 0,05 400 0,36 641
10 0.0166 0.08 4,61 0,00132 0,00132 0,076 0,076 400 0,412 56011 0,0086 0.08 4,61 0,00068 0,00068 0,039 0,039 400 0,338 6831 2 0,022 0.08 4,61 0,00176 0,00176 0,1 0,1 400 0,46 5021 3 0,024 0.08 4,61 0,0019 0,0019 0,11 0,11 400 0,479 4811 4 0,016 0.08 4,61 0,00128 0,00128 0,073 0,073 400 0,4 568,1 5 0.006 0.08 4,61 0,00048 0,00048 0,027 0,027 400 0,31 734,1 6 0,022 0.08 4,61 0,00176 0,00176 0,1 0,1 400 0,46 5021 7 0,052 0.08 0,387 0,00416 0,00416 0,02 0,02 400 0,659 3501 8 0.043 0.08 1.15 0,00344 0,00344 0,049 0,049 400 0,358 643,
Ch ng VII:La chon thiet b bao ve
Ta chon CB lam thiet b bao ve cho day da n va thiet b. CBg om co hai loai chn h la CB hieu ch n h c va CB khon h hieuch n h c.
CB khong hieu ch n h cy Thong soUdmCB : 380 660IdmCB : 1A - 125AICU :5KA 40KA
y C 3 ieu kien la chon
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UdmCB u U LUOI Idm C B u I maxlv ICU u I N M 3
CB hieu ch n h cy Thong soU dm C B : 380 660
IdmCB : 125A - 8000AICU :40KA 145KAIr : dong c t i ncua CB
Ir = Kr *Io=Kr *k0*IdmCB Kr = 0.8 ; 0.85 ; 0.88 ; 0.9 ; 0.93 ; 0.95 ; 0.98 ; 1
Io : dong ch n h n h tho cua CBIo = Ko* IdmCB Ko = 0.5 ; 0.63 ; 0.7 ;0.8; 0.9;1
Im (I sd ) : dong cat t, do ng cat co thi g ian INM(1) Im (I sd ) = Km * Ir Idn Km = 2 ; 3 ; 4 ; 5 ; 6 ; 8 ;10
Ii : dong cat t c th i c a CB Ii = Ki * IdmCB INM(3) Ki = 1.5 ; 2 ; 3 ; 4 ; 6 ; 8 ; 10 ; 11
y C 6 i ukien la chon CB hieu ch n h cU dm C B u U L UO I IdmCB u I maxlv IC U u I N M 3 I maxlv e Ir e Icpdd*khc Idn e Im e I N M 1 Ii e I N M 3
I: La chon CB cho tu phan phoi phan xn g :Dong ien tn h toan tu phan phoi pha n xng sau bu :IlvM A X =
luoi
dmMB A
U
S
*3=
38.0*3630 = 957 (A)
Vi tu phan phoi pha n xng ta chon loai CB hieu chn h cco ma hieu CB: C1 001 N co cac tho ng so n h sau:
ImCB = 1000(A) lvMAX=957A
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UmCB = 415(V) >Uluoi=380(V)ICU =150(KA) (3) N M I u =IK1=14823(A
Vi cac thong so c hieu ch n h n h sau :
Ta co : Ilvmax
ko*k
r *I
dmcbe
Icpdd*khc
dmcb
lv
I I max ko*kr
dmcbhc
c pdd
I k I
*
1000957
e ko*kr e 62,0*10001680
0,957 e ko*kr e 2,7Ch n ko=10,957 e kr e 2,7Chon kr =0,98
Ir = ko*kr *Idmcb =0,98*1*1000 =980(A)Io=Ir /kr = )(100098,0
980 A!
Ta co : Idn km*Ir e I(1)nm
Ir I dn km
r
nm
I
I )1(
98014 87
e km e 9802178
1,5e
kme
2,2 Ch n k m=2Im = km*Ir =2*980=1960 (A)
Ta co : ki*Idmcbe I(3)nm
Kiedmcb
nm
I
I )3(
km e 100014 823
kie 14,8
Ch n k i=11Ii = ki*Idmcb =11*1000=11000 (A)
Vay CB:1 001 Na chon th o m n v i cc i u ki nco cac he so sau: Io =1 000(A),ko = 1 Ir =980(A),kr = 0,98
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Ii=11 000(A),Ki=11 Im=1 960(A),km=2 ImCB = 1 000(A) UmCB = 41 5(V) ICU =1 50(KA)
II: La chon CB cho cac tu ong lc1 :Tu ong lc 1 , I
lvmax= 224,5 A
Vi tu ong lc 1 n ay ta chon loai CB:NS250N hieu ch n h cco thong so n h sau: ImCB = 250(A)> Ilvmax
UmCB = 415(V) >Uluoi=380(v)
ICU =36(KA) (3) N M
I u =3216(A)V ay la CB hieu chn h c n en ta ca n phai kiem tra va hieu
ch n h lai cac thong so cua CB
Ta co : Ilvmax ko*kr *Idmcbe Icpdd*khc
dmcb
lv
I I max ko*kr
dmcbhc
c pdd
I k I
*
2505,224
e ko*kr e 62,0*2504 00
0,89 e ko*kr e 2,5Ch n ko=10,89 e 1*kr e 2,5Chon kr =0,95Ir = ko*kr *Idmcb =0,95*1*250 =237,5(A)
Io=Ir /kr = )(25095,05,237
A!
Ta co : Idn km*Ir e I(1)nm
Ir I dn km
r
nm
I I )1(
5,23775,356
e km e 5,237879
1,5 e kme 3,7 Ch n k m=3
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Im = km*Ir =3*237,5=712,5(A)
Ta co : ki*Idmcbe I(3)nm
Kiedm cb
nm
I
I )3(
km e 2503216
kie 12,8 Ch n k i=11Ii = ki*Idmcb =11*250=2750 (A)
Vay CB: NS250N a chon th o m n v i cc i u ki n vco cac he sosau:
Io =250(A),ko = 1 Ir =237,5(A),kr = 0,95 Ii=2750(A),Ki=11 Im=721 ,5(A),km=3 ImCB = 250(A) UmCB = 41 5(V) ICU =36(KA)
2:Tu ong lc 2 , I tt = 172,2AVi tu ong lc 2 n ay ta chon loai CB:NS250N hieu ch n h cco thong so n h sau: ImCB = 250(A)> Ilvmax
UmCB = 415(V) >Uluoi=380(v)
ICU =36(KA) (3) N M I u =10265(A)V ay la CB hieu chn h c n en ta ca n phai kiem tra va hieu
ch n h lai cac thong so cua CB
Ta co : Ilvmax ko*kr *Idmcb e Icpdd
*khc
dmcb
lv
I I max ko*kr
dm cbhc
c pdd
I k I
*
2502,172
e ko*kr e 62,0*250298
0,68 e ko*kr e 1,8Ch n ko=0,8
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0,68 e 0,8*kr e 1,8Chon kr =0,98Ir = ko*kr *Idmcb =0,8,98*250 =196(A)
Io=Ir /kr = )(20098,0
196 A!
Ta co : Idn km*Ir e I(1)nm
Ir I dn km
r
nm
I
I )1(
1966,394
e km e 196879
2 e km e 4,48 Ch n k m=3Im
= km*Ir =3*196=588(A)
Ta co : ki*Idmcbe I(3)nm
Kiedm cb
nm
I
I )3(
km e 25010265
kie 41 Ch n k i=11Ii = ki*Idmcb =11*250=2750 (A)
Vay CB: NS250N a chon th o m n v i cc i u ki nco cac he so sau: Io =200(A),ko = 0,8 Ir =1 96(A),kr = 0,98 Ii=2750(A),Ki=11 Im=588(A),km=3 ImCB = 250(A) UmCB = 41 5(V) ICU =36(KA)
3:Tu ong lc 3 , I tt = 229,6AVi tu ong lc 3 n ay ta chon loai CB:NS250N hieu ch n h cco thong so n h sau: ImCB = 250(A)> Ilvmax
UmCB = 415(V) >Uluoi=380(v)
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ICU =36(KA) (3) N M I u =15288(A)V ay la CB hieu chn h c n en ta ca n phai kiem tra va hieu
ch n h lai cac thong so cua CB
Ta co : Ilvmax ko*kr *Idmcbe Icpdd*khc
dmcb
lv
I I max ko*kr
dmcbhc
c pdd
I k I
*
2506,229
e ko*kr e 62,0*2504 00
0,91 e ko*kr e 2,58Ch n ko=10,91 e 1*kr e 2,58Chon kr =0,98Ir = ko*kr *Idmcb =1*0,98*250 =245(A)
Io=Ir /kr = )(25098,024 5
A!
Ta co : Idn km*Ir e I(1)nm
Ir I dn km
r
nm
I
I )1(
24 58,34 3
e km e 24 51033
1,4 e kme 4,21 Ch n k m=3Im = km*Ir =3*245=733(A)
Ta co : ki*Idmcb e I(3)nm
Ki edm cb
nm
I
I )3(
km e 25015288
kie 61 Ch n k i=11Ii = ki*Idmcb =11*250=2750 (A)
Vay CB: NS250N a chon th o m n v i cc i u ki nco cac he so sau: Io =250(A),ko = 1
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Ir =245(A),kr = 0,98 Ii=2750(A),Ki=11 Im=733(A),km=3 ImCB = 250(A) UmCB = 41 5(V)
ICU =36(KA) 4:Tu ong lc 4 , I tt = 171A
Vi tu ong lc 4 n ay ta chon loai CB:NS250N hieu ch n h cco thong so n h sau: ImCB = 250(A)> Ilvmax
UmCB = 415(V) >Uluoi=380(v)
ICU =36(KA) (3) N M I u =7333(A)V ay la CB hieu chn h c n en ta ca n phai kiem tra va hieu
ch n h lai cac thong so cua CB
Ta co : Ilvmax ko*kr *Idmcbe Icpdd*khc
dmcb
lv
I I max ko*kr
dmcbhc
c pdd
I k I
*
250171
e ko*kr e 62,0*250298
0,684 e ko*kr e 1,9Ch n ko=0,90,684 e 0,9*kr e 1,9Chon kr =0,8Ir = ko*kr *Idmcb =0,9*0,8*250 =180(A)
Io=Ir /kr = )(2259,0180
A!
Ta co : Idn km*Ir e I(1)nm
Ir
I dn km r
nm
I
I )1(
1808,307
e km e 180890
1,7 e kme 4,9 Ch n k m=3Im = km*Ir =3*180=540(A)
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Ta co : ki*Idmcbe I(3)nm
Kiedmcb
nm
I
I )3(
km e250
7333
kie 29 Ch n k i=11Ii = ki*Idmcb =11*250=2750 (A)
Vay CB: NS250N a chon th o m n v i cc i u ki nco cac he so sau: Io =225(A),ko = 0,9 Ir =1 80(A),kr = 0,8 Ii=2750(A),Ki=11 Im=540(A),km=3 ImCB = 250(A) UmCB = 41 5(V) ICU =36(KA)
III: La chon CB cho cac thi t b
1 : K t qu l a ch n CB cho cc thit b TDL1
STT
Tenthiet b KHMB
Idm (A)
Xi(m)
Yi(m)
I(3)NM (A) Lo i CB
U mCB(V)
ImCB(A)
ICU(KA)
1 Motor cau 74 10,8 4,5 6,7 1805,7
C120H 415 16 15
2
Bmncnonghap
73 10,2 9 6,7 1589,5
C120H 415 16 15
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3Q uatcong
nghiep75 1,63 5,2 15,2
461
C60N 415 2 6
4 Motor cau 74 10,8 5,38 20 1528,4
C120H 415 16 15
5 Motor cau 74 10,8 5,37 24,3
1367
C120H 415 16 15
6
Bmncnonghap
73 10,2 6,6 29,1 1056,4
C120H
415 16 15
7 Motor cau 74 10,8 5,37 33,5
1096
C120H 415 16 15
8 Motor cau 74 10,8 5,55 38,7 987,
7
C120H 415 16 15
9
Bmncnonghap
73 10,2 6,41 43,5
787
C120H415 16 15
10
Bmncnonghap
73 10,2 15,1 43,5
865
C120H 415 16 15
11 Maychiec bia 72 40,8 16,3 27,4
6633
C120H 415 50 15
12 Maychiec bia 72 40,8 22,5 27,4
6817
C120H 415 50 15
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13 Maygap rong 67 22,8 30,8 30,1
3013
C120H 415 25 15
14 Maygap rong 67 22.8 30,8 23.4
3013
C120H 415 25 15
15 Motor cau 74 10,8 40 6,7
2119
C120H 415 16 15
16 Motor
cau74 10,8 30,4 6.7
2119
C120H 415 16 15
2:K t qu l a ch n Cb cho cc thit b TDL2
STTTenthiet
bKHMB Idm
(A)Xi
(m) Yi(m)
I(3)NM (A)
Lo i CBU mCB(V)
ImCB(A)
ICU(KA)
1Q uatcongnghiep
75 1,63 79 52,1538
C60N 415 2 6
2 Motor cau 74 10,8 76,3 60,2 940 C120H 415 16 153 Motor cau 74 10,8 71,1 60,7 1056 C120H 415 16 15
4Bmnc
nonghap73 10,2 66 60,2
850
C120H 415 16 15
5 Motor cau 74 10,8 58 60,2 734 C120H 415 16 15
6Mayxucchai
68 106 75 39,111799
C120H 415 125 15
7 May 68 106 77,6 22 10986
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xucchai
C120H 415 125 10
8Q uatcongnghiep
75 1,63 79 14,8461
C60N 415 2 6
3:K t qu l a ch n CB cho cc thit b T L 3
STT Ten thietb KHMBIdm (A)
Xi(m)
Yi(m)
I(3)NM (A)
Lo iCB U mCB
(V)
ImCB(A)
ICU(KA)
1May gap
rong 67 22,8 67 32,5 3084 C120H 415 25 15
2May gap
rong 67 22.8 67 24,3 3607 C120H 415 25 15
3 Motor cau 74 10,8 76,3 6,7 1353 C120H 415 16 15
4 Motor cau 74 10,8 71,1 6,7 1595 C120H 415 16 15
5 Motor cau 74 10,8 54,8 6,7 4543C120H 415 16 15
6 Motor cau 74 10,8 49,2 6,7 4543 C120H415 16 15
7May chiet
bia 72 40,8 54,5 28 6861 C120H 415 50 15
8May chiet
bia 72 40,8 61,4 28 6011 C120H 415 50 15
9motor vi
tren 70 46 56 14,8 8817C120H
415 50 15
10
Bm nc
nong hap 73 10,2 68 6,7 7975 C120H 415 16 1511
Bm ncnong hap 73 10,2 44 6,7 9209 C120H 415 16 15
12May gap
day 69 22,8 10,5 10,5 7843 C120H 415 25 15
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4:K t qu l a ch n CB cho cc thit b TDL4
STT Ten thietb KHMBIdm (A)
Xi(m)
Yi(m)
I(3)NM (A)
Lo iCB U mCB
(V)
ImCB(A)
ICU(KA)
1 Motor cau 74 10,8 5,5 47,8 1112 C120H 415 16 15
2 Motor cau 74 10,8 5,5 52,1 1230 C120H 415 16 15
3 Motor cau 74 10,8 13,9 52,1 1368 C120H 415 16 15
4 Motor cau 74 10,8 14 47,81531 C120H 415 16 15
5 Motor cau 74 10,8 21,5 49,7 1878C120H 415 16 15
6 Motor cau 74 10,8 26,7 49,7 733 C120H415 16 15
7 Motor cau 74 10,8 5,2 60,2 3538 C120H 415 16 15
8 Motor cau 74 10,8 25,3 60,7 3147 C120H 415 16 15
9 Motor cau 74 10,8 26,3 60,7 3514C120H 415 16 15
10 Motor cau 74 10,8 36,8 60,7 2590 C120H 415 16 15
11 Motor cau 74 10,8 42,3 60,7 13134C120H 415 16 15
12 Motor cau 74 10,8 47,5 60,2 2044 C120H 415 16 15
13
Bmncnonghap 73 10,2 52,7 60,2 1879
C120H 415 16 15
14
Bmncnonghap 73 10,2 32 50,6 2679
C120H 415 16 15
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15
Bmncnonghap 73 10,2 10,4 60,2 5598
C120H415 16 15
16
Bm
ncnonghap 73 10,2 31,5 60,2 2044
C120H 415 16 15
17Motor vid i 71 46 47,5 60,7 6365 C120H 415 50 15
18may gapday 69 22,8 50 41,6 3656
C120H 415 25 15
5: l a ch n CB cho sinh ho t v chiu sng
a) L a ch n CB cho sinh ho t
Ta c IlvM A X =luoi
m S
U
S
*3=
38.0*3187 = 284 (A)T ra bang 8.8 trang 49 sach
HCC ta la chon c : Icpdd =298 (A) S= 185(mm2); Ro=0,0991 .
v Ilvmax > 125A n n ta ch n CB cho Sin h ho t l lo i CB hi u ch n h u c co ma hi u CB:NS400N v i thong so n h sau:
ImCB = 400(A)> IlvmaxUmCB = 415(V) >Uluoi=380(v)
ICU =36(KA) (3) N M I u =14823(A)=IK1V ay la CB hieu chn h c n en ta ca n phai kiem tra va hieu
ch n h lai cac thong so cua CB
Ta co : Ilvmax ko*kr *Idmcb e Icpdd*khc
dmcb
lv
I I max ko*kr
dm cbhc
c pdd
I k I *
4 00284
e ko*kr e 62,0*4 00298
0,71 e ko*kr e 1,2Ch n ko=0,9
K m/;
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0,71 e 0,9*kr e 1,2Chon kr =0,95Ir = ko*kr *Idmcb =0,95*0,9*400 =342(A)
Io=Ir /kr = )(36095,0
34 2 A!
Ta co : Idn km*Ir e I(1)nm
kmr
nm
I
I )1(
km e 34 22178
km e 6,3 Ch n k m=6
Im
= km*Ir =6*342=2052(A)
Ta co : ki*Idmcbe I(3)nm
Kie !dm cb
nm
I
I )3(
dm cbk
I
I )3(1
km e4 00
14 823
kie 37 Ch n k i=11Ii = ki*Idmcb =11*400=4400 (A)
Vay CB: NS400N a chon th o m n v i cc i u ki n vco cac he sosau:
Io =360(A),ko = 0,9 Ir =342(A),kr = 0,95 Ii=4400(A),Ki=11 Im=2052(A),km=6 ImCB = 400(A) UmCB = 41 5(V) ICU =36(KA)
b> l a ch n CB chi u sang
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Ta c IlvM A X =luoi
mC S
U
S
*3=
38.0*381 =123(A) 125(A)nn ta ch n lo i Cb khng
hi u ch nh lCB:C1 20H co cac tho ng so:ImCB = 125(A)> IlvmaxUmCB = 415(V) 4 00( )
luoiU V u !
ICU =15(KA) (3) N M I u = 14823 (A)6: l a ch n CB cho my pht d phng
Smpdp=Smba=630kva
Nn ta ch n CB cho my pht d phng gi ng thong s nhta ch n cho MBA n TPPPX cthong s v c hi u ch nh nhsau
CB:1 001 N Io =1 000(A),ko = 1 Ir =980(A),kr = 0,98 Ii=11 000(A),Ki=11 Im=1 960(A),km=2 ImCB = 1 000(A) UmCB = 41 5(V) ICU =1 50(KA)
7: l a ch n CB cho t b
y Ilvmax =)(5313*38,0
50*73
7 A
U
Q
luoi
dm
!!
Vi t bta chon loai CB hieu chn h c co ma hieu CB: C801 N co cac tho ng so n h sau:
ImCB = 800(A) lvMAX=531AUmCB = 415(V) >Uluoi=380(V)ICU =150(KA) (3) N M I u =IK1=14823(A
Vi cac thong so c hieu ch n h n h sau :
Ta co : Ilvmax ko*kr *Idmcbe Icpdd
*khc
dmcb
lv
I I max ko*kr
dm cbhc
c pdd
I k I
*
800531
e ko*kr e 62,0*800864
0,66 e ko*kr e 1,7Ch n ko=0,7
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0,85 e kr e 2,4Chon kr =0,98Ir = ko*kr *Idmcb =0,98*0,7*800 =548(A)
Io=Ir /kr = )(56098,0
54 8 A!
Ta co : Idn km*Ir e I(1)nm
Ir I dn km
r
nm
I
I )1(
54 814 87
e km e 54 82178
2,7 e kme 3,9 Ch n k m=3Im
= km*Ir =3*548=1644 (A)
Ta co : ki*Idmcbe I(3)nm
Kiedm cb
nm
I
I )3(
km e 80014 823
kie 18 Ch n k i=11Ii = ki*Idmcb =11*800=8800 (A)
Vay CB:801 N a chon th o m n v i cc i u ki nco cac he so sau: Io =560(A),ko = 0,7 Ir =548(A),kr = 0,98 Ii=8800(A),Ki=11 Im=1 644(A),km=3 ImCB = 800(A) UmCB = 41 5(V) ICU =1 50(KA)
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CHNG VIIITHIET KE BAO VE AN TOAN IEN CHO NGI
I.Khai quat he ve thong noi at phan xng : M uc ch xay dng he thong noi at:
y Chong giaty Chong hoa hoan do ieny Cung cap ien lien tucy Chong qua apy Chong nhieu ien
Moi s o noi eu co u iem va khuyet iem rieng cua notuy thuoc he thong x nghiep ma ta la chon sao cho phu hp.
Tnh chat lam viec cua phan xng ang xet la : cong nhanlam viec phan xng luon tiep xuc may moc, thiet b lam viec vvay oi hoi an toan, v the trng hp co s co cham vo xay ra batbuot phai ngat ien neu khong se gay nguy hiem cho cong nhan.
e n gian va kinh te cho lap at ta chon he thong bao vetheo s o TN-C-S, tuy nhien s o nay cung con nhieu han che.
N hng ieu kien bat buoc: S o TN yeu cau trung tnh pha ha ap cua MBA nguon,vo cuatu phan phoi,vo cua tat ca cac phan t khac trong mangva cac vat dan t nhien phai c noi at chung.
oi vi mot tram co phan o lng c thc hien pha ha ap,can co bien phap cach ly co the nhn thay c pha aunguon ha the.
Day PEN khong bao gi c cat trong bat k trng hpnao.Viec ieu khien va bao ve cac may cat mach trongmang TN c sap xep nh sau:
-loai 3 cc khi mach co day PEN;
-loai 4 cc (3 pha +day trung tnh)khi mach co day trung tnhrieng biet vi day PE.Chon s o TN-C-Se thiet ke an toan cho xng ckh.
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S o TN- C -S :
-S o TN-C-S la s ket hp gia hai s o TN-C va TN-S,v vay mata tan dung c nhng u khuyet iem cua chung e bao ve antoan cho phan xng.
-Khi dung chung mot li , s o TN-S luon s dung sau s o TN-Cva iem phan gia day PE khoi day PEN thng la iem au cuali.
-T tu ien chnh en cac tu phan phoi ta s dung mang kieu TN-C,t tu phan phoi en cac thiet b ta s dung mang kieu TN-S