Cc phn t bo v v tn hiu ho ngha ca bo v v tn hiu ho* Cc phn t bo v v tn hiu ho c vai tr rt to ln:m bo qu trnh lm vic an ton cho ngi v my mc, thit b. Qu trnh lm vic c th xy ra s c hoc ch lm vic xu cho ngi v my mc, thit b, ng thi c th bo hiu cho ngi vn hnh bit tnh trng lm vic ca h thng KT x l.* Chc nng ca cc thit b bo v v tn hiu ho:Ngng h thng (my mc) khi s c nguy him trc tip n ngi, thit b, my mc: U < Uquy nh , U > Ucp , I > Icp , ....Khi qu ti hoc s c cha nguy him n thit b, my mc th thit b bo v v tn hiu ho phi bo cho ngi vn hnh bit s l kp thi.Bo m khi ng, hm, o chiu ..., mt cch bnh thng, ngha l phi m bo sao cho: I < Icp, to < tocp ,...Cc dng bo v:Bo v ngn mch:- Bo v ngn mch l bo v cc s c c th gy nn h hng cch in, hoc h hng cc c cu ca thit b, my mc (khi ngn mch s gy nn nhit tng nhanh gy chy hoc sc t ng tng mnh gy va p, ...).- Cc thit b bo v thng dng: cu ch, aptmat, rle dng in cc i, cc khu bo v ngn mch bng bn dn, in t, ...- Dng tc ng ca cu ch:Idc = Ik / ( (7-18)Trong :Idc l dng tc ng ca dy chy c chn.Ik l dng khi ng ca ng c, ph ti c bo v.( l h s xt n qun tnh nhit.( = 2,5 i vi ng c khi ng bnh thng.( = (1,6 ( 2) i vi ng c khi ng nng.+ Cm t cu ch trn dy trung tnh, mch ni t, v t dy ch th v my s c in p cao nguy him. Dng cu ch bo v ngn mch th n gin, r tin, nhng tc ng khng chnh xc, dng tc ng ph thuc vo thi gian, thay th lu, khng bo v c ch lm vic 2 pha.- Dng chnh nh ca aptmat:Ic = (1,2 ( 1,3).Ik ; (7-19)+ Aptmat tc ng ri th c th ng li nhanh, ct c dng ln, bo v c ch lm vic dng 2 pha (khi b mt 1 trong 3 pha).- Dng rle dng in cc i (RM) bo v ngn mch phi chnh nh dng tc ng cho ph hp vi dng ngn mch.Thng t rle dng cc i trn 3 pha ca ng c khng ng b 3 pha, hoc t trn 1 cc i vi ng c mt chiu. Tip im ca RM l loi khng t phc hi.+ V d dng cu ch v aptmat bo v ngn mch:

+ V d dng rle dng cc i bo v ngn mch:

Bo v nhit:- Nhm trnh qu ti lu di, nu khng th kh c, thit b, ng c s pht nng qu nhit cho php.- Thng dng rle nhit, aptmt c bo v nhit, phn t bo v qu ti bng bn dn, bo v qu ti cho ph ti di hn.- Cc tip im rle nhit (RN) l loi khng t phc hi, sau khi rle nhit tc ng th phi n reset bng tay. Phi chn rle nhit c c tnh pht nng gn vi c tnh pht nng ca thit b, ng c cn c bo v (hnh 7-13).+ Dng chnh nh ca rle nhit, aptmat:Ic = (1,2 ( 1,3)Im (7-20)Trong : Im l dng nh mc ca ng c, ph ti.+ V d dng rle nhit v aptmat bo v qu ti di hn:

- Dng rle dng cc i (RI) bo v qu ti cho ph ti ngn hn hoc ngn hn lp li. Khi ph ti lm vic trong thi gian ngn, s pht nng ca ph ti khng ph hp vi c tnh ca rle nhit, nn rle nhit khng tc ng kp, bi vy phi dng rle dng cc i tc ng nhanh.+ V d dng rle dng cc i bo v qu ti ngn hn:

- Dng chnh nh ca rle dng cc i bo v qu ti:Ic.RI = (1,4 ( 1,5)Im (7-21)- Thng dng 1 rle dng cc i bo v ngn mch (RM) v 2 rle dng cc i bo v qu ti (RI). Tip im ca rle dng cc i bo v qu ti l loi t phc hi (hnh 7-14).Bo v im khng v cc tiu:- Nhm trnh lm vic vi in p ngun thp hoc mt p ngun, v trnh t khi ng li khi in p ngun phc hi.- Thng dng cc rle in p (RA), cng tc t (CTT), khi ng t (KT), bo v m khng v cc tiu.- Chnh nh in p ht, nh ca rle in p, cng tc t:Uh.RA > Ung.st.cp (7-22)Unh.RA ( Ung.st.cp (7-23)Trong :Uh.RA l in p ht ca rle in p, hay ca cng tc t, khi ng t.Unh.RA l in p nh ca RA, CTT, KT.Ung.st.cp = 85%Ung.m l in p ngun st cho php.Nguyn l lm vic v bo v ca s hnh 6 - 20:t cng tc xoay KC v tr 0 th tip m KC1 s kn, KC2 h; ng cu dao CD, nuin p lm vic t gi tr cho php (Ung > 85%Ung.m) th RA tc ng, n t duy tr thng qua tip im RA(1-3) ca n.Quay cng tc KC n v tr 1 tri (T) th K c in, lm cho ng c quay. Khi in p Ung ( 85%Ung.m th RA s nh lm K mt in v ng c cng c loi khi li in, trnh cho ng c khi b t nng qu nhit cho php (v in p thp s dn n dng tng qu dng cho php ca ng c).Khi ng c ang lm vic, nu mt in ngun th khi c in li, ng c vn khng t khi ng li c, v khi KC vn v tr 1 tri v KC1 vn h, RA mt in khi mt in p ngun, do khi c in li th K vn khng c in.+ V d dng rle in p (RA) bo v m khng v cc tiu:

Bo v thiu v mt t trng:- Nhm bo v thiu v mt kch t ng c. Khi in p hay dng kch t ng c b gim, gy ra tc ng c cao hn tc cho php, hoc dng in ng c ln hn dng cho php, dn n h hng cc phn ng hc ca my, lm xu iu kin chuyn mch, ...- Dng rle dng in, rle in p, ... bo v thiu v mt t trng.+ V d dng rle dng in, rle in p bo v thiu v mt t trng (hnh 7-16)Nguyn l bo v: khi in p th rle thiu t trng RTT s ng kn tip im ca n, KC t v tr gia nn tip im KC1 kn, RA tc ng. Quay KC sang v tr 1 (T) th cho ng c lm vic bnh thng.Khi in p st qu gi tr cho php, hoc dng kch t gim thp n gi tr: Ikt. = Inh.RTT , Inh.RTT ( Ikt.min.cp , nn RTT nh lm K mt in, loi ng c ra khi li in bo v ng c.

Bo v lin ng:- Nhm bo m s lm vic an ton cho mch (bo m nghim ngt mt trnh t lm vic hp l gia cc thit b, trnh thao tc nhm).- Cc thit b bo v lin ng bng c kh nh: cc nt n kp, cc cng tc hnh trnh kp, ... V cc phn t bo v lin ng in nh:Cc tip im kho cho ca cc cng tc t, rle, lm vic cc ch khc nhau. V d:

Khi khi ng thun, n nt MT th T c in, ng in cho ng c quay, cn tip im thng kn ca MT m ra khng cho N c in, m bo khng b ngn mch mch stato. Khi T c in th tip im thng kn ca T m ra, m bo cho N khng th c in nu nh khng may c ngi tc ng vo nt MN.Khi ang quay thun, mun o chiu, n nt MN th T s mt in v N s c in, qu trnh o chiu din ra bnh thng. Nu khng may trong qu trnh quay thun, tip im ca T mch stato b dnh th tip im ca T mch ca cun dy N s khng kn li c, nn mc d n MN nhng N vn khng th c in c, trnh c s ngn mch bn pha stato nu nh c T v N u tc ng.Nh vy cc lin ng c v in trong s bo m cho s hot ng bnh thng, ng trnh t lm vic t ra, trnh thao tc nhm.Tn hiu ho- Khi xut hin ch lm vic xu nhng cha cn phi dng my th thit b bo v s hot ng lm cho cc thit b tn hiu bo cho ngi vn hnh bit x l kp thi.- Khi tn hiu bo m khng c x l kp thi th thit b bo v s tc ng nh ch s lm vic ca h thng truyn ng in.- Thit b tn hiu ho: m thanh: chung, ci, ...; nh sng: n, mu, ...; C bo: rle tn hiu, ...V d:

S hnh 6-23 ang hot ng bnh thng. Nu nh qu ti th rle nhit s tc ng, lm RA ri n K mt in, loi ng c ra khi tnh trng nguy him, ng thi ng tip im ca n lm n sng ln, bo cho ngi vn hnh bit x l, sau khi x l xong, ngi vn hnh n reset ca RN th mi c th vn hnh li c.Cn nu b ngn mch trong ng c th rle bo v dng cc i RM tc ng, loi ngay ng c khi tnh trng nguy him, ng thi ng tip im ca n lm cho chung Chg ku ln, bo cho ngi vn hnh bit x l kp thi, sau khi x l xong, ngi vn hnh n reset ca RM th mi c th vn hnh li c.Cu hi n tp1. Da vo nhng c s no ngi ta a ra cc nguyn tc iu khin t ng theo cc thng s thi gian, tc , dng in, v hnh trnh, v.v ?2. Phn tch ni dung ca nguyn tc iu khin t ng theo thi gian, tc , dng in, hnh trnh ? Gii thch nguyn l lm vic ca s minh ha cho mi nguyn tc trn?3. Ti sao c th xy ra cc s c trong h thng truyn ng in t ng ? cch khc phc s c nh th no ?4. Phn tch bo v ngn mch, bo v qu ti, bo b im khng v cc tiu, bo v thiu hoc mt t trng, bo v lin ng ? Gii thch nguyn l bo v ca cc mch in hnh tng ng vi mi bo v trn ?5. Tn hiu ha l g ? Cc mch tnh hiu ha c tc dng g trong h thng truyn ng in t ng ?