1. DOng co kh0ng ddng bO rdto ldng s6c 3 pha c6 c6ng sudtOinn muc Pdm = 28 kW, di6n dp
dinh mrrc Udm = 380 V, tdn sd 50 Hz, ddu Y, 6 cqc, h0 s6 cdng sudt dinh mrlc cosqdm = 0,88, t6c d6 dinh mfc ndm = 980 v/ph. (Tdn hao ddng + tdn hao s5t stato; ld2,2kw, tdn hao
co (do ma sdt, quat 9i6...) la 1,1 kW, khi d6ng co ldm vi6c 6 tai dinh mr.rc. Tinh, lrlc tAi dinh muc:
a/ h6 so trudt sdm (0,5 d)
b/ tdn hao d6ng r6to pCu2dm (0,5 d)
c/ hi6u sudt qdm (0,5 d)
d/ dong di6n stato lldm (0,5 d)
e/tdn so dong di6n rOtof2 (1d)
2. Mdy phdt di6n d6ng b6 cung cdp di6n cho hd ti6u thu mdt c6ng sudt S = 2500+j3000 kVA, 6 di6n 6p 6,3 kV.
a/ Tinh tdng tdn hao tr6n dudng dAy vd trong mdy phdt, biSt rdng di6n tr6 c0a drlong ddy / pha ld rd = 0,15 Q, di6n trd dAy qudn phdn tlng/ pha c0a m6y phdt lir rU = 0,045 CI (1d). b/ Ndu ddt th6m m6t dQng cd dien ddng b6 ldm vi6c trong chd d0 quei kich tr.t, ph6t ra c6ng sudt phAn khdng, vdi S = 30- j3000 kVA, thi tdng tdn hao tr6n s6 ld bao nhi6u ? (2d)
3. M6y biSn dp 1 pha c6 cdc sd lieu sau :c6ng sudt dinh mrlc Sdm = 1OO kVA, tdn sd OO
Hz, di6n aip dinh mrlc U1/U2 =7200/480 V, co cdcthdng sdtinh bing ohms: R,cao dp = 3,06 X,cao 6p = 6,05
R,ha dp = 0,014 X,ha aip = 0,027
XM, cao dp = 17 809
RM, cao Ap =21 4OO
Mdy bidn 6p cung cdp di6n cho tAi ti6u thu dong dinh mfc, 6 di6n 6p dinh mfc 480 V, tai co cosg =O,75 tr5. VC sd dd mach trjdng dtJdng cOa mdy biSn dp vd tfnh :
a/ di6n tr6 vir di6n khiing tuong dudng c0a mdy biSn 6p nhin trJ phia cao ap (1 d)
b/ tdng tr6 nhin vdo mdy biSn 6p trt phia cao 6p (k€i cA tdng trd cOa tAi) (1 d)
c/ di6n dp phia c4o 6p cria mdy bidn dp khi d6 (1 d)
d/tdng trd nhinvdo mdy bidn dp trJ phfa cao dp khi kh6ng tai (1 d)
4. D6ng cd di6n 1 chidu kich trl song song, cong sudt dinh m0c Pdm = 95 kw, di6n 6p dinh mrrc Udm'= 220V, dong tirou thtl dinh mfc ldrn = 470 A, ddng kich ttJ dinh mrlc iktdm = 4,25 A, ndm = 500 v/ph. Di6n tr6 day qudn phdn fng ru = 0,0125 Q. Tinh :
a/ hi6u sudt 1O,s 0;
b/ tdn hao ddng trong mdy, tdn hao khi kh6ng tAi vA ddng khOng tai (0,5 d)
c/ moment c0a d6ng cd (1 d)
d/ di6n trd didu chinh Rdc cdn thidt mdc ndi tidp tr6n phdn rlng Od dOng cd quay. vdi n = ndm, lrJ = lUdm va tr} thOng giAm di 40 % (1 d).
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oE rnl Hgc ri rY rnuAr slEr'r 2 (s1.12.1eee)Thdi gian ldm bdi : 120 phrtL Sinh vi6n duOc ph6p s0 duno tAi tiQu.