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Background: one of my co-worker found that a utility client is using two CT's (say same CT's 600/5A at the same side of a breaker) in series to supply current to relays. Because this is not a normal practise, they were wondering what is the current going through the relays at norminal situation: 5A or 10A? They cannot agree with each other and then they come to me for advices. 0 l3 _- l* _1 k* X6 L- a: `( }7 P0 @3 z8 ~+ w3 _
My response: Becasue the CT's are in series at primary side, the primary currents of CT's are the same. And the CT ratios are the same. So the seconday side current should be 5A, even they are connected in series at secondary side. You can consider these two CT's are only one CT with primary and secondary side coils doubled. Also I tried to give a reason why the client use two CT's as one: to increase the CT class, say from C400 to C800, to avoid saturation. Even though I think this answer should be acceptable, I still didn't forget to add one word: this is my understanding without any proving. Now I need you help to confirm this answer is correct. 6 }3 q5 l! S& t3 s/ Q) m% Y( {* f2 h" q2 C9 q
I don't think above is difficult because those two CT's are identical. I go another step, making things a little bit challenging. My question is what is the secondary current if two CT's with different ratios, say 1200/5A and 600/5A are connected in series to supply current to relays, assuming the primary current is 1200A? . T' M) S }6 A0 U2 S- i4 g, a
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First, it seems nothing wrong to do it this way. Then you will get 5A for 1200/5A CT and 10A for 600/5A CT. But you cannot have two currents if you put them in series. So what is the current? or they cannot be connected this way? Why?9 K3 r. E2 X2 m; O p9 Q
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Any answer, suggestion or attention are appreciated.
本帖最后由 zxygedi 于 2010-5-3 02:02 编辑 7 D* Q9 l8 ]" a5 l0 a) ^6 i
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This is my opinion: ! ]8 V% l, f0 C" e6 J& U1. Two CTs with the same ratio in series connection can acquire 5A secondary current. Furthermore, the secondary load can be doubled. But I don't think it can avoid saturation.( H7 ^( X* _8 N* B( l9 t2 P
3 D! [& U0 w d4 g$ R" e2. Two CTs with different ratio should not be connected in series! . @9 b$ ]+ b) K$ r6 R2 R' `) U Inorder to explain the problem, I drew the following diagram. 5 M; O# q, {2 Y2 q, @ G% U , F% s+ z- X. S9 X# g
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t: B- e6 P, _We can regard a CT as a Current Control Current Source (CCCS). We know that the ideal current sources can't be connected in series. And the internal impedance of an ideal current source is infinite. Though a CT is not actually an ideal current source, its internal impedance is very large. 2 }4 o! u2 W, w) Z5 e
4 t0 S0 m! Q3 ~In this case, the following parameters is supposed. CT1: 1200/5 A; CT2: 600/5 A; I1 = 1200 A; ZL = 1 ohm; Z1 = Z2 = 10000 ohm. Calculating results show that I6 is approximately 2.5 A. So U2 is 25000 V. It's too high. The insulation of the secondary wire of CT2 will be punctured, I guess. The situation of CT1 is about the same. : p; k/ j [' ] * h; q0 q! Z* L$ F/ uBy the way, for this reason I just said, connecting two CTs with the same ratio in series is not recommended, because their ratio can't be precisely equal. 2 O! z$ t' N9 J; x' D- z6 ?4 T; B5 x( Z2 A
Sorry for my bad English. It's too late. I got go sleeping.
! C7 r4 S' `9 c( r, l, }& S ) @( B1 ], a, F" [2 Q8 l I'll come back to this topic with further question/comment soon. Thanks for your patience. ! [# W$ F# s* f ( d+ `0 Y5 v4 Q0 m+ _1 m& WBy the way can you tell me where the model you used comes from (which paper, book or website)? I would like to understand the model first. I believe the correct model is the key to the questions. Thanks!
回复 3#MonkeyTang8 B# y' q1 k' l, o. i) @# A) N
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The modal came to myself. I don't know whether it is correct or not.
“Two CTs with the same ratio in series connection can acquire 5A secondary current. Furthermore, the secondary load can be doubled. But I don't think it can avoid saturation。”8 E+ a! N7 _6 u# @5 B
相同变比的CT串联后二次负载能力提高了两倍，在相同短路电流的情况下两个CT串联就不容易饱和了。
回复 2#zxygedi ) q8 G' P4 H3 R# R * b' q7 G" n# O$ s/ g- d 3 `" t4 u$ a* b$ B Everything looks good except that the assumption of Z1 = Z2 = 10000 ohms. The calculation is good. I6=2.5A. But also I5 = -2.5A. As I mentioned in my last post, I5 and I6 are approximately the exciation currents (here you ignored the internal impedance. It should be ok). So how can the excitation current is negative? The explanation is that Z1 and Z2 are not fixed when the exciatation current is changing. The question now are how the excitation impedances Z1 and Z2 changing?
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狗仔卡
发表于 2010-5-2 12:47:26
提升卡
置顶卡
沉默卡
喧嚣卡
变色卡
显身卡
). 这些有待考证研究.