The ratio question of CT's in series
Need helps: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.
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.
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?
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?
Any answer, suggestion or attention are appreciated. 本帖最后由 zxygedi 于 2010-5-3 02:02 编辑
This is my opinion:
1. 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.
2. Two CTs with different ratio should not be connected in series!
Inorder to explain the problem, I drew the following diagram.
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.
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.
By 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.
Sorry for my bad English. It's too late. I got go sleeping. 本帖最后由 MonkeyTang 于 2010-5-3 08:46 编辑
回复 2# zxygedi
I'll come back to this topic with further question/comment soon. Thanks for your patience.
By 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# MonkeyTang
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。”
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 5# zy7239
两个CT串联为啥就不容易饱和还是想不明白。我最近正打算研究一下《电流互感器和电压互感器选择及计算导则》，对CT的了解太肤浅了，不行啊。
我解释一下,不知是否能解释清楚：
1、CT的二次电压U=I2*ZL
2.CT的饱和时它的磁通密度B大于某个值，一次侧电流无法正确传变到二次侧。
3、B=(I2*ZL)/(4.44fAN),从公式中可看出，若在B\I2\A\N等条件确定的情况下,CT二次负载ZL也就确定了。
4、当两个CT串联时，那不就是他们的负载可以承受2ZL。 谢谢楼上两位的参与. 到目前为止, 我们基本上可以断定: 同规格同变比的CT串联后可以提高抗饱和能力至2倍(约). 这点现场应用及zy7239的计算都说明了这点. 不过有不同看法的仍可以提出来讨论.
至于不同变比的CT的串联, 我觉得还得从变压器的基本原理入手, 搞了几天也没个结果. 不过参照一本介绍CT原理及应用的书, 觉得zxygedi的模型很有道理, 模型中的Z1及Z2可理解为励磁阻抗, 可变, I3及I4为恒流源, 取决于变比及一次侧电流. 变比不同时它们也不同, 但继电器电流I2同时流过两CT. 所以恒流源的差异只能由流过励磁阻抗的励磁电流I5及I6来平衡.
基于上述分析, 不同变比的CT的串联问题可以归结到励磁电流的作用及影响. 励磁电流I5及I6的正常值范围是多少? 这种情况下它们会不会超出正常值范围? 会不会引起CT饱和? 猜想很可能会. 猜测理由是对偶原理: 不同变比的PT不能并接, 否则会引起很大的环流及烧毁PT; 同理, 不同变比的CT不能串接, 否则..., (产生高的环压?::smile::). 这些有待考证研究. 回复 2# zxygedi
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? 回复 6# zxygedi
对于一个C400类CT, 其额定电流是5安培, 额定负载是4欧姆, 当流过CT的电流为额定值的20倍时, CT输出电压应为400伏. 用实际短路电流, 继电器及二次侧导线电阻校核时, CT输出电压不能高于400伏, 否则有饱和的可能.
两个相同的CT串联后, 这个电压将是两个CT的输出电压, 各个CT将只承担其一半, 也就是如果输出电压是400V的话, 一个CT就会饱和, 而两个CT时, 一个CT只承担了200V, 仅是饱和值的一半. 这是从应用角度解释了两个CT串联后为何不易饱和的原因.
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