When is Inrush Current Considered? Direct Start Engine, 5kW, 3 Phase, 0.8 Factor, 400V, Bm 10A

Question

Hello to all forum members! I have a question like this: We have a direct start engine and some current consumed there: Let it be 5 kW. 3 phase. factor 0.8. U-400V The engine will take 9 and mper that rating plate , a le boot more. Let's take 6 times a factor of 4. So 13.5 and inrush current...

zdzichra · Answer

First of all, it depends on how long this section of duct is. The longer the section, the more heat is released on it. The recommendations are general, no one plays in specific cases.

tomek1944 · Answer

I do not know how long, I want to know when I, as an ordinary electrician, should take into account the starting current when choosing a cable. I just don't know what it depends on, engine switching cycles. ? Do I always have to fulfill the relationship (bm * 1.6

zdzichra · Answer

Yes, because the purpose of the cartridge is to protect the cable

Strumien swiadomosci swia · Answer

This will count the average current for the period.

zdzichra · Answer

Yes, and in a moment it will come out that this motor needs to be turned on more often and what? New cables? I always count the cables against the starting current, I take it as a long-term load and I know that the topic will not come back in the form of burned wires or anything else. A bit more expensive (thicker cables) but restful sleep - priceless

Miniax · Answer

Interesting to me, but where did my colleague get the starting current of 13.5 A for a 5 kW motor with direct starting? Just as I do not understand how the author wants to protect this engine with such a fuse, at best, after a few starts, such BMa will burn out. First of all, you need to know what the motor overload protection is and what is the short-circuit protection ... However, when it comes to the frequency of starts, the author should take into account whether the start-up is light or heavy, because it has a significant impact on the load on the cables and protections. I recommend that you read the document, these are the basics: A colleague seriously for the 5.5 kW engine uses 25 mm ^ 2 cables?

tomek1944 · Answer

Interestingly enough, where did my colleague get the starting current of 13.5 A for a 5 kW motor with direct starting? Just as I do not understand how the author wants to protect this engine with such a fuse, at best, after a few starts, such BMa will burn out. First of all, you need to know what the motor overload protection is and what is the short-circuit protection ... However, when it comes to the frequency of starts, the author should take into account whether the start-up is light or heavy, because it has a significant impact on the load on the cables and protections. I recommend that you read the document, these are the basics: Since when does the fuse protect the engine? And how do I put on a delayed cartridge that is supposed to burn? I've seen this reading. If we have the rated current of this motor 9A, how do we calculate the starting current? I think the rated current times the times (let it be an induction motor 6-8) By a starting factor of 2-4 depending on the size of the engine - but 5 kW is a motor. And I count the starting current. 9 * 6/4 = 13.5 Amps I choose Fuses higher degree, i.e. 16, and I choose the cable. How can you move the engine every now and then with heavy starting? Dtr prohibits it. In many cases, in high-power engines, starting is given every hour as many times as possible. We are talking about a 5 kW engine all the time, it will not have a heavy start.

Krzysztof Kamienski · Answer

Who said that? Automatic washmachine ? Is a mobile crane? The word is not about the nature of the driving machine, that is, the nature of the load mechanical - pump, fan, compressor or other devil? Buddy, even an engine in a toy car may start hard or not, depending on the construction of the transmission and the power selected for the weight of the vehicle. Engine power has nothing to do with it. as you do so far.

Anonymous · Answer

9A * kr / kr =? Typical motor 5.5kW 1440rpm, rated current 10A at 400V, starting multiplicity of 7.3, starting current 73A ... This value flows through the windings until the rotor reaches half speed, around 720rpm, only after that the current value starts to decrease ... If you accelerate something for 5 seconds, then 70A flows for at least half of this time ... These are to be endured by the selected protections and cables. 1mm2 to 5.5kW?

tomek1944 · Answer

And did you forget about the starting rate? Added after 3 [minutes]: We make the start-up dependent on the machine. I write about the engine, you about the machine. The electrode is a forum of scientists themselves and I am asking for a simple thing, even the formula is not known to calculate the start-up. The engine itself is simple that it will have a light start. And I'm not asking for the type of boot. only when you take this current into account whether the motor is turned on once a day or every 20 minutes, you know that I will not choose the same wire

djlukas · Answer

My friend, the specificity of the drive, i.e. what and when to be driven, tells us how often the start-up will be made and we secure the entire installation in this respect.

It is normal that when the specifications of the machine come out that it will be turned on once a day, one start-up is a problem for us. I'm not talking about the voltage drop and the motor getting stuck with too thin wires.

tomek1944 · Answer

So when to consider starting up, always?

Miniax · Answer

Always ;) Seeing is not understanding. If a colleague had understood, this topic would not have arisen - the appendix I provided earlier contains all the necessary information on: overload protection, short-circuit protection and selection of wires. What more information does your colleague need? A colleague does not understand the definition of the inrush rate. The starting factor is a characteristic of the engine given in its specification: For example, this engine has a start-up rate of 9, therefore it consumes more than 93 A when starting. The phrase every now and then is a relative concept. In the document I attached earlier, the alpha factor is given, depending on the frequency and severity of the start. I can see that my friend has information about the new laws of physics, because I have never heard that a small engine could not have a hard start. Yes, but it is necessary to know the type of drive operation and the frequency of starts, because while the starting current for a given motor is always the same, the starting time may be different depending on its operating conditions, and that is important.

tomek1944 · Answer

For example, this engine has a start-up rate of 9, therefore it consumes more than 93 A when starting. I don't understand why you don't divide this current by the inrush factor.

Miniax · Answer

I do not understand where the colleague got the division by the starting factor? As already mentioned, the starting factor is the ratio of the starting current to the rated current of the motor and it is a design feature of the motor. There is no division there. The starting current for a given motor is ALWAYS the same and it is exactly for a squirrel-cage motor: rated current x inrush factor = inrush current. This is the simplest possible formula. Even one of my colleagues had previously uploaded a nice current-torque graph during start-up. If a colleague does not believe it, I recommend the oscilloscope, motor and measurement.

When is Inrush Current Considered? Direct Start Engine, 5kW, 3 Phase, 0.8 Factor, 400V, Bm 10A

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