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

tomek1944 15171 17
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Kiedy należy uwzględniać prąd rozruchowy przy doborze kabla i zabezpieczenia dla silnika trójfazowego z rozruchem bezpośrednim?

Prąd rozruchowy należy uwzględniać zawsze przy doborze zabezpieczenia i kabla silnika, bo zabezpieczenie ma chronić przewód, a nie tylko pracę silnika. Startowy współczynnik jest cechą silnika podaną w danych katalogowych i prąd rozruchowy liczy się jako prąd znamionowy × współczynnik rozruchu; dla przykładu silnik 5,5 kW, 10 A i współczynnik 7,3 daje około 73 A przy starcie [#16859221][#16859363][#16858371] Trzeba też brać pod uwagę rodzaj napędu i częstotliwość startów, bo to decyduje o czasie rozruchu i obciążeniu całej instalacji [#16859156][#16859221] Długość odcinka kabla ma znaczenie, bo im dłuższy przewód, tym większe wydzielanie ciepła [#16857243]
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  • #1 16857221
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    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.
    I put on a Bm 10 A and choose a cable that will last 10 A.
    Suppose 1 mm 2 .
    In the book by Mr. Markiewicz there was such a relationship that the fuse * 1.6 is to be smaller than the short-term load capacity of the cable, i.e. 145 percent of the long-term load capacity:
    Let us assume that our cable has a long-term load capacity of 10 A , i.e. 1.6x10
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  • #2 16857243
    zdzichra
    Level 32  
    Posts: 1531
    Help: 167
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    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.
  • #3 16857257
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    zdzichra wrote:
    First of all, it depends on how long this episode is. The longer the episode, the more heat is released on it. The recommendations are general, no one plays in detailed cases. The guidelines are made for it to work.


    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
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  • #4 16857309
    zdzichra
    Level 32  
    Posts: 1531
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    Rate: 303
    tomek1944 wrote:
    Do I always have to fulfill the relationship (bm * 1.6
  • #5 16857323
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    zdzichra wrote:
    tomek1944 wrote:
    Do I always have to fulfill the relationship (bm * 1.6
  • #6 16857386
    zdzichra
    Level 32  
    Posts: 1531
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    Yes, because the purpose of the cartridge is to protect the cable
  • #7 16857416
    Strumien swiadomosci swia
    Level 43  
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    tomek1944 wrote:
    And now, when I attach it once a day, it is known that I will choose a thin wire, even this 1 mm2, and if I would attach it every 2 minutes, what then?


    This will count the average current for the period.
  • #8 16857450
    zdzichra
    Level 32  
    Posts: 1531
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    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 :D
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  • #9 16857756
    Miniax
    Electrician specialist  
    Posts: 1438
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    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:

    http://www.zue.pwr.wroc.pl/download/lab_urzadzen/12.pdf


    zdzichra wrote:
    Yes, and in a moment it will turn out that this engine needs to be turned on more often and what? New wires? 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 burnt wires or anything. A little more expensive (thicker wires) but a good night's sleep - priceless :D


    A colleague seriously for the 5.5 kW engine uses 25 mm ^ 2 cables?
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  • #10 16858200
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    [quote = "Miniax"] 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:

    http://www.zue.pwr.wroc.pl/download/lab_urzadzen/12.pdf

    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.
  • #11 16858224
    Krzysztof Kamienski
    Level 43  
    Posts: 21874
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    tomek1944 wrote:
    We are talking about a 5 kW engine all the time, it will not have a heavy start.
    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.
  • #12 16858371
    Anonymous
    Anonymous  
  • #13 16858686
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    wieslaw.siczek wrote:
    tomek1944 wrote:
    The motor will draw 9 amps rated but start amps more.
    Let's take 6 times a factor of 4.
    So 13.5 amps inrush current.


    tomek1944 wrote:
    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

    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 ...

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

    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?




    And did you forget about the starting rate?
    Added after 3 [minutes]:
    Krzysztof Kamienski wrote:
    tomek1944 wrote:
    We are talking about a 5 kW engine all the time, it will not have a heavy start.
    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.



    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
  • #14 16859156
    djlukas
    Level 27  
    Posts: 1606
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    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.
  • #15 16859159
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    djlukas wrote:
    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.


    So when to consider starting up, always?
  • #16 16859221
    Miniax
    Electrician specialist  
    Posts: 1438
    Help: 173
    Rate: 289
    tomek1944 wrote:

    Since when does the fuse protect the engine? And how do I put on a delayed cartridge that is supposed to burn?


    Always ;)

    tomek1944 wrote:

    I've seen this reading.


    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?
    tomek1944 wrote:

    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.


    A colleague does not understand the definition of the inrush rate.

    The starting factor is a characteristic of the engine given in its specification:

    http://www.motors.celma.pl/Offer/2SIE112M2A,1525

    For example, this engine has a start-up rate of 9, therefore it consumes more than 93 A when starting.

    tomek1944 wrote:

    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.


    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.
    tomek1944 wrote:

    We are talking about a 5 kW engine all the time, it will not have a heavy 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.

    tomek1944 wrote:


    So when to consider starting up, always?


    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.
  • #17 16859298
    tomek1944
    Level 7  
    Posts: 9
    Rate: 2
    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.
  • #18 16859363
    Miniax
    Electrician specialist  
    Posts: 1438
    Help: 173
    Rate: 289
    tomek1944 wrote:
    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.


    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.
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Topic summary

✨ The discussion revolves around the considerations for inrush current when selecting cables for a direct start 5 kW, 3-phase motor operating at 400V with a power factor of 0.8. The initial query highlights the calculated inrush current of 13.5 A based on a starting factor of 4. Forum members debate the importance of accounting for inrush current in cable selection, emphasizing that the cable must withstand both long-term and short-term load capacities. The conversation also touches on the significance of the motor's starting characteristics, the frequency of starts, and the appropriate fuse ratings to protect the motor and cables. Various opinions are shared regarding the adequacy of a 1 mm² cable for this application, with some suggesting thicker cables for safety and reliability.
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FAQ

TL;DR: A typical 5.5 kW, 400 V motor can draw about 73 A at direct-on-line start (≈7.3× rated). “Rated current × inrush factor = inrush current.” [Elektroda, 3065572, post #16858371]

Why it matters:** Correctly accounting for inrush prevents nuisance trips, overheated cables, and burned starters in real plants. This FAQ is for electricians, designers, and maintainers sizing protection and wiring for 3‑phase motors.

Quick Facts

When do I need to consider inrush current for a three‑phase motor?

Always include inrush. It dictates cable heating, voltage drop, and whether protective devices trip during starts. One expert notes that the start current is fixed for a given motor, while start time varies with the load and frequency of starts. Ignoring either leads to nuisance trips or overheated wiring during acceleration. Plan for the motor’s starting duty, not just its steady state. [Elektroda, Miniax, post #16859221]

How do I estimate inrush current from nameplate data?

Use the motor’s rated current multiplied by its inrush (starting) factor. Quote: “Rated current × inrush factor = inrush current.” If the rated current is 10 A and the factor is 7.3, expect about 73 A at a DOL start. This current persists near its peak until roughly half synchronous speed, then decays as the rotor accelerates. [Elektroda, 3065572, post #16858371]

What is the inrush (starting) factor?

It is the ratio of starting current to rated current and is a motor characteristic from its design. In practice, you multiply rated current by this factor to get the start current. Expert reminder: “The starting factor is the ratio of the starting current to the rated current.” Use the manufacturer’s datasheet when available. [Elektroda, Miniax, post #16859363]

Does cable sizing depend on start frequency or just steady load?

It depends on both. Frequent or heavy starts raise thermal stress and can exceed a cable’s short‑time capacity even if steady current is within limits. Designers account for how often the machine starts and how hard it accelerates to avoid overheating or voltage sag that stalls the motor. [Elektroda, Miniax, post #16859221]

Can I size the cable by averaging current over time?

No. Averaging hides thermal and magnetic stresses from start pulses. Short, high currents still heat copper and insulation and can trigger protection. A forum expert cautions that treating starts as mere averages is misleading; start bursts matter. [Elektroda, Strumien swiadomosci swia, post #16857416]

What protects the motor vs. what protects the cable?

Short‑circuit protection (fuse/MCB) must protect the cable first. Overload protection (thermal relay or protective function) protects the motor against prolonged overcurrent. Don’t expect a small breaker to survive repeated DOL starts without proper coordination. [Elektroda, zdzichra, post #16857386]

Will a 10 A breaker survive starting a 5 kW motor?

Unlikely with DOL. One practitioner warns that a small B‑curve device may fail after a few starts due to high inrush. Expect nuisance trips or damage unless you coordinate curve, rating, and motor inrush or use soft‑starting. [Elektroda, Miniax, post #16857756]

Is 1 mm² cable acceptable for a 5–5.5 kW DOL motor?

No. With ≈73 A inrush typical for 5.5 kW, 1 mm² is undersized. It risks voltage drop, overheating, stalled starts, and insulation damage. The forum challenges such sizing directly in context of a 5.5 kW, 400 V example. Choose a larger cross‑section and verify thermal/voltage drop. [Elektroda, 3065572, post #16858371]

How long does high inrush last during DOL starting?

Peak current persists until about half speed, then it ramps down as torque rises and slip falls. For a multi‑second start, the cable and protection must endure near‑peak current for a significant fraction of that time. Coordinate trip curves accordingly. [Elektroda, 3065572, post #16858371]

Does motor size alone tell me if the start is heavy or light?

No. The driven machine defines the starting load. Pumps, fans, and compressors have different torque‑speed profiles. Even small motors can have heavy starts with unfavorable gearing or load inertia. Evaluate the mechanical load, not just kW. [Elektroda, Krzysztof Kamienski, post #16858224]

Quick how‑to: estimate DOL inrush and pick protection?

  1. Read rated current from the motor plate.
  2. Multiply by the manufacturer’s inrush factor to get start current.
  3. Select protection that won’t trip on inrush and still protects the cable; then size the cable for thermal stress and voltage drop. [Elektroda, Miniax, post #16859363]

What happens if I ignore inrush when selecting fuses?

You risk immediate trips or cumulative damage. A practitioner notes small B devices can fail after only a few DOL starts. Replacements and downtime cost more than correct coordination and proper cable sizing. [Elektroda, Miniax, post #16857756]

How does start frequency affect protection coordination?

More starts per hour raise thermal accumulation in both cable and protective devices, lengthening cool‑down times. You must consider the ‘severity’ of starts and their frequency to set suitable device curves and sizes. [Elektroda, Miniax, post #16859221]

Why do some designers size cables “against the starting current”?

They aim to avoid future rework when processes demand more frequent starts. Slightly larger conductors reduce thermal stress and voltage drop, supporting reliable operation if duty intensifies later. [Elektroda, zdzichra, post #16857450]

Is there a simple rule to compute inrush for a squirrel‑cage motor?

Yes. Inrush current equals rated current times the motor’s starting factor. This succinct relation is repeatedly emphasized by experts in the thread and aligns with practical measurement. [Elektroda, Miniax, post #16859363]
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