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  1. #1
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    Default Question about currents - idle, load, inrush

    Hi guys,

    Got a hydraulic guillotine in at work that we had to run to ensure it works before sending it out to the customer, and a few things created a bit of discussion in the office (that I'd really like to come in and say "Eureka! This is the reason why!" haha).
    Before anyone frets, all the electrical work was performed by a licensed and competent electrician, though I'm not sure if this particular line is his forte (or he just wasn't in a mood to try explaining anything).

    It's a decent unit, good for 12mm steel. In the course of getting it running, we had to hire a 100kva genset as the building power isn't enough to handle the required draw.
    On startup, the sparkie's clamp meter read about 50a inrush, with room for error (apparently it's a bit of a slow unit), though this does indicate the written requirement of a 63a circuit is certainly not kidding.
    Anyway, after the inrush, the idling draw was 10a, so motor running but hydraulics not engaged.

    Now, the bit where I'm hoping to get some clarification - unfortunately when we got to the stage of pumping the oil/giving it a proper run, the sparky was long gone and hence could not provide us with the handy clamp reading to let us know what the amp draw under load was. And thus, here I am.

    I grabbed some details off the plate on the motor (I really am too keen to be a smartypants), specs are 420v 50Hz 15kW cos0.86 A26

    Bits and pieces I've picked up over time has suggested a loose rule of thumb of inrush current being around 3 to 7 times the load current, which to me seems right if the idle load is the one to go by (the 10a reading), but there's the 26a reading on the motor plate and I'm just really out of my depth.

    This is a very rare size of equipment for us, and I'd love to sate my curiosity whilst it remains. Also, whether a 100kva generator was suitable or vast overkill

    Thanks in advance, and also apologies if I've done my thing of structuring my sentences a bit odd

  2. #2
    BobL is offline Member: Blue and white apron brigade
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    Default

    50A inrush for a 15kW motor sounds light on to me.
    I see that or more with a 3HP dust extractor, but that is of course under load while starting.
    It also depends on the response action time of the clamp meter if it was too slow it might not catch the virus.

  3. #3
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    The equation for 3 phase power is:
    P = √(3) * V * I * p.f

    Rearranging it gives:
    I = P / (√(3) * V * p.f)
    I = 15000 / (√(3) * 420 * 0.86) = 23.98 Amps.

    That's in theory. The reality is that how much it draws at full load in that machine depends on a number of factors - supply voltage and actual mechanical load being the main ones. Its quite possible in that machine, it may only ever draw say 20A FLC, because it has been oversized for the application. Its also possible, if someone puts a full width piece of 16mm plate in that machine, that the motor draws in excess of 30A (or more). Whether the machine will physically break is another matter, but assuming it doesn't, the motor is quite capable of delivering that current for a period of time. Not at all good for it though, as it will overheat the windings. A longer run at that amperage, or repeated short bursts at that load will result in a destroyed motor (and if its a crappy motor, that time frame becomes a lot shorter!)

    I'd agree with Bob, 50A inrush at first glance sounds a bit low - but like him I note you say the motor can run without the hydraulics going, which means that the inrush current is only used to get the rotor in the motor going. That may cause a lower figure than if the motor was started from stop with a load connected (ie. hydraulics going, or a big fan etc), and certainly will cause the peak current figure to be reached and then drop back to the running figure much faster. The other factor on inrush current is whether that motor starts DOL (direct online) or has for example a star/delta starting arrangment, or a VFD etc - its getting towards the size of motor where you might expect to start seeing those arrangements.

    In fairly vague terms, not intended to be 100% accurate but to attempt to convey the concept (as I can't exactly remember the specifics, and a quick google has been unproductive). I'm sure someone will be along to correct me. The torque (and thus current draw) produced by an electric motor are kind of determined by whats called 'slip' - the difference between the speed of the rotor poles (output shaft, effectively) and the speed of the rotating magnetic field in the stator (generated by the electricity you supply). The magnetic field will ALWAYS rotate at just slightly higher than the motor plate rpm spec, and the poles of the rotor are dragged around by this field. There will always be a small amount of slip which means that even unloaded, the rotor never quite matches the speed of the field. The rotor can not possibly ever match the speed of the magnetic field, as if it was moving at the same speed as the field, there would be no magnetic pull applied to the rotor.

    When you start up a motor that is stopped, the rotor is obviously stationary, but the magnetic field is instantly rotating around at the rated rpm. This means that the magnetic fields pull on the rotor (and vice versa) at this time is enormous, and it pulls a massive amount of current in order to get the rotor to match speed to the magnetic field, which drops off as the rotors speed approaches the speed of the magnetic field and decreases the magnetic pull.

    The same effect occurs when you have a motor running without load, like your machine. The rotor is happily spinning away at the speed of the magnetic field, and then you hit the button to engage the hydraulics. Suddenly you've effectively applied the brakes to the rotor, and it begins to slow down. As it slows down, the amount of slip increases, in turn increasing the magnetic pull between the rotor and field, and increasing current. The point of all this gibberish, is that when a motor is idling with no load, it might be running at (number plucked out of air) 0.2% slip,and with a current to suit. If you then apply a constant load to that motor, the amount of slip increases, say to 3% (again, number plucked out of air to demonstrate concept), and the current required to sustain the magnetic field likewise increases.

    For any given load (including idle!), the amount of slip for that load will never actually vary. At the previous example of 3% slip, there is a given amount of magnetic pull present, which is what's required to drive whatever is connected to the motor. If the rotor caught up to the magnetic field, then the magnetic pull would be less, and insufficient to actually drive the connected load - thus the rotor speed would drop. Likewise, if the rotor speed drops with that given load, the magnetic field increases, accelerates the load, and catches back up...

  4. #4
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    The magnetic field will ALWAYS rotate at just slightly higher than the motor plate rpm spec,
    This means that the rotor will keep falling back until it moves back into another pole (?) and keep dropping back poles?

    The rotor can not possibly ever match the speed of the magnetic field, as if it was moving at the same speed as the field, there would be no magnetic pull applied to the rotor.
    This is something I have wondered about.

    Dean

  5. #5
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    Quote Originally Posted by Oldneweng View Post
    This means that the rotor will keep falling back until it moves back into another pole (?) and keep dropping back poles?
    Basically, yes. It's the difference between the nameplate speed (say 1425 rpm) and the theoretical speed ((2 x frequency x 60 seconds/minute)/no. of motor poles. 1500rpm in this case)

    At a previous place we had a 3 phase motor on a hydraulic pump that steady state would draw 235A. Start up current was around 850A.

    Michael

  6. #6
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    That is some excellent information thank you Jekyll and Hyde

    Now, a second question to throw out there - When I got to work yesterday, I noted the plate I missed the other day, which states that the total trip current is 37.02A, and the fusing current is 63A.
    Which of those values is the one that determines the size of the breaker/outlet required? Again, I wasn't present for the discussion/decision process, but the fusing value is the reason why the large (I felt it to be oversized) genset was hired, but after seeing that plate I'm tentatively theorising (stabbing in the dark) it relates to the fuse size required between the supply poles and the power board?

    Thanks again

  7. #7
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    I'm not really sure what that second plate is referring to exactly, is it attached to the machine or the motor? And are there other electrical items running in the machine beyond the motor discussed?

    My gut instinct here is that for the motor alone, on the date from the original post, a 32A D-curve breaker should be all that's needed... Looking at http://www.hagerelectro.com.au/files...HINFO_MCBS.PDF on page 4, a 32A D curve breaker at 100A would take between 3 and 10 seconds to trip, at 150A would take between 1.5 and 4 seconds to trip, and at 200A between 1 to 3 seconds. Should allow enough time for the inrush current to be over and done with (after all, that is the idea of D curve breakers!). Happy to be corrected on this though, not exactly my area of expertise.

    Generator sizing is also not something I know much about, but the brief investigations I've done into it previous (mostly thinking of running welders) seems to suggest you often 'should' use a generator far bigger than you expected. How much bigger seems to depend greatly on where you look

  8. #8
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    I think of it this way - the current has to come from somewhere. Even if the steady state draw is only say 5A, if the starting current is say 17A, that has to come from somewhere. On mains, the line will supply that and because of the delays associated with fusing, the breaker will almost but not quite go. On a generator though there is no reserve, so the power (current) would have to be there.

    Michael

  9. #9
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    sounds like a 63a circuit is the go.
    If we assume that the control system of the gillo has a motor protection then a 63a breaker should be OK..but it can be 80A providing cabling is suitable. As Mr Jeckle says and providing no motor protection is built in to existing control system then fit a D curve breaker for motor protection.

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