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Starter motor current draw

  1. Jul 20, 2017 #1
    hello, if i use a 12v 1.2kw starter motor on 24v. how the power will change and how much current will it consume.
     
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  3. Jul 20, 2017 #2

    Averagesupernova

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    That depends on how fast you allow it to spin up. I can tell you from experience that 6 volt starters have been used for years on systems that were updated to 12 volts.
     
  4. Jul 20, 2017 #3

    CWatters

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    1.2kW will be the maximum power the starter can deliver (under some conditions specified by the maker). The actual power and current drawn will depend on the load (how difficult it is for the engine to turn over).

    A 12V starter on 24V will probably try to turn around twice as fast. How much power it takes to turn the load twice as fast is anyone's guess. Could be anywhere from double to four times the power. Say 2.4kW to 4.8kW. That would equate to current of 100 to 200A.

    Google will tell you what might happen..

    https://www.steelsoldiers.com/showthread.php?31476-Converting-a-12v-starter-to-24v

    https://www.steelsoldiers.com/showt...NEED-HELP!&s=bf043bc6013ac56b871864ed9e25743d

     
  5. Jul 31, 2017 #4
    In my opinion, if the starter it is an D.C. motor provided with permanent magnet poles then at start-0 rpm-the current will be double than at 12 V but the torque will be also double and then the starting time will be a half of the previous. In this case the temperature will rise from-let's say 100oC to 150oC. It depends on how many times per an hour you need to start it and if the pause duration will be enough to get cooling
     
  6. Jul 31, 2017 #5

    jim hardy

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    Here's mt simplistic thinking:

    Consider a series wound starter at stall:

    Current with 24 volts applied will be twice what it would be with 12 volts applied.
    So armature current and field current(they're one and the same) will both get doubled.
    Doubling field current will about double flux.
    Locked rotor torque , being product of armature current and flux (and a proportionality constant) , will quadruple.
    So,
    Initial power input to the starter motor also quadruples (2X volts and 2X amps) and the engine being started will be accelerated by the starter noticeably more quickly.


    Since torque of a series machine is proportional to square of current through it
    If we then consider the engine being started to be a constant torque load,
    24V cranking speed will settle at whatever RPM gives same current as 12V cranking speeed, probably just a little over double RPM .

    Same current X twice the volts = twice the power delivered to starter .

    Power at any other RPM will fall someplace between 2X of a stalled permanent magnet machine and 4X of a stalled series machine.

    Actual number of course depends largely on the condition of your battery, cables and starter brushes.

    old jim
     
  7. Jul 31, 2017 #6
    For starting an IC engine, in most cases, I don't think the starter motor ever reaches steady state, unless the IC engine is balky. If the IC engine starts promptly, the starter motor is likely still accelerating when the greater speed of the IC engine causes it to disengage.

    Old Jim hit on the key point, I think, when he observed that the torque is proportional to the square of the current through the motor. The motor impedance is the same at a particular speed, no matter what the applied voltage, so the current should double. This would give four times the electromagnetic torque and correspondingly much faster acceleration. It will also give rise to much more rapid heating, which is also proportional to the square of the current.
     
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