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My Solenoid heated more than it should have

  1. Aug 8, 2017 #1
    physfoum.jpg
    Please help me understand. Attached is a pic of some quick calculations
    My solenoids self inductance is 0.145mH. The coil is 117 grams of copper. I calculated with 300 amps on the picture but I only put 170A through it the other day. I had it hooked up for 2 seconds touching it at the same time, it got luke warm just from the current running through P=I^2R but then I unhooked it and as soon as I disconnected the coil interestingly it went from luke warm to insanely hot from the magnetic field collapsing and cycling through the coil. According to the math I did in the picture the energy from the magnetic field should've been way under 6.5 Joules which wouldn't raise the temperature much for 117 grams of copper according to its specific heat capacity. But it felt like it went from 35C to 90 plus C. How do I explain this because the math doesn't explain it here.
     
  2. jcsd
  3. Aug 8, 2017 #2
    Thinking about it more the heat must've been just hiding underneath the 5th layer and took a while to show itself which made me think it came from the collapsed magnetic field because it was delayed but in actuality it was from the current running through not the magnetic field collapse.
     
  4. Aug 8, 2017 #3

    rbelli1

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    That and for very fast temperature changes you can get quite a lot of ΔT at the surface of your fingers before you actually feel the heat.

    BoB
     
  5. Aug 9, 2017 #4

    davenn

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    would love to know how you got those figures
    You DONT put xxx Amps of current through anything, that isn't the way it works
    A load will draw xxx Amps when a given voltage is applied to a given resistance ( of the load)

    1) ... what voltage did you put across the coil ?
    2) ... what resistance is the coil ?
    3) ... what is the wire gauge used to wind the coil

    I see none of this info presented
     
  6. Aug 10, 2017 #5
    A little over 26 Volts, resistance of the coil is 0.12 ohms, the IGBT shouldn't have added a lot to this. AWG 16 wound in 5 layers across 4 cm solenoid, could've wound it better so only got 145 turns. 170A was a conservative figure, I actually thought my resistance was higher but i measured it by running 0.3 Volts through the coil and seeing how many amps ran through with my power supply and it came out to 0.12 ohm so if that is accurate I was running over 200 amps.
     
  7. Aug 10, 2017 #6

    davenn

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    and that only applies if your 26V source can actually supply that current
    170A running through 16AWG wire would fuse the wire in seconds, so this assumes that the current is very much lower

    you also didn't state what the the IGBT type number was and also I doubt that it could pass 170A

    so, where is you 26V being sourced from ?
    and what is the the IGBT ?

    ohhh and you don't run voltage through anything :wink:


    Dave
     
  8. Aug 10, 2017 #7
    They can supply the current, Gens Aces Lipo 4400mAh 60C constant discharge 120C peak. It's for a coilgun so I'm only trying to light up for a few ms. This time it went wrong but I disconnected pretty quickly between 1 to 2 seconds. I don't remember the IGBT number but it was rated high enough. I'll be going with something even beefier next an SOT-227 packaging MOSFET rated for 480A next and gonna use a heatsink for it too this time.
    The copper magnet wire is rated for 155C but I was getting worried it got pretty hot.
    And the voltage yeah believe me I ran that sucker through with my own hands lol
     
  9. Aug 11, 2017 at 10:54 AM #8
    What does matter on temperature rise it is not the magnetic field energy but heat losses. For 2 seconds the loss heat will produce only copper conductor temperature rise since the heat evacuation is negligible[adiabatic phenomenon].
    So I^2*Ro(1+a*(T-To))*dt=TCAP*Vol*dT
    TCAP is the thermal capacity per unit volume , in J/(cm3 · °C) TCAP=3.4 for copper.
    I^2*Ro*dt=TCAP*Vol*dT/(1+a*(T-To)) and by integrating from T=Ti to Tf
    we get[eventually]:
    Tf=(exp(Ro*I^2/TCAP/Vol*t*a)*(1+a*(Ti-To))-1)+To
    If Ro=0.12 ohm at To=20 oC and Ti=ambient temp.=30 a=0.00393
    Tf=2270 oC for 300 A and 310 oC for 170 A [2 sec.]
    You may use the IEEE 80/2013 formula 37 in order to check the current:
    I=Amm^2*sqrt(TCAP/10^4/tc/ar/ror*ln((Ko+Tm)/(Ko+Ta)))
    tc=2 sec ar=0.00393 ror=1.724 Ko=234 Ta=30 oC and Tm=Tf
    Amm^2=1.31 mm^2 [16 awg].
     
  10. Aug 11, 2017 at 11:41 AM #9

    jim hardy

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    No, it got insanely hot as the heat produced inside the wire conducted itself out through the insulation arriving where you can feel it.


    Back of the envelope calculation: (Babadaq's formula is a lot better, this method gives just a hobby level sanity check)
    Using your current number ,
    170 amps 2 X 0.12 ohms = 3468 Joules per second
    and using your voltage number
    26 volts 2 / 0.12 ohms = 5633 Joules per second.
    How fast does that heat up your mass of copper ?

    I'd unwind the coil and see if the insulation near the center has melted.

    BTW that's quite a battery.
    upload_2017-8-11_11-58-49.png
    http://www.gensaceusa.com/98p-60c-4400-6s1p-6666.html
     
  11. Aug 11, 2017 at 3:12 PM #10
    However, if the voltage is limited to 26 V d.c. then the current will be 26/R(temp)-in my opinion not more than 115A since the resistance rises to 0.224 ohm at 262 oC
     
  12. Aug 11, 2017 at 6:06 PM #11

    jim hardy

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    The writing in his picture is too dim for me to read. So i didn't know what he was calculating - temperature, power, or resistance .
    Your differential equation is the scholarly answer to his dilemma. Nice Job !
    My simple algebra is just a first step that might get an amateur started in the right direction.

    old jim
     
  13. Aug 11, 2017 at 11:47 PM #12
    Yeah that battery is a beast huh hehe got it for half off too. I actually have a second one, was planning to run them in series haha. I'll probably limit the current to the first coil though as it's gonna be on for the longest. This is meant for a multi stage coil gun btw.

    As for the heat melting the insulation it should be fine. The guy who brought up the variable resistance with rising temperature has a good point. It's too late for hard math right now lol so I just pulled 4000J out of the air per second a little less than if the resistance was at room temperature. Given my 117g coil and heat capacity of cu of 0.385 that would give a temp rise for the coil of 88 degrees per second. Could've had that sustain for 1.8 seconds until it would get to the copper magnet wire's limit of 155 C. I think there's magnet wire rated for close to 200 that I've seen, should probably upgrade to that although the coil wasn't meant to stay on that long.
     
    Last edited: Aug 11, 2017 at 11:56 PM
  14. Aug 12, 2017 at 9:29 AM #13

    jim hardy

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    There exists glass insulated high temperature wire for aerospace and steel mill applications.
     
  15. Aug 15, 2017 at 1:49 PM #14

    sophiecentaur

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    Make sure that you keep count of the Ahrs you are taking from it for these experiments or you can damage it and lose its capacity. Batteries shouldn't be taken below HALF their nominal capacity. (take no more than 10Ahr from a 20Ahr battery) Manufacturers are naughty in the way the specify their batteries. Your "Beast" could soon be a bit feeble if you aren't nice to it. :frown:
     
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