System Protection - compute for fuse size

Discussion in 'General Electronics Chat' started by ritchell77, Jun 16, 2008.

  1. ritchell77

    Thread Starter New Member

    Jun 16, 2008
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    i'm supposed to put fuses as system protection for my 4.5KW heating elements @230Vac. i had it connected at star (3 heating elements each heaters) and i have 3 heaters.i rated my fuse at 25Amp.do you think that ampere is enough?
    thanks!
     
  2. nanovate

    Distinguished Member

    May 7, 2007
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    How many amps? Amperage is not the only rating to consider.
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    Well, 4.5kW at 230VAC = 19.56 Amperes (approximately)
    So, a 25A slow-blow fuse might be sufficient. You may need to bring them up one at a time, or two, then the last one, as the initial surge current will be considerably more than once the elements have reached operating temperature.
     
  4. nanovate

    Distinguished Member

    May 7, 2007
    665
    1
    You should also derate based on ambient/operating temp of the fuse.

    As Sgt Wookie mentioned you should also look at the inrush current.

    Are you using NEC or IEC?
     
  5. ritchell77

    Thread Starter New Member

    Jun 16, 2008
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    Thank you both for the response!
    I am using IEC. My conflict now is when I did an actual measurement of each heating elements resistance I got 34ohms each? How can I possibly arrive to a 4.5KW heating element?Do you think that there's a problem on my heating element?It may not have been rated as to what it says?I mean to be at 230V and 4.5KW? If I will consider the actual resistance values that i got on every element then, I'm supposed to use a lower ampere rating for my fuse right? (With 34ohms I think I can consider 15-20Amps fuse).I would have love to use lower ampere rating simply because I have space constraint. With lower ampere rating of fuse, I can use a fuse-terminal block which is not so big.
     
  6. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    Were they at room temperature when you measured them? Resistance varies with temperature. If you know the material they are made from, you can look up the temperature coefficient of resistivity and estimate the resistance at working temperature.

    Also, were the individual elements removed and measured by themselves? Or were they measured while assembled in the unit?
     
  7. ritchell77

    Thread Starter New Member

    Jun 16, 2008
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    yes they were measured by themselves since i had them disconnected and yes they are at room temperature.shouldn't my heating element be rated at 1.8kw with the measured 34ohms instead of 4.5kw and my fuse should then be allowed to be as low as 10Amp?thanks!
     
  8. thingmaker3

    Retired Moderator

    May 16, 2005
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    If they are rated for 4.5KW and you can only account for 1.8KW, then you are missing some part of the puzzle. If you use too small a fuse, the fuses will blow.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Well, P = Esquared/R
    So if E=230 and R=34, P = 1555.9 Watts (approximately) per element when cold.
    You have three elements, so 1555.9 x 3 = 4667.7 Watts
    That is your load when cold.
    I = E/R, so since E=230 and R=34/3, your inrush current (when cold) will be 20.3 Amperes.
    So, you could get away with a 20A slow-blow fuse. Alternatively, a pair of 10A slow blow fuses would work, but you wouldn't gain much.
     
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