Delta-connected heater current calculation

Discussion in 'General Electronics Chat' started by Girish Bodhale, Jun 14, 2016.

  1. Girish Bodhale

    Thread Starter New Member

    Jun 14, 2016
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    Dear Forum, Its my first post on heaters, also am working for the first time on heaters.
    I have 12 No's, 230v 750w single phase heater. i have divided the heaters in 2 bunches of 6 heaters each, so made 2 delta connection of it. Now please do guide me to find out how much current will be delivered by each phase (R, Y, B) when heaters are powered on. Please see the attachment for the heater connections.
     
  2. #12

    Expert

    Nov 30, 2010
    16,261
    6,768
    R=Esquared/P
    R=70.5333...ohms
    2 heaters in series = 141.0666 ohms
    I = 415 volts/141.0666 ohms
    I=2.942 amps in each pair of heaters
    P=IsquaredR
    P=610.44 watts per heater
     
  3. Girish Bodhale

    Thread Starter New Member

    Jun 14, 2016
    2
    0



    so as per the above calculations my current per phase will be 2.942*2= 5.884 (as am using two delta)
     
  4. #12

    Expert

    Nov 30, 2010
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    I went as far as I knew how to do this. If that is deficient, other people here can finish the calculations.
     
  5. BR-549

    Well-Known Member

    Sep 22, 2013
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    It appears the voltage across the heaters is close to the heater rated voltage, so I would assume the current draw will be close to total power rating and should plan for that occurrence.

    9000 W / 415 volts......~21 amps. It won't be that high because the voltage isn't quite up to snuff for the power rating.
     
  6. gerty

    AAC Fanatic!

    Aug 30, 2007
    1,153
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    I believe you forgot the magic 3ø number 1.732.
    AMPS=WATTS÷(1.732 x VOLTS x PF)

    So, roughly 9000/717.95 = 12.5 amps
     
  7. #12

    Expert

    Nov 30, 2010
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    My beliefs include the idea that there are 415 volts RMS at the ends of each (2 resistor) load. By estimating the resistance of the loads, I can find the current and power in each load. The instant you try to add them, you hit the phase relationship problem. I have never worked in that area, so I'm stuck. I could look it up and educate myself, but so can the Thread starter.
     
  8. BR-549

    Well-Known Member

    Sep 22, 2013
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    You can calculate further, and get a closer number, but it won't be correct either. The voltage on 3 phase is never balanced.

    I wouldn't waste the time. Turn the switch on, measure the current, adjust thermal overloads on starter relay.

    Get back to work.
     
  9. Kjeldgaard

    Member

    Apr 7, 2016
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    When I, through time, has had a number of tasks to the control and measurement of three-phase systems, I dare to describe my calculation of this task.

    Each of the twelve heating elements has a resistance of: U^2 / P = 230^2 / 750 = 70.5 Ohm

    Each of the three groups heaters consists of two series-connected heaters, parallel with further two series-connected heaters, so we can reduce it to three heaters of 70.5 Ohm, nicely symmetrical between the three phases.

    The quick solution, when it is symmetrical / resistive load, is the current draw from each of the three phases: U / R = 415 / 70.5 = 5.9 A.

    Proof of the above short calculation, may be looking at the overall power consumption. Each of the three groups of four heaters consumer U^2 / R = 415^2 / 70.5 = 2443 W. The total power will then be 3 * 2443W = 7329W, and with symmetrical load, each phase should deliver a third = 2443W, and this gives a current of: P / U = 2443/415 = 5.9 A.
     
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