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Single Phase Controlled Rectifier Problem

  1. Nov 16, 2017 #1
    • Thread moved from the technical forums, so no Homework Template is shown
    Question
    Consider a heating system with strongly inductive characteristic and with an electronic control of temperature performed by varying the firing angle of a single-phase controlled bridge.
    a) Determine the value of resistance to to ensure a maximum power of 7500W if this system is powered with 220V (effective).
    b) Determine the fire angle to a power of 3000W.

    My attempt of solution:
    a)
    If the system is strongly inductive so the current is constant, then we should use average voltage. So, maximum power is when the average voltage is maximum and this is possible when fire angle = 0 degree. Then:
    quw8aKp.png
    And the resistance is:
    VcAStaF.png
    b) Using the same idea:
    QHUw6Ez.png
    The problem is:

    Many friends (and maybe the professor) solved this problem on a exam using 220 V as voltage to find the resistance when maximum power, but I don't think this is totally correct because the problem tells us that the system is strongly inductive, so Irms = Iaverage = I constant, then, to find the voltage we should use average voltage and not effective as many other did. So, What do you think ? If I'm incorrect, why ? I'm a bit upset because I studied a lot for this exam and I was confident that the correct was average voltage when constant current.
     
  2. jcsd
  3. Nov 19, 2017 #2

    Baluncore

    User Avatar
    Science Advisor

    We assume that the 220V is AC RMS. We assume that conduction is symmetrical on both polarities of the AC supply.

    If an SCR bridge is used the current through the heater will be DC. If it is DC it will be hard to phase control since there will be no zero current to turn-off with the SCRs.

    If a TRIAC is used the current will reverse twice per cycle. So the control is with a TRIAC and the current through the load is AC. Now, just how big is the “strongly inductive characteristic”? The power factor will be close to 0.9, and so the inductive component of the impedance XL, will be about 10% of the resistance. you can compute R and L to get say a 0.9 phase angle.

    If we ignore the unknown inductance, W = 7500 = V2 / R.
    For 100% duty cycle, the resistance required will be R = V2 / 7500 = 6.543 ohm

    Energy is the integral of v^2 / R, while the switch is conducting. Looking at the first half cycle we need less than 50% duty.
    It will need to turn on at about 95° and conduct until 180°. I leave you to solve for R and XL.
     
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