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Voltage drop along cable

  1. Oct 20, 2008 #1
    I have a 45kW VSD drive with pf of .9 and FL amp of 75A, I have a 350m run from the MCC to the VSD itself and I'm trying to calculate what size cable I need so that I don't exceed the specified voltage drop of 3%. The line-line voltage where the motor is to be placed is 400V and the cable impedance is say .2ohms per km.

    I was looking in the australian standards and their voltage formula is as follows:
    Vdrop = (sqrt(3) * I(line) * Distance * Z(cable))/1000
    V(% drop) = Vdrop/400

    What I want to know is, why are they multiplying by sqrt(3), when you already have the line value Current. I thought sqrt(3) was only used to convert between phase/line values of current or voltage in delta/star configurations. The motor has a delta winding and I'm running 4C (3C+E) out to it.

    From what I understand it should be:
    Vdrop = I(line) * Distance * Z(cable)
    So;
    Vdrop = (75 * 350 * .2) / 1000
    Vdrop = 5.25V

    V(%) = (Vdrop/V) * 100
    V(%) = (5.25 / 400) *100
    V(%) = 1.31%

    Thanks
     
  2. jcsd
  3. Oct 21, 2008 #2

    stewartcs

    User Avatar
    Science Advisor

    It appears they are approximating the voltage drop from line-to-line of a three phase system, not line-to-neutral.

    The general approximation for a line-to-neutral voltage drop is:

    VD = IRcos(phi) + IXsin(phi)

    where,

    VD = voltage drop in one conductor, one way
    I = current flowing in one conductor
    R = line resistance for one conductor in ohms
    X = line reactance for one conductor in ohms
    phi = angle whose cosine is the load power factor

    Note that I is usually assumed to be the load carrying capacity of the conductor.

    The line-to-line voltage drop for a three-phase system is then found by multiplying by sqrt(3).

    CS
     
  4. Oct 26, 2008 #3
    You have given cable Z of 0.2 ohm/km which is value for 95mm2
    This cable has 3ph volt drop of 0.4mV/Am
    Hence VD = 0.4 *75*350/1000 = 10.5v
    Volt drop usually is % of voltage at MCC which usually 415v
    3% of 415V is 12.45v hence you are below this
     
  5. Oct 27, 2008 #4
    %V = ( phase voltage drop / rated phase voltage)* 100

    %V (in symetrical system) = ( sqrt (3) * phase voltage drop / sqrt (3) * rated phase voltage) * 100


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