Feeder with transformer and loads

Discussion in 'Homework Help' started by Jess_88, Sep 4, 2011.

  1. Jess_88

    Thread Starter Member

    Apr 29, 2011
    Hey guys :)
    I'm having a little trouble with this problem


    for the circuit given above I need to determine

    a) Compute ~IL load1 (magnitude and phase).
    b) Compute IL load2 (magnitude and phase).
    c) Determine the total load current (magnitude and phase).
    d) Compute the voltage at the sending end of the feeder Vsend (magnitude and phase) if the voltage at the load ~(receiving) end is maintained at Vload 2400<0° V .
    e) What is the real power Psend and reactive power Qsend at the sending end of the feeder?
    f)Calculate the voltage regulation VR

    Two single-phase loads are supplied through a 35 kV feeder whose impedance is Zf=(Rf+jXf)=(100+j300) and a  VH / VL  =35000V/2400V single-phase transformer whose equivalent (series-branch) impedance is Z L =( R Low + jX Low ) =(0.2+j1.0) referred to its low-voltage (secondary) side. Note H and L stand for high side and low side, respectively.

    I feel like I must be looking at this question wrong... seems a maybe a bit 'easy' to calculate the currents.

    for parts a,b,c would I do the following

    current in Load2
    S = VI
    I = S/V

    current in Load1
    could I just use I = P/V???
    or would I work out the power at load2 and use that to determine the current in load 1

    part 'c' I just add them together right?

    Last edited: Sep 4, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    To find the current in each load you divide the apparent power (S) by the voltage Vload.

    No you can't just divide P1/Vload to give Iload1. You must use the apparent power for this case as well.

    Also remember the two currents Iload1 & Iload2 are complex values so they must be added accordingly to derive the total current Iload - i.e. not just as a simple algebraic addition.
  3. Jess_88

    Thread Starter Member

    Apr 29, 2011
    ok I think I have it.

    I = P/(Vcosθ)

    θ = cos^(-1) (0.6)

    I = 69.44 < -53.13

    I = S/V

    θ = cos^(-1)(0.3)

    I = 41.66 <72.54

    I1 + I2 = 56.42< -16.262

    Referring Xeq and Req to primary (x(35k/2400)^(2))

    Zeq = 42.534 + j212.673

    I(send) = (1/a)*IL
    = [1/(35k/2400)]*(54.164<-16.262)
    =3.714<-16.262 A

    voltage drop across lines
    Vd = Isend*(Zf + Zeq)

    = 1976.286<58.2

    Vsend = VH + Vd
    = 35000<0 + 1976.286<58.2
    = 36 080.53<2.668 V

    = (36 080.53<2.668 V)(3.714<16.262 A)
    =126 755.62 + j43 472.31

    P = 126 755.62 W
    Q = 43 472.31 VAR

    VL = 2400<0 + (56.41<-16.26)(1.02<80.25)

    VH = VL*(35k/2400)
    =35 364.5<1.22 V

    VR = [(VH - 35k)/35k]*100

    Hows that look?
  4. shadow3

    New Member

    Sep 15, 2013
    Sorry, but I am just confused... I have a homework that asks me to calculate the voltage at the sending end of a feeder.... Is it the voltage Vsend? (in this circuit). thanks