ask the resistance/phase of the load meter (wattmeter)

Discussion in 'Homework Help' started by cupcake, Oct 20, 2010.

  1. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    I was asked to calculate the value of the resistance/phase of the load in balanced 3-phase purely resistive load. it stated that there are two different ways to obtain the solution.

    W1 = VL IL cos (30-theta)
    W2 = VL IL cos (30+theta)

    where voltage line= 110, Is = 0.86, W1=77 and W2=76.5, where W1 and W2 is wattmeter

    W1+W2=153.5

    theta=tan-1 ( * (W1-W2)/(W1+W2))

    thus, I obtained theta = 0.810 where cos theta (pf) = 0.999

    from this, how I calculated the resistance? is it only R=V/I = 110/0.86 = 127.91? or using the wattmeter result?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    In brief -

    (1) The two watt-hour meter method of power measurement in a balanced three phase system states that the total 3-phase (real) power is the sum of the two watt meter readings.

    (2) A purely resistive load will always have unity power factor.


    So Ptot=W1+W2=153.5W as you state.

    Power per phase P_phase=153.5/3=51.17W (it's a balanced 3-phase load)

    Hence each load resistor dissipates 51.17W with a line voltage of 110V applied.

    You are not told whether the load is connected in star or delta. It seems the question implies the effective load resistance from line to neutral is required - but I'm not 100% sure.

    In delta connection it would be 236.46 Ω from line-to-line
     
  3. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    it's state there are two different ways to obtain the load resistance.. what's the other approach?

    btw, the load is connected in delta connection.
     
  4. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    I have another question, why the resistive load bank and the reactive load banks are effectively (connected) in parallel with respect to the supply? is there any reason behind it?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    What does this question relate to with respect to the content of your original post?
     
  6. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    I just want to know why, that question doesn't relate with my original post...
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    OK - but you need to provide a schematic of what you are referring to. As it stands the question unfortunately makes no sense.
     
  8. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    this is the schematic diagram..see the attachment..
     
  9. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    and also, during the experiment, I found out that the sum of wattmeter readings (W1+W2), is nearly the same in all the cases (purely resistive load, inductive load and capacitive load). why is it so? is it because the power factor of the reactive load?
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I believe the experiment shown in your schematic is intending to show that the true power observation will be almost constant even though the effective load power factor is varying.

    The experiment is "contrived" to the extent that the additional reactive components (whether capacitive or inductive) are placed in parallel with the same resistive load bank. If the reactive elements were being added in series with constant resistor values (per phase) then both the real power and power factor would vary.

    I would expect to see small variations in the real power in the experiment as shown. Real capacitors and inductors also have some losses which would add to the power losses in the resistive load bank.

    This would appear to be what you observed.
     
  11. cupcake

    Thread Starter Member

    Sep 20, 2010
    73
    0
    ok, last one, in Balanced 3-phase resistive-inductive load, wattmeter reading W1 is greater than W2, but in Balanced 3-phase resistive-capacitive load, wattmeter reading W1 is less than W2. why is it so?
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The wattmeter signs would depend upon the effective load power factor value and whether it is leading or lagging case.

    It's also possible to have both positive values for W1 & W2 for either a leading or lagging condition at the same power factor.

    You would have to analyse each condition based on the actual circuit values.
     
Loading...