1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluating indication of the wattmeter in three-phase system

  1. Feb 10, 2015 #1
    • Member warned about posting without the HW template
    Hi all, I am new to this forum and I hope you can help me.

    I have the following three-phase system where I have to evaluate the indication of the wattmeter W. Available data is: V = 380 V, R = 12 Ohm, R' = 120 Ohm.

    IMG_0087.jpg

    Can you make me understand the process for obtaining W?
     
  2. jcsd
  3. Feb 10, 2015 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    You have 5 energy sinks here. Three identical resistors, one in each phase---that's fortunate, it maintains balance! There's a fourth resistor, and a 0.8pf load of Q=600VAr .

    It looks like the wattmeter measures the power in R', in addition to the power in one phase of a balanced load.

    What are your thoughts?
     
  4. Feb 10, 2015 #3
    My thoughts are that I don't know how to calculate V' or I'.

    I am completely in trouble.

    I would like to understand how to proceed step by step to solve the problem.
     
    Last edited: Feb 10, 2015
  5. Feb 10, 2015 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Since you know V, you should be able to calculate the line current going into that box on the right.
     
  6. Feb 10, 2015 #5
    OK, the line current going into the box should be the following:

    Since
    Q = sqrt(3) * V * I * sin(phi)
    cos(phi)
    = 0,8 => phi = 36,87°

    Then
    I = Q / ( sqrt(3) * V * sin(phi) ) = 1,52

    I = I1 = I2 = I3 because the load is balanced without considering R'.

    Is it correct? If so, how should I proceed?
     
  7. Feb 10, 2015 #6

    NascentOxygen

    User Avatar

    Staff: Mentor

    Once line current is known, you can determine the voltage across R. You'll need to start drawing a phasor diagram.
     
  8. Feb 10, 2015 #7
    OK, maybe voltage across R should be the following:

    Since
    R = R1 = R2 = R3
    I = I1 = I2 = I3


    Then
    Vr = V1 = V2 = V3 = R * I = 12 Ohm * 1,52 A = 18,24 V

    Now I should be able to use Kirchhoff for tensions in the mesh where the voltmeter V is present, this way (V12 refers to the voltage of the three-phase system):

    V12 - V1 - V + V2 = 0

    Since V1 and V2 are equal:
    V12 = V => V12 = 380 V
    and since the three-phase system is symmetrical:
    V12 = V23 = V31 = 380 V

    Is it correct? If so, how should I proceed?
     
  9. Feb 10, 2015 #8

    NascentOxygen

    User Avatar

    Staff: Mentor

    Time to start putting this on a phasor diagram, I think.
     
  10. Feb 13, 2015 #9
    No too much help.
     
  11. Feb 13, 2015 #10

    NascentOxygen

    User Avatar

    Staff: Mentor

    Are you able to draw a phasor diagram? Decide which phasor you will use as a reference, then draw others in relation to it. This is the type of problem where a phasor diagram is essential, IMO.

    Can you explain in words how you will go about finishing this problem?

    If you have not learnt about phasor diagrams, I expect you will ask further questions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluating indication of the wattmeter in three-phase system
Loading...