• Support PF! Buy your school textbooks, materials and every day products Here!

Evaluating indication of the wattmeter in three-phase system

  • Thread starter pizzico85
  • Start date
  • #1
5
0
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?
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
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?
 
  • #3
5
0
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:
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
Since you know V, you should be able to calculate the line current going into that box on the right.
 
  • #5
5
0
Since you know V, you should be able to calculate the line current going into that box on the right.
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?
 
  • #6
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
Once line current is known, you can determine the voltage across R. You'll need to start drawing a phasor diagram.
 
  • #7
5
0
Once line current is known, you can determine the voltage across R. You'll need to start drawing a phasor diagram.
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?
 
  • #8
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
Time to start putting this on a phasor diagram, I think.
 
  • #9
5
0
No too much help.
 
  • #10
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
No too much help.
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.
 

Related Threads on Evaluating indication of the wattmeter in three-phase system

  • Last Post
Replies
11
Views
780
  • Last Post
Replies
11
Views
750
  • Last Post
Replies
1
Views
503
  • Last Post
Replies
1
Views
532
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
658
Replies
0
Views
9K
Replies
1
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
2
Views
1K
Top