Evaluating indication of the wattmeter in three-phase system

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The discussion revolves around evaluating the wattmeter indication in a three-phase system with given voltage and resistance values. The user seeks guidance on calculating the wattmeter reading, particularly focusing on the line current and voltage across resistors in a balanced load scenario. They have calculated the line current as 1.52 A and the voltage across each resistor as 18.24 V, using the known parameters and Kirchhoff's laws. The importance of drawing a phasor diagram is emphasized for visualizing the relationships between voltages and currents in the system. Further clarification on phasor diagrams and their application in solving the problem is requested.
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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?
 
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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?
 
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:
Since you know V, you should be able to calculate the line current going into that box on the right.
 
NascentOxygen said:
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?
 
Once line current is known, you can determine the voltage across R. You'll need to start drawing a phasor diagram.
 
NascentOxygen said:
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?
 
Time to start putting this on a phasor diagram, I think.
 
No too much help.
 
  • #10
pizzico85 said:
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 learned about phasor diagrams, I expect you will ask further questions.
 
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