# Can we assume this 3ph. system is balanced?

Discussion in 'Homework Help' started by ElectricMagician, Aug 5, 2012.

1. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
Hello everybody,

I have 2 questions regarding part b of the attached problem.

1. Can we assume the system is balanced and so |Vab| = |Vbc| = |Vca| ?

2. Can we assume that W1=Vab * IaA * cos (θvab -θiaA) and similarly for W2 with Vbc and Icc?

Thank you very much.

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2. ### WBahn Moderator

Mar 31, 2012
17,768
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Upon what basis would such an assumption be made?

3. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
Actually, that was my question, if there were any grounds for the assumption that the system is balanced, since the solution would be straightforward from there.

Otherwise, we're left with only one equation: √3(W1+W2) = √3VLIL*cos(θz)

which cannot be broken down to reflect different line voltages currents or loads since we do not know how they're related.

Thank you

4. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
Since you are given three different loads, it seems unlikely that there is any basis to assume the system is balanced.

How are W1 and W2 related to the line voltages and currents?

5. ### mlog Member

Feb 11, 2012
276
36
I don't think you have enough information to assume the loads are balanced.

The problem doesn't ask you this question, does it?

Last edited: Aug 6, 2012
6. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
W1 = |VAB||IaA| cos (θVABiaA)
W2 = |VBC||IcC| cos (θVBCicC)

What got me stuck is that we have 2 equations and so many unknown variables (|VAB|, IaA, IcC, θVAB, θVBC, θIaA, θIcC)

7. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
But it asks for the per phase P and Q, pf, and the "phase impedance"
So all the above variables are needed.

Doesn't the "phase impedance" wording above suggest that all loads are equal?

8. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
But don't you have a whole bunch of other relationships to work with. Namely, what is the relationship between each phase voltages and that phase currents? What is the relationship between the phase voltages? What is the relationship between the phase currents?

9. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
Why would it suggest that?

10. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
But they keep generating/requiring new unknowns

Phase voltage = Line Voltage. The magnitude is known to be 200V
Phase Voltage/ Phase current = Phase impedance

We cannot determine the relationship between phase current and line current because we cannot assume the loads are equal.

Phase voltages are equal.
Phase currents are not necessarily equal or related, since we did not assume the loads are equal.

That's why I'm stuck with this problem, since I'm assuming that the system is not balanced.

11. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
Actually, after reading it again, I don't think it signifies anything.

I guess I was just trying to ease things for myself there.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The relationship for two wattmeter indication of reactive power only seems to hold for balanced loads. On that basis one assumes the alternative [unbalanced loads] renders the problem unsolvable. If the latter [unbalanced loading] is true then why ask the question in the first place?

Last edited: Aug 7, 2012

Jul 26, 2012
57
0
Makes sense.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Can you therefore solve the problem on the basis of balanced loading?

15. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
It's still problematic because of voltage imbalance

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I'm puzzled that in the case of a balanced 3 phase load with a balanced 3 phase source you anticipate a voltage imbalance condition. How would this occur in your opinion?

17. ### ElectricMagician Thread Starter Member

Jul 26, 2012
57
0
By voltage imbalance I meant that we were not sure that the sources are balanced, since the problem doesn't state so.

I asked him today and he said we can assume the problem is balanced, which makes it a lot easier.

Thanks everybody for your generous help.

18. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
If you have a balanced source driving a balanced load, then doesn't symmetry result in the only difference between what the two wattmeters read being due to the (presumed tiny) influence of the imbalance due to the presence of one of the wattmeter coils in the other wattmeter's path?

It just seems odd that the problem carefully sets up the notation for different phase voltages and different phase currents and, most telling, different phase loads and that you would then be expected to assume that everything is neatly balanced.

And, if it does only work if things are neatly balanced, then this method would seem to be pretty useless in practice.

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If by a balanced load you mean a purely resistive 3-phase load then yes the difference in the watt-meter readings would be very small. In the case of a balanced 3 phase load with a reactive power component as well as real power then there will be a difference in the two readings consistent with the equation noted in the OP's original problem.

With respect to explicitly stating [in a question or problem] that a three phase supply is balanced [in both magnitude & phase] my feeling is that this is normally implicit in questions regarding 3-phase systems. If the contrary condition required consideration then the questioner would presumably state the individual line voltages and / or phase discrepancies. The original question in this case states [in part b] that the line voltage is 220V which suggests there is no variation between individual phase voltages - otherwise any discrepancies would be noted.

For my "two pennies worth", the use of the two watt-meter method to find the reactive power is virtually useless anyway. Balanced loads aren't commonplace and who would remember to include the √3 factor even if the loads were balanced. To me it just seems to be one of those mathematical curiosities that electrical engineering professors find fascinating for some reason.

With respect to the two watt-meter method for power measurement [actually adding two meter readings together], I would suspect that with the advent of modern electronics based measuring instruments the concept itself is probably also just an interesting but largely outdated curiosity .....????? A curiosity for most folk - apart from those instrument manufacturers who build the concept transparently into their devices. So it's still worth understanding the principle.

Last edited: Aug 8, 2012