# power analysis concepts etc.

Discussion in 'Mathematics and Physics' started by PG1995, Sep 3, 2017.

1. ### PG1995Active Member

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There is something fundamentally wrong in this problem. Now I find that the solution I posted in post #24 is wrong too although it matches with the given answer. The reason being that I wrote that Q=3VpIpsinθ=3*V_AB*I_AB*sinθ but it should have been Q=3VpIpsinθ=3*V_an*I_AB*sinθ where V_an=2400<-30.

2. ### RatchitWell-Known Member

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See the attachment where I get the same results using VL and IL or VP and IP.

OK, visualize a delta configuration. Put a voltage across one of the 3 impedances. Now, isn't the voltage across the impedance (VP) the same as the line voltage (VL)? Certainly the currents will be different by a factor of Sqrt(3), but the voltages are the same for a delta configuration. For a wye configuration, the phase voltage and line voltage will differ, but the current will be the same. Are you sure you interpreted the text correctly?

The power triangle follows the impedance triangle. Since the impedance has a positive angle, the reactive power should also. I should have known better. The attachment shows the same answer worked 3 different ways using Vl=VP and a positive angle.

Ratch

3. ### RatchitWell-Known Member

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The above attachment shows the correct answer.

Ratch

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5. ### PG1995Active Member

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Thank you.

Do you think that I'm interpreting the text wrongly? I don't think so because in the section on balanced Y-Δ configuration, it clearly says that phase voltage and line voltage are different. In this text, which, by the way, I use the most, phase voltage is defined as line to neutral.

But I think that mentioned text is confusing because I checked three other texts on that and phase and line voltages for Y-Δ configuration are taken the same. In one text, phase voltage, Van, for the alternator and phase voltage, Vab for the load are taken differently. In the other, phase and line voltage are considered equal. The text which the problem under discussion is taken from also considers phase and line voltages equal.

Thanks.

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• ###### y-delta_phase_neutral_voltage2.jpg
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6. ### RatchitWell-Known Member

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Yes.

I read the attached texts carefully, and they say that the line voltage equals the phase voltage for a delta configuration. Did you visualize a delta configuration? By the geometry of the loads, how can it be anything else? The line voltages and phase voltages are not equal in a wye configuration, but the line and phase currents are equal. When the texts talk about unequal line and phase voltages, they refer to a wye configuration. Also, notice how the correct answers appear when VL = VP in a delta configuration. Do you still have a hard time understanding the line and phase relationships in a delta configuration?

Ratch

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7. ### PG1995Active Member

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Thank you.

I agree with you but the original text is clearly misleading. It could have defined phase voltage for the alternator and phase voltage for load separately at least as was done in this text. The original text at the very beginning clearly defines phase and line voltage as differently. So, I don't agree with you where you say that I was interpreting the original text wrongly.

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9. ### RatchitWell-Known Member

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As long as we agree that line and phase voltages are equal for a delta configuration, that is what really matters. The quality of the text narrative is a matter between you and the publishers.

Ratch

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10. ### PG1995Active Member

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Something is still confusing me. You wrote sqrt(3)*V_l*I_l*cos(-36.87)=sqrt(3)*4157*150*cos(-36.87)=864 kW. Why is so? When line current, I_l should lag corresponding phase current, I_AB, by 30°. In my humble opinion, I_l, should be 150∠-66.87°. Please check calculation on phase current and line current in part (a) here. Thank you.

11. ### RatchitWell-Known Member

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It would help if you disclosed the post number where that occurred. Anyway, apparent
power does not care what the phase of the current is with respect to voltage. It is proportional to the square of the absolute value of current. Current phase does not enter into the apparent power calculations. Apparent power is V I*, where I* is the conjugate of I. Multiplying by I and dividing by I gives V I I*/I = V I^2/I = Z I^2 , where the phase of the current is not considered.

Ratch

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12. ### PG1995Active Member

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Thank you.

This is from your post #42.

We are calculating real power, not apparent power so the phases of current and voltage do matter. I'm sorry if I'm missing something obvious. Thanks.

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13. ### RatchitWell-Known Member

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Once you have apparent power, you can easily calculate real power by multiplying by the cosine of the impedance triangle or reactive power by multiplying by the sine of the impedance triangle. In any case, apparent power does not care about the current phase in relation to the voltage or other current. The current phase angle vs. voltage might or might not match the impedance triangle angle depending on the configuration, but the impedance triangle is the gold standard for the power triangle. Neglecting that fact is what got me the wrong answer the first time I did the last problem.

Ratch

Last edited: Oct 7, 2017
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14. ### PG1995Active Member

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Thank you.

Sorry, I was missing a point about power calculation.

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