# Power for 3 Phase Y Conn. Load

Discussion in 'Homework Help' started by jegues, Sep 29, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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See figure attached for problem statement as well as my attempt.

I'm having trouble finding the power dissapated across the transmission line.

I thought this would be,

Where VL and IL are the line voltage and current across the transmission line.

Where,

The solution gets the answer by doing the following,

Can someone explain my confusion or clarify things?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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There are several ways of coming up with an answer but having determined the line current, I would have thought the simplest way of expressing the transmission line power loss would be

or using your values

Sorry that was per phase

it should obviously read

BTW. Your formula would work fine as well provided you convert the voltage drop across the line impedance to a line-to-line equivalent voltage drop.

Last edited: Sep 29, 2011
3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
Where does your formula come from?

I can see you did P = IV, where V = I/R, but wheres does the 3 infront come from?

Will the math show how it comes about?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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No I didn't use P=IV - rather I used P=I^2R

The RMS magnitude of the current in any line conductor is |I_line|.

This current flows in the line resistance R_line of each of the 3-phase conductors.

So the total power loss in the transmission line for all 3 conductors is

3*|I_line|^2*R_line

The phase angle of the current doesn't matter in this approach since current and voltage in the line resistances (as lumped elements) are always in phase.

5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
I mentioned after that V= I/R

Hence, P=I^2R.

Thank you for your other comments, this is much more clear now.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I guess you meant V=I*R ....