# 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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43
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,

$P_{TL} = \sqrt{3}V_{L}I_{L}cos\phi, \quad \text{Where, } \quad cos\phi \quad \text{is the power factor across the transmission line}$

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

$I_{L} = 150 \angle -36.87^{o}$

$V_{L} = I_{L} \cdot Z_{TL}$

Where,

$Z_{TL} = (2.5 + j10.2) \Omega$

The solution gets the answer by doing the following,

$Re \left{ 3 I_{L}^{2} \cdot Z_{TL} \right}$

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

$P_{TL}=3|{I_L}|^2R_{line}$

or using your values

$P_{TL}=3*150^2*2.5=56.25kW$

Sorry that was per phase

it should obviously read

$P_{TL}=3*150^2*2.5=168.75kW$

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
735
43
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
735
43
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
5,448
783
I guess you meant V=I*R ....