# Power for 3 Phase Y Conn. Load

#### jegues

Joined Sep 13, 2010
733
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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#### t_n_k

Joined Mar 6, 2009
5,455
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

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

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

Sorry that was per phase

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

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:

#### jegues

Joined Sep 13, 2010
733
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

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

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

Sorry that was per phase

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

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.
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?

#### t_n_k

Joined Mar 6, 2009
5,455
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?
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.

#### jegues

Joined Sep 13, 2010
733
No I didn't use P=IV - rather I used P=I^2R
I mentioned after that V= I/R

Hence, P=I^2R.

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

#### t_n_k

Joined Mar 6, 2009
5,455
I mentioned after that V= I/R

Hence, P=I^2R.
I guess you meant V=I*R ....