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?

 

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.

 

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.

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?

 

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.

 

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.

 

I mentioned after that V= I/R

Hence, P=I^2R.

I guess you meant V=I*R ....