Transmission Line - power sent. (Shorebrook PE Exam Question 30)

[akyip]

Hey guys,

Typically when you are asked to calculate for the max power that can be sent on a transmission line given Vs, Vr, and theta, this should really be the 3-phase power, right?

The reason I ask is that in Shorebrook PE Power Exam question 30, the solution only uses the 1-phase power.

Generally speaking, in a transmission line, max power should really be the 3-phase power, correct? So: Pmax, line 3-ph = 3 * (Vs * Vr / X) * sin(theta)

Whenever I am asked to calculate the max power possible sent in synchronous generators, it usually refers to the 3-phase max power. So I would think the same concept applies to transmission lines...

Thanks for any input on this!

 

[Orchid PE]

If you are using L-L voltages, you don't need to multiply by 3. If you are using phase voltages, you will need to multiply by 3.

 

[BebeshKing PE]

I have a concern with this problem, though. The problem was asking the "MAXIMUM" power transmitted. should we just use Pmax=(Vs)(Vr)/X ,where phase angle should be 90 degrees during maximum power?

 

[Orchid PE]

I have a concern with this problem, though. The problem was asking the "MAXIMUM" power transmitted. should we just use Pmax=(Vs)(Vr)/X ,where phase angle should be 90 degrees during maximum power?

I get what you're saying. However, the question essentially asks what is the maximum power that can be transmitted at 30°.

You are correct though that the most amount of power is transmitted when the phase angle is 90°.

 

[DLD PE]

If you are using L-L voltages, you don't need to multiply by 3. If you are using phase voltages, you will need to multiply by 3.

This.  For transmission lines, I would always assume (unless specifically stated) that voltages given are line voltages.