Timeline for How do you determine power flow direction in a transmission line?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2011 at 20:00 | comment | added | Alan Rominger | Wait, power would be of the form $A*sin(2 \omega t - \theta)+B$ if we take $\omega$ to be the frequency of the circuit, or the frequency of the current and voltage. I don't want to leave my comment with an incorrect equation. | |
| Oct 3, 2011 at 16:40 | comment | added | Alan Rominger | @CarlBrannen Yes, you are expressing the correct physical picture. The instantaneous power delivered in any given line will be a sinusoid with the form $A*sin(\omega t-\theta)+B$, and the average, $B$, will be 1/3rd of the total power. | |
| Oct 3, 2011 at 15:52 | comment | added | Carl Brannen | @Zassou; With 3-phase, you can do the "experiment" on any single one of the three lines. You'll get the power flow for that line. Assuming things are balanced, you will get 1/3 of the total power. | |
| Oct 3, 2011 at 14:34 | comment | added | Alan Rominger | @RonMaimon There is no power flowing back to the plant. The plant produces power and that power is delivered through the lines. Now, current does flow back to the plant through the lines. However, in 3-phase power there is no return line. The sum of the current in all lines is always zero, so there is no net current flowing through the entire system of wires either! This is the magic of 3-phase power. | |
| Oct 3, 2011 at 12:40 | history | edited | Marty Green | CC BY-SA 3.0 |
added 1113 characters in body
|
| Oct 3, 2011 at 6:52 | comment | added | Martin Gales | @Carl: I agree, this is actually a good answer. But the question itself asks for a physics experiment. There is no physics experiments suggested in this answer. | |
| Oct 3, 2011 at 4:15 | comment | added | Carl Brannen | I believe this is the correct answer. Basically you get a direction when you measure the current. The voltage either leads or lags the current and that defines the power flow. (Uh, not sure which.) | |
| Oct 1, 2011 at 15:48 | comment | added | Ron Maimon | @Martin: The "reactive power" is just the power returning to the plant from the returning line. This is not an objection. | |
| Oct 1, 2011 at 7:40 | comment | added | Martin Gales | @Ron: Not quite. The mentioned power meter measures only the active power. The Poynting vector includes the whole power flow(active + reactive). The active power is transmitted for ex. to your computer. The reactive power can be compared to the foam in a glass of beer : it is not the real stuff, but there is no way to avoid it and the glass must be oversized unless you will have overflow. | |
| Sep 30, 2011 at 17:09 | comment | added | Ron Maimon | @Martin--- you can determine all these quantities away from the wire by field measurements. This is equivalent to the Poynting vector business you want. | |
| Sep 30, 2011 at 6:32 | comment | added | Martin Gales | @Marty: The power meter is not a right answer. The question is to determine power flow direction without a direct connection to the line. | |
| Sep 29, 2011 at 15:07 | comment | added | Ron Maimon | Yes, you are right. I was worried about the fact that there is power returning to the plant, but you get that too. | |
| Sep 29, 2011 at 10:14 | history | answered | Marty Green | CC BY-SA 3.0 |