Power through a line
The key idea
Everything that a power flow computes reduces to one picture. Two voltage phasors sit with an impedance between them. The angle between the phasors decides how much active power crosses, and in which direction. The difference in their lengths decides the reactive power. The impedance keeps the difference between the two ends, and that difference is the loss.
The idea
Every branch in a network has two ends. Call them the sending end, at voltage Vs, and the receiving end, at voltage Vr. Each voltage is a phasor, so each one has a magnitude and an angle. The branch impedance, R + jX, sits between the two ends. That is the entire problem.
The difference between the two ends forces the current: I = (Vs − Vr) / (R + jX). Two identical phasors give no flow at all. Every number that a study reports comes out of that one subtraction, including the flow, the loading, and the loss.
You can make the phasors differ in two ways, and each way has a different effect. You can rotate one phasor relative to the other, which opens an angle difference δ. You can also make one phasor longer than the other, which opens a magnitude difference. On a transmission line the two differences act almost independently. That independence is the most useful fact in this lesson.
Take the angle first. On a transmission line the reactance is much larger than the resistance. Active power then comes out as P ≈ VsVr·sin δ / X. The magnitudes barely matter here, because both are near 1.0 per unit and they multiply together. Active power follows the angle, and the leading end sends it.
Tip the sending end forward by 10°, and megawatts appear. Tip the sending end backward, and the same line imports power instead. Nothing physical changes in the line. Only the order of the two angles changes.
Now take the magnitudes. Reactive power comes out as Q ≈ (Vs² − VsVr·cos δ) / X. For small angles the cosine is nearly 1, so this result is mostly Vs(Vs − Vr) / X. Reactive power follows the magnitudes, and the higher-voltage end sends it. Raise a generator's terminal voltage by one percent, and the generator exports vars. Its megawatts hardly move.
The angle and the magnitude are two separate controls with almost no crosstalk. Engineers call this effect decoupling. It is why voltage control and dispatch are separate problems. It is also why a solver can use the split to converge faster. An operator who must move MW changes the governor setting. An operator who must move Mvar changes the excitation.
The impedance keeps part of the flow. The resistance turns |I|²R into heat, so less active power arrives than leaves. The reactance absorbs |I|²X of reactive power, and this absorption rises faster than the loading rises.
A heavily loaded line therefore pulls the voltage down at both ends. The reactive flows at the two ends of a busy line often point in opposite directions. Both ends supply the reactive power that the line absorbs.
Distribution breaks this clean split. On a low-voltage feeder, R and X are comparable, and R is sometimes the larger of the two. The two couplings then mix. Active power now moves the voltage noticeably. A street of rooftop solar that exports on a sunny afternoon pushes the local voltage up. The same export on a transmission line would be almost invisible.
Try it
Start with equal magnitudes and δ at +10°. Drag the angle through zero and watch both power arrows turn around. Then set δ back to zero and move the sending voltage instead.
every value in per unit
P sent
1.736
P received
1.736
P loss
0.000
Q sent
0.152
Q received
−0.152
Q absorbed
0.304
Active power crosses from the sending bus to the receiving bus. Reactive power enters the line from both ends, and the line keeps it.
This is a single-phase per-unit model. The receiving end holds 1.00 pu ∠0°, and X is 0.1 pu. The model has no shunt capacitance, so nothing on this line generates vars. Move the angle, and P moves. Move the sending voltage, and Q moves. Turn up R/X to model a distribution feeder, and the two controls are no longer separate.
Why it matters
- Flow direction is a consequence, not a setting. Nobody tells the power which way to go along a branch. Dispatch and voltage set the angles, and the angles set the direction. A power flow study solves for exactly that result.
- Voltage support is a reactive power problem. A shortage of vars collapses the magnitude gradient, and the voltage sags. Capacitors, generator excitation and tap changers all change the magnitude difference, not the angle.
- Losses grow with the square of the current. Double the flow, and the loss quadruples. The same MW is therefore cheaper to move at a higher voltage, where the per-unit current is smaller.
- On a feeder, the two couplings mix. R is comparable to X, so active power exports raise the voltage. The voltage limits then constrain how much generation a feeder can host.
The math, if you want itOptional — the page reads completely without it
Nothing here is an approximation. The whole branch is Ohm's law and S = V·I*:
the exact relations
I = Vs − VrR + jX · Ss = Vs·I* · Sr = Vr·I*
Now set R to zero. This step is a fair approximation on transmission, where X is many times R. The two flows then separate into the classic pair:
active power, R ≈ 0
P ≈ Vs·Vr·sin δX
reactive power at the sending end, R ≈ 0
Qs ≈ Vs² − Vs·Vr·cos δX
P depends on δ through a sine. For small angles P is roughly proportional to the angle, and P reverses sign with the angle. Q depends on δ through a cosine, which is flat near zero. Small angle changes therefore barely move Q, and the magnitudes control it.
The impedance keeps exactly the difference between the two ends, because Ss − Sr = (Vs − Vr)·I* = |I|²·Z:
what the impedance keeps
Ploss = |I|²·R · Qloss = |I|²·X
Both losses are positive whenever current flows, so a series branch can only consume reactive power. Real lines also have shunt capacitance, which generates reactive power in proportion to the voltage squared. A lightly loaded transmission line is therefore a net var source. This model has no shunt branch, so it cannot show that effect.
See it in Phasor
In Phasor, every branch result is this calculation. The study reports P and Q at both ends of each branch, plus the loss between them. A flow that leaves one bus at 12 MW arrives at the next bus at 11.8 MW. The difference is exactly what the impedance takes. A sign change tells you that the angle order flipped.