Power flow
Introducido| English | Significado |
|---|---|
| power flow | The calculation of voltages and real and reactive power at every bus of a power network in steady state. |
| slack bus | slack bus |
| reactive compensation | A voltage tool. |
Will the voltage hold?
- "Will the voltage at the end of this feeder stay above 0.95 pu?" Grid planners answer that question thousands of times a day.
- They use a power flow 潮流计算. The two-bus version fits on one page, and it shows why capacitors fix voltage.
Two buses
- Bus 1 is the slack bus 平衡节点: held at $1.0\angle 0°$ pu, it supplies whatever the losses turn out to be.
- A line $Z = R + jX$ feeds a load $S = P + jQ$ at bus 2. The load current depends on $V_{2}$, and $V_{2}$ depends on the current: a loop.
The slack bus has a fixed real power, set before the power flow is solved.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
The slack bus fixes the voltage and angle. Its power is whatever the loads plus the losses need, and it is known only after solving.
Guess, correct, repeat
- Guess $V_{2}$, then update $V_{2} \leftarrow V_{1} - Z \cdot \mathrm{conj}(S/V_{2})$ until it stops changing.
- The conjugate matters: $S = VI^{*}$, so the current is $I = \mathrm{conj}(S/V)$.
What happens if the conjugate is left out of the iteration?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
With $S = VI^{*}$ the current is $\mathrm{conj}(S/V)$. Without the conjugate the iteration uses the wrong current.
The quick estimate
- Near 1 pu the voltage drop is about $\Delta V \approx \dfrac{PR + QX}{|V_{2}|}$.
- On most lines $X$ is much larger than $R$, so the reactive flow $Q$ causes most of the drop. That is why reactive compensation 无功补偿 is a voltage tool.
A line has R = 0.05 pu and X = 0.2 pu and carries P = 0.5 pu and Q = 0.2 pu, with V near 1 pu. Estimate the voltage drop in pu.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$PR + QX = 0.025 + 0.040 = 0.065$ pu. The $QX$ term is the larger, although $Q$ is less than half of $P$.
Worked example: the base case
- $Z = 0.02 + j0.10$ pu, load $0.8 + j0.4$ pu. The iteration settles at $V_{2} = 0.937\angle -4.4°$ pu, a drop of 0.063.
- The estimate gives $(0.016 + 0.040)/0.937 = 0.060$: close, and $QX = 0.040$ is most of it. The line uses $0.954^{2} \times 0.02 = 0.018$ pu of real power.
With Z = 0.02 + j0.10 pu and load 0.8 + j0.4 pu, the receiving voltage settles at about
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Iterating $V_{2} = 1.0 - Z\,\mathrm{conj}(S/V_{2})$ settles at 0.937 pu, −4.4°. The estimate gives 0.060 against the exact drop of 0.063.
In the base case the line current is 0.954 pu and R = 0.02 pu. What real power does the line lose, in pu? Give 3 decimal places.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$|I|^{2}R = 0.954^{2} \times 0.02 = 0.018$ pu, so the slack bus sends 0.818 pu for a 0.8 pu load.
Worked example: adding capacitors
- Capacitors at bus 2 cut the net load to $0.8 + j0.1$ pu. Now $V_{2} = 0.970\angle -4.6°$ pu: the drop falls from 0.063 to 0.030.
- The angle hardly moves, because $P$ carries the angle. The capacitors lift $V_{2}$ as much as halving the line impedance would, at far lower cost.
Cutting the reactive load at that bus from 0.4 to 0.1 pu mainly
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
El $QX$ term falls from 0.040 to 0.010 and the magnitude follows. The angle depends on $P$, which did not change.
Three common slips
- Dropping the conjugate. The iteration then diverges, or quietly settles on a wrong answer.
- Overloading the slack bus. It sets the angle reference and covers losses; the loads must still balance.
- Trusting the estimate at large angles. It ignores the $(PX - QR)/V$ part, which matters past about 15°.
Match each term to what it mainly affects.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Magnitude follows Q, angle follows P, and the line losses grow with the square of the current.
Recap
- Iterate $V_{2} = V_{1} - Z\,\mathrm{conj}(S/V_{2})$ until it settles.
- Check with $\Delta V \approx (PR + QX)/V$.
- Reactive flow causes the voltage problem; capacitors are the cheap fix.