Power flow
Introduit| English | Sens |
|---|---|
| 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.
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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?
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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.
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$PR + QX = 0.025 + 0.040 = 0.065$ 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
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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.
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$|I|^{2}R = 0.954^{2} \times 0.02 = 0.018$ 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
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Le $QX$ 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.
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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.