Power flow
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| slack bus | slack bus |
The loop and its conjugate
- Slack bus at $1.0\angle 0°$, line $Z = R + jX$, load $P + jQ$ at bus 2. Each pass computes the current the present $V_{2}$ implies: $V_{2} = V_{1} - Z\,\mathrm{conj}(S/V_{2})$. The conjugate is physics, not decoration.
- Linearised: $\Delta V \approx (PR + QX)/|V_{2}|$ — reactive flow through reactance eats the voltage.
$Z = 0.02 + j0.10$, load $0.8 + j0.4$: converges to $V_{2} = 0.937\angle -4.4°$, current $0.954\angle -31°$, sending power $0.818$ pu ✓ — and $QX = 0.040$ of the 0.063 drop dwarfs $PR = 0.016$.
{exercise:uni-ee0961-v2}
With Z = 0.02 + j0.10 pu and load 0.8 + j0.4 pu, the receiving voltage converges to about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Iterating V2 = 1.0 − Z·conj(S/V2) settles at 0.937 pu, −4.4°; the (PR+QX)/V estimate gives 0.060 against the exact 0.063.
Dropping the conjugate in the iteration formula would —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
With S = V·I*, the current is conj(S/V); skipping the conjugate feeds the iteration the wrong complex current and it drifts to nonsense.
Capacitors are a voltage tool
- Cut the reactive load to 0.1 pu: $V_{2}$ rises to $0.970\angle -4.6°$ — the drop halves while the angle barely moves. Magnitude belongs to Q, angle to P.
{exercise:uni-ee0962-comp}
Cutting the reactive load from 0.4 to 0.1 pu at that bus mainly —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The QX term falls from 0.040 to 0.010 and the magnitude follows; the angle belongs to P·X, which is untouched.