Synchronous machines
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| power angle | góc công suất |
| static margin | static margin |
| excitation | excitation |
Locked to the grid
- The rotor turns at exactly $n_{s} = 120f/p$, no slip. Its EMF has the transformer form $E = 4.44 f N \Phi k_{w}$.
- Power crosses through the power angle $\delta$: $P = 3VE\sin\delta/X_{s}$ — a sine-shaped torsion spring.
{exercise:uni-ee0941-pmax}
That generator (V = 11 kV line, E = 12 kV line, Xs = 5 Ω/phase) has a maximum power of about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Per phase V = 6351, E = 6928; 3·V·E/Xs = 26.4 MW at δ = 90°. Line values divided by √3 first — the classic trap.
Margin arithmetic
- 11 kV bus, $X_{s} = 5\ \Omega$/phase, E = 12 kV line: $P = 26.4\sin\delta$ MW. At $\delta = 30°$, P = 13.2 MW with a 100% static margin — the cushion operators keep.
Schedule 10 MW: $\delta = 22.3°$ and the machine exports $Q = 0.22$ MVAr over-excited. Under-excite to E = 10 kV and the same 10 MW needs $\delta = 27.0°$ while Q swings to −4.6 MVAr: the excitation dial owns the VARs.
{exercise:uni-ee0942-margin}
Running at δ = 30°, the machine could roughly —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
P(30°) = 13.2 MW against a 26.4 MW peak: the static margin is (26.4 − 13.2)/13.2 = 100%.
At the same 10 MW schedule, lowering the excitation from 12 kV to 10 kV mainly changes —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The turbine fixes P (δ adjusts to 27.0°); excitation moves Q, here from exporting 0.22 MVAr to absorbing 4.6 MVAr.