Synchronous machines
Diperkenalkan| English | Makna |
|---|---|
| synchronous machine | synchronous machine |
| power angle | sudut daya |
| excitation | excitation |
| infinite bus | infinite bus |
| static margin | static margin |
Every power station
- Almost every watt from a power station leaves through a synchronous machine 同步电机. It is a magnet spun by a turbine and locked to the grid.
- It delivers power through an angle, like a spring that twists. This lesson measures that spring.
Locked to the grid
- The rotor is an electromagnet that locks to the rotating field, so it turns at exactly $n_{s} = 120f/p$, with no slip.
- Its induced EMF is $E = 4.44\, f N \Phi k_{w}$, where $k_{w}$ (the winding factor) accounts for how the coils are laid out.
A 2-pole generator feeds a 50 Hz grid. At what speed does it turn, in rpm?
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$n_{s} = 120 \times 50/2 = 3000$ rpm, exactly, with no slip.
The power angle
- The power angle 功率角 $\delta$ is how far the rotor's EMF leads the bus voltage. Per phase, only the synchronous reactance $X_{s}$ lies between them.
- The power delivered is $P = \dfrac{3VE}{X_{s}}\sin\delta$. It peaks at 90°; past that the machine loses step.
Per phase, V = 6000 V, E = 7000 V and X_s = 6 Ω. What is the largest power the three-phase machine can deliver, in MW?
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$P_{max} = 3VE/X_{s} = 3 \times 6000 \times 7000/6 = 21$ MW, reached at $\delta = 90°$.
If the power angle is pushed past 90°, the machine can lose synchronism.
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Past 90° more angle gives less power, so the spring cannot hold the rotor and it slips a pole.
Excitation and reactive power
- The reactive power is $Q = \dfrac{3V(E\cos\delta - V)}{X_{s}}$. Raising the field current, the excitation 励磁, raises $E$ and makes $Q$ positive.
- The turbine sets $P$; the excitation sets $Q$. Grid operators adjust both every day.
A generator's field current is raised while the turbine power stays the same. Mainly what changes?
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More excitation raises $E$, and $Q = 3V(E\cos\delta - V)/X_{s}$ rises. Real power is set by the turbine.
Worked example: rating and margin
- An infinite bus 无穷大母线 at 11 kV line (6351 V per phase), $E = 12$ kV line (6928 V per phase), $X_{s} = 5\ \Omega$. Then $P = 26.4\sin\delta$ MW.
- At 30°, $P = 13.2$ MW. The static margin 静态裕度 is $(26.4 - 13.2)/13.2 = 100\%$: the load could double before the spring lets go.
A generator runs at δ = 20°. What is its static margin, in %? Give the nearest whole number.
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The margin is $(P_{max} - P)/P = 1/\sin 20° - 1 = 1/0.342 - 1 = 1.92$, so 192%.
Worked example: P and Q together
- At 10 MW, $\sin\delta = 10/26.4$ and $\delta = 22.3°$. Then $Q = +0.23$ MVAr: over-excited, it exports a little reactive power.
- Lower the excitation to $E = 10$ kV line and keep 10 MW: $\delta = 27.0°$ and $Q = -4.6$ MVAr. Now it absorbs reactive power.
Match each control to what it sets.
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Two independent dials, P and Q, and a speed fixed by the grid.
Three common slips
- Line voltages in a per-phase formula. Divide by $\sqrt{3}$ first.
- "More excitation, more power." The turbine sets $P$; excitation moves $Q$.
- Electrical and mechanical angles. With $p$ poles, the mechanical angle is $2\delta/p$.
A 4-pole machine runs at a power angle of 30° (electrical). What is the angle in mechanical degrees?
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Mechanical angle $= 2\delta/p = 2 \times 30/4 = 15°$.
Recap
- Speed is locked: $n_{s} = 120f/p$.
- $P = \dfrac{3VE}{X_{s}}\sin\delta$, with the limit at 90°.
- The turbine owns the angle; the field owns the VARs.