DC machines
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| shunt motor | shunt motor |
| series motor | series motor |
| field weakening | field weakening |
Two equations, one machine
- Force $F = BIl$ and EMF $e = Blv$ wrap into the pair $E = k\Phi\omega$ and $T = k\Phi I_{a}$, joined by the identity $T\omega = E I_{a}$: the same air-gap power in both languages.
- The shunt motor (field in parallel) holds nearly constant speed; the series motor (field in series) makes torque $\propto I_{a}^{2}$.
Shunt motor, 220 V, $R_{a} = 0.2\ \Omega$, $I_{a} = 50$ A at 1500 rpm: $E = 210$ V, $k\Phi = 1.337$, $T = 66.9$ N·m, and $T\omega = EI_{a} = 10.5$ kW ✓.
{exercise:uni-ee0921-rating}
A shunt motor at 220 V, Ra = 0.2 Ω, draws Ia = 50 A at 1500 rpm (157.1 rad/s). The EMF is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
E = V − Ia·Ra = 220 − 10 = 210 V; then kΦ = 210/157.1 = 1.337 and T = 66.9 N·m.
Starting is resistance's job
- At switch-on $E = 0$, so the inrush would be $220/0.2 = 1100$ A — twenty times rated. A starter limiting current to 100 A needs $R_{start} = 2.0\ \Omega$, cut out in steps.
{exercise:uni-ee0922-start}
Starting that motor from rest with no starter, the armature current would be about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
At standstill E = 0, so only Ra limits current: 220/0.2 = 1100 A, twenty times rated — hence the 2.0 Ω starter.
Field weakening for speed
- Cut the flux 20% ($k\Phi = 1.070$) at the same load torque: current rises to 62.5 A, EMF settles at 207.5 V and the speed climbs to 193.9 rad/s = 1852 rpm — a fifth less flux, 23% more speed.
- Check: $T\omega = 12.96$ kW against $EI_{a} = 12.97$ kW ✓. Above base speed this trade is standard.
{exercise:uni-ee0923-weaken}
With the field weakened 20% at constant load torque, the speed —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Ia must rise to 62.5 A for the same torque, E settles at 207.5 V and ω = 207.5/1.070 = 1852 rpm.