Máquinas de corriente continua
Introducido| English | Significado |
|---|---|
| armature | armature |
| commutator/ˈkɒmjuːteɪtə/ | commutator |
| shunt motor | shunt motor |
| series motor | series motor |
| field weakening | field weakening |
Torque on command
- A cordless drill, a train's axle motor and a car's starter are all DC machines. Each gives torque you can command with a current.
- The whole machine follows from two lines of physics you already know.
Force and EMF
- A conductor of length $l$ in a field $B$ carrying current $I$ feels a force $F = BIl$. Moving at speed $v$, it makes an EMF $e = Blv$.
- A DC machine puts many such conductors on a rotor, the armature 电枢. The commutator 换向器, a rotating switch, keeps the torque in one direction.
Match each part to its job.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
The field makes the flux, the armature carries the current, and the commutator keeps them working together.
The two machine equations
- Collect the geometry and the flux into one constant $k\Phi$: $E = k\Phi\omega$ and $T = k\Phi I_{a}$, with $\omega$ in rad/s.
- Multiply them and $T\omega = E I_{a}$: mechanical power and electrical power are the same air-gap power.
A DC motor has kΦ = 2 V·s and draws 30 A of armature current. What torque does it make, in N·m?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$T = k\Phi I_{a} = 2 \times 30 = 60$ N·m.
Shunt or series
- A shunt motor 并励电机 has its field in parallel, so the flux is fixed and the speed $\omega = \dfrac{V - I_{a}R_{a}}{k\Phi}$ barely sags with load.
- A series motor 串励电机 has its field in series, so the flux follows $I_{a}$ and torque grows as $I_{a}^{2}$. Unloaded, it races.
A series motor can safely run with no load on its shaft.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Its flux comes from the armature current. With no load the current, and so the flux, is small, and the speed rises until something fails.
Worked example: the rating point
- A shunt motor on 220 V with $R_{a} = 0.2\ \Omega$ draws 50 A at 1500 rpm (157.1 rad/s). $E = 220 - 50 \times 0.2 = 210$ V, so $k\Phi = 1.337$ V·s and $T = 66.9$ N·m.
- Check: $T\omega = 66.9 \times 157.1 = 10.5$ kW and $E I_{a} = 210 \times 50 = 10.5$ kW.
A shunt motor on 240 V has R_a = 0.5 Ω and draws 40 A. What is its back EMF, in V?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$E = V - I_{a}R_{a} = 240 - 40 \times 0.5 = 220$ V.
Worked example: starting
- At standstill $E = 0$, so only $R_{a}$ limits the current: $220/0.2 = 1100$ A, about twenty times rated.
- A starter resistance holding the current to 100 A is $220/100 - 0.2 = 2.0\ \Omega$, cut out in steps as $E$ builds.
The same 240 V motor (R_a = 0.5 Ω) must start at no more than 80 A. What starter resistance is needed, in Ω?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
At standstill $E = 0$, so the total resistance must be $240/80 = 3\ \Omega$; take away $R_{a}$ to leave 2.5 Ω.
Worked example: field weakening
- Cut the flux 20% to $k\Phi = 1.070$ with the same 66.9 N·m load. Now $I_{a} = 62.5$ A, $E = 207.5$ V and $\omega = 193.9$ rad/s, which is 1852 rpm.
- Field weakening 弱磁 trades flux for speed and pays in armature current. Check: $T\omega = 12.96$ kW $= E I_{a}$.
A shunt motor's field flux is reduced while the load torque stays the same. What happens?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Less flux needs more current for the same torque ($I_{a} = T/k\Phi$), and the speed $\omega = E/k\Phi$ rises. That is field weakening.
Three common slips
- rpm in the equations. $E = k\Phi\omega$ needs rad/s; rpm is wrong by a factor of $60/2\pi$.
- Unloaded series motor. With no current there is no flux to limit the speed.
- Direct start. At standstill, only the armature resistance limits the current.
Convert 3000 rpm to rad/s. Give 1 decimal place.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$3000 \times 2\pi / 60 = 314.2$ rad/s. Use this, never rpm, in $E = k\Phi\omega$.
Recap
- $E = k\Phi\omega$, $T = k\Phi I_{a}$, and $T\omega = E I_{a}$.
- Shunt holds speed; series gives big torque at low speed.
- Start through a resistance; weaken the field for more speed.