DC machines
Diperkenalkan| English | Makna |
|---|---|
| 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.
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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?
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$T = k\Phi I_{a} = 2 \times 30 = 60$ 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.
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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?
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$E = V - I_{a}R_{a} = 240 - 40 \times 0.5 = 220$ 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?
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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?
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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.
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$3000 \times 2\pi / 60 = 314.2$ 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.