Induction machines
مُقدَّم| English | المعنى |
|---|---|
| squirrel-cage | squirrel-cage |
| rotating field | A magnetic field that turns at a steady speed, produced by three-phase currents in the stator windings. |
| synchronous speed | synchronous speed |
| slip | الانزلاق |
| breakdown torque | breakdown torque |
| wound rotor | wound rotor |
Seven motors in ten
- Around seven of every ten industrial motors are squirrel-cage 鼠笼式 induction machines. They have no brushes and only one moving part.
- The secret is a field that rotates by itself. This lesson follows the rotor as it chases that field.
The rotating field
- Three coils 120° apart, fed with currents 120° apart in time, make a rotating field 旋转磁场 of constant size.
- It turns at the synchronous speed 同步转速 $n_{s} = \dfrac{120f}{p}$ rpm, with $f$ in Hz and $p$ the number of poles.
What is the synchronous speed of a 2-pole motor on 60 Hz, in rpm?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
$n_{s} = 120 \times 60 / 2 = 3600$ rpm.
Slip
- The rotor chases the field but never quite catches it. The shortfall is the slip 转差率 $s = \dfrac{n_{s} - n}{n_{s}}$.
- The rotor bars see the field slide past at frequency $sf$, so every rotor quantity follows from the slip.
The field rotates, the rotor chases
Move the slip and watch the stator field, the rotor and their relative speed; read the rotor frequency off the gap between them.
A 6-pole, 50 Hz motor runs at 960 rpm. What is its slip? Give 2 decimal places.
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
$n_{s} = 120 \times 50/6 = 1000$ rpm, so $s = (1000 - 960)/1000 = 0.04$.
A 50 Hz motor runs at a slip of 0.05. What frequency do the rotor currents have, in Hz?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
The rotor sees $sf = 0.05 \times 50 = 2.5$ Hz, not 50 Hz.
Where the power goes
- In the equivalent circuit the rotor appears as $R_{2}'/s = R_{2}' + R_{2}'\dfrac{1-s}{s}$. The first part is rotor copper loss; the second is the shaft.
- So slip divides the air-gap power $P_{ag}$: copper gets $sP_{ag}$ and the shaft gets $(1-s)P_{ag}$.
10 kW crosses the air gap of a motor running at 3% slip. How much is lost as rotor copper loss, in W?
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
Copper takes $sP_{ag} = 0.03 \times 10\,000 = 300$ W; the shaft gets the other 9.7 kW.
Worked example: the nameplate
- A 4-pole, 50 Hz motor: $n_{s} = 120 \times 50/4 = 1500$ rpm. Running at 1440 rpm, $s = 60/1500 = 0.04$.
- The rotor frequency is $sf = 2$ Hz, and $R_{2}'/s = 25R_{2}'$: near synchronous speed the rotor looks almost purely resistive.
The torque–slip curve
- Torque rises with slip, peaks at the breakdown torque 最大转矩, then falls. The peak sits where $R_{2}'/s$ equals $\sqrt{R_{th}^{2} + (X_{th} + X_{2}')^{2}}$.
- More rotor resistance moves the peak towards starting without changing its height. The wound rotor 绕线转子 sells exactly this.
Doubling the rotor resistance doubles the breakdown torque.
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
The peak height does not depend on $R_{2}'$. More rotor resistance only moves the peak to a larger slip, which raises the starting torque.
Worked example: the torque chain
- With $V_{th} = 231$ V, $R_{th} = 0.5\ \Omega$, $X_{th} + X_{2}' = 2.4\ \Omega$, $R_{2}' = 0.4\ \Omega$ and $s = 0.04$: $T = 87.8$ N·m.
- $P_{ag} = 13.8$ kW, rotor copper loss 552 W, shaft 13.2 kW. Breakdown: 172 N·m at $s = 0.163$; locked rotor, 62 N·m.
In the torque chain, T = 87.8 N·m at ω_s = 157.1 rad/s and s = 0.04. What is the shaft power?
الصعوبة 1/5 · 100 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
$P_{ag} = T\omega_{s} = 13.8$ kW, and the shaft gets $(1 - 0.04) \times 13.8 = 13.2$ kW. The other 552 W heats the rotor.
Three common slips
- rpm in the torque formula. $\omega_{s}$ must be in rad/s.
- "More slip, more torque." Torque peaks at $s_{max}$ and falls on both sides of it.
- 50 Hz in the rotor. The rotor frequency is $sf$: 2 Hz here.
Match each operating point to its slip.
الصعوبة 2/5 · 140 عملة + مكافأة صغيرة عند الإنجاز الأول اليوم
At s = 0 the bars see no moving field, so there is no current and no torque; a motor always runs a little below synchronous speed.
Recap
- $n_{s} = 120f/p$; slip $s = (n_{s} - n)/n_{s}$; rotor frequency $sf$.
- Copper takes $sP_{ag}$; the shaft takes $(1-s)P_{ag}$.
- Rotor resistance moves the torque peak, not its height.