CN1038553A - Dual-shaft ac governor motor - Google Patents

Abstract

The present invention proposes a kind of speed-regulating theory of alternating current machine and according to the stepless speed-regulating motor of this Design Theory, it is simple in structure, and control is convenient.

The present invention makes two parts with the stator of motor, and wherein the rotatable α angle of one part changes rotor electromotive force by regulating the α angle, thereby reaches the speed governing purpose.

Description

Dual-shaft AC governor motor

In motor driving system, speed regulating control is a kind of method of often using, and the present invention proposes a kind of new speed-regulating theory and according to the buncher of this Design Theory.

Governing system of the present invention is compared with governing system in the past, has simple in structurely, easy to operate, and mechanical property is good, the power factor height, and characteristics such as efficient height are a kind of comparatively ideal speed regulating methods.

One, basic principle

By the Electrical Motor principle as can be known, the rotor electromotive force E of AC asynchronous motor ₂Determine rotating speed of motor, changed E ₂Can change rotating speed of motor.

E ₂Change can be undertaken by number of ways, as in rotor loop, adding an adjustable electromotive force E _f, by changing E _fChange E ₂Size, can reach the speed governing purpose.

Because E ₂Frequency change with motor speed, by mathematical principle as can be known, have only the identical phasor of frequency just stackable, so E _fMust and E ₂Same frequency, E _fWant and E ₂Be consistent, must under similarity condition, produce, as when stator terminal adds the identical electromotive force of two frequencies, will producing two identical electromotive forces of frequency simultaneously at same epitrochanterian winding, but the just addition of these two electromotive forces.

Two electromotive forces that frequency is identical have produced, and how to change E with it ₂Size? by mathematical principle as can be known, change the angle of two phasors, can make these two phasors and size change.As Fig. 1, work as α ₁＜α ₂The time, E _{2 α 1}＞E _{2 α 2}

How to change the α angle, by Electrical Motor as can be known, the rotor electromotive force of AC asynchronous motor, all the time follow the stator electromotive force, if the stator and the stator winding of alternating current motor are made two parts, wherein a part is fixing, and another part is adjustable, the rotatable α angle in adjustable part relatively fixed part, the rotor winding of motor is corresponding stator winding respectively.If the stator fixedly electromotive force responded to respectively on the rotor winding of winding and adjustable winding is E ₂₁And E ₂₂Then the combined potential of rotor loop is:

₂＝

₂₁+ ₂₂

If change the α angle between fixing and adjustable stator, then E ₂₁And E ₂₂Between the α angle also change α degree, then E thereupon ₂Amplitude also change E ₂Changed, rotating speed of motor will change thereupon, therefore, changes the α angle and can change rotating speed of motor.

Two, basic structure

Fig. 2 is the principle assumption diagram of this motor.

Wherein 1. being the fixed stator of motor, 2. is the adjustable stator of motor, 3. be to control 2. the relatively 1. control device of rotation, and 4. be rotor axis of electric, 5. be the fixed stator winding, common name X ₁₁, 6. be the adjustable stator winding.Common name X ₂₁, 7. be the rotor winding, common name X ₂, 8. be tachogenerator, 9. be motor base.A, B, C are the three phase mains inlet wires.

The stator of this motor is winding X fixedly ₁₁With adjustable winding X ₂₁Difference respective rotor winding X ₂, and at X ₂On induce E respectively ₁₂And E ₂₂The electromotive force that two class frequencys are identical.1. and the α angle 2. by 3. regulating, 5. 6. between with 1. 2. changing the phase difference that produces the α angle, 8. be the device that is used as closed-loop control.

Fig. 3 is the stator and rotor circuit theory diagrams of this motor.

Because this motor is equivalent to two AC induction motor mechanical axis and links to each other respectively with electric axis, so claim dual-axle motor

Three, mechanical property

For the electric current of the simple circuit of trying to achieve motor and secondary level circuit, must list the equivalent circuit diagram of this motor, for the ease of analyzing theoretically, suppose that the fixed stator winding of motor is consistent with the parameter of adjustable stator winding, the label of Cai Yonging is here:

U: network voltage.

I _1,2: fixed stator winding X ₁₁With adjustable stator winding X ₂₁Electric current.

i ₁₂: fixed stator winding X ₁₁With adjustable stator winding X ₂₁Respectively at rotor winding X ₂The electric current of last induction.

r ₁, X ₁: the resistance of stator and leakage reactance.

r ₂, X ₂: the resistance of rotor and leakage reactance

X ₀: the reactance of useful flux

S, Sm: revolutional slip and breakdown slip.

e ^ja。Stator is relative stator phase angle 1. 2., is as the criterion with electrical degree.

Because 6. 5. winding be to obtain electromotive force from same electrical network, therefore, their voltage equates, but 5. 6. winding have the phase difference of α angle relatively, so, 5. can be write as U, Ue corresponding to winding with winding voltage 6. ^Ja

According to above parameter can be listed as equivalent circuit such as Fig. 4

If use:

Z ₁＝r ₁+j（X ₀+X ₁）

Z ₂＝r ₂/S+j（X ₀+X ₂）

The simple whole impedance of expression motor when revolutional slip is S and secondary level whole impedance, the voltage equation of this motor stator loop and rotor loop just has following form so.

U＝I ₁Z ₁-i ₁ jX ₀①

Ue ^ja＝I ₂e ^jaZ ₁+i ₁ jX ₀②

2i ₁Z ₂-I ₁jX ₀+I ₂e ^jajX ₀＝0 ③

Separating above-mentioned equation 3. 2. 1. gets

$I_1=\frac{{\mathrm{2Z}}_1{\mathrm{<figure-callout\; id="2"\; label="Z"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">Z</figure-callout>}}_2{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}^2{}_0{\mathrm{(1+e}}^{\mathrm{j\; a}})}{{\mathrm{2Z}}_1{\mathrm{(Z}}_1{\mathrm{<figure-callout\; id="2"\; label="Z"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">Z</figure-callout>}}_2{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}^2{}_0)}\mathrm{<figure-callout\; id="4"\; label="U"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">U</figure-callout>\; ④}$

$I_2=\frac{{\mathrm{2Z}}_1{\mathrm{<figure-callout\; id="2"\; label="Z"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">Z</figure-callout>}}_2{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}^2{}_0{\mathrm{(1+e}}^{\mathrm{-\; j\; a}})}{{\mathrm{2Z}}_1{\mathrm{(Z}}_1{\mathrm{<figure-callout\; id="2"\; label="Z"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">Z</figure-callout>}}_2{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}^2{}_0)}\mathrm{<figure-callout\; id="5"\; label="U"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">U</figure-callout>\; ⑤}$

$i_1=\frac{{\mathrm{jX}}_0{\mathrm{(1-e}}^{\mathrm{j\; a}}\mathrm{)U}}{{\mathrm{2(Z}}_1{\mathrm{<figure-callout\; id="2"\; label="Z"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">Z</figure-callout>}}_2{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}^2{}_0)}⑥$

According to above parameter draw phasor diagram such as Fig. 5.

The torque characteristics of each stator of this motor is with the three-phase alternating-current motor torque characteristics.Motor under specified situation, α=0 o'clock, then the breakdown torque of each stator is:

$M_m=\frac{{\mathrm{<figure-callout\; id="2"\; label="3U"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">3U</figure-callout>}}^2}{{\mathrm{2\sigma }}_1{\mathrm{(<figure-callout\; id="1"\; label="X"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}_2{\sigma }_1)}$

Wherein:

${\mathrm{<figure-callout\; id="1"\; label="\sigma "\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">\sigma </figure-callout>}}_1=\frac{{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1{\mathrm{+X}}_2}{X_0}$

Order: X=X ₁+ X ₂σ ₁

Then: Mm=(3U ²)/(2X σ ₁)

Breakdown slip correspondingly

$S_m=\frac{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">r</figure-callout>}}_2{\sigma }_1}{{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1{\mathrm{+<figure-callout\; id="2"\; label="X"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">X</figure-callout>}}_2{\mathrm{<figure-callout\; id="1"\; label="\sigma "\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">\sigma </figure-callout>}}_1}=\frac{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">r</figure-callout>}}_2{\sigma }_1}{X}$

4., 5. can try to achieve the moment that each stator of this motor produces according to formula:

The total torque of this motor is:

${\mathrm{M=<figure-callout\; id="1"\; label="M"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">M</figure-callout>}}_1\mathrm{+<figure-callout\; id="2"\; label="M"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">M</figure-callout>}{}_2\mathrm{=2M}{}_m\frac{\mathrm{1-cosa}}{\frac{S}{S_m}+\frac{S_m}{S}}⑨$

From formula 9. as can be seen, the torque of this motor is relevant with the α angle, and when α=0, M=0 is when α=180 °

$\mathrm{M=}\frac{{\mathrm{4M}}_m}{\frac{S}{S_m}+\frac{S_m}{S}}$

That is to say that the angle when between stator winding is zero, the electromotive force that electromotive force is an equal and opposite in direction, direction is opposite that two stators that make are responded in the rotor winding, these two electromotive forces are cancelled each other, the combined potential that makes rotor loop is zero, and therefore, motor torque is zero.When the angle between stator winding is 180 °, the direction unanimity of the electromotive force that two stators that make are responded on the rotor winding.After these two electromotive force stacks, form bigger electromotive force, therefore, torque just strengthens.

When the stator winding parameter of motor not simultaneously, the parameter difference here is meant that the number of pole-pairs of two stators is identical, the impedance parameter difference.

Then the total torque of motor is:

${\mathrm{M=<figure-callout\; id="1"\; label="M"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">M</figure-callout>}}_1\mathrm{+<figure-callout\; id="2"\; label="M"\; filenames="881036854\_DRIMG1.png,881036854\_DRIMG2.png"\; state="\{\{state\}\}">M</figure-callout>}{}_2\mathrm{=2M}{}_m\frac{\mathrm{K-cosa}}{\frac{S}{S_m}+\frac{S_m}{S}}\mathrm{⑩K=}\frac{M_2}{{\mathrm{<figure-callout\; id="1"\; label="M"\; filenames="881036854\_DRIMG1.png"\; state="\{\{state\}\}">M</figure-callout>}}_1}\mathrm{＞1}$

Formula 10. be dual-axle motor when two stator moments differ K times, the mechanical characteristic equation formula of dual-axle motor.

By formula 7. 8. as can be known, when some α angle, M is a negative value, and at this moment, this stator is introduced mechanical energy and done generating state from the outside.When this mainly shows the dual-axle motor low-speed motion.

Fig. 6 is the characteristic curve of this motor.

As can be seen from the figure the characteristic of this motor is harder, and this is because its mechanical property is equivalent to the synthetic of two alternating current machine characteristics.

Four, the ideal no-load speed of motor

Because the speed governing of this motor is equivalent to seal in the additional electromotive force E of a same frequency in an AC motor rotor _fTandem control, according to the theory of tandem control as can be known:

E _f＝SE ₂₀

In the formula: E ₂₀Locking rotor electromotive force during for this motor α=180 °

If:

｜E ₂₁｜＝K｜E ₂₂｜

E ₂₁With E ₂₀Between angle be (β)/(X)

Then: E ₂₂With E ₂₀Angle be (X-1)/(X) β

As shown in Figure 7: alpha+beta=180 °

As β=0, then E _f=0

｜E ₂₀｜＝｜E ₂₁｜+｜E ₂₂｜＝（K+1）｜E ₂₂｜

When β ≠ 0

｜E _f｜＝｜E ₂₀｜-（｜E ₂₁｜cos (β)/(X) +｜E ₂₂｜cos (X-1)/(X) β）

＝（K+1）｜E ₂₂｜-（K｜E ₂₂｜cos (β)/(X) +｜E ₂₂｜cos (X-1)/(X) β）

＝｜E ₂₂｜〔（K+1）-Kcos (β)/(X) +cos (X-1)/(X) β）

＝S｜E ₂₀｜

＝S（K+1）｜E ₂₂｜

Then:

S（K+1）＝（K+1）-（Kcos (β)/(X) +cos (X-1)/(X) β）

$\frac{n_O\mathrm{-n}}{n_O}\mathrm{(K+1)=(K+1)-(Kcos}\frac{\beta }{\alpha }\mathrm{+cos}\frac{\mathrm{X-1}}{X}\mathrm{\beta \; )n=}\frac{n_O}{\mathrm{K+1}}\mathrm{(Kcos}\frac{\beta }{X}\mathrm{+\; cos}\frac{\mathrm{X-1}}{X}\mathrm{\beta \; )\; ⑾}$

β＝±180～0°

When K=1, X must equal 2.

Then: n=n ₀Cos (β)/2 (12)

N in the formula (11) is meant dual-axle motor when the different beta angle, the ideal no-load speed of motor.In (11) (12).When β=0, n=n ₀, the rotating speed maximum, during β=180 °, the rotating speed minimum, β changes, and rotation speed n also changes thereupon.

The asynchronous mechanical characteristics of motor curve of the β that drawn among Fig. 6.

The positive negative value at α angle, the 2. leading or 1. decision that lags behind by stator.Be meant in advance 2. in 1. the place ahead of rotating magnetic field.Otherwise, lag behind, when leading, α is for just, and the positive negative value of α influences the power factor of motor.Usually, E ₁₁＞E ₂₁, α is timing, rotor power factor height.

Five, the starting of motor and control

Because the rotor electromotive force E of this motor ₂Scalable, therefore, the directly bringing onto load starting of this motor need not be adopted step-down or rotor-resistance starting, particularly K value hour, when K value than greatly the time, then need rotor crosstalk resistance or reduced voltage starting.During this electric motor starting, can earlier the α angle be transferred to minimum, then, increase the α angle gradually, reach desired speed to motor and end.

The speed governing of this motor can be adopted manually or control automatically, by 3. making 2. relatively 1. rotation of stator, installs 3. available handle or remote-control motor and forms.

Six, other structure of this motor

According to the operation principle of this motor, can make two series connection windings to the rotor winding of this motor, it is the fixedly winding and the adjustable winding of corresponding stator respectively, and the same rotor winding of the effect of this structure is identical, as shown in Figure 8.

The stator winding of this motor also can be made sleeve shaped, wherein fixes for one group, and another group relative fixed winding is adjustable, and this structure also has the speed governing effect.

But sometimes, motor is not required speed governing, only require that normal speed is lower, can in the stator slot of motor,, just can reach the purpose that reduces rated speed according to this theory with two cover windings under certain α angle.At this moment, the desirable unloaded fast cotype (11) of motor.

If the stator winding separated into two parts of motor is with at present in a stator slot, wherein first directly connects power supply.Second portion is drawn, and receives on the adjustable stator by brush, connects power supply again, by design make shifting brush then second portion winding and first winding produce the effect that differs α angular phase difference, then this structure also has the adjustable speed effect.

If the rotor of this motor is fixed, then this motor will become a three-phase induction regulator, and this voltage regulator voltage output is adjustable continuously, the fricton-tight contact of voltage regulator, and it is very big that power can be done.

Claims (6)

1, a kind of alternating current machine stepless speed regulation theoretical and according to the alternating stepless speed regulating machine of this Design Theory is characterized in that:

(1), 1. the stator of motor make, 2. two parts, its 2. relatively 1. rotatable α angle,

(2), winding is 5. 6. with 1. 2. rotation and induce two groups of electromotive forces at the rotor winding on 7. respectively.

(3), the rotor combined potential changes with α.

2, motor as claimed in claim 1, it is characterized in that: stator is made socket type, wherein the rotatable α angle of one part.

3, motor as claimed in claim 1 is characterized in that: the rotor winding is in series by two parts.

4, motor as claimed in claim 1 is characterized in that: following two cover windings in same stator slot, and wherein, a cover directly connects power supply, another set ofly connects power supply by brush, and shifting brush makes two cover windings produce the effect that differs the α angular phase difference.

5, a kind of low speed AC motor is characterized in that: stator winding is made two parts and is differed under the α angle in same stator slot.

6, a kind of alternating current impression voltage regulator is characterized in that: former limit is made up of two cover windings, but does not wherein have connection between former, the secondary winding of an another set of relatively phase shift of cover.

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