CN1570390A - Whirlpool compressor - Google Patents

Abstract

One kind of new vortex compressor, its vortex root line based on three base circle (Three Basic Cricle, is called TBC) the theory to extend the revision line. This line inside revises involute the section E1J1, inside by flank host involute section J2E2, the connection tangential path E2E1, to revise involute section J1k1, revision great-circle segmental arc K1N1, the revision small circle segmental arc NK2, flank to revise involute section K2J2 to be composed. Revision circular arc center of a circle O1 and O2 in do not bind me first base circle tangent on. This invention has solved the PMP line turbulence root section (namely high pressure region spot) the intensity and the compression ratio mutually limits, the vortex compressor compression ratio cannot largely enhance tangential point, thus while enhances the vortex compressor compression ratio, the vortex root intensity also has the enhancement.This invention involves the vortex compressor is suitable for the air conditioning, the refrigeration, aspect and so on vacuum pump and gas loading.

Description

Scroll compressor

Technical field

The present invention relates to a kind of scroll compressor.This scroll compressor is used for aspects such as air-conditioning, refrigeration, vacuum pump and gas compression.

Background technique

Scroll compressor is the nearly two a kind of brand-new compressors that occur in the world during the last ten years, has advantages such as mechanism is simple, efficient, low noise.For obtaining compression ratio efficiently, each state has all made extensive work to the correction of scroll compressor vortex root molded lines.The extensive swirl compressor spinning disk vortex root (being the position, zone of high pressure) that adopts in various countries is revised a kind of Perfect MeshIndustrial molded lines (being called for short the PMP molded lines) that meshes fully that molded lines is the exploitation of Mitsubishi heavy industry at present.It is characterized in that the position, zone of high pressure is formed by two sections circular arc corrections, and revise the circular arc center of circle all on the tangent line of crossing main vortex filament basic circle.

Though the PMP molded lines can be with the residual gas emptying, avoided repeated compression, but analyze it and revise principle as can be known: for obtaining high compression ratio, the exhaust angle is increased, but the main vortex filament section of being corrected is reduced, only the vortex filament wall thickness increases to some extent near exhaust port, and remaining part vortex filament wall thickness is constant, thereby can not guarantee at root section enough intensity is arranged, otherwise, root section intensity is improved, and can only be cost to reduce compression ratio then.Therefore, make and cause the scroll compressor compression ratio significantly not improve, power is restricted, and makes how most of scroll compressors use under air conditioning condition.

Summary of the invention

The objective of the invention is to limit the shortcoming that the scroll compressor compression ratio can not significantly improve mutually in order to solve PMP molded lines vortex root section (being the position, zone of high pressure) intensity and compression ratio.The present invention relates to a kind of scroll compressor, proposed a kind of based on the theoretical extension correction molded lines of three basic circles (Three Basic Circle is called for short TBC).

The scroll compressor that the present invention relates to, its whirlpool dish vortex root section (being the position, zone of high pressure) molded lines is based on the extension correction molded lines of " three basic circles " molded lines (being called for short the TBC molded lines) revised theory, and this molded lines is by the outside main involute section J ₂E ₂, connect straightway E ₂E ₁, inboard main involute section E ₁J ₁, the inboard involute section J that revises ₁K ₁, revise great circle segmental arc K ₁N revises roundlet segmental arc NK ₂, involute section K is revised in the outside ₂J forms.Wherein:

(1) the main involute section J in the outside ₂E ₂Equation is:

Wherein, x ₀=0, y ₀=0, α=α, α is the initial angle of the main involute section of survey outside forming,

${\mathrm{\θ}}_{\min }=\frac{3\mathrm{\π}}{2}+\mathrm{\ψ}-\mathrm{\α},$ ψ is three basic circle center of circle place straight line and X-axis angles.

(2) connect straightway E ₂E ₁Equation is:

Straightway E ₂E ₁End points E ₂, E ₁Respectively with involute J ₂E ₂, E ₁J ₁On E ₂And E ₁Point is a same point, so coordinate figure is known, then E ₂E ₁The parametrization equation be:

$y=\mathrm{LN}(x_{E_1},y_{E_1},x_{E_2},y_{E_2},x)$

(3) inboard main involute section E ₁J ₁Equation is:

Wherein, x ₀=0, y ₀=0, α=-α ,-α is the initial angle of the main involute section of survey in forming,

${\mathrm{\θ}}_{\min }=\frac{3\mathrm{\π}}{2}+\mathrm{\ψ}+\mathrm{\α},$ ψ is three basic circle center of circle place straight line and X-axis angles.

(4) the inboard involute section J that revises ₁K ₁(the auxiliary basic circle in left side forms) equation is:

Wherein, x ₀=-Dcos ψ, y ₀=-Dsin ψ, α=α ₀₁, ${\mathrm{\α}}_{01}=\frac{3\mathrm{\π}}{2}+\mathrm{\ψ}-\frac{L_1}{a},$ α ₀₁For forming the inboard involute section J that revises ₁K ₁Initial angle, form by the auxiliary basic circle in left side, D be the left and right auxiliary basic circle center of circle respectively with the distance in the basic circle center of circle, center, θ=α _K1, ${\mathrm{\α}}_{K1}=\mathrm{\π}-{\mathrm{\α}}_{01}+\mathrm{\ψ}+{\mathrm{\δ}}^{\′},$ ${\mathrm{\δ}}^{\′}=\frac{\mathrm{\π}}{2}-\mathrm{\δ},$ δ is the inboard involute section J that revises ₁K ₁The angle of spread.

(5) revise great circle segmental arc K ₁The N equation is:

Revise great circle segmental arc K ₁The end points K of N ₁For revising involute section J in the inboard ₁K ₁On point, center of circle x ₀₁, y ₀₁Point coordinates is:

γ is the angle of two correction circular arc center of circle place straight lines and place, three basic circle centers of circle straight line,

For the auxiliary basic circle distance of center circle in center basic circle and left side from, revise great circle segmental arc K ₁N and correction roundlet segmental arc NK ₂Tangent at the N point.

$x_N=\frac{(R_1-R_2)\cos {\mathrm{\γ}}_P}{2},$ ? $y_N=\frac{(R_1-R_2)\sin {\mathrm{\γ}}_P}{2}$

R ₁, R ₂Be two correction radius of arc;

(6) revise roundlet segmental arc NK ₂Equation is:

Revise great circle segmental arc NK ₂End points K ₂For revising involute section K in the outside ₂J ₂On point, center of circle x ₀₂, y ₀₂Point coordinates is:

Revise great circle segmental arc K ₁N and correction roundlet segmental arc NK ₂Tangent at the N point.

(7) involute section K is revised in the outside ₂J ₂(the auxiliary basic circle in right side forms) equation is:

Wherein, x ₀=Dcos ψ, y ₀=Dsin ψ, α=α ₀₂, ${\mathrm{\α}}_{02}=\frac{\mathrm{\π}}{2}+\mathrm{\ψ}-\frac{L_2}{a},$ α ₀₂For forming

Involute section K is revised in the outside ₂J ₂Initial angle, form by the auxiliary basic circle in right side, D be the left and right auxiliary basic circle center of circle respectively with the distance in the basic circle center of circle, center, θ=α _K2, α _K2=δ '+ψ-α ₀₂, ${\mathrm{\δ}}^{\′}=\frac{\mathrm{\π}}{2}-\mathrm{\δ},$ δ is the inboard involute section J that revises ₁K ₁The angle of spread.

The novel scroll compressor that the present invention relates to, vortex root section (being the position, zone of high pressure) molded lines is completed into by the extension correction molded lines based on " three basic circles " molded lines (being called for short the TBC molded lines) revised theory.Revise circular arc center of circle O ₁And O ₂All not on the tangent line of crossing main vortex filament basic circle, overcome the shortcoming of PMP molded lines, the benefit of bringing thus is:

＜1 〉. when β ↓, $R_1=\stackrel{\‾}{K_1{O}_1}\↓$ The time, involute J ₁K ₁Section vortex filament wall thickness still can extend internally into wall thickness to be increased, and scroll molded line root intensity is improved.

＜2 〉. because the section wall thickness increases near this theoretical molded lines root, a pair of enclosed cavity volume that makes a pair of swirl chamber mesh instantaneous formation reduces, thereby can improve compression ratio.

Description of drawings

Fig. 1 is scroll molded line root " extension deformation correction " principle schematic;

Fig. 2 is scroll molded line root " the extension deformation correction " principle schematic that additional straight line correction is arranged;

Fig. 3 is a Japan Patent molded lines PMP schematic representation;

Fig. 4 is for after " extending the deformation correction " through the scroll molded line root, to Japanese molded lines contrast schematic representation;

When Fig. 5 is PMP molded lines Moving plate, the engagement of quiet dish, the exhaust cavity area;

When Fig. 6 is scroll molded line root " extension deformation correction " molded lines Moving plate, the engagement of quiet dish, the exhaust cavity area;

Embodiment

Basic thought that " extension deformation correction " is theoretical such as following explanation:

As shown in Figure 1, main vortex filament basic circle O, the center of circle is now introduced radius and is a at the O point, apart from the O point be D about two from basic circle O ' ₁And O ' ₂, the center of circle of three master and slave basic circles point-blank, three centers of circle place straight line and X-axis angle are ψ.The inside and outside main vortex filament that is generated by this main basic circle is involute J ₁E ₁And J ₂E ₂, its initial angle be respectively α and-α, main Base radius is a, existing sets inside and outside vortex filament starting point J ₁And J ₂Do not change with the β angle, but the fixed point in Fig. 1.To Fig. 1

J ₁The beginning angle of point is: ${\mathrm{\α}}_1=\frac{3\mathrm{\π}}{2}+\mathrm{\ψ}-\mathrm{\α}----\left(1\right)$

Outwards launch to generate main vortex filament J thus ₁E ₁Section

J ₂The beginning angle of point is: ${\mathrm{\α}}_2=\frac{\mathrm{\π}}{2}+\mathrm{\ψ}+\mathrm{\α}----\left(2\right)$

Outwards launch to generate main vortex filament J thus ₂E ₂Section

From J ₁And J ₂Respectively with two from the basic circle involute J that extends internally ₁K ₁Section and J ₂K ₂Section, Fig. 1 middle conductor J ₁C ₁Around from basic circle O ' ₁Turning over angle δ as pure rolling cuts from basic circle O ' ₁In Q ₁Point, line segment J ₂C ₂Turning over equal angular cuts from basic circle O ' ₂In Q ₂The point.Owing to introduced extension deformation involute J ₁K ₁And J ₂K ₂, therefore, revise circular arc center of circle O in the Figure of abstract ₁And O ₂All not on the tangent line of crossing main vortex filament basic circle, by O ₁And O ₂Remake the circular arc correction, the center of circle of having broken through the vortex root molded lines correction circular arc of stipulating in the Japan Patent can only be in the restriction on the main vortex filament basic circle tangent line.

The present invention is based on the extension correction molded lines of " three basic circles " molded lines (being called for short the TBC molded lines) revised theory, and it specifically derives as follows:

Revise the theoretical calculation and the derivation of molded lines:

1, the no additional straight line segment correction of introducing parameter D, ψ is known by Fig. 1: δ=π-β, $\mathrm{\δ}+{\mathrm{\δ}}^{\′}=\frac{\mathrm{\π}}{2},$ So ${\mathrm{\δ}}^{\′}=\mathrm{\β}-\frac{\mathrm{\π}}{2}$ ；

By this correction thought and involute character as can be known: curved section J ₁K ₁Section and J ₂K ₂The section initial angle is respectively:

It begins angle (corresponding K ₁And K ₂Point) and end angle (corresponding J ₁And J ₂Point) be respectively:

Be correct engagement, should have:

$R_1+R_2=\stackrel{\‾}{O_1{O}_2}=2\stackrel{\‾}{O_{1}O}=2\stackrel{\‾}{O_{2}O}$

By right-angled triangle OO among Fig. 1 ₁B ₁Have:

${\left(\frac{R_1+R_2}{2}\right)}^2={\stackrel{\‾}{O_1{B}_1}}^2+{\stackrel{\‾}{{\mathrm{OB}}_1}}^2----\left(7\right)$

And ${\mathrm{\ΔOO}}_1{B}_1\≅\mathrm{\Δ}{\mathrm{OO}}_2{B}_2,$ B ₁, B ₂Being respectively the O point does Elongation line reaches

The intersection point of the vertical line of elongation line.

$R_2=\stackrel{\‾}{Q_1{K}_1}+\stackrel{\‾}{Q_1{B}_1}-\stackrel{\‾}{O_1{B}_1}----\left(8\right)$

$R_2=\stackrel{\‾}{Q_2{K}_2}+\stackrel{\‾}{Q_2{B}_2}-\stackrel{\‾}{O_2{B}_2}----\left(9\right)$

Wherein: $\stackrel{\‾}{O_1{B}_1}=\stackrel{\‾}{O_2{B}_2},\stackrel{\‾}{Q_1{B}_1}=\stackrel{\‾}{Q_2{B}_2}----\left(10\right)$

Formula (8)+formula (9):

${R_1+R}_2=\stackrel{\‾}{Q_1{K}_1}+\stackrel{\‾}{Q_2{K}_2}+2\stackrel{\‾}{Q_1{B}_1}-2\stackrel{\‾}{O_1{B}_1}----\left(11\right)$

Order $S=\stackrel{\‾}{Q_1{K}_1}+\stackrel{\‾}{Q_2{K}_2}+2\stackrel{\‾}{Q_1{B}_1}----\left(12\right)$

So formula (11) can be expressed as:

$R_1+R_2=S-2\stackrel{\‾}{O_1{B}_1}---\left(13\right)$

Substitution formula (7), try to achieve:

$O_1{B}_1=\frac{S}{4}-\frac{{\stackrel{\‾}{{\mathrm{OB}}_1}}^2}{S}----\left(14\right)$

Character by Fig. 1 and involute is known:

By figure Q among Fig. 1 ₁O ₁O ' ₁OB ₁Shown in:

$\stackrel{\‾}{O_1^{\′}O}=D,$ ? $\stackrel{\‾}{O_1^{\′}{Q}_1}=a$

To Δ O ' ₁Q ₁O, You Yu Xian theorem gets:

$\stackrel{\‾}{{\mathrm{OQ}}_1}=\sqrt{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}}$

And You Zheng Xian theorem gets:

$\sin \mathrm{\σ}=\frac{D}{\stackrel{\‾}{{\mathrm{OQ}}_1}}{\sin \mathrm{\δ}}^{\′}$

So: $\stackrel{\‾}{Q_1{B}_1}=\stackrel{\‾}{{\mathrm{OQ}}_1}\sin \mathrm{\σ}=D\sin {\mathrm{\δ}}^{\′}----\left(16\right)$

$\stackrel{\‾}{{\mathrm{OB}}_1}=\sqrt{{\stackrel{\‾}{O{Q}_1}}^2-{\stackrel{\‾}{Q_1{B}_1}}^2}---\left(17\right)$

Thus: can be by formula (8), (9) obtain R ₁, R ₂As shown in Figure 1:

$\mathrm{\γ}={\mathrm{\δ}}^{\′}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{OB}_1}}\right)=\mathrm{\β}-\frac{\mathrm{\π}}{2}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{\mathrm{OB}}_1}}\right)----\left(18\right)$

Gu Tuoniejiao is: θ ^*=2 π-γ (19)

2, introduce parameter D, ψ has a correction of additional straight line segment

If the O that obtains by Fig. 1 ₁, O ₂Corresponding to the O among Fig. 2 _1P, O _2P, R among Fig. 1 then ₁, R ₂Corresponding to

$R_1^{\′}=\stackrel{\‾}{K_1{O}_{1P}},$ $R_2^{\′}=\stackrel{\‾}{K_2{O}_{2P}},$ Now will $\stackrel{\‾}{O_{1P}{O}_{2P}}$ Line segment turns over several angle clockwise around the O point, makes O _1PPoint goes to O ₁Point, O _2PPoint goes to O ₂Point, O _1PPoint is to O ₁Dot spacing then has from being ε:

Make among Fig. 1 In Fig. 2 be Then among Fig. 2:

$\stackrel{\‾}{O_1{B}_1}=\stackrel{\‾}{O_{1P}{B}_1}+\mathrm{\ϵ}----\left(21\right)$

Among Fig. 2:

$\stackrel{\‾}{O_1{O}^{\′}}=\frac{(R_1+R_2)}{2}----\left(22\right)$

$\stackrel{\‾}{{\mathrm{OO}}_1}=\sqrt{{\stackrel{\‾}{{\mathrm{OB}}_1}}^2+{\stackrel{\‾}{O_1{B}_1}}^2}----\left(23\right)$

$\mathrm{\η}=\sin \left(\frac{\stackrel{\‾}{O_1{O}^{\′}}}{\stackrel{\‾}{{\mathrm{OO}}_1}}\right)----\left(24\right)$

Know by no straight line correction Chinese style (18): γ → γ _P

Then: $\mathrm{\γ}=\frac{\mathrm{\π}}{2}-\mathrm{\η}+{\mathrm{\γ}}_P+\mathrm{\ψ}----\left(25\right)$

The drop out of gear angle gets: θ ^*=2 π-γ (26)

3, the constraint conditio of parameter D and ψ

Parameter D and ψ and main vortex filament design basic parameter have certain restrictive condition.If the value of D and ψ is excessive, do not satisfy constraint conditio, revising circular arc for two then can not be correctly tangent.

The constrained procedure derivation of equation is as follows:

When $y_{Q_2}\≥y_{K_2}$ And $y_{O_2}\≥y_{K_2}$ The time, the smooth connection of molded lines.

Right $y_{O_2}\≥y_{K_2},$ Promptly $\stackrel{\‾}{{\mathrm{OO}}_2}\cos (\mathrm{\γ}+\mathrm{\ψ})\≥\stackrel{\‾}{{\mathrm{OO}}_2}\cos (\mathrm{\γ}+\mathrm{\ψ})+R_2\sin (\mathrm{\δ}-\mathrm{\ψ})----\left(27\right)$

The following formula abbreviation gets: R ₂(δ-ψ)≤0 (28) for sin

Analytic expression (28), R ₂Can only be for just, so

Sin (δ-ψ)＜0 o'clock, might form molded lines, because R ₂And δ, ψ is also relevant.

By formula (12) $R_1+R_2=S-2\stackrel{\‾}{O_1{B}_1}----\left(29\right)$

By formula (8), (9) R ₁-R ₂=α (π-2 α) (30)

Simultaneous formula (29), (30): $R_2=\frac{1}{2}[S-2\stackrel{\‾}{O_1{B}_1}-a(\mathrm{\π}-2\mathrm{\α})]----\left(31\right)$

In the formula (31):

$S=\stackrel{\‾}{Q_1{K}_1}+\stackrel{\‾}{Q_2{K}_2}+2\stackrel{\‾}{Q_1{B}_1}$

$=L_1-\mathrm{a\δ}+L_2-\mathrm{a\δ}+2D\sin {\mathrm{\δ}}^{\′}$

$=a(\frac{3\mathrm{\π}}{2}+\mathrm{\ψ}-\mathrm{\α})-D-\mathrm{a\δ}+a(\frac{\mathrm{\π}}{2}+\mathrm{\ψ}+\mathrm{\α})-D-\mathrm{a\δ}+2D\sin {\mathrm{\δ}}^{\′}$

$=2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D(\sin {\mathrm{\δ}}^{\′}-1)----\left(32\right)$

$O_1{B}_1=\frac{S}{4}-\frac{{\stackrel{\‾}{{\mathrm{OB}}_1}}^2}{S}$

$=\frac{S}{4}-\frac{{\stackrel{\‾}{{\mathrm{OQ}}_1}}^2-{\stackrel{\‾}{Q_1{B}_1}}^2}{S}$

$=\frac{S}{4}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{S}$

$=\frac{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D(\sin {\mathrm{\δ}}^{\′}-1)}{4}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}$

$=\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D(\sin {\mathrm{\δ}}^{\′}-1)}{2}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}----\left(33\right)$

Formula (32), (33) substitution formula (31):

$R_2=\frac{1}{2}[S-2\stackrel{\‾}{O_1{B}_1}-a(\mathrm{\π}-2\mathrm{\α})]$

$=\frac{1}{2}\{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)-2[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D({\sin \mathrm{\δ}}^{\′}-1)}{2}$

$=\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D(\sin {\mathrm{\δ}}^{\′}-1)}]-a(\mathrm{\π}-2\mathrm{\α})\}---\left(34\right)$

So, right $y_{O_2}\≥y_{K_2}$

R ₂＞0 and sin (δ-ψ)＜0 o'clock, can form smooth molded lines;

Right $y_{Q_2}\≥y_{K_2},$ Promptly

$D\sin \mathrm{\ψ}+a\sin ({\mathrm{\δ}}^{\′}+\mathrm{\ψ})\≥\stackrel{\‾}{{\mathrm{OO}}_2}\cos (\mathrm{\γ}+\mathrm{\ψ})+R_2\sin (\mathrm{\δ}-\mathrm{\ψ})----\left(35\right)$

In the formula (35),

${\stackrel{\‾}{{\mathrm{OO}}_2}}^2={\stackrel{\‾}{{\mathrm{OB}}_1}}^2+{\stackrel{\‾}{O_1{B}_1}}^2$

$={\stackrel{\‾}{{\mathrm{OQ}}_1}}^2-{\stackrel{\‾}{{Q_{1}B}_1}}^2+{\stackrel{\‾}{O_1{B}_1}}^2$

$=a^2+D^2+2\mathrm{aD}{\cos \mathrm{\δ}}^{\′}-D^2\mathrm{si}{n}^2{\mathrm{\δ}}^{\′}$

$+[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D(\sin {\mathrm{\δ}}^{\′}-1)}{2}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}{]}^2----\left(36\right)$

Formula (36) substitution formula (35):

$D\sin \mathrm{\ψ}+a\sin ({\mathrm{\δ}}^{\′}+\mathrm{\ψ})\≥$

$\{a^2+D^2+2\mathrm{aD}{\cos \mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}$

$+[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D(\sin {\mathrm{\δ}}^{\′}-1)}{2}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}{]}^2{\}}^{\frac{1}{2}}\cos (\mathrm{\γ}+\mathrm{\ψ})$

$+\frac{1}{2}\{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D(\sin {\mathrm{\δ}}^{\′}-1)-2[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D({\sin \mathrm{\δ}}^{\′}-1)}{2}$

$-\frac{a^2+D^2+2\mathrm{aD}{\cos \mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}]-a(\mathrm{\π}-2a)\}\sin (\mathrm{\δ}-\mathrm{\ψ})----\left(37\right)$

The γ angle can be got by formula (18) in the formula (37).Promptly

$\mathrm{\γ}={\mathrm{\δ}}^{\′}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{\mathrm{OB}}_1}}\right)=\mathrm{\β}-\frac{\mathrm{\π}}{2}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{\mathrm{OB}}_1}}\right)$

So obtaining formula (37) is the judgement formula.As parameter a, δ, δ ', ψ one timing, the scope of D can be decided, and the gained scope can form the restrictive condition of smooth closed curve.Then whether molded lines can smooth closure can be judged.

In sum, the constraint conditio of D and ψ is

When

$R_2=\frac{1}{2}[S-2\stackrel{\‾}{O_1{B}_1}-a(\mathrm{\π}-2\mathrm{\α})]$

$=\frac{1}{2}\{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)-2[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D({\sin \mathrm{\δ}}^{\′}-1)}{2}$

$-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}]-a(\mathrm{\π}-2\mathrm{\α})\}>0$

And sin (δ-ψ)＜0 o'clock and

$D\sin \mathrm{\ψ}+a\sin ({\mathrm{\δ}}^{\′}+\mathrm{\ψ})\≥$

$\{a^2+D^2+2\mathrm{aD}{\cos \mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}$

$+[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D(\sin {\mathrm{\δ}}^{\′}-1)}{2}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}{]}^2{\}}^{\frac{1}{2}}\cos (\mathrm{\γ}+\mathrm{\ψ})$

$=\frac{1}{2}\{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)-2[\frac{a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+D({\sin \mathrm{\δ}}^{\′}-1)}{2}$

$-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\δ}}^{\′}-D^2{\sin }^2{\mathrm{\δ}}^{\′}}{2a(\mathrm{\π}+\mathrm{\ψ}-\mathrm{\δ})+2D({\sin \mathrm{\δ}}^{\′}-1)}]-a(\mathrm{\π}-2\mathrm{\α})\}\sin (\mathrm{\δ}-\mathrm{\ψ})$

Wherein $\mathrm{\γ}={\mathrm{\δ}}^{\′}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{\mathrm{OB}}_1}}\right)=\mathrm{\β}-\frac{\mathrm{\π}}{2}+\mathrm{\ψ}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\‾}{O_1{B}_1}}{\stackrel{\‾}{{\mathrm{OB}}_1}}\right)$ The time, can form smooth molded lines.

More than be the parameter D and the ψ constrained procedure derivation of equation.

Be another D and the ψ constrained procedure derivation of equation below, can be used as the check formula:

Revise the circular arc distance of center circle from being for two $L_{O_1{O}_2}=\sqrt{{(x_{O_1}-x_{O_2})}^2+{(y_{O_1}-y_{O_2})}^2}----\left(38\right)$

When $L_{O_1{O}_2}<(R_1+R_2)$ The time, the smooth connectivity of molded lines is destroyed, and in such cases, it is too big to revise circular arc.

When $L_{O_1{O}_2}=(R_1+R_2)$ The time, two correction circular arcs are tangent, and this moment, molded lines was only used the circular arc correction.

When $L_{O_1{O}_2}>(R_1+R_2)$ The time, must further revise with straightway, in this case its point of contact N ₁, N ₂Coordinate is determined by following formula:

4, each point coordinates of deriving

Point on the involute, try to achieve by following equation:

J ₁The point:

x ₀=0, y ₀=0, θ=α ₁[formula (1)], α=α; Substitution formula (40);

J ₂The point:

x ₀=0, y ₀=0, θ=α ₂[formula (2)], α=-α; Substitution formula (40);

K ₁The point:

x ₀=-Dcos ψ, y ₀=-Dsin ψ, θ=α _K1[formula (5)], α=α ₀₁[formula (4)]; Substitution formula (40);

K ₂The point:

x ₀=Dcos ψ, y ₀=Dsin ψ, θ=α _K2[formula (6)], α=α ₀₂[formula (4)]; Substitution formula (40);

x ₀₁, y ₀₁The point:

x ₀₂, y ₀₂The point:

Determine by formula (23);

γ _PDetermine by formula (8);

x _N1, y _N1The point:

x _N2, y _N2The point:

$\stackrel{\&OverBar;}{N_1{N}_2}=\sqrt{{(x_{N2}-x_{N1})}^2+{(y_{N2}-y_{N1})}^2}----\left(45\right)$

When ε=0, x _N1=x _N2=x _N, y _N1=y _N2=y _N

Calculate accumulated error for reducing, then:

$x_N=\frac{(R_1-R_2)\cos {\mathrm{\&gamma;}}_P}{2},$ ? $y_N=\frac{(R_1-R_2)\sin {\mathrm{\&gamma;}}_P}{2}----\left(46\right)$

Can form the novel scroll molded line that gets by top derivation.

Referring to Fig. 1: main vortex filament basic circle O, the center of circle is now introduced radius and is a at the O point, apart from the O point be D about two from basic circle O ' ₁And O ' ₂, the center of circle of three master and slave basic circles point-blank, three centers of circle place straight line and X-axis angle are ψ.The inside and outside main vortex filament that is generated by this main basic circle is involute J ₁F ₁And J ₂E ₂, its initial angle be respectively α and-α, main Base radius is a, existing sets inside and outside vortex filament starting point J ₁And J ₂Do not change with the β angle, but the fixed point in Fig. 3.

From J ₁And J ₂Respectively with two from the basic circle involute J that extends internally ₁K ₁Section and J ₂K ₂Section, line segment J ₁C ₁Around from basic circle O ' ₁Turning over angle δ as pure rolling cuts from basic circle O ' ₁In Q ₁Point, line segment J ₂C ₂Turning over equal angular cuts from basic circle O ' ₂In Q ₂The point.Owing to introduced extension deformation involute J ₂K ₁And J ₂K ₂, therefore, revise circular arc center of circle O ₁And O ₂All not on the tangent line of crossing main vortex filament basic circle, by O ₁And O ₂Remake the circular arc correction.

In Fig. 1, the center of circle of three master and slave basic circles point-blank, three centers of circle place straight line and X-axis angle be ψ and about the condition that should satisfy from basic circle and main basic circle center of circle distance D be:

When

$R_2=\frac{1}{2}[S-2\stackrel{\&OverBar;}{O_1{B}_1}-a(\mathrm{\&pi;}-2\mathrm{\&alpha;})]$

$=\frac{1}{2}\{2a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+2D({\sin \mathrm{\&delta;}}^{\&prime;}-1)-2[\frac{a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+D({\sin \mathrm{\&delta;}}^{\&prime;}-1)}{2}$

$-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\&delta;}}^{\&prime;}-D^2{\sin }^2{\mathrm{\&delta;}}^{\&prime;}}{2a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+2D({\sin \mathrm{\&delta;}}^{\&prime;}-1)}]-a(\mathrm{\&pi;}-2\mathrm{\&alpha;})\}>0$

And sin (δ-ψ)＜0 o'clock and

$D\sin \mathrm{\&psi;}+a\sin ({\mathrm{\&delta;}}^{\&prime;}+\mathrm{\&psi;})\&GreaterEqual;$

${\{a}^2+D^2+2\mathrm{aD}\cos {\mathrm{\&delta;}}^{\&prime;}-D^2{\sin }^2{\mathrm{\&delta;}}^{\&prime;}$

$+[\frac{a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+D(\sin {\mathrm{\&delta;}}^{\&prime;}-1)}{2}-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\&delta;}}^{\&prime;}-D^2{\sin }^2{\mathrm{\&delta;}}^{\&prime;}}{2a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+2D({\sin \mathrm{\&delta;}}^{\&prime;}-1)}{]}^2{\}}^{\frac{1}{2}}\cos (\mathrm{\&gamma;}+\mathrm{\&psi;})$

$+\frac{1}{2}\{2a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+2D({\sin \mathrm{\&delta;}}^{\&prime;}-1)-2[\frac{a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+D({\sin \mathrm{\&delta;}}^{\&prime;}-1)}{2}$

$-\frac{a^2+D^2+2\mathrm{aD}\cos {\mathrm{\&delta;}}^{\&prime;}-D^2{\sin }^2{\mathrm{\&delta;}}^{\&prime;}}{2a(\mathrm{\&pi;}+\mathrm{\&psi;}-\mathrm{\&delta;})+2D({\sin \mathrm{\&delta;}}^{\&prime;}-1)}]-a(\mathrm{\&pi;}-2\mathrm{\&alpha;})\}\sin (\mathrm{\&delta;}-\mathrm{\&psi;})$

Wherein $\mathrm{\&gamma;}={\mathrm{\&delta;}}^{\&prime;}+\mathrm{\&psi;}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\&OverBar;}{O_1{B}_1}}{\stackrel{\&OverBar;}{{\mathrm{OB}}_1}}\right)=\mathrm{\&beta;}-\frac{\mathrm{\&pi;}}{2}+\mathrm{\&psi;}-{\mathrm{tg}}^{-1}\left(\frac{\stackrel{\&OverBar;}{O_1{B}_1}}{\stackrel{\&OverBar;}{{\mathrm{OB}}_1}}\right)$ The time, can form smooth molded lines.

The scroll compressor vortex root section that the present invention relates to (being the position, zone of high pressure) molded lines is by forming based on the extension correction molded lines of " three basic circles " molded lines (being called for short the TBC molded lines) revised theory, and the scroll compressor of forming with the PMP molded lines of same ratio size compares as Fig. 5 and Fig. 6.Can know, under identical air inflow, the scroll compressor that the present invention relates to is obviously little than Japanese PMP molded lines compressor at the exhaust cavity area, and this just illustrates that compression ratio has had raising, and also increase at root vortex filament thickness, overcome the shortcoming of PMP molded lines compressor fully.

Claims (1)

1 one kinds of scroll compressors is characterized in that swirl compressor spinning disk vortex root molded lines is by the outside main involute section J ₂E ₂, connect straightway E ₂E ₁, inboard main involute section E ₁J ₁, the inboard involute section J that revises ₁K ₁, revise great circle segmental arc K ₁N revises roundlet segmental arc NK ₂, involute section K is revised in the outside ₂J ₂Form; Wherein:

(1) the main involute section J in the outside ₂E ₂Pattern curve satisfies equation:

Wherein, x ₀=0, y ₀=0, α=α, α is the initial angle of the main involute section of survey outside forming, ${\mathrm{\&theta;}}_{\min }=\frac{3\mathrm{\&pi;}}{2}+\mathrm{\&psi;}-\mathrm{\&alpha;},$ ψ is three basic circle center of circle place straight line and X-axis angles;

(2) connect straightway E ₂E ₁Pattern curve satisfies equation:

Straightway E ₂E ₁End points E ₂, E ₁Respectively with involute J ₂E ₂, E ₁J ₁On E ₂And E ₁Point is a same point, so coordinate figure is known, then E ₂E ₁The parametrization equation be:

$y=\mathrm{LN}(x_{E_1},y_{E_1},x_{E_2},y_{E_2},x)$

(3) inboard main involute section E ₁J ₁Pattern curve satisfies equation:

Wherein, x ₀=0, y ₀=0, α=-α ,-α is the initial angle of the main involute section of survey in forming,

${\mathrm{\&theta;}}_{\min }=\frac{3\mathrm{\&pi;}}{2}+\mathrm{\&psi;}+\mathrm{\&alpha;},$ ψ is three basic circle center of circle place straight line and X-axis angles;

(4) the inboard involute section J that revises ₁K ₁(the auxiliary basic circle in left side forms) pattern curve satisfies equation:

Wherein, x ₀=-Dcos ψ, y ₀=-Dsin ψ, α=α ₀₁, ${\mathrm{\&alpha;}}_{01}=\frac{3\mathrm{\&pi;}}{2}+\mathrm{\&psi;}-\frac{L_1}{a},$ α ₀₁For forming the inboard involute section J that revises ₁K ₁Initial angle, form by the auxiliary basic circle in left side, D be the left and right auxiliary basic circle center of circle respectively with the distance in the basic circle center of circle, center, θ=α _K1, α _K1=π-α ₀₁+ ψ+δ ', ${\mathrm{\&delta;}}^{\&prime;}=\frac{\mathrm{\&pi;}}{2}-\mathrm{\&delta;},$ δ is the inboard involute section J that revises ₁K ₁The angle of spread;

(5) revise great circle segmental arc K ₁The N pattern curve satisfies equation:

Revise great circle segmental arc K ₁The end points K of N ₁For revising involute section J in the inboard ₁K ₁On point, center of circle x ₀₁, y ₀₁Point coordinates is:

γ is the angle of two correction circular arc center of circle place straight lines and place, three basic circle centers of circle straight line, For the auxiliary basic circle distance of center circle in center basic circle and left side from, revise great circle segmental arc K ₁N and correction roundlet segmental arc NK ₂Tangent at the N point;

$x_N=\frac{(R_1-R_2)\cos {\mathrm{\&gamma;}}_P}{2},$ $y_N=\frac{(R_1-R_2)\sin {\mathrm{\&gamma;}}_P}{2}$

R ₁, R ₂Be two correction radius of arc;

(6) revise roundlet segmental arc NK ₂Pattern curve satisfies equation:

Revise great circle segmental arc NK ₂End points K ₂For revising involute section K in the outside ₂J ₂On point, center of circle x ₀₂, y ₀₂Point coordinates is:

Revise great circle segmental arc K ₁N and correction roundlet segmental arc NK ₂Tangent at the N point;

(7) involute section K is revised in the outside ₂J ₂(the auxiliary basic circle in right side forms) pattern curve satisfies equation:

Wherein, x ₀=Dcos ψ, y ₀=Dsin ψ, α=α ₀₂, ${\mathrm{\&alpha;}}_{02}=\frac{\mathrm{\&pi;}}{2}+\mathrm{\&psi;}-\frac{L_2}{a},$ α ₀₂Revise involute section K for forming the outside ₂J ₂Initial angle, form by the auxiliary basic circle in right side, D be the left and right auxiliary basic circle center of circle respectively with the distance in the basic circle center of circle, center, θ=α _K2, α _K2=δ '+ψ-α ₀₂, ${\mathrm{\&delta;}}^{\&prime;}=\frac{\mathrm{\&pi;}}{2}-\mathrm{\&delta;},$ δ is the inboard involute section J that revises ₁K ₁The angle of spread.

Images